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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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To apply the method of separation of variables to this problem we assume that
and substitute for u in the differential equation (19). This yields
or
r-------+ r— =---------= X, (21)
where X is the separation constant. Thus we obtain the two ordinary differential equations
r2R" + rR' - XR = 0, (22)
In this problem there are no homogeneous boundary conditions; recall, however, that solutions must be bounded and also periodic in â 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 ê < 0, let ê = — j2, where j > 0. Then Eq. (23) becomes 0" — j2© = 0, and consequently
0(6) = c1ej6 + c2e—j6. (24)
Thus 0(6) can be periodic only if c1 = c2 = 0, and we conclude that ê cannot be
negative.
If ê = 0, then Eq. (23) becomes 0" = 0, and thus
0(6) = c1 + c2Q. (25)
For 0(6) to be periodic we must have c2 = 0, so that 0(6) is a constant. Further, for ê = 0, Eq. (22) becomes
r2 R" + rR' = 0. (26)
This equation is of the Euler type, and has the solution
R(r) = k1 + k2 ln r. (27)
The logarithmic term cannot be accepted if u(r, 6) is to remain bounded as r ^ 0;hence
k2 = 0. Thus, corresponding to ê = 0, we conclude that u(r,6) must be a constant, that is, proportional to the solution
u^(r,e) = 1. (28)
Finally, if ê > 0, we let ê = /j2 where ë > 0. Then Eqs. (22) and (23) become
r2 R" + rR' — j2 R = 0 (29)
and
0" + j2© = 0, (30)
respectively. Equation (29) is an Euler equation and has the solution
R(r) = k1r j + k2r —j, (31)
while Eq. (30) has the solution
0(0) = c1 sin j6 + c2cos j6. (32)
In order that 0 be periodic with period 2n it is necessary that j be a positive integer n. With j = n it follows that the solution r—j 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, 6) = rn cos n6, vn (r, 6) = rn sinn6, n = 1, 2,.... (33)
These functions, together with u0(r, 6) = 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,
OO
t(r, 6) = -0 + ^2rn(cn cosn6 + kn sinn6). (34)
2 n = 1
10.8 Laplace’s Equation
611
The boundary condition (18) then requires that
c “
è(à,â) = -2 + ^an(cn cosne + kn sinne) = f (â) (35)
2 n=1 n n
for 0 < â < 2n. The function f may be extended outside this interval so that it is
periodic with period 2n, and therefore has a Fourier series of the form (35). Since
the extended function has period 2n, we may compute its Fourier coefficients by integrating over any period of the function. In particular, it is convenient to use the original interval (0, 2n); then
1 f2n
anc = — I f (â) cos ne dâ, n = 0, 1, 2,...; (36)
n J0
1 f2ÿ
ank = —1 f (â) sinïª dâ, n = 1,2,.... (37)
Ï J0
With this choice of the coefficients, Eq. (34) represents the solution of the boundary value problem of Eqs. (18) and (19). Note that in this problem we needed both sine and cosine terms in the solution. This is because the boundary data were given on
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