Em um conjunto aberto
U
⊂
R
n
{\displaystyle U\subset \mathbb {R} ^{n}}
[ 1]
Δ
φ
=
f
{\displaystyle \Delta \varphi =f}
onde,
f
:
U
→
R
{\displaystyle f:U\to \mathbb {R} }
Δ
{\displaystyle \Delta }
Δ
φ
:=
∑
i
=
1
n
∂
2
φ
∂
x
i
2
{\displaystyle \Delta \varphi :=\sum _{i=1}^{n}{\frac {\partial ^{2}\varphi }{\partial x_{i}^{2}}}}
Aqui, a incógnita
φ
{\displaystyle \varphi }
U
⊂
R
n
{\displaystyle U\subset \mathbb {R} ^{n}}
R
.
{\displaystyle \mathbb {R} .}
∇
2
{\displaystyle \nabla ^{2}}
Δ
=
∇
⋅
∇
{\displaystyle \Delta =\nabla \cdot \nabla }
∇
{\displaystyle \nabla }
gradiente . Quando
f
≡
0
{\displaystyle f\equiv 0}
equação de Laplace .
Em duas dimensões, i.e. no espaço euclidiano
R
2
{\displaystyle \mathbb {R} ^{2}}
[ 2] coordenadas cartesianas ):
∂
2
φ
∂
x
2
+
∂
2
φ
∂
y
2
=
f
(
x
,
y
)
{\displaystyle {\frac {\partial ^{2}\varphi }{\partial x^{2}}}+{\frac {\partial ^{2}\varphi }{\partial y^{2}}}=f(x,y)}
Em coordenadas polares
(
r
,
θ
)
{\displaystyle (r,~\theta )}
∂
2
g
∂
r
2
+
1
r
∂
g
∂
r
+
1
r
2
∂
2
g
∂
θ
2
=
h
(
r
,
θ
)
{\displaystyle {\frac {\partial ^{2}g}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial g}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}g}{\partial \theta ^{2}}}=h(r,\theta )}
Para obter esta equação faz-se as mudanças de variáveis
x
=
r
cos
θ
{\displaystyle x=r{\text{cos }}\theta }
y
=
r
sen
θ
{\displaystyle y=r{\text{sen }}\theta }
g
(
r
,
θ
)
=
φ
(
r
cos
θ
,
r
sen
θ
)
{\displaystyle g(r,\theta )=\varphi (r{\text{cos }}\theta ,r{\text{sen }}\theta )}
h
(
r
,
θ
)
=
ϕ
(
r
cos
θ
,
r
sen
θ
)
{\displaystyle h(r,\theta )=\phi (r{\text{cos }}\theta ,~r{\text{sen }}\theta )}
Em três dimensões, i.e. no espaço euclidiano
R
3
{\displaystyle \mathbb {R} ^{3}}
∂
2
φ
∂
x
2
+
∂
2
φ
∂
y
2
+
∂
2
φ
∂
z
2
=
f
(
x
,
y
,
z
)
{\displaystyle {\partial ^{2}\varphi \over \partial x^{2}}+{\partial ^{2}\varphi \over \partial y^{2}}+{\partial ^{2}\varphi \over \partial z^{2}}=f(x,y,z)}
Em coordenadas cilíndricas
(
ρ
,
θ
,
z
)
{\displaystyle (\rho ,~\theta ,~z)}
1
ρ
∂
∂
ρ
(
ρ
∂
g
∂
ρ
)
+
1
ρ
2
∂
2
g
∂
θ
2
+
∂
2
g
∂
z
2
=
h
(
ρ
,
θ
,
z
)
{\displaystyle {1 \over \rho }{\partial \over \partial \rho }\left(\rho {\partial g \over \partial \rho }\right)+{1 \over \rho ^{2}}{\partial ^{2}g \over \partial \theta ^{2}}+{\partial ^{2}g \over \partial z^{2}}=h(\rho ,\theta ,z)}
Pode-se obter esta fazendo as mudanças de variáveis
x
=
r
cos
θ
{\displaystyle x=r{\text{cos }}\theta }
y
=
r
sen
θ
{\displaystyle y=r{\text{sen }}\theta }
z
=
z
{\displaystyle z=z}
g
(
r
,
θ
,
z
)
=
φ
(
r
cos
θ
,
r
sen
θ
,
z
)
{\displaystyle g(r,\theta ,z)=\varphi (r{\text{cos }}\theta ,r{\text{sen }}\theta ,z)}
h
(
r
,
θ
,
z
)
=
ϕ
(
r
cos
θ
,
r
sen
θ
,
z
)
{\displaystyle h(r,\theta ,z)=\phi (r{\text{cos }}\theta ,~r{\text{sen }}\theta ,z)}
Em coordenadas esféricas
(
r
,
ϕ
,
θ
)
{\displaystyle (r,\phi ,\theta )}
1
r
2
∂
∂
r
(
r
2
∂
g
∂
r
)
+
1
r
2
sin
ϕ
∂
∂
ϕ
(
sin
ϕ
∂
g
∂
ϕ
)
+
1
r
2
sin
2
ϕ
∂
2
g
∂
θ
2
=
h
(
r
,
ϕ
,
θ
)
{\displaystyle {1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial g \over \partial r}\right)+{1 \over r^{2}\sin \phi }{\partial \over \partial \phi }\left(\sin \phi {\partial g \over \partial \phi }\right)+{1 \over r^{2}\sin ^{2}\phi }{\partial ^{2}g \over \partial \theta ^{2}}=h(r,\phi ,\theta )}
Para resolver uma equação de Poisson podem-se utilizar vários métodos como, por exemplo, uma função de Green ou métodos numéricos como o método das diferenças finitas (MDF), o método dos elementos finitos (MEF) ou o Element Free-Gallerkin Method (EFGM).
