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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?

user122965
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2 Answers2

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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.

Gyu Eun Lee
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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.

Hamed
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