Chap . 5 Series Solutions of ODEs Special Functions
This chapter introduces you to higher functions that are needed in physics and its engineering applications. Just as you could not do problems in calculus without knowing trigonometric and exponential functions, you could not solve important models involving ordinary or partial differential equations without knowing Bessel functions and Legendre polynomials, to mention just the two most prominent classes of these functions. Although these and the many other special functions of practical interest have quite different properties and serve very distinct purposes, it is most remarkable that these functions are accessible by the same instrument, namely, power series , (whose simplest examples you know from calculus), perhaps multiplied by a fractional power or a logarithm. You should see and learn which of the many functions and series are of importance in applications and what methods are available for investigating properties through series and for discovering relations between functions, in addition to the calculation of values by series and representing the results graphically by curves. Such an overview will help you to find your way through the vast number of books and articles in journals related to special functions when you need help in solving specific engineering or other problems. Your CAS knows all the functions that you will ever need, but it is your task to establish a general orientation in this wide field of relationships and formulas, in order to find and select what functions and relations you need for specific tasks and situations.
Sec . 5.1 Power Series Method
Problem Set 5.1. Page 170
5. Terminating power series . The ODE 2 + x y′ = y can be solved by separating variables, dy dx = , ln |y| = ln |x + 2 | + c, y = c̃ x + 2. y x + 2 The solution is a polynomial, a terminating power series, and you will see why, and how termination occurs. Substitute the power series for y and y′ into the given ODE to get 2 2 2 + xa1 + 2a2 x + 3a3 x + ⋯ = a0 + a1 x + a2 x + ⋯. Write the left side as a power series by multiplying out and ordering, 2 2a1 + 4a2 x + 6a3 x + ⋯ 2 3 + a1 x + 2a2 x + 3a3 x + ⋯ 2 = 2a1 + 4a2 + a1 x + 6a3 + 2a2 x + ⋯. Now compare powers on both sides:
a0 a0 = 2a1 hence a1 = 2
a1 = 4a2 + a1 a2 = 0
a2 = 6a3 + 2a2 a3 = 0, etc.
and you see how it happens that the power series terminates after two terms. a0 remains arbitrary, so that x you have obtained a general solution of your first-order ODE, namely, y = a0 1 + 2 .
13. Verhulst ODE , initial value . y′ = y − y2 = y1 − y has the general solution (see (9) on p. 31 in Sec. 1.5 with A = B = 1) y = 1/Ce−t + 1, Chap. 5 Series Solutions of ODEs. Special Functions 41
and the initial value 1/2 gives the solution (C = 1) y = 1/e−t + 1. 1 1 You see that you can introduce y0 = 2 into the power series, obtaining y = 2 + a1 x + ⋯, so that also 1 1 − y = 2 − a1 x − ⋯ in the other factor on the right. Hence by substitution, 2 3 4 a1 + 2a2 x + 3a3 x + 4a4 x + 5a5 x + ⋯
1 2 3 1 2 3 = + a1 x + a2 x + a3 x + ⋯ a1 x a2 x a3 x ⋯ . 2 2 − − − −
0 1 1 1 1 Compare like powers on both sides. x gives a1 = 2 − 4 = 4 . Then x gives 1 1 1 2a2 = a0 a1 + a1 a0 a1 = + = 0, hence a2 = 0. − − 8 4 − 8 x2 gives
2 1 1 3a3 = a0 a2 + a + a0 1 a0 = , hence a3 = . − 1 − − 16 − 48 x3 gives 1 1 4a4 = a3 + a3 = 0, hence a4 = 0. − 2 2 x4 gives 1 1 1 5a5 = a4 a1 a3 + a2 a2 + a3 a1 + a4 = , 2 − − − − − 4 96 hence a5 = 1/480, as given on p. A13. The value s1 = 0.73125 (the sum of the coefficients just calculated!) is exact to 3D, and the value exact to 5D is 1/e−1 + 1 = 0.73106. This calculation shows that for a nonlinear ODE, working with power series is inconvenient . Indeed, for ODEs these series find their main application in connection with linear ODEs; and all the ODEs we shall now consider in this chapter will be linear.
