Showing the solution is unique or not for Poisson problem Consider the following problem:
$$\nabla^2u=u+\exp(xyz)$$
in the sphere $x^2+y^2+z^2<1$ (call this $V$) and
$$\frac{\partial u}{\partial n}+u=5$$
on the surface of $S$ of the volume $V$.
Question: How do we show the solution $u(x,y,z)$ is unique or not?
My attempt: The above is in general a Poisson problem $\nabla^2u=f(x,y,z)$ in some three-dimensional domain $D$ with surface $S$. The boundary condition in general takes the form of $\partial u/\partial n=g(x,y,z)$ which is a Neumann condition.
Suppose that there are two such solutions, say $u_1$ and $u_2$. We have $$\nabla^2u_1=u_1+\exp(xyz),\quad\nabla^2u_2=u_2+\exp(xyz)\;\;\text{in}\;\;V.$$
Similarly, we have
$$\frac{\partial u_1}{\partial n}+u_1=5,\quad \frac{\partial u_2}{\partial n}+u_2=5\;\;\text{on}\;\;S.$$
We use Green's first identity with $u=v=w$, that is
$$\int_V\nabla w\cdot\nabla w+w\nabla^2w\;dV=\int_S w\frac{\partial w}{\partial n}\;dA,$$
where we have define $w=u_1-u_2$ so that 
$$\nabla^2w=w\;\text{in}\;V\;\text{and}\;\frac{\partial w}{\partial n}=-w\;\text{on}\;S.$$
Thus Green's first identity gives:
$$\int_V\nabla w\cdot\nabla w+w^2\;dV=\int_S -w^2\;dA.\qquad(1)$$
The aim is to show that $w=0$ but I am not sure how to continue from $(1)$.
I'd appreciate any help or hint. Thank you.
 A: Given the expression
$\displaystyle \int_V\nabla w\cdot\nabla w+w^2\;dV=\int_S -w^2\;dA, \tag 1$
observe the left-hand side is non-negative whilst the right is non-positive; the only way this can bind is with
$w = 0. \tag 2$
Note Added in Edit, Wednesday 1 May 2019 1:05 PM PST:  Here we show how a Green identity gives rise to (1).
Green's first identity for two functions $u$ and $v$ defined on the region $V$ is
$\displaystyle \int_V (\nabla u \cdot \nabla v + u\nabla^2 v) \; dV = \int_{\partial V} u \nabla v \cdot \vec{dS}, \tag 3$
where $dV$ is the volume element on $V$ and $\vec{dS}$ is the (normal) area element on $\partial V$; this follows from the well-known differential identity
$\nabla \cdot (u \nabla v) = \nabla u \cdot \nabla v + u \nabla^2 v, \tag 4$
which integrated over $V$ yields
$\displaystyle \int_V (\nabla u \cdot \nabla v + u \nabla^2 v) \; dV = \int_V \nabla \cdot (u \nabla v) \; dV = \int_{\partial V} u \nabla v \cdot \vec{dS}, \tag 5$
where we have used the divergence theorem in the second equality; with
$u = v = w \tag 6$
we obtain
$\displaystyle \int_V (\nabla w \cdot \nabla w + w \nabla^2 w) \; dV = \int_V \nabla \cdot (w \nabla w) \; dV = \int_{\partial V} w \nabla w \cdot \vec{dS}; \tag 7$
now with
$\nabla^2 w = w \tag 8$
and
$\dfrac{\partial w}{\partial \vec n} = -w \; \text{on} \; \partial V = S \tag 9$
as in the text of the question itself, we have
$\displaystyle \int_{\partial V} w\nabla w \cdot \vec{dS} = \int_{\partial V} -w\dfrac{\partial w}{\partial \vec n}dA = \int_{\partial V}-w^2dA; \tag{10}$
combining this with (7) we obtain
$\displaystyle \int_V (\nabla w \cdot \nabla w + w \nabla^2 w) \; dV = \int_{\partial V}-w^2dA, \tag{11}$
which is (1).  End of Note.
