# How do you prove uniqueness of a partial differntial equation solution with boundry conditions? What approach should be taken for such proofs?

The PDE looks like this

$-$$\Delta$$u$ $=$ $f(x)$ with boundary conditions and $x \in$ $\mathbb R^n$.

(I didn't include boundary conditions because I would like to figure this problem on my own after I received some help understanding what to do in general for PDE uniqueness proofs.)

My ideas:

Suppose $u_1$$(x) and u_2(x) are smooth solutions to the PDE. NTS u_1(x) = u_2(x) <=> u_1(x) - u_2(x) = 0 Afterwards I'm not sure how to continue. Will I need to use the boundary conditions in some way for the proof? ## 2 Answers The question as stated cannot be answered without additional information, such as information about the boundary. Two standard approaches to uniqueness of C^2 solutions to Poisson's equation are the maximum principle and energy estimates, both of which rely on a bounded domain and a reasonably regular boundary. Without information on the boundary of the domain you will need a new method. Moreover, in most situations uniqueness fails rather explicitly without boundary conditions. For example, in the case of the homogeneous equation \Delta u = 0, with no assumptions on the boundary there is always the trivial solution, while for the inhomogeneous equation -\Delta u = f you can always construct a new solution by adding a solution of the homogeneous equation if you don't care about the boundary conditions. In fact, for the homogeneous equation you will have an infinite-dimensional vector space of solutions (say, a subspace of C^2), and for the inhomogeneous equation you can get an infinite-dimensional affine subspace. Finally, even if you specify boundary conditions uniqueness may fail. For example, for the homogeneous equation \Delta u = 0 on the half-space \mathbb{R}_+^n, you will obtain uniqueness only for bounded solutions of the boundary-value problem. (This argument is actually pretty nice: if u is bounded, harmonic, and u=0 on the boundary of the half-space, you can reflect u across the boundary to obtain a bounded harmonic function.) If on an unbounded domain you do not specify behavior at infinity, you may lose uniqueness: for example, for the boundary-value problem$$ \begin{cases} \Delta u = 0 & |x|\geq 1\\ u=0 & |x|=1 \end{cases}$$in$\mathbb{R}^n\setminus B_1(0)$, in dimension$n=2$the functions$u(x) = a\log|x|$,$a$any constant, are solutions. Adding the assumption that$u(x)$is bounded restores the uniqueness for$n=2$. In$n=3$, the functions$u(x) = a(1-|x|^{-1})$are solutions for any constant$a$. So again we lose uniqueness to the BVP; asserting that$u(x)$tends to a limit as$|x|\to\infty$restores the uniqueness. We have two major uniqueness theorems for this type of PDE. 1) The solution to$\Delta u=f$is uniquely determined by its value at the boundary. Hint: Construct a function$w$such that$\Delta w =0$. This is called a Laplace equation. It is well-known that solutions to Laplace equation (called Harmonic functions) do not admit any local minima or maxima, use this to prove. 2) Take the Poisson equation$\Delta u=f$. Then$\nabla u$is uniquely determined on$S$if$\int_{\partial S}\nabla u\cdot da$is known. Hint for proof: Use the identity$\nabla\cdot (g\nabla g) = g\Delta g + (\nabla g)^2$with some nice choice for$g\$.

Remark: In general we do not have nice uniqueness theorems for PDEs like we do for ODEs. The uniqueness, if there is such a thing, is proven case by case.