A Some Special Functions and Their Properties

The main purpose of this appendix is to introduce several special functions and to state their basic properties that are most frequently used in the theory and applications of ordinary and partial differential equations. The subject is, of course, too vast to be treated adequately in so short a space, so that only the more important results will be stated. For a fuller discussion of these topics and of further properties of these functions the reader is referred to the standard treatises on the subject.

A-1 Gamma, Beta, and Error Functions

The gamma function (also called the factorial function) is deﬁned by a deﬁnite integral in which a variable appears as a parameter ∞ Γ (x)= e−ttx−1 dt, x > 0. (A-1.1) 0 The integral (A-1.1) is uniformly convergent for all x in [a, b] where 0 0. Integrating (A-1.1) by parts, we obtain the fundamental property of Γ (x) ∞ − −t x−1 ∞ − −t x−2 Γ (x)= e t 0 +(x 1) e t dt 0 =(x − 1)Γ (x − 1) for x − 1 > 0. Then we replace x by x +1to obtain the fundamental result Γ (x +1)=xΓ(x). (A-1.2) In particular, when x = n is a positive integer, we make repeated use of (A-1.2) to obtain Γ (n +1)=nΓ (n)=n(n − 1)Γ (n − 1) = ··· = n(n − 1)(n − 2) ···3 · 2 · 1Γ (1) = n!, (A-1.3) where Γ (1) = 1.

L. Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, DOI 10.1007/978-0-8176-8265-1, c Springer Science+Business Media, LLC 2012 690 A Some Special Functions and Their Properties

We put t = u2 in (A-1.1) to obtain ∞ Γ (x)=2 exp −u2 u2x−1 du, x > 0. (A-1.4) 0 1 Letting x = 2 , we ﬁnd √ 1 ∞ π √ Γ =2 exp −u2 du =2 = π. (A-1.5) 2 0 2 Using (A-1.2), we deduce √ 3 1 1 π Γ = Γ = . (A-1.6) 2 2 2 2

5 7 2n+1 Similarly, we can obtain the values of Γ ( 2 ),Γ( 2 ),...,Γ( 2 ). The gamma function can also be deﬁned for negative values of x by the rewritten form of (A-1.2)as Γ (x +1) Γ (x)= ,x=0 , −1, −2,.... (A-1.7) x For example, 1 Γ ( 1 ) 1 √ Γ − = 2 = −2 Γ = −2 π, (A-1.8) 2 − 1 2 2 3 Γ (− 1 ) 4√ Γ − = 2 = π. (A-1.9) − 3 2 2 3 We differentiate (A-1.1) with respect to x to obtain d ∞ d e−t Γ (x)=Γ (x)= tx dt dx dx t 0 ∞ d e−t ∞ = exp(x log t) dt = tx−1(log t)e−t dt. (A-1.10) 0 dx t 0 At x =1, this gives ∞ Γ (1) = e−t log tdt= −γ, (A-1.11) 0 where γ is called the Euler constant and has the value 0.5772. The graph of the gamma function is shown in Figure A.1. The volume, Vn, and the surface area, Sn, of a sphere of radius r in n-dimensional space Rn are given by { 1 }n n { 1 }n n−1 Γ ( 2 ) r 2 Γ ( 2 ) r Vn = n ,Sn = n . Γ ( 2 +1) Γ ( 2 )

dVn Thus, dr = Sn. A-1 Gamma, Beta, and Error Functions 691

Fig. A.1 The gamma function.

2 4 3 In particular, when n =2, 3,..., we get V2 = πr , S2 =2πr; V3 = 3 πr , 2 S3 =4πr ;etc. Using (A-1.2) and (A-1.5), we obtain the following results: πmr2m 2πmr2m−1 V = ,S= , 2m m! 2m (m − 1)! 2(2π)mr2m+1 22m+1m!πmr2m V = ,S = . 2m+1 1.3.5 ···(2m +1) 2m+1 (2m)!

Legendre Duplication Formula

Several useful properties of the gamma function are recorded below for reference without proof. We begin with 1 √ 22x−1Γ (x)Γ x + = πΓ(2x). (A-1.12) 2 In particular, when x = n (n =0, 1, 2,...), √ 1 π (2n)! Γ n + = . (A-1.13) 2 22n n! The following properties also hold for Γ (x):

Γ (x)Γ (1 − x)=π cosec πx, x is a noninteger, (A-1.14) ∞ Γ (x)=px exp(−pt) tx−1 dt, (A-1.15) 0 ∞ Γ (x)= exp xt − et dt. (A-1.16) −∞ 692 A Some Special Functions and Their Properties √ x+ 1 Γ (x +1)∼ 2π exp(−x)x 2 for large x, (A-1.17) √ n+ 1 n! ∼ 2π exp(−n)x 2 for large n. (A-1.18)

The latter formulas are known as Stirling approximation of Γ (x +1)for large x and of n! for large n. The incomplete gamma function, γ(x, a), is defined by the integral x γ(a, x)= e−tta−1 dt, a > 0. (A-1.19) 0 The complementary incomplete gamma function, Γ (a, x), is defined by the integral ∞ Γ (a, x)= e−t ta−1 dt, a > 0. (A-1.20) x Thus, it follows that γ(a, x)+Γ (a, x)=Γ (a). (A-1.21) The beta function, denoted by B(x, y), is defined by the integral t B(x, y)= tx−1(1 − t)y−1 dt, x > 0,y>0. (A-1.22) 0 The beta function B(x, y) is symmetric with respect to its arguments x and y, that is, B(x, y)=B(y, x). (A-1.23) This follows from (A-1.22) by the change of variable 1 − t = u, that is, 1 B(x, y)= uy−1(1 − u)x−1 du = B(y, x). 0 If we make the change of variable t = u/(1 + u) in (A-1.22), we obtain another integral representation of the beta function ∞ ∞ B(x, y)= ux−1(1 + u)−(x+y) du = uy−1(1 + u)−(x+y) du. (A-1.24) 0 0 Putting t = cos2 θ in (A-1.22), we derive π/2 B(x, y)=2 cos2x−1 θ sin2y−1 θdθ. (A-1.25) 0 Several important results are recorded below for ready reference without proof: 1 1 B(1, 1) = 1,B, = π, (A-1.26) 22 x − 1 B(x, y)= B(x − 1,y), (A-1.27) x + y − 1 A-1 Gamma, Beta, and Error Functions 693

Fig. A.2 The error function and the complementary error function.

Γ (x)Γ (y) B(x, y)= , (A-1.28) Γ (x + y) 1+x 1 − x πx B , = π sec , 0

The error function, erf(x), is deﬁned by the integral 2 x erf(x)=√ exp −t2 dt, −∞

erf(−x)=− erf(x), (A-1.31) d 2 erf(x) = √ exp −x2 , (A-1.32) dx π erf(0) = 0, erf(∞)=1. (A-1.33)

The complementary error function, erfc(x), is deﬁned by the integral 2 ∞ erfc(x)=√ exp −t2 dt. (A-1.34) π x Clearly, it follows that

erfc(x)=1− erf(x), (A-1.35) erfc(0) = 1, erfc(∞)=0. (A-1.36)

The graphs of erf(x) and erfc(x) are shown in Figure A.2. 1 erfc(x) ∼ √ exp −x2 for large x. (A-1.37) x π

Closely associated with the error function are the Fresnel integrals, which are deﬁned by 694 A Some Special Functions and Their Properties

Fig. A.3 The Fresnel integrals C(x) and S(x).

x πt2 x πt2 C(x)= cos dt and S(x)= sin dt. (A-1.38) 0 2 0 2 These integrals arise in diffraction problems in optics, in water waves, in elasticity, and elsewhere. Clearly, it follows from (A-1.38) that

C(0) = 0 = S(0), (A-1.39) π C(∞)=S(∞)= , (A-1.40) 2 d πx2 d πx2 C(x) = cos , S(x)=sin . (A-1.41) dx 2 dx 2

It also follows from (A-1.38) that C(x) has extrema at the points where x2 = (2n+1), n =0, 1, 2, 3,..., and S(x) has extrema at the points where x2 =2n, n = 1, 2√, 3,.... The largest maxima occur ﬁrst and are found to be C(1) = 0.7799 and S( 2) = 0.7139. We also infer that both C(x) and S(x) are oscillatory about the line y =0.5. The graphs of C(x) and S(x) for non-negative real x are shown in Figure A.3. We prove further properties of the Gamma and the Beta functions. We ﬁrst prove that π/2 1 sin2p−1 x cos2q−1 xdx= B(p, q). (A-1.42) 0 2 We put sin2 x = t so that the left hand side of the above integral becomes 1 π/2 sin2p−2 x cos2q−2 x · 2 cos x sin xdx 2 0 1 1 1 = tp−1(1 − q)q−1 dt = B(p, q). (A-1.43) 2 0 2 We next prove that A-1 Gamma, Beta, and Error Functions 695 1 1 Γ (2p)Γ =22p−1Γ (p) Γ p + . (A-1.44) 2 2 We have π/2 1 π π/2 1 sin2p 2xdx= sin2p θdθ=2 sin2p θdθ, (2x = θ). 0 2 0 0 2

1 Putting q = 2 in (A-1.42) with x =2θ gives 1 1 π/2 π/2 B p + , =2 sin2p xdx=2 sin2p 2θdθ 2 2 0 0 π/2 =22p+1 sin2p θ cos2p θdθ 0 1 1 =22pB p + ,p+ , 2 2 which is, using (A-1.28) and (A-1.2),

Γ (p + 1 )Γ ( 1 ) Γ (p + 1 )Γ (p + 1 ) Γ (p + 1 )Γ (p + 1 ) 2 2 =22p 2 2 =22p 2 2 , Γ (p +1) Γ (2p +1) 2pΓ(2p) or equivalently, 1 1 Γ (2p) Γ =22p−1Γ (p)Γ p + . 2 2 We next deﬁne (1 − t )ntx−1 if 0 ≤ t ≤ n, f(n, t)= n (A-1.45) 0 if t ≥ n.

Using x n lim 1 − = e−x, n→∞ n we obtain, for ﬁxed t, t n lim f(n, t) = lim 1 − tx−1 = e−t tx−1. (A-1.46) n→∞ n→∞ n Hence, for x>0, ∞ ∞ Γ (x)= e−ttx−1 dt = lim f(n, t) dt n→∞ 0 0 n t n = lim 1 − tx−1 dt, →∞ n 0 n 696 A Some Special Functions and Their Properties

t which is, putting n = z, 1 = lim nx (1 − z)nzx−1 dz n→∞ 0 zx 1 1 z = lim nx (1 − z)n · + n (1 − z)n−1 dz n→∞ x x 0 0 n 1 = lim nx · (1 − z)n−1zx dz n→∞ x 0 n · (n − 1) ···1 1 = lim nx zn+x−1 dz →∞ · ··· − n x (x +1) (n + x 1) 0 nx n! = lim . (A-1.47) n→∞ x(x +1)···(x + n) This is the celebrated Gauss formula. We next prove that, for 0

A random variable X with values in (0, 1) has the Beta distribution if its density function is, for some p, q>0, 1 f(x)= xp−1(1 − x)q−1, 0 0 and q>0, qp f(x)= xp−1e−qx. (A-1.55) Γ (p)

A-2 Bessel and Airy Functions

The Bessel function of the ﬁrst kind of order v (non-negative real number) is denoted by Jv(x) and deﬁned by ∞ (−1)rx2r J (x)=xv . (A-2.1) v 22r+vr! Γ (r + v +1) r=0 This series is convergent for all x. 698 A Some Special Functions and Their Properties

The Bessel function y = Jv(x) satisﬁes the Bessel equation x2y + xy + x2 − v2 y =0. (A-2.2)

When v is not a positive integer or zero, Jv(x) and J−v(x) are two linearly independent solutions so that

y = AJv(x)+BJ−v(x) (A-2.3) is the general solution of (A-2.2), where A and B are arbitrary constants. However, when v = n, where n is a positive integer or zero, Jn(x) and J−n(x) are no longer independent, but are related by the equation

n J−n(x)=(−1) Jn(x). (A-2.4)

Thus, when n is a positive integer or zero, equation (A-2.2) has only one solution given by ∞ (−1)r x n+2r J (x)= . (A-2.5) n r!(n + r)! 2 r=0

A second solution, known as Neumann’s or Webber’s solution, Yn(x) is given by

Yn(x) = lim Yv(x), (A-2.6) v→n where (cos vπ)J (x) − J− (x) Y (x)= v v . (A-2.7) v sin vπ Thus, the general solution of (A-2.2)is

y(x)=AJn(x)+BYn(x), (A-2.8) where A and B are arbitrary constants. In particular, from (A-2.5),

∞ (−1)r x 2r J (x)= , (A-2.9) 0 (r!)2 2 r=0 ∞ (−1)r x 2r+1 J (x)= . (A-2.10) 1 r!(r + 1)! 2 r=0 Clearly, it follows from (A-2.9) and (A-2.10) that − J0(x)= J1(x). (A-2.11)

Bessel’s equation may not always arise in the standard form given in (A-2.2), but more frequently as x2y + xy + k2x2 − v2 y =0 (A-2.12) A-2 Bessel and Airy Functions 699 with the general solution

y(x)=AJv(kx)+BYv(kx). (A-2.13)

The recurrence relations are recorded below for easy reference without proof: v J (x)= J (x) − J (x), (A-2.14) v+1 x v v v J − (x)= J (x)+J (x), (A-2.15) v 1 x v v 2v J − (x)+J (x)= J (x), (A-2.16) v 1 v+1 x v − Jv−1(x) Jv+1(x)=2Jv(x). (A-2.17)

We have, from (A-2.5),

∞ (−1)r2−(n+2r) xnJ (x)= x2n+2r. n r!(n + r)! r=0 Differentiating both sides of this result with respect to x and using the fact that 2(n + r)/(n + r)! = 2/(n + r − 1)!, it turns out that

∞ − r −(n+2r+1) d n ( 1) 2 2n+2r−1 n x J (x) = x = x J − (x). (A-2.18) dx n r!(n + r − 1)! n 1 r=0 Similarly, we can show d x−nJ (x) = −x−nJ (x). (A-2.19) dx n n+1 The generating function for the Bessel function is

∞ 1 1 exp x t − = tnJ (x). (A-2.20) 2 t n n=−∞

The integral representation of Jn(x) is 1 π Jn(x)= cos(nθ − x sin θ) dθ. (A-2.21) π 0 The following are known as the Lommel integrals: a xJn(px)Jn(qx) dx 0 a = pJ (qa)J (pa) − qJ (pa) J (qa) ,p= q, (A-2.22) (q2 − p2) n n n n 700 A Some Special Functions and Their Properties

Fig. A.4 Graphs of y = J0(x), J1(x),andJ2(x). and a 2 2 2 a 2 − n 2 xJn(px) dx = Jn (pa)+ 1 2 2 Jn(pa) . (A-2.23) 0 2 p a ± 1 When n = 2 , 2 2 J 1 (x)= sin x, J− 1 (x)= cos x. (A-2.24) 2 πx 2 πx A rough idea of the shape of the Bessel functions when x is large may be obtained − 1 from equation (A-2.2). Substitution of y = x 2 u(x) eliminates the ﬁrst derivative, and hence, gives the equation 4n2 − 1 u + 1 − u =0. (A-2.25) 4x2 For large x, this equation approximately becomes

u + u =0. (A-2.26)

This equation admits the solution u(x)=A cos(x + ε), that is, A y = √ cos(x + ε). (A-2.27) x

This suggests that Jn(x) is oscillatory and has an inﬁnite number of zeros. It also →∞ ± 1 tends to zero as x . The graphs of Jn(x) for n =0, 1, 2 and for n = 2 are shown in Figures A.4 and A.5, respectively. An important special case arises in particular physical problems when k2 = −1 in equation (A-2.12). we then have the modiﬁed Bessel equation x2y + xy − x2 + v2 y =0, (A-2.28) with the general solution

y = AJv(ix)+BYv(ix). (A-2.29) A-2 Bessel and Airy Functions 701

Fig. A.5 Graphs of J 1 (x) and J− 1 (x). 2 2

Fig. A.6 Graphs of y = Y0(x), Y1(x),andY2(x).

We now deﬁne a new function −v Iv(x)=i Jv(ix), (A-2.30) and then use the series (A-2.1)forJv(x) so that ∞ ∞ (−1)r ix v+2r 1 x v+2r I (x)=i−v = . v r!Γ (r + v +1) 2 r!Γ (r + v +1) 2 r=0 r=0 (A-2.31)

Similarly, we can find the second solution, Kv(x), of the modified Bessel equation (A-2.28). Usually, Iv(x) and Kv(x) are called modified Bessel functions and their properties can be obtained in a similar way to those of Jv(x) and Yv(x). The graphs of Y0(x), Y1(x), and Y2(x) are shown in Figure A.6. We state a few important infinite integrals involving Bessel functions which arise frequently in the application of Hankel transforms. ∞ (2b)vΓ (v + 1 ) 1 exp(−at)J (bt)tv dt = √ 2 ,v>− , (A-2.32) v v+ 1 π(a2 + b2) 2 2 0 ∞ v 3 v+1 2a(2b) Γ (v + 2 ) exp(−at)Jv(bt)t dt = √ ,v>−1, (A-2.33) 2 2 v+ 3 0 π(a + b ) 2 702 A Some Special Functions and Their Properties

Fig. A.7 The Airy function.

∞ v 2 − 2 2 v+1 b − b − exp a t Jv(bt)t dt = 2 v+1 exp 2 ,v>1, 0 (2a ) 4a (A-2.34) ∞ 2 2 − 2 2 1 −b + c bc exp a t Jv(bt)Jv(ct)tdt = 2 exp 2 Iv 2 , 0 2a 4a 2a v>−1, (A-2.35) ∞ 2μ−v−1 2μ−v−1 2 Γ (μ) 1 −1 t Jv(t) dt = − , 0 <μ< ,v> . 0 Γ (v μ +1) 2 2 (A-2.36)

The Airy function, y = Ai(x), is the ﬁrst solution of the differential equation

y − xy =0. (A-2.37)

The second solution is denoted by Bi(x). Then these functions are expressed in terms of the Bessel and modiﬁed Bessel functions in the form x 2 3/2 2 3/2 1 x Ai(x)= I− 1 x − I 1 x = K 1 (ξ), (A-2.38) 3 3 3 3 3 π 3 3 x 2 3/2 2 3/2 x iπ Bi(x)= I− 1 x + I 1 x = Re e 6 H 1 (−iξ) , 3 3 3 3 3 3 3 (A-2.39)

2 3/2 where ξ = 3 x . The integral representation of Ai(x) is 1 ∞ 1 Ai(x)= cos t3 + xt dt. (A-2.40) π 0 3 The graph of y = Ai(x) is shown in Figure A.7 using the values of Ai(x) at x =0 and x →∞: 1 1 Ai(0) = √ Bi(0) = =0.355028, 3 33/2Γ ( 2 ) 3 Ai(x), Bi(x) → [0, ∞]asx →∞. A-3 Legendre and Associated Legendre Functions 703

Similarly, the graph of y = Bi(x) can be drawn. The integral representation of Bi(x) is 1 ∞ t3 t3 Bi(x)= exp xt − +sin xt + dt. (A-2.41) π 0 3 3 A slightly more general integral representation of Ai(ax) is 1 ∞ t3 Ai(ax)= cos xt + 3 dt. (A-2.42) πa 0 3a When a =1, this reduces to (A-2.40). The general power series solution of the Airy equation (A-2.37) is given by x3 x6 x3n y(x)=a 1+ + + ···+ + ··· 0 2 · 3 2 · 3 · 5 · 6 2 · 3 ···(3n − 1)(3n) x4 x7 x3n+1 + a x + + + ···+ + ··· 1 3 · 4 3 · 4 · 6 · 7 3 · 4 ···(3n)(3n +1) (A-2.43) ∞ x3n = a 1+ 0 3 · 4 ···(3n − 4)(3n − 3)(3n − 1)(3n) n=1 ∞ x3n+1 + a x + , (A-2.44) 1 3 · 4 ···(3n − 3)(3n − 2)(3n)(3n +1) n=1 where a0 and a1 are arbitrary constants of integration. Finally, the asymptotic representations of Ai(x) and Bi(x) are given by 1 2 Ai(x) ≈ √ exp − x3/2 as x → +∞, 1/4 (A-2.45) 2 πx 3 1 3 Bi(x) ≈ √ exp x3/2 as x →∞. (A-2.46) πx1/4 2

A-3 Legendre and Associated Legendre Functions

The Legendre polynomials, Pn(x), are deﬁned by the Rodrigues formula

n 1 d n P (x)= x2 − 1 . (A-3.1) n 2nn! dxn The ﬁrst seven Legendre polynomials are

P0(x)=1,

P1(x)=x, 704 A Some Special Functions and Their Properties 1 P (x)= 3x2 − 1 , 2 2 1 P (x)= 5x3 − 3x , 3 2 1 P (x)= 35x4 − 30x2 +3 , 4 8 1 P (x)= 63x5 − 70x3 +15x , 5 8 1 P (x)= 231x6 − 315x4 + 105x2 − 5 . 6 16 The generating function for the Legendre polynomials is ∞ − 1 2 2 n 1 − 2xt + t = t Pn(x). (A-3.2) n=0 This function provides more information about the Legendre polynomials. For example,

n Pn(1) = 1,Pn(−1) = (−1) , (A-3.3) 1 · 3 · 5 ···(2n − 1) (2n − 1)!! P (0) = (−1)n =(−1)n , (A-3.4) 2n 2nn! (2n)!!

P2n+1(0) = 0,n=0, 1, 2,..., (A-3.5) dn (2n)! P (−x)=(−1)nP (x), P (x)= , (A-3.6) n n dxn n 2nn! where the double factorial is deﬁned by

(2n − 1)!! = 1 · 3 · 5 ···(2n − 1) and (2n)!! = 2 · 4 · 6 ···(2n).

The graphs of the ﬁrst four Legendre polynomials are shown in Figure A.8. The recurrence relations for the Legendre polynomials are

(n +1)Pn+1(x)=(2n +1)xPn(x) − nPn−1(x), (A-3.7) − Pn+1(x) Pn−1(x)=(2n +1)Pn(x), (A-3.8) 2 1 − x P (x)=nP − (x) − nxP (x), (A-3.9) n n 1 n − 2 − 1 x Pn(x)=(n +1)xPn(x) (n +1)Pn+1(x). (A-3.10)

The Legendre polynomials, y = Pn(x), satisfy the Legendre differential equation 1 − x2 y − 2xy + n(n +1)y =0. (A-3.11)

If n is not an integer, both solutions of (A-3.11) diverge at x = ±1. The orthogonal relation is 1 2 Pn(x)Pm(x) dx = δnm. (A-3.12) −1 (2n +1) A-3 Legendre and Associated Legendre Functions 705

Fig. A.8 Graphs of y = P0(x), P1(x), P2(x),andP3(x).

The associated Legendre functions are deﬁned by

m m+n m d 1 m d n P m(x)= 1 − x2 2 P (x)= 1 − x2 2 x2 − 1 , (A-3.13) n dxm n 2nn! dxm+n where 0 ≤ m ≤ n. Clearly, it follows that

0 Pn (x)=Pn(x), (A-3.14) (n − m)! P m(−x)=(−1)n+mP m(x),P−m(x)=(−1)m P m(x). (A-3.15) n n n (n + m)! n

m The generating function for Pn (x) is

m ∞ − 2 2 (2m)!(1 x ) m r 1 = Pr+m(x)t . (A-3.16) m − 2 m+ 2 2 m!(1 2tx + t ) r=0 The recurrence relations are

m m − m (2n +1)xPn (x)=(n + m)Pn−1(x)+(n m +1)Pn+1(x), (A-3.17) 1 d − 2 1 − x2 2 P m(x)=P m+1(x) − (n + m)(n − m +1)P m 1(x). (A-3.18) dx n n n

m The associated Legendre functions Pn (x) are solutions of the differential equation m2 1 − x2 y − 2xy + n(n +1)− y =0. (A-3.19) (1 − x2)

This reduces to the Legendre equation when m =0. 706 A Some Special Functions and Their Properties

Listed below are a few associated Legendre functions with x = cos θ:

1 1 − 2 2 P1 (x)= 1 x =sinθ, 1 P 1(x)=3x 1 − x2 2 = 3 cos θ sin θ, 2 2 − 2 2 P2 (x)=3 1 x =3sin θ, 3 1 3 1 2 − − 2 2 2 − P3 (x)= 5x 1 1 x = 5 cos θ 1 sin θ, 2 2 2 − 2 2 P3 (x)=15x 1 x = 15 cos θ sin θ, 3 − 2 3/2 3 P3 (x)=15 1 x =15sin θ.

The orthogonal relations are 1 m m 2 · ( + m)! Pn (x)Pt (x) dx = δn, (A-3.20) − (2 +1) ( − m)! 1 1 − 2 −1 m (n + m)! 1 x Pn (x)Pn(x) dx = − δm. (A-3.21) −1 m(n m)!

A-4 Jacobi and Gegenbauer Polynomials

(α,β) The Jacobi polynomials, Pn (x),ofdegreen are deﬁned by the Rodrigues formula (−1)n dn P (α,β)(x)= (1 − x)−α(1 + x)−β (1 − x)α+n(1 + x)β+n , (A-4.1) n 2nn! dxn where α>−1 and β>−1. When α = β =0, the Jacobi polynomials become Legendre polynomials, that is, (0,0) Pn(x)=Pn (x),n=0, 1, 2,.... (A-4.2) On the other hand, the associated Laguerre functions arise as the limit α (α,β) − 2x Ln(x) = lim Pn 1 . (A-4.3) β→∞ β

(α,β) The recurrence relations for Pn (x) are

2(n + 1)(α + β + n + 1)(α + β +2n)P (α,β)(x) n+1 2 − 2 (α,β) =(α + β +2n +1) α β + x(α + β +2n + 2)(α + β +2n) Pn (x) − (α,β) 2(α + n)(β + n)(α + β +2n +2)Pn−1 (x), (A-4.4) where n =1, 2, 3,..., and A-4 Jacobi and Gegenbauer Polynomials 707

(α,β−1) − (α−1,β) (α,β) Pn (x) Pn (x)=Pn−1 (x). (A-4.5) The generating function for Jacobi polynomials is ∞ (α+β) −1 − −α −β (α,β) n 2 R (1 t + R) (1 + t + R) = Pn (x)t , (A-4.6) n=0

2 1 where R =(1− 2xt + t ) 2 . (α,β) The Jacobi polynomials, y = Pn (x), satisfy the differential equation 1 − x2 y + (β − α) − (α + β +2)x y + n(n + α + β +1)y =0. (A-4.7)

The orthogonal relation is 1 0 if n = m, − α β (α,β) (α,β) (1 x) (1 + x) Pn (x)Pm (x) dx = (A-4.8) −1 δn if n = m, where 2α+β+1Γ (n + α +1)Γ (n + β +1) δ = . (A-4.9) n n!(α + β +2n +1)Γ (α + β + n +1) − 1 When α = β = v 2 , the Jacobi polynomials reduce to the Gegenbauer polynomi- v als, Cn(x), which are deﬁned by the Rodrigues formula

n n (−1) v− 1 d v+n− 1 Cv(x)= 1 − x2 2 1 − x2 2 . (A-4.10) n 2nn! dxn

v The generating function for Cn(x) of degree n is ∞ −v 1 1 − 2xt+ t2 = Cv(x) tn, |t| < 1, |x|≤1,v>− . (A-4.11) n 2 n=0 The recurrence relations are

v − v − v (n +1)Cn+1(x) 2(v + n)xCn(x)+(2v + n 1)Cn−1(x)=0, (A-4.12) v − v+1 v+1 (n +1)Cn+1(x) 2vCn (x)+2vCn−1(x)=0, (A-4.13) d Cv(x) =2vCv+1(x). (A-4.14) dx n n+1

v The differential equation satisﬁed by y = Cn(x) is 1 − x2 y − (2v +1)xy + n(n +2v)y =0. (A-4.15)

The orthogonal property is 1 v− 1 − 2 2 v v 1 x Cn(x)Cm(x) dx = δnδnm, (A-4.16) −1 708 A Some Special Functions and Their Properties where 21−2vnΓ (n +2v) δ = . (A-4.17) n n!(n + v)[Γ (v)]2 1 When v = 2 , the Gegenbauer polynomials reduce to Legendre polynomials, that is,

1 2 Cn (x)=Pn(x). (A-4.18)

The Hermite polynomials can also be obtained from the Gegenbauer polynomials as the limit −n/2 v √x Hn(x)=n! lim v Cn . (A-4.19) v→∞ v 1 Finally, when α = β = 2 , the Gegenbauer polynomials reduce to the well-known Chebyshev polynomials, Tn(x), which are deﬁned by a solution of the second order difference equation

un+2 − 2xun+1 + un =0, |x|≤1, (A-4.20)

u(0) = u0 and u(1) = u1. (A-4.21)

The generating function for Tn(x) is

∞ (1 − t2) = T (x)+2 T (x)tn, |x|≤1,t<1. (A-4.22) (1 − 2xt+ t2) 0 n n=1 The ﬁrst seven Chebyshev polynomials of degree n of the ﬁrst kind are

T0(x)=1,

T1(x)=x, 2 T2(x)=2x − 1, 3 T3(x)=4x − 3x, 4 2 T4(x)=8x − 8x +1, 5 3 T5(x)=16x − 20x +5x, 6 4 2 T6(x)=32x − 48x +18x − 1.

The graphs of the ﬁrst four Chebyshev polynomials are shown in Figure A.9. The Chebyshev polynomials y = Tn(x) satisfy the differential equation 1 − x2 y − xy + n2y =0. (A-4.23)

It follows from (A-4.22) that Tn(x) satisﬁes the recurrence relations

Tn+1(x) − 2xTn(x)+Tn−1(x)=0, (A-4.24) − Tn+m(x) 2Tn(x)Tm(x)+Tn−m(x)=0, (A-4.25) − 2 − 1 x Tn(x)+nxTn(x) nTn−1(x)=0. (A-4.26) A-4 Jacobi and Gegenbauer Polynomials 709

Fig. A.9 Chebyshev polynomials y = Tn(x).

The parity relation for Tn(x) is

n Tn(−x)=(−1) Tn(x). (A-4.27)

The Rodrigues formula is √ n 2 1 n π(−1) (1 − x ) 2 d n− 1 T (x)= · 1 − x2 2 . (A-4.28) n n − 1 n 2 (n 2 )! dx

The orthogonal relation for Tn(x) is ⎧ ⎪ 1 ⎨ 0 if m = n, − 1 1 − x2 2 T (x) T (x) dx = π if m = n, m n ⎪ 2 (A-4.29) −1 ⎩ π if m = n =0.