Sec . 5.2 Theory of the Power Series Method
Problem Set 5.2. Page 176
7. Radius of convergence . A power series in powers of x may converge for all x (this is the best possible case) or within an interval with the center x0 as midpoint (in the complex plane: within a disk with center x0) or only at the center (the practically useless case). In the second case the interval of convergence has length 2R, where R is called the radius of convergence (it is a radius in the complex case, as has just been said) and is given by (7a) or (7b) of Sec. 5.2. Here it is assumed that the limits in these formulas exist. This will be the case in most applications. (For help when this is not the case, see Sec. 15.2.) The convergence radius is important whenever you want to use series for computing values, exploring properties of functions represented by series, or proving relations between functions, tasks of which you will gain a first impression in Secs. 5.3-5.7 and corresponding problems. In Prob. 7 you may set 2 m x − 1 = t. Then you have a power series in t with coefficients of absolute value |am| = 1/4 . Hence the root in (7a) is 1/4, so that the radius of convergence of the power series in t is 4. That is, the series converges for |t| < 4. This implies |x − 1| = |t| < 2. Hence the given series has the convergence radius 2. Confirm this by using (7b). The quotient in (7b) is
|am+1/am| = 1/4. This leads to the same result as before. Note that the problem is special; in general, the sequences of those roots and quotients in (7) will not be constant, that is, the terms of such a sequence of quotients (or roots) will not be all the same. 42 Ordinary Differential Equations (ODEs) Part A
17. General solution . Whittaker functions . A general solution of a second-order ODE involves two arbitrary constants. In power series solutions these are two coefficients which are left arbitrary and to which the other coefficients are recursively related. Usually these are a0 and a1. In simpler cases the coefficients a2, a4, ⋯ are then related to a0, and the coefficients a3, a5, ⋯ of the odd powers are related to a1. For instance, this is the situation in the case of a power series solution of y′′ + y = 0
in which you obtain y = a0 cos x + a1 sin x. However, it may very well happen that each of the two functions of a basis of solutions involves both even and odd powers. This is the case for our present ODE y′′ − y′ + x2 y = 0 2 as you will see. If you ask your CAS, it will tell you that solutions of this ODE are Mk,mix and 2 x/2 Wk,mix , i = −1 , multiplied by e / x . From this you conclude that these functions are special functions that have been investigated, so that if you need to know some of their properties, you need not start from scratch but can make a search in the literature, notably, in Ref [GR1] in Appendix 1 of AEM, where you will find on p. 1045 in the Index of Notations that Mκ,μz and Wκ,μz are confluent hypergeometric functions , called Whittaker functions (W suggests Whittaker) and which are discussed on p. 505 of Ref. [GR1]. You obtain the coefficients of the power series as given on p. A13 by substituting the usual expressions for y′′, y′ and y, obtaining
2 3 4 5 2a2 + 3 ∙ 2a3 x + 4 ∙ 3a4 x + 5 ∙ 4a5 x + 6 ∙ 5a6 x + 7 ∙ 6a7 x + ⋯ 2 3 4 5 − a1 − 2a2 x − 3a3 x − 4a4 x − 5a5 x − 6a6 x − ⋯ 2 3 4 5 + a0 x + a1 x + a2 x + a3 x + ⋯ = 0. Now equate the sum of the coefficients of each occurring power of x to 0:
0 1 x : a2 = a1, a1 arbitrary 2
1 1 1 x : 2a2 = 3 ∙ 2a3, a3 = a2 = a1 3 6
2 1 1 1 x : a0 3a3 + 12a4 = a0 a1 + 12a4 = 0, a4 = a0 + a1 − − 2 − 12 24
3 1 1 x : a1 4a4 + 20 a5 = a1 4 a0 + a1 + 20a5 − − − 12 24 1 5 1 1 = a0 + a1 + 20a5 = 0 a5 = a0 a1. 3 6 − 60 − 24
With these coefficients depending on both a0 and a1, which remain arbitrary, you can now make up the two series as given on p. A13. A revised edition of [GR1] will appear in 2006 as the Digital Library of Mathematical Functions (http://dlmf.nist.gov).
