Bessel Functions and Their Applications to Solutions of Partial Diﬀerential Equations

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Math 456 Lecture Notes: Bessel Functions and their Applications to Solutions of Partial Diﬀerential Equations

Vladimir Zakharov June 3, 2009 1 Gamma Function

Gamma function Γ(s) is deﬁned as follows:

Z ∞ Γ(s) = e−tts−1dt (1) 0

As far as: 1 ∂ ts−1 = ts (2) s ∂t

By plugging (2) into (1) we get

Z ∞ Z ∞ −t d s −t s ∞ −t s sΓ(s) = e t dt = e t |0 + e t dt (3) 0 dt 0 or sΓ(s) = Γ(s + 1) (4)

Then Γ(1) = 1 and Γ(2) = 1. By induction we obtain: Γ(n + 1) = n! (5)

Then 1 Γ(s) = Γ(s + 1) s 1 Γ(s) → if s → 0 s 1 Γ(s − 1) = Γ(s + 1) s(s − 1) 1 Γ(s − n) = Γ(s + 1) (6) (s − n)(s − n + 1) ··· s

Γ(s) has holes at all negative integral values of s.

To ﬁnd asymptotic behavior of Gamma-function as s → ∞, we use so called ”Laplace Method.”

2 Z ∞ Z ∞ Γ(s + 1) = e−ttsdt = e−φ(t,s)dt (7) 0 0

φ(t, s) = t − s ln(t)

Function φ(t, s) has a minimum at t = s. Indeed:

∂φ s = 1 − = 0 if t = s (8) ∂t t

Near this minimum: 1 φ = φ (s) + φ00(s)τ 2 + ··· , τ = t − s o 2 φo(s) = s − s ln(s) 1 φ00(s) = s

Now we will replace in (7) φ(t, s) to its approximate value (8) and go from integration by t to integration by τ. Whithout loss of accuracy we can consider that −∞ < τ < ∞. Then

Z ∞ 2 −φ (s) − τ Γ(s + 1) ≈ e o e 2s dτ (9) −∞ ss e−φo(s) = l √ R ∞ −y2 √ Now we replace τ = 2sy and remember that −∞ e dy = π. We end up with the following answer:

√ sn Γ(s + 1) ≈ 2πs (10) l √ nn n! ≈ 2πn l

3 This is the Stirling approximation for n = 5, where n! = 120. The Stirling approximation gives 5! ≈ 118.045. The accuracy of the Stirling approximation is reasonable. We accept without proof: π Γ(x)Γ(1 − x) = (11) sin(πx)

2 1 1 √ where Γ ( 2 ) = π so Γ( 2 ) = π.

2 Bessel Equation Appears

Let us try to solve the diﬀusion equation

ut = χ∆u (12) inside the disk of radius a in polar coordinates:

1 ∂ ∂u 1 ∂2u ∆u = r + (13) r ∂r ∂r r2 ∂θ2

We impose boundary conditions u(r = a) = 0 with initial data u(t = 0) = φ(r, θ). In polar coordinates the previous equation becomes:

1 ∂ ∂u 1 ∂2u u = χ r + (14) t r ∂r ∂r r2 ∂θ2 Partial solutions to this equation can be found of the following form:

u(r, θ, t) = einθe−χtk2 R(r) (15)

The radial part R(r) satisﬁes the equation

1 ∂ ∂R n2 r + k2 − R = 0 (16) r ∂r ∂r r2

2 2 2 2 k can take discrete values k = k1, ··· , kN , ··· Corresponding radial functions RN (r) satisfy the Dirichlet condition RN (a) = 0. By change of variables z = kr we have:

4 1 ∂ ∂R s2 z + 1 − R = 0 (17) z ∂z ∂z z2 where we have replaced n2 = s2, assuming that s is an arbitrary real number. The previous equation is the Bessel equation. At z → 0 it becomes the equipotent equation: 1 ∂ ∂R s2 z − R = 0 (18) z ∂z ∂z z2 which can be solved explicitly:

s −s R = C1z + C2z (19)

One can seek a solution of (17) in the form z s R = F (z, s) (20) 2

F satisﬁes the equation: 2s + 1 F 00 + F 0 + F = 0 (21) z

The solution of equation (21) can be found in the form of series:

∞ 2k X z F = Ck (22) k=0 2

After diﬀerentiating by z, the ﬁrst term in (22) vanishes. One can see that:

∞ 2(k−1) 00 2s + 1 0 X [(2k − 1)2k + 2k(2s + 1)] z F + F = Ck (23) z k=1 4 2

Let us replace k → k + 1. Now:

∞ 2k 00 2s + 1 0 X z F + F = Ck+1(k + 1)(k + s + 1) (24) z k=0 2

5 By substitution we will ﬁnally get:

Ck+1(k + 1)(k + s + 1) + Ck = 0 (25) or C C = − k (26) k+1 (k + 1)(k + s + 1)

This can then be solved as follows: (−1)k (−1)k C = = (27) k Γ(k + 1)Γ(k + s + 1) k!Γ(k + s + 1)

In particular for integral s = n: (−1)k C = (28) k k!(n + k)!