The Chebyshev polynomials of the second kind, Un(x), are deﬁned by

− 1 2 2 −1 Un(x)= 1 − x sin (n + 1) cos x , −1 ≤ x ≤ 1. (A-4.30)

The generating function for Un(x) is ∞ 2 −1 n 1 − 2xt+ t = Un(x)t , |x| < 1, |t| < 1. (A-4.31) n=0

The ﬁrst seven Chebyshev polynomials Un(x) are given by

U0(x)=1, U1(x)=2x, 2 U2(x)=4x − 1, 710 A Some Special Functions and Their Properties

3 U3(x)=8x − 4x, 4 2 U4(x)=16x − 12x +1, 5 3 U5(x)=32x − 32x +6x, 6 4 2 U6(x)=64x − 80x +24x − 1.

The differential equation for y = Un(x) is 1 − x2 y − 3xy + n(n +2)y =0. (A-4.32)

The recurrence relations are − Un+1(x) 2xUn(x)+Un−1(x)=0, (A-4.33) − 2 − 1 x Un(x)+nxUn(x) (n +1)Un−1(x)=0. (A-4.34)

The parity relation is n Un(−x)=(−1) Un(x). (A-4.35) The Rodrigues formula is √ n n π(−1) (n +1) d n+ 1 U (x)= 1 − x2 2 . (A-4.36) n 1 1 n n+1 − 2 2 dx 2 (n + 2 )!(1 x )

The orthogonal relation for Un(x) is 1 1 2 2 π 1 − x Um(x) Un(x) dx = δmn. (A-4.37) −1 2

A-5 Laguerre and Associated Laguerre Functions

The Laguerre polynomials Ln(x) are deﬁned by the Rodrigues formula dn L (x)=ex xn e−x , (A-5.1) n dxn where n =0, 1, 2, 3,.... The ﬁrst seven Laguerre polynomials are

L0(x)=1,

L1(x)=1− x, 2 L2(x)=2− 4x + x , 2 3 L3(x)=6− 18x +9x − x , 2 3 4 L4(x)=24− 96x +72x − 16x + x , 2 3 4 5 L5(x) = 120 − 600x + 600x − 200x +25x − x , 2 3 4 5 6 L6(x) = 720 − 4320x + 5400x − 2400x + 450x − 36x + x . A-5 Laguerre and Associated Laguerre Functions 711

The generating function is ∞ xt (1 − t)−1 exp = tn L (x). (A-5.2) 1 − t n n=0 In particular, Ln(0) = 1. (A-5.3) The orthogonal relation for the Laguerre polynomial is ∞ −x 2 e Lm(x)Ln(x) dx =(n!) δnm. (A-5.4) 0 The recurrence relations are

(n +1)Ln+1(x)=(2n +1− x)Ln(x) − nLn−1(x), (A-5.5) − xLn(x)=nLn(x) nLn−1(x), (A-5.6) − Ln(x)=Ln−1(x) Ln−1(x). (A-5.7)

The Laguerre polynomials, y = Ln(x), satisfy the Laguerre differential equation xy +(1− x)y + ny =0. (A-5.8) The associated Laguerre polynomials are defined by dm Lm(x)= L (x) for n ≥ m. (A-5.9) n dxm n m The generating function for Ln (x) is ∞ xz (1 − z)−(m+1) exp − = Lm(x)zn, |z| < 1. (A-5.10) 1 − z n n=0 It follows from this that (n + m)! Lm(0) = . (A-5.11) n n!m! The associated Laguerre function satisfies the recurrence relation m − m − m (n +1)Ln+1(x)=(2n + m +1 x)Ln (x) (n + m)Ln−1(x), (A-5.12) d x Lm(x)=nLm(x) − (n + m)Lm (x). (A-5.13) dx n n n−1 m The associated Laguerre function, y = Ln (x), satisfies the associated Laguerre differential equation xy +(m +1− x)y + ny =0. (A-5.14) m The Rodrigues formula for Ln (x) is exx−m dn Lm(x)= e−xxn+m . (A-5.15) n n! dxn m The orthogonal relation for Ln (x) is ∞ −x m m m (n + m)! e x Ln (x)Ll (x) dx = δnl. (A-5.16) 0 n! 712 A Some Special Functions and Their Properties A-6 Hermite Polynomials and Weber–Hermite Functions

The Hermite polynomials Hn(x) are deﬁned by the Rodrigues formula dn H (x)=(−1)n exp x2 exp −x2 , (A-6.1) n dxn where n =0, 1, 2, 3,.... The ﬁrst seven Hermite polynomials are

H0(x)=1,

H1(x)=2x, 2 H2(x)=4x − 2, 3 H3(x)=8x − 12x, 4 2 H4(x)=16x − 48x +12, 5 3 H5(x)=32x − 16x + 120x, 6 4 2 H6(x)=64x − 480x + 720x − 120.

The generating function is

∞ tn exp 2xt − t2 = H (x). (A-6.2) n! n n=0

It follows from (A-6.2) that Hn(x) satisﬁes the parity relation

n Hn(−x)=(−1) Hn(x). (A-6.3)

Also, it follows from (A-6.2) that (2n)! H (0) = 0,H(0) = (−1)n . (A-6.4) 2n+1 2n n! The recurrence relations for Hermite polynomials are

Hn+1(x) − 2xHn(x)+2nHn−1(x)=0, (A-6.5) Hn(x)=2xHn−1(x). (A-6.6)

The Hermite polynomials, y = Hn(x), are solutions of the Hermite differential equation y − 2xy +2ny =0. (A-6.7) The orthogonal property of Hermite polynomials is ∞ √ 2 n exp −x Hn(x)Hm(x) dx =2 n! πδmn. (A-6.8) −∞

With repeated use of integration by parts, it follows from (A-6.1) that A-7 Mittag-Lefﬂer Function 713 ∞ 2 m exp −x Hn(x)x dx =0,m=0, 1,...,(n − 1), (A-6.9) −∞ ∞ √ 2 n exp −x Hn(x)x dx = πn!. (A-6.10) −∞

The WeberÐHermite function,orsimplyHermite function, x2 y = h (x)=exp − H (x) (A-6.11) n 2 n satisﬁes the Hermite differential equation y + λ − x2 y =0,x∈ R, (A-6.12) where λ =2n +1.Ifλ =2 n +1, then y is not ﬁnite as |x|→∞. { }∞ The Hermite functions hn(x) 0 form an orthogonal basis for the Hilbert space L2(R) with weight function 1. They satisfy the following fundamental properties: − hn(x)+xhn(x) 2nhn−1(x)=0, − hn(x) xhn(x)+hn+1(x)=0, − 2 hn(x) x hn(x)+(2n +1)hx =0, ˜ n F hn(x) = hn(k)=(−i) hn(k).

The normalized WeberÐHermite functions are given by 2 −n/2 − 1 − 1 x ψ (x)=2 π 4 (n!) 2 exp − H (x). (A-6.13) n 2 n

Physically, they represent quantum-mechanical oscillator wave functions. The graphs of these functions are shown in Figure A.10.

A-7 Mittag-Lefﬂer Function

Another important function that has widespread use in fractional calculus and fractional differential equation is the Mittag-Leffler function. The Mittag-Leffler function is an entire function defined by the series

∞ zn E (z)= ,α>0. (A-7.1) α Γ (αn +1) n=0 The graph of the Mittag-Leffler function is shown in Figure A.11. The generalized Mittag-Leffler function, Eα,β(z), is defined by

∞ zn E (z)= ,α,β>0. (A-7.2) α,β Γ (αn + β) n=0 714 A Some Special Functions and Their Properties

Fig. A.10 The normalized WeberÐHermite functions.

Fig. A.11 Graph of the Mittag-Lefﬂer function Eα(x).

Also the inverse Laplace transform yields m!sα−β L−1 = tαm+β−1E(m) ±atα , (A-7.3) (sα+¯ a)m+1 α,β where dm E(m)(z)= E (z). (A-7.4) α,β dzm α,β A-8 The Jacobi Elliptic Integrals and Elliptic Functions 715

Obviously,

z Eα,1(z)=Eα(z),E1,1(z)=E1(z)=e . (A-7.5)

A-8 The Jacobi Elliptic Integrals and Elliptic Functions

The parametric equation of an ellipse is given by

x = a sin θ, y = b cos θ, (a>b), 0 ≤ θ ≤ φ. (A-8.1)

Using the arclength formula from calculus, the length of the elliptic arc (A-8.1)is ds2 = dx2 + dy2 = a2 cos2 θ + b2 sin2 θ dθ2 a2 − b2 = a2 1 − sin2 θ dθ2 = a2 1 − m2 sin2 θ dθ2, (A-8.2) a2

2− 2 1 a b 2 where e = m =( a2 ) < 1 is the eccentricity of the ellipse. Consequently, (A-8.2) gives the length of the elliptic arc s φ s = ds = a 1 − m2 sin2 θdθ. (A-8.3) 0 0 This integral cannot be evaluated in terms of elementary functions. Because of its origin, it is called an elliptic integral. In general, there are three classes of elliptic integrals, called elliptic integrals of the ﬁrst, second and third kinds, and they deﬁned by φ dθ F (φ, m)= , (A-8.4) 2 0 1 − m2 sin θ φ E(φ, m)= 1 − m2 sin2 θdθ, (A-8.5) 0 φ dθ Π(φ, m, n)= , (m = n), (A-8.6) 2 2 0 (1 + n2 sin θ) 1 − m2 sin θ where the parameter φ is called the amplitude, φ = am(F, m) so that am(0,m)=0 π and m (0

sn(u, m)=sinφ, (A-8.10) 1 cn(u, m) = cos φ = 1 − sn2 u 2 , (A-8.11) 1 1 dn(u, m)= 1 − m2 sin2 φ 2 = 1 − m2 sn2 u 2 (A-8.12) so that

sn(−u)=− sn u, cn(−u)=cn u, dn(−u)=dn u, (A-8.13) sn(0) = 0, and cn(0) = dn(0) = 1. (A-8.14)

The following limiting results also hold lim sn(u, m)=sinu, lim cn(z,m) = cos u, → → m 0 m 0 (A-8.15) lim dn(z,m)=1, m→0 lim sn(u, m) = tanh u, lim cn(u, m)=sechu, → → m 1 m 1 (A-8.16) lim dn(u, m)=sechu. m→1 Making reference to Dutta and Debnath (1965) without proof, we state the following basic properties of the Jacobi elliptic functions: sn2 u + cn2 u =1, dn2 u + m2 sn2 u =1, dn2 u − m2 cn2 u =1− m2, (A-8.17) d d sn u = cn u dn u, cn u = − sn u dn u, (A-8.18) du du and d dn u = −m2 sn u cn u. (A-8.19) du Putting sn(u, m)=x in the ﬁrst result in (A-8.18) gives the differential equation dx = 1 − x2 1 − m2x2 , (A-8.20) du so that it leads to the Legendre normal form for the sn-function sn(u,m) dx u = . (A-8.21) 2 2 2 0 (1 − x )(1 − m x ) A-8 The Jacobi Elliptic Integrals and Elliptic Functions 717

Similarly, it follows from (A-8.18)Ð(A-8.19) that the Legendre normal forms for cn and dn functions are 1 dx u = , (A-8.22) 2 2 2 2 cn(u,m) (1 − x )(m + m x ) 1 dx u = , (A-8.23) 2 2 2 dn(u,m) (1 − x )(x − m ) √ where m = 1 − m2 is called the complementary modulus of the elliptic integral. The complete elliptic integrals are then given by π π K(m)=F ,m = K m ,E(m)=E ,m = E m . (A-8.24) 2 2

The limiting values of K(m) and E(m) are given as follows: π π lim K(m)=K(0) = , lim E(m)=E(0) = , (A-8.25) m→0 2 m→0 2 lim K(m)=K(1) = ∞, lim E(m)=E(1) = 1. (A-8.26) m→1 m→1 Finally, the addition theorems for sn, cn, and dn functions are sn u cn v dn v + sn v cn u dn u sn(u + v)= , (A-8.27) (1 − m2 sn2 u sn2 v) cn u cn v − sn u dn u sn v dn v cn(u + v)= , (A-8.28) (1 − m2 sn2 u sn2 v) dn u dn v − m2 sn u cn u sn v cn v dn(u + v)= , (A-8.29) (1 − m2 sn2 u sn2 v) where sn(u +4K)=sn u, cn(u +4K)=cn u, and dn(u +2K)=dn u so that sn and cn are periodic functions of period 4K, and dn is a periodic function of period 2K. B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

The main purpose of this appendix is to discuss Fourier series and generalized functions, and to state their basic properties that are most frequently used in the theory and applications of ordinary and partial differential equations. The subject is, of course, too vast to be treated adequately in so short a space, so that only the more important results will be stated. Included are basic properties of Fourier and Laplace transforms which are used in ﬁnding solutions of ordinary and partial differential equations. For a fuller discussion of these topics and of further properties of these functions the reader is referred to the standard treatises on the subjects including Debnath and Bhatta (2007).

B-1 Fourier Series and Its Basic Properties

If f(x) is a periodic function of period 2π defined in (−π, π), then f(x) can be represented as an infinite series in terms of trigonometric functions in the form ∞ 1 f(x)= a + (a cos nx + b sin nx). (B-1.1) 2 0 n n n=1 This is known as the Fourier Series. If we assume that the infinite series is term-by- term integrable on (−π, π), then ∞ π π 1 f(x) dx = a0 + (an cos nx + bn sin nx) dx = πa0 − − 2 π π n=1 so that 1 π a0 = f(x) dx. (B-1.2) π −π Multiplying both sides of (B-1.1)bycos mx and integrating the resulting series from −π to π gives

L. Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, DOI 10.1007/978-0-8176-8265-1, c Springer Science+Business Media, LLC 2012 720 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms π f(x) cos mx dx −π ∞ π 1 = a0 + (an cos nx + bn sin nx) cos mx dx = πan,m= n. − 2 π n=1 Thus, 1 π an = f(x) cos nx dx. (B-1.3) π −π Similarly, multiplying (B-1.1)bysin mx and integrating over (−π, π) gives 1 π bn = f(x)sinnx dx. (B-1.4) π −π

The Fourier coefﬁcients an and bn given by (B-1.2)Ð(B-1.4) are known as the EulerÐ Fourier formulas. If f(x) is an even function of x deﬁned on [−π, π], then 1 π a = f(x) cos nx dx n π −π 2 π = f(x) cos nx dx, n =0, 1, 2,..., (B-1.5) π 0 1 π bn = f(x)sinnx dx =0,n=1, 2, 3,.... (B-1.6) π −π Hence, the Fourier series of an even function f(x) can be written as

∞ 1 f(x)= a + a cos nx, (B-1.7) 2 0 n n=1 where an are given by (B-1.5). Similarly, if f(x) is an odd function of x on (−π, π), the Fourier series of an odd function is given by ∞ f(x)= bn sin nx, (B-1.8) n=1 where an =0for all n, and bn are given by 1 π b = f(x)sinnx dx n π −π 2 π = f(x)sinnx dx, n =1, 2, 3,.... (B-1.9) π 0 B-1 Fourier Series and Its Basic Properties 721

Fig. B.1 The function f(x) and its extension.

Fig. B.2 Graph of f(x)=x2 and its extension.

Example B-1.1. The Fourier series of f(x) (see Figure B.1) 0 if − π

∞ π 2 (−1)n f(x)= − cos(2n − 1)x + sin nx . (B-1.10) 4 π(2n − 1)2 n n=1

Example B-1.2. The Fourier series of the (see Figure B.2) function

f(x)=x2, −π ≤ x ≤ π with f(x ± 2nπ)=f(x),n=1, 2, 3,..., 722 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Fig. B.3 The triangular wave function and its extension. is ∞ π2 (−1)n x2 = +4 cos nx. (B-1.11) 3 n2 n=1

Since f(x) is even, bn =0for all n ≥ 1, it turns out that 1 π 2 π 2π2 a = x2 dx = x2 dx = , 0 π π 3 −π 0 π π 1 2 2 2 an = x cos nx dx = x cos nx dx π −π π 0 4 = (−1)n,n≥ 1. n2

Example B-1.3. The triangular wave function (see Figure B.3) −x if − π ≤ x<0, f(x)=|x| = x if 0 ≤ x<π, with f(x ± 2πn)=f(x) has the Fourier cosine series representation

∞ π 4 1 f(x)= − cos(2n − 1)x. (B-1.12) 2 π (2n − 1)2 n=1

In this case, f(x) is even, thus, bn =0for all n ≥ 1, and 1 π 2 π a = |x| dx = xdx= π, 0 π π −π 0 1 π 2 π an = |x| cos nx dx = x cos nx dx π −π π 0 2 = (−1)n − 1 , πn2 and so 0 if n is even, a = n − 4 πn2 if n is odd. B-1 Fourier Series and Its Basic Properties 723

Fig. B.4 The sawtooth wave function.

Example B-1.4. The sawtooth wave function (see Figure B.4)

f(x)=x, −π

∞ sin nx f(x)=2 (−1)n+1 . (B-1.13) n n=1

In this case, f(x) is odd and hence, an =0for all n ≥ 0, and π π 1 2 2 n+1 bn = x sin nx dx = x sin nx dx = (−1) . π −π π 0 n Example B-1.5. The Fourier series representation of the square wave function (see Figure B.5) deﬁned by ⎧ ⎪ ≤ ⎨⎪ 1 if 0

Obviously, f(x) is odd, and hence, an =0for all n ≥ 0, and bn is given by 1 π 2 π bn = sgn x sin nx dx = sin nx dx π −π π 0 2 1 − (−1)n 0 if n is even, = = 4 π n nπ if n is odd. 724 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Fig. B.5 The square wave function and its extension.

Fig. B.6 The triangular wave function and its extension.

Example B-1.6. The triangular wave function (see Figure B.6) f(x) on [−π, π] is given by π + x if − π ≤ x ≤ 0, f(x)= π − x if 0 ≤ x ≤ π. Since f(x) is even, b =0for n ≥ 1, and n 1 π 2 π a0 = f(x) dx = (π − x) dx = π, π −π π 0 π 2 − 2 1 − − n an = (π x) cos nx dx = 2 1 ( 1) π 0 π n 2 0 if n is even, = · πn2 2 if n is odd. Thus, the Fourier series for f(x) is ∞ a f(x)= 0 + a cos nx 2 n n=1 ∞ π 4 1 = + cos(2n − 1)x. (B-1.15) 2 π (2n − 1)2 n=1 B-1 Fourier Series and Its Basic Properties 725

If f(x) is a periodic function of period 2l and is deﬁned on [−l, l], then the Fourier representation of f(x) is

∞ a nπx nπx f(x)= 0 + a cos + b sin , (B-1.16) 2 n l n l n=1 where a0, an, and bn are given by the EulerÐFourier formulas: 1 l a0 = f(x) dx, (B-1.17) l − l 1 l nπx an = f(x) cos dx, (B-1.18) l − l l 1 l nπx bn = f(x)sin dx. (B-1.19) l −l l If f(x) is an even function of period 2l deﬁned on [−l, l], then

∞ a nπx f(x)= 0 + a cos , (B-1.20) 2 n l n=1 where bn =0, n ≥ 1, and 2 l nπx an = f(x) cos dx, n =0, 1, 2,.... (B-1.21) l 0 l If f(x) is an odd function of period 2l, then

∞ nπx f(x)= b sin , (B-1.22) n l n=1 where an =0, n ≥ 0, and 2 l nπx bn = f(x)sin dx. (B-1.23) l 0 l The functions f(x) in all Examples B-1.1ÐB-1.6 can be deﬁned on [−l, l] and the corresponding Fourier series can be obtained directly by calculating Fourier co- πt efﬁcients an and bn, or by using the transformation x = l . In Example B-1.1,the Fourier series on [−l, l] is

∞ l l 2 (2n − 1)πx (−1)n nπx f(x)= − cos + sin . 4 π π(2n − 1)2 l n l n=1 (B-1.24)

In Example B-1.2, the Fourier series on [−l, l] is given by 726 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

∞ l2 4l2 (−1)n nπx f(x)= + cos . (B-1.25) 3 π2 n2 l n=1 In Example B-1.3, the Fourier series on [−l, l] is ∞ l 4l 1 xπ f(x)= − cos (2n − 1) . (B-1.26) 2 π2 (2n − 1)2 l n=1 When l = π, this reduces to (B-1.12). In Example B-1.4, the Fourier series on [−l, l] is ∞ 2l (−1)n+1 nπx f(x)= sin . (B-1.27) π n l n=1 In Example B-1.5, the Fourier series of sgn(x) on [−l, l] is ∞ 4 1 xπ sgn(x)= sin (2n − 1) . (B-1.28) π (2n − 1) l n=1 In Example B-1.6, the Fourier series on [−l, l] is ∞ l 4l 1 πx f(x)= + cos(2n − 1) . (B-1.29) 2 π2 (2n − 1)2 l n=1 It is sometimes convenient to represent a function f(x) by a Fourier series in complex form. This expansion can easily be derived from the Fourier series (B-1.1), that is, ∞ a f(x)= 0 + (a cos nx + b sin nx) 2 n n n=1 ∞ a 1 b = 0 + a einx + e−inx + n einx − e−inx 2 2 n 2i n=1 ∞ a 1 1 = 0 + (a − ib )einx + (a + ib )e−inx 2 2 n n 2 n n n=1 ∞ ∞ inx −inx inx = c0 + cne + c−ne = cne , n=1 n=−∞ where π a0 1 c0 = = f(x) dx, 2 2π −π 1 1 π cn = (an − ibn)= f(x)(cos nx − i sin nx) dx 2 2π −π 1 π = f(x)e−inx dx, 2π −π B-1 Fourier Series and Its Basic Properties 727 1 1 π c− = (a + ib )= f(x)(cos nx + i sin nx) dx n 2 n n 2π −π π 1 inx = f(x)e dx = cn. 2π −π Thus, we obtain the Fourier series of f(x) in complex form ∞ inx f(x)= cne , −π

Thus, (B-1.32) is known as the Parseval formula for a complex Fourier series. We next consider the nth partial sum a n s (x)= 0 + (a cos kx + b sin kx), (B-1.33) n 2 k k k=1 − π 2 of the Fourier series for f(x) in (B-1.1) defined on [ π, π].If −π f (x) dx exists and is finite, then π 2 0 ≤ f(x) − sn(x) dx −π π π π 2 − 2 = f (x) dx 2 f(x)sn(x) dx + sn(x) dx. (B-1.34) −π −π −π It follows from the definition of the Fourier coefficients (B-1.2)Ð(B-1.4) and the orthogonality of the cosine and sine functions that π π n a0 f(x)sn(x) dx = f(x) + (ak cos kx + bk sin kx) dx − − 2 π π k=1 πa2 n = 0 + π a2 + b2 . 2 k k k=1 728 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Similarly, it turns out that 2 π π n 2 a0 sn(x) dx = + (ak cos kx + bk sin kx) dx −π −π 2 k=1 π 2 n π π a0 2 2 2 2 = dx + ak cos kx dx + bk sin kx dx − 4 − − π k=1 π π πa2 n = 0 + π a2 + b2 . 2 k k k=1 Consequently, (B-1.34) reduces to π 2 0 ≤ f(x) − sn(x) dx −π π 2 n 2 − πa0 2 2 = f (x) dx + π ak + bk . (B-1.35) − 2 π k=1 This leads to the inequality 2 n π a0 2 2 ≤ 1 2 + ak + bk f (x) dx, (B-1.36) 2 π − k=1 π and since the right-hand side of (B-1.36) is independent of n, it follows in the limit as n →∞that ∞ 2 π a0 2 2 ≤ 1 2 + ak + bk f (x) dx. (B-1.37) 2 π − k=1 π This is known as the Bessel inequality for a Fourier series. Since the left-hand side of (B-1.36) is non-decreasing and is bounded above, the series ∞ a2 0 + a2 + b2 (B-1.38) 2 k k k=1 converges. Thus, the necessary condition for the convergence of the series (B-1.38) is that

lim ak = 0 and lim bk =0. (B-1.39) k→∞ k→∞ That is, π π lim f(x) cos kx dx =0, lim f(x)sinkx dx =0. (B-1.40) →∞ →∞ k −π k −π These results are known as the RiemannÐLebesgue Lemma. B-1 Fourier Series and Its Basic Properties 729

The Fourier series is said to converge in the mean to f(x) when π 2 lim f(x) − s (x) dx =0. (B-1.41) →∞ n n −π If the Fourier series converges in the mean to f(x), then

∞ 2 π a0 2 2 1 2 + an + bn = f (x) dx. (B-1.42) 2 π − n=1 π This is called Parseval’s relation, and it is one of the fundamental results in the theory of Fourier series. This relation can formally be derived from the convergence of the Fourier series to f(x) on [−π, π]. In other words, if

∞ 1 f(x)= a + (a cos nx + b sin nx), (B-1.43) 2 0 n n n=1 where a0, an, and bn are given by (B-1.2)Ð(B-1.4), we multiply by (B-1.43)by 1 − π f(x) and integrate the resulting expression from π to π to obtain 1 π f 2(x) dx π −π a π = 0 f(x) dx 2π −π ∞ 1 π 1 π + an f(x) cos nx dx + bn f(x)sinnx dx . (B-1.44) π − π − n=1 π π Replacing all integrals on the right hand side of (B-1.44) by the Fourier coefﬁcients gives the Parseval relation (B-1.42). If two (2π)-periodic integrable functions f(x) and g(x) deﬁned on [−π, π] have the Fourier series expansions

∞ 1 f(x)= a + (a cos nx + b sin nx), (B-1.45) 2 0 n n n=1 ∞ 1 g(x)= α + (α cos nx + β sin nx), (B-1.46) 2 0 n n n=1 where a0, an, and bn are given by (B-1.1)Ð(B-1.3), and α0, αn, and βn are given by results similar to those of (B-1.1)Ð(B-1.3), then the following Parseval’s relations hold π ∞ 1 2 1 2 2 2 f(x)+g(x) dx = (a0 + α0) + (an + αn) +(bn + βn) , π − 2 π n=1 (B-1.47) 730 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms π ∞ 1 2 1 2 2 2 f(x) − g(x) dx = (a0 − α0) + (an − αn) +(bn − βn) . π − 2 π n=1 (B-1.48)

Subtracting (B-1.48) from (B-1.47) yields

∞ 1 π 1 f(x)g(x) dx = a0α0 + (anαn + bnβn). (B-1.49) π − 2 π n=1

This is a general Parseval relation for the product function f(x)g(x). When f = g, (B-1.49) reduces to (B-1.42). In the context of the complex Fourier series expansion (B-1.30)ofa(2π)- periodic function f(x), we can replace the Fourier coefﬁcient cn by f(n) so that π 1 −inx cn = f(n)= e f(x) dx. (B-1.50) 2π −π

The concept of convolution of two (2π)-periodic integrable functions f and g on R arises naturally, so that we deﬁne their convolution (f ∗ g)(x) on [−π, π] by 1 π (f ∗ g)(x)= f(x − t)g(t) dt. (B-1.51) 2π −π

Physically, the convolution (f ∗ g)(x) represents an integral output of the two functions f and g in contrast to the ordinary pointwise product (output) f(x)g(x). Clearly, the convolution is commutative, that is, 1 π (f ∗ g)(x)= f(ξ)g(x − ξ) dξ =(g ∗ f)(x). (B-1.52) 2π −π Another interpretation of a convolution is that it represents an average (or mean) value. In particular, if g =1in (B-1.52), then f ∗ g is constant and is equal to 1 π (f ∗ 1)(x)= f(ξ) dξ. (B-1.53) 2π −π

This means that (f ∗ 1)(x) represents the average value of f(x) on [−π, π]. In addition to commutativity, the convolution satisﬁes the following algebraic and analytic properties for any constant α and β:

f ∗ (αg + βh)=α(f ∗ g)+β(f ∗ h) (Distributive), (B-1.54)

(f ∗ g) ∗ h = f ∗ (g ∗ h) (Associative), (B-1.55)

(f ∗ g)(x) is continuous (Continuity), (B-1.56)

f∗ g(n)=f(n)g(n)(Convolution). (B-1.57) B-1 Fourier Series and Its Basic Properties 731

To prove (B-1.57), we use the deﬁnition 1 π f∗ g(n)= (f ∗ g)(x)e−inx dx 2π −π 1 π 1 π = e−inx f(t)g(x − t) dt dx 2π −π 2π −π 1 π 1 π = e−intf(t) e−in(x−t)g(x − t) dx dt, x − t = s 2π −π 2π −π 1 π 1 π = e−intf(t) dt g(s)e−ins ds 2π −π 2π −π = f(n)g(n).

The nth partial sum of a complex Fourier series (B-1.30)is n n π ikx ikx 1 −ikt sn(x)= cke = e f(t)e dt 2π − k=−n k=−n π 1 π n = f(t) eik(x−t) dt 2π −π k=−n 1 π = f(t)Dn(x − t) dt, (B-1.58) 2π −π = Dn(x) ∗ f(x), (B-1.59) where Dn(x) is called the Dirichlet kernel deﬁned by n n ikx ikx −ikx Dn(x)= e =1+ e + e k=−n k=1 n =1+ 2 cos kx. (B-1.60) k=1 We next make the following observations regarding the genesis of the convolution in the theory of Fourier. The convolution (Dn ∗f)(x) arises in the nth partial sum sn(x) of the Fourier series of f(x). Thus, the problem of understanding sn(x) reduces to that of (Dn ∗ f)(x). It follows from (B-1.60) that 1 x 1 x x Dn(x)sin = sin +sin (cos x + cos 2x + ···+ cos nx) 2 2 2 2 2 1 x 1 3x x 5x 3x = sin + sin − sin + sin − sin 2 2 2 2 2 2 2 1 1 + ···+ sin n + x − sin n − x 2 2 1 1 = sin n + x. 2 2 732 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Fig. B.7 Graph of Dn(x = θ) against (x = θ).

