Show uniqueness in Dirichlet problem in unit disk

Consider the solution of the Dirichlet problem in the unit disk, i.e. solving Laplace equation there with some known function on the boundary. The solution, obtained via separation of variables, can be expressed as $$u(r,\theta)=\frac{a_0}{2}+\sum_{n=1}^\infty r^n(a_n\cos{n\theta}+b_n\sin{n\theta}).$$ I am now trying to show this solution is unique. To that end, I'm trying to show that if the known function at the boundary is identically ##0##, i.e. ##u(1,\theta)\equiv0##, then the only solution must be ##0##. The author of the text merely refers to a section of the book that contains the following two results:

Theorem 4.3 Suppose that ##f## is piecewise continuous and that all its Fourier coefficients are ##0##. Then ##f(t)=0## at all points where ##f## is continuous.

Corollary 4.1 If two continuous functions ##f## and ##g## have the same Fourier coefficients, then ##f=g##.

I'm confused over how or if these results apply here. If ##u(1,\theta)\equiv0##, then $$0=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos{n\theta}+b_n\sin{n\theta}.$$ Does this mean the Fourier series converges pointwise to ##0## for all ##\theta##? If so, I don't think I can use any of the theorems above. I am stuck.

I believe I'm overthinking this, or thinking too fast!

The idea is to expand the function ##u(1,\theta)=g(\theta)\equiv 0## into a Fourier series. Clearly all its coefficients are ##0##, i.e. ##a_n=b_n=0## for all ##n##. The theorem assures us that $$u(r,\theta)\equiv0,$$ is the only solution. Why? Well, suppose there was another non-zero (continuous) solution, then it too would have all its coefficients equal to ##0## by the boundary condition. But the theorem would force it to be equal to ##0## identically. Thus ##u(r,\theta)\equiv0## is the only solution to the boundary condition ##u(1,\theta)\equiv0##.

For the second part, suppose there are two solutions ##u,v## to the problem with boundary condition ##u(1,\theta)=v(1,\theta)=g(\theta)##. Then we apply the result we just proved to ##w(r,\theta)=u(r,\theta)-v(r,\theta)## to get ##w(r,\theta)\equiv0##, i.e. ##u(r,\theta)=v(r,\theta)##.

IIRC, uniqueness is obtained through the Maximum priciple, i.e., extrema are reached only in the boundary. Existence , I believe, is harder, though it depends on your perspective, interests.

You can also do it this way. If you have two solutions, then their difference will be a solution with zero boundary conditions. Multiply the equation by ##u## and integrate

##\int u\Delta u=0##

Use one of those integration formulas (basicaly integration by parts) to get

##\int \nabla u\cdot \nabla u =0##

This implies that the function has zero gradient, thus is constant, and from the boundary condition it follows that it is zero.

martinbn said:

You can also do it this way. If you have two solutions, then their difference will be a solution with zero boundary conditions. Multiply the equation by ##u## and integrate

##\int u\Delta u=0##

Use one of those integration formulas (basicaly integration by parts) to get

##\int \nabla u\cdot \nabla u =0##

This implies that the function has zero gradient, thus is constant, and from the boundary condition it follows that it is zero.

This argument is preferable, since it also covers solutions (if any) which cannot be obtained by separation of variables in polar coordinates.

Well, Max, min, will be reached at the boundary, forcing the function to be identically zero.

WWGD said:

Well, Max, min, will be reached at the boundary, forcing the function to be identically zero.

Yes, but you need to prove it or take it for granted.