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Title basically says everything. Prove that if $u\in C^{2}(\mathbb{R}^{n}_{+})\cap C(\bar{\mathbb{R}^{n}_{+}})$ is a bounded solution of the BVP

$$\left\{\begin{array} -\Delta u=0&\text{in}\;\mathbb{R}^{n}_{+}\\ u=g&\;\text{on}\;\partial\mathbb{R}^{n}_{+}, \end{array}\right.$$ then it is unique.

Various tools I have in mind are maximum principle, mean value formulas, Liouville's theorem, "energy" functionals, and Harnack's inequality, uniqueness of Green's function, Hopf's lemma, etc....but in all my scratch work to prove the problem, I keep running into technical difficulties in working with the boundary at infinity.


I arrived at a proof by using a result from Evans exercise #2.5.10 (Schwarz reflection principle):

Proof. Consider the ball $B_{R}(0)$ and suppose $u_{R}$ is a bounded solution to the above problem, but posed on the domain $B_{R}^{+}(0):=\{x:x\in B_{R}(0), x_{n}>0\}.$ Let $v_{R}$ be another bounded solution and define $w_{R}:v_{R}-u_{R}.$ Then $w_{R}=0$ on $\partial B^{+}_{R}(0)\cap\{x_{n}=0\}$, and the Schwarz reflection principle states that the odd extension of $w_{R}$ to $B^{-}_{R}(0)$ is harmonic in all of $B_{R}(0)$ (the proof of this is trivial if we assume $w\in C^{2}(\bar{B^{+}_{R}(0)})$ by using the mean-value formulas, and only a little more difficult under the present assumptions by using Poisson's formula for the ball). Now, $u_{R},v_{R}$ both being bounded implies $w_{R}$ is also bounded. Sending $R\to\infty$, we find that $w:=\lim_{R\to\infty}w_{R}$ is a bounded and harmonic in $\mathbb{R}^{n}$, from which it follows $w_{R}\equiv0$ by Liouville's theorem and the fact that $w=0$ on the hyperplane $x_{n}=0.$

Let $u_{1}$ be a bounded solution to the problem above. Suppose $u_{2}$ is another solution and define $$w:=u_{1}-u_{2}.$$ Then $w=0$ on $\partial\mathbb{R}^{n}_{+}.$

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2 Answers 2

up vote 1 down vote accepted

Added: For the half space $H$ there is a nice direct argument, using the following result from Chapter 2 of Harmonic Function Theory by Sheldon Axler, Paul Bourdon, and Wade Ramey.

Corollary 2.2. Suppose that $u$ is a continuous bounded function on $\bar H$ that is harmonic on $H$. If $u=0$ on $\partial H$, then $u\equiv 0$ on $\bar H$.

Proof: For $x\in \mathbb{R}^{n-1}$ and $y<0$ define $u(x,y)=-u(x,-y)$, thereby extending $u$ to a bounded continuous function defined on all of $\mathbb{R}^n$. Clearly $u$ satisfies the local mean-value property specified in Theorem 1.24, so $u$ is harmonic on $\mathbb{R}^n$. Liouville's Theorem (2.1) now shows that $u$ is constant on $\mathbb{R}^n$.


The uniqueness can be proved using Brownian motion $(B_t)$. Let $D$ be your domain and define $T_D=\inf(t>0: B_t\notin D)$ to be the first time when Brownian motion exits the domain. For $x\in D$, the process $u(B_{t\wedge T_D})$ is a bounded $\mathbb{P}_x$-martingale, so $u(x)=\mathbb{E}_x(u(B_{t\wedge T_D}))$. Since $u$ is bounded and $\mathbb{P}_x(T_D<\infty)$, we have no difficulty in letting $t\to\infty$ and concluding that $$u(x)=\mathbb{E}_x(g(B_{T_D})).$$

A similar argument is given in Proposition 2.6 in Chapter 4 of Brownian motion and classical potential theory by Sidney C. Port and Charles J. Stone. There they approximate $D$ from within by relatively compact regular open sets $D_n$, use uniqueness on each $D_n$, and take limits to obtain uniqueness on $D$.

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Slick...but unfortunately I have not yet studied stochastic processes (next year!), and am therefore not really in a position to understand your proof very well. :[ –  Taylor Martin Feb 1 '13 at 5:52
    
I imagine that it could be turned into a purely analytical argument. But as a probabilist, this way of thinking is very natural to me. –  Byron Schmuland Feb 1 '13 at 5:59
    
@JTian I've added a more analytic proof. Hope it helps! –  Byron Schmuland Feb 1 '13 at 6:28
    
I came up with the strictly analytic proof by using the Schwarz (odd) reflection principle for harmonic functions on a ball (see my edit). But I want to accept your answer since your proof is distinct and provides a helpful alternative. –  Taylor Martin Feb 1 '13 at 6:44
    
I see you that you added a proof which is essentially the same as the one I came up with. Thank you! –  Taylor Martin Feb 1 '13 at 6:44
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Here's a standard technique in proving uniqueness of such boundary value problems. If $u_1$ and $u_2$ are two such solutions, consider $v = u_1 - u_2$. What BVP will it solve? Can you show that $v = 0$? Note that there is an explicit formula for the Laplace equation on the half-space.

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I'm familiar with the technique, but all that I am able to show so far is the obvious fact that $v:=0$ on $\partial\mathbb{R}^{n}_{+}$. On a bounded domain $U$ (assuming it is connected), this would be adequate and the max/min principle shows immediately $v\equiv0$ in $U$. But $\partial\mathbb{R}^{n}_{+}$ is just a hyper-plane. How do you deal with the boundary at infinity? And the representation formula you're referring to is the one corresponding to Green's function for the half-space? Even so, there are unbounded counter-examples not given by this formula. –  Taylor Martin Feb 1 '13 at 2:53
    
Note that I am considering $u_{1}$ and $u_{2}$ to be two bounded solutions. But all I know about $v:=u_{1}-u_{2}$ is that it vanishes on $\partial\mathbb{R}^{n}_{+}$ and is bounded in $\partial\mathbb{R}^{n}_{+}$. I don't know anything else about $u_{1}$ or $u_{2}$, so it's hard to see how you can conclude anything else about $v$ using this difference method. Unless one of the tools I mentioned in the original post can somehow be modified on this unbounded domain to show $v\equiv0$...... –  Taylor Martin Feb 1 '13 at 2:57
    
$v$ is a bounded solution to $\Delta v = 0$, with $u = 0$ on $\partial \mathbb{R}_{+}^n$. By the uniqueness of the Green's function representation formula, then you can conclude that $v \equiv 0$. –  Christopher A. Wong Feb 1 '13 at 10:14
    
I don't disagree that the Green's function representation of a solution with zero boundary data yields a identically zero solution (this is obvious from the definition of $P_{\mathbb{R}^{n}}[0]$). But how do you conclude that $v$ is represented this way? How do you preclude other representation formulas for bounded solutions with zero boundary data? –  Taylor Martin Feb 1 '13 at 21:26
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