Thus, the exact form of the Dirichlet kernel is

1 sin(n + 2 )x Dn(x)= x . (B-1.61) sin 2

This reveals that the denominator of the Dirichlet kernel Dn(x) vanishes at the points x =2πm which are removable points of discontinuity. Furthermore, it follows from (B-1.60) that Dn(x) is an even function with period 2π and satisﬁes π Dn(x) dx =2π. (B-1.62) −π

The graph of Dn(x = θ) isshowninFigureB.7. It looks very similar to that of the diffusion kernel as shown in Figure 1.6 in Chapter 1 except for its symmetric oscillatory trail in both positive and negative θ-axes. It follows from (B-1.58), by putting t−x = ξ and noting that Dn(ξ) is even, that 1 π−x s (x)= f(x + ξ)D (ξ) dξ n 2π n −π−x 1 π = f(x + ξ)Dn(ξ) dξ. (B-1.63) 2π −π We close this section by stating the Pointwise Convergence Theorem:Iff(x) is a piecewise smooth and periodic function with period 2π on [−π, π], then, for any x in (−π, π), n a0 lim sn = lim + (ak cos kx + bk sin kx) n→∞ n→∞ 2 k=1 1 = f(x +0)+f(x − 0) , (B-1.64) 2 B-1 Fourier Series and Its Basic Properties 733 or equivalently,

∞ a 1 0 + (a cos kx + b sin kx)= f(x+) + f(x−) , (B-1.65) 2 k k 2 k=1 where 1 π a = f(x) cos kx dx, k =0, 1, 2, 3,..., (B-1.66) k π −π 1 π bk = f(x)sinkxdx, k =1, 2, 3,.... (B-1.67) π −π We refera to Myint-U and Debnath (2007) for its proof. Another remarkable feature of Fourier series of a function at its ordinary points of discontinuity deals with the behavior of its nth partial sums sn(x) as n →∞.At points where f(x) is continuous, the nth partial sums sn(x) approach smoothly the value f(x) as n →∞. However, for the functions f(x)=x in Example B-1.4 or f(x)=sgnx in Example B-1.5, the graphs of their partial sums exhibit a large error in the neighborhood of points of discontinuity at x =0and x = ±π independent of the number of terms in their partial sums. In other words, the partial sums do not converge smoothly to the mean value. Instead, they overshoot the mark at each end of the jumps of the function. The explanation of this phenomenon was ﬁrst provided by J.W. Gibbs (1839Ð1903) in 1899, who showed that overshooting was not the result of computational errors. This feature is typical for the Fourier series of a function at the points of discontinuity, and is now universally known as the Gibbs phenomenon. One of the most effective and useful applications of Fourier series deals with the summation of inﬁnite series in closed form which is one of the major problems in mathematics. We shall use Examples B-1.1ÐB-1.6 to derive the sums of many important numerical series. Substituting x =0in (B-1.10) gives the numerical series ∞ π 2 1 0=f(0) = − . 4 π (2n − 1)2 n=1 Hence, ∞ 1 π2 = , (2n − 1)2 8 n=1 or equivalently, 1 1 1 1 π2 + + + + ···= . (B-1.68) 12 32 52 72 8 This can be used to obtain another numerical series 1 1 1 1 1 π2 S = + + + + ···+ + ···= . (B-1.69) 12 22 32 42 n2 6 734 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

In fact, 1 1 1 1 1 1 1 1 S = + + + ··· + + + + + ··· 12 32 52 4 12 22 32 42 π2 1 = + S. 8 4

− 1 π2 Thus, S(1 4 )= 8 , which gives (B-1.69). Putting x =0in (B-1.11) leads to the numerical series

∞ π2 (−1)n 0= +4 , 3 n2 n=1 or equivalently,

∞ (−1)n+1 π2 = . n2 12 n=1 Thus,

1 1 1 1 π2 − + − + ···= . (B-1.70) 12 22 32 42 12 π Substituting x = 2 in the series (B-1.13)gives ∞ π (−1)n+1 nπ =2 sin 2 n 2 n=1 sin π sin 2π sin 3π sin 4π =2 2 − 2 + 2 − 2 + ··· 1 2 3 4 1 1 1 =2 1 − + − + ··· . 3 5 7

Therefore, 1 1 1 π 1 − + − + ···= . (B-1.71) 3 5 7 4 This is celebrated Leibniz series for π discovered by Leibniz in 1673. It is also known as the Gregory series independently discovered by James Gregory (1638Ð1675) in around 1670. π Putting x = 4 in (B-1.13) gives another numerical series π 1 1 1 1 1 1 1 1 1 1 = √ 1+ − − + + −··· − 1 − + − + ··· . 8 2 3 5 7 9 11 2 3 5 7 In view of (B-1.71), this leads to the following series B-1 Fourier Series and Its Basic Properties 735 1 1 1 1 1 π 1+ − − + + −···= √ . (B-1.72) 3 5 7 9 11 2 2 Subtracting (B-1.70) from (B-1.69) yields 1 1 1 1 π2 + + + + ···= . (B-1.73) 22 42 62 82 24 In Example B-1.2, the Fourier series for f(x)=x2 isgivenby(B-1.11). It follows from the Parseval relation (B-1.42) that ∞ ∞ π 4 4 1 4 2π 2 2π 16 x dx = + an = + 4 , π − 9 9 n π n=1 n=1 or equivalently, ∞ 2π4 2π4 16 = + . 5 9 n4 n=1 Thus, ∞ 1 π4 = . (B-1.74) n4 90 n=1 If we apply the Parseval formula (B-1.42) to Example B-1.3, we can derive the following numerical series ∞ 1 π4 = . (B-1.75) (2n − 1)4 96 n=1 A similar calculation can be used for the Fourier series of f(x)=x(π − x), 0

∞ nπx f(x)= b sin , (B-1.80) n l n=1 where 2 l nπx b = f(x)sin dx n l l 0 2 l/2 2hx nπx l 2h(l − x) nπx = sin dx + sin dx , l 0 l l l/2 l l which is, integrating by parts, l/2 l/2 4h − l nπx − nπx = 2 x cos cos dx l nπ l 0 l 0 l l nπx nπx +(l − x) cos + cos dx l l l/2 l/2 8h nπ = sin . (B-1.81) n2π2 2

Thus, when h =1and l = π, the Fourier sine series on the interval 0

π Thus, the Fourier sine series for f(x)=x on 0

π Putting x = 2 in (B-1.83) gives the numerical series π2 1 1 1 1 = + + + + ··· . (B-1.84) 8 12 32 52 72 π The Fourier sine series (B-1.83)forx can be integrated from x to 2 term-by-term so that 1 π2 4 1 1 − x2 = cos x − cos 3x + cos 5x −··· . (B-1.85) 2 4 π 33 53

Substituting x =0into (B-1.85) gives the numerical series B-1 Fourier Series and Its Basic Properties 737

1 1 1 1 π3 − + − + ···= . (B-1.86) 13 33 53 73 32 Integrating (B-1.85) with respect to x from 0 to x leads to 1 π2 x3 4 1 1 x − = sin x − sin 3x + sin 5x −··· . (B-1.87) 2 4 3 π 34 54

π Putting x = 2 in (B-1.87) yields the numerical series 1 1 1 1 π π3 π4 + + + + ···= · = . (B-1.88) 14 34 54 74 8 12 96 π Integrating (B-1.87) again from x to 2 gives π2 π 2 1 π 4 − x2 − − x4 16 2 24 2 4 1 1 = cos x − cos 3x + cos 5x −··· . (B-1.89) π 35 55 Substituting x =0in (B-1.89) gives the numerical series

1 1 1 1 5π5 − + − + ···= . (B-1.90) 15 35 55 75 1536 Integrating (B-1.89) again from 0 to x yields a new numerical series π4 π2 π4 1 x − x3 − x + x5 64 48 384 120 4 1 1 = sin x − sin 3x + sin 5x −··· . (B-1.91) π 36 56

π Putting x = 2 in (B-1.91) leads to another numerical series 1 1 1 π6 1+ + + + ···= . (B-1.92) 36 56 76 960 We consider applications of Fourier series to differential equations. As a simple application, we discuss the periodic solution of a non-homogeneous simple harmonic oscillator

x¨ + ω2x = f(t), (B-1.93) where the forcing function f(t) has a Fourier series expansion

∞ 1 f(t)= A + (A cos nt + B sin nt), (B-1.94) 2 0 n n n=1 the coefﬁcients A0, An, and Bn are known. 738 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

We seek a uniformly convergent Fourier series solution of (B-1.93) in the form

∞ 1 x(t)= a + (a cos nt + b sin nt). (B-1.95) 2 0 n n n=1 Differentiating (B-1.95) term by term and assuming the series for x˙(t) and x¨(t) converge uniformly, substituting in (B-1.93)gives

∞ ∞ a −n2 + ω2 a cos nt + 0 ω2 + −n2 + ω2 b sin nt n 2 0 n n=1 n=1 ∞ 1 = A + (A cos nt + B sin nt). (B-1.96) 2 0 n n n=1 Consequently, equating the coefﬁcients, we obtain A A B a = 0 ,a= n ,b= n . 0 ω2 n ω2 − n2 n ω2 − n2 Thus, the solution of (B-1.93) is given by ∞ A (A cos nt + B sin nt) x(t)= 0 + n n (B-1.97) 2ω2 ω2 − n2 n=1 provided ω = n. However, if ω2 = n2 for some integer n, the solution becomes unbounded. This phenomenon is known as resonance. If damping is included in equation (B-1.93), the solution will remain bounded. However, when ω2 = n2, we can write

g(t)=f(t) − Aω cos ωt + Bω sin ωt,

2 2 and then we can solve separately x¨ + ω x = g(t) and x¨ + ω x = Aω cos ωt + Bω sin ωt. The second equation gives rise to the non-periodic solutions (At+B) cos ωt and (At+B)sinωt. The sum of the solutions of the two equations gives a solution of the original equation. Equation (B-1.93) can be solved by the use of integrating factor in the form t τ x(t)=exp(−iωt) exp(2iωτ) eiωxf(x) dx dτ. (B-1.98) 0 0 We next apply the method of Fourier series to solve the damped harmonic oscillator governed by x¨ + kx˙ + ω2x = f(t), (B-1.99) where kx˙ (k>0) is the damping term. If f(t) is periodic with period 2π and has the same Fourier series expansion (B-1.94), then we can ﬁnd a Fourier series solution of (B-1.95) by differentiating and substituting in (B-1.99) so that equation (B-1.99) takes the form B-1 Fourier Series and Its Basic Properties 739

∞ ω2a −n2a + nkb + ω2a cos nt + 0 n n n 2 n=1 ∞ 2 2 + −n bn − nkan + ω bn sin nt n=1 ∞ 1 = A + (A cos nt + B sin nt). (B-1.100) 2 0 n n n=1

A0 Equating the coefﬁcients from both sides gives a0 = ω2 and ω2 − n2 a + nkb = A , n n n 2 2 −nkan + ω − n bn = Bn.

Solving for an and bn gives An nk B ω2 − n2 n 1 2 − 2 − an = = ω n An nkBn , (B-1.101) ω2 − n2 nk D −nk ω2 − n2 2 2 ω − n An −nk B n 1 2 − 2 bn = = ω n Bn + nkAn , (B-1.102) ω2 − n2 nk D −nk ω2 − n2 where 2 D = ω2 − n2 + n2k2. (B-1.103)

Thus, the coefﬁcients a0, an, and bn of the Fourier series solution (B-1.95)ofthe damped simple harmonic equation (B-1.99)aregivenby(B-1.101)Ð(B-1.102), and the solution represents the periodic response. There may be a transient response de- pending on the initial conditions, and the transient solution will eventually decay as t →∞. Because of the presence of the damping term, there will be no resonance when n = ω. We close this section by adding an example of an application of Fourier series to a simple boundary value problem d2u = λu = f(x), 0

∞ nπx f(x)= b sin , 0

∞ ∞ n2π2 nπx nπx − + λ B sin = b sin , 0

Consequently, the coefﬁcients Bn are given by − n2π2 1 B = λ − b , (B-1.108) n l2 n

nπ 2 provided λ =( l ) . nπ 2 If λ =(l ) for all or some values of positive integer n, then Bn cannot be determined, and hence no solution exists unless Bn ≡ 0. Thus, the Fourier series solution of the boundary value problem is ∞ l2b nπx u(x)= n sin , (B-1.109) λl2 − n2π2 l n=1 where the zero denominator must be handled separately.

B-2 Generalized Functions (Distributions) The most widely known example of a generalized function is the Dirac delta function δ(x) which was ﬁrst introduced by P.M.M. Dirac in 1920s as a mathematical device in the formulation of quantum mechanics. The Dirac delta function δ(x) is deﬁned by ∞ δ(x) = 0 for all x =0 , and δ(x) dx =1. (B-2.1) −∞

In the ordinary sense, δ(x) cannot be considered as a function because if a function vanishes everywhere except at a single point x =0, then its integral over any interval must be zero, so that integrating it over an interval including the origin cannot be equal to one. However, from the physical point of view, the delta function is the natural mathematical quantity which can be used to describe many of the ab- stractions which arise in physical sciences. For example, the mass-density function B-2 Generalized Functions (Distributions) 741

Fig. B.8 The sequence of delta functions, δn(x). is zero everywhere except at x =0, where it is infinite because a finite mass is con- centrated in zero length, and it is so infinitely large here that the integral is non-zero even through the integrand is positive over an infinitesimally small region only. This makes sense physically, though it is mathematically absurd. So the delta function δ(x) can be used as the mass-density function ρ(x) describing the mass distribution per unit length of a rod at a point x. Similarly, the point charge, the impulsive force, the point dipole, and the frequency response of an undamped harmonic oscillator are all aptly represented by the delta function or other generalized functions. In general, the generalized functions play a major role in Fourier analysis and in the theory of partial differential equations. The function f(x)=1has no Fourier √transform in the ordinary Fourier transform theory, but it has Fourier transform 2πδ(k) in the generalized function theory. Thus, the generalized functions remove difficulties which existed in the classical Fourier analysis. There is a beautiful analogy with the way in which the use of complex numbers helps in the solution of quadratic equations as within the realm of real numbers only certain quadratic equations have no solutions. Sometimes, the Dirac delta function δ(x) is defined by its fundamental property

∞ f(x)δ(x − a) dx = f(a), (B-2.2) −∞ where f(x) is any continuous function in any interval containing the point x = a. Although there are no ordinary functions which have the properties required of the delta function, we may approximate δ(x) in one dimension by a sequence of ordinary functions n δ (x)= exp −nx2 ,n=1, 2, 3,.... (B-2.3) n π 742 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Clearly, δn(x) → 0 as n →∞for any x =0 and δn(0) →∞as n →∞as showninFigureB.8. Also, for all n =1, 2, 3,..., ∞

δn(x) dx =1 −∞ and ∞ ∞

lim δn(x) dx = δ(x) dx =1, n→∞ −∞ −∞ as expected. Thus, the delta function can be considered as the limit of a sequence of ordinary Gaussian functions, and we write n δ(x) = lim exp −nx2 . (B-2.4) n→∞ π As mentioned in Chapter 1, the major problem of ﬁnding the solution of inhomogeneous partial differential equations deals with the construction of the Green’s function in each case. This problem becomes easier by the use of a generalized function together with the methods of Fourier and Laplace transforms. In spherical polar coordinates (r, θ, φ), the Laplacian is 1 d dψ ∇2ψ(r)= r2 , (B-2.5) r2 dr dr

1 ∇2 which is independent of θ and φ.Ifψ(r)= r , then ψ(r)=0for all r =0.At r =0, ∇2ψ(r) does not exist. By the divergence theorem (or Gauss’ theorem) 1 1 1 ∇2 dV = div ∇ dV = ∇ · ndS, (B-2.6) V r V r S r where V is the volume of a sphere of radius a and center at r =0so that ∇ 1 d 1 − eˆr =ˆer = 2 , (B-2.7) r r=a dr r r=a a where eˆr is the direction of the outward normal n to the surface of the sphere. It follows from (B-2.6) that ∇2 1 − 1 · − dV = 2 eˆr nds= 4π. (B-2.8) V r a S ∇2 1 Thus, the Laplacian ( r ) is a function of r which has the following fundamental properties: (i) It vanishes for r =0 . (ii) It is not deﬁned at r =0. (iii) And its integral over any sphere with center at r =0is −4π. B-2 Generalized Functions (Distributions) 743

All of these lead to the result 1 −4π if V contains r =0, ∇2 dV = (B-2.9) V r 0 if V does not contain r =0.

This may be written in the compact form 1 ∇2 = −4πδ(r), (B-2.10) r where δ(r) is the Dirac delta function with the property 1 if V contains r =0, δ(r) dV = (B-2.11) V 0 otherwise.

This result is a special case of the general definition of the vector form of the delta function f(a) if V contains the point a, f(r)δ(r − a) dr = (B-2.12) V 0 otherwise, obtained by setting f(r)=1and a = 0 in (B-2.12). For the results (B-2.11) and (B-2.12) to be valid, it is certainly sufficient that the functions be continuous and infinitely differentiable everywhere. Result (B-2.12) reduces to (B-2.2) in one dimension. We would not go into great detail, but refer to the famous books of Lighthill (1958) and Jones (1982) for the introduction to the subject of generalized functions. A good function, g(x), is a function in C∞(R) that decays sufficiently rapidly so that g(x) and all of its derivatives decay to zero faster than |x|−N as |x|→∞for all N>0. In other words, suppose that for each positive integer N and n, lim xN g(n)(x)=0, (B-2.13) |x|→∞ then g(x) is called a good function. Usually, the class of good functions is represented by S. The good functions play an important role in Fourier analysis because the inversion, convolution, and differentiation theorems as well as many others take simple forms with no problem of convergence. The rapid decay and infinite differentiability properties of good functions lead to the fact that the Fourier transform of a good function is also a good function. Good functions also play an important role in the theory of generalized functions. A good function of bounded support is a special type of good function that also plays an important part in the theory of generalized functions. Good functions also have the following important properties. The sum (or difference) of two good functions is also a good function. The product and convolution of two good functions are good functions. The derivative of a good function is a good function; xn g(x) is a good 744 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms function for all non-negative integers n whenever g(x) is a good function. A good function belongs to Lp (a class of pth power Lebesgue integrable functions) for every p in 1 ≤ p ≤∞. The integral of a good function is not necessarily good. However, if φ(x) is a good function, then the function g defined for all x by x g(x)= φ(t) dt −∞ ∞ is a good function if and only if −∞ φ(t) dt exists. Good functions are not only continuous, but are also uniformly continuous on R and absolutely continuous on R. However, a good function cannot necessarily be represented by a Taylor series expansion on every interval. As an example, consider a good function of bounded support exp[−(1 − x2)−1] if |x| < 1, g(x)= 0 if |x|≥1. The function g is infinitely differentiable at x = ±1, as it must be in order to be good. It does not have a Taylor series expansion in every interval because a Taylor expansion based on the various derivatives of g at any point x satisfying |x| > 1 would lead to zero value for all x. For example, exp(−x2), x exp(−x2), (1 + x2)−1 exp(−x2), and sech2x are good functions, while exp(−|x|) is not differentiable at x =0, and the function (1 + x2)−1 is not a good function as it decays too slowly as |x|→∞. A sequence of good functions, {fn(x)}, is called regular if, for any good function g(x), ∞

lim fn(x)g(x) dx (B-2.14) n→∞ −∞

1 exists. For example, fn(x)= n φ(x) is a regular sequence for any good function φ(x), since ∞ 1 ∞ lim fn(x)g(x) dx = lim φ(x)g(x) dx =0. n→∞ −∞ n→∞ n −∞ Two regular sequences of good functions are equivalent if, for any good function g(x), the limit (B-2.14) exists and is the same for each sequence. A generalized function, f(x), is a regular sequence of good functions, and two generalized functions are equal if their deﬁning sequences are equivalent. General- ized functions are, therefore, only deﬁned in terms of their action on integrals of good functions if ∞ ∞

f,g = f(x)g(x) dx = lim fn(x)g(x) dx = lim fn,g (B-2.15) −∞ n→∞ −∞ n→∞ for any good function g(x), where the symbol f, g is used to denote the action of the generalized function f(x) on the good function g(x),orf, g represents the number that f associates with g.Iff(x) is an ordinary function such that B-2 Generalized Functions (Distributions) 745

(1 + x2)−N f(x) is integrable on (−∞, ∞) for some N, then the generalized function f(x) equivalent to the ordinary function is deﬁned as any sequence of good functions {fn(x)} such that, for any good function g(x), ∞ ∞

lim fn(x)g(x) dx = f(x)g(x) dx. (B-2.16) n→∞ −∞ −∞

For example, the generalized function equivalent to zero can be represented by either { φ(x) } { φ(x) } of the sequences n and n2 . The unit function, I(x), is defined by ∞ ∞ I(x)g(x) dx = g(x) dx (B-2.17) −∞ −∞ for any good function g(x). A very important and useful good function that defines { − x2 } the unit function is exp( 4n ) . Thus, the unit function is the generalized function that is equivalent to the ordinary function f(x)=1. The Heaviside function, H(x), is defined by ∞ ∞ H(x)g(x) dx = g(x) dx. (B-2.18) −∞ 0 The generalized function H(x) is equivalent to the ordinary unit function 0 if x<0, H(x)= (B-2.19) 1 if x>0, and since generalized functions are defined through the action on integrals of good functions, the value of H(x) at x =0does not have significance here. The sign function, sgn(x), is defined by ∞ ∞ 0 sgn(x)g(x) dx = g(x) dx − g(x) dx (B-2.20) −∞ 0 −∞ for any good function g(x). Thus, sgn(x) can be identified with the ordinary function −1 if x<0, sgn(x)= (B-2.21) +1 if x>0.

In fact, sgn(x)=2H(x) − I(x), which can be seen as follows: ∞ ∞ sgn(x)g(x) dx = 2H(x) − I(x) g(x) dx −∞ −∞ ∞ ∞ =2 H(x)g(x) dx − I(x)g(x) dx −∞ −∞ ∞ ∞ =2 g(x) dx − g(x) dx 0 −∞ 746 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms ∞ 0 = g(x) dx − g(x) dx. 0 −∞ The following results are also true xδ(x)=0, (B-2.22) δ(x − a)=δ(a − x). (B-2.23)

Result (B-2.23) shows that δ(x) is an even function. Clearly, the result x 1 if x>0, δ(y) dy = = H(x) −∞ 0 if x<0 shows that d H(x)=δ(x). (B-2.24) dx The Fourier transform of the Dirac delta function is ∞ 1 1 F δ(x) = √ e−ikxδ(x) dx = √ . (B-2.25) 2π 2π −∞

Hence, 1 1 ∞ δ(x)=F −1 √ = eikx dk. (B-2.26) 2π 2π −∞ This is an integral representation of the delta function extensively used in quantum mechanics. Also, (B-2.26) can be rewritten as 1 ∞ δ(k)= eikx dx. (B-2.27) 2π −∞

The Dirac delta function, δ(x), is deﬁned so that for any good function g(x), ∞ δ, g = δ(x)g(x) dx = g(0). (B-2.28) −∞

Derivatives of generalized functions are defined by the derivatives of any equivalent sequences of good functions. We can integrate by parts using any member of the sequences and assuming g(x) vanishes at infinity. We can obtain this definition as follows: ∞ f ,g = f (x)g(x) dx −∞ ∞ ∞ − − = f(x)g(x) −∞ f(x)g (x) dx = f,g . (B-2.29) −∞

The derivative of a generalized function f is the generalized function f defined by B-2 Generalized Functions (Distributions) 747 f ,g = − f,g (B-2.30) for any good function g. The differential calculus of generalized functions can easily be developed with locally integrable functions. To every locally integrable function f, there corresponds a generalized function (or distribution) defined by ∞ f,φ = f(x)φ(x) dx, (B-2.31) −∞ where φ is a test function on R → C with bounded support (φ is infinitely differentiable and such that its derivatives of all orders exist and are continuous). The derivative of a generalized function f is the generalized function f defined by f ,φ = − f,φ (B-2.32) for all test functions φ. This definition follows from the fact that ∞ f ,φ = f (x)φ(x) dx −∞ ∞ ∞ − − = f(x)φ(x) −∞ f(x)φ (x) dx, = f,φ −∞ which was obtained from integration by parts and using the fact that φ vanishes at infinity. It is easy to check that H(x)=δ(x),for ∞ ∞ H,φ = H(x)φ(x) dx = − H(x)φ(x) dx −∞ −∞ ∞ − − ∞ = φ (x) dx = φ(x) 0 = φ(0) = δ, φ . 0 Another result is ∞ δ,φ = − δ(x)φ(x) dx = −φ(0). −∞

It is easy to verify that f(x)δ(x)=f(0)δ(x). We next deﬁne |x| = xsgn(x) and calculate its derivative as follows. We have d d d dx |x| = xsgn(x) = x sgn(x) +sgn(x) dx dx dx dx d = x 2H(x) − I(x) +sgn(x) dx =2xδ(x)+sgn(x)=sgn(x), (B-2.33) which is true since sgn(x)=2H(x) − I(x) and xδ(x)=0. 748 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Similarly, we can show that d sgn(x) =2H(x)=2δ(x). (B-2.34) dx Theorem B-2.1. The Fourier transform of a good function is a good function.

Proof. The Fourier transform of a good function f(x) exists and is deﬁned by 1 ∞ F f(x) = F (k)=√ e−ikxf(x) dx. (B-2.35) 2π −∞

Differentiating F (k) n times and integrating N times by parts, we get − N ∞ N (n) ≤ ( 1) √1 −ikx d − n F (k) N e N ( ix) f(x) dx (−ik) 2π −∞ dx ∞ N ≤ 1 √1 d n N N x f(x) dx. |k| 2π −∞ dx

Evidently, all derivatives tend to zero as fast as |k|−N as |k|→∞for any N>0, and hence F (k) is a good function.

Theorem B-2.2. If f(x) is a good function with the Fourier transform (B-2.35), then the inverse Fourier transform is given by 1 ∞ f(x)=√ eikxF (k) dk. (B-2.36) 2π −∞ Proof. For any >0,wehave ∞ ∞ 2 1 2 F e− x F (−x) = e−ikx− x eixtf(t) dt dx. 2π −∞ −∞

Since f is a good function, the order of integration can be interchanged to obtain ∞ ∞ 2 1 2 F e− x F (−x) = f(t) dt e−i(k−t)x− x dx, 2π −∞ −∞ which is, by a similar calculation as that used in Example 1.7.1 in Chapter 1, 1 ∞ (k − t)2 = √ exp − f(t) dt. 4π −∞ 4 Using the fact that 1 ∞ (k − t)2 √ exp − dt =1, 4π −∞ 4 we can write B-2 Generalized Functions (Distributions) 749

2 F e− x F (−x) − f(k) · 1 1 ∞ (k − t)2 = √ f(t) − f(k) exp − dt. (B-2.37) 4π −∞ 4 Since f is a good function, we have f(t) − f(k) ≤ max f (x) . t − k x∈R

It follows from (B-2.37) that

2 F e− x F (−x) − f(k) 1 ∞ (t − k)2 ≤ √ maxf (x) |t − k| exp − dt x∈R 4 4π −∞ ∞ 1 2 = √ maxf (x)4 |α|e−α dα → 0 4π x∈R −∞

→ t−√k as 0, where α = 2 . Consequently, 1 ∞ f(k)=F F (−x) = √ e−ikxF (−x) dx 2π −∞ 1 ∞ = √ eikxF (x) dx 2π −∞ 1 ∞ ∞ = eikx dx e−iξxf(ξ) dξ. 2π −∞ −∞

Interchanging k with x, this reduces to the celebrated Fourier integral formula 1 ∞ ∞ f(x)= eikx e−ikξf(ξ) dξ dk. (B-2.38) 2π −∞ −∞

Hence, the theorem is proved.

Example B-2.1. The Fourier transform of a constant function c is √ F{c} = 2πcδ(k). (B-2.39)

In the ordinary sense, c ∞ F{c} = √ e−ikx dx 2π −∞ is not a well deﬁned integral; it diverges. However, it can be treated as a generalized { − x2 } function, namely taking c = cI(x) and considering exp( 4n ) as an equivalent sequence to the unit function I(x) gives 750 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

Fig. B.9 Graph of the function fa(x).

x2 c ∞ x2 F c exp − = √ exp −ikx − dx, 4n 2π −∞ 4n which, by Example 1.7.1,is √ √ n = c 2n exp −nk2 = 2πc exp −nk2 , √ √ π = 2πcδn(k) → 2πcδ(k)asn →∞, { } { n − 2 } since δn(k) = π exp( nk ) is a sequence equivalent to the delta function deﬁned by (B-2.4).

Example B-2.2. Show that 1 F e−axH(x) = √ ,a>0. (B-2.40) 2π(ik + a)

We have, by deﬁnition, 1 ∞ 1 F e−axH(x) = √ exp −x(ik + a) dx = √ . 2π 0 2π(ik + a) Example B-2.3. By considering the function (see Figure B.9)

−ax ax fa(x)=e H(x) − e H(−x),a>0, (B-2.41)

ﬁnd the Fourier transform of sgn(x). In Figure B.9, the vertical axis (y-axis) represents fa(x) and the horizontal axis represents the x-axis. B-3 Basic Properties of the Fourier Transforms 751

We have, by deﬁnition, 1 0 1 ∞ F fa(x) = −√ exp (a − ik)x dx + √ exp −(a + ik)x dx 2π −∞ 2π 0 1 1 1 2 (−ik) = √ − = · . 2π a + ik a − ik π a2 + k2 In the limit as a → 0,f (x) → sgn(x) and then a 2 1 F sgn(x) = · . (B-2.42) π ik

B-3 Basic Properties of the Fourier Transforms

The Fourier transform of f(x), F{f(x)} = F (k), is deﬁned by (1.7.1) and its inverse F −1{F (k)} = f(x) is deﬁned by (1.7.2). We state the following properties of the Fourier transform: (a) (Shifting) F f(x − a) = e−iakF f(x) , (B-3.1) (b) (Scaling) 1 k F f(ax) = F , (B-3.2) |a| a (c) (Conjugate) F f(−x) = F f(x) , (B-3.3) (d) (Translation) F eiaxf(x) = F (k − a), (B-3.4) (e) (Duality) F f(x) = F (k), and F F (x) = f(−k), (B-3.5) (f) (Composition) ∞ ∞ F (k)g(k)eikx dk = f(ξ)G(ξ − x) dξ, (B-3.6) −∞ −∞ where F (k)=F f(ξ) and G(ξ)=F g(k) .

(g) (RiemannÐLebesgue Lemma) If F{f(x)} = F (k), then lim F (k)=0. (B-3.7) |k|→∞ 752 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

(h) (Poisson Summation Formula)

∞ √ ∞ 2π nπ f(2πn)= F . (B-3.8) 2a a n=−∞ n=−∞

When a = π,

∞ ∞ 1 f(2πn)=√ F (n). (B-3.9) n=−∞ 2π n=−∞

When 2a =1,formula(B-3.8) becomes ∞ √ ∞ f(n)= 2π F (n). (B-3.10) n=−∞ n=−∞

The convolution (f ∗ g)(x) of two integrable function f(x) and g(x) is deﬁned by (1.7.13) and the convolution Theorem 1.7.1 states that F (f ∗ g)(x) = F (k)G(k), (B-3.11) or equivalently, F −1 F (k)G(k) =(f ∗ g)(x). (B-3.12)

Proof. We have 1 ∞ F (f ∗ g)(x) = √ e−ikx (f ∗ g)(x) dx, 2π −∞ which is, by deﬁnition of the convolution, 1 ∞ ∞ = e−ikx f(x − ξ)g(ξ) dξ dx 2π −∞ −∞ 1 ∞ ∞ = e−ikξg(ξ) dξ e−ik(x−ξ)f(x − ξ) dξ 2π −∞ −∞ 1 ∞ ∞ = e−ikξg(ξ) dξ e−ikηf(η) dη, (x − ξ = η) 2π −∞ −∞ = F (k)G(k).

