Poisson Equation方程求解--转载Poisson Equation

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概述

Poisson Equation

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1 Poisson's Equation
- 1.1 Definition
- 1.2 Description
- 1.3 Solution to Case with 4 Homogeneous Boundary Conditions
  - 1.3.1 Step 1: Separate Variables
  - 1.3.2 Step 2: Translate Boundary Conditions
  - 1.3.3 Step 3: Solve Both SLPs
  - 1.3.4 Step 4: Solve Non-homogeneous Equation
- 1.4 Solution to General Case with 4 Non-homogeneous Boundary Conditions
  - 1.4.1 Step 1: Decompose Problem
  - 1.4.2 Step 2: Solve Subproblems
  - 1.4.3 Step 3: Combine Solutions

Poisson's Equation[edit]

Definition[edit]

$nabla^2 u = f rightarrow frac{partial^2 u}{partial x_1^2} + frac{partial^2 u}{partial x_2^2} + frac{partial^2 u}{partial x_3^2} = f ~.$

Description[edit]

Appears in almost every field of physics.

Solution to Case with 4 Homogeneous Boundary Conditions[edit]

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

$begin{align} u(0,y)&=&0 \ u(l,y)&=&0 \ u(x,0)&=&0 \ u(x,m)&=&0 \end{align}$

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[edit]

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[edit]

As in the solution to the Laplace equation, translation of the boundary conditions yields:
$begin{alignat}{2} x(0) & = & 0\ x(l) & = & 0\ y(0) & = & 0\ y(m) & = & 0end{alignat}$

Step 3: Solve Both SLPs[edit]

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

$left .begin{alignat}{2} x''-mu x & = & 0 \ x(0) & = & 0 \ x(l) & = & 0end{alignat}right } x_m(x) = sin frac{(m+1)pi x}{l},m=0,1,2,cdots$

$left .begin{alignat}{2} y''-mu y & = & 0 \ y(0) & = & 0 \ y(m) & = & 0end{alignat}right } y_n(y) = sin frac{(n+1)pi y}{m},n=0,1,2,cdots$

Step 4: Solve Non-homogeneous Equation[edit]

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

$begin{align} u(x,y) & := sum_{m,n=0}^infty a_{mn}x_m(x)y_n(y) \ & = sum_{m,n=0}^infty a_{mn}sin frac{(m+1)pi x}{l}sin frac{(n+1)pi y}{m}end{align}$

We substitute this into the Poisson equation and solve:

$begin{align}f(x,y)& = u_{xx}+u_{yy} \& = sum_{m,n=0}^infty left { a_{mn}left lbrack -frac{(m+1)^2pi^2}{l^2} right rbrack sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m} right } +left { a_{mn}left lbrack -frac{(n+1)^2pi^2}{m^2} right rbrack sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m} right } \& = sum_{m,n=0}^infty underbrace{left [ -a_{mn} left ( frac{(m+1)^2pi^2}{l^2}+frac{(n+1)^2pi^2}{m^2} right ) right ]}_{a_{mn}} sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m}end{align}$ $begin{align}a_{mn}&=frac{intlimits_0^m int limits_0^l f(x,y) sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m} dx dy}{intlimits_0^m sin^2 frac{(n+1)pi y}{m} dy intlimits_0^l sin^2 frac{(m+1)pi x}{l} dx } \&=frac{4}{lm}intlimits_0^m int limits_0^l f(x,y) sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m} dx dyend{align}$ $a_{mn}=-frac{4}{lmleft [ frac{(m+1)^2pi^2}{l^2} + frac{(n+1)^2pi^2}{m^2} right ]}intlimits_0^m int limits_0^l f(x,y) sin frac{(m+1)pi x}{l} sin frac{(n+1)pi y}{m} dx dy; m,n=0,1,2,cdots$

Solution to General Case with 4 Non-homogeneous Boundary Conditions[edit]

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

$begin{align} u(x,0)&=f_1 \ u(x,m)&=f_2 \ u(0,y)&=f_3 \ u(l,y)&=f_4end{align}$

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

Step 1: Decompose Problem[edit]

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.

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:

$begin{cases}u_{xx}+u_{yy}=0 \u(x,0)=f_1 \u(x,m)=f_2 \u(0,y)=f_3 \u(l,y)=f_4end{cases}$
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:

$begin{cases}u_{xx}+u_{yy}=f(x,y) \u(x,0)=0 \u(x,m)=0 \u(0,y)=0 \u(l,y)=0end{cases}$

Step 2: Solve Subproblems[edit]

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.

Step 3: Combine Solutions[edit]

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 $u_2(x,y)$ :
$u(x,y)=u_1(x,y)+u_2(x,y)$