Sec . 5.3 Legendre ’s Equation . Legendre Polynomials Pnx
Note well that Legendre’s equation involves the parameter n, so that (1) is actually a whole family of ODEs, with basically different properties for different n. In particular, for integer n = 0,1,2,⋯ one of the series (6) or (7) reduces to a polynomial, and it is remarkable that these “simplest” cases are of particular interest in applications. Chap. 5 Series Solutions of ODEs. Special Functions 43
Problem Set 5.3. Page 180
5. Legendre functions for n = 0. The power series and Frobenius methods were instrumental in establishing large portions of the very extensive theory of special functions (see, for instance, Refs. [GR1], [GR10] in Appendix 1 of AEM), as needed in engineering, physics, and other areas, simply because many special functions appeared first in the form of power series solutions of differential equations. In general, this concerns properties and relationships between higher transcendental functions. The point of Prob. 5 is an illustration that sometimes such functions may reduce to elementary functions known from calculus. If you set n = 0 in (7), it was observed in the problem that y2x becomes 1 2 ln 1 + x/1 − x. In this case, the answer suggests using 1 1 ln 1 + x = x x2 + x3 + ⋯ . − 2 3 − Replacing x by −x and multiplying by −1 on both sides gives 1 1 1 − ln 1 − x = ln = x + x2 + x3 + ⋯ . 1 − x 2 3 Addition of these two series and division by 2 verifies the last equality sign in the formula of Prob. 5. You are requested to obtain this result directly by solving the Legendre equation (1) with n = 0, that is, 1 − x2y′′ − 2xy′ = 0 or 1 − x2z′ = 2xz, where z = y′. Separation of variables and integration gives
dz 2x 2 2 = dx , ln |z| = −ln |1 − x | + c, z = C1/1 − x . z 1 − x2 y is now obtained by another integration, using partial fractions 1 1 1 1 = − . 1 − x2 2 x + 1 x − 1 This gives 1 1 x + 1 y = z dx = C1ln x + 1 − ln x − 1 + c = C1 ln + c. ∫ 2 2 x − 1 Since y1x in (6), Sec. 5.3, reduces to 1 if n = 0, you can now readily express your solution obtained in terms of the standard functions y1 and y2 in (6) and (7), namely,
y = cy 1x + C1y2x.
7. ODE . Set x = az , thus z = x/a, and apply the chain rule, according to which d d dz 1 d d2 1 d2 = = and = . dx dz dx a dz dx 2 a2 dz 2 Substitution now gives d2 y 1 dy 1 a2 − a2 z2 − 2az + nn + 1y = 0. dz 2 a2 dz a The factors a cancel and you are left with 1 − z2y′′ − 2zy′ + nn + 1 y = 0.
Hence solutions are Pnz = Pnx/a and Qnx/a, as claimed in Appendix 2, p. A13.
Sec . 5.4 Frobenius Method
To write down the indicial equation, e.g. in order to find out what case an ODE corresponds to, you often need not consider any series, just write the ODE in standard form, with bx and cx as in (1); then you see bx and cx and can take their constant terms b0 and c0 and use (4). For instance, the ODE in Prob. 5 is x2 y′′ + 4xy′ + x2 + 2 y = 0. 44 Ordinary Differential Equations (ODEs) Part A
Writing it in the form (1), you have 4 x2 + 2 y′′ + y′ + y = 0. x x2 2 You see that bx = 4 and cx = x + 2. Hence you have b0 = 4, c0 = 2, so that the indicial equation is rr − 1 + 4r + 2 = r2 + 3r + 2 = r + 2 r + 1 = 0, and the roots are −2 and −1, as given on p. A13.
