Let
U
⊆
C
{\displaystyle U\subseteq \mathbb {C} }
u
:
U
→
R
{\displaystyle u\colon U\to \mathbb {R} }
harmonic if it is twice differentiable and satisfies
Δ
u
:=
∂
2
u
∂
x
2
+
∂
2
u
∂
y
2
=
0
{\displaystyle \Delta u:={\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}=0}
have.
The real part of a holomorphic function is harmonic, as follows from the Cauchy-Riemann-Differential equation . Interestingly, the converse also holds: every harmonic function is the real part of a holomorphic function.
Connection to Holomorphic Functions
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Let
U
⊆
C
{\displaystyle U\subseteq \mathbb {C} }
u
∈
C
2
(
U
,
R
)
{\displaystyle u\in C^{2}(U,\mathbb {R} )}
Δ
u
=
0
{\displaystyle \Delta u=0}
There exists
v
∈
C
2
(
U
,
R
)
{\displaystyle v\in C^{2}(U,\mathbb {R} )}
u
+
i
v
{\displaystyle u+iv}
(2).
⟹
{\displaystyle \implies }
Cauchy-Riemann-Differential equation , we have:
∂
2
u
∂
x
2
+
∂
2
u
∂
y
2
=
∂
∂
x
(
∂
u
∂
x
)
+
∂
∂
y
(
∂
u
∂
y
)
=
∂
∂
x
(
∂
v
∂
y
)
+
∂
∂
y
(
−
∂
v
∂
x
)
=
∂
2
v
∂
x
∂
y
−
∂
2
v
∂
y
∂
x
=
0.
{\displaystyle {\begin{array}{rl}{\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}&={\frac {\partial }{\partial x}}\left({\frac {\partial u}{\partial x}}\right)+{\frac {\partial }{\partial y}}\left({\frac {\partial u}{\partial y}}\right)\\&={\frac {\partial }{\partial x}}\left({\frac {\partial v}{\partial y}}\right)+{\frac {\partial }{\partial y}}\left(-{\frac {\partial v}{\partial x}}\right)\\&={\frac {\partial ^{2}v}{\partial x\partial y}}-{\frac {\partial ^{2}v}{\partial y\partial x}}\\&=0.\end{array}}}
since partial derivatives commute.
1.
⟹
{\displaystyle \implies }
g
:=
∂
u
∂
x
−
i
∂
u
∂
y
{\displaystyle \textstyle g:={\frac {\partial u}{\partial x}}-i{\frac {\partial u}{\partial y}}}
Cauchy-Riemann-Differential equation ,
g
{\displaystyle g}
holomorphic .since
U
{\displaystyle U}
f
:
U
→
C
{\displaystyle f\colon U\to \mathbb {C} }
g
{\displaystyle g}
u
(
z
0
)
=
f
(
z
0
)
{\displaystyle u(z_{0})=f(z_{0})}
z
0
∈
U
{\displaystyle z_{0}\in U}
f
=
u
1
+
i
v
1
{\displaystyle f=u_{1}+iv_{1}}
∂
u
1
∂
x
−
i
∂
u
1
∂
y
=
f
′
=
g
=
∂
u
∂
x
−
i
∂
u
∂
y
{\displaystyle {\frac {\partial u_{1}}{\partial x}}-i{\frac {\partial u_{1}}{\partial y}}=f'=g={\frac {\partial u}{\partial x}}-i{\frac {\partial u}{\partial y}}}
so is
u
1
−
u
{\displaystyle u_{1}-u}
u
1
(
z
0
)
=
f
(
z
0
)
=
u
(
z
0
)
{\displaystyle u_{1}(z_{0})=f(z_{0})=u(z_{0})}
u
1
=
u
{\displaystyle u_{1}=u}
v
:=
v
1
{\displaystyle v:=v_{1}}
Translation and Version Control
edit
This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Harmonische_Funktion
