This is a Complex Analysis problem Suppose u1 and u2 are har

This is a Complex Analysis problem:

Suppose u_1 and u_2 are harmonic in a bounded domain D and continuous on the closure of D. Suppose u_1 = u_2 on the boundary of D. Use the maximum principle for harmonic functions to show that u_1 = u_2 in D. (This is uniqueness of solutions of the Dirichlet problem.)

Solution

It’s easy to show that there is only one function that can satisfy Poisson’s equation with given Dirichlet boundary data. Suppose that both u1 and u2 are C 2 and satisfy Poisson’s equation with forcing function f and boundary data h. Then the function v = u2 u1 satisfies deltav = 0 in D and v = 0 on D.

We will show that v = 0 in D, so u1 = u2 in D. Take the equation 4v = 0 and multiply both sides by v to obtain v 4v = 0.

By Green’s first identity

Z D v 4 v dV = Z D v v n dA Z D |v| 2 dV.

But since v = 0 on D and since 4v = 0 this becomes Z D |v| 2 dV = 0. (4) But |v| 2 is clearly non-negative on D; if the integral is zero then we must have v 0 on D, so that v is constant. But since v = 0 on D we conclude that v 0 throughout D. Thus u1 = u2 in D. Now consider the same situation but with Neumann data. The same computations which led to equation (4) still work, and we conclude that v = u2 u1 is constant. But since we now have only v n = 0 on D, rather than v = 0, we can’t conclude that v 0 on D, only that v = c, so that u2 = u1 + c. And indeed, it’s easy to see that if 4u1 = f on D with u1 n = g then u2 = u1 + c satisfies the same conditions for any choice of c. So when dealing with Neumann boundary conditions we obtain uniqueness only up to an additive constant. We can obtain uniqueness by specifying one additional condition, e.g., the value of the solution u at a given point in D, or, for example, the condition that

Z D u(x) dx = 0.

Thus, for example, if two solutions with the same Neumann data differ by a constant, say u2 = u1 + c, and both satisfy the additional condition (5) we can immediately deduce that c = 0, so u1 u2.