Problem Set 5.4. Page 187
1. Basis of solutions . Substitute y, y′, and y′′, given by (2) and (2*) in Sec. 5.4, into the differential equation xy ′′ + 2y′ − xy = 0. This gives ∞ ∞ ∞ m+r−1 m+r−1 n+r+1 ∑ m + rm + r − 1 am x + ∑ 2m + r am x − ∑ an x = 0. m=0 m=0 n=0 The first two series have the same general power, and you can take them together. In the third series set n = m − 2 to get the same general power. n = 0 then gives m = 2. You obtain ∞ ∞ m+r−1 m+r−1 ∑ m + r m + r + 1 am x − ∑ am−2 x = 0.(A) m=0 m=2 For m = 0 this gives the indicial equation rr + 1 = 0. The roots are r = 0 and −1. They differ by an integer. This is Case 3 of Theorem 2 in Sec. 5.4. Consider the larger root r = 0. Then (A) takes the form ∞ ∞ m−1 m−1 ∑ m m + 1 am x − ∑ am−2 x = 0. m=0 m=2
m = 1 gives 2a1 = 0. This implies a3 = a5 = ⋯ = 0, as is seen by taking m = 3,5,⋯ . Furthermore,
m = 2 gives 2 ∙ 3a2 − a0 = 0, hence a0 arbitrary, a2 = a0/3!
m = 4 gives 4 ∙ 5a4 − a2 = 0, hence a4 = a2/4 ∙ 5 = a0/5!
and so on. Since you want a basis and a0 is arbitrary, you can take a0 = 1. Recognize that you then have the Maclaurin series of
y1 = sinh x/x.
Now determine an independent solution y2. Since in Case 3 one would have to assume a term involving the logarithm (which may turn out to be zero), reduction of order (Sec. 2.1) seems to be simpler. This begins by writing the equation in standard form (divide by x):
′′ 2 ′ y + x y − y = 0. In (2) of Sec. 2.1 you then have p = 2/x, −∫ p dx = −2ln |x| = ln 1/x2, hence exp −∫ p dx = 1/x2. 2 2 Insertion of this and y1 into (9) and cancellation of a factor x gives
2 cosh x U = 1/sinh x, u = ∫ U dx = −coth x, y2 = uy 1 = − x .
7. Basis of solutions . Try the substitution x + 3 = t. Can you see why?
19. Hypergeometric equation . Your task is to find the values of the parameters a, b, c of the given hypergeometric ODE 1 1 x1 x y′′ + 2x y′ y = 0. − 2 − − 4 The first coefficient x1 − x has the same form as in (15) on p. 188, so that you can immediately Chap. 5 Series Solutions of ODEs. Special Functions 45
compare the coefficients of y′, 1 2x = c a + b + 1 x (A) 2 − − and the coefficients of y, 1 ab = . (B) − − 4 From (B) you have b = 1/4a. Substitute this into (A), obtaining from the terms in x and from the constant terms 1 1 a + + 1 = 2 and c = . 4a 2 2 1 1 1 Hence a + 1/4a = 1 or a − a + 4 = 0, so that a = 2 and b = 2 . Consequently, a first solution is 1 1 1 1 F 2 , 2 , 2 ; x. A second solution is given by (17) on p. 188 of AEM with 1 − c = 2 , that is,
1/2 3 y2x = x F 1, 1, ; x , 2 1 1 3 because a − c + 1 = 2 − 2 + 1 = 1, b − c + 1 = 1, and 2 − c = 2 .