This completes the proof. It is often convenient to delete the factor √1 in the deﬁnition of convolution 2π (1.7.13) as this factor does not have any effect on the properties of the convolution. The convolution has the following algebraic properties:

f ∗ g = g ∗ f (Commutative), (B-3.13) f ∗ (g ∗ h)=(f ∗ g) ∗ h (Associative), (B-3.14) B-3 Basic Properties of the Fourier Transforms 753

(αf + βg) ∗ h = α(f ∗ h)+β(g ∗ h) (Distributive), (B-3.15) f ∗ δ = f = δ ∗ f (Identity), (B-3.16) where α and β are any two constants, and δ(x) is the Dirac delta function. ∗ ∗ (f g)(x)=(f g)(x), (B-3.17) x (f ∗ g)(x) = xf(x) ∗ g(x)+ f(x) ∗ xg(x) , (B-3.18) 1 f(ax + b) ∗ g(ax + c)= h(ax + b + c), (B-3.19) |a| where h(x)=(f ∗ g)(x). In particular,

f(x + b) ∗ g(x + c)=h(x + b + c). (B-3.20)

If a1, a2, b1, b2 are any constants and f1 ∗ g1, f1 ∗ g2, f2 ∗ g1, and f2 ∗ g2 exist, then

(a1f1 + a2f2) ∗ (b1g1 + b2g2)=a1b1(f1 ∗ g1)+a1b2(f1 ∗ g2)

+ a2b1(f2 ∗ g1)+a2b2(f2 ∗ g2). (B-3.21)

To prove (B-3.14), we have ∞ f ∗ (g ∗ h) (x)= f(x − ξ)(g ∗ h)(ξ) dξ −∞ ∞ ∞ = f(x − ξ) dξ g(ξ − t)h(t) dt −∞ −∞ ∞ ∞ = f(x − ξ)g(ξ − t) dξ h(t) dt, (ξ − t = η) −∞ −∞ ∞ ∞ = f(x − t − η)g(η) dη h(t) dt, −∞ −∞ ∞ = (f ∗ g)(x − t)h(t) dt = (f ∗ g) ∗ h (x). −∞

To prove (B-3.18), we apply the Fourier transform to the left hand side so that ∞ F x(f ∗ g)(x) = e−ikxx(f ∗ g)(x) dx −∞ ∞ ∞ = xe−ikx dx f(x − ξ)g(ξ) dξ −∞ −∞ ∞ ∞ = (x − ξ + ξ)e−ikx dx f(x − ξ)g(ξ) dξ −∞ −∞ ∞ ∞ = (x − ξ)e−ikx dx f(x − ξ)g(ξ) dξ −∞ −∞ 754 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms ∞ ∞ + e−ikx dx f(x − ξ)ξg(ξ) dξ −∞ −∞ ∞ ∞ = e−ikx dx (x − ξ)f(x − ξ)g(ξ) dξ −∞ −∞ ∞ ∞ + e−ikx dx f(x − ξ)ξg(ξ) dξ −∞ −∞ ∞ = e−ikx xf(x) ∗ g(x) dx −∞ ∞ + e−ikx f(x) ∗ xg(x) dx −∞ = F xf(x) ∗ g(x) + F f(x) ∗ xg(x) . We next apply F −1 to obtain result (B-3.18). To prove (B-3.19), we apply the Fourier transform to its left-hand side and use the convolution theorem so that F f(ax + b) ∗ g(ax + c) = F f(ax + b) F g(ax + c) 1 ikb k 1 ikc k = exp F · exp G |a| a a |a| a a 1 ik(b + c) k = · exp H , since H(k)=F (k)G(k) |a|2 a a 1 = F h(ax + b + c) . |a| We next apply F −1 to both sides to obtain (B-3.19). Theorem B-3.1 (General Parseval’s Relation). If F{f(x)} = F (k) and F{g(x)} = G(k), then ∞ ∞ f(x)g(x) dx = F (k)G(k) dk. (B-3.22) −∞ −∞ Proof. We proceed formally to obtain ∞ 1 ∞ ∞ ∞ f(x)g(x) dx = dx eikxF (k) dk e−ilxG(l) dl 2π −∞ −∞ −∞ −∞ ∞ ∞ 1 ∞ = F (k) dk G(l) dl ei(k−l)x dx −∞ −∞ 2π −∞ ∞ ∞ = F (k) dk G(l)δ(k − l) dl −∞ −∞ ∞ = F (k)G(k) dk. −∞ Thus, the proof is complete. B-3 Basic Properties of the Fourier Transforms 755

In particular, if f(x)=g(x), then ∞ ∞ f(x)f(x) dx = F (k)F (k) dk, −∞ −∞ or equivalently, ∞ ∞ |f(x)|2 dx = |F (k)|2 dk. (B-3.23) −∞ −∞

This is well known as the Parseval relation. The function space L2(R) of all complex-valued Lebesgue square integrable functions with the inner product (f,g) deﬁned by ∞ (f,g)= f(x)g(x) dx (B-3.24) −∞ is a complex Hilbert space with the norm f2 deﬁned by

∞ 1 2 2 f2 = (f,f)= f(x) dx . (B-3.25) −∞

In terms of this norm, the Parseval relation (B-3.23) takes the form

f2 = F 2 = Ff2. (B-3.26)

This means that the Fourier transform is a unitary transformation on the Schwartz space S(R) which consists of the set of all inﬁnitely differentiable functions f so that f and all its derivatives are rapidly decreasing in the sense that sup |x|mf (n)(x) < ∞ for every m, n ≥ 0. x∈R

Physically, the quantity f2 is a measure of energy and F 2 represents the power spectrum of f. The following examples of the Fourier transforms are useful in applied mathematics (see Debnath and Bhatta 2007). If χ[−a,a](x) is the characteristic function of [−a, a] deﬁned by 1 if |x| a, then 2 sin ak F χ − (x) = F (k)= . (B-3.28) [ a,a] a π k

If |x| |x| f(x)= 1 − H 1 − , (B-3.29) a a 756 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms then − a ak 2 ak F f(x) = √ sin2 . (B-3.30) 2π 2 2 The following analytic properties of the convolution also hold: d (f ∗ g)(x) = f ∗ g (x)= f ∗ g (x), (B-3.31) dx 2 d ∗ ∗ ∗ 2 (f g)(x) = f g (x)= f g (x), (B-3.32) dx (f ∗ g)(m+n)(x)= f (m) ∗ g(n) (x), (B-3.33) ∞ ∞ ∞ (f ∗ g)(x) dx = f(ξ) dξ g(η) dη, (B-3.34) −∞ −∞ −∞ b x−a (f ∗ χ[a,b])(x)= f(x − ξ) dξ = f(η) dη, (B-3.35) a x−b 1 ∞ F f(x)g(x) =(F ∗ G)(k)=√ F (k − ξ)G(ξ) dξ, (B-3.36) 2π −∞ f(x)=(f ∗ δ)(x)=F −1 F (k) . (B-3.37)

To prove (B-3.31), we have d ∞ ∞ (f ∗ g)(x)= f(x − ξ)g(ξ) dξ = f (x − ξ)g(ξ) dξ dx −∞ −∞ = f ∗ g (x).

To prove (B-3.33), we ﬁrst apply the Fourier transform to the left-hand side and then use the convolution Theorem 1.7.1 so that F (f ∗ g)m+n(x) =(ik)m+nF (f ∗ g)(x) =(ik)m+nF (k)G(k) = (ik)mF (k) (ik)nG(k) = F f (m)(x) F g(n)(x) = F f (m) ∗ g(n) (x) .

The use of the inverse Fourier transform proves the result (B-3.33). To prove that the integral of the convolution satisﬁes (B-3.34), we have ∞ ∞ ∞ (f ∗ g)(x) dx = f(x − ξ)g(ξ) dξ dx −∞ −∞ −∞ ∞ ∞ = g(ξ) f(x − ξ) dξ dξ, (x − ξ = η) −∞ −∞ ∞ ∞ = g(ξ) dξ f(η) dη. −∞ −∞

We next prove (B-3.36) which is a dual result of (B-3.11) as follows. We have B-3 Basic Properties of the Fourier Transforms 757 1 ∞ F f(x)g(x) = √ e−ikxf(x)g(x) dx, 2π −∞ which is, by replacing g(x) by its inverse Fourier transform formula, ∞ ∞ 1 1 = √ e−ikxf(x) dx√ eik xG k dk 2π −∞ 2π −∞ ∞ ∞ 1 1 = √ G k dk · √ e−ix(k−k )f(x) dx 2π −∞ 2π −∞ 1 ∞ = √ F k − k G k dk 2π −∞ =(F ∗ G)(k).

Thus, the results (B-3.11) and (B-3.36) represent a duality since the Fourier transform of the convolution product of two functions is equal to the ordinary product of their Fourier transforms, and the Fourier transform of the ordinary product of two functions is equal to the convolution product of their Fourier transforms. If the diffusion kernel function Gt(x) is 1 x2 Gt(x)=√ exp − , (B-3.38) 4πκt 4κt then

(Gt ∗ Gs)(x)=Gt+s(x). (B-3.39) To prove (B-3.39), we apply the Fourier transform without the factor √1 .This 2π F{ − 2 } π − k2 means that exp( ax ) = a exp( 4a ). Consequently, F (G ∗ G )(x) = F G (x) F G (x) t s t s =exp −κk2t exp −κk2s 2 =exp−k κ(t + s) = F Gt+s(x) . The inverse Fourier transform gives the result. 1 −| | ∗ If g(x)= 2a H(a x ), then (f g)(x) is the average value of f(x) on [x − a, x + a]. Using the value of g(x) gives ∞ 1 a (f ∗ g)(x)= f(x − ξ)g(ξ) dξ = f(x − ξ) dξ −∞ 2a − a 1 x+a = f(η) dη. 2a x−a We close this section by adding an application of the convolution to the solution of the wave equation in Example 1.7.2. The solution of the Fourier transform U(k, t)=F{u(x, t)} of the transformed wave equation (1.7.17) is given by 758 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

U(k, t)=A cos ckt + B sin ckt, (B-3.40) where dU U(k, 0) = F (k)and = G(k). (B-3.41) dk t=0 Consequently, the use of (B-3.41)gives G(k) U(k, t)=F (k) cos ckt + sin ckt. (B-3.42) ck Using the results 2 F δ(x − ct)+δ(x + ct) = cos(ckt), π 2 sin ckt F χ − (x) = F H ct −|x| = , [ ct,ct] π k applying the inverse Fourier transform to (B-3.42) gives the solution 1 sin ckt u(x, t)=F −1 F (k) cos ckt + F −1 G(k) . c k Application of the convolution Theorem 1.7.1 gives the solution π u(x, t)=f(x) ∗ δ(x − ct)+δ(x + ct) 2 1 π + g(x) ∗ χ − (x) (B-3.43) c 2 [ ct,ct] 1 ∞ = f(x − ξ) δ(ξ − ct)+δ(ξ + ct) dξ 2 −∞ 1 ∞ + g(x − ξ)χ[−ct,ct](ξ) dξ 2c −∞ 1 1 ct = f(x − ct)+f(x + ct) + g(x − ξ) dξ, (x − ξ = α) 2 2c − ct 1 1 x+ct = f(x − ct)+f(x + ct) + g(α) dα. (B-3.44) 2 2c x−ct This is identical with the d’Alembert solution of the wave equation.

B-4 Basic Properties of Laplace Transforms

The Laplace transform of a continuous or piecewise continuous function f(t) for t>0 is denoted by L{f(t)} = f(s), and it deﬁned by (1.9.1) and the inverse B-4 Basic Properties of Laplace Transforms 759

Laplace transform, L−1{f(s)} = f(t), is deﬁned by (1.9.2). In this section, some basic properties of the Laplace transform are presented. For more information, the reader is referred to Debnath and Bhatta (2007): L eatf(t) = f(s − a)(shifting), (B-4.1) 1 b s L f(at + b) = exp s f (scaling), (B-4.2) |a| a a dn L tnf(t) =(−1)n f(s),n≥ 1, (B-4.3) dsn f(t) ∞ L = f(s) ds, (B-4.4) t s t f(s) L f(τ) dτ = . (B-4.5) 0 s

If L{f(t)} = f(s), then f(s) t L−1 = f(τ) dτ, (B-4.6) s 0 t t1 t L−1 f(s) − 2 = f(τ) dτ = (t τ)f(τ) dτ. (B-4.7) s 0 0 0 In general, t t1 t2 tn−1 −1 f(s) L = ··· f(τ) dτ dt ··· dt − sn 1 n 1 0 0 0 0 t (t − τ)n−1 = − f(τ) dτ. (B-4.8) 0 (n 1)! If f(t) is a periodic function with period T and L{f(t)} exists, then T −1 L f(t) = 1 − e−sT e−stf(t) dt. (B-4.9) 0 For example, if f(t) is a square wave function with period 2a deﬁned by

f(t)=H(t) − 2H(t − a)+2H(t − 2a) − 2H(t − 3a)+··· , (B-4.10) then the Laplace transform of f(t) is 1 as f(s)= tanh . (B-4.11) s 2

Clearly, the graph of f(t) shows that f(t)=1if 0

Theorem B-4.1 (Convolution Theorem). If f(t) ∗ g(t) is the Laplace Convolution of f(t) and g(t) deﬁned by t f(t) ∗ g(t)=(f ∗ g)(t)= f(t − τ)g(τ) dτ, (B-4.12) 0 and if L{f(t)} = f(s) and L{g(t)} = g(s), then L f(t) ∗ g(t) = L f(t) L g(t) = f(s)g(s), (B-4.13) or equivalently, L−1 f(s)g(s) = f(t) ∗ g(t). (B-4.14)

To prove the convolution theorem, we have ∞ t L f(t) ∗ g(t) = e−st f(t − τ)g(τ) dτ dt, 0 0 ∞ t = e−s(t−τ)f(t − τ) · e−sτ g(τ) dτ dt, 0 0 which is, reversing the order of integration, ∞ ∞ = e−s(t−τ)f(t − τ) dt e−sτ g(τ) dτ, 0∞ τ ∞ = e−sxf(x) dx e−sτ g(τ) dτ, (t − τ = x) 0 0 = L f(t) L g(t) = f(s)g(s).

In case of the Laplace transform, we proved the convolution theorem (B-4.12) and now we prove the dual result 1 L f(t)g(t) = f¯(s) ∗ g¯(s), (B-4.15) 2πi where ∞ ∞ 1 c1+i 1 c2+i f(t)=√ estf¯(s) ds, g(t)=√ estg¯(s) ds, 2πi c1−i∞ 2πi c2−i∞ c1,c2 > 0. B-4 Basic Properties of Laplace Transforms 761

We have, by deﬁnition and replacing g(t) by its inverse Laplace transform, ∞ L f(t)g(t) = e−stf(t)g(t) dt 0 ∞ ∞ 1 c2+i = e−stf(t) dt · eztg¯(z) dz 0 2πi c2−i∞ ∞ ∞ 1 c2+i = g¯(z) dz e−t(s−z)f(t) dt 2πi − ∞ c2 i 0 c2+i∞ 1 ¯ = f(s − z)¯g(z) dz, Re(s − z) ≥ c1 2πi c2−i∞ 1 = f¯(s) ∗ g¯(s), Re s ≥ c + c . 2πi 1 2 Like the above duality property for the Fourier transform, results (B-4.12) and (B-4.16) represent the duality property for the Laplace transform. The following properties of the Laplace convolution (B-4.12) also hold: L (f ∗ g)(t) = L f ∗ g (t) + f(0)g(s)=sf(s)g(s), (B-4.16) where the prime denotes the derivative with respect to t. Similarly, L (f ∗ g)(t) = L f ∗ g (t) + g(0)f(s)=sf(s)g(s). (B-4.17)

Results similar to (B-4.16)Ð(B-4.17) can be proved for the second and the higher derivatives of (f ∗ g)(t). The Duhamel formulas follow from (B-4.16) and (B-4.17)as t L−1 sf(s)g(s) = f(0)g(t)+ g(t − τ)f (τ) dτ, (B-4.18) 0 or equivalently, t L−1 sf(s)g(s) = g(0)f(t)+ f(t − τ)g(τ) dτ. (B-4.19) 0

p−1 −t If fp(t)=t e , t>0, then direct differentiation gives − − fp(t)=(p 1)fp−1(t) fp(t). (B-4.20)

It also follows that fp(t) ∗ fq(t) exists for all p, q>0 and satisﬁes the following identities ∗ (fp fq)(t)=B(p, q)fp+q(t) (B-4.21) (f ∗ f ) (t)=B(p, q) (p + q − 1)f − (t) − f (t) (B-4.22) p q p+q 1 p+q ∗ ∗ − − ∗ (fp fq) (t)=fp fq(t)= (p 1)fp−1(t) fp(t) fq(t), (B-4.23)

=(p − 1)B(p − 1,q)fp+q−1(t) − B(p, q)fp+q(t), (B-4.24) where B(p, q) are the Beta function of p and q. 762 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms

To prove (B-4.21), we have t (fp ∗ fq)(t)= fp(t − τ)fq(τ) dτ 0 t = (t − τ)p−1e−(t−τ)τ q−1e−τ dτ, (t − τ = tu) 0 1 = e−ttp+q−1 up−1(1 − u)q−1 du, 0 −t p+q−1 = B(p, q)e t = B(p, q)fp+q(t).

Then ∗ − − (fp fq) (t)=B(p, q)fp+q(t)=B(p, q) (p + q 1)fp+q−1(t) fp+q(t) .

The Laplace convolution (B-4.12) also satisﬁes the properties similar to (B-3.13). In particular, L ∗ ∗···∗ ··· (f1 f2 fn)(t) = f 1(s)f 2(s) f n(s). (B-4.25) n L f ∗n(t) = f(s) , (B-4.26) where f ∗n is the nth convolution product deﬁned by

f ∗n(t)=(f ∗ f ∗···∗f)(t). (B-4.27)

In general, mathematical operations such as addition, f(x)+g(x), multiplication, f(x)g(x), and composition, f(g(x)), for two functions f(x) and g(x) form a new function or an ordinary output. On the other hand, the convolution f(x) ∗ g(x) represents the integral output of two functions f and g, and it plays a central role in the subjects, such as Fourier series, Fourier transforms, number theory, harmonic analysis, probability theory, and almost any integral transform. We have already stated the convolution Theorem 1.7.1 for the Fourier transform, and the convolution The- orem B-4.1 for the Laplace transform of two functions. Obviously, these theorems can be generalized to obtain a relation between the n-fold convolution of n functions and the product of the transforms of these functions. We next establish a nice connection between the Weierstrass transform and the two-sided Laplace transform of f(ξ) on −∞ <ξ<∞ deﬁned by ∞ L f(ξ) = f(s)= e−sξf(ξ) dξ (B-4.28) −∞ which is the Fourier transform of f(ξ) for s = ik so that √ √ L f(ξ) = f(s)= 2πF f(ξ) (ik)= 2πF(ik). (B-4.29)

− ξ2 Thus, the two-sided Laplace transform of f(ξ)=exp( 4a ) is given by B-4 Basic Properties of Laplace Transforms 763 2 √ 2 ξ − ξ L exp − = 2πF e 4a (ik = s) 4a √ = 4πa exp as2 . (B-4.30)

In Example 1.7.6 in Chapter 1, the solution of the Cauchy problem for the diffusion equation is given by ∞ u(x, t)= f(ξ)G(x − ξ) dξ = f(x) ∗ G(x), (B-4.31) −∞ where the diffusion kernel G(x) is given by 1 x2 G(x)=√ exp − . (B-4.32) 4πκt 4κt The above results (B-4.31)Ð(B-4.32) are used to introduce the Weierstrass transform of f(ξ) with positive parameter t in the form ∞ W f(ξ) (x)=F (x)= f(ξ)G(x − ξ) dξ, (B-4.33) −∞ where G(x), or, more precisely, Gt(x), is the kernel of the Weierstrass transform deﬁned by (B-4.33). We next expand the kernel Gt(x − ξ) to express the Weierstrass transform (B-4.33)off(ξ) in terms of the two-sided Laplace transform (B-4.28) so that (B-4.33) reduces to F (x)=W f(ξ) (x) 2 exp(− x ) ∞ x ξ2 = √ 4κt exp − − ξ exp − f(ξ) dξ 4πκt −∞ 2κt 4κt 2 exp(− x ) ξ2 x = √ 4κt L exp − f(ξ) − , (B-4.34) 4πκt 4κt 2κt where L is the two-sided Laplace transform (B-4.28). We use (B-4.32) with κt = a and (B-4.30) to derive the inverse Weierstrass transform so that ∞ 2 2 1 ξ eas = √ e−sξ exp − dξ. (B-4.35) 4πa −∞ 4a

d Replacing s by the operator D = dx and using the fact that ∞ (−ξ)kDkf(x) e−ξDf(x)= = f(x − ξ), (B-4.36) k! k=0 equation (B-4.35) becomes 764 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms ∞ 2 2 1 ξ eaD f(x)=√ e−ξDf(x) exp − dξ 4πa −∞ 4a 1 ∞ ξ2 = √ f(x − ξ)exp − dξ 4πa −∞ 4a 1 ∞ (x − ξ)2 = √ exp − f(ξ) dξ 4πa −∞ 4a ∞ = G(x − ξ)f(ξ) dξ = F (x). (B-4.37) −∞

This gives the formula for the inverse Weierstrass transform

2 f(x)=e−aD F (x),x∈ R. (B-4.38)

The celebrated RiemannÐLiouville fractional integral of order α of a function f(t) is usually deﬁned by t −α −α 1 − α−1 D f(t)=0 Dt f(t)= (t x) f(x) dx, Re α>0. (B-4.39) Γ (α) 0 Clearly, D−α is a linear integral operator, and (B-4.39) can be expressed in terms of the convolution product

D−αf(t)=f(t) ∗ g(t), (B-4.40)

− tα 1 L{ } −α where g(t)= Γ (α) and g(s)= g(t) = s . Using the Laplace convolution theorem to (B-4.40)gives L D−αf(t) = L f(t) ∗ g(t) = L f(t) L g(t) = s−αf(s), (B-4.41) or equivalently, D−αf(t)=L−1 s−αf(s) . (B-4.42)

This can be used to evaluate the fractional integral of a given function f(t).For example, if f(t)=tβ, then Γ (β +1) Γ (β +1) D−αtβ = L−1 = tα+β. (B-4.43) sα+β+1 Γ (α + β +1)

Another consequence of (B-4.39)forthefractional derivative, D−αf(t), is that it can be deﬁned as the solution φ(t) of the integral equation

D−αφ(t)=f(t) (B-4.44) so that its Laplace transform gives B-4 Basic Properties of Laplace Transforms 765

φ(s)=sαf(s). (B-4.45)

Consequently, the inverse Laplace transform yields φ(t)=D−αf(t)=L−1 sαf(s) (B-4.46) −α−1 t t ∗ 1 − −α−1 = − f(t)= − (t x) f(x) dx. (B-4.47) Γ ( α) Γ ( α) 0 This is known as the Cauchy integral formula for the fractional derivative of f(t). In fact, (B-4.46) can often be used to obtain the fractional derivative of f(t).For example, if f(t)=tβ then Γ (β +1) Γ (β +1) Dαtβ = L−1 = t(β−α). (B-4.48) sβ−α+1 Γ (β − α +1

In general, the Laplace convolution integral equation is of the form t f(t)=h(t)+λ g(t − τ)f(τ) dτ, (B-4.49) 0 where λ is a given constant parameter, f(t) is the unknown function, h(t) and g(t) are given functions. Applications of the Laplace transform to (B-4.49) combined with the convolution theorem yields

h(s) f(s)= . (B-4.50) 1 − λg(s)

The inverse Laplace transform gives the formal solution of f(t) in the form h(s) f(t)=L−1 . (B-4.51) 1 − λg(s)

For example, the Abel integral equation of the ﬁrst kind in the form t (t − τ)α−1f(τ) dτ = g(t), (B-4.52) 0 or equivalently,

tα−1 ∗ f(t)=g(t), (B-4.53) can be solved by application of the Laplace transform so that 1 1 f(s)= sαg(s)= s sα−1g(s) . (B-4.54) Γ (α) Γ (α)

This leads to the solution 766 B Fourier Series, Generalized Functions, and Fourier and Laplace Transforms t 1 1 d − −α f(t)= − (t τ) g(τ) dτ. (B-4.55) Γ (α) Γ (1 α) dt 0 The Abel integral equation of the second kind is given by λ t f(t)+ (t − τ)α−1f(τ) dτ = g(t),α>0, (B-4.56) Γ (α) 0 where λ is a real or complex parameter and g(t) is a given function. Application of the Laplace transform to (B-4.56) leads to the solution sα sα−1 f(s)= g(s)= s · · g(s) , (B-4.57) sα + λ sα + λ so that the inverse Laplace transform gives the solution of (B-4.56)as t d α f(t)= Eα,1 −λτ g(t − τ) dτ, (B-4.58) dt 0 where the Mittag-Lefﬂer function, Eα,β(z), is given by

∞ zn E (z)= ,α,β>0. (B-4.59) α,β Γ (nα + β) n=0 C Answers and Hints to Selected Exercises

1.15 Exercises

2 − dy 1. (a) A =4, B =5, C =1, B 4AC =9> 0. Hyperbolic, dx = √ B± B2−4AC 1 − 1 2A =1, 4 . Integrating gives y x = c1 and y = 4 x + c2, 1 − 8 − − 1 − − 8 uξη = 3 (uη 3 ); α = ξ + η, β = ξ η, uαα uββ = 3 (uα uβ 3 ). − 2 − dy − 1 − (b) A =2, B = 3, C =1, B 4AC =1> 0. Hyperbolic, dx = 2 , 1.

Integrating gives x +2y = c1, x + y = c2; uξη = η − ξ. 2 − 2 − 1 (c) Hyperbolic, ξ = y x = c1, η = y x = c2; uξη + η uξ =0. (d) Hyperbolic for y<0 and elliptic for y>0. Parabolic for y =0. √ √ − − − − 1 − 2 For the hyperbolic case, ξ = x +2 y, η = x 2 y, y = 4 (x c) , and 1 − uξη = 2(ξ−η) (uη uξ). √ 1 For the elliptic case, α = x, β =2 y; uαα + uββ = β uβ. (e) Elliptic for y>e−x, parabolic for y = e−x, and hyperbolic for y

(f) Hyperbolic for x<0 and elliptic for x>0. 3 ± − 3/2 − ± 2 − 3/2 For x<0; ξ, η = 2 y ( x) , (y c)= 3 ( x) (cubic parabolas). 1 − uξη = 6(ξ−η) (uξ uη). 3 − 3/2 1 For x>0, α = 2 y, β = x ; uαα + uββ +(3β )uβ =0, where α and β √ − √1 satisfy the Beltrami equations βx = xαy, βy = x αx. (g) Hyperbolic for |x| < 2|y|, elliptic for |x| > 2|y|, and parabolic for |x| =2|y|.

L. Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, DOI 10.1007/978-0-8176-8265-1, c Springer Science+Business Media, LLC 2012 768 C Answers and Hints to Selected Exercises

(h) Hyperbolic in the ﬁrst and third quadrants, elliptic in the second and fourth

quadrants.

(i) Elliptic for |x| a.

2 2 (j) ξ = y , η = x ; 2ξη(uξξ + uηη)+ηuξ + ξuη =0.

dy 2. (d) Elliptic for all x and y. In this case, the characteristics equations are dx = ±i sech2 x so that the characteristics are y ∓ i tanh x = constant so that α = y,

β = tanh x. Thus, the canonical form of the given equation is uαα + uββ =

2 −1 2β(1 − β ) uβ. dy ± 2 (f) Hyperbolic for all x and y. The characteristic equations are dx = sech x so that the characteristics are y ∓ tanh x = const., ξ = y + tanh x, and η =

y − tanh x. Thus, the canonical form of the given equation is uξη =(η − ξ) ×

2 −1 (uξ − uη){4 − (ξ − η) } .

dy 2 (g) dx = cosec y; ξ = x + cos y, η = y; uηη =sin η cos ηuξ. 2 − 2 2 − 2 2 4 4. (a) A = u , B =2uxuy, C = u . Hence, B 4AC =4(uxuy + u ) > 0 for all u(x, y). Hyperbolic for all u(x, y). |∇ | 2 2 − 2 − − 2 (b) u = ux + uy. A =1 ux, B = 2uxuy, C =1 uy. Hence, 2 − − 2 2 − 2 | | B 4AC = 4+4(ux + uy)= 4+4(Δu) > 0, =0,or< 0 for Δu > 1, =1,or,< 1. nπ − 2 nπx 5. αn =( ), n =0, 1, 2, 3,...; un(x, t)=an exp( αnkt) cos( ), and ! ∞ nπx u(x, 0) = a0 + n=1an cos( ) is the cosine Fourier series, where a0 = 1 2 nπx 0 f(x) dx and an = 0 f(x) cos( ) dx. 6. For λ = −α2, u(x, y)=(A cos αx + B sin αx)(C cosh αy + D sinh αy), nπ C − A =0, and α = a , n =1, 2, 3,...; D = tanh αb, and u(x, y)= ! ∞ nπx nπ(b−y) nπb −1 2 a × n=1an sin( a )sinh a , where an =(sinha ) ( a ) 0 f(x) nπx sin( a ) dx. 7. Hint: We assume u(x, t)=X(x)T (t) =0 and substitute in the given equation

2 − 1 T X 4 to obtain XT (t)+c TX (x)=0, or equivalently, c2 T = X = λ . 1.15 Exercises 769

We have X = λ4X, T = −λ4c2T , where λ4 is a separation constant. Thus,

X(x)=A cosh λx + B cos λx + C sinh λx + D sin λx,

T (t)=E cos(λ2ct)+F sin(λ2ct),B= −A, D = −C,

X(x)=A(cosh λx − cos λx)+C(sinh λx − sin λx).

From conditions X( )=0=X( ), we obtain cos λ − cos λ sinh α − sin α =0 =⇒ cos λ cosh λ =1. sinh λ +sinλ cosh α − cos α

8. We seek a nontrivial separable solution u(x, t)=X(x)T (t) so that sin α =0, ! nπ ∞ { nπ 2 } α =( ), n =1, 2, 3, 4,...; sinh α =0. u(x, t)= n=1[an cos ( ) ct + { nπ 2 } nπx bn sin ( ) ct ]sin( ). 4 − 4 9. an =0for all even n, an =(n2π2 ), n =1, 5, 9,..., and an = ( n2π2 ), n =3, 7, 11,.... ! ∞ − n2π2κt nπx Thus, u(x, t)= n=1an exp( 2 )sin( ). 11. Application of the Fourier transform, F{u(x, t)} = U(k, t), to the problem

d2U 4 gives dt2 + k U =0, U(k, 0) = F (k), and Ut(k, 0) = 0. Thus, the solution of the transformed system is U(k, t)=F (k) cos(k2t). Consequently, the inverse Fourier transform together with the Fourier convolution yields the so- ∞ lution u(x, t)=F −1{F (k) cos(k2t)} = √1 f(x − ξ)h(ξ,t) dξ, where 2π −∞ 2 F −1{cos(k2t)} = √1 cos( x − π )=h(x, t). 2t 4t 4 12. Apply the joint Fourier and Laplace transforms (1.9.12) to obtain U(k, s)=

(s+α)F (k)+G(k) (s2+sα+c2k2) . Application of the joint inverse transforms combined with the convolution of the Fourier transform gives the solution for u(x, t). ∞ √1 { − } 13. u(x, t)= −∞ A(k)exp[i(kx + ωt)] + B(k)exp[i(kx ωt)] dk, where √ 2π 2 2 2 1 1 1 − ω = c k + a , and A(k)= 2 [F (k)+ iω G(k)], and B(k)= 2 [F (k) 1 iω G(k)]. 770 C Answers and Hints to Selected Exercises

2 2 15. Hint: F −1 cos (atk2) = √1 [cos( x ) ± sin( x )]. sin 2 at 4at 4at 17.

