2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.