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

ISBN 0-471-31999-6

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By an argument similar to the one above we conclude that both sides of the last equation must be equal to the same constant 5. This leads to the two equations r2R" + rR' + (Or2 - 5)R = 0 and 0" + 50 = 0. Since the

circular membrane is continuous, we must have 0 (2n) = 0(0), which requires 5 = p2, p. a non-negative integer. The condition 0(2n) = 0(0) is also known as the periodicity condition. Since we also desire solutions which vary periodically in time, it is clear that the separation constant O should be positive,

O = X2. Thus we arrive at the three equations r2R" + rR' + (X2r2 - p2)R = 0, 0" + p20 = 0, and T" + X2a2T = 0.

230

Section 10.8

Section 10.8, Page 611

1a. Assuming that u(x,y) = X(x)Y(y) leads to the two O.D.E. X" - OX = 0, Y" + GY = 0. The B.C. u(0,y) = 0, u(a,y) = 0 imply that X(0) = 0 and X(a) = 0. Thus nontrivial solutions to X" - GX = 0 which satisfy these boundary conditions are possible only if O = -(nn/a) 2, n = 1,2...; the corresponding solutions for X(x) are proportional to sin(nnx/a). The B.C. u(x,0) = 0 implies that Y(0) = 0. Solving Y" - (nn/a) 2Y = 0 subject to this condition we find that Y(y) must be proportional to sinh(nny/a). The fundamental solutions are then un(x,y) = sin(nnx/a)sinh(nny/a), n = 1,2,..., which satisfy Laplace's equation and the homogeneous B.C. We

1b. Substituting for g(x) in the equation for cn we have

so cn = [4a sin(nn/2)]/[n2n 2sinh(nnb/a)]. Substituting these values for cn in the above series yields the desired solution.

assume that u(x,y) = ^ cnsin(nnx/a)sinh(nny/a), where

n=1

the coefficients cn are determined from the B.C.

assume

u(x,b) = g(x) = ^ cnsin(nnx/a)sinh(nnb/a). It follows

n=1

1c.

Section 10.8

231

1d.

2. In solving the D.E. Y" - X2Y = 0, one normally writes Y(y) = c1sinhXy + c2coshXy. However, since we have Y(b) = 0, it is advantageous to rewrite Y as Y(y) = d1sinhX(b-y) + d2coshX(b-y), where d1, d2 are also arbitrary constants and can be related to c1, c2 using the appropriate hyperbolic trigonometric identities. The important thing, however, is that the second form also satisfies the D.E. and thus Y(y) = d1sinhX(b-y) satisfies the D.E. and the homogeneous B.C. Y(b) = 0. The rest of the problem follows the pattern of Problem 1.

3a. Let u(x,y) = v(x,y) + w(x,y), where u, v and w all

satisfy Laplace's Eq., v(x,y) satisfies the conditions in Eq. (4) and w(x,y) satisfies the conditions of Problem 2.

4. Following the pattern of Problem 3, one could consider

adding the solutions of four problems, each with only one non-homogeneous B.C. It is also possible to consider adding the solutions of only two problems, each with only two non-homogeneous B.C., as long as they involve the same variable. For instance, one such problem would be uxx + uyy = 0, u(x,0) = 0, u(x,b) = 0, u(0,y) = k(y),

u(a,y) = f(y), which has the fundamental solutions un(x,y) = [cnsinh(nnx/b) + dncosh(nnx/b)]sin(nny/b).

Assuming u(x,y) = un(x,y) and using the B.C.

n=1

u(0,y) = k(y) we obtain dn = (2/b) k(y)sin(nny/b)dy.

J0

Using the B.C. u(a,y) = f(y) we obtain

cnsinh(nna/b) + dncosh(nna/b) = (2/b)Ibf(y)sin(nny/b)dy, which

0

can be solved for cn, since dn is already known. The second problem, in this approach, would be uxx + uyy = 0, u(x,0) =

h(x), u(x,b) = g(x), u(0,y) = 0 and u(a,y) = 0. This has the

fundamental solutions

232

Section 10.8

un(x,y) = [ansinh(nny/a) + bncosh(nny/a)]sin(nnx/a, so that u(x,y) = un(x,y). Thus u(x,0) = h(x) gives

n=1

bn = (2/an ah(x)sin(nnx/a)dx and u(x,b) = g(x) gives Jo

ansinh(nnb/a) + bncosh(nnb/a) = (2/a) ag(x)sin(nnx/a)dx, which

Jo

can be solved for an since bn is known.

5. Using Eq.(20) and following the same arguments as

presented in the text, we find that R(r) = k1rn + k2r-n and 0(8) = c1sinn0 + c2cosn0, for n a positive integer, and u0(r,8) = 1 for n = 0. Since we require that u(r,8) be bounded as r ^ ^, we conclude that k1 = 0. The fundamental solutions are therefore un(r,8) = r-ncosn8, vn(r,8) = r-nsinn8, n = 1,2,... together with u0(r,8) = 1. Assuming that u can be expressed as a linear combination of the fundamental solutions we have

u(r,8) = c0/2 + r-n(cncosn8 + knsinn8). The B.C.

n=1

requires that

u(a,8) = c0/2 + a-n(cncosn8 + knsinn8) = f(8) for

n=1

0 < 8 < 2n. This is precisely the Fourier series representation for f(8) of period 2n and thus

a-ncn = (1/n) Pf(8)cosn8d8, n = 0,1,2,... and

0

a-nkn = (1/n) [2f(8)sinn8d8, n = 1,2... .

0

7. Again we let u(r,8) = R(r)0(8) and thus we have

r2R" + rR' - OR = 0 and 0" + O0 = 0, with R(0) bounded and

the B.C. 0(0) = 0(a) = 0. For o < 0 we find that

0(0) = 0, so we let O = X2 (X2 real) and thus

0(8) = c^osX8 + c2sinX8. The B.C. 0(0) = 0 ^ ci = 0 and

the B.C. 0(a) = 0 ^ X = nn/a, n = 1,2,... .

Substituting these values into Eq.(31) we obtain R(r) = k1rnn/a + k2r-nn/a. However k2 = 0 since R(0) must be bounded, and thus the fundamental solutions are un(r,8) = rnn/asin(nn8/a). The desired solution may now be formed using previously discussed procedures.

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