Partial differential equations/Poisson Equation

Poisson's Equation

Definition

${\mathrm{\nabla }}^{2}u=f\Rightarrow \frac{{\partial }^{2}u}{\partial x_1^2}+\frac{{\partial }^{2}u}{\partial x_2^2}+\frac{{\partial }^{2}u}{\partial x_3^2}=f.$

Description

Appears in almost every field of physics.

Solution to Case with 4 Homogeneous Boundary Conditions

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

$\begin{array}{ccc}u(0,y)& =& 0\\ u(L,y)& =& 0\\ u(x,0)& =& 0\\ u(x,M)& =& 0\\ \end{array}$

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: $u_{xx}+u_{yy}=0.$

Step 1: Separate Variables

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

Step 2: Translate Boundary Conditions

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

Step 3: Solve Both SLPs

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

$\begin{array}{ccc}{X}^{″}-\mu X& =& 0\\ X(0)& =& 0\\ X(L)& =& 0\end{array}\}{X}_m(x)=\sinh \frac{m\pi x}{L},m=1,2,3,\cdots$

$\begin{array}{ccc}{Y}^{″}+\mu Y& =& 0\\ Y(0)& =& 0\\ Y(M)& =& 0\end{array}\}{Y}_n(y)=\sin \frac{n\pi y}{M},n=1,2,3,\cdots$

Step 4: Solve Non-homogeneous Equation

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

$\begin{array}{cc}u(x,y)& :=\sum _{m,n=1}^{\mathrm{\infty }}{a}_{mn}{X}_m(x)Y_n(y)\\ & =\sum _{m,n=1}^{\mathrm{\infty }}{a}_{mn}\sinh \frac{m\pi x}{L}\sin \frac{n\pi y}{M}\end{array}$

We substitute this into the Poisson equation and solve:

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

$\begin{array}{cc}{A}_{mn}& =\frac{\underset{0}{\overset{M}{\int }}\underset{0}{\overset{L}{\int }}F(x,y)\sinh \frac{m\pi x}{L}\sin \frac{n\pi y}{M}dxdy}{\underset{0}{\overset{M}{\int }}{\sin }^2\frac{n\pi y}{M}dy\underset{0}{\overset{L}{\int }}{\sinh }^2\frac{m\pi x}{L}dx}\\ & =\frac{8\pi }{LM}\frac{m}{(\sinh (2\pi m)-2\pi m)}\underset{0}{\overset{M}{\int }}\underset{0}{\overset{L}{\int }}F(x,y)\sinh \frac{m\pi x}{L}\sin \frac{n\pi y}{M}dxdy\end{array}$

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

Solution to General Case with 4 Non-homogeneous Boundary Conditions

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

$\begin{array}{cc}u(x,0)& =f_1\\ u(x,M)& =f_2\\ u(0,y)& =f_3\\ u(L,y)& =f_4\end{array}$

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:
   
   $\{\begin{array}{c}{u}_{xx}+u_{yy}=0\\ u(x,0)=f_1\\ u(x,M)=f_2\\ u(0,y)=f_3\\ u(L,y)=f_4\end{array}$
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:
   
   $\{\begin{array}{c}{u}_{xx}+u_{yy}=F(x,y)\\ u(x,0)=0\\ u(x,M)=0\\ u(0,y)=0\\ u(L,y)=0\end{array}$

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 $u_1(x,y)$ and the homogeneous Poisson sub-problem $u_2(x,y)$:
$u(x,y)=u_1(x,y)+u_2(x,y)$