Sec . 5.5 Bessel ’s Equation . Bessel Functions Jνx
We begin with a guide through the present long section. Bessel’s equation x2 y′′ + xy′ + x2 − ν2 y = 0(1) involves a given parameter ν ≥ 0. Hence the solutions yx depend on ν, and one writes yx = Jνx. Integer values of ν appear often in applications, so that they deserve a special notation ν = n. We treat these first. The Frobenius method, just as the power series method, leaves a0 arbitrary. This would make formulas and numeric values involving an arbitrary constant. To avoid this, a0 is assigned a definite value (that depends on n). A relatively simple series (11) is obtained by choosing 1 a0 = . (9) 2n n! From integer n we now turn to arbitrary ν. The choice of (9) makes it necessary to generalize the factorial function n! to noninteger ν. This is done on p. 192 by the gamma function , leading to (20), a power series times a single power xν. Pages 193-194 concern the problem of finding a second linearly independent solution. Formula (24) is the backbone of the formalism for Bessel functions. Finally, on p. 196 we show that special parametric values may lead to elementary functions. This is generally true for special functions; see, for instance, Problem Set 5.4.
Problem Set 5.5. Page 197
5. Reduction to Bessel ’s equation . This ODE x2 y′′ + xy′ + λ2 x2 − ν2 y = 0 is particularly important in applications. The second parameter λ slips into the independent variable if ∙∙ you set z = λx, hence by the chain rule y′ = dy /dx = dy /dz dz /dx = λẏ, y′′ = λ2 y , where ∙ = d/dz . Substitute this to get ∙∙ z2/λ2 λ2 y + z/λλẏ + z2 − ν2y = 0.
λ cancels. A solution is Jνz = Jνλx.
13. Reduction to Bessel ’s equation . The transformation is perhaps suggested to some extent by the presence of x6 in the ODE 46 Ordinary Differential Equations (ODEs) Part A
x2 y′′ + xy′ + 9x6 − ν2 y = 0. Use the chain rule to get y′ = dy /dz dz /dx = 3x2 ẏ = 3z2/3 ẏ y′′ = 9x4 y∙∙ + 6xẏ = 9z4/3 y∙∙ + 6z1/3 ẏ. Substitute this into the given ODE to obtain 2/3 4/3 ∙∙ 1/3 1/3 2/3 2 2 z 9z y + 6z ẏ + z ∙ 3z ẏ + 9z − ν y = 0. Simplify this, taking the two ẏ-terms together, ∙∙ 9z2 y + 6z + 3z ẏ + 9z2 − ν2 y = 0. 3 3 Divide by 9 to see that solutions are y1 = Jνz = Jνx and y2 = J−νz = J−νx ; cf. p. A14.
15. Reduction to Bessel ’s equation . It is common that simple-looking ODEs often may have nonelementary solutions. This is the case for xy′′ + y = 0. Use the first given transform and the chain rule to obtain y = x1/2 u 1 y′ = x1/2 u′ + x−1/2 u 2 1 y′′ = x1/2 u′′ + x−1/2 u′ x−3/2 u. − 4 Substitute this into the given ODE, obtaining 1 x3/2 u′′ + x1/2 u′ x−1/2 u + x1/2 u = 0. − 4 Now switch from x to z = 2x1/2, as given. Denote derivatives with respect to z by dots and use 1 ∙ −1/2 −1/2 dz /dx = 2 2 x = x . Calculate u′ = x−1/2 u̇
∙∙ 1 u′′ = x−1 u x−3/2 u̇ . − 2 Substitute this into the previous ODE, obtaining
∙∙ 1 1 x1/2 u u̇ + x1/2 u̇ x−1/2 x−1/2 u + x1/2 u = 0. − 2 − 4 Simplify this and introduce z = 2x1/2, hence x1/2 = z/2, finding
1 ∙∙ 1 1 1 z u + u̇ + z z−1 u = 0. 2 2 2 − 2 Multiply this by 2z to see that this is Bessel’s equation with n = 1, so that a solution is u = J1z, and 1/2 1/2 1/2 y = x u = x J12x .