3 Bessel Function

The Bessel function Js(z) is deﬁned by the series:

s ∞ k 2k z X (−1) z Js(z) = (29) 2 k=0 k!Γ(s + k + 1) 2

This series converges for all z on the complex plane, thus Js(z) is the entire function. If z → 0, then z s 1 J (z) → (30) s 2 Γ(s + 1)

2 If s is not an integer, then J−s(z) is the second solution of the Bessel equation. Now: z −s 1 J (z) → (31) −s 2 Γ(−s + 1)

Js(z) is regular at z → 0, while J−s(z) is singular. So, let s → ∞. By the use of the Stirling Formula we get: 1 lz s Js(z) → √ (32) 2πs 2s

6 What happens if s → −n? Notice that (−s + 1)Γ(−s + 1) = Γ(−s + 2). Hence (31) can be re-written as follows: z −s −s + 1 J (z) → (33) −s 2 Γ(−s + 2)

Let s → 1, then (33) tends to zero and the ﬁrst term in (29) vanishes. All the other terms are ﬁnite thus: −1 ∞ k 2k z X (−1) z J−1(z) = (34) 2 k=1 k!Γ(k) 2 By replacing k → k + 1, we get ∞ k 2k z X (−1) z J−1(z) = − (35) 2 k=0 (k + 1)!Γ(k + 1) 2

Now (k + 1)! = (k + 1)k! and (k + 1)Γ(k + 1) = Γ(k + 2), hence:

∞ k 2k z X (−1) z J−1(z) = − (36) 2 k=0 k!Γ(k + 2) 2

On the right hand side of this equation we have −J1(z), hence

J−1(z) = −J1(z) (37)

In a similar way, we obtain:

n J−n(z) = (−1) Jn(z) (38)

For integral n, Jn and J−n are linearly dependent, and we must construct a second solution of the Bessel equation by another way. Let us calculate the following derivative: ∞ k 2k d −s 1 d X (−1) z z Js(z) = s dz 2 dz k=0 k!Γ(s + k + 1) 2 ∞ k 2(k−1) 1 X (−1) z z = s 2k 2 k=1 k!Γ(s + k + 1) 4 2 ∞ k 2k z X (−1) z = − s+1 2 k=0 k!Γ(s + k + 2) 2

7 By multiplying by zs we get: d zs z−sJ (z) = −J (z) (39) dz s s+1 or s J (z) = J (z) − J 0(z) (40) s+1 z s s

Let us consider now the following function:

∞ k 2(s+k) s s X (−1) z z Js = 2 (41) k=0 k!Γ(s + k + 1) 2 After diﬀerentiating by z and multpling by z−s, we obtain:

s−1 ∞ k 2k −s d s z X (−1) (s + k) z z (z Js) = dz 2 k=0 k!Γ(s + k + 1) 2 z s−1 ∞ (−1)k z 2k = X 2 k=0 k!Γ(s + k) 2 = Js−1

Finally, d z−s (zsJ (z)) = J (z) (42) dz s s−1 s 0 J (z) = J + J (z) (43) s−1 z s s By combining (40) and (43) we get 2s J (z) + J (z) = J (z) (44) s+1 s−1 z s 4 Bessel Functions of Half-integral Index

Let us introduce the function g deﬁned as follows:

8 s 1 1 Js = 1 g = g (45) z 2 z After plugging into the Bessel equation (17), one realizes that g satisﬁes the equation:

s2 − 1 g00 + 1 − 4 g = 0 (46) z2 1 Let s = z . Then,

g00 + g = 0 (47) as far as Js is regular at z → ∞

1 g = c sin(z) Js → cz 2

To ﬁnd c, we remember the asymptotics of the Bessel functions at z → 0.

1 z 2 1 J 1 (z) → 2 3 2 Γ( 2 )

√ 3 1 1 1 π Γ = Γ + 1 = Γ = 2 2 2 2 2

s 2z J 1 (z) → 2 π Finally, s 2 J 1 (z) = sin(z) (48) 2 πz

According to (40), all J 1 (z) are expressed through a combination of power and n+ 2 trigonometric functions. In particular,

s s ! 1 d −1 2 1 d sin(z) 2 1 1 2 2 2 J 3 (z) = −z (z J 1 (z)) = − z = 3 sin(z) − 1 cos(z) 2 dz 2 π dz z π z 2 z 2

9 or

s 2 sin(z) J 3 (z) = − cos(z) 2 πz z Separating this procedure we get:

s 2 3 3 J 5 (z) = − 1 sin(z) − cos(z) 2 πz z2 z Now, let z → ∞ in equation (40). One can put approximately:

0 Js+1(z) ≈ −Js(z) Then, we have:

s s 2 2 J 3 → − cos(z) J 5 → − sin(z) 2 πz 2 π s 2 J 7 → cos(z) ··· 2 πz

One can see that J 1 (z) has an inﬁnite amount of zeros on the real axis. The same n+ 2 statement is correct for all Bessel functions.

