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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.
Download (direct link): elementarydifferentialequations2001.pdf
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where X is the separation constant. Thus we obtain the two ordinary differential equations
X" - XX = 0, (6)
Y" + XY = 0. (7)
FIGURE 10.8.1 Dirichlet problem for a rectangle.
10.8 Laplace’s Equation
607
If we now substitute for u from Eq. (5) in each of the homogeneous boundary conditions, we find that
X (0) = 0 (8)
and
Y (0) = 0, Y (b) = 0. (9)
We will first determine the solution of the differential equation (7) subject to the boundary conditions (9). However, this problem is essentially identical to one encountered previously in Sections 10.1, 10.5, and 10.7. We conclude that there are nontrivial solutions if and only if X is an eigenvalue, namely,
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) = cosh(nn x/b) + k2 sinh(nn x/b), (11)
and the boundary condition (8) then requires that kx = 0. Therefore X(x) must be proportional to sinh (nn x/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
OO OO
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 ~b~ = f (ó). (!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 fb nny
cn sinh = by f (Ó) sin dy. (15)
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 ? = e?/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) = (y’ 0 - y - 1
fW )2 — y, 1 - y - 2.
By evaluating cn from Eq. (15) we find that
8 sin (nn/2)
n2n 2 sinh(3nn/2)
(16)
(17)
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.
cn =
FIGURE 10.8.3 Level curves of u(x, y) for Example 1
10.8 Laplace’s Equation
609
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,6) = f (â), (18)
where f is a given function on 0 < â < 2n (see Figure 10.8.4). In polar coordinates Laplace’s equation takes the form
urr + 1 ur + ~2 uââ = 0. (19)
To complete the statement of the problem we note that for u(r, â) to be single-valued,
it is necessary that u be periodic in â with period 2n. Moreover, we state explicitly that u(r, â) must be bounded for r < a, since this will become important later.
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