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.
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 $\displaystyle v(x,y) = -\tfrac{1}{5}\sin (x) \sin (2y)$ then $\displaystyle v_{xx} + v_{yy} = \sin(x)\sin(2y)$.
We will try to find a function $\displaystyle w(x,y)$ so that $\displaystyle u=w+v$ would solve the equation you want to ask. This would require $\displaystyle (w+v)_{xx} + (w+v)_{yy} = \sin(x)\sin(2y) \implies w_{xx}+w_{yy} = 0$. We also want $\displaystyle u(x,0) = \sin(3x)$ therefore $\displaystyle w(x,0)+v(x,0) = \sin(3x) \implies w(x,0) = \sin(3x)$. We also want $\displaystyle u(x,\pi) = 0$ therefore $\displaystyle w(x,\pi) + v(x,\pi) = 0 \implies w(x,\pi) = 0$. We also want $\displaystyle u(0,y) = 0$ therefore $\displaystyle w(0,y) + v(0,y) = 0 \implies w(0,y) = 0$. We also want $\displaystyle u(\pi,y) = \sin(y)$ therefore $\displaystyle w(\pi,y) + v(\pi,y) = \sin (y) \implies w(\pi,y) = \sin (y)$.
Now we can find the solution. Solve $\displaystyle w_{xx}+w_{yy}=0$ subject to the boundary value problem $\displaystyle w(x,0) = \sin(3x),w(x,\pi)=0,w(0,y)=0,w(\pi,y)=\sin(y)$. This is a standard exercise with seperation of variables. One you get the solution $\displaystyle w(x,y)$ then the solution to the original problem would be $\displaystyle w(x,y) - \tfrac{1}{5}\sin (x)\sin (2y)$.
Solve the two problems
$\displaystyle w_{xx} + w_{yy} = 0, $
$\displaystyle w(x,0) = \sin(3x),w(x,\pi)=0,w(0,y)=0,w(\pi,y)=0 $
and
$\displaystyle w_{xx} + w_{yy} = 0, $
$\displaystyle w(x,0) = 0,w(x,\pi)=0,w(0,y)=0,w(\pi,y)=\sin(y) $
and then add the two solutions.