Diese Lerneinheit startet mit einem Beispiel für das Lösen einer Differentialgleichung mit
γ
′
(
t
)
=
γ
(
t
)
n
{\displaystyle \gamma {\,}'(t)={\gamma (t)}^{n}}
n
∈
N
{\displaystyle n\in \mathbb {N} }
γ
:
[
a
,
b
]
→
C
{\displaystyle \gamma :[a,b]\to \mathbb {C} }
Vektorfeld
f
(
z
)
=
z
n
{\displaystyle f(z)=z^{n}}
z
a
∈
C
{\displaystyle z_{a}\in \mathbb {C} }
z
a
=
γ
(
a
)
{\displaystyle z_{a}=\gamma (a)}
Ein Punkt
z
a
{\displaystyle z_{a}}
Vektorfeld ausgesetzt, das durch einen holomorphe Funktion
f
:
G
→
C
{\displaystyle f:G\to \mathbb {C} }
Die folgende Abbildung zeigt Vektorfeld von einzelnen Punkten
z
∈
C
{\displaystyle z\in \mathbb {C} }
f
(
z
)
=
z
2
{\displaystyle f(z)=z^{2}}
z
∈
G
{\displaystyle z\in G}
z
{\displaystyle z}
vector field for z square Wenn man versucht
γ
′
(
t
)
=
γ
(
t
)
2
{\displaystyle \gamma {\,}'(t)=\gamma (t)^{2}}
holomorphen Weg
γ
:
[
a
,
b
]
→
C
{\displaystyle \gamma :[a,b]\to \mathbb {C} }
γ
(
a
)
=
z
a
{\displaystyle \gamma (a)=z_{a}}
Vektorfeld
f
(
z
)
=
z
3
{\displaystyle f(z)=z^{3}}
γ
(
t
)
:=
z
a
{\displaystyle \gamma (t):=z_{a}}
t
∈
[
a
,
b
]
{\displaystyle t\in [a,b]}
meromorphen Funktion
γ
U
:
U
→
C
{\displaystyle \gamma _{_{U}}:U\to \mathbb {C} }
[
a
,
b
]
⊂
U
{\displaystyle [a,b]\subset U}
γ
(
t
)
:=
γ
U
(
t
)
{\displaystyle \gamma (t):=\gamma _{_{U}}(t)}
t
∈
[
a
,
b
]
{\displaystyle t\in [a,b]}
[
a
,
b
]
{\displaystyle [a,b]}
Die Lösung der Differentialgleichung mit Anfangswert
γ
(
a
)
=
z
a
∈
C
{\displaystyle \gamma (a)=z_{a}\in \mathbb {C} }
γ
(
t
)
=
z
a
1
−
z
a
⋅
(
t
−
a
)
{\displaystyle \gamma (t)={\frac {z_{a}}{1-z_{a}\cdot (t-a)}}}
Die Lösung hat eine Pol bei
t
=
1
+
a
z
a
{\displaystyle t={\tfrac {1+a}{z_{a}}}}
Damit ist
γ
G
{\displaystyle \gamma _{_{G}}}
meromorphe Funktion auf
G
:=
C
∖
{
1
+
a
z
a
}
{\displaystyle G:=\mathbb {C} \setminus \left\{{\tfrac {1+a}{z_{a}}}\right\}}
γ
:
[
a
,
b
]
→
C
{\displaystyle \gamma :[a,b]\to \mathbb {C} }
[
a
,
b
]
{\displaystyle [a,b]}
t
o
:=
1
+
a
z
a
{\displaystyle t_{o}:={\tfrac {1+a}{z_{a}}}}
[
a
,
b
]
{\displaystyle [a,b]}
γ
{\displaystyle \gamma }
t
o
∉
[
a
,
b
]
{\displaystyle t_{o}\notin [a,b]}
γ
{\displaystyle \gamma }
t
o
=
1
+
a
z
a
∈
[
a
,
b
]
{\displaystyle t_{o}={\tfrac {1+a}{z_{a}}}\in [a,b]}
Für
z
a
:=
1
{\displaystyle z_{a}:=1}
[
a
,
b
]
:=
[
1
,
3
]
{\displaystyle [a,b]:=[1,3]}
t
0
:=
1
+
a
z
a
=
2
{\displaystyle t_{0}:={\tfrac {1+a}{z_{a}}}=2}
[
a
,
b
]
{\displaystyle [a,b]}
t
0
{\displaystyle t_{0}}
[ Bearbeiten ] Über die Trennung der Variablen kann man die Differentialgleichung lösen:
γ
′
(
t
)
=
γ
(
t
)
2
{\displaystyle \gamma {\,}'(t)=\gamma (t)^{2}}
Man betrachtet dazu die folgende Gleichung nach Äquivalenzumformung:
1
γ
(
t
)
2
⋅
γ
′
(
t
)
=
1
(
∗
)
{\displaystyle {\frac {1}{\gamma (t)^{2}}}\cdot \gamma {\,}'(t)=1\quad (\ast )}
Dabei ist es wesentlich, dass
γ
(
t
)
2
{\displaystyle \gamma (t)^{2}}
γ
(
t
)
{\displaystyle \gamma (t)}
t
∈
[
a
,
b
]
{\displaystyle t\in [a,b]}
Durch Integration über die Gleichung
(
∗
)
{\displaystyle (\ast )}
∫
a
b
1
γ
(
t
)
2
⋅
γ
′
(
t
)
d
t
=
∫
a
b
1
d
t
(
∗
∗
)
{\displaystyle \int _{a}^{b}{\frac {1}{\gamma (t)^{2}}}\cdot \gamma {\,}'(t)\,dt=\int _{a}^{b}1\,dt\quad (\ast \ast )}
Auf der linken Seite der Gleichung erhält man die Defintion des Wegintegrals .
