# Math Help - non-homogeneous PDE

1. ## non-homogeneous PDE

Can someone help me solve this non-homogeneous PDE (Poisson's equation)?

uxx + uyy = sin(x)sin(2y), 0 < x < pi, 0 < y < pi
with the conditions:
u(x,0) = sin(3x), u(x,pi) = 0, u(0,y) = 0, and u(pi,y) = sin(y)
Any help is greatly appreciated.

2. Originally Posted by PvtBillPilgrim
Can someone help me solve this non-homogeneous PDE (Poisson's equation)?

uxx + uyy = sin(x)sin(2y), 0 < x < pi, 0 < y < pi
with the conditions:
u(x,0) = sin(3x), u(x,pi) = 0, u(0,y) = 0, and u(pi,y) = sin(y)
Any help is greatly appreciated.
Let $v(x,y) = -\tfrac{1}{5}\sin (x) \sin (2y)$ then $v_{xx} + v_{yy} = \sin(x)\sin(2y)$.

We will try to find a function $w(x,y)$ so that $u=w+v$ would solve the equation you want to ask. This would require $(w+v)_{xx} + (w+v)_{yy} = \sin(x)\sin(2y) \implies w_{xx}+w_{yy} = 0$. We also want $u(x,0) = \sin(3x)$ therefore $w(x,0)+v(x,0) = \sin(3x) \implies w(x,0) = \sin(3x)$. We also want $u(x,\pi) = 0$ therefore $w(x,\pi) + v(x,\pi) = 0 \implies w(x,\pi) = 0$. We also want $u(0,y) = 0$ therefore $w(0,y) + v(0,y) = 0 \implies w(0,y) = 0$. We also want $u(\pi,y) = \sin(y)$ therefore $w(\pi,y) + v(\pi,y) = \sin (y) \implies w(\pi,y) = \sin (y)$.

Now we can find the solution. Solve $w_{xx}+w_{yy}=0$ subject to the boundary value problem $w(x,0) = \sin(3x),w(x,\pi)=0,w(0,y)=0,w(\pi,y)=\sin(y)$. This is a standard exercise with seperation of variables. One you get the solution $w(x,y)$ then the solution to the original problem would be $w(x,y) - \tfrac{1}{5}\sin (x)\sin (2y)$.

3. Well, the PDE you give at the end has two nonhomogeneous conditions. So how do you solve it?

4. Originally Posted by PvtBillPilgrim
Well, the PDE you give at the end has two nonhomogeneous conditions. So how do you solve it?
Solve the two problems

$w_{xx} + w_{yy} = 0,$
$w(x,0) = \sin(3x),w(x,\pi)=0,w(0,y)=0,w(\pi,y)=0$

and

$w_{xx} + w_{yy} = 0,$
$w(x,0) = 0,w(x,\pi)=0,w(0,y)=0,w(\pi,y)=\sin(y)$

and then add the two solutions.