5 Integral Representation

Let us study the integral:

Z π 1 iz sin(θ)−inθ An(z) = e dθ (49) 2π −π To evaluate this integral, we use the Taylor expansion of the exponent:

∞ 1 ∞ 1 z p eiz sin(θ) = X (iz sin(θ))p = X (eiθ − e−iθ)p (50) p=0 p! p=0 p! 2 Now, notice that the integral:

10 Z π 1 iθ −iθ p −inθ Ip,π = (e − e ) e dθ = 0 if p < 0 (51) 2π −π Then, we denote p = n + q. The integrand in (51) can be presented in the form:

1 (eiθ − e−iθ)n+qe−inθ = (1 − e−2iθ)n(eiθ − e−iθ)q 2π Suppose that q is odd (q = 2k + 1). All terms in the ﬁrst parentheses are even powers of e−iθ, while all terms in the second parentheses are odd powers (positive or negative) on e−iθ. As a result, the integrand is a linear combination of odd powers of e−iθ. Thus the integral is zero, and we can put q = 2k. We obtain the following intermediate result:

n ∞ k z X 1 z An(z) = Ik,n (52) 2 n=0 (n + 2k)! 2 Where

Z π 1 iθ −iθ n+2k −inθ Ik,n = (e − e ) e dθ (53) 2π −π

To calculate Ik,n, we use the binomial expansion in the parentheses. In this expansion, we are interested only in the single term proportional to einθ. All other terms after multiplication to (53) and integration over θ are cancelled. Hence,

(n + 2k)! (−1)k(n + 2k)! (eiθ − e−iθ)n+2k ≈ (eiθ)n+k(−e−iθ)k = einθ k!(n + k)! k!(n + k)! and

(−1)k(n + 2k)! I = (54) n,k k!(n + k)! By plugging (54) into (52), we get ﬁnally:

n ∞ k k z X (−1) z An(z) = = Jn(z) 2 k=0 k!(n + k)! 2

We obtained the integral representation for Jn(z):

11 Z π 1 iz sin(θ)−inθ Jn(z) = e dθ (55) 2π −π This result is correct for positive n. Let us notice that,

n Jn(−z) = (−1) Jn(z) (56) Bessel functions of even order are even functions on z, while functions of odd order are odd. Now, we can ﬁnd An(z) at negative n. Let us change simultaneously the signs on z and n.

Z π 1 −iz sin(θ)+inθ A−n(−z) = e dθ 2π −π Now by replacing θ → −θ, we restore the previous result. Hence,

A−n(−z) = An(z) = Jn(z) n A−n(z) = Jn(−z) = (−1) Jn(z) (57) Finally, for all integrals on

n An(z) = (−1) Jn(z) notice also that Jn is real. Then (55) can be rewritten as follows: 1 Z π Jn(z) = cos(z sin θ − nθ)dθ (58) 2π −π Now, look at eiz sin(θ). This is a periodic function which can be expanded in the Fourier series. Apparently,

∞ ∞ iz sin(θ) X inθ X inθ n −inθ e = Jn(z)e = J0(z) + Jn(z)(e + (−1) e ) (59) n=−∞ n=1 The separating of imaginary and real parts in (59) gives us:

∞ X cos(z sin(θ)) = J0(z) + 2 J2k(z) cos(2kθ) (60) k=1 ∞ X sin(z sin(θ)) = 2 J2k+1(z) sin((2k + 1)θ) k=−∞

12 By introducing t = eiθ, one can transform (59) to the following expansion:

∞ z (t− 1 ) X n e 2 t = Jn(z)t (61) n=−∞

z (t− 1 ) This means that F (z, t) = e 2 t is a ’generating function’ for the entire community of Bessel functions of integral orders.

6 Asymptotic behavior at z → ∞

To ﬁnd the asymptotic behavior of the Bessel functions at z → ∞, we will use the device similar to the one used for the derivation of the Stirling formula. We present integral (55) in the form:

Z π 1 iΦ(z,θ) Jn = e dθ (62) 2π −π

Φ(z, θ) = z sin(θ) − nθ (63) If z → ∞, the integrand is the fast oscillation function everywhere except the two dΦ points where dθ = 0. These points are deﬁned by the equation:

z cos(θ) = n at z → ∞ π cos(θ) → 0 θ → ± 2

± π The contributions of points θ = ± 2 give complex conjugated results. Hence, it is π π enough to study the neighbourhood of the point θ = 2 . Let us introduce θ = 2 + τ. For small τ, nπ 1 Φ(z, θ) ≈ z − − zτ 2 (64) 2 2 Integral (62) can be replaced approximately by the following integral:

Z ∞ 1 i(z− πn − π ) −iz τ 2 Jn(z) =

13 s 2 1 −πi τ = y, √ = e 4 iz i Then, √ Z iπ ∗∞ 2 i(z− πn − π ) 4 −y2 J (z) = √

Figure 1: Contour of Integration. Integration is going in the complex plane along the straight line turned by 45◦ with respect to the real axis. This is demonstrated in Fig. 1. However, the contour of integration can be turned back and returned to the real axis (To justify this fact, we need to use some elemetns of complex analysis. But, this is true.) In other words, the integral in (65) can be replaced by the integral R ∞ −y2dy √ −∞ e = π. We end up with the following result: s 2 πn π J (z) → cos z − − (66) n πz 2 4 We derived this expression only for integral n. In fact, this is correct for all s. To prove this, we have to use a more sophisticated integral respresenation for Js(z) which is valid not only for integrals. In general, s 2 πs π J (z) → cos z − − (67) s πz 2 4 In particular,

s s 2 π 2 J 1 → cos z − → sin(z) 2 πz 2 πz This is the unique Bessel function coinciding with its own asymptotic behavior.