′ 27. Integration . Proceed as suggested on p. A14. Formula (24a) with ν = 1 is xJ1 = xJ0, integrated xJ1 = ∫ xJ0 dx , and gives (with integration by parts similar to that in Example 2 in the text)
2 ∫ x J0 dx = ∫ xxJ0 dx = xxJ1 − ∫ 1 ∙ xJ1 dx (C) 2 = x J1 − ∫ xJ1 dx .
′ Now use (24b) with ν = 0, that is, J0 = −J1, to obtain on the right side of (C) 2 x J1 − x−J0 − ∫ 1 ∙ −J0 dx
2 = x J1 + xJ0 − ∫ J0 dx . Chap. 5 Series Solutions of ODEs. Special Functions 47
Your CAS will perhaps give you some other functions. However, it seems essential that everything is expressed in terms of Jn with several n because Bessel functions and their integrals are usually computed recursively from J0 and J1; similarly for fractional values of the parameter ν.
Sec . 5.6 Bessel Functions of the Second Kind Yνx
These Bessel functions are introduced to have a basis of solutions of Bessel’s equation for all values of ν; recall that integer ν = n gave trouble (linear dependence of Jn and J−n; Theorem 2 on p. 193). We discuss the case ν = n = 0 first and in detail on pp. 198-200. For general ν we give just the main results (on pp. 200-202) without going into all the details.
Problem Set 5.6. Page 202
9. Reduction to Bessel ’s equation . The given ODE x2 y′′ − 5xy′ + 9x6 − 8 y = 0 could be transformed to Bessel’s equation in two steps, transforming the independent variable by setting z = x3 and the dependent variable (the unknown solution of the ODE) by setting y = x3 u. You can also combine the two transformations, as we shall now explain. This is done by the chain rule. You need dz /dx = 3x2. Transform y′ to get (denoting derivatives with respect to z by dots)
dy d 3 2 3 du 2 5 2 = x u = 3x u + x ∙ 3x = 3x u̇ + 3x u dx dx dz where ∙ = d/dx . From this you obtain for the second derivative d2y d = 3x5 u̇ + 3x2 u dx 2 dx 4 7 ∙∙ 2 2 = 15x u̇ + 9x u + 6xu + 3x u̇ ∙ 3x = 9x7 u∙∙ + 24x4 u̇ + 6xu. Substitute these derivatives into the given ODE to get 9x9 u∙∙ + 24x6 u̇ + 6x3 u − 15x6 u̇ − 15x3 u + 9x6 − 8 x3 u = 0. Collect terms and divide by 9:
∙∙ x9 u + x6 u̇ + x9 − 9x3 u = 0. Divide by x3 and set x3 = z: ∙∙ z2 u + zu̇ + z2 − 9 u = 0. 3 This is Bessel’s equation with parameter n = 3. A solution is u = J3z = J3x . Hence 3 3 3 3 3 y = x u = x J3x , and a second linearly independent solution is x Y3x .
11. Hankel functions . Show linear independence, starting from 1 2 c1 Hν + c2 Hν = 0. Insert the definitions of the Hankel functions:
c1 Jν + iYν + c2 Jν − iYν = c1 + c2 Jν + ic1 − ic2 Yν = 0.
Since Jν and Yν are linearly independent, their coefficients must be zero, that is (divide the second coefficient by i)
c1 + c2 = 0
c1 − c2 = 0. 1 2 Hence c1 = 0 and c2 = 0, which means linear independence of Hν and Hν on any interval on which 48 Ordinary Differential Equations (ODEs) Part A
these functions are defined. (The proof on p. A14 also uses the linear independence of Jν and Yν.