Pode-se obter uma solução clássica para a equação de Poisson em
R
n
{\displaystyle \mathbb {R} ^{n}}
−
Δ
φ
=
g
{\displaystyle -{\Delta }\varphi =g}
supondo
g
∈
C
c
2
(
R
n
,
R
)
{\displaystyle g\in C_{c}^{2}(\mathbb {R} ^{n},\mathbb {R} )}
g
{\displaystyle g}
continuamente diferenciável com suporte compacto . Neste caso, a solução é dada por:
φ
(
x
)
=
∫
R
n
Φ
(
x
−
y
)
g
(
y
)
d
y
{\displaystyle \varphi (x)=\int _{\mathbb {R} ^{n}}\Phi (x-y)g(y)dy}
onde,
Φ
:
R
n
−
0
→
R
{\displaystyle \Phi :\mathbb {R} ^{n}-{0}\to \mathbb {R} }
equação de Laplace .[ 1]
Mostraremos, primeiro, que
φ
∈
C
2
(
R
n
,
R
)
.
{\displaystyle \varphi \in C^{2}(\mathbb {R} ^{n},\mathbb {R} ).}
φ
(
x
)
=
∫
R
n
Φ
(
y
)
g
(
x
−
y
)
d
y
{\displaystyle \varphi (x)=\int _{\mathbb {R} ^{n}}\Phi (y)g(x-y)dy}
Como
g
∈
C
c
2
(
R
n
,
R
)
{\displaystyle g\in C_{c}^{2}(\mathbb {R} ^{n},\mathbb {R} )}
∂
φ
∂
x
i
(
x
)
=
lim
h
→
0
φ
(
x
+
h
⋅
e
i
)
−
φ
(
x
)
h
=
∫
R
n
Φ
(
y
)
[
lim
h
→
0
g
(
x
+
h
e
i
−
y
)
−
g
(
x
−
y
)
h
]
d
y
=
∫
R
n
Φ
(
y
)
∂
g
∂
x
i
(
x
−
y
)
d
y
{\displaystyle {\frac {\partial \varphi }{\partial x_{i}}}(x)=\lim _{h\to 0}{\frac {\varphi (x+h\cdot e_{i})-\varphi (x)}{h}}=\int _{\mathbb {R} ^{n}}\Phi (y)\left[\lim _{h\to 0}{\frac {g(x+he_{i}-y)-g(x-y)}{h}}\right]dy=\int _{\mathbb {R} ^{n}}\Phi (y){\frac {\partial g}{\partial x_{i}}}(x-y)dy}
e, de forma análoga, temos
∂
2
φ
∂
x
i
2
=
∫
R
n
Φ
(
y
)
∂
g
2
∂
x
i
2
(
x
−
y
)
d
y
{\displaystyle {\frac {\partial ^{2}\varphi }{\partial x_{i}^{2}}}=\int _{\mathbb {R} ^{n}}\Phi (y){\frac {\partial g^{2}}{\partial x_{i}^{2}}}(x-y)dy}
o que mostra que
φ
∈
C
2
(
R
n
,
R
)
.