14 7 Zeros of Bessel function

It is clear from (67) that the Bessel function Js(z) has an inﬁnite amount of zeros for s the half axis 0 < z < ∞. Let us denote these zeros as aN , where N = 1, 2, ...∞. From (67), one can conclude that the distance between two neighboring zeros tends to π.

s s aN+1 − aN → π as N → ∞ (68) The first five Bessel functions of integral order are plotted on Figure 1. The first five of each are presented in Table 1. Notice that:

5 5 a5 − a4 = 3.2377

While:

0 0 a5 − a4 = 3.1394

Both values are close to π. The derivatives of Bessel functions have the following asymptotic behavior: s 2 sπ π J 0(z) → − sin z − − (69) s zπ 2 4 0 s The derivatives of Js(z) also have an inﬁnite amount of zeros bN . Again:

s s bN+1 − bN → π if N → ∞ (70)

15 Table 1: ROOTS of the FUNCTION Jn(x) are given in the following table. zero J0(x) J1(x) J2(x) J3(x) J4(x) J5(x) 1 2.4048 3.8317 5.1336 6.3802 7.5883 8.7715 2 5.5201 7.0156 8.4172 9.7610 11.0647 12.3386 3 8.6537 10.1735 11.6198 13.0152 14.3725 15.7002 4 11.7915 13.3237 14.7960 16.2235 17.6160 18.9801 5 14.9309 16.4706 17.9598 19.4094 20.8269 22.2178

Table 2: The ROOTS of its DERIVATIVES are given in the following table. 0 0 0 0 0 0 zero J0(x) J1(x) J2(x) J3(x) J4(x) J5(x) 1 3.8317 1.8412 3.0542 4.2012 5.3175 6.4156 2 7.0156 5.3314 6.7061 8.0152 9.2824 10.5199 3 10.1735 8.5363 9.9695 11.3459 12.6819 13.9872 4 13.3237 11.7060 13.1704 14.5858 15.9641 17.3128 5 16.4706 14.8636 16.3475 17.7887 19.1960 20.5755

0 Zeros of the ﬁrst ﬁve Jn(z) are represented in Table 2.

16 Figure 1

Bessel Function of the First Kind J 0.6 1 J 2 J 3 J 0.4 4 J5

0.2

0 x 2 4 6 8 10

-0.2

-0.4

16a Let us rewrite the Bessel equation as follows:

d s2 zJ 0 + zJ − J = 0 (71) dz s s z s By multiplying by 2zJ 0, we get

d (z2J 02 − s2J 2) + 2z2JJ 0 = 0 dz s s

d d 2z2JJ 0 = z2 J 2 = z2J 2 − 2zJ 2 dz dz Finally

d h i 2zJ 2 = z2J 02 + (z2 − s2)J 2 (72) s dz s s

Integrating (72) with respect to z from 0 to aN , we get:

Z aN 2 1 2 02 1 2 2 zJs (z)dz = aN Js (aN ) = aN Js±1(aN ) (73) 0 2 2 The last part of equation (72) follows from equations (43) and (44). In virtue of (43), 0 Js(aN ) = Js−1(aN ). In virtue of (44), Js+1(aN ) = −Js−1(aN ), thus:

2 2 02 Js+1(aN ) = Js−1(aN ) = Js (aN ) (74) n From Table 1, one can see that the ﬁrst zero a0 grows with n. The following statement is correct: The number of zeros of Js(z) on the interval s 1 0 < z < m + + π (75) z 4 is exactly m. Putting m = 1 into (75) we get: 3 s as < + π (76) 1 4 2 5 5 For s = 5, we get a1 < 10.35. In reality a1 = 8.7715. We see that this estimate is rather accurate.

