Fourier Series and Partial Differential Equations Lecture Notes

Chapter 5. Laplace’s equation in the plane 53
Figure 20
In the figure, R is the shadowed region. It has two ‘holes’ and ∂R is the union of three
simple closed curves oriented as shown.
5.3.1 Uniqueness for the Dirichlet problem
We consider uniqueness of solutions to the Dirichlet problem, working in Cartesian coordinates.
Theorem 5.4 Consider the BVP
∂2T ∂x2 + ∂2T = 0 in R, T = f on ∂R, (5.50)
∂y2 where f is a prescribed function and R is a bounded and connected region as in the
statement of Green’s theorem. Then the BVP has at most one solution.
Proof. Let S also be a solution, so that
∂2S ∂x2 + ∂2S = 0 in R, S = f on ∂R. (5.51)
∂y2 Then the difference W := T −S in a solution of the BVP
Consider the identity
W
∂2W ∂x2 + ∂2W = 0 in R, W = 0 on ∂R. (5.52)
∂y2
∂2W ∂x2 + ∂2W ∂y2 2 2 ∂W ∂W
+ + =
∂x ∂y
∂
W
∂x
∂W
+
∂x
∂
W
∂y
∂W
. (5.53)
∂y
Integrate both sides over R and appeal to Laplace’s equation and Green’s theorem to find
that
∂W 2
2
∂W
∂
+ dxdy = W
R ∂x ∂y
R ∂x
∂W
+
∂x
∂
W
∂y
∂W
dxdy,
∂y
= −W ∂W ∂W
dx+W
∂y ∂x dy
. (5.54)
∂R
Since W = 0 on ∂R the line integral must vanish and so
∂W 2
2
∂W
+ dxdy = 0. (5.55)
∂x ∂y
R
This is possible only if ∂W/∂x = 0, ∂W/∂y = 0 in R. Hence W is constant and since
W = 0 on ∂R the constant can only equal zero. Hence T = S and the solution is
unique.