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!

Hint. Try
$$
u(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1.
$$
Find a solution, and then show uniqueness.