17 8 Orthogonality and Fourier-Bessel series

s Let Js(z) be the Bessel function of order (or index) s. Let aN be its zeros such that s Js(aN ) = 0. Suppose that 0 < r < a is an interval on the real azis. We consider now (s) r s the set of the function RN (r) = Js( R aN ). This is the set of functions against the weight r. In other words

a Z (s) (s) RN (r)RM (r) rdr = 0 if N 6= M (77) 0 To prove this fact, we ﬁrst mention that

s s RN (r) = Js(aN ) = 0 (78) Then it is easy to check that these functions satisfy the equations

1 ∂ ∂Rs s2 a r N + k − Rs = 0, k = N (79) r ∂r ∂r N r2 N N a 1 ∂ ∂Rs s2 a r M + k − Rs = 0. k = M (80) r ∂r ∂r M r2 M M a

By multiplying these equations by rRM and rRN respectively and subtracting the results, we get ∂ ∂Rs ∂ ∂Rs Rs r N − Rs r M = (k2 − k2 )rR R (81) M ∂r ∂r N ∂r ∂r M M N M The left hand side can be rewritten as the following: ∂ r[R ,R ] = (k2 − k2 )rR R (82) ∂r M N M M N M ∂R δR [R ,R ] = R M − R N (83) M N M ∂r N δr

[RM ,RN ]|r=a = 0 (84) 2 2 Then if kM 6= kM , integration from zero to a leads to the condition of (77). Notice that s s r s we could replace functions RN (r) by Ren(r) = Js( a bN ). They satisfy the condition 0 Ren(a) = 0. In this case, equation (8.8) is again satisfied. The Wronskian [RM ,RN ] is s zero at r = a. Hence function ReN (r) satisfies the orthogonality condition (77). Suppose that f(r), 0 < r < a, is some real or complex function defined on the s interval (0, r). We can represent this function as a linear combination of RN (r).

18 Let

∞ X s f(r) = fN RN (r) (85) N=1 s By multiplying this to rRM (r) and integrating, we get

Z a Z a 1 s 1 s r fN = s2 f(r)rRN (r)dr = s2 f(r)rJN aN dr λN 0 λN 0 a Here

Z a 2 Z aN 2 5 1 a 2 1 2 2 λN = rRN (r)dr = 2 zJz (z)dz = a Js±1(aN ) (86) 0 2 aN 0 2 s r s Now one important remark, all functions RN (r) → ( 2a aN ) at r → 0. It means that the series (78) reasonably converges if the function f(r) behaves at r → as

f(r) → crs (87) If the asymptotics of (80) holds, the conditions for the convergence of the series are very similar to corresponding conditions for the standard Fourier series. In particular, if f(a) = 0 |f 0(r)| < C, where C is some arbitrary constant, this series converges absolutely and uniformly on 0 < r < a. A function f(r, θ) deﬁned in the disk 0 < r < a can be expanded in this disk in the Bessel - Fourier series. First, present f(r, θ) as a Fourier series in angles.

∞ X inθ f(r, θ) = fn(r)e (88) n=−∞ Z 2π 1 −inθ fn(r) = f(r, θ)e dθ (89) 2π 0

What is asymptotic of fn(r) if r → 0? Let us return to the Cartesian coordinates (x = r cos θ, y = r sin θ). Let fn(θ) be presented as follows: 1 1 f = f (r) + rf (θ) + r2f (θ) + ··· + Rn−1f (θ) (90) n 0 1 2 2 r n−1

f1(θ) = fx cos θ + fy sin θ

All other fn(θ) are trigonometric polynomials of order n − 1. Apparently:

19 Z 2π −inθ fk(θ)e dθ = 0 if k < n 0

n Hence fn(r) → Pnr as r → 0, with Pn some constant, and functions fn(r) are god for expansion in series of Fourier function of order n. Finally

∞ ∞ X X inθ raN f(r, θ) = fnN e Jn (91) n=−∞ N=1 a Z 2π Z a 1 −inθ raN fnN = 2 e dθ rJn f(r, θ)dr (92) 2πλnM 0 0 a

In particular, if f(x, y) = δ(x − x0)δ(y − y0) = r0δ(θ − θ0)δ(r − r0), 1 r −inθ0 0 fnN = 2 r0e JN aN (93) 2πλnM a Series (84) are especially good and fast converging if f(r, θ) satisﬁes the Dirichlet condition f(a, θ) = 0. If this function satisﬁes the Neumann condition (fr(a, θ) = 0), one can use the better following set of orthogonal functions:

r r 1,J b , ··· J b , ··· n a 1 n a n 9 Application of the Fourier - Bessel series to solutions of PDEs in circular domains

In this chapter, we apply Bessel function to solution of boundary problems for some basic equations of mathematical physics. We will solve equations inside the circle of radius a, 0 < r < a, with zero boundary conditions on the circle. First, we will start with the Poisson equation

1 ∂ ∂u 1 ∂2 ∆U = r + = f(r, θ) (94) r ∂r ∂r r2 ∂θ2

U|r=a = 0

20 We expand f(r, θ) in the Bessel - Fourier series

∞ ∞ n X X inθ raN f(r, θ) = fnN e Jn (95) N=1 n=−∞ a Z 2π Z a n 1 −inθ raN fnN = 2 e dθ rJn f(r, θ)dr (96) 2πλnM 0 0 a A solution of the Poisson equation can be found in the form of a similar series

∞ ∞ n X X inθ raN U(r, θ) = UnN e Jn (97) N=1 n=−∞ a

The coeﬃecients UnN and fnN are connected by a simple relation a2 UnN = − n 2 fnN (98) (aN )

Suppose that f(x, y) − δ(x − x0)δ(y − y0) = r0δ(r − r0)δ(θ − θ0) and x0 = r0 cos θ0, y0 = r0 sin θ0. Here we used the relation

dxdy = rdrdθ (99)

