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644 CHAPTER 18 The Potential Equation

u = f 2

u = g 1 D u = g 2

u = f 1

0 0 u = f 2 0
u = g 2
0 D 0 0 D D u = g 1 D
0 0
0
0 0 0
u = f 1
4
FIGURE 18.2 u(x, y) = u j (x, y).
j=1
With this choice of coefficients, the solution can be written
∞ ∞
2 sinh(nπy/L)
u(x, y) = f (ξ)sin(nπξ/L)dξ sin(nπx/L) .
L 0 sinh(nπ K/L)
n=1
EXAMPLE 18.1
Suppose, in the problem just solved, L = K = π and f (x) = x(π − x). Compute the numbers
2 π 4
n
x(π − ξ)sin(nξ)dξ = (−1) .
π 0 πn 3
The solution is
∞
4 sinh(ny)
n
u(x, y) = (−1) sin(nx) .
πn 3 sinh(nπ)
n=1
If nonzero boundary data is prescribed on all four sides of D, define four Dirichlet problems
in each of which the data is nonzero on just one side (Figure 18.2). Each of these problems can
be solved by the separation of variables. The sum of the solutions of these four problems is the
solution of the original problem.

SECTION 18.2 PROBLEMS

In each of Problems 1 through 5, solve the Dirichlet 3. u(0, y) = u(1, y) = 0for0 ≤ y ≤ 4and
problem for the given rectangle and boundary conditions. u(x,4) = x cos(πx/2),u(x,0) = 0for0 ≤ x ≤ 1
4. u(0, y) = sin(y),u(π, y) = 0for 0 ≤ y ≤ π and
1. u(0, y) = u(1, y) = 0for0 ≤ y ≤ π and
u(x,0) = x(π − x),u(x,π) = 0for 0 ≤ x ≤ π
u(x,π) = 0,u(x,0) = sin(πx) for 0 ≤ x ≤ 1.
2. u(0, y) = y(2 − y),u(3, y) = 0for0 ≤ y ≤ 2and 5. u(0, y) = 0,u(2, y) = sin(y) for 0 ≤ y ≤ π and
u(x,0) = u(x,2) = 0for0 ≤ x ≤ 3 u(x,0) = 0,u(x,π) = x sin(πx) for 0 ≤ x ≤ 2

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October 14, 2010 15:27 THM/NEIL Page-644 27410_18_ch18_p641-666