Schritt 2: Definition des Wegintegrals [ Bearbeiten ] Die Integration der Gleichung
(
∗
∗
)
{\displaystyle (\ast \ast )}
∫
γ
1
z
2
d
z
=
∫
a
b
1
γ
(
t
)
2
⋅
γ
′
(
t
)
d
t
=
∫
a
b
1
d
t
=
b
−
a
{\displaystyle \int _{\gamma }{\frac {1}{z^{2}}}\,dz=\int _{a}^{b}{\frac {1}{\gamma (t)^{2}}}\cdot \gamma {\,}'(t)\,dt=\int _{a}^{b}1\,dt=b-a}
Die Stammfunktion von
f
^
(
z
)
=
z
−
2
{\displaystyle {\widehat {f}}(z)=z^{-2}}
F
^
(
z
)
=
1
−
1
⋅
z
−
1
=
−
1
z
{\displaystyle {\widehat {F}}(z)={\tfrac {1}{-1}}\cdot z^{-1}=-{\tfrac {1}{z}}}
[ Bearbeiten ] Das Integral
∫
γ
f
^
(
z
)
d
z
=
F
^
(
γ
(
b
)
)
−
F
^
(
γ
(
a
)
)
{\displaystyle \int _{\gamma }{\widehat {f}}(z)\,dz={\widehat {F}}{\big (}\gamma (b){\big )}-{\widehat {F}}{\big (}\gamma (a){\big )}}
F
^
{\displaystyle {\widehat {F}}}
b
−
a
=
∫
γ
1
z
2
d
z
=
−
1
γ
(
b
)
+
1
γ
(
a
)
{\displaystyle b-a=\int _{\gamma }{\frac {1}{z^{2}}}\,dz=-{\frac {1}{\gamma (b)}}+{\frac {1}{\gamma (a)}}}
Schritt 4 - Endpunkt des Wegintegrals als Variable [ Bearbeiten ] Man betrachtet nun das Wegintegral über das Intervall
[
a
,
t
]
{\displaystyle [a,t]}
t
>
a
{\displaystyle t>a}
γ
(
a
)
=
z
a
∈
C
{\displaystyle \gamma (a)=z_{a}\in \mathbb {C} }
t
−
a
=
∫
γ
1
z
2
d
z
=
−
1
γ
(
t
)
+
1
z
a
{\displaystyle t-a=\int _{\gamma }{\frac {1}{z^{2}}}\,dz=-{\frac {1}{\gamma (t)}}+{\frac {1}{z_{a}}}}
Man erhält den Weg
γ
(
t
)
=
z
a
1
−
z
a
(
t
−
a
)
{\displaystyle \gamma (t)={\tfrac {z_{a}}{1-z_{a}(t-a)}}}
−
1
z
a
+
t
−
a
=
−
1
γ
(
t
)
⟺
1
z
a
−
z
a
(
t
−
a
)
z
a
=
=
1
γ
(
t
)
⟺
1
−
z
a
(
t
−
a
)
z
a
=
1
γ
(
t
)
{\displaystyle {\begin{array}{rclc}-{\frac {1}{z_{a}}}+t-a&=&-{\frac {1}{\gamma (t)}}&\Longleftrightarrow \\{\frac {1}{z_{a}}}-{\frac {z_{a}(t-a)}{z_{a}}}=&=&{\frac {1}{\gamma (t)}}&\Longleftrightarrow \\{\frac {1-z_{a}(t-a)}{z_{a}}}&=&{\frac {1}{\gamma (t)}}&\\\end{array}}}
Um die Spur des Weges in Geogebra zu zeichnen, zerlegt man
γ
(
t
)
=
γ
1
(
t
)
+
i
⋅
γ
2
(
t
)
{\displaystyle \gamma (t)=\gamma _{1}(t)+i\cdot \gamma _{2}(t)}
γ
1
:
[
a
,
b
]
→
R
{\displaystyle \gamma _{1}:[a,b]\to \mathbb {R} }
γ
2
:
[
a
,
b
]
→
R
{\displaystyle \gamma _{2}:[a,b]\to \mathbb {R} }
R
2
{\displaystyle \mathbb {R} ^{2}}
[ Bearbeiten ] In der Definition der Lösung
γ
(
t
)
{\displaystyle \gamma (t)}
γ
(
t
)
=
z
a
1
−
z
a
⋅
(
t
−
a
)
=
z
a