1 −inθ r0 n fnN = 2 e r0Jn aN (100) 2πλnN a Now,

ran r ∞ ∞ J ( N ) r an 0 X X in(θ−θ0) n a 0 N f(r, θ) = G(r, r0, θ − θ0) = e 2 Jn 2π n=−∞ N=1 λnN a

G(r, r0, θ − θ0) is the Green function for the Poisson equation. In other words:

Z 2π Z a U(r, θ) = dθ0 G(r, r0, θ − θ0)f(r0, θ0)dr0 (101) 0 0 Now we will solve the diﬀusion equation

∂u 1 ∂ ∂u 1 ∂2 = κ r + ∂r r ∂r ∂r r2 ∂θ2

21 inside of the circle 0 < r < a with zero boundary condition u|r=a = 0 and initial value u|t=0 = f(r, θ). Again, we must perform Fourier - Bessel expansion on (95) and (96). The solution is given by the expression:

∞ ∞ X X r aN 2 −inθ n −κ( a ) t U(r, θ, t) = fnN e Jn aN e (102) N=1 n=−∞ a The Green function of the the diﬀusion equation on the circle can be written as follows:

∞ ∞ r0 X X 1 r r0 1 aN 2 i(θ−θ0) n n −κ( a ) τ G(r, r0, θ − θ0, τ) = 2 e Jn aN Jn aN √ e 2π N=1 n=−∞ λnN a a 4kκτ (103) A solution of the uniform diﬀusion equation inside the circle ∂U = κ∆U + G(r, θ, τ) (104) ∂t

U|t=0 = f(r, θ),U|r=a = 0 is expressed through this Green function by the use of the standard formula

Z 2π Z a U(r, θ, t) = dφ0 f(r0, θ0)G(r, r0, θ − θ0, t)dr0 (105) 0 0

Z t Z 2π Z a + dτ dφ0 g(r0, θ0, τ)G(r, r0, θ − θ0, t − τ)dr0 0 0 0 The solution of the forced wave equation

∂2U = c2∆U + g(r, θ, t) ∂t2

U|t=0 = A(r, θ) Ut|t=0 = B(r, θ) is presented through the Green function:

22 r ∞ ∞ 1 r r 0 X X i(θ−θ0) n 0 G(r, r0, θ − θ0, τ) = 2 e Jn aN J aN sin wnN (106) 2π N=1 n=−∞ λnN wnN a a by the standard formula again:

Z 2π Z a U(r, θ, t) = dθ0 A(r0, θ0)Gt(r, r0, θ − θ0, t)dr0+ (107) 0 0

Z 2π Z a + dθ0 B(r0, θ0)G(r, r0, θ − θ0, t)dr0+ 0 0

Z t Z 2π Z a dτ dθ0 g(r0, θ0, τ)G(r, r0, θ − θ0, t − τ)dr0 0 0 0 c n Where wnN are eigenfrequencies (wnN = a aN ). Using Table 1, one can order the eigenfrequencies as follows.

wq+1 > wq, q1 = 1, ··· , ∞

c w = w a0 = 2.4048w , w = 1 0 1 0 0 a

1 w2 = w0a1 = 3.8317w0

2 w3 = w0a1 = 5.1336w0

0 w4 = w0a2 = 5.5201w0

0 w5 = w0a3 = 6.3802w0 If the Dirichlet boundary condition is replaced by the Neumann boundary condition. r r n In this case instead of Jn( a aN ) there should be Jn( a bN ). We must now correct the 2 value for λnm:

Z a 2 Z bN 2 2 r n a 2 λnm = Jn bN rdr = 2 Jn(z)zdz (108) 0 a bN 0

23 10 Bessel Functions of Second and Third Kinds

Bessel functions of the second kind, also called Neumann or Weber’s Functions, are deﬁned as solutions of the Bessel equation with the following asymptotics: s 2 πs π N (z) → sin z − − (109) s πz 2 4

If s is not an integer, one can present Ns(z) as a linear combination of Js and J−s

Ns(z) = AJs + BJ−s (110) To ﬁnd A, B we have to solve the following equation:

sin(φ) = A cos(φ) + B cos(φ + πs) πs π φ = z − − 2 4 As far as

cos(φ + πs) = cos(φ) cos(πs) − sin(φ) sin(πs) we get

A + B cos(πs) = 0 1 = −B sin(πs)

Finally,

cos(πs)J − J N = s −s (111) s sin(πs) n n What is going on if s → n? As far as cos(πn) = (−1) , J−n = (−1) Jn, both the numerator and denominator in (111) tend to zero and we should use L’hopital’s Rule. As a result:

cos(πn) ∂ J − ∂ J | 1 ∂ ∂ N (z) = ∂s s ∂s −s s=n = J − (−1)n J | (112) s π cos(πn) π ∂s s ∂s −s s=−n

