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We have $\Delta u=f$ in $D$, and $\dfrac{\partial u}{\partial n}+au=h$ on boundary of D, where $D$ is a domain in three dimension and $a$ is a positive constant. $\dfrac{\partial u}{\partial n}=\triangledown u\cdot n$ ($n$ is normal vector).

My thoughts: Suppose there are $u_1$ and $u_2$, satisfis the above equations. Let $w=u_1-u_2$, then we have $\Delta w=0$ in $D$, and $\dfrac{\partial w}{\partial n}=-aw$ on boundary of D. Maximum modulus principle may be useful but I don't know where to put it in. And energy method seems not helpful in this question.

Any help would be appreciated!

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$\newcommand{div}{\nabla \cdot} \newcommand{grad}{\nabla}$ Start by taking $w$ as in your question so that $\Delta w = 0$ in $D$ and $\frac{\partial w}{\partial n} = -aw$ on $\partial D$ as you note.

Also note the following identity for the divergence operator for a scalar field $\phi$ and a vector field $\bf F$ $$\div (\phi {\bf F}) = \phi \div {\bf F} + \grad \phi \cdot {\bf F}$$

Then consider $$\iiint_D \div(w \grad w) dV = \iiint_D w \Delta w + (\grad w)^2 dV = \iiint_D (\grad w)^2 dV \geq 0$$ since $\Delta w = 0$ in $D$.

But also by the divergence theorem $$\iiint_D \div(w \grad w) dV = \iint_{\partial D} w \grad w \cdot {\bf dS} = \iint_{\partial D} w\grad w \cdot {\bf n} ds = \iint_{\partial D} -aw^2 ds \leq 0$$

Hence $$\iiint_D (\grad w)^2 dV = 0 \implies \grad w = 0 \mbox{ in } D$$ This gives us that $w$ is a constant so that solutions are unique up to a constant.

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  • $\begingroup$ Can we prove the uniqueness in a stronger sense such that $w=0$? By the second equation we have $w=0$ on the boundary. However, we can't use maximum principle since $D$ might not be bounded and even so, the function may not be able to be extended continuously to its closure. Is it right? $\endgroup$ – jack Jan 11 '16 at 20:52
  • $\begingroup$ To clarify, if $w=0$ on the boundary and $\nabla w = 0$ inside, then it immediately follows (assuming $w$ is differentiable) that $w=0$ everywhere. $\endgroup$ – user76284 May 7 '16 at 17:20

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