{\displaystyle \varphi \in C^{2}(\mathbb {R} ^{n},\mathbb {R} ).}
e
i
{\displaystyle e_{i}}
i
{\displaystyle i}
base canônica do
R
n
{\displaystyle \mathbb {R} ^{n}}
Mostraremos, agora,
−
Δ
φ
=
g
{\displaystyle -\Delta \varphi =g}
Φ
{\displaystyle \Phi }
singularidade em
0
{\displaystyle 0}
ε
>
0
{\displaystyle \varepsilon >0}
[ 1]
(1)\quad
Δ
φ
(
x
)
=
∫
B
(
0
,
ε
)
Φ
(
y
)
Δ
x
g
(
x
−
y
)
d
y
+
∫
R
n
−
B
(
x
,
ε
)
Φ
(
y
)
Δ
x
g
(
x
−
y
)
d
y
{\displaystyle \Delta \varphi (x)=\int _{B(0,\varepsilon )}\Phi (y)\Delta _{x}g(x-y)dy+\int _{\mathbb {R} ^{n}-B(x,\varepsilon )}\Phi (y)\Delta _{x}g(x-y)dy}
Aqui,
B
(
0
,
ε
)
{\displaystyle B(0,\varepsilon )}
bola de centro
0
{\displaystyle 0}
ε
{\displaystyle \varepsilon }
(2)\quad
|
∫
B
(
0
,
ϵ
)
Φ
(
y
)
Δ
x
g
(
x
−
y
)
d
y
|
≤
{
C
|
|
D
2
g
|
|
∞
ε
2
|
ln
ε
|
,
n
=
2
C
|
|
D
2
g
|
|
∞
ε
2
,
n
≥
3
{\displaystyle \left|\int _{B(0,\epsilon )}\Phi (y)\Delta _{x}g(x-y)dy\right|\leq \left\{{\begin{array}{ll}C||D^{2}g||_{\infty }\varepsilon ^{2}|\ln \varepsilon |&,n=2\\C||D^{2}g||_{\infty }\varepsilon ^{2}&,n\geq 3\end{array}}\right.}
Aqui,
|
|
⋅
|
|
∞
{\displaystyle ||\cdot ||_{\infty }}
norma
L
∞
(
R
n
)
{\displaystyle L_{\infty }(\mathbb {R} ^{n})}
integrado por partes , o que nos fornece:
(3)\quad
∫
R
n
−
B
(
x
,
ϵ
)
Φ
(
y
)
Δ
x
g
(
x
−
y
)
d
y
=
−
∫
R
n
−
B
(
0
,
ε
)
D
Φ
(
y
)
⋅
D
y
g
(
x
−
y
)
d
y
+
∫
∂
B
(
0
,
ε
)
Φ
(
y
)
∂
g
∂
ν
(
x
−
y
)
d
S
(
y
)
{\displaystyle \int _{\mathbb {R} ^{n}-B(x,\epsilon )}\Phi (y)\Delta _{x}g(x-y)dy=-\int _{\mathbb {R} ^{n}-B(0,\varepsilon )}D\Phi (y)\cdot D_{y}g(x-y)dy+\int _{\partial B(0,\varepsilon )}\Phi (y){\frac {\partial g}{\partial \nu }}(x-y)dS(y)}
Aqui,
∂
g
∂
ν
{\displaystyle {\frac {\partial g}{\partial \nu }}}
derivada normal de
g
{\displaystyle g}
(4)\quad
|
∫
∂
B
(
0
,
ε
)
Φ
(
y
)
∂
g
∂
ν
(
x
−
y
)
d
S
(
y
)
|
≤
{
C
|
|
D
g
|
|
∞
ε
|
ln
ε
|
,
n
=
2
C
|
|
D
g
|
|
∞
ε
,
n
≥
3
{\displaystyle \left|\int _{\partial B(0,\varepsilon )}\Phi (y){\frac {\partial g}{\partial \nu }}(x-y)dS(y)\right|\leq \left\{{\begin{array}{ll}C||Dg||_{\infty }\varepsilon |\ln \varepsilon |&,n=2\\C||Dg||_{\infty }\varepsilon &,n\geq 3\end{array}}\right.}
Se integrarmos por partes o penúltimo termo de (3) novamente, vemos que:
(5)\quad
−
∫
R
n
−
B
(
0
,
ε
)
D
Φ
(
y
)
⋅
D
y
g
(
x
−
y
)
d
y
=
∫
R
n
−
B
(
0
,
ε
)
Δ
Φ
(
y
)
g
(
x
−
y
)
d
y
−
∫
∂
B
(
0
,
ϵ
)
∂
Φ
∂
ν
(
y
)
g
(
x
−
y
)
d
S
(
y
)
{\displaystyle -\int _{\mathbb {R} ^{n}-B(0,\varepsilon )}D\Phi (y)\cdot D_{y}g(x-y)dy=\int _{\mathbb {R} ^{n}-B(0,\varepsilon )}\Delta \Phi (y)g(x-y)dy-\int _{\partial B(0,\epsilon )}{\frac {\partial \Phi }{\partial \nu }}(y)g(x-y)dS(y)}
Aqui, o penúltimo termo é nulo, pois
Δ
Φ
≡
0
{\displaystyle \Delta \Phi \equiv 0}
R
n
−
B
(
0
,
ε
)
{\displaystyle \mathbb {R} ^{n}-B(0,\varepsilon )}
(6)\quad
∫
∂
B
(
0
,
ϵ
)
∂
Φ
∂
ν
(
y
)
g
(
x
−
y
)
d
S
(
y
)
=
1
n
α
(
n
)
ε
n
−
1
∫
∂
B
(
x
,
ε
)
g
(
y
)
d
S
(
y
)
→
g
(
x
)
quando
ε
→
0
{\displaystyle \int _{\partial B(0,\epsilon )}{\frac {\partial \Phi }{\partial \nu }}(y)g(x-y)dS(y)={\frac {1}{n\alpha (n)\varepsilon ^{n-1}}}\int _{\partial B(x,\varepsilon )}g(y)dS(y)\to g(x)\quad {\text{quando}}\quad \varepsilon \to 0}
pois, notemos que o termo a direita deste símbolo de igualdade é a média de
g
{\displaystyle g}
B
(
x
,
ε
)
{\displaystyle B(x,\varepsilon )}
−
Δ
φ
=
g
{\displaystyle -\Delta \varphi =g}
A equação de Poisson em domínios limitados deve ser complementada com condições de contorno .