24 To study the behavior of Nn(z) at z → 0 one can use the series representation of (29)

s n k k z X (−1) z Js(z) = 2 k=0 k!Γ(s + k + 1) 2 Now,

z s·ln z = e 2 2

s s d z z s·ln z z z = ln e 2 = ln ds 2 2 2 2 In the same way, Line is Cut Off in Notes Thus, Nn(z) can be presented as follows: z πN (z) = 2 ln J (z) − Nf (z) (113) n 2 n n (1) (2) 1 In (113), Nfn(z) = Nn (z) + Nn (z) appears as a result of differentiating Γ(±s+k+1) by s. Finally, we put s = n, then Nfn(z) is expanision only over integer powers of n, both positive and negative. When s → −n, the first n terms in the expansion (29) tend to zero. However, their derivatives by s are not zero. They can easily be calculated. The last term in Js vanishes as s → −n, which corresponds to k = n − 1. This term gives the following contribution to J−s(z)

1 z n−2 −s + n z n−2 = (n − 1)!Γ(−s + n) 2 (n − 1)!Γ(−s + n + 1) 2

Diﬀerentiating the above expression by s at s = n gives the following contribution to (1) Nn (z):

1 z n−2 (n − 1)! 2

25 Collecting all similar terms (k ≤ n−1) together, we end up with the following explicit (1) expression for Nn (z)

n−1 2k−n (1) X (n − k − 1)! z Nn (z) = (114) k=0 k! 2 (1) Nn (z) has singularities as z → ∞. The most singular term is: 2n N (1)(z) → (n − 1)! (115) n z ∞ 2k+n (2) X k z Nn (z) = ak(−1) (116) k=0 2 is the fast converging power series in (116) 1 a = ψ(k + 1) + ψ(k + n + 1) (117) k k!(n + k)! Here, ψ(s) is a new special function

d Γ0(s) ψ(s) = ln(Γ(s)) = ds Γ(s)

Z ∞ Γ0(s) = e−t ln(t)ts−1dt (118) 0 Through the use of integration by parts, one can ﬁnd the explicit value of ψ(s) in all −t d −t integral points can be [Line Cut Oﬀ Notes]. Remembering that e = − dt e , we get from (118)

Γ0(s) = (s − 1)Γ0(s − 1) + Γ(s − 1)

By dividing by Γ(s) and replacing s → s = 1, we end up with the diﬀerence equation for ψ

1 ψ(s + 1) = ψ(s) + s Z ∞ ψ(1) = e−t ln(t)dt = −c (c ≈ 0.5772) (119) 0

26 c is the so-called Euler constant. Finally, we get for any integral point. 1 ψ(k + 1) = −c + 1 + ··· + (120) k To ﬁnish with the Neumann functions, we present the asymptotic behavior of the ﬁrst two at z → 0

z N (z) → 2 ln + c + 0(z2) 0 2 2 N (z) → − + 0(z) · ln(z) (121) 1 z

Functions Nn are even if n is even and odd if Nn is odd.

n Nn(−z) = (−1) Nn(z) (122) The Bessel functions of third kind are also know as the Hankel functions, which are deﬁned as follow:

(1) Hs (z) = Js(z) + iNs(z)

(2) Hs (z) = Js(z) − iNs(z) (123) At z → ∞ they have the following amymptotics s (1) 2 i(z− πs − π ) H (z) → e 2 4 s πz s (2) 2 −i(z− πs − π ) H (z) → e 2 4 (124) s πz (2) ¯ (1) apparently, Hs (z) = Hs (z). The Neumann functions are plotted in Figure 2.

11 Modiﬁed Bessel Functions

Modiﬁed Bessel functions are solutions of the modiﬁed Bessel equation

1 d dR s2 z − 1 − R = 0 (125) z dz dz z2

27 This equation appears if we perform the transformation z → iz. In other words, the modiﬁed Bessel functions are Bessel functions of imaginary argument. However, one muse be careful performing the transform z → iz because we have to observe not only asymptotics at z → 0, but also asmyptotics at z → ∞. One solution Ys(z) of equation (125) is deﬁned by the series

s ∞ 2k z X 1 z Is(z) = (126) 2 k=0 k!Γ(k + s + 1) 2 At z → 0 zs I (z) ≈ (127) s zsΓ(s + 1) At z → ∞ s 2 I (z) → ez (128) s πz

Is is the real function of real argument. They are connected with Bessel functions of the ﬁrst kind by the relation:

− π si Is(z) = e 2 Js(iz) (129) In particular, n In(z) = −i Jn(iz) (130) Modiﬁed Bessel functions of second kind are deﬁned by the relation

πi π si (1) k (z) = e 2 H (iz) (131) s 2 s They have asymptotics at r π k (z) ≈ e−z, z → ∞ (132) s 2z

Both Is(z) and ks(z). Modiﬁed Bessel functions of the First and Second kind are plotted on Figures 3 and 4.

