solving this pde with weird B.C.s

Question

$$ \nabla^2 u = 1\;\;\;;\;\;\; 0<x<\pi\,,\,0<y<\pi\\ u_y(x,0)=u(x,\pi) = u(y,0)=u(y,\pi) = 0 $$

I think this is some version of the poisson's equation. I am not particularly sure on how to solve with its boundary conditions.

Answer 1

I presume you mean $\nabla^2 u= 1$. Let $v= u- x^2/2$. Then $\nabla^2 v= \nabla^2 u- 1= 1- 1= 0$.

I would now try "separating variables"- let $v= X(x)Y(y)$. $\nabla^2 u= Y\frac{d^2X}{dx^2}+ X\frac{d^2Y}{dy^2}= 0$. $Y\frac{d^2X}{dx^2}= -Y\frac{d^2Y}{dy^2}$. $\frac{1}{X}\frac{d^2X}{dx^2}= -\frac{1}{Y}\frac{d^2Y}{dy^2}$. Since the left side depends only on x and the right side only on x, in order to be equal for all x and y they must be the same constant. We have $\frac{1}{X}\frac{d^2X}{dx^2}= \alpha$ so $\frac{d^2X}{dx^2}= \alpha X$ and $\frac{1}{Y}\frac{d^2Y}{dy^2}= -\alpha$ so $\frac{d^2Y}{dy^2}= -\alpha Y$.

The conditions are $X(x)Y'(0)= X(\pi)Y(y)= X(x)Y(\pi)= X(0)Y(y)$.

Are you sure those are correct? X(x)Y'(0), and $X(x)Y(\pi)$ are functions of x only while X(0)Y(y) and $X(\pi)Y(0)$ are functions of y only. In order that those are the same for all x and y, they must be constants. But if $X(x)Y(\pi)$ is a constant for all x, so is X(x). Similarly, if $X(\pi)Y(y)$ is a constant then so is Y(y). The only way I can see out of that bind is if $Y'(0)= 0$, $X(\pi)= 0$, $Y(\pi)= 0$, and $X(0)= 0$.

But in that case the solution is v(x, y)= 0 for all x and y so that $u(x, y)= x^2/2$.