# Partial differential equations/Poisson Equation

## Poisson's Equation

### Definition

${\displaystyle \nabla ^{2}u=f\Rightarrow {\frac {\partial ^{2}u}{\partial x_{1}^{2}}}+{\frac {\partial ^{2}u}{\partial x_{2}^{2}}}+{\frac {\partial ^{2}u}{\partial x_{3}^{2}}}=f~.}$

### Description

Appears in almost every field of physics.

### Solution to Case with 4 Homogeneous Boundary Conditions

Let's consider the following example, where ${\displaystyle u_{xx}+u_{yy}=F(x,y),(x,y)\in \lbrack 0,L\rbrack \times \lbrack 0,M\rbrack ~.}$  and the Dirichlet boundary conditions are as follows:

{\displaystyle {\begin{aligned}u(0,y)&=&0\\u(L,y)&=&0\\u(x,0)&=&0\\u(x,M)&=&0\\\end{aligned}}}

In order to solve this equation, let's consider that the solution to the homogeneous equation will allow us to obtain a system of basis functions that satisfy the given boundary conditions. We start with the Laplace equation: ${\displaystyle u_{xx}+u_{yy}=0~.}$

#### Step 1: Separate Variables

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:
${\displaystyle X''-\mu X=0}$
${\displaystyle Y''+\mu Y=0}$

#### Step 2: Translate Boundary Conditions

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}}}

#### Step 3: Solve Both SLPs

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)=\sin {\frac {(m+1)\pi x}{L}},m=0,1,2,\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+1)\pi y}{M}},n=0,1,2,\cdots }

#### Step 4: Solve Non-homogeneous Equation

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

{\displaystyle {\begin{aligned}u(x,y)&:=\sum _{m,n=0}^{\infty }a_{mn}X_{m}(x)Y_{n}(y)\\&=\sum _{m,n=0}^{\infty }a_{mn}\sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\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=0}^{\infty }\left\{a_{mn}\left\lbrack -{\frac {(m+1)^{2}\pi ^{2}}{L^{2}}}\right\rbrack \sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}\right\}+\left\{a_{mn}\left\lbrack -{\frac {(n+1)^{2}\pi ^{2}}{M^{2}}}\right\rbrack \sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}\right\}\\&=\sum _{m,n=0}^{\infty }\underbrace {\left[-a_{mn}\left({\frac {(m+1)^{2}\pi ^{2}}{L^{2}}}+{\frac {(n+1)^{2}\pi ^{2}}{M^{2}}}\right)\right]} _{A_{mn}}\sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}\end{aligned}}}  {\displaystyle {\begin{aligned}A_{mn}&={\frac {\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}dxdy}{\int \limits _{0}^{M}\sin ^{2}{\frac {(n+1)\pi y}{M}}dy\int \limits _{0}^{L}\sin ^{2}{\frac {(m+1)\pi x}{L}}dx}}\\&={\frac {4}{LM}}\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}dxdy\end{aligned}}}

${\displaystyle a_{mn}=-{\frac {4}{LM\left[{\frac {(m+1)^{2}\pi ^{2}}{L^{2}}}+{\frac {(n+1)^{2}\pi ^{2}}{M^{2}}}\right]}}\int \limits _{0}^{M}\int \limits _{0}^{L}F(x,y)\sin {\frac {(m+1)\pi x}{L}}\sin {\frac {(n+1)\pi y}{M}}dxdy;m,n=0,1,2,\cdots }$

### Solution to General Case with 4 Non-homogeneous Boundary Conditions

Let's consider the following example, where ${\displaystyle u_{xx}+u_{yy}=F(x,y),(x,y)\in \lbrack 0,L\rbrack \times \lbrack 0,M\rbrack ~.}$  and the boundary conditions are as follows:

{\displaystyle {\begin{aligned}u(x,0)&=f_{1}\\u(x,M)&=f_{2}\\u(0,y)&=f_{3}\\u(L,y)&=f_{4}\end{aligned}}}

The boundary conditions can be Dirichlet, Neumann or Robin type.

#### Step 1: Decompose Problem

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:

${\displaystyle {\begin{cases}u_{xx}+u_{yy}=F(x,y)\\u(x,0)=0\\u(x,M)=0\\u(0,y)=0\\u(L,y)=0\end{cases}}}$

#### Step 2: Solve Subproblems

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

#### Step 3: Combine Solutions

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 ${\displaystyle u_{2}(x,y)}$ :
${\displaystyle u(x,y)=u_{1}(x,y)+u_{2}(x,y)}$