12 Applications of the Modiﬁed Bessel Function

The modiﬁed Bessel functions are commonly used for solutions to many diﬀerent applied problems. Let us consider the cylinder of radius R and length 2a 0 < r < R −a < z < a

28 Figure 2

Bessel Function of the Second Kind

0.4

0.2

0 2 4 6 8 10

-0.2

-0.4

-0.6

-0.8

-1

A Bessel function of the second kind Nn(x) is a solution to the BESSEL DIFFERENTIAL EQUATION which is singular at the origin. Bessel functions of the second kind are also called NEUMANN FUNCTIONS or WEBER FUNCTIONS. The above plot shows Nn(x) for n = 1 , 2 , ... , 5

28a Figure 3

Modi ed Bessel Function of the First Kind

3.0

2.5

2.0 I 1 I2 I3 1.5 I4 I5 1.0

0.5

0 x 1 2 3 4 5

A function In(x) which is one of the solutions to the MODIFIED BESSEL DIFFERENTIAL EQUATION and is closely related to the BESSEL FUNC- TION OF THE FIRST KIND Jn(x). The above plot In(x) for n = 1, 2, ... , 5.

28b Figure 4

Modi ed Bessel Function of the Second Kind

0 1 2 3 4 5

The function Kn(x) which is one of the solutions to the MODIFIED BESSEL DIFFERENTIAL EQUATION. The above plot shows Kn(x) for n = 1, 2, ... , 5.

28c Let us solve the Laplace equation:

1 ∂ ∂u 1 ∂2u ∂2u ∆u = r + + = 0 (133) r ∂r ∂r r2 ∂θ2 ∂z2 with the ”lateral” boundary conditions:

u|z=a = 0, u|z=−a = 0

u|r=R = f(z, θ) − a < z < a

f(z, θ + 2π) = f(z, θ)

To solve equation (133), we will use separation of variables and look for solutions in the form: πm u = einθ cos z R (r) (134) 2a n,m

Rn,m satisﬁes the following equation: 1 ∂ ∂R n2 πm r − k2 + R = 0, k2 = (135) r ∂r ∂r m z2 m 2a Equation (135) has the following solution satisfying the condition:

In(kmz) R|r=R = 1,R = (136) In(kmR) The solution of the Laplace equation is:

∞ ∞ X X inθ πm In(kmz) u = fn,me cos z (137) n=−∞ m=1 2a In(kmR) fn,m - coeﬃcient of double Fourier series to be found from the condition:

∞ ∞ X X inθ πm f(z, θ) = fn,me cos z (138) n=−∞ m=1 2a They are:

Z 2π Z a 1 −inθ πm fn,m = dθ f(θ, z)e cos z dz (139) 2πa 0 −a 2a

29 Let us now solve the Laplace equation (133) outside the inﬁnite cylinder (0 < r < R and −∞ < z < ∞). Again, the boundary condition is:

u|r=R = f(z, θ) − ∞ < z < ∞

f(z, θ + 2π) = f(z, θ) To solve the problem, we must specify the boundary condition in inﬁnity:

∂u → 0 at r → ∞ ∂r Now, we should perform the Fourier transformations in the z-direction:

Z ∞ u(k, θ, r) = u(z, θ, r)e−ikzdz −∞ Z ∞ f(k, θ) = f(z, θ)e−ikzdz (140) −∞ and realize the expansion in the Fourier series in angles, thus:

X inθ f(k, θ) = fn(k)e

Z 2π 1 −inθ fn(k) = f(k, θ)e dθ (141) 2π 0 The radial part of the solution satisﬁes the modiﬁed Bessel equation: 1 ∂ ∂R n r − k2 + R = 0 (142) r ∂r ∂r r It should be taken as follows:

K (kr) R(k, n) = n Kn(kR) The solution is given by the inverse Fourier Transform:

Z ∞ ∞ 1 X Kn(kr) −ikz+inθ u(z, r, θ) = dt fn(k) e (143) 2π −∞ n=−∞ Kn(kR)

30 13 Heavy Chain

The future Bessel function appeared in mathematics in 1732 when Danie Bernoully solved the problem on oscillations of the hung, heavy chain. Let the chain of length l and linear density ρ be hung such that it can move only in one direction. Let the coordinate x be taken such that x = 0 at the free end of the chain. Then the deviation from equilibrium state u = u(x, t) satisﬁes the equation:

∂2u ∂2u ∂u = g x + (144) ∂t2 ∂x2 ∂x

u|x=l = 0

Use separation of variables:

u = X(x)T (t) where T (t) = sin(ωt + ϕ) leads to the equation:

ω2 xX00 + X0(x) + X(x) = 0 (145) g with the boundary condition:

X(l) = 0,X(0) < ∞

By introducing the new variable:

sx y = 2ω g

We transform (145) to the Bessel equation:

d2X 1 dX + + X = 0 (146) dy2 y dy This is the equation for Bessel functions of zero order. Thus, the solution is:

31 sx u = AJ 2ω (147) 0 g

The characteristic frequency ωk can take consequence of discrete values (ωk, k = 1, 2, ··· , ∞). They can be found from the boundary condition:

s l u(l) = 0 J 2ω = 0 0 k y Hence,

s l 2ω = a0 J (a0) = 0 k g k 0 k 1rg ω = a0 k 2 l k

0 an - zeros of the Bessel function J0.