P ∞ sin ωt φ(x, z, t)=− exp ikx + |k|z dk, ω2 = g|k|, 2π ω −∞ √ P ∞ Pt g gt2 η(x, t)= cos ωtexp(ikx) dk ≈ √ · cos 3/2 2π −∞ 2 2π x 4x for gt2 4x.

19.

iP exp(εt) ∞ (Uk − iε)exp(|k|z + ikx) φ(x, z, t)= dk, 2πρ (Uk − iε)2 − g|k| −∞ P exp(εt) ∞ |k| exp(ikx) dk η(x, t)= 2 . 2πρ −∞ (Uk − iε) − g|k|

23. Using the Fourier transform, U(k, y)=F{u(x, y)}, gives the solution of the

transformed system in the form U(k, y)=F (k) cos(k2y). The inverse Fourier

transform together with the Fourier convolution gives the solution u(x, y)=

∞ 2 √ 1 − ξ − π 4πy −∞ f(x ξ) cos( 4y 4 ) dξ. 24.

∞ ∞ 1 1 u(x, y, t)= F (k, ) cos c k2 + 2 2 t 2π −∞ −∞ × exp i(kx + y) dk d .

25. Application of the joint Laplace and Fourier transform gives U¯(k, s)=

(s + κk2)−1Q¯(k, s). The use of the inverse Laplace transform combined with

the convolution theorem yields

− 2 ∗ U(k, t)=exp κk t Q(k, t) t = exp −κk2(t − τ) Q(k, τ) dτ. 0 1.15 Exercises 771

Application of the inverse Fourier transform and its convolution theorem gives

t u(x, t)= F −1 P (k, t − τ)Q(k, τ) dτ 0 t = p(x, t − τ) ∗ q(x, τ) dτ 0 1 t ∞ = √ p(x − ξ,t − τ)q(ξ,τ) dξ dτ 2π −∞ 0 t ∞ 2 1 − 1 (x − ξ) √ − 2 − = (t τ) dτ q(ξ,τ)exp − dξ, 4πκ 0 −∞ 4κ(t τ)

2 F −1{ − 2 } √ 1 − x where p(x, τ)= exp( κk τ) = exp( 4κτ ). 2κτ 1 ∞ { G(k) } ikt 2 2 27. u(x, t)= 2π −∞ F (k) cos(xα)+ α sin(xα) e dk, where α = α (k)= 1 2 − − c2 (k iak b).

28. Hint: Seek a solution of the form ψ(x, y, t)=φn(x, t)sin(nπy) with ψ0(x, y)=

2 ∂ ∂ φn − 2 ψ0n(x)sin(nπy), so that φn(x, t) satisﬁes the equation ∂t[ ∂x2 α φn]+

∂φn 2 2 2 β ∂x =0, where α =(nπ) +κ . Apply the Fourier transform of φn(x, t) with ∞ respect to x, and use Ψ (k, 0) = F{ψ (x)}. φ (x, t)= √1 Ψ (k, 0) × n 0n n 2π −∞ n exp[i{kx − ω(k)t}] dk, where ω(k)=−βk(k2 + α2)−1.

29. Divide the ﬁrst equation by L and the second equation by C to obtain the third R − 1 G − 1 and the fourth equations It + L I = L Vx, and Vt + C V = C Ix. Differentiate

the third equation with respect to time t and replace Vt and Vx on the right- hand side to derive a second order equation telegraph equation for I in the form − 2 2 −1 R G Itt c Ixx +(p + q)It + pqI =0, where c =(LC) , p = L , and q = C . Similar calculation can be used to obtain the same equation for V . Thus, u = I

or V satisﬁes the above telegraph equation with coefﬁcients c2, p, and q, or with

coefﬁcients c2, a, and b.

− − 30. (a) V (x, t)=V erfc( √x ), κ = a 1 =(RC) 1. I(x, t)= Vo √ 1 × 0 2 κt R πκt − x2 exp( 4κt ). √ √ 1 −x b { x a bt } 1 −x b { x a − bt } (b) V (x, t)= 2 e erfc 2 t + a + 2 e erfc 2 t a . 772 C Answers and Hints to Selected Exercises

− kx − x − x 31. V (x, t)=V0 exp( c )f(t c )H(t c ). 32. u(x, t)=x cos ωt. k − πct πx 33. u(x, t)= (πc)2 [1 cos( a )] sin( a ). ω 1 z 34. u(z,t)U exp[iωt − ( ) 2 (1 + i)z], u(z,t)=Uerfc( √ ). 2ν 2 νt 35. u(z,t)=Ut[(1 + 2ζ2)erfc(ζ) − √2ζ exp(−ζ2)], ζ = √z . π 2 νt 36. a − 1 q(z,t)= eiωt e λ1zerfc ζ − it(2Ω + ω) 2 2 1 + eλ1zerfc ζ + it(2Ω + ω) 2 b − − 1 + e iωt e λ2zerfc ζ − it(2Ω − ω) 2 2 1 + eλ2zerfc ζ + it(2Ω + ω) 2 ,

i(2Ω±ω) 1 { } 2 where λ1,2 = ν . Thus,

1 ν 2 q(z,t) ∼ a exp(iω − λ z)+b exp(−iωt − λ z),δ = . 1 2 1,2 |2Ω ± ω|

˜ Q 43. (b) Hint: f(k)=(πak )J1(ak). ∞ ˜ 2 45. u(r, t)= 0 k f(k) cos(btk )J0(kr) dk. − r2 ˜ a2 − a2k2 If f(r)=exp( a2 ), then f(k)= 2 exp( 4 ). Using the self-reciprocity of the Hankel transform gives the solution 1 a2k2 u(r, t)= a2H exp − cos btk2 (r) 2 0 4 = a2Ω(t)exp −a2Ω(t)r2 a2 cos 4btΩ(t)r2 +4bt sin 4btΩ(t)r2 ,

where Ω(t)=(a4 +16b2t2)−1.

48. Hint: The solution of the dual integral equations

∞

kJ0(kr)A(k) dk = u0, 0 ≤ r ≤ a, 0 ∞ 2 k J0(kr)A(k) dk =0,a

2u0 sin(ak) is given by A(k)=( π ) k2 . 1.15 Exercises 773

49. Hint: See Debnath (1994, pp. 103Ð105). 1 ∞ −1 − 50. u(r, z)=(πa ) 0 k J1(kr)J0(kr)exp( kz) dk. √ −1 2 2 − 1 2 2 1 2 2 52. Hint: L [(s + a ) 2 exp{−k(s + a ) 2 }]=H(t − k)J0(a t − k ). Application of the joint Laplace and Hankel transforms

∞ ∞ −st u¯˜(k, z, s)= rJ0(kr) dr e u(r, z, t) dt 0 0

to the given equation and the general boundary condition uz(r, 0,t)=f(r, t)= H(a − r)g(t) gives

d2u¯˜ 1 a − s2 + k2c2 u¯˜ =0, u¯˜ (k, 0,s)= J (ak)¯g(s). dz2 c2 z k 1

The bounded solution of the equation is

z u¯˜(k, z, s)=A¯˜(k, s)exp − s2 + c2k2 c ac g¯(s)J (ak) with A¯˜(k, s)=− √ 1 . k s2 + c2k2 √ Thus, u¯˜(k, z, s)=−( ac ) √J1(ak) g¯(s)exp[− z s2 + c2k2]. k s2+c2k2 c The inverse Laplace transform gives

ac −1 g(s) z 2 2 2 u˜(k, z, t)=− J1(ak)L √ exp − s + c k k 2 2 2 c s + c k t ac z z2 − − − 2 − = J1(ak) g(t τ)H τ J0 ck τ 2 dτ. k 0 c c

The inverse Hankel transform leads to the ﬁnal solution

∞ t z u(r, z, t)=(−ac) J0(kr)J1(ak) dk g(t − τ)H τ − 0 0 c " # 2 2 2 × J0 k c τ − z dτ. 774 C Answers and Hints to Selected Exercises

This, for g(t)=δ(t), becomes ∞ z 2 2 2 u(r, z, t)=(−ac)H t − J0(kr)J1(ak)J0 k c t − z dk. c 0 ∞ −1 sinh kz 53. u(r, z)=b 0 k ( cosh ka )J1(bk)J0(kr) dk. − 1 sin(ak) H 2 − 2 2 − 54. Hint: 0[(a r ) H(a r)] = k , and

√ √ −1 √ L−1 s( s − a) exp −k s k √ =exp −ak − a2t erfc √ − a t . 2 t

55. Hint: Use the joint Hankel and Laplace transform method.

57. Use the Hankel transform and derive

∞ ∞ 1 r0(t) u(r, z, t)= k exp(kz)J0(kz) αp(α, τ) ρ 0 0 0 × J0(kα) dα cos ω(t − τ) dτ ,

where ω2 = gk.

59. Hint: Use the joint Laplace and Fourier transform.

∞ W sin αt 1 G(x, t)= exp(ikx) dk, α = a2k4 + ω2 2 , 2πm −∞ α

2 EI 2 κ where a = m and ω = m . 61. Hint: Set

∞ ∞ mπx nπy G(x, y; ξ,η)= a sin sin mn a b m=1 n=1

in the original equation to ﬁnd amn from mπ 2 nπ 2 ab mπξ nπη + a =sin sin . a b 4 mn a b 1.15 Exercises 775

62. Apply the joint Laplace and double Fourier transform.

2 2 63. Hint: Gtt − c Gxx + d G = δ(x)δ(t). The joint Laplace and Fourier transform gives

1 1 1 G¯˜(k, s)=√ ,α= c2k2 + d2 2 . 2π (s2 + α2)

The inverse transforms give ∞ − 1 2 2 2 1 2 d 2 d G(x, t)= k + 2 sin ct k + 2 exp(ikx) dk 2πc −∞ c c 1 d = J c2t2 − x2 H ct −|x| . 2c 0 c

nπ 2 64. Hint: Eigenvalues are λn =( ) , n =0, 1, 2, 3,.... Eigenfunctions are nπx nπx Xn(x)=An cos( )+Bn sin( ). Thus, ∞ ∞ u(x, t)= Xn(x)Tn(t)= un(x, t) n=0 n=0 ∞ 1 nπx nπx nπkt = a + a cos + b sin exp − , 2 0 n n n=1 ∞ a nπx nπx f(x)=u(x, 0) = 0 + a cos + b sin , 2 n n n=1

where 1 nπξ an = f(ξ) cos dξ, n =0, 1, 2,..., − 1 nπξ bn = f(ξ)sin dξ, n =1, 2, 3,.... −

65. (b) Initial data tend to zero as n →∞, but the solution

un(x, y) →∞ as n →∞. 776 C Answers and Hints to Selected Exercises

66. (a) Hint: λ =0, u(x)=A+Bx with the boundary conditions leads to the trivial

solution.

If λ<0, u(x)=A cosh kx + B sinh kx(λ = −k2). The ﬁrst boundary con-

dition gives A =0, and the second boundary condition yields k = − tanh k.

The graphical representation of y = −k and y = tanh k shows that there is

no intersection of these curves for k>0. Hence, there are no negative eigen-

values. For λ = k2 > 0, u(x)=A cos kx + B sin kx. The boundary con-

ditions give A =0, and B =0 , k = − tan k. The graphical representation

of y = −k and y = tanh k shows that there is an inﬁnite number of values

k = kn (n =1, 2, 3,...) so that there is an inﬁnite number of eigenvalues 2 ≈ 1 2 2 λ = kn 4 (2n +1) π for large n. The corresponding eigenfunctions are

un(x)=sinknx, where n =1, 2, 3,....

2 2 (b) λn = kn =(nπ) , un(x)=An cos nπx, n =0, 1, 2,.... 2 2 { } (c) λn = kn =(2nπ) , un(x)= sin 2πnx, cos 2πnx , (n =0, 1, 2,...) are eigenfunctions. To each eigenvalue, there correspond two eigenfunctions.

Eigenvalues are degenerate.

(d) For 1 − 4λ =0,or> 0, only trivial solutions. − − 2 1 n2π2 For 1 4λ = k < 0, we obtain eigenvalues λn = 4 (1 + a2 ) and eigen- −x nπ functions un(x)=Bne sin knx, (kn = a ,n=1, 2, 3,...). 67. Hint: Multiply the equation by p(x) so that a2(x) p(x)u + p(x)u + q(x)+λρ(x) u =0,

where p(x)= p(x)a1(x) , q(x)= p(x)a0(x) , and ρ(x)= p(x) . a2(x) a2(x) a2(x) 69. Hint: λ = k2 > 0. The general solution is

y(x)=A + Bx + C cos kx + D sin kx.

The ﬁrst boundary conditions at x =0give A = C =0. 1.15 Exercises 777

The remaining boundary conditions at x = a give

Ba + D sin ka =0,k2D sin ak =0.

nπ For nontrivial solutions, B = C, D =0, k = a . n2π2 nπx Thus, λn = a2 , and yn(x)=Dn sin( a ). n2π2 The critical buckling loads are Pn = EI( a2 ). The Euler load is the largest π2 load which the beam can withstand before possible buckling: P1 = a2 (EI). πx The corresponding fundamental bucking mode is y1(x)=D1 sin( a ). 71. X +2bX + k2X =0, T˙ + k2T =0, where −k2 is a separation constant.

m = −b ± iα, α2 = k2 − b2 > 0.

X(x)=exp(−bx)(A cos αx + B sin αx). Using boundary conditions, we ob- nπ tain A =0, sin αa =0, B =0, α =( a ), n =1, 2, 3,.... Hence, T (t)=C exp(−k2t). Thus, the ﬁnal solution is

∞ − 2 − 2 2 2 u(x, t)= an exp knt bx sin αnx, kn = b + αn. n=1

2 λ 2 72. (a) Eigenvalues are k =(a ) and eigenfunctions are λr R(r)=AJn ,Jn(ka)=0, a λct λct T (t)=B cos + C sin ,Θ(θ)=D cos nθ + E sin nθ, a a ∞ ∞ λ r λ ct u(r, θ, t)= (a cos nθ + b sin nθ)J mn cos mn , mn nm n a a n=0 m=0

where λmn are positive roots of Jn(x)=0.

∞ ∞ λ r f(r, θ)= (a cos nθ + b sin nθ)J mn , where nm nm n a n=0 m=0 a π 1 λm0r ≥ a0n = 2 2 f(r, θ)J0 rdrdθ and n 1, πa J1 (λn0) 0 −π a 778 C Answers and Hints to Selected Exercises π a 2 λmnr (anm,bnm)= 2 2 f(r, θ)Jn πa Jn+1(λmn) −π 0 a × (cos nθ, sin nθ)rdrdθ.

{ πc ≥ ≥ } The set of frequencies is a λmn : n 0,m 1 . (b) Since f is independent of θ,soisu. The solution is

∞ λ r λ ct u(r, t)= a J m cos m ,λ= λ , m 0 a a m m0 m=1 ∞ λ r f(r)= a2 − r2 = u(r, 0) = a J m m 0 a m=1

8a2 so that am = 3 . λmJ1(λm) 73. (a) Hint: Use u(r, z)=R(r)Z(z) so that

r2R + rR + k2r2R =0,R(a)=0; Z = k2Z, Z(0) = 0.

2 2 The eigenvalues of this SturmÐLiouville problem are kn =(λn/a) with the

λnr corresponding eigenfunctions Rn(r)=J0( a ), where λn are the positive roots

of J0(x)=0. !∞ zλn rλn zλn Thus, Zn(z)=sinh( a ). u(r, z)= n=1anJ0( a )sinh( a ), and g(r)= !∞ hλn rλn hλn u(r, h)= n=1 an sinh( a )J0( a ), where bn = an sinh( a )= a rλn 0 rg(r)J0( a ) dr. hλn rλn a rλn 2cosech( a ) (b) an = cosech( ) rJ0( ) dr = . a 0 a λnJ1(λn) Hence, the solution u(r, z) follows.

74. (a) Hint: Multiply the left-hand side of the wave equation by 2ut, and the resulting expression follows from performing the indicated differentiations on the

right-hand side of the differential equality.

(b) 2∇2 − 2 ··· 2ut c nu utt =2c (utux1 )x1 +(utux2 )x2 + +(utuxn )xn − c2 u2 + u2 + ···+ u2 + u2 . x1 x2 xn t t 1.15 Exercises 779

can be obtained.

75. Hint: (a) In the context of a stretched string, −d2u can be interpreted as an

additional spring force normal to the string.

Multiply the KleinÐGordon equation by ut to obtain

2 2 ututt − c utuxx + d utu =0, 1 ∂ c2 ∂ ∂ 1 u2 + u2 − c2 (u u )+ d2 u2 =0. 2 ∂t t 2 ∂x x ∂x t x 2

2 wtt = c wxx,a≤ x ≤ b,

w(x, 0) = 0 = wt(x, 0); w(orwx)=0at x = a, b.

Using (b), E(t)=E0 =0, and noting w(x, 0) = 0 implies wx(x, 0) = 0. 1 b 2 2 2 2 2 Therefore, E(t)= 2 a (wt + c wx + d w ) dx =0. Since the integrand is positive, w ≡ 0.

76. (a) Hint: Seek a separable solution u(r, t)=R(r)T (t) so that

r2R +2rR + k2r2R =0,T + c2k2T =0, 1 R(r)=√ A1J 1 (kr)+B1Y 1 (kr) , r 2 2 1 T (t)=C cos(ckt)+D sin(ckt)= (A sin kr + B cos kr). r

We assume that R(r) is ﬁnite at r =0, and hence, B =0.

nπ The eigenvalues are k = kn = a , n =1, 2,.... ! ∞ 1 nπr u(r, t)= n=1(an cos cknt + bn sin cknt) r sin( a ).

The initial conditions u(r, 0) = f(r) and ut(r, 0) = g(r) can be satisﬁed by the Fourier series expansion of rf(r) and rg(r) on (0,a).

8a3 (b) an =0, b2n =0, b2n−1 = cπ4(2n−1)4 . 780 C Answers and Hints to Selected Exercises

77. We seek a separable solution u(r, z)=R(r)Z(z) so that r2R +rR −k2r2R =

2 2 n2π2 0, and Z + k Z =0with Z(0) = Z(h)=0. k = h2 and Z(z)= nπz nπr an sin( h ). We assume R(r) is ﬁnite at r =0, and hence R(r)=anI0( h ), ! ∞ nπr nπz and thus, the solution is u(r, z)= n=1 anI0( h )sin( h ).Wehavef(z)= ! ∞ nπa nπz nπa −1 2 h nπz n=1 anI0( h )sin( a ) so that an =[I0( h )] h 0 f(z)sin( h ) dz. !∞ a0 78. (a) Hint: u(1,θ)= 2 + n=1(an cos nθ + bn sin nθ) = 1 + cos 2θ.

a0 =2, a2 =1, an =0, n =0 ,2.

2 bn =0for all n, u(r, θ)=1+r cos 2θ. 1 π | | 2 − 0 π (b) a0 = π −π 2θ dθ = π [ −πθdθ + 0 θdθ]=2π. 1 π | | − 8 an = π −π 2θ cos θdθ= πn2 for odd n, and 0 for even n. 1 π | | bn = π −π 2θ sin nθ dθ =0. ! − 8 ∞ −2 2n+1 Thus, u(r, θ)=π π n=0(2n +1) r cos(2n +1)θ. !∞ (c) Hint: n=1(an cos nθ + bn sin nθ) = 2 cos 2θ.

a2 =1, an =0, n =2 ; bn =0for all n.

a0 2 Therefore, u(r, θ)= 2 + r cos 2θ.

a0 (d) u(r, θ)= 2 + r(cos θ +sinθ). 79. (a) Hint: Seek a separable solution u(r, θ)=R(r)Θ(θ) =0 so that

r2R(r)+rR(r) − λ2R =0, Θ(θ)+λ2Θ(θ)=0.

Since u(r, θ) is periodic with period 2π,soisΘ(θ).

Eigenvalues: λn = λ = n, n =1, 2,.... The solution of the CauchyÐEuler equation for R(r) is

1 n =0: R (r)= (A + B ln r), 0 2 0 0 n −n n ≥ 1: Rn(r)=Anr + Bnr ,Θn(θ)=Cn cos nθ + Dn sin nθ, ∞ 1 u(r, θ)= (a + b ln r)+ a rn + b r−n cos nθ 2 0 0 n n n=1 n −n + cnr + dnr sin nθ. 1.15 Exercises 781

The Fourier coefﬁcients for r = a are 1 π (a + b ln a)= f(θ) dθ, 0 0 π −π 1 π a an + b a−n = f(θ) cos nθ dθ, n n π −π π n −n 1 cna + dna = f(θ)sinnθ dθ. π −π

The process is similar, when r = b gives the Fourier coefﬁcients.

(b) Use the same method described in Exercise 76(a) to ﬁnd

r =1: a0 =1,an + bn =0,n≥ 1; c1 + d1 =1,

cn + bn =0,n>1. b r =2: b = −1, 2a + 1 =1, 2na +2−nb =0 for n>1, 0 1 2 n n 2 2 2nc +2−nd =0,n≥ 1; a = ,b= − , n n 1 3 1 3 1 4 c = − ,d= , 1 3 1 3 1 2 1 u(r, θ)= (1 − ln r)+ r − r−1 cos θ − r − 4r−1 sin θ. 2 3 3

tions.

r =1: b0 =0,an − bn =0,n≥ 1; cn − dn =0,n≥ 1, 1 3 1 3 r =2: b =0,a− b = ,c− d = . 0 1 4 1 4 1 4 1 4

Thus, a1 = b1 =1and c1 = d1 = −1. − − a0 1 − 1 Hence, the solution is u(r, θ)= 2 +(r + r ) cos θ (r + r )sinθ. 1 2 2 ∇2 81. Deﬁne v(x, y)= 4 (x + y ) so that φ =1and assume w = u + v so that 2 ∇ w =0, w = v on ∂D, and w(0, 0) = u(0, 0). Use the result min∂Dv ≤

u(0, 0) ≤ max∂Dv. 782 C Answers and Hints to Selected Exercises

a2 r2 Max v = 4 and the minimum of v = 4 , where r is determined from the 2 2 2 x y circle x + y = r which touches ∂D: a + b =1, (x>0,y>0). r2 r2 2 2 2 2 2 −1 Since (x, y)=(4 , 4 ), one has r = a b (a + b ) . 2 82. Hint: We take v = u1 − u2 where u1 and u2 are solutions of ∇ u =0on D. ∇2v =0for x ∈ D and v(x)=0on ∂D.

By the minÐmax principle, v attains its maximum and minimum on ∂D so that

0 ≤ v ≤ 0 for x ∈ D, and hence, v =0on D.

2 91. Hint: At the leading order ∇ u0 =0,u0(1,θ)=sinθ. 1 − 1 Seek a separable solution u0 = f(r)sinθ so that f + r f r2 f =0,f(1) = 1. The solution is f(r)=Ar + Br−1. For the bounded solution, B ≡ 0. Hence

2 2 f(r)=r, and u0 = r sin θ.AtO(ε ), u2 satisﬁes the equation ∇ u2 =

−r sin θ, u2(1,θ)=0. For the separable solution u2 = g(r)sinθ, g(r) satis- 1 − 1 − r fies g + r g r2 g = A ,g(1) = 0. Using the variation of parameters, the 1 − 3 bounded solution is g(r)= 8 (r r ). 92. Hint: u =0satisfies the given equation and the far field boundary condition, but

not the boundary condition at r =1. We need a boundary layer near r =1so that

we can deﬁne r =1+εr with r = O(1) in the boundary layer for ε 1.At

r −r the leading order urr − u =0, which gives solution u = A(θ)e + B(θ)e . This matches with the far ﬁeld if A ≡ 0 and satisﬁes the condition at r =0

when B =1. Thus the inner solution and a composite solution are valid at the

leading order for r ≥ a. Hence the solution follows.

93. Hint: Seek a separable solution u(x, t)=X(x)T (t) =0 , where X(x) and

T (t) satisfy X + λX =0, 0

nπ 2 λκT =0,t>0. The eigenvalues are λn =( ) and the eigenfunctions are nπx Xn(x)=An cos( ), n =0, 1, 2,.... Thus, the solution is

∞ nπx κn2π2 u(x, t)=a + a cos exp − t , 0 n 2 n=1 1.15 Exercises 783

∞ nπx 1 f(x)=u(x, 0) = a + a cos , where a = f(x) dx, 0 n 0 n=1 0 2 nπx an = f(x) cos dx, n =1, 2, 3,.... 0

1 2 − n − (b) a0 = 2 , an = n2π2 [( 1) 1]. 96. The solution follows from (1.9.15), when q(ξ,τ)=sin(kξ − ωτ). ω In case (a), when c = k ,

−1 u(x, t)= k2c2 − ω2 sin(kx − ωt) −1 +(kc − ω) 2kc ω2 − k2c2 sin(kt + kct) −1 +(kc + ω) 2kc ω2 − k2c2 sin(kt − kct).

Thus, the solution consists of three sinusoidal waves that propagate with differ- ent amplitudes and with velocities ±c and the phase velocity (ω/k).

ω In case (b), when c = k , the solution is given by

1 1 t u(x, t)= sin(x − t) − sin(x + t)+ cos(x − t). 4 4 2

Thus, the solution consists of two harmonic waves that propagate with velocities ±1 and another harmonic wave whose amplitude grows linearly with time t.

98. Use the joint Laplace and ﬁnite Hankel transform of order one deﬁned by

a ∞ −st V¯ (km,s)= rJ1(rkm) dr e v(r, t) dt, 0 0

where V¯ (km,s) is the Laplace transform of V (km,t), and km are the roots of

the equation J1(akm)=0. The solution of the transformed system is 784 C Answers and Hints to Selected Exercises

2 ¯ ¯ a νkmΩf(s)J1(akm) V (km,s)= 2 . (s + νkm)

(See Myint-U and Debnath 2007, Chapter 12, p. 507.)

The solution is

∞ k J (rk ) t v(r, t)=−2νΩ m 1 m f(t − τ)exp −νk2 τ dτ. J (ak ) m m=1 1 m 0

When f(t) = cos ωt, the solution is given by

∞ k J (rk ) t v(r, t)=−2νΩ m 1 m cos ω(t − τ)exp −νk2 τ dτ. J (ak ) m m=1 1 m 0

When ω =0, the solution becomes

∞ J (rk )exp(−νtk2 ) v(r, t)=rΩ − 2Ω 1 m m . k J (ak ) m=1 m 2 m

In the limit as t →∞, the transients die out and the ultimate steady-state is

attained as the rigid body rotation about the axis of the cylinder.

99. (b) The Cauchy data tend to zero as n →∞. But, for t>0, the solution

un(x, t) →∞as n →∞for certain values of x and t. So, the problem is ill-posed.

100. (b) The initial data tend to zero as n →∞. However, for y =0 , the solution

un(x, y) →∞as n →∞. Thus, the problem is not well-posed. 101. (b) For y =0 , the amplitude of the solution tends to inﬁnity as n →∞due √ to the factor sinh(ny), even though the initial data uy(x, 0) = exp(− n) ×

sin nx → 0 as n →∞. However, the solution un(x, y) and all of its derivatives tend to zero as n →∞uniformly throughout the half-strip in the (x, y) plane.

102. Without the last term in the ﬁrst equation, the equation is known as the Eu-

ler equation. The second equation is the continuity equation for a certain vir- 2.9 Exercises 785

tual ﬂuid. The square of the amplitude a2 plays the role of density ρ and

aV (x) plays the role of pressure. The second term on the right-hand side of

the ﬁrst equation plays the role of dispersion. Thus, the transformation of the

Schrödinger equation into two equations in ﬂuid dynamics is usually referred

to as the hydrodynamic analogy of quantum mechanics.

103. (b) uxx = −n cos nx cosh ny, uyy = n cos nx cosh ny, and hence, uxx +

uyy =0. The boundary data is changed by only a small amount as u(0,y) → 0 as n →∞. Yet the solution u(x, y) is changed from zero by a large amount. 1 →∞ Along the line x =0, the solution is u(0,y)= n cosh ny as y →∞. Thus, a small change of the boundary data produces a large change in

the solution. So, the problem is ill-posed.

2 2 2 2 104. (a) We have ut = κn exp(κn t)sinnx and uxx = −n exp(κn t)sinnx,

and hence, ut + κuxx =0. 1 → →∞ →∞ →∞ (b) Thus, u(x, 0) = n sin nx 0 as n , yet u(x, t) as n for any positive t. Hence, the problem is ill-posed. 1 2− 4 105. (b) It is easy to verify that u(x, t)= n sin nx exp(κn δn )t is the solution of the equation. It also satisﬁes the initial condition. The solution is well behaved

for large n, and also it is bounded for all n for any ﬁnite t. However, the solution

is unstable as n → 0. For small δ, the negative diffusion equation is obviously

ill-posed.

2.9 Exercises t2 − t2 − t2 · · 1. (i) Hint: 0=δ t (T V ) dt = t (δT δV ) dt = t (mr˙ δr˙ + F δr) dt. 1 1 1 (ii) 0=δ t2 ( 1 mx˙ 2 − 1 mω2x2) dt,orx¨ + ω2x =0. t1 2 2 2. Hint: Apply the variational principle t2 1 δ (T − V ) dt =0, where T = ρ y˙2 dx and 2 t1 0 1 V = EIy2 dx. 2 0 786 C Answers and Hints to Selected Exercises

Thus, ρy¨ + EIy(iv) =0.

3. ηt +6ηηx + ηxxx =0. 10. (a) We have ∇4u =0(biharmonic equation).

2 2 2 (b) utt − α ∇ u + β u =0(KleinÐGordon equation).

