Can Two Solutions of a PDE be Equal with Given Boundary Conditions?

[Euge]

I'm stepping in for Ackbach temporarily until he returns. Here is this week's POTW:

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Suppose $D$ is a compact domain of $\Bbb R^3,$ $F : D \to \Bbb R^3$ is continuous, and $\phi_1, \phi_2 : D \to \Bbb R$ are $C^2$-solutions of the PDE $\nabla^2 \phi = F$. If $\dfrac{\partial \phi_1}{\partial n} = \dfrac{\partial \phi_2}{\partial n}$ on the boundary $\partial D$ and $\phi_1(x_0) = \phi_2(x_0)$ for some $x_0\in \partial D,$ show that $\phi_1 = \phi_2$.

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[Euge]

Hi MHB community,

There was a misprint in the original statement. The hypothesis added is $\phi_1(x_0)= \phi_2(x_0)$ for some $x_0\in \partial D$. I'm giving an extra week for users to submit solutions.

 

[Euge]

No one answered this week’s problem. You can read my solution below.

Spoiler

Let $\phi := \phi_1 - \phi_2$. Then $\phi$ is a harmonic function on $D$, and the divergence theorem gives

$$\iint_{\partial D} \phi \frac{\partial \phi}{\partial n}\, dS = \iiint_D \operatorname{div}(\phi\nabla \phi)\, dV = \iiint_D (\lvert \nabla \phi\rvert^2 + \phi \Delta \phi)\, dV = \iiint_D \lvert \nabla \phi\rvert^2\, dV$$

As $\phi = 0$ on $\partial D$, the surface integral above is zero. Hence $\iiint_D \lvert \nabla \phi\rvert^2\, dV = 0$. Since $\lvert \nabla \phi\rvert^2$ is nonnegative and continuous on $D$ it follows that $\nabla \phi = 0$. Connectedness of $D$ implies $\phi$ is constant. By hypothesis $\phi_1(x_0)=\phi_2(x_0)$, whence $\phi \equiv 0$. Thus $\phi_1 \equiv \phi_2$.