Sec . 5.7 Sturm -Liouville Problems Orthogonal Functions
Example 2 concerns cosine and sine functions. These functions constitute the simplest and most important set of orthogonal functions because they lead to Fourier series (Chap. 11), whose invention was perhaps the most important progress ever made in applied mathematics. Indeed, these series turned out to be fundamental for both ODEs and PDEs (Chaps. 2 and 12). Example 5 on p. 207 is remarkable inasmuch as you need no boundary conditions in dealing with Legendre polynomials. Example 6 and Theorem 2 concern Bessel functions, infinitely many for every fixed n. So in (13), n is fixed. The smallest n is n = 0. Then (13) concerns J0. Eq. (13) then takes the form R xJ0km0 x J0kj0 x dx = 0 j ≠ m, both integer . ∫0
If n = 0 were the only possible value of n, you could simply write km and kj instead of km0 and kj0; write it down for youself to see what (13) then looks like. Recall that km0 is related to the zero αm0 of J0 by km0 = αm0/R. In applications (vibrating drumhead in Sec. 12.9, for instance) the number R can have any value depending on the problem (in Sec. 12.9 it is the radius of the drumhead); this is the reason for introducing the arbitrary k near the beginning of the example; it gives us the flexibility needed in practice.
Problem Set 5.7. Page 209
1. Case 3 of Theorem 1. In this case the proof runs as follows. By assumption, pa = 0 and pb ≠ 0. The starting point of the proof is (9), as before. Since pa = 0, you see that (9) reduces to ′ ′ pb ynbymb − ymbynb and you have to show that this is zero. Now from (2b) you have ′ l1ynb + l2ynb = 0 ′ l1ymb + l2ymb = 0.
At least one of the two coefficients is different from zero, by assumption; say, l2 ≠ 0. Now multiply the first equation by ymb and the second by −ynb and add, obtaining ′ ′ l2ynbymb − ymbynb = 0.
Since l2 is not zero, the expression in the brackets must be zero. But this expression is identical with that in the brackets in the first line of (9), which we have written above. The second line of (9) is zero because of the assumption pa = 0. Hence (9) is zero, and from (8) you obtain the relationship (6) to be proved. (For l1 ≠ 0 the proof is similar. Supply the details; this will show you whether you really understand the present proof.)
3. Change of x. Equate ct + k to the endpoints of the given interval and solve for t to get the new interval on which you can prove orthogonality.
9. Sturm -Liouville problem . The given equation and boundary conditions do constitute a Sturm-Liouville problem. The equation is of the form (1) with p = 1, q = 0, r = 1. The interval in which solutions are sought is given by a = 0 and b = L as its endpoints. In the boundary conditions, k1 = 1, k2 = 0, l1 = 0, l2 = 1. You first have to find a general solution. In Prob. 9 it is y = A cos kx + Bsin kx where k = λ .(A) You obtain eigenvalues and functions by using the boundary conditions. The first condition gives y0 = A = 0. Differentiation of the remaining part of equation (A) gives y′x = kB cos kx , hence y′L = kB cos kL = 0, Chap. 5 Series Solutions of ODEs. Special Functions 49
thus cos kL = 0. This yields kL = knL = 2n + 1π/2, where n = 0,1,⋯; these are the positive values for which the cosine is zero. You need not consider negative values of n because the cosine is even, so that 2 you would get the same eigenfunctions. The eigenvalues are λ = λn = kn . The corresponding eigenfunctions are 2n + 1πx yx = y x = sin k x = sin . n n 2L The figure shows the first few eigenfunctions assuming that L = 1. All of them start at 0 and have a horizontal tangent at the other end of the interval from 0 to 1. This is the geometric meaning of the boundary conditions. y1 has no zero in the interior of this interval. Its graph shown corresponds to 1/4 of the period of the cosine. y2 has one such zero (at 2/3), and its graph shown corresponds to 3/4 of that period. y3 has two such zeros (at 0.4 and 0.8). y4 has three, and so on.
