Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$$
my try:  I know this is  screened Poisson equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
I  only find this poisson equation one of the solution
$$u(x,y)=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
because 
$$\dfrac{\partial u}{\partial x}=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(-\dfrac{2x}{a^2}\right)$$
$$\dfrac{\partial^2 u}{\partial x^2}=c\cdot\dfrac{b^2}{a^2+b^2}$$
and 
$$\dfrac{\partial^2 u}{\partial y^2}=c\cdot\dfrac{a^2}{a^2+b^2}$$
so
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c\cdot\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=c$$
and when
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Longrightarrow u=0$$
and this solution is uniqueness? and How prove it? Thank you
Thank you very much!
 A: You found one solution $u_0$ of the problem. Then for any solution $u$ of the problem the function $v:=u-u_0$ would satisfy
$$\Delta v=0,\qquad v=0 \quad {\rm on}\quad \partial \Omega,$$
where $$\Omega:=\left\{(x,y)\Biggm| {x^2\over a^2}+{y^2\over b^2}\leq1\right\}$$
is an elliptical domain. Now a harmonic function which vanishes on the boundary of a  compact domain has to vanish identically therein, as it cannot have a local extremum in the interior. So $v=0$, or $u=u_0$, which shows that your solution is unique.
A: Hint. Try
$$
u(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1.
$$
Find a solution, and then show uniqueness.
A: The shape that your boundary conditions describe is in fact an ellipse. So if you switch to an elliptic coordinate system your pde should separate out nicely.
A: Let
$$v(x,y)=u(x,y)+\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
Then $\Delta v=0$ in $\mathbb R^2$ and $v(x,y)=0$ on the boundary of the ellipse $U=\big\{\frac{x^2}{a^2}+\frac{y^2}{b^2}<1\big\}$. Thus, as the Dirichlet problem
$$
\Delta v=0 \,\,\text{in $U$} \quad\text{and} \quad  v=0 \,\,\text{on $\partial U$},
$$
has unique solution $v\equiv 0$, then $v$ vanishes in $U$ as well. But $v$ is a harmonic function in $\mathbb R^2$, and harmonic functions are real analytic,  and thus it is expressible as a power series around $(0,0)$
$$
v(x,y)=\sum_{m,n=0}^\infty \frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}\frac{x^my^n}{m!n!}=0,
$$
since $\frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}=0$, for all $m,n$.
