# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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X = (nn/b)2, n = 1, 2,...; (10)

and Y(y) is proportional to the corresponding eigenfunction sin(nny/b). Next, we substitute from Eq. (10) for X in Eq. (6), and solve this equation subject to the boundary condition (8). It is convenient to write the general solution of Eq. (6) as

X(x) = kx cosh(nnx/b) + k2 sinh(nnx/b), (11)

and the boundary condition (8) then requires that k1 = 0. Therefore X(x) must be proportional to sinh(nnx/b). Thus we obtain the fundamental solutions

nnx nny

u„ (x, y) = sinh ——sin —-—, n = 1, 2,.... (12)

n b b

These functions satisfy the differential equation (1) and all the homogeneous boundary conditions for each value of n.

To satisfy the remaining nonhomogeneous boundary condition at x = a we assume, as usual, that we can represent the solution u(x, y) in the form

CO CO

nn x nny

u (x > y) = 2^ cnun(x > y) = 2^ cn sinh~ sin~ • (13)

n=1 n=1

The coefficients cn are determined by the boundary condition

nn a nny

u(a, y) = 2_^ cn sinh b sin = f (y)• (!4)

n = 1

Therefore the quantities cn sinh(nn a/b) must be the coefficients in the Fourier sine series of period 2b for f and are given by

nn a 2 rb nny

cn sinh ~b~ = b J0 f (y) sin ~b~ dy. (!5)

Thus the solution of the partial differential equation (1) satisfying the boundary conditions (4) is given by Eq. (13) with the coefficients cn computed from Eq. (15).

From Eqs. (13) and (15) we see that the solution contains the factor sinh(nnx/b)/sinh(nna/b). To estimate this quantity for large n we can use the approximation sinh f = ef/2, and thereby obtain

sinh (nn x/b) ^ 2exp(nn x/b)

= exp[—nn(a — x )/b].

sinh(nn a/b) 2 exp (nn a/b)

Thus this factor has the character of a negative exponential; consequently, the series (13) converges quite rapidly unless a — x is very small.

608

Chapter 10. Partial Differential Equations and Fourier Series

EXAMPLE

1

To illustrate these results, let a = 3, b = 2, and

f(y) = I2—y, 0 i y i 1: (16)

By evaluating cn from Eq. (15) we find that

8 sin(nn/2)

c„ = —z—n----------------• (17)

n n sinh(3nn/2)

Then u(x, y) is given by Eq. (13). Keeping 20 terms in the series we can plot u versus x and y, as shown in Figure 10.8.2. Alternatively, one can construct a contour plot showing level curves of u (x, y); Figure 10.8.3 is such a plot, with an increment of 0.1 between adjacent curves.

FIGURE 10.8.2 Plot of u versus x and y for Example 1.

FIGURE 10.8.3 Level curves of u(x, y) for Example 1.

10.8 Laplace’s Equation

Dirichlet Problem for a Circle. Consider the problem of solving Laplace’s equation in a circular region r < a subject to the boundary condition

u(a,0) = f (0), (18)

where f is a given function on 0 < d < 2n (see Figure 10.8.4). In polar coordinates Laplace’s equation takes the form

urr + 1 ur + 1 u 00 = 0 (19)

To complete the statement of the problem we note that for u(r,0) to be single-valued, it is necessary that u be periodic in 0 with period 2n. Moreover, we state explicitly that u(r,0) must be bounded for r < a, since this will become important later.

To apply the method of separation of variables to this problem we assume that

u(r, 0) = R(r)©(0), (20)

and substitute for u in the differential equation (19). This yields

R"© + 1 R'© + 1 R@” = 0,

r r2

or

2 R" R' &"

r + r— =-----------------= X, (21)

R R © v '

where X is the separation constant. Thus we obtain the two ordinary differential equations

r2R" + rR' - XR = 0, (22)

©" + X© = 0. (23)

In this problem there are no homogeneous boundary conditions; recall, however, that solutions must be bounded and also periodic in 0 with period 2n. It is possible to show

(Problem 9) that the periodicity condition requires that X must be real. We will consider

in turn the cases in which X is negative, zero, and positive.

FIGURE 10.8.4 Dirichlet problem for a circle.

610

Chapter 10. Partial Differential Equations and Fourier Series

If X < 0, let X = —j2, where j > 0. Then Eq. (23) becomes ©" — j2© = 0, and consequently

Thus ©(9) can be periodic only if c1 = c2 = 0, and we conclude that X cannot be negative.

If X = 0, then Eq. (23) becomes ©" = 0, and thus

For ©(9) to be periodic we must have c2 = 0, so that ©(9) is a constant. Further, for X = 0, Eq. (22) becomes

The logarithmic term cannot be accepted if u(r, 9) is to remain bounded as r ^ 0;hence k2 = 0. Thus, corresponding to X = 0, we conclude that u(r,9) must be a constant, that is, proportional to the solution

Finally, if X > 0, we let X = j2 where j > 0. Then Eqs. (22) and (23) become

In order that © be periodic with period 2n it is necessary that i be a positive integer n. With i = n it follows that the solution r-i in Eq. (31) must be discarded since it becomes unbounded as r ^ 0. Consequently, k2 = 0 and the appropriate solutions of Eq. (19) are

un (r,9) = rn cos n9, vn (r,9) = rn sin n9, n = 1, 2,.... (33)

These functions, together with u0(r,9) = 1, form a set of fundamental solutions for the present problem.

In the usual way we now assume that u can be expressed as a linear combination of the fundamental solutions; that is,

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