editar
Diz que a equação de Poisson tem condições de contorno de Dirichlet quando a função incógnita
φ
{\displaystyle \varphi }
△
φ
=
f
,
x
∈
D
φ
=
g
,
x
∈
∂
D
.
{\displaystyle {\begin{array}{rclcl}\triangle \varphi &=&f,\quad &x\in D\\\varphi &=&g,&x\in \partial D\end{array}}.}
Como consequência do princípio do máximo forte para funções harmônicas , mostra-se que se
D
{\displaystyle D}
φ
∈
C
2
(
D
)
∩
C
(
D
¯
)
{\displaystyle \varphi \in C^{2}(D)\cap C({\bar {D}})}
g
∈
C
(
∂
D
)
{\displaystyle g\in C(\partial D)}
[ 1]
A unicidade de solução também é garantida mesmo que
D
{\displaystyle D}
D
{\displaystyle D}
∂
D
∈
C
1
{\displaystyle \partial D\in C^{1}}
φ
,
φ
~
∈
C
2
(
U
)
{\displaystyle \varphi ,~{\tilde {\varphi }}\in C^{2}(U)}
u
=
φ
−
φ
~
{\displaystyle u=\varphi -{\tilde {\varphi }}}
△
u
=
0
,
x
∈
D
u
=
0
,
x
∈
∂
D
.
{\displaystyle {\begin{array}{rclcl}\triangle u&=&0,\quad &x\in D\\u&=&0,&x\in \partial D\end{array}}.}
Agora, usando de integração por partes, obtemos:
∫
D
‖
D
u
‖
2
d
x
=
−
∫
D
u
Δ
u
d
x
=
0
{\displaystyle \int _{D}\|D\,u\|^{2}\,dx=-\int _{D}u\Delta u\,dx=0}
o que implica que
D
u
=
0
{\displaystyle D\,u=0}
u
{\displaystyle u}
u
=
0
{\displaystyle u=0}
∂
D
{\displaystyle \partial D}
u
=
0
{\displaystyle u=0}
D
¯
{\displaystyle {\bar {D}}}
φ
=
φ
~
{\displaystyle \varphi ={\tilde {\varphi }}}
editar
Diz que a equação de Poisson tem condições de contorno de Neumann quando a derivada normal da função incógnita
φ
{\displaystyle \varphi }
△
φ
=
f
,
x
∈
D
∂
∂
η
φ
=
g
,
x
∈
∂
D
.
{\displaystyle {\begin{array}{rclcl}\triangle \varphi &=&f,\quad &x\in D\\{\frac {\partial }{\partial \eta }}\varphi &=&g,&x\in \partial D\end{array}}.}
↑ a b c d Evans, Lawrence C. (2010). Partial Differential Equations 2 ed. [S.l.]: American Mathematical Society. ISBN 978-0821849743 ↑ Figueiredo, Djairo (1987). Análise de Fourier e equações diferenciais parciais 2 ed. [S.l.]: IMPA. ISBN 8524400269 
![{\displaystyle {\frac {\partial \varphi }{\partial x_{i}}}(x)=\lim _{h\to 0}{\frac {\varphi (x+h\cdot e_{i})-\varphi (x)}{h}}=\int _{\mathbb {R} ^{n}}\Phi (y)\left[\lim _{h\to 0}{\frac {g(x+he_{i}-y)-g(x-y)}{h}}\right]dy=\int _{\mathbb {R} ^{n}}\Phi (y){\frac {\partial g}{\partial x_{i}}}(x-y)dy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41aa93cccdaade5c35f3c470a7cdffdee709c887)
