I had read Poisson equation theory . To understand that theory I wanted to solve some basic Poisson equation .
So I constructed following problem
$u_{xx}+u_{yy}=1$
$u(0,y)=u(x,0)=u(x,1)=u(1,y)=0$.
I had studied green representation formula. But for this simple looking region say unit square. How to solve this PDE?
Any Help/Hint will be useful.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[10px,#ffd]{\mrm{u}_{xx} + \mrm{u}_{yy} = 1\,,\quad \mrm{u}\pars{0,y} =\mrm{u}\pars{x,0} =\mrm{u}\pars{x,1} = \mrm{u}\pars{1,y} = 0}: {\Large ?}}$
$\ds{\mrm{u}\pars{x,y} \equiv \sum_{n = 1}^{\infty}a_{n}\pars{x}\sin\pars{n\pi y}}$ already satisfies the boundary condition $\ds{\mrm{u}\pars{x,0} = \mrm{u}\pars{x,1} = 0}$. In addittion, $\ds{\mrm{u}\pars{x,y}}$ satisfies the above differential equation. Namely, $$ \sum_{n = 1}^{\infty}\bracks{a_{n}''\pars{x} - \pars{n\pi}^{2}a_{n}\pars{x}}\sin\pars{n\pi y} = 1 $$ Multiply both members by $\ds{\sin\pars{n\pi y}}$ and integrate over $\ds{y \in \pars{0,1}}$: \begin{align} &{1 \over 2}\bracks{a_{n}''\pars{x} - \pars{n\pi}^{2}a_{n}\pars{x}} = {1 - \pars{-1}^{n} \over n\pi} \\[2mm] \implies & a_{n}\pars{x} = -2\,{1 - \pars{-1}^{n} \over \pars{n\pi}^{3}} + b_{n}\sinh\pars{n\pi\bracks{{1 \over 2} - x}} + c_{n}\cosh\pars{n\pi\bracks{{1 \over 2} - x}} \end{align}
The general solution is reduced to \begin{align} \mrm{u}\pars{x,y} & = -2\sum_{n = 1}^{\infty}{1 - \pars{-1}^{n} \over \pars{n\pi}^{3}}\,\sin\pars{n\pi y} \\[2mm] & + \sum_{n = 1}^{\infty}\braces{b_{n}\sinh\pars{n\pi\bracks{{1 \over 2} - x}} + c_{n}\cosh\pars{n\pi\bracks{{1 \over 2} - x}}}\sin\pars{n\pi y} \end{align}
\begin{align} &\mrm{u}\pars{0,y} = 0 \implies \\[2mm] & -2\sum_{n = 1}^{\infty}{1 - \pars{-1}^{n} \over \pars{n\pi}^{3}}\,\sin\pars{n\pi y} + \sum_{n = 1}^{\infty}\bracks{b_{n}\sinh\pars{n\pi \over 2} + c_{n}\cosh\pars{n\pi \over 2}}\sin\pars{n\pi y} = 0 \\[5mm] & \mrm{u}\pars{1,y} = 0 \implies \\[2mm] & -2\sum_{n = 1}^{\infty}{1 - \pars{-1}^{n} \over \pars{n\pi}^{3}}\,\sin\pars{n\pi y} + \sum_{n = 1}^{\infty}\bracks{-b_{n}\sinh\pars{n\pi \over 2} + c_{n}\cosh\pars{n\pi \over 2}}\sin\pars{n\pi y} = 0 \end{align} Adding and subtracting both members of this equations yields $\ds{b_{n} = 0}$ and $$ c_{n} = 2\,{1 - \pars{-1}^{n} \over \pars{n\pi}^{3}} \mrm{sech}\pars{n\pi \over 2} $$
Finally, \begin{align} \mrm{u}\pars{x,y} & = \bbox[10px,#ffd]{-\,{4 \over \pi^{3}} \sum_{n = 0}^{\infty}{\sin\pars{\bracks{2n + 1}\pi y} \over \pars{2n + 1}^{3}}} \\[2mm] & \bbox[10px,#ffd]{+ {4 \over \pi^{3}}\sum_{n = 0}^{\infty} {\mrm{sech}\pars{\bracks{n + 1/2}\pi} \over \pars{2n + 1}^{3}} \cosh\pars{\bracks{n + {1 \over 2}}\pi\bracks{1 - 2x}} \sin\pars{\bracks{2n + 1}\pi y}} \end{align}