⋅
(
1
−
z
a
⋅
(
t
−
a
)
)
¯
(
1
−
z
a
⋅
(
t
−
a
)
)
⋅
(
1
−
z
a
⋅
(
t
−
a
)
)
¯
=
z
a
⋅
(
1
−
z
a
¯
⋅
(
t
−
a
)
)
(
1
−
z
a
⋅
(
t
−
a
)
)
⋅
(
1
−
z
a
¯
⋅
(
t
−
a
)
)
=
z
a
−
|
z
a
|
2
⋅
(
t
−
a
)
(
1
−
z
a
⋅
(
t
−
a
)
−
z
a
¯
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
=
z
a
−
|
z
a
|
2
⋅
(
t
−
a
)
1
−
2
⋅
R
e
(
z
a
)
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
{\displaystyle {\begin{array}{rcl}\gamma (t)&=&{\frac {z_{a}}{1-z_{a}\cdot (t-a)}}={\frac {z_{a}\cdot {\overline {(1-z_{a}\cdot (t-a))}}}{(1-z_{a}\cdot (t-a))\cdot {\overline {(1-z_{a}\cdot (t-a))}}}}\\&=&{\frac {z_{a}\cdot (1-{\overline {z_{a}}}\cdot (t-a))}{(1-z_{a}\cdot (t-a))\cdot (1-{\overline {z_{a}}}\cdot (t-a))}}\\&=&{\frac {z_{a}-|z_{a}|^{2}\cdot (t-a)}{(1-z_{a}\cdot (t-a)-{\overline {z_{a}}}\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}}}\\&=&{\frac {z_{a}-|z_{a}|^{2}\cdot (t-a)}{1-2\cdot {\mathfrak {Re}}(z_{a})\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}}}\\\end{array}}}
[ Bearbeiten ] Mit
γ
(
a
)
=
z
a
=
z
a
,
1
+
i
⋅
z
a
,
2
{\displaystyle \gamma (a)=z_{a}=z_{a,1}+i\cdot z_{a,2}}
γ
(
t
)
=
z
a
−
|
z
a
|
2
⋅
(
t
−
a
)
1
−
2
⋅
R
e
(
z
a
)
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
=
z
a
,
1
−
|
z
a
|
2
⋅
(
t
−
a
)
⏞
∈
R
+
i
⋅
z
a
,
2
⏞
∈
R
1
−
2
⋅
R
e
(
z
a
)
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
⏟
∈
R
{\displaystyle {\begin{array}{rcl}\gamma (t)&=&{\frac {z_{a}-|z_{a}|^{2}\cdot (t-a)}{1-2\cdot {\mathfrak {Re}}(z_{a})\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}}}\\&=&{\frac {\overbrace {z_{a,1}-|z_{a}|^{2}\cdot (t-a)} ^{\in \mathbb {R} }+i\cdot \overbrace {z_{a,2}} ^{\in \mathbb {R} }}{\underbrace {1-2\cdot {\mathfrak {Re}}(z_{a})\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}} _{\in \mathbb {R} }}}\\\end{array}}}
[ Bearbeiten ] Damit erhält man nun die Realteilfunktion
γ
1
{\displaystyle \gamma _{1}}
γ
2
{\displaystyle \gamma _{2}}
z
a
{\displaystyle z_{a}}
γ
1
(
t
)
=
z
a
,
1
−
|
z
a
|
2
⋅
(
t
−
a
)
⏞
∈
R
1
−
2
⋅
R
e
(
z
a
)
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
γ
2
(
t
)
=
z
a
,
2
⏞
∈
R
1
−
2
⋅
R
e
(
z
a
)
⋅
(
t
−
a
)
+
|
z
a
|
2
⋅
(
t
−
a
)
2
{\displaystyle {\begin{array}{rcl}\gamma _{_{1}}(t)&=&{\frac {\overbrace {z_{a,1}-|z_{a}|^{2}\cdot (t-a)} ^{\in \mathbb {R} }