1
0.5
0 0.2 0.4 0.6 0.8 1 x
-0.5
-1
Sec . 5.7. Prob . 9. First four eigenfunctions of the Sturm-Liouville problem with L = 1
Sec . 5.8 Orthogonal Eigenfunction Expansions
Example 1 shows the beginning of a discussion that will be continued in the chapter on Fourier series (Chap. 11). In Example 2 the first two terms have the largest coefficients in absolute value, much larger than those of any further terms, because a1P1x + a3P3x resembles sin πx very closely (make a sketch, using Fig. 104 on p. 180). Also notice that a2 = a4 = ⋯ = 0 because sin πx is an odd function. Example 3 shows a Fourier-Bessel series in terms of J0, with arguments of the terms determined by the location of the zeros of J0. The next series would be in terms of J1, the second next in terms of J2, and so on. The last part of the section concerns a concept of convergence that is suitable in connection with orthogonal expansions and is quite different from the convergence considered in calculus. The basics are given in the section, and more details (not needed here) belong to special courses of higher level; see, for instance, Ref. [GR7] in Appendix 1.
Problem Set 5.8. Page 216
1. Fourier -Legendre series . In Example 2 of the text we had to determine the coefficients by integration. In the present case this would be possible, but the method of undetermined coefficients is much simpler. The given function fx = 7x4 − 6x2
is of degree 4, hence you need only P0, P1,⋯,P4. Since f is an even function, you actually need only P0, P2, P4. Write 4 2 fx = a4 P4x + a2 P2x + a0 P0x = 7x − 6x . 4 1 4 2 Begin by determining a4 so that the x -terms on both sides agree. Since P4x = 8 35x − 30x + 3 (see Sec. 5.3), you have the condition 50 Ordinary Differential Equations (ODEs) Part A
35 7 ∙ 8 a = 7, hence a = = 1.6. 8 4 4 35 Calculate the remaining function 4 2 4 2 f1x = fx − 1.6P4x = 7x − 6x − 1.6/835x − 30x + 3 2 2 = −6x + 6x − 0.6 = −0.6 = −0.6P0.
You thus obtain the answer fx = 1.6P4x − 0.6P0x shown p. A15.
5. Even function . Pm with odd m contains only odd powers, so it is odd, Pm−x = −Pmx because a sum of odd functions is odd (proof?). Let f be even, f−x = fx. Then the integrand g = fPm in am with odd m is odd because g−x = f−xPm−x = −fxPmx = −gx. Now let p = −x, then dp = −dx and p = −1⋯0 corresponds to x = 1⋯0, (so that 0 0 ∫ gp dp = ∫ g−x −dx −1 1 1 = g−x dx ∫0 1 = −gx dx ∫0 so that the integral of g from −1 to 1 is zero, which makes am = 0.
7. Fourier -Legendre series . The coefficients are given by (7) in the form 2m + 1 1 am = ∫ sin 2πx Pmx dx . 2 −1 For even m these coefficients are zero (why?). For odd m the answer to this CAS experiment will have the coefficients − 0.4775, − 0.6908, 1.844, − 0.8236, ⋯ which decrease rather slowly. Nevertheless, it is interesting to see in the figure that convergence to sin 2πx seems rather rapid. The figure shows the partial sums up to and including P3, P5, P7.
Sec . 5.8. Prob . 7. Partial sums as mentioned Chap. 5 Series Solutions of ODEs. Special Functions 51
15. Fourier -Legendre series . Here you are asked to develop a Bessel function, J0, into a Fourier-Legendre series. The argument is α0,2x, where α0,2 is the second positive zero of J0. This has the effect that J0 is zero at x = 1 and has another zero between 0 and 1 (as well as infinitely many further zeros greater than 1). Your CAS will evaluate 2m + 1 1 am = ∫ J0α0,2x Pmx dx 2 −1 and give you the values (except for a roundoff error) 0.1212, 0, − 0.7950, 0, 0.9595, 0, − 0.3359, 0, 0.0553,⋯.
The figure shows the partial sums of the Fourier-Legendre series of the given fx up to and including P4, P6, P8. Convergence again seems better than you would expect in the light of the slow decrease of the size of the coefficients. This is as in the previous problem.
Sec . 5.8. Prob . 15. Partial sums as mentioned