## Partial Differential Equation Question

Question:
Consider the problem

$\displaystyle \displaystyle u_{xx} + u_{yy} = y \ 0 \leq x \leq L \ \ 0 \leq y \leq H$

with boundary conditions $\displaystyle u(x,0) = 1, u(x, H) = 0, u(0,y) = 0$ and $\displaystyle \displaystyle \frac{\partial u}{\partial x} (L,y) = 0$

This problem is best solved by making the substitution $\displaystyle u(x,y) = v(y) + w(x,y)$ and choosing $\displaystyle v(y)$ so that $\displaystyle w(x,y)$ satisfies a homogenous equation.

Find an explicit expression for $\displaystyle v(y)$ and state the precise boundary conditions and equation which $\displaystyle w(x,y)$ must satisfy. Do not find an explicit expression for $\displaystyle w$

Let $\displaystyle u(x,y) = v(y) + w(x,y)$

$\displaystyle \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 v}{\partial y^2} = y$

Choose $\displaystyle \frac{\partial^2 v}{\partial y^2} = y$
Then

$\displaystyle V' = \frac{1}{2} y^2 + A$ and $\displaystyle V = \frac{1}{6}y^3 + Ay + B$

$\displaystyle V(0) = 1$ so $\displaystyle B = 1$

$\displaystyle V(H) = 0$ so $\displaystyle \frac{1}{6} H^3 + AH + 1 = 0$ and therefore $\displaystyle A = - \frac{1}{H} - \frac{1}{6} H^2$

$\displaystyle V = \frac{1}{6}y^3 - \frac{y}{H} - \frac{1}{6}H^2 y + 1$
and then

$\displaystyle u(x,0) = v(0) + w(x,0) = 1$ so $\displaystyle w(x,0) = 0$

$\displaystyle u(x,H) = v(H) + w(x,H) = 0$ so $\displaystyle w(x,H) = 0$