∇
2
u
=
f
⇒
∂
2
u
∂
x
1
2
+
∂
2
u
∂
x
2
2
+
∂
2
u
∂
x
3
2
=
f
.
{\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~.}
Appears in almost every field of physics.
Solution to Case with 4 Homogeneous Boundary Conditions
edit
Let's consider the following example, where
u
x
x
+
u
y
y
=
F
(
x
,
y
)
,
(
x
,
y
)
∈
[
0
,
L
]
×
[
0
,
M
]
.
{\displaystyle u_{xx}+u_{yy}=F(x,y),(x,y)\in \lbrack 0,L\rbrack \times \lbrack 0,M\rbrack ~.}
Dirichlet boundary conditions are as follows:
u
(
0
,
y
)
=
0
u
(
L
,
y
)
=
0
u
(
x
,
0
)
=
0
u
(
x
,
M
)
=
0
{\displaystyle {\begin{aligned}u(0,y)&=&0\\u(L,y)&=&0\\u(x,0)&=&0\\u(x,M)&=&0\\\end{aligned}}}
u
x
x
+
u
y
y
=
0
.
{\displaystyle 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
)
.
{\displaystyle u(x,y)=X(x)Y(y)~.}
Laplace equation yields:
X
″
−
μ
X
=
0
{\displaystyle X''-\mu X=0}
Y
″
+
μ
Y
=
0
{\displaystyle Y''+\mu Y=0}
Step 2: Translate Boundary Conditions
edit
As in the solution to the Laplace equation, translation of the boundary conditions yields:
X
(
0
)
=
0
X
(
L
)
=
0
Y
(
0
)
=
0
Y
(
M
)
=
0
{\displaystyle {\begin{alignedat}{2}X(0)&=&0\\X(L)&=&0\\Y(0)&=&0\\Y(M)&=&0\end{alignedat}}}
Step 3: Solve Both SLPs
edit
Because all of the boundary conditions are homogeneous, we can solve both SLPs separately.
X
″
−
μ
X
=
0
X
(
0
)
=
0
X
(
L
)
=
0
}
X
m
(
x
)
=
sinh
m
π
x
L
,
m
=
1
,
2
,
3
,
⋯
{\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 }
Y
″
+
μ
Y
=
0
Y
(
0
)
=
0
Y
(
M
)
=
0
}
Y
n
(
y
)
=
sin
n
π
y
M
,
n
=
1
,
2
,
3
,
⋯
{\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 }
Step 4: Solve Non-homogeneous Equation
edit
Consider the solution to the non-homogeneous equation as follows:
u
(
x
,
y
)
:=
∑
m
,
n
=
1
∞
a
m
n
X
m
(
x
)
Y
n
(
y
)
=
∑
m
,
n
=
1
∞
a
m
n
sinh
m
π
x
L
sin
n
π
y
M
{\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}}}
F
(
x
,
y
)
=
u
x
x
+
u
y
y
=
∑
m
,
n
=
1
∞
{
a
m
n
[
m
2
π
2
L
2
]
sinh
m
π
x
L
sin
n
π
y
M
}
+
{
a
m
n
[
−
n
2
π
2
M
2
]
sinh
m
π
x
L
sin
n
π
y
M
}
=
∑
m
,
n
=
1
∞
[
a
m
n
(
m
2
π
2
L
2
−
n
2
π
2
M
2
)
]
⏟
A
m
n
sinh
m
π
x
L
sin
n
π
y
M
{\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}}}
A
m
n
=
∫
0
M
∫
0
L
F
(
x
,
y
)
sinh
m
π
x
L
sin
n
π
y
M
d
x
d
y
∫
0
M
sin
2
n
π
y
M
d
y
∫
0
L
sinh
2
m
π
x
L
d
x
=
8
π
L
M
m
(
sinh
(
2
π
m
)
−
2
π
m
)
∫
0
M
∫
0
L
F
(
x
,
y
)
sinh
m
π
x
L
sin
n
π
y
M
d
x
d
y
{\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}}}
a
m
n
=
8
π
L
M
m
(
sinh
(
2
π
m
)
−
2
π
m
)
[
(
m
+
1
)
2
π
2
L
2
−
(
n
+
1
)
2
π
2
M
2
]
∫
0
M
∫
0
L
F
(
x
,
y
)
sinh
m
π
x
L
sin
n
π
y
M
d
x
d
y
;
m
,
n
=
1
,
2
,
3
,
⋯
{\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 }
Solution to General Case with 4 Non-homogeneous Boundary Conditions
edit
Let's consider the following example, where
u
x
x
+
u
y
y
=
F
(
x
,
y
)
,
(
x
,
y
)
∈
[
0
,
L
]
×
[
0
,
M
]
.
{\displaystyle u_{xx}+u_{yy}=F(x,y),(x,y)\in \lbrack 0,L\rbrack \times \lbrack 0,M\rbrack ~.}
u
(
x
,
0
)
=
f
1
u
(
x
,
M
)
=
f
2
u
(
0
,
y
)
=
f
3
u
(
L
,
y
)
=
f
4
{\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}}}
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:
{
u
x
x
+
u
y
y
=
0
u
(
x
,
0
)
=
f
1
u
(
x
,
M
)
=
f
2
u
(
0
,
y
)
=
f
3
u
(
L
,
y
)
=
f
4
{\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}}}
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:
{
u
x
x
+
u
y
y
=
F
(
x
,
y
)
u
(
x
,
0
)
=
0
u
(
x
,
M
)
=
0
u
(
0
,
y
)
=
0
u
(
L
,
y
)
=
0
{\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
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.
f(x,y)=x+3*y-2
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
)
{\displaystyle u_{1}(x,y)}
u
2
(
x
,
y
)
{\displaystyle u_{2}(x,y)}
u
(
x
,
y
)
=
u
1
(
x
,
y
)
+
u
2
(
x
,
y
)
{\displaystyle u(x,y)=u_{1}(x,y)+u_{2}(x,y)}
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e480e226689cd9ff8755d8ef0ad9e805106b43d)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ec9d5779c72d9cd651736e508fe675e94bfc2d0)
