Partial differential equations/Poisson Equation

Consider the solution to the Poisson equation as ${\displaystyle u(x,y)=X(x)Y(y)~.}$ Separating variables as in the solution to the Laplace equation yields:
${\displaystyle X''-\mu X=0}$
${\displaystyle Y''+\mu Y=0}$

As in the solution to the Laplace equation, translation of the boundary conditions yields:
${\displaystyle {\begin{alignedat}{2}X(0)&=&0\\X(L)&=&0\\Y(0)&=&0\\Y(M)&=&0\end{alignedat}}}$

Because all of the boundary conditions are homogeneous, we can solve both SLPs separately.

${\displaystyle \left.{\begin{alignedat}{2}X''-\mu X&=&0\\X(0)&=&0\\X(L)&=&0\end{alignedat}}\right\}X_{m}(x)=\sinh {\frac {m\pi x}{L}},m=1,2,3,\cdots }$

${\displaystyle \left.{\begin{alignedat}{2}Y''+\mu Y&=&0\\Y(0)&=&0\\Y(M)&=&0\end{alignedat}}\right\}Y_{n}(y)=\sin {\frac {n\pi y}{M}},n=1,2,3,\cdots }$

Consider the solution to the non-homogeneous equation as follows:

${\displaystyle {\begin{aligned}u(x,y)&:=\sum _{m,n=1}^{\infty }a_{mn}X_{m}(x)Y_{n}(y)\\&=\sum _{m,n=1}^{\infty }a_{mn}\sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}\end{aligned}}}$

We substitute this into the Poisson equation and solve:

${\displaystyle {\begin{aligned}F(x,y)&=u_{xx}+u_{yy}\\&=\sum _{m,n=1}^{\infty }\left\{a_{mn}\left\lbrack {\frac {m^{2}\pi ^{2}}{L^{2}}}\right\rbrack \sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}\right\}+\left\{a_{mn}\left\lbrack -{\frac {n^{2}\pi ^{2}}{M^{2}}}\right\rbrack \sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}\right\}\\&=\sum _{m,n=1}^{\infty }\underbrace {\left[a_{mn}\left({\frac {m^{2}\pi ^{2}}{L^{2}}}-{\frac {n^{2}\pi ^{2}}{M^{2}}}\right)\right]} _{A_{mn}}\sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}\end{aligned}}}$

${\displaystyle {\begin{aligned}A_{mn}&={\frac {\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}dxdy}{\int \limits _{0}^{M}\sin ^{2}{\frac {n\pi y}{M}}dy\int \limits _{0}^{L}\sinh ^{2}{\frac {m\pi x}{L}}dx}}\\&={\frac {8\pi }{LM}}{\frac {m}{\left(\sinh(2\pi m)-2\pi m\right)}}\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}dxdy\end{aligned}}}$

${\displaystyle a_{mn}={\frac {8\pi }{LM}}{\frac {m}{\left(\sinh(2\pi m)-2\pi m\right)\left[{\frac {(m+1)^{2}\pi ^{2}}{L^{2}}}-{\frac {(n+1)^{2}\pi ^{2}}{M^{2}}}\right]}}\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sinh {\frac {m\pi x}{L}}\sin {\frac {n\pi y}{M}}dxdy;m,n=1,2,3,\cdots }$

For the Poisson equation, we must decompose the problem into 2 sub-problems and use superposition to combine the separate solutions into one complete solution.

1. The first sub-problem is the homogeneous Laplace equation with the non-homogeneous boundary conditions. The individual conditions must retain their type (Dirichlet, Neumann or Robin type) in the sub-problem:
   
   ${\displaystyle {\begin{cases}u_{xx}+u_{yy}=0\\u(x,0)=f_{1}\\u(x,M)=f_{2}\\u(0,y)=f_{3}\\u(L,y)=f_{4}\end{cases}}}$
2. The second sub-problem is the non-homogeneous Poisson equation with all homogeneous boundary conditions. The individual conditions must retain their type (Dirichlet, Neumann or Robin type) in the sub-problem:
   
   ${\displaystyle {\begin{cases}u_{xx}+u_{yy}=F(x,y)\\u(x,0)=0\\u(x,M)=0\\u(0,y)=0\\u(L,y)=0\end{cases}}}$

Depending on how many boundary conditions are non-homogeneous, the Laplace equation problem will have to be subdivided into as many sub-problems. The Poisson sub-problem can be solved just as described above. f(x,y)=x+3*y-2

The complete solution to the Poisson equation is the sum of the solution from the Laplace sub-problem ${\displaystyle u_{1}(x,y)}$ and the homogeneous Poisson sub-problem ${\displaystyle u_{2}(x,y)}$:
${\displaystyle u(x,y)=u_{1}(x,y)+u_{2}(x,y)}$