2 (d) utt + α uxxxx =0(elastic beam equation).

d (e) dx (pu )+(r + λs)u =0(SturmÐLiouville equation). 2 ∇2 − 11. ( 2m ) ψ +(E V )ψ =0. ∗ − 2 2 T 14. utt c uxx = F (x, t), c = ρ . 18. Hint: (a) Use the vector identity ·∇ ∇ 1 · − × ∇× (u )u = ( 2 u u) u ( u). (b) Use the fact that u · (u × ω)=0to get the desired result. (c) Use ∇·u =0and add the following result: 1 · p ∇· ∂ 1 · ∇· ( 2 u u + ρ + Ω)( u)=0to the equation in (b) to ﬁnd ∂t( 2 u u)+ 1 · p ·∇ ·∇ ∇· · [u( 2 u u + ρ + Ω)] = 0, where (u )φ +(u )φ = (φu) ∂Ω Adding a zero contribution ∂t gives the energy equation. The ﬁrst term represents the rate of change of the total energy (kinetic and po-

tential), and the second term describes the energy ﬂow carried by the velocity combined with the contribution from the rate of working of the pressure forces. 19. Hint: Follow Example 2.4.2 to obtain

2 2 2 2 2 2 1+q + r px + 1+p + r qy + 1+p + q rz

− 2(pqpy + qrqz + rprx)=0.

− 2 − − − uy 2 1 A (y1 y) 22. (a) uy uxy 2 =0,(b)y = 2 . 1+y A (y1−y) 24. (a) Hint: For a conservative system, T + V = C.

Putting V = C − T in (2.4.25) gives the principle of least action. 2.9 Exercises 787

(b) The principle of least action asserts that time action is stationary for any

conservative system. − 1 2 − 27. (a) In one dimension, the Lagrangian is L = T V = 2 mx˙ V (x) so that ∂L − ∂V ∂L ∂L − d ∂L ∂x = ∂x and ∂x˙ = mx˙. The EulerÐLagrange equation is ∂x dt ( ∂x˙ )=0, − ∂V or equivalently, mx¨ = ∂x = F . This is Newton’s law of motion. − 1 2 2 − 2 2 (b) The Lagrangian L = T V = 2 m(˙x +˙y ) V (x + y ).TheEulerÐ d ∂L ∂L d ∂L ∂L Lagrange equations are dt ( ∂x˙ )= ∂x and dt ( ∂y˙ )= ∂y . In this case, these − ∂V − ∂V equations give mx¨ = ∂x and my¨ = ∂y . In this formulation, conserved quantities are not obvious.

y˙ =˙r sin θ + rθ˙ sin θ. Thus, x2 + y2 = r2 and x˙ 2 +˙y2 =˙x2 + r2θ˙2. Thus, 1 2 2 ˙2 − 2 the Lagrangian becomes L = 2 m(˙r + r θ ) V (r ). The EulerÐLagrange equations are d ( ∂L)= ∂L,or, dpr = mr¨ = mrθ˙2 − ∂V , d ( ∂L)= ∂L,or, dt ∂r˙ ∂r dt ∂r dt ∂θ˙ ∂r dpθ = d (mr2θ˙)=0because ∂L = − ∂V + mrθ˙2, ∂L =0, and p = ∂L = dt dt ∂r ∂r ∂θ˙ r ∂r˙ mr˙, p = ∂L = mr2θ˙. Since dpθ =0, one gets mr2θ˙ = constant which shows θ ∂θ˙ dt that the angular momentum is conserved and the radial equation of motion is ˙2 − ∂V mr¨ = mrθ ∂r . The ﬁrst term in the radial equation on the right-hand side represents the centrifugal force, while the second term gives the dynamical radial

force.

−brn 28. In the linearized limit, f(r)=ae − a ≈−abrn so that the nonlinear Toda

lattice equation mr¨ =2f(rn)−f(rn−1)−f(rn+1) becomes mr¨n = ab(rn+1 +

rn−1 − 2rn). Substituting the traveling wave solution into the linearized Toda 2 − 2 k lattice equation gives mω cos θ =2ab cos θ(1 cos k)=4ab sin 2 . Thus, 2 4ab 2 k the dispersion relation is ω = m sin 2 , where k corresponds to the discrete wavenumber that is similar to the continuous wavenumber. 788 C Answers and Hints to Selected Exercises 3.6 Exercises

2. (a) xp − yq = x − y,(d)yp − xq = y2 − x2.

3. (a) u = f(y),(b)u = f(bx − ay),(c)u = f(ye−x), − −1 x2−y2 (d) u = f(y tan x),(e)u = f( x ), x+u 2 − 2 x+u 2 − 2 (f) y = C1, (u + y) x = C2, f( y , (u + y) x )=0, 2 2 x + y = C1,y(u − y)=C2, dx − dy du d(u−y) −1 2 2 (g) We have y2 = xy = xu−2xy = x(u−y) , u = y + y f(x + y ), (h) u +logx = f(xy),(i)f(x2 + u2,y3 + u3)=0. − 3 2 − 2 5. (a) u =sin(x 2 y),(b)u =exp(x y ), y − y 3 − 1 2 (c) u = ⎧xy + f( x ), u = xy +2 ( x ) ,(d)u =sin(y 2 x ). ⎨⎪ 1 2 − 2 − 2 2 y +exp[ (x y )] for x>y, (e) u = ⎪ ⎩ 1 2 − 2 − 2 2 x +exp[ (y x )] for x

− 1 u = 1+(x + y)2 e 2y − (x + y)2 2 .

dy − y d y 1 y (i) dx x =1, dx ( x )= x which implies that x = C1 exp( x ). u+1 u+1 − y x = C2, f( x ,xexp( x )) = 0. x2+1 1 Initial data imply x = C1 and = C2. Hence C2 = C1 + . x C1 u+1 − y 1 y 2 − y y − x = x exp( x )+ x exp( x ). Thus, u = x exp( x )+exp(x ) 1. √ 6. We ﬁnd u2 − 2ut +2x =0, and hence, u = t ± t2 − 2x.

x 7. u(x, y)=exp(x2−y2 ). 3.6 Exercises 789

y z x−y d(x−y) dz x−z 8. (a) u = f( x , x ), (b) Hint: u1 = xy = C1, x2−y2 = z(x+y) gives u = z = x−y x−y C2, u = f( xy , z ).(c)φ =(x + y + z)=C1. dx ( dy ) dz dx + dy + dz ( z ) y ( z ) x y z d log(xyz) Hint: y−z = z−x = x−y = 0 = 0 , ψ = xyz = C2, u = f(x + y + z,xyz) is the general solution.

2 2 2 2 (d) Hint: xdx+ ydy=0, x + y = C1, zdz= −(x + y )ydy= −C1ydy,

2 2 2 2 2 2 2 2 2 2 z +(x + y )y = C2, u = f(x + y ,z +(x + y )y ).

− − − x 1dx y 1dy z 1dz d(log xyz) 2 2 2 (e) y2−z2 = z2−x2 = y2−x2 = 0 . Thus, u = f(x + y + z ,xyz) is a general solution. − x2 − x3 − x3 − x2 9. (a) Hint: y 2 = C1, u = xy 3 + C2,φ(u xy + 3 ,y 2 )=0. − x3 − x2 − x3 − x2 2 u = xy 3 + f(y 2 ), u = xy 3 +(y 2 ) . − 1 3 − x2 5 (b) u = xy 3 x + y 2 + 6 . x+u 2 − − 2 2 − 2u − − 2 − 2 − 11. y = C1, u (x y) = C2, u y (x y) y (x y)=0. 2 − Thus, u = y +(x y), y>0. τ 2 (2x−2y+y2) 12. (a) x = 2 + τs+ s, y = τ +2s, u = τ + s = 2(y−1) . τ 2 2 − 2 − 2 (b) x = 2 + τs + s , y = τ +2s, u = τ + s, (y s) =2x s ,asetof parabolas.

1 2 (c) x = 2 (τ + s) , y = u = τ + s. 13. Hint: The initial curve is a characteristic, and hence, no solution exists.

xy x2−y2+1 1 2 14. (a) u =exp(x+y ),(b)u = sin[( 2 ) ], dx dy du − − (c) x = −y = −1 gives xy = c1 and u ln y = c2 = f(xy). Hence, 2x √ xy 1 y 2 t − 2 1 ln(3x)=f(3x ).Or,f(t)=2 3 ln( 3t). Thus, u =2(3 ) + 2 log( 3x ), dx dy du x xdx+ydy du (e) d = y = x2+y2 gives y = c1, and x2+y2 = x2+y2 yields 2u = 2 2 2 2 x 2 − x + y + c2.Or,2u = x + y + f( y ). Hence, f(x)=x 1. Thus, 2u = 2 2 x2 − x + y + y2 1. dx dy du 2 − 2 − 2 − 2 (f) y2 = xy = x gives x y = c1 and u ln y = c2 = f(x y ). Hence, f(t)=t +1. Thus, u =lny +(x2 − y2)+1. 790 C Answers and Hints to Selected Exercises

2 dx dy du dy c1x (g) x = y = xy gives x = c1 and du = ydx =(c1x)dx, u = 2 + c2 = xy xy y xy y 2 + c2. u = 2 + f( x ). Hence, f(1) = 0. Thus, u = 2 + f( x ), where f is an arbitrary function such that f(1) = 0.

xdx ydy udu xdx+ydy+udu 2 2 2 15. x(u2−y2) = xy2 = −xu2 = 0 , and hence, x + y + u = c1. dy − du 2 2 2 And y = u gives uy = c2. Thus, x + y + u = f(uy), and hence, 3u2 = f(u2). Thus, 3uy = u2 + x2 + y2.

τ 2 16. (a) x(s, τ)=τ, y(s, τ)= 2 + aτs + s, u(s, τ)=τ + as. τ = x, s =(1+ −1 − 1 2 −1{ 1 2 } ax) (y 2 x )a, and hence, u(x, y)=x+as =(1+ax) x+a(y + 2 x ) , − 1 singular at x = a . √ u2 − 2 − 2 (b) y = 2 + f(u x), 2y = u +(u x) , u(0,y)= y. d(x+y+u) d(y−u) d(u−x) − 2 17. (a) Hint: 2(x+y+u) = −(y−u) = −(u−x) , (x + y + u)(y u) = c1 and 2 (x + y + u)(u − x) = c2.

dx dy dx du dx u(u2+a) (b) Hint: x = −y . Hence, xy = a. xu(u2+a) = x4 . So, du = x3 giving x4 = u4 +2au2 + b and, thus, x4 − u4 − 2u2xy = b. dx dy dy 2 − − 2 (c) x+y = x−y = 0 (exact equation). u = f(x 2xy y ). 2 − 2 − 1 2 2 − 2 (d) f(x y ,u 2 y (x y )) = 0. (e) f(x2 + y2 + z2,ax+ by + cz)=0.

dx dy dz x y 2 2 2 −1 y 18. Hint: x = y = z , and hence, z = c, z = d. x +y = a and z = tan ( x ) 2 2 2 2 −1 d a a give (c + d )z = a and z = b tan ( c ). c =(z ) cos θ, d =(z )sinθ, and z = b tan−1(tan θ)=bθ. Thus, the curves are xbθ = az cos θ and ybθ=

az sin θ. (dx−2dy) du { − 2 2} 19. Hint: 9u = −3(x−2y) . F x + y + u, (x 2y) +3u =0. Thus, (x − 2y)2 +3u2 =(x + y + u)2.

20. F (x2 + y, yu)=0, (x2 + y)4 = yu. − dz (dx+dy+dz) 2 21. Hint: x y + z = c1, −(x+y+z) = 8z , and hence, 8z +(x + 2 2 2 y + z) = c2. F {(x − y + z), 8z +(x + y + z) } =0. 2 2 − 2 2 2 2 c1 + c2 =2a ,or(x y + z) +(x + y + z) +8z =2a . 3.6 Exercises 791

22. F (x2 + y2 + z2,y2 − 2yz − z2)=0. √ (a) y2 − 2yz − z2 =0, two planes are y =(1± 2)z.

(b) x2 +2yz +2z2 =0, a quadric cone with vertex at the origin.

23. Use the Hint of 17(c). dx dy − d2x dt = x + y, dt = x y, dt2 =2x. √ dx 2 2 dx ( dt ) =2x + c. When x =0=y, dt = 2x. √ 2 u =lnx + x2 − 2xy +2y.

3 24. (a) a = f(x + 2 y). ct − a cy (b) x = at + c1, y = bt, u = c2e , c2 = f(c1), u(x, y)=f(x b y)exp( b ). x − c (c) u = f( 1−y )(1 y) . 1 2 1 −1 − 2 (d) x = 2 t + αst + s, y = t, and u = y + 2 α(αy +1) (2x y ). √ 26. (a) Hint: (f )2 =1− (g)2 = λ2; f (x)=λ and g(y)= 1 − λ2. √ √ 2 2 f(x)=λx + c1 and g(y)=y 1 − λ + c2. u(x, y)=λx + y 1 − λ + c. (b) Hint: (f )2 +(g)2 = f(x)+g(y) or (f )2 − f(x)=g(y) − (g)2 = λ. Hence, (f )2 = f(x)+λ and (g)= g(y) − λ.

√df √dg Or, f+λ = dx and g−λ = dy.

x+c1 2 − y+c2 2 f(x)+λ =( 2 ) and g(y) λ =( 2 ) .

x+c1 2 y+c2 2 u(x, y)=( 2 ) +( 2 ) . (c) Hint: (f )2 + x2 = −g(y)=λ2. √ 2 2 2 Or, f (x)= λ − x , and g(y)=−λ y + c2. Putting x = λ sin θ, we obtain √ 1 2 −1 x x 2 − 2 f(x)= 2 λ sin ( λ )+ 2 λ x + c1, √ 1 2 −1 x x 2 − 2 − 2 u(x, y)= 2 λ sin ( λ )+ 2 λ x λ y +(c1 + c2). (d) Hint: x2(f )2 = λ2 and 1 − y2(g)2 = λ2. √ 2 Or, f(x)=λ ln x + c1 and g(y)= 1 − λ ln y + c2. 1 · 1 · 27. (a) Hint: v =lnu gives vx = u ux, vy = u uy. 792 C Answers and Hints to Selected Exercises

2 ux 2 2 uy 2 x ( u ) + y ( u ) =1. 2 2 2 2 2 2 2 2 Or, x vx + y vy =1gives x (f ) + y (g ) =1. x2{f (x)}2 =1− y2(g)2 = λ2. √ 2 Or, f(x)=λ ln x + c1 and g(y)= 1 − λ (ln y)+c2. √ 2 Thus v(x, y)=λ ln x + 1 − λ (ln y)+lnc, (c1 + c2 =lnc). √ u(x, y)=cxλy 1−λ2 .

(b) Hint: v = u2 and v(x, y)=f(x)+g(y) may not work.

Try u = u(s), s = λxy, so that ux = u (y) · (λy) and uy = u (s) · (λx).

2 1 du 2 Consequently, 2λ ( u ds ) =1. Or, 1 du = √1 1 . Hence, u(s)=c exp( √s ). u ds 2 λ 1 λ 2 Thus, u(x, y)=c exp( √xy ). 1 2

1 √ux 1 √uy 4 2 2 2 28. Hint: vx = 2 u , vy = 2 u . This gives x (f ) + y (g ) =1. Or, x4(f )2 =1− y2(g)2 = λ2.

Or, x4(f )2 = λ2 and y2(g)2 =1− λ2. √ − λ − 2 Hence, f(x)= x + c1 and g(y)= 1 λ ln y + c2. √ − λ − 2 2 Thus, u(x, y)=( x + 1 λ ln y + c) . 2 v2 ux uy vx y 29. Hint: vx = u , vy = u . x2 + y2 =1, and v = f(x)+g(y). 2 (f ) − 1 2 2 Or, x2 =1 y2 (g ) = λ . √ f (x)=λx, and g(y)= 1 − λ2y. √ λ 2 1 2 − 2 Or, f(x)= 2 x + c1, and g(y)= 2 y 1 λ + c2. 2 √ λ 2 y − 2 v(x, y)= 2 x + 2 1 λ + c =lnu. 2 √ λ 2 y − 2 u(x, y)=c exp[ 2 x + 2 1 λ ],c1 + c2 =lnc. x2 λ x2 e = u(x, 0) = ce 2 , which gives c =1and λ =2.

6x+3at2+5at3 31. (b) v(x, t)=x + at, u(x, t)= 6(1+2t) . − x dt dx dv 32. (a) We have vt avx =0, v(x, 0) = e . So, 1 = −a = 0 gives dv =0,

or, v = const. = c1.Also,dx + adt =0yields x + at = const. = c2. Thus, 3.6 Exercises 793

v = f(x + at), and ex = v(x, 0) = f(x).Or,v = ex+at. The ﬁrst equation

−x at dt dx du becomes ut + uux = e v = e . This gives 1 = u = eat . − eat eat Or, u a = c1. Thus, dx = udt =( a + c1)dt. eat eat − eat eat − teat Or, x = a2 + c1t + c2 = a2 +(u a )t + c2 = a2 + ut a + c2. − t at − eat − eat − t at − eat Or, x ut + a e a2 = c2. Thus, u(x, t) a = f(x ut + a e a2 ). − 1 at − 1 at − t at − 1 at Hence, u a e = c1 and u(x, t) a e = f(x ut + a e a2 e ). − 1 − 1 1 − 1 Using the initial condition gives x a = f(x a2 ),orf(x)=(x + a2 a ). 1 at − t at − 1 at 1 − 1 Hence, u(x, t)= a e +(x ut + a e a2 e + a2 a ). −1 1 t − 1 at 1 − 1 Thus, u(x, t)=(1+t) [x +(a + a a2 )e +(a2 a )]. √ √dx dy du − 1 − 33. (a) We have x = u = −u2 leads to 2 x u = C1 and y = ln(C2u). √ √ −1 At (x0, 0), u =1so that C1 =2 x0 − 1 and C2 =1. Thus, u =2( x − √ √ √ −y −1 x0)+1and u = e , y =lnu =ln[2( x − x0)+1].

dx dy du 1 y 1 (b) We have ux2 = e−y = −u2 so that x =lnu + C1 and e = u + C2. −y At (x0, 0), u =1, that is, C2 =0and u = e .

1 1 1 1 1 Hence, C1 = . Thus, ln u =( − ) and y = − . x0 x x0 x0 x ∂nu ∂nu ∂u ∂u − ∂u 34. For any function u(ξ,τ),wehave ∂xn = ∂ξn , ∂t = μ ∂τ c ∂ξ , 2 2 utt = μ uττ − 2μcuξ,τ + c uξξ,etc.

n dτ dξ Thus, the equation assumes the form uτ +(αu )uξ =0.Wehave 1 = αun = du du dξ n n 0 or dτ =0and dτ = αu . Thus, u = const. = C1 and ξ = αu τ + C2. Then the general solution is f(u, ξ − αunτ)=0,oru = φ(ξ − αunτ) which

is a Riemann simple wave. Hence, u(ξ,0) = φ(ξ)=u0 sin kξ. Thus, u(ξ,τ)=

u n n u0 sin[kξ − ( ) (αku τ)]. u0 0 dx dy du dy x−y 35. (a) We have x+y = x−y = 0 . So, u = const. = c1. Hence, dx = x+y .Thisis 2 2 an exact equation and hence, f(x, y)=c2,or,x − 2xy − y = c2. Thus, the general solution is u = f(x2 − 2xy − y2), f is an arbitrary function. 794 C Answers and Hints to Selected Exercises

dx dy du dy (b) We have 1 = −(ax+by) = 0 . Thus, u = const. = c1. Hence, dx + by = 2 2 −ax,givingy(b y − abx + a)=c2. Hence, u = f(y(b y − abx + a)) is the general solution. dx dy du 2 − 2 du 2 − 2 (c) We have x−1 = y−1 = x2−y2 . Hence, x y = c1. dx = x(x y )=c1x 1 2 − 1 2 2 − 2 − 1 2 2 − 2 giving u = 2 c1x + c2. Thus, u 2 x (x y )=c2 and u 2 x (x y )= f(x2 − y2). dx dy du 2 − 2 (e) We ﬁnd that y = −x = 0 . This gives x y = C1 and u = C2. So, u = f(x2 + y2). Hence, y = f(a2 + y2)=f(t), a2 + y2 = t. Therefore, √ f(t)= t − a2. Thus, u = f(x2 + y2)= x2 + y2 − a2. dx dy du 1 u2 − (f) We ﬁnd that x = x+y = udu,or, dx = xu ,or 2 log x = c1. Thus, dy x+y y y − u2 − dx = x , and hence, x =logx + c2. So, x log x = f( 2 log x). Thus, − − − − −t y x log x = f( log x). Putting t = log x, f(t)=t e . Thus, x = u2 − − u2 2 x exp( 2 ). dt dx du 36. (a) We have, 1 = −x = u gives t +lnx = C1 and xu = C2. 1 Or, g(xu, t +lnx)=0,or,u = x h(t +lnx) is the general solution, where h is 1 an arbitrary function. Hence, u(x, 0) = f(x)= x h(ln x),or,h(ln x)=xf(x), that is, h(x)=exf(ex). Thus, u(x, t)=etf(xet).

4.6 Exercises

1. (a) 16u =(x +4y)2,(b)u = α(x + y). √ 2. (a) a log u = x + ay + b,(b)2 u(1 + a2)=x + ay + b,

1 1 2 − 2 2 (d) u = ay + 4 (x + b) ,(g)u = a log x +(4 a ) log y + b, 1 2 2 1 2 (h) u = a(1 + x ) + 2 (ay) + b. − xy 3. (a) log u = x + y 1,(c)u = 2(y−2) which is singular at y =2, (e) u(x, y)=x + y, (f) 4u =(x + y +2)2. (i) u =exp(−act)f(x − ct). Use equation (4.2.19)

x(τ,0) = x0(τ)=τ, t(τ,0) = 0, u(τ,0) = f(τ). 4.6 Exercises 795

4. Hint: At s =0, p(t, x)=q(t, s)=1, then, p(t, s)=q(t, s)=1.

x(t, s)=(t − 2)es +2, y(t, s)=1+es,

u(t, s)=(t − 1)es +2, u(x, y)=x + y − 1.

8. Hint: x(0,s)=0, y(0,s)=s, u(0,s)=−s.

x(t, s)=− sinh t, y(t, s)=s cosh t,

1 − s − 2 2 u(t, s)= 2 (1 + cosh t),u(x, y)= y(x +1) . 10. Hint: Characteristics are x − 2ct = s where s is a parameter. The points on

the characteristic lines travel with speed 2c, whereas points on the wave proﬁle

u = c(x − ct) move with speed c. The wave proﬁle u is not constant on the

characteristic lines.

1 y 11. (a) u = 2 xy + f( x ) where f is an arbitrary function such that f(1) = 0. (b) Characteristics are xy = const. (a family of hyperbolas).

y xy 1 a 2 u = a log y + f(xy), u = 2 log( 3x )+2( 3 ) . 1 t t (c) At t =0, p =0, q = s ; sp = tanh( s ), sq = sech( s ). − 2 t t 2 2 t − 2 x(s, t)= 2s tan( s ) sech( s ), y(s, t)=2s sech ( s ) s . t 2 − 2 2 2 u(s, t)=2s sech( s ), (u 2y) =4(x + y ). (d) u(x, y)=x − y,(e)u(x, t)=x2 tan t.

dx 1 dp 1 dq 12. Hint: Write the characteristic strip equations dt = Fp,.... Since p dt = q dt ,

p and q are proportional to each other. The initial data p0(s) and q0(s) are

2 2 1 the solutions of p0(s)=0and (p0 + q0) 2 = tan θ so that p0(s)=0

and q0(s) = tan θ. Since p0(s)=0, the constant of proportionality must be zero, and so p(t, s)=0. We then ﬁnd x(t, s)=s and z(t, s) = cos(θ − t). dq − 2 − − Then dt = [q + q cos(θ t)] gives q(t, s) = tan(θ t) y(t, s)= − sin(θ − t)+2sinθ, or equivalently, (y − 2sinθ)2 + z2 =1, which rep-

resents a cylinder.

− 2 1 dz ± 1 1 z 2 14. Hint: ( dξ )= z ( 1+a2 ) , 2 1 x+ay d(1 − z ) 2 = ±d( √ )=±d(x cos θ + y sin θ). a2+1 796 C Answers and Hints to Selected Exercises

dp p 15. (a) Hint: dq = q gives p = c1q. dx − dp du − dt +(u a) dt + p dt =0,ordx +(u a) dp + pdu=0, or d(x +(u − a)p)=0gives x + p(u − a)=const.

Similarly, y + q(u − a)=const.

These equations and F =0give the integral surfaces 2 2 − 2 − (x−ct) x + y +(u a) 1=0, u(x, t)= 1+t(x−ct) . dx 2 dy 2 du − 2 (b) Hint: dt =2pu , dt =2qu , dt = 2(1 u ), dp − 2p dq − 2q dt = u , and dt = u . d − 2 d − 2 (˙pu +˙up)= dt (up)= 2pu , dt (uq)= 2qu . dx 2 − d dy − d Consequently, dt =2pu = dt (up) and dt = dt (uq), and hence, (x − a)2 +(y − b)2 = u2(p2 + q2)=1− u2, where a and b are con-

stants.

2 2 2 (c) (x − x0) +(y − y0) =(u − u0) , where x0, y0, and u0 are constants. This

is a three-parameter⎧ family of solutions. Use equation (4.3.1). ⎪ ⎨ x2 if y =0, 16. (a) u(x, y)= √ ⎩⎪ − (2y2) 1(1 + 2xy − 1+4xy) if y =0 . ⎧ ⎪ ⎨ x if y =0, (b) u(x, y)= √ ⎩⎪ − (2y) 1( 1+4xy − 1) if y =0 .

5.5 Exercises

1. (a) u(x, t)=sinξ, ξ = x − t sin ξ. x →∞ → (b) u(x, t)= 1−t , u as t 1. 1 2 n−1 2. Hint: (u)t +(2 u )x =0. Multiplying it by nu gives the ﬁrst result. − − 1 − 1 4. ξ = x t exp( ξ ) for x>0; u(x, t)=exp( ξ ). 6. The characteristic diagram in the (x, t)-plane shows that the initial signal f(x)

is focused along the characteristics to a region near x =0as t increases. 5.5 Exercises 797

− t2 7. u = t + A, x = ut 2 + B, where A and B are constants. √ 8. Hint: Shock t =2(x − 1) from (1, 0) to (2, 2) and shock x = 2t from (2, 2). √ ≥ x u =0for x<0, and, when x 0, u = t to the left of t = x and x = 2t, u =0to the right of shocks, and u =1in a triangular region bounded by t =0,

t = x, and t =2(x − 1).

9. u = −1 to the left of the line t + x =0, and u =1to the right of the line

x − t =1. Between these lines, u =(2x − 1)/(1 + 2t). ≤ ≤ 1 1 10. Characteristics from 0 x t, t =0intersect at ( 2 , 2 ). Characteristics from ≤ ≥ ≥ 1 x 0, x 1 also intersect in the region t 2 , and hence, no single-valued 1 1 solution. Shock parallel to the t-axis from ( 2 , 2 ). We ﬁnd u(x, t)=(2x−1)/(1−2t) in the triangular region bounded by x−t =0, x + t =1 t =0 , . ⎧ ⎨⎪ +1 to the left of shock, u(x, t)=⎪ ⎩ −1 to the right of shock.

11. When a<0, shock along the line at =2x, −a to the left of shock, 2a to the

right. When a>0, −a to the left of the line x = −at and 2a to the right of the

x line x =2at. Between these lines, u =(t ). √ 2 dt dx du 15. Hint: Here c(u)=u , f(x)= x, 1 = c(u) = 0 . Thus, du =0gives u(x, t) = const. on the characteristics. √ u(x, t)=f(ξ)= ξ. √ ξ = x − tF (ξ)=x − tc(f(ξ)) = x − tc( ξ)=x − tξ.

1 x x 2 Thus, ξ = 1+t . Hence, u(x, t)=(1+t ) . 1 3 3 dt dx du 16. c(u)=u , f(x)=x ,wehave 1 = u3 = 0 . u(x, t) = const. on the characteristics.

1 Thus, u(x, t)=f(ξ)=ξ 3 .

1 ξ = x − tF (ξ)=x − tc(f(ξ)) = x − tc(ξ 3 )=x − tξ.

1 x x 3 Hence, ξ = 1+t and u(x, t)=(1+t ) . 798 C Answers and Hints to Selected Exercises

dt dx du du du 17. c(u)=u, f(x)=x; 1 = u = 2t gives dt = u and dt =2t. Hence, u = t2 + A on the characteristic curve x = x(t) joining (x, t) and (ξ,0). dx − 2 2 − 2 2 Also, dt =[u(x, t) t ]+t =[u(ξ,0) 0] + t = ξ + t with ξ = x(0). t3 The equation of the characteristic curve is x(t)=ξt + 3 + B with x(0) = ξ. t3 x(t)=ξ(1 + t)+ 3 . u(x(0), 0) = u(ξ,0) = f(ξ)=ξ. 2 2 3x−t3 Thus, A = ξ, u(x, t)=t + ξ = t + 3(1+t) . dt dx du 18. Hint: 1 = 3t = u , u(x, 0) = f(x) = cos x. dx du 3 2 t Or, dt =3t and u = dt.Or,x = 2 t + A and u(x, t)=Be . − 3 2 B = u(ξ,0) = cos ξ, with x(0) = ξ. Then A = ξ and x 2 t = ξ. t t − 3 2 Consequently, u(x, t)=e cos ξ = e cos(x 2 t ). dt dx du 19. Hint: 2 = 1 = 0 , and hence u(x, t) = const. on the characteristic curve dx 1 t dt = 2 , or equivalently, x = 2 + A. t If x(0) = ξ, then x = ξ + 2 . ⎧ ⎪ ⎨ sin ξ if 0 ≤ ξ ≤ π, u =(x, t)=const. = u(ξ,0) = ⎩⎪ 0 otherwise. ⎧ ⎨⎪ − t ≤ − 1 ≤ sin(x 2 ) if 0 x 2 t π, Or equivalently, u(x, t)= ⎩⎪ 0 otherwise. dt dx du 20. Hint: 1 = 1 = x gives x = t + A, and hence, x = t + ξ,ifx(0) = ξ. du 1 2 dt = x(t)=t + ξ, and hence, u(x, t)= 2 t + ξt + B. − 1 2 − Or, B = u(x, t) 2 t ξt = u(ξ,0) = sin ξ. 1 2 − − If x>t, u(x, t)= 2 t + t(x t)+sin(x t). If (x, t) lies above the line x = t, characteristic passing through this point

never reaches the x-axis, but it reaches the t-axis so that the boundary condi-

tion u(0,t)=t can be used (the initial condition cannot be used in this case, du − − x

With η = t − x, the solution is given by 1 2 − − − 1 − 2 1 2 − u(x, t)= 2 t (t x)t +(t x)+ 2 (t x) = 2 x + t x. t2 On x = t, u(x, t)= 2 . dx dt du dx 2 du dt 21. We have x2 = u = 1 = dz gives dz = x , dz =1, dz = u, whence − −1 1 2 x =(a z) , u = z + b, t = 2 z + bz + c, when a, b, and c are constants. The characteristic equation originating from the initial curve at x = s, t =1− s 1 2 − s(z =0)is x =(1−sz ), u = z(b =0), t = 2 z +1 s. For each x, t, the solution u depends on the parameters s and z. Eliminating 1 2 − 1 2 − s and z gives the answer as follows: s = 2 z +1 t =(2 u +1 t) and x x 1 2 − s = 1+zx =(1+ux ). Thus, ( 2 u +1 t)(1 + ux)=x.

22. The Darcy law for water is Vd = −KHx, K is the hydraulic conductivity and p v2 ≈ p 2 H is the hydraulic head, H = gρ + z + 2g gρ + z for small (v /2g), p is the pressure, g is the acceleration to gravity, and ρ is the density. The Darcy law

describes the assumption that water ﬂows slowly in the soil pores for the head

loss to be proportional to the velocity.

The characteristic equations associated with BuckleyÐLeverett equation are

dt dx ds ds 1 = c = 0 . Or equivalently, dt =0, and hence, s = const. and it is a dx conserved quantity. Hence, dt = c is the speed at which the interface between ∂F water and hydrocarbon moves. Thus, c = ∂s =0when s =0and s =1, and c attains its maximum when 0

known property of hydrocarbons as non-wetting liquids, the mobility of which

is very low for small water saturation.

6.12 Exercises

1. Hint: Use ht +(uh)x =0. 800 C Answers and Hints to Selected Exercises ⎧ ⎪ ⎪ 200 if ξ<−ε, ⎨⎪ 2. ρ(x, t)=f(ξ)= 100(1 − ξ ) if − ε<ξ<ε, ⎪ ε ⎪ ⎩⎪ 0 if ξ>ε, ⎧ ⎪ − − ⎪ ξ 60t if ξ< ε, ⎨⎪ x = ξ +[60− 3 f(ξ)]t = (60t − ε)( ξ ) if − ε<ξ<ε, 5 ⎪ ε ⎪ ⎩⎪ ξ +60t if ξ>ε, ⎧ ⎪ ⎨ 200 if x<−(60t + ε), ρ(x, t)=⎪ ⎩ − x − 100(1 60t+ε ) if (60t + ε)

1 (b) For t< 15 , we eliminate ξ from the above solution to obtain ⎧ ⎪ − ⎪ 25(3 + x 15t ) if − 1+30t ≤ x ≤ 15t, ⎨⎪ 1−15t − ρ(x, t)=⎪ 25(3 − x 15t ) if 15t ≤ x ≤ 1+30t, ⎪ x+15t ⎩⎪ 50 if |x − 30t|≥1.

1 − The solution ρ(x, t) for t = 15 (1 ε) can be found from the result given in 1 Exercise 3(a). Use results in 3(a) to ﬁnd ρ and x for t = 15 . ⎧ ⎪ ⎨ 25(3 −|ξ|) if |ξ|≤1, ρ = and ⎩⎪ 50 if |ξ>1|, 6.12 Exercises 801 ⎧ ⎪ ⎪ 2+ξ if ξ>1orξ<−1, ⎨⎪ x = 1+2ξ if 0 ≤ ξ ≤ 1, ⎪ ⎪ ⎩⎪ 1 if − 1 ≤ ξ<0.

1 Use these results to draw a graph of ρ(x, 15 ).

4. Hint: Use u = x − c0t and v = t as independent variables. 12. Hint: The equation in matrix form is ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2 ⎜ ucρ 0 ⎟ ⎜ px ⎟ ⎜ 100⎟ ⎜ pt ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1 u 0 ⎟ ⎜ u ⎟ + ⎜ 010⎟ ⎜ u ⎟ =0. ⎝ ρ ⎠ ⎝ x ⎠ ⎝ ⎠ ⎝ t ⎠

00u Sx 001 St

The eigenvalues are the roots of the equation 2 u − λcρ 0 dx 1 u − λ 0 =0 which gives λ = = u, u ± c. ρ dt 00u − λ

13. Hint: du + λ2,1 dv =0along the characteristics C1 and C2, respectively. a(du)2 +2bdudv+ c(dv)2 =0, which corresponds to the characteristics

dx = λdt, a(dx)2 − 2bdxdt+ c(dt)2 =0.

− du+ 1 14. Hint: Along the characteristics C+: η = x ct = const. and dt + 2 au+ =0 − 1 − with the solution u+(t)=u+(0) exp( 2 at) with η = x ct and u−(t)= − 1 u−(0) exp( 2 at) with ξ = x + ct. dρ ∂ρ dx · ∂ρ 15. Hint: ρ = ρ(x(t),t), dt = ∂t + dt ∂x, and then use equations (6.3.1) and (6.3.2).

16. Hint: λ1,2 = u ± c, L1,2 =(1, ±2).

∂ ∂ dx ( ∂t + λ1 ∂x)(u +2c)=Hx along C+: dt = u + c, ∂ ∂ − dx − ( ∂t + λ2 ∂x)(u 2c)=Hx along C−: dt = u c. 802 C Answers and Hints to Selected Exercises

18. Hint: See Jeffrey and Taniuti (1964, pp. 72, 73).

19. Hint: Equate xsr and xrs to derive the EulerÐPoissonÐDarboux equation.

∂ρ 22. Hint: The first equation can be written as ∂t + div(ρu, ρv, ρw)=0. The first and second equations can be combined to find the result ⎛ ⎞ ⎛ ⎞ ⎜ ρu ⎟ ⎜ p + ρu2 ρuv ρuw ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ⎜ ⎟ ⎜ ⎟ ⎜ ρv ⎟ +div⎜ ρuv p + ρv2 ρvw ⎟ =0. ∂t ⎝ ⎠ ⎝ ⎠ ρw ρuw ρvw p + ρw2

The above two results lead to the desired conservation form. ⎛ ⎞ ⎛ ⎞ c u ⎜ 0 0 ⎟ ⎜ ⎟ ⎜ a ⎟ ⎜ a ⎟ ⎜ ⎟ ⎜ ⎟ 25. For (i), A = ⎜ −100⎟ and B = ⎜ 0 ⎟. ⎝ ⎠ ⎝ ⎠ 001 −v − w ⎛ ⎞ ⎛ ⎞ c c u ⎜ − 0 ⎟ ⎜ ⎟ ⎜ a a ⎟ ⎜ a ⎟ ⎜ ⎟ ⎜ ⎟ For (ii), A = ⎜−1 − c c 0 ⎟ and B = ⎜ u ⎟, ⎝ a a ⎠ ⎝ a ⎠ 1 − v − w 002 2 2 and thus, for both cases (i) and (ii), ⎛ ⎞ ⎜ v ⎟ ⎜ ⎟ ⎜ ⎟ U = ⎜ w ⎟ . ⎝ ⎠ u

26. Characteristics drawn from the negative x-axis are vertical lines x = const.,

and along these characteristics, ρ = ρ0 and ρ(x, t)=ρ0 for x<0, t>0. Characteristics drawn from points x>asatisfy x˙(t)=c(ρ(x, 0), 0) =

c(0) = ρm.Ifx>a+ ρmt, t>0, ρ(x, t)=0.

ρ ρ If 0

27. (a) This system of two equations is known to be hyperbolic. The ﬁrst equa-

tion is already in the one-dimensional divergence form because it becomes ρt +

(ρu)x =0. The one-dimensional form of the second equation is not in divergence form, but it is multiplied by ρ and added to u times the ﬁrst equation to

2 obtain the conservation form (ρu)t +(ρu + p)x =0.

∂ρ ∂u (c) We have ρt +(ρu)x =0, or equivalently, ∂u ∂t +(ρxu + ρux)=0. dρ · ∂u dρ ∂u ∂u 1 ∂p Thus, du ∂t +(ρ + u du )( ∂x)=0. It follows that ∂t +(uux)+ ρ ∂x =0,or ∂u 1 ∂p equivalently, ∂t +(uux + ρ ∂x)=0. ∂u c2(ρ) dρ ∂u Thus, ∂t +(u + ρ du ) ∂x =0.

28. We have xvt +wx =0and close the system by using vx = wt. Thus, the Tricomi

1 v 0 x equation becomes Ut + AUx =0, where U = and A = . This has w −10 2 − 1 eigenvalues λ = x for x =0. When x<0, the eigenvalue are real and distinct, and the system is hyperbolic.

When x>0, the system is elliptic, and the system is parabolic when x =0.

29. The ﬁrst equation is in conservation form: ht +(uh)x =0. Multiply the second equation by h and add the result to u times the ﬁrst equation to obtain the

conservation form.

7.9 Exercises

5. Hint: ∞ ∞ u2 I˙4 = xut dx = − x∂x + Ku˜ dx −∞ −∞ 2 ∞ ∞ ∞ 1 2 1 = u + Ku˜ dx = I2(u)+ dx K(x − s)u(s, t) ds −∞ 2 2 −∞ −∞ 1 = I (u)+c(0)I (u). 2 2 1

− 1 − Integrating with respect to t gives I4(u) 2 tI2(u) tc(0)I1(u)=const. 804 C Answers and Hints to Selected Exercises √ 6. Hint: (a) Multiply the given equation by t to get

√ √ − 2 tu t + t uxx 3u x =0.

(b) Multiply the given equation by 2tu to obtain

2 − 2 − 3 tu t + t 2uuxx ux 4u x =0.

7. Hint: Multiply equation (2.7.67)byu and follow the procedure described by Benney (1974) and Miura (1974) to obtain

1 3 1 4 2 2 u + u + wu uz + u hx =0, 3 t 4 x

1 3 1 4 2 1 3 − 1 3 − or equivalently, ( 3 u )t +(4 u + u h)x +(3 u w)z 3 u wz 2huux =0. Multiply equation (2.7.67)byh to obtain

1 (hu) − uh + huu + hwu + h2 =0. t t xx z 2 x

This equation is added to the previous equation to generate 1 3 2 1 4 1 2 1 3 − hu + u + u h + h + h + u w z huux 3 t 4 2 x 3

+ hwuz − huux − uht =0.

h Write m = 0 udzso that ht + mx =0, and then 1 1 1 hu + u3 + hu2 + um + u4 + h2 3 3 2 x x 1 + w hu + m + u3 =0. 3 z

This gives the answer. 8.14 Exercises 805

∞ ∞ 8. Hint: (a) ( −∞ udx)y =0, and hence, −∞ udx= c(t). Under suitable conditions, c(t) is a constant, and hence, it can be set equal to

zero. To prove (c), we ﬁnd ∞ ∞ − 2 ∞ udx + uxx 3u −∞ +3 vdx =0. −∞ t −∞ y

Both (b) and (c) represent the conservation of momentum in both x- and y-

directions. H ∂H 10. It follows from the deﬁnition of that ∂J = ω, which means that the frequency of the modulated wave is determined by the partial derivative of H with respect

∂k ∂ω to the wave action J. Thus, ∂t + ∂x =0reduces to the Hamiltonian form ∂k ∂ ∂H L ∂t + ∂x( ∂J )=0. We next replace ω by J in the Whitham equation to obtain ∂J ∂ ∂H ∂H − ∂L ∂t + ∂x( ∂k )=0because ∂k = ∂k . Thus, the above equation describes ∂H conservation of the wave action J, and ∂k represents the density of its ﬂux. ∞ The quantity I = −∞Jdxis conserved in a localized wave packet. It is noted that variables J and k are introduced to describe the modulated wave, and they

play the role similar to the canonical variables of the action-angle in classical

mechanics.

8.14 Exercises

1. (a) α = β =1and γ =2,(b)β =2α,

2. In D1 and for c ≥ 2, u(ξ)=A exp(m1ξ)+B exp(m2ξ), where m1, m2 = √ 1 − ± 2 − 2 [ c1 c 4].InD2, u(ξ)=1+A1 exp(n1ξ)+B1 exp(n2ξ), where n1, √ 1 − ± 2 n2 = 2 [ c c +4]. 1 To determine A, B, A1, and B1, we use the facts that (i) u(0) = 2 , (ii) the values of u (0) in D1 and D2 are equal, and (iii) u(−∞)=1and u(∞)=0.InD1, 806 C Answers and Hints to Selected Exercises

for ξ>0,wehave

1 2 2 u(ξ)= √ c − 4+2c − c +4 exp(m1ξ) 4 c2 − 4 2 2 + c − 4 − 2c + c +4 exp(m2ξ) .

− 1 In D2,forξ<0, we ﬁnd u(ξ)=1 2 exp(n1ξ).

When c =2, we ﬁnd solutions in D1 and in D2.

1 − − √1 − In D1,forξ>0, u(ξ)= 2 exp( ξ)+(1 )ξ exp( ξ), and in D2,forξ<0, √ 2 − 1 − u(ξ)=1 2 exp[( 2 1)ξ]. 8. Hint: ψ˜ = aβ+1ψ, and set β =0to ﬁnd the similarity solutions.

9. Hint: The boundary conditions are u(η =0)=0and u(η →∞)=1,

v(η =0)=vω.

c2 2 11. v + 4 (η v +5ηv +3v)=0. η γ −1 − α 1 β − Hint: ηt = ( β ) t , ut = β t (γv αηv), γ −2 2γ γ γ β { α 2 2 α − α − } utt = t ( β ) η v + β ( β + β +1)ηv +(β 1)( β )v , γ − 1 γ −2 ux = t β 2 v , uxxxx = t β v . 13. Hint:

a a ˙ a − Q(t)=2 uxuxt dx =2[uxut]0 2 uxxut dx 0 0 a = −2 uxx uxx + F (u) dx 0 a a − 2 2 = 2 uxx dx +2 uxF (u) dx. 0 0

Using the boundary condition gives the Poincaré inequality

a a 2 ≥ 2 2 uxx dx π ux dx 0 0

and 9.14 Exercises 807 a a a 2 ≤ 2 ≤ 2 uxF (u) dx ux F (u) dx A ux dx, 0 0 0 Q˙ (t) ≤ 2 A − π2 Q(t)=μQ(t).

Integrating from 0 to a gives the inequality.

14. See Murray (1997), p. 357.

2 20. Combining ut +(ρu)x =0with (ρu)t +(ρu )x =0gives ut + uux =0.

1 2 dt Or in other words, ut +(2 u )x =0. The characteristic equations are 1 = dx du dx u = 0 , which leads to u = const. for dt = u. Thus, u is invariant, and ρ 1 2 1 2 and ρu are conserved quantities. Writing U = u and F = 2 u = 2 U gives

Ut + Fx =0. Or equivalently, Ut + FU Ux =0, that is, Ut + cUx =0, where

∂F c = ∂U = U = u is the propagation velocity. Differentiating the inviscid

Burgers equation ut + uux =0with respect to x gives (ux)t + u(ux)x =

− 2 dt dx dux ux. Thus, the characteristic equations for ux are = = − 2 , that is, 1 u ux d 2 dx 1 ux = −u for = u. Integrating gives − = −t + A, where A is a dt x dt ux 1 constant. When t = t0, (ux)=u , and A = t0 − . Thus, the solution becomes 0 u0 − −1 (ux)(t)=u0[1 + (t t0)u0] .

9.14 Exercises

dE 20. Hint: Add two equations in 20(d) and integrate the result to show dt =0.

21. Hint: If u0(x)=(c1 − γ)exp(−|x − x1|)+(c2 − γ)exp(−|x − x2|), x ∈ R

with c1 <γ, c2 >γand x1

m0 = u0 − u0xx =2c1δ(x − x1)+2c2δ(x − x2),

where δ(x) is the Dirac delta function.

22. Hint: u(x, t) is a solution of 808 C Answers and Hints to Selected Exercises T (uφt + F (u)φx) dx dt + u(x, 0)φ(x, 0) dx =0, 0 R R

∈ ∞ × R 1 2 ∗ 3 2 for any T>0 and φ C0 ([0,T] ) with F (u)= 2 u + p ( 2 u )+γu, where p(x) is deﬁned in 21(a). (See Guo 2009.)

10.9 Exercises

10. We seek the solution ψ(x, t)=Ψ(x)e−iωt where ω =(E/) is the frequency

of the De Broglie wave and the function Ψ(x) satisﬁes the equation

2mE Ψ (x)+ (1 − x)Ψ =0. 2

2 1 Making the change of variable z = −(1 − x)(2mE/ ) 3 reduces the equation

into the form Ψ − zΨ =0. The bounded solution at |z|→∞can be expressed

in terms of the Airy function

∞ s3 Ψ(z)=CAi(z)=C cos sz + ds, 0 3

where C is a constant. Thus, the solution describing the behavior of the electron

is ψ(z,t)=CAi(z)exp(−iωt). The function |ψ|2 describes the probability of

the electron at a point z.Forz>0, the probability drops to zero suddenly as if

there were an obstacle reﬂecting the electron at z =0which is referred to as the

turning point. For z<0, the solution ψ(z,t) represents a nonuniform standing

wave. Using the asymptotic expansion of Ai(z) for |z|→∞, in the form ⎧ ⎪ ⎨⎪ 1 − 2 3/2 →∞ √ 1 exp( 3 z ), when z , ∼ πz 4 Ai(z) ⎪ ⎩⎪ 1 2 | |3/2 π →−∞ √ 1 cos( 3 z + 4 ), when z . πz 4 12.10 Exercises 809

The solution for ψ(z,t) can be written as the superposition of two progressive

waves in the form C 2 π ψ(z,t) ∼ √ exp −i ωt + |z|3/2 − 4 3 4 z 2 π +exp −i ωt − |z|3/2 + . 3 4

This represents the De Broglie wave for the particle in free space.

11.14 Exercises

u 6. Hint: φ(x, t) = tan( 4 ), x + Ut x − Ut φ(x, t) ∼±U exp ±√ ∓ U exp ∓√ as t →∓∞. 1 − U 2 1 − U 2

1 γ γ 1 Note that ∓U exp(θ)=± exp{θ + √ }, where = 1(1 − U) 2 log U. 2 1−U 2 2

−1 a1+a2 1 14. (b) u4 = 4 tan [( ) tan{ (u2 − u3)}] for any solutions u2,u3. a1−a2 4 u tan r =exp(θ ),θ=(a x + a−1t + ε ),r=2, 3. 4 r r r r r

15. Hint: Use ux(x, t) instead of u(x, t). 22. v − c2v − a2v =0. tt xx 1 2 − x2 −| | 24. u(x, t)= 2c J0[d t c2 ]H(ct x ).

12.10 Exercises

1. (a) The continuity and the momentum equations for the inviscid Burgers equa- ρ tion can be written in the matrix form Ut + AFx =0by deﬁning U = = ρu ρu u u1 , F = F1 = = 2 . 2 2 u2 F2 ρu u2/u1

∂F1 ∂F1 ∂u1 ∂u2 01 01 Thus, A = = u2 2 2u2 = 2 . ∂F2 ∂F2 −( ) −u 2u u1 u1 ∂u1 ∂u2 810 C Answers and Hints to Selected Exercises The eigenvalues of A are the roots of 0=|A − λI| = −λ 1 = λ2 − −u2 2u−λ 2uλ + u2. Thus, A has two equal eigenvalues (λ = u, u) which are not distinct.

Hence, the system is not hyperbolic. ∂F1 ∂F1 ∂u1 ∂u2 01 01 ∂P (b) A = = ∂P u2 2 2u2 = 2 2 , where c = .The ∂F2 ∂F2 −( ) c −u 2u ∂u1 ∂u1 u1 u1 ∂u1 ∂u2 eigenvalues are real and distinct roots of |A − λI| =0. −λ 1 Or, |A − λI| = =0, that is, λ = λ1, λ2 = u ∓ c. c2−u2 2u−λ The corresponding eigenvectors are 1 and 1 , and a new matrix B can u−c u+c be deﬁned by B = 11, and hence, B−1 = 1 c+u −1 with B−1AB = u−cu+c 2c c−u 1 − diagonal matrix, D = u c 0 = λ1 0 . 0 u+c 0 λ2 −1 1 u − The Riemann invariants are dR = B dU, and hence, dR1 =(2 + 2c )dU1 1 1 − u 1 R1 2 dU2 and dR2 =(2 2c )dU1 + 2 dU2, where the matrix R = . R2

Fx = S. Or equivalently, Ut + AUx = S, where U, F , and S are n × 1 matrices given by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ u1 ⎥ ⎢ F1 ⎥ ⎢ S1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ u2 ⎥ ⎢ F2 ⎥ ⎢ S2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ U = ⎢ . ⎥ ,F= ⎢ . ⎥ ,S= ⎢ . ⎥ , ⎢ . ⎥ ⎢ . ⎥ ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ ⎣ . ⎦

un Fn Sn

A is the n × n Jacobian matrix of F : ⎡ ⎤ ∂F ∂F ∂F ⎢ 1 1 ··· 1 ⎥ ⎢ ∂u1 ∂u2 ∂un ⎥ ⎢ ⎥ ⎢ ∂F2 ∂F2 ··· ∂F2 ⎥ ⎢ ∂u ∂u ∂u ⎥ A = ⎢ 1 2 n ⎥ . ⎢ . . . . ⎥ ⎢ . . .. . ⎥ ⎣ ⎦ ∂Fn ∂Fn ··· ∂Fn ∂u1 ∂u2 ∂un 12.10 Exercises 811

A hyperbolic system of conservation laws is a system of conservation laws with the following properties: (i) The components of F and S are functions of components of U and possibly of x and t,butF does not contain any derivative of

U with respect to x and t, and (ii) the matrix A has n real and distinct eigenvalues which are roots of |A − λI| =0. The theory of hyperbolic systems of conservation laws with many examples of applications is available in many books including Defermos (2000), Holden and Risebro (2002), Jeffrey (1976),

Lax (1973, 2006), Serre (2000), Sharma (2010), and Zheng (2001). Bibliography1

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Abdelkader, 417 Asano, 153, 625, 633, 636, 645 Abel integral equation of the ﬁrst kind, 765 Associated Laguerre function, 706, 710, 711 Abel integral equation of the second kind, Associated Laguerre polynomial, 711 766 Associated Legendre function, 486, 703, 705 Abel’s formula, 86 Associated SL problem, 92, 94 Ablowitz, 152, 155, 417, 425, 506, 542 Associative, 730, 752 Ablowitz and Ladik (AL) equation, 199 Asymptotically stable, 401 Absorption isotherm, 300 Average value, 730, 757 Acoustic radiation potential, 9, 65 Averaged conservation equation, 589 Acrivos, 153 Averaged Hamiltonian, 379 Action, 198 Averaged Lagrangian, 592, 594 Action integral, 165 Averaged variational principle, 593 Addition theorems for elliptic functions, 717 Axisymmetric acoustic radiation problem, Adiabatic invariant, 588 65 Airy, 427 Axisymmetric biharmonic equation, 67 Airy differential equation, 570 Axisymmetric CauchyÐPoisson problem, 69 Airy function, 10, 38, 156, 468, 702 Axisymmetric diffusion equation, 64, 135 Airy integral, 38 Axisymmetric Dirichlet problem, 141 AKNS method, 425, 506, 508 Axisymmetric initial value problem, 133 Axisymmetric wave equation, 62, 79 Alfvén wave velocity, 358 Amick, 182 Bäcklund transformations, 494Ð498, 530, Ammerman, 152, 397 605, 606 Amplitude dispersion, 181 Balmforth, 155 Amplitude parameter, 183 Barenblatt, 416Ð418 Angular frequencies, 24 Barone, 151 Anharmonic forces, 515 Ben-Jacob, 155 Antikinks, 431 Benguria, 374 Antisoliton, 597, 600, 605 Benjamin, 153, 173, 527, 547 Aoki, 152, 397 BenjaminÐBonaÐMahony (BBM) equation, Arecchi, 151 154, 357, 528 Aris, 152 BenjaminÐFeir (modulational) instability, Arnold, 152, 283, 397 552, 554 Aronson, 415 BenjaminÐFeir side-band instability, 674

L. Debnath, Nonlinear Partial Differential Equations for Scientists and Engineers, DOI 10.1007/978-0-8176-8265-1, c Springer Science+Business Media, LLC 2012 848 Index

BenjaminÐOno (BO) equation, 153 BurgersÐHuxley equation, 155 Benney, 152, 153, 550 Burton, 321 Benton, 416 Berezin, 462, 463 Camassa, 518 Bernoulli equation, 175, 182 Camassa and Holm (CH) equation, 157, 518, BernoulliÐEuler equation, 137 522 Bespalov, 153 Canonical form, 12 Bessel, 85, 105, 110, 697 Canonical transformation, 241 Bessel function, 61, 105, 106, 697 Canosa, 152, 397, 417 Bessel’s equation, 85, 105, 110, 698 Capillary solitary wave, 456 Bessel’s inequality, 91, 728 Capillary waves, 190 Beta distribution, 697 Case, 416 Beta function, 692, 694 Cauchy conditions, 6 Bhatnagar, 636 Cauchy data, 210, 214, 331 Bianchi theorem, 497 Cauchy integral formula, 765 Biharmonic equation, 10, 67, 126, 133 Cauchy principal value, 153 Bilinear expansion of Green’s function, 99 Cauchy problem, 6, 7, 17, 18, 36, 42, 49, 50, Birkhoff, 404 52, 126, 134, 209, 210, 222Ð224, 234, Birkhoff theorem, 546 254Ð256, 274 Blackstock, 416 CauchyÐEuler equation, 84 Blasius problem, 131, 142 CauchyÐPoisson problem, 58, 68, 128, 135 Bleakney, 321 CauchyÐRiemann equations, 495 BohrÐSommerfeld rule, 493 Caustic, 569 Boltzmann constant, 438 Cavalli-Sforva, 152 Boltzmann distribution, 438, 638 Chaotic solutions, 346 Boltzmann nonlinear diffusion problem, 406 Characteristic base curve, 207 Boltzmann transformation, 408 Characteristic curves, 12, 207, 316, 435 Bona, 154, 357, 508, 528 Characteristic direction, 207, 229 Bond number, 187, 453 Characteristic equation, 12, 207, 230 Bore, 328 Characteristic function, 755 BornÐInfeld equation, 193 Characteristic shock, 341 Bound state, 478, 605 Characteristic strip, 231 Boundary-value problem, 6, 36, 51, 61, 96, Characteristic strip equations, 231 118, 134, 141, 142, 180, 394, 402, Characteristic velocity, 370 417, 633, 739 Characteristics, 12, 207, 327 Boussinesq, 425, 427, 436 Charpit equations, 230Ð235 Boussinesq equation, 10, 152, 194, 357, 428, Chebyshev polynomial, 708, 709 434Ð436, 451, 452, 467, 513, 651 Chen, 152, 155 Brauner, 416 Chézy formula, 298 Breakers, 299 Chiu, 416 Breather solution, 602, 604 Chromatographic models, 300 Bright soliton, 543 Chu, 153, 378, 549 Brillouin, 569 Circular frequency, 24 Britton, 417 Clairaut equation, 233, 237 Brunt–Väisälä frequency, 358 Classical solution, 3 BuckleyÐLeverett equation, 281 Claude, 517 Burgers, 151 Cnoidal wave, 458, 461, 462 Burgers equation, 151, 296, 382, 630, 637, Cole, 386, 396 638 ColeÐHopf transformation, 387 Index 849

Collins, 618 371, 432, 460, 464, 539, 541, 542, Column isotherm, 300 580, 584, 598, 622 Commutative, 36, 730, 752 Debye, 428 Commutator, 499 Debye length, 439 Compact breather, 514 Deep water waves, 178 Compactons, 509Ð511, 513, 514 Defocusing branch, 509 Complementary error function, 693 Degasperis and Procesi (DP) equation, 157, Complete elliptic integral, 461, 715, 717 518, 525 Complete integral, 203 Delta function, 746 Complete set, 89 Density, 284, 469 Complete solution, 203 Depassier, 374 Complex amplitude, 360 Derivative expansion method, 650 Composition, 751 Deterministic chaos, 342, 345 Compression wave, 262, 302 Developable surface, 238 Concentric KdV equation, 155 Dielectric constants, 567 Conjugate, 751 Differential equation for Conjugate variables, 240 Airy function, 702 Conservation law, 277, 374, 469Ð473, 547, associated Laguerre function, 711 621, 670, 671 associated Legendre function, 705 Conservation of energy, 164, 165, 192, 480 Bessel function, 697, 698 Constant of motion, 469 Chebyshev polynomials, 708, 710 Continuity, 730 Gegenbauer polynomials, 707 Continuous spectrum, 51, 479 Hermite function, 713 Convolution, 36, 521, 730, 762 Hermite polynomials, 712 Convolution theorem, 36, 752, 760 Jacobi polynomials, 707 Laguerre polynomials, 711 Couette flow, 121 Legendre polynomial, 704 Courant, 336 modified Bessel function, 700 Cramer, 637 Weber-Hermite function, 713 Crandall, 491, 529 Diffusion coefficient, 306 Crank, 409, 415 Diffusion equation, 8, 27, 29, 42, 43, 48, 49, Crighton, 416 54, 55, 64, 71Ð73, 109, 112, 114, 117, Cylindrical KdV equation, 155, 529 122, 132, 137, 382, 387 Cylindrical wave equation, 103 Diffusion kernel function, 757 Diffusion length, 348 D’Alembert solution, 18, 37, 332, 434, 758 Diffusive wave, 383, 386 Damped harmonic oscillator, 738 Dipole solution, 393 Darcy law, 347 Dirac, 381, 535, 623, 740 Dark soliton, 543, 575 Dirac delta function, 34, 36, 60, 70, 740, Davey, 153, 156, 550 741, 746 DaveyÐStewartson equation, 156, 663, 668, Dirichlet condition, 6 670 Dirichlet kernel, 731, 732 Davis, 153, 612 Dirichlet problem, 31, 39, 47, 139, 141, 163 De Broglie wave, 45, 358, 577 Discontinuous initial data, 267, 274 De Leonardis, 604 Discontinuous solution, 261, 270, 275Ð277 De Vries, 152, 428 Dispersion relation, 45, 69, 177, 186, 187, Debnath, 52, 55, 58, 60, 62, 68, 69, 83, 88, 246, 355Ð359, 370, 439, 444, 569, 115, 119Ð124, 128, 129, 132, 135, 595, 631, 637, 661, 672 142, 152, 173, 182Ð185, 324, 369, for Alfvén gravity wave, 358 850 Index

for cnoidal wave, 461 Enz, 151 for de Broglie wave, 45 EPDiff equation, 524 for electromagnetic waves, 358 Equation for ion-acoustic wave, 439 Ablowitz and Ladik (AL), 199 for water waves, 69, 178, 357, 426, 540 associated Legendre, 486 Dispersion relation for de Broglie wave, 45, averaged conservation, 589 358 axisymmetric wave, 62 Dispersion relation for the gravity-capillary BenjaminÐBonaÐMahony (BBM), 154, waves, 187 357, 528 Dispersion relation for water waves, 69, 178, BenjaminÐOno (BO), 153 357, 426, 540 Bernoulli, 175, 182 Dispersive wave, 357 BernoulliÐEuler, 137 Dissipative wave equation, 622 Bessel, 85, 105, 110, 698 Distribution, 740, 747 biharmonic, 10, 67, 126, 133 Distributive, 730, 753 BornÐInfeld, 193 Dodd, 503, 504 Boussinesq, 10, 152, 194, 357, 428, Domain of dependence, 18 434Ð436, 451, 452, 467, 513, 651 Drazin, 489, 492, 601 BuckleyÐLeverett, 281 Dressler, 310, 311 Burgers, 151, 296, 382, 630, 637, 638 Duality, 751, 761 BurgersÐHuxley, 155 Duhamel formula, 145, 761 Camassa and Holm (CH), 157, 518, 522 Dunbar, 417 CauchyÐEuler, 84 Duncan et al., 618 CauchyÐRiemann, 495 Dutta, 460, 464, 539, 541, 598 characteristic, 12, 207, 230 Dynamical equation, 587 characteristic strip, 231 Charpit, 230Ð235 Effective mode, 546 Clairaut, 233, 237 Eigenfunctions, 24, 28, 30, 83, 84, 112 concentric KdV, 155 Eigenvalues, 22, 23, 28, 30, 83, 84, 316, 492 continuity, 284, 307, 438, 445 Eikonal equation, 247, 248, 252, 569 cylindrical KdV, 155, 529 Eilbeck, 618 DaveyÐStewartson, 156, 663, 668, 670 Einstein, 149, 425 Degasperis and Procesi (DP), 157, 518, Ekman layer, 132 525 Elastic particles, 604 diffusion, 8, 27, 29, 42, 43, 48, 49, 54, 55, Electrical susceptibility, 568 64, 71Ð73, 109, 112, 114, 117, 122, Electromagnetic wave equation, 10 132, 137, 382, 387 Electron plasma frequency, 439 dynamical, 587 Elliptic equation, 13, 14 eikonal, 247, 248, 252, 569 Elliptic functions, 616, 715 electromagnetic wave, 10 Elliptic integral, 460, 539, 715 elliptic, 13, 14 Energy, 26, 45, 101, 107, 191, 755 Euler, 173, 174, 183, 194, 440, 446, 447, Energy density, 364, 366 451 Energy equation, 192, 197 EulerÐDarboux, 20 Energy ﬂux, 192, 364, 366 EulerÐLagrange, 159, 160, 162, 168, 171, Energy inequality, 107 194, 195, 197 Energy integral, 101, 140 EulerÐPoincaré (EP), 524 Engelbrecht, 636 EulerÐPoissonÐDarboux, 324, 351 Entropy criterion, 277 evolution, 499, 608, 624, 627 Envelope soliton, 430, 542, 543 Fisher, 152, 396 Index 851

FokkerÐPlanck, 226 NavierÐStokes, 307 fractional diffusion, 114, 118, 122 nonlinear diffusion, 382, 406, 408, 409 fractional partial differential, 114 nonlinear evolution, 500, 502 fractional Schrödinger, 124 nonlinear KleinÐGordon (KG), 151, 554 fractional wave, 116, 118, 122, 123 nonlinear partial differential, 150, 203, Gardner, 471 362, 650 GelfandÐLevitanÐMarchenko (GLM), 483 nonlinear reactionÐdiffusion, 412 generalized NLS, 564, 645 nonlinear Schrödinger (NLS), 153, 378, GinzburgÐLandau (GL), 154 503, 505, 529, 550, 552, 555, 558, Hamilton, 163 562, 574, 577, 654, 661, 663, 668, 671 HamiltonÐJacobi, 238Ð240, 243, 244 nonlinear wave, 151, 328, 350 Hamilton’s canonical, 524 OttoÐSudanÐOstrovsky, 508 Harry Dym, 525 Painlevé, 467 heat, 8, 53 parabolic, 13, 14, 383 Helmholtz, 9, 65, 75, 127, 144 paraxial, 571 Hermite differential, 712 Peregrine, 640 Hodgkin and Huxley, 155 Poisson, 9, 73, 195 hyperbolic, 12, 322, 323, 351, 370 PrandtlÐBlasius, 419, 420 hypergeometric, 323 quasi-linear partial differential, 3, 194, internal wave, 358 202, 378 internalÐinertial wave, 358 reactionÐdiffusion, 397, 412 inviscid derivative Burgers, 521 Riccati, 475 Johnson, 155, 447, 449 roll wave, 309 KadomtsevÐPetviashvili (KP), 155, 444, Rossby wave, 130, 358 445, 473, 529 scattering, 608 Kawahara, 376 Schrödinger, 9, 44, 124, 195, 253, 477, KdVÐBurgers, 372, 508, 624, 637 486, 504, 507, 541, 544, 564 kinematic wave, 151, 259, 284 shallow water, 327, 348, 433 KleinÐGordon (KG), 9, 126, 138Ð140, sine-Gordon (SG), 151, 506, 594, 595, 151, 171, 357, 579, 582, 584, 585, 598, 605, 620, 621 592, 619, 622 sinh-Gordon, 506, 621 KortewegÐde Vries (KdV), 10, 38, 152, telegraph, 9, 126, 130, 132, 144, 145, 349 357, 428, 438, 440, 443, 452, 458, Toda lattice, 157, 158 468, 475, 505, 530, 531, 638, 658, 660 traffic flow, 286, 287 KPP, 398 Tricomi, 14, 351 KuramotoÐSivashinsky, 376 water wave, 173, 184, 185, 432, 434 Lagrange, 163, 164 wave, 4, 17, 22, 36, 49, 50, 52, 58, 62, Laplace, 8, 9, 39, 41, 47, 59, 138, 146, 78Ð80, 92, 101Ð104, 115, 122, 140, 663 151, 167Ð170, 444, 613 Lax, 499, 501, 530, 531 Whitham, 157, 193, 364, 372, 375Ð378, linear partial differential, 1Ð3, 10 531, 549, 593 linear scattering, 608 YangÐMills, 158 linear wave, 4, 8, 17, 22, 36, 49, 50, 52, Zakharov, 562 58, 62, 78Ð80, 102, 109, 441, 444 Error function, 693 Lorenz, 343 Estabrook, 495, 497, 503 Maxwell, 438, 567, 613 Euler constant, 690 minimal surface, 163 Euler equation, 173, 174, 183, 194, 440, modified KdV (mKdV), 153, 505, 528, 446, 447, 451 530, 635 EulerÐDarboux equation, 20 852 Index

EulerÐFourier formulas, 720, 725 Fractional StokesÐEkman problem, 142 EulerÐLagrange equation, 159, 160, 162, Fractional unsteady Couette flow, 121 168, 171, 194, 195, 197 Fractional wave equation, 116, 118, 122, 123 EulerÐLagrange variational problem, 159, Freeman, 155, 498, 529 160 Frenkel, 151 EulerÐPoincaré (EP) equation, 524 Fresnel integrals, 693 EulerÐPoissonÐDarboux equation, 324, 351 Friedrichs, 336 Evolution equation, 499, 608, 624, 627 Froude number, 187, 310 Expansive wave, 261, 275 Fulton, 153 Function Factorial function, 689 Airy, 10, 38, 156, 468, 702 Fan, 291, 301 associated Laguerre, 706, 710, 711 Fan centered, 291, 292 associated Legendre, 486, 703, 705 Feigenbaum constant, 345, 346 Bessel, 61, 105, 106, 697 Feir, 371, 547 beta, 692, 694 Ferguson, 546 characteristic, 755 Fermi, 428, 547 complementary error, 693 FermiÐPastaÐUlam recurrence, 428, 547 diffusion kernel, 757 Fife, 152, 415, 417 Dirac delta, 34, 36, 740, 741 Finite Hankel transform, 123, 686 eigen, 24 Fisher, 152, 396 error, 693 Fisher equation, 152, 396 factorial, 689 Fission, 490 gamma, 689 FitzHugh, 155 Gaussian, 742 Flesch, 618 generalized, 740, 744, 747 Flood wave, 297Ð299 good, 743 Flux, 284, 469 Hamilton, 163, 243 Flux transport, 412 Hamilton principle, 238 Focusing branch, 509 Focusing instability, 574 Heaviside, 745 FokkerÐPlanck equation, 226 Hermite, 713 Forcing term, 356 hypergeometric, 323 Fordy, 504 incomplete gamma, 692 Fourier, 381 Jacobi Epsilon, 616 Fourier cosine transforms, 679 Jacobi theta, 395 Fourier integral formula, 749 Laguerre, 710, 711 Fourier integral solution, 49 Legendre, 703 Fourier method, 5, 22 Mittag-Leffler, 114, 713, 766 Fourier series, 719Ð731, 733, 736, 739 modified Bessel, 700 Fourier sine transforms, 677 normal probability density, 696 Fourier transform, 34, 46, 360, 675, 751 ordinary Gaussian, 742 FourierÐBessel series, 111 phase, 360 Fractional Blasius problem, 142 Riemann, 79, 322 Fractional derivative, 55, 764 sawtooth wave, 723 Fractional diffusion equation, 114, 118, 122 smooth, 204 Fractional partial differential equations, 114 special, 689 Fractional Rayleigh problem, 119 square wave, 723, 759 Fractional Schrödinger equation, 124 theta, 464 Fractional Stokes problem, 119 triangular wave, 390, 722, 724 Index 853

WeberÐHermite, 712, 713 Gibbs, 733 Wright, 115 Gibbs phenomenon, 733 Functional, 159Ð162 GinzburgÐLandau (GL) equation, 154 Fundamental harmonic, 25 Glacier flow, 305 Fundamental parameters, 183, 438 Goldstein, 243 Fundamental period, 25 Good function, 743 Fundamental solution, 42, 43, 389, 408 Gordon, 9 Gorschkov, 649 Galvin, 430 GorterÐMellink law, 411 Gamma distribution, 697 GravityÐCapillary surface waves, 187, 452 Gamma function, 689 Green function, 42, 43, 70Ð74, 76Ð82, 97, Gardner, 430, 472, 474, 635 137 Gardner equation, 471 Green signal problem, 292 Gardner transformation, 471 Gregory, 734 GardnerÐMorikawa transformation, 628, Griffith, 321 633, 634 Grimshaw, 527 Gauss formula, 696 Grindrod, 417 Gaussian function, 742 Group velocity, 45, 178, 189, 357, 662 Gazdag, 417 Growth rate, 396 Gegenbauer polynomial, 707, 708 Growth term, 259 GelfandÐLevitanÐMarchenko (GLM) Gurney, 415 equation, 483 Gurtin, 415 General integral, 204 General Parseval’s relation, 754 General solution, 4, 204 Haberman, 152 Generalized Fourier coefficients, 89, 93 Hadamard example, 7 Generalized Fourier series, 89 Hagstrom, 417 Generalized function, 740, 744, 747 Hahn, 612 Generalized Mittag-Leffler function, 713 Haight, 287 Generalized NLS equation, 564, 645 Hamilton equations, 163 Generalized RankineÐHugoniot shock Hamilton function, 163, 243 condition, 340 Hamilton principle, 165, 168 Generalized Riemann invariants, 339 Hamilton principle function, 238 Generalized simple wave, 338Ð340 Hamilton variational principle, 649 Generalized solitary wave, 457, 458 HamiltonÐJacobi equation, 238Ð240, 243, Generalized solution, 3, 257, 271, 272, 275 244 Generating function for Hamiltonian, 44, 163, 173, 375, 524 associated Laguerre function, 711 Hamiltonian density, 604 associated Legendre function, 705 Hamilton’s canonical equation, 524 Bessel function, 698 Hammack, 542 Chebyshev polynomials, 708 Hankel transform, 61, 446, 683 Gegenbauer polynomials, 707 Harry Dym equation, 525 Hermite polynomials, 712 Hasegawa, 570 Jacobi polynomials, 707 Hasimoto, 555 Laguerre function, 711 Hassan, 650 Legendre polynomial, 704 Heat equation, 8, 53 Geometrical optics approximation, 585 Heaviside function, 745 Ghez, 415 Heisenberg, 149, 535 Gibbon, 151, 503, 563 Helmholtz equation, 9, 65, 75, 127, 144 854 Index

Hermite differential equation, 712 Jacobi elliptic function, 461, 616, 716 Hermite function, 713 Jacobi elliptic integrals, 715 Hermite polynomial, 712 Jacobi Epsilon function, 616 Hilbert, 201, 355, 579, 623 Jacobi polynomial, 706, 707 Hilbert transform, 153 Jacobi principle of least action, 245, 248 Hirota, 430, 617 Jacobi theta function, 395 Hodgkin and Huxley equation, 155 Jeffrey, 508, 569, 636, 655 Hodograph transformation, 336 Johnson, 417, 445, 489, 661 Holm, 522 Johnson equation, 155, 447, 449 Hopf, 386 Jones, 360, 743 Hoppensteadt, 402 Josephson, 151 Howard, 413 Jost solutions, 481 Hydraulic jump, 299, 328 Jump condition, 257, 270, 274 Hyman, 509Ð511, 513, 515 Jump discontinuity, 270 Hyperbolic equation, 12, 322, 323, 351, 370 Hypergeometric equation, 323 Kadomtsev, 155 Hypergeometric function, 323 KadomtsevÐPetviashvili (KP) equation, 155, 444, 445, 473, 529 Ichikawa, 153 Kako, 152 Identity, 36, 753 Kakutani, 153, 625, 636 Ilichev, 376 Kaliappan, 413 Ill posed, 7, 146, 147, 370 Kametaka, 417 Incomplete gamma function, 692 Karpman, 153, 462, 509, 563 Index of refraction, 248 Kaup, 506, 509 Infeld, 563 Kawahara, 636, 650, 655 Initial value problem, 17, 128, 133, 219, Kawahara equation, 376 259, 313, 331, 359 KdVÐBurgers equation, 372, 508, 624, 637 Inner product, 87, 89, 755 Keller, 417 Inoue, 655 Kelley, 153 Instantons, 158 Kerr effect, 570 Integrability condition, 230 Kinematic wave equation, 151, 259, 284 Integral equation, 765 Kinetic energy, 26, 45, 163, 164, 167, 190, Integral surface, 4, 206 196, 524 Interaction strength, 546 Kink, 431 Internal wave equation, 358 Kirchhoff transformation, 409 InternalÐinertial wave equation, 358 Kivshar, 515, 516 Inverse Fourier transform, 34, 46 Klein, 9 Inverse Hankel transform, 61 KleinÐGordon (KG) equation, 9, 126, Inverse Laplace transform, 51 138Ð140, 151, 171, 357, 579, 582, Inverse scattering transform, 475, 476, 478 584, 585, 592, 619, 622 Inverse Weierstrass transform, 763 KleitzÐSeddon law, 299 Inviscid derivative Burgers equation, 521 Kodama transformation, 522 Ion plasma frequency, 439 Kolmogorov, 152, 397 Ion-acoustic wave, 438, 440, 638 Konno, 577 Isentropic ﬂow, 314, 351 Kontorova, 151 Isothermal curves, 40 Kopell, 413 Korteweg, 152, 428 Jacobi dn function, 541, 716 KortewegÐde Vries (KdV) equation, 10, 38, Jacobi sn function, 460, 716 152, 357, 428, 438, 440, 443, 452, Index 855

458, 468, 475, 505, 529Ð531, 638, Levi-Civita, 426 658, 660 Lie group, 404, 466 KPP equation, 398 Lifshitz, 493 Kramers, 569 Lighthill, 151, 152, 194, 283, 286, 369, 416 Kriess, 416 Lighthill criterion, 566 Kronecker delta, 24, 240 Linear operator, 3, 70 Krushkal, 153 Linear partial differential equation, 1Ð3, 10, Kruskal, 428, 429, 525, 547 105 Kulikovskii, 636 Linear scattering equation, 608 KuramotoÐSivashinsky equation, 376 Linear superposition principle, 5, 6, 43, 74 Kynch, 151, 303, 348 Linear wave equation, 4, 8, 17, 22, 36, 49, 50, 52, 58, 62, 78Ð80, 102, 109, 441, Lagrange bracket, 242, 243 444 Lagrange equation, 163, 164 Liu, 152, 155 Lagrange identity, 86 Local frequency, 361, 362, 367 Lagrangian, 163, 168, 172, 198, 367 Local Mach number, 335 Lagrangian density, 168, 604, 621 Local wavenumber, 361, 362, 367 Lagrangian functional, 169 Localized modes, 516 Lagrangian stability, 546 Logan, 417 Laguerre differential equation, 711 Logistic map, 345 Laguerre function, 710, 711 Lommel integrals, 699 Laguerre polynomial, 710, 711 Long wavelength parameter, 183 Lake, 153, 547 Lorenz, 342, 417 Lamb, 151, 612 Lorenz attractor, 342 Lamé constants, 127 Lorenz equation, 343 Landau, 493 Lorenz system, 342 Langmuir isotherm model, 301 Luke, 171 Langmuir soliton, 559 Laplace equation, 8, 9, 39, 41, 47, 59, 138, Ma, 576 146, 663 MacCamy, 415 Laplace spherical harmonics, 105 Mach lines, 335 Laplace transform, 51, 59, 680, 758 Mach number, 252, 335 Laplacian, 44, 73, 79, 102, 109, 742 Madsen, 152 Lardner, 416 Magnetic permeability, 567 Larson, 416, 417 Mainardi, 115, 118 Lattice solitary waves, 618 Makhankov, 430 Law of conservation of energy, 164, 165, Manning formula, 298 480 Manoranjan, 417 Lax, 257, 277, 340, 811 Marchenko, 376 Lax entropy criterion, 277 Matrix operators, 502 Lax equation, 499, 501, 530, 531 Matsumoto, 655 Lax pair, 499, 530 Maximum principle, 108 Legendre duplication formula, 691 Maxon, 156 Legendre functions, 703 Maxwell equation, 438, 567, 613 Legendre polynomial, 487, 703, 704, 708 Maxworthy, 430 Leibniz, 734 McCall, 612 Leibniz series, 734 McKean, 417 Leibovich, 153 Mean value property, 34 LeVeque, 341 Mei, 153, 378, 549 856 Index

Mersenne law, 25 Nonlinear evolution equation, 500, 502 Method of characteristics, 208, 333 Nonlinear group velocities, 370 Method of multiple scales, 656, 661 Nonlinear KleinÐGordon (KG) equation, Minimal surface equation, 163 151, 554 Mitchell, 417 Nonlinear lattices, 615 Mittag-Leffler function, 114, 713, 766 Nonlinear partial differential equation, 150, Miura, 470, 475, 528 203, 362, 650 Miura transformation, 470 Nonlinear reactionÐdiffusion equation, 412 Modes, 25 Nonlinear Schrödinger (NLS) equation, 153, Modified Bessel equation, 700 378, 503, 505, 529, 550, 552, 555, Modified Bessel function, 701 558, 562, 574, 577, 654, 661, 663, Modified KdV (mKdV) equation, 153, 505, 668, 671 528, 530, 635 Nonlinear shallow water waves, 354 Modulational instability, 554, 564 Nonlinear superposition principle, 497, 607 Mollenauer, 570 Nonlinear wave equation, 151, 328, 350 Moment of instability, 436 Normal dispersion, 575 Momentum equation, 312 Normal modes, 24 Monge axis, 207 Normal probability density function, 696 Monge cone, 208, 229 Novick-Cohen, 376 Monochromatic waves, 568 Nozaki, 636 Monoclinal flood wave, 299 Numerical series, 734, 736, 737 Montroll, 413 Nwogu, 152 Morikawa, 628 Nye, 151, 305 Multiple Fourier transform, 46, 47 Munier, 409 Oikawa, 636 Murray, 413Ð417 Okubo, 408 Olver, 173 N-soliton solution, 486, 762 Ono, 153, 550, 625, 633, 636 n-th convolution product, 762 Order, 2 N-wave solution, 392 Ordinary Gaussian functions, 742 Nagumo, 155 Ordinary output, 762 Naumkin, 376, 377 Orthogonal relation, 706, 707, 711 NavierÐStokes equation, 307 Orthogonal set, 89 Nayfeh, 153 Ostrovsky, 508, 624, 646 Neumann boundary-value problem, 6 OttoÐSudanÐOstrovsky equation, 508 Neumann conditions, 6 Outi, 542 Newell, 604 Overtones, 25 Newman, 413Ð416 Nigmatullin, 115 p-mode, 517 Nisbet, 415 Pack, 321 Nishihara, 637 Page, 517 Nodes, 25 Painlevé equation, 467 Non-periodic solutions, 738 Parabolic equation, 13, 14, 382, 383 Nondispersive wave, 178, 357, 452 Paraxial equation, 571 Nonlinear diffusion equation, 382, 406, 408, Parity relation, 709, 712 409 Parker, 416 Nonlinear Dirichlet problem, 163 Parseval formula, 727 Nonlinear dispersion relation, 181, 516, 540, Parseval relation, 91, 729, 755 573, 620 Partial differential equation, 2 Index 857

Particular solution, 204 RankineÐHugoniot condition, 270, 340 Pasta, 428, 547 Ravindran, 152, 153 Pattle, 413 Rayleigh, 427 Pawlik, 153 Rayleigh formula, 190 Peakon solitary traveling wave, 523 Rayleigh problems, 119 Peakons, 519, 522, 524 Rayleigh quotient, 545 PeierlsÐNabarro potential, 517 RayleighÐBenard convection problem, 344 Pelinovsky, 624, 646 Reaction term, 259, 397 Penel, 416 ReactionÐdiffusion equation, 397, 412 Peregrine, 152, 527, 576 Recurrence relation, 699, 706, 711, 712 Peregrine equation, 640 Redekopp, 152 Period-doubling, 346 Reductive perturbation method, 627 Perring, 600, 608 Reflected wave, 479 Petviashvili, 155 Reflection coefficient, 479 Phase function, 360 Refractive index, 246, 568, 570 Phase velocity, 177, 245, 661 Refractive wave, 261, 275 Phillips, 577 Resonance, 738 Piston problem, 319, 325 Reynolds number, 386, 388, 393, 395 Planck constant, 9, 44, 124, 358, 494 Riabouchinsky, 434 Plane waves, 102 Riccati equation, 475 Plateau problem, 162, 197 Richards, 151, 286 Platzman, 416 Riemann, 311, 385 Poincaré, 1, 227 Riemann function, 79, 322 Poincaré return map, 345 Riemann initial-value problem, 313 Poincaré theorem, 241 Riemann invariants, 312, 317Ð320, 325, 329, Point of inflection, 414 330 Pointwise Convergence Theorem, 732 RiemannÐLebesgue Lemma, 728, 751 Poisson bracket, 240 RiemannÐLiouville fractional integral, 764 Poisson equation, 9, 73, 195 Ripples, 189 Poisson integral formula, 34, 39, 43 Robin boundary-value problem, 6 Poisson kernel, 34 Robin condition, 6 Poisson summation formula, 752 Rodin, 416 Polarization, 568 Rodrigues formula, 703, 707, 709Ð712 Polytropic gas, 319 Roll wave equation, 309 Potential, 507, 550, 608 Rosales, 468 Potential energy, 26, 45, 165, 168, 190, 196 Rosenau, 417, 509, 511, 513, 515, 518 Power spectrum, 755 Roskes, 153, 550 PrandtlÐBlasius equation, 419, 420 Rossby wave equation, 130, 358 Prasad, 152, 153, 636 Rowland, 563 Principle of least action, 198 Rowlands, 153, 563 Prolongation structure method, 503 Ruled surface, 238 Properties of the convolution, 756 Russell, 425, 426

Quasi-linear partial differential equation, 3, s-mode, 517 194, 202, 378 Saddle point, 401 Saffman, 526 Radian frequency, 24 Sagdeev, 440 Raizer, 416 Sandri, 650 Rankine, 182, 371 Sandusky, 517 858 Index

Sanuki, 564 Skyrme, 151 Saric, 153 Sleeman, 155 Satsuma, 155 Slobodkina, 636 Sawtooth wave function, 723 Smoller, 341, 417 Scaling, 751, 759 Smooth function, 204 Scattering data, 475 sn function, 460 Scattering equation, 608 Solitary wave, 429, 430, 436, 457, 540Ð542, Scattering state, 476, 479 600 Schiff, 151 Solution Schimizu, 153 complete, 203 Schmidt, 563 discontinuous, 261, 275Ð277 Schneider, 115 general, 4, 204 Schrödinger equation, 9, 44, 124, 195, 253, generalized, 3, 257, 271, 272, 275 477, 486, 504, 507, 541, 544, 564, 670 particular, 204 Schwartz space, 755 Schwarz inequality, 545 singular, 204, 249 Scott, 151, 155, 416, 619 weak, 3, 257, 271Ð274, 341 Secular (unbounded) term, 652 Sorensen, 152 Seddon, 299 Source solution, 389 Segur, 155, 425, 488, 506 Source term, 259, 397 Self-focusing instability, 570 Sparrow, 420 Self-induced transparency (SIT), 612 Special functions, 689 Seliger, 376 Spectrum, 83 Sen, 136, 637 Speed of shallow water waves, 178, 428, Separation of variables, 21, 22, 29, 32, 598 431, 434 Separatrix, 401 Spherical waves, 103 SerretÐFrenet formulas, 556 Square wave function, 723, 759 Shabat, 502, 535, 542, 544, 545, 547, 550, Stable node, 400 576 Standing wave, 25 Shallow water equation, 327, 348, 433 Stationary phase method, 60, 69, 391 Shallow water speed, 431, 459 Stationary point, 60, 67, 360, 391 Shallow water waves, 178, 327 Stewartson, 153, 154, 156, 550 Shifting, 751, 759 Stirling approximation, 692 Shigesada, 415 Stokes, 355, 426 Shishmarev, 376, 377 Stokes’ analysis, 182 Shock condition, 270, 271, 340 Stokes expansion, 182, 369 Shock wave, 205, 264, 270, 288, 321 Stokes problem, 119, 131 Sievers, 516, 517 Stokes wave, 179, 552 Sign function, 745 Similarity solution, 404, 406, 409 StokesÐEkman problem, 131, 142 Similarity transformation, 406 Strange attractor, 345 Similarity variable, 389, 406 Stretched variable, 632 Simple harmonic oscillator, 244, 737 Stretching transformation, 404 Simple wave, 311, 314, 318 Strong convergence, 90 Sine-Gordon (SG) equation, 151, 506, 594, Stuart, 154 595, 598, 605, 620, 621 SturmÐLiouville systems, 82Ð85, 138, 139, Single-soliton solution, 604, 610 501 Singular solution, 204, 249 Sturrock, 650 Sinh-Gordon equation, 506, 621 Su, 635 Index 859

Superposition principle Two-sided Laplace transform, 762 linear, 5, 6 Two-soliton solution, 488, 497, 498 nonlinear, 497 Supersonic flow, 334, 341 Ulam, 428, 547 Surface area of a sphere in n-dimension, 690 Undular bore, 461 Surface wave problem, 128, 134 Unitary transformation, 755 Symmetric, 692 Van Saarloos, 155 Van Wijngaarden, 153 Takeno, 516, 517 Vanden-Broeck, 527 Talanov, 153 Variational principle, 159, 171, 172, 193, Tang, 417 196 Taniuti, 153, 625 Varma, 153 Tappert, 153 Velocity of sound, 312 Telegraph equation, 9, 126, 130, 132, 144, Velocity potential, 175 145, 349 Viecelli, 156 Thermal conductivity, 54 Volume of a sphere in n-dimension, 690 Theta function, 464 Von Neumann, 149, 201, 227 Three-soliton solution, 488 Vortex filament, 555 Thyagaraja, 544 Vorticity, 174 Toda, 152, 158, 615 Wadati, 152, 577 Toda lattice equation, 157, 158 Wahlquist, 495, 497, 503 Toda lattice soliton, 158, 615, 618 Walsh, 396, 416 Toland, 182 Wang, 155 Traffic flow, 286, 287 Ward, 158 Traffic flow equation, 286, 287 Washimi, 153, 625 Traffic signal problem, 294 Water wave equation, 173, 184, 185, 432, Transform 434 finite Hankel, 686 Watson, 650 Fourier, 34, 46, 360, 675, 751 Wave Fourier cosine, 679 Alfvén gravity, 358 Fourier sine, 677 capillary solitary, 456 Hankel, 61, 446, 683 cnoidal, 458, 461 Hilbert, 153 compression, 262, 302 inverse scattering, 475, 476 de Broglie, 45, 358, 577 Laplace, 51, 59, 680, 762 diffusive, 383, 386 Weierstrass, 763 dispersive, 357 Transition layer, 386 electromagnetic, 10 Transition region, 271 expansive, 261, 275 Translation, 751 flood, 297, 298 Transmission coefficient, 479 generalized simple, 338Ð340 Transmitted wave, 479 generalized solitary, 457, 458 Transverse vibration of string, 22 gravity, 358 Traveling wave solution, 50, 384 gravityÐcapillary, 187, 452 Triangular wave function, 390, 722, 724 incident, 486 Tricomi equation, 14, 351, 354 inertial, 358 Trochoid, 180 internal, 358 Trullinger, 604 internal-inertial, 358 Turin, 412 ion-acoustic, 438, 440, 638 860 Index

kinematic, 151, 259 Weiland, 153 N, 392 Well posed problem, 7, 18, 147 nondispersive, 178, 357 Wentzel, 569 reﬂected, 479 West, 413 refractive, 261, 275 Whitham, 151, 171, 173, 193, 286, 299, 359, roll, 309 371, 377 Rossby, 130, 358 Whitham averaged Lagrangian, 368, 592 shallow water, 178, 640 Whitham averaged variational principle, shock, 205, 264, 270, 288, 321 367, 368, 378 simple, 275, 311, 314, 318, 320 Whitham equation, 157, 193, 364, 366, 368, solitary, 429, 430, 436, 457, 540Ð542 372, 375Ð378, 531, 549, 593 standing, 25 Whitham rule, 290, 392 Stokes, 179, 552 Wigner, 579 transmitted, 479 Wiljelmsson, 153 WKBJ method, 569 water, 176, 357 Wright function, 115 Wave diffraction, 572 Wronskian, 480 Wave energy, 546 Wyss, 115 Wave equation, 4, 8, 17, 22, 36, 49, 50, 52, 58, 62, 78Ð80, 92, 101Ð104, 115, 122, Yajima, 152 126, 140, 151, 167Ð170, 444, 613 YangÐMills equation, 158 Wave front, 245 Yuen, 153, 546 Wave propagation velocity, 285 Wavepacket, 570 Zabusky, 152, 428Ð430, 462, 469, 547 Weak solution, 3, 257, 271Ð274, 341, 519 Zakharov, 152, 173, 502, 535, 542, 544, 545, Webber, 417 547, 550, 576 Webber solution, 698 Zakharov equation, 562 WeberÐHermite function, 712, 713 ZakharovÐShabat (ZS) scheme, 502 Wei, 625 Zel’dovich, 416, 418 Weidman, 152 Zeppetella, 417 Weierstrass transform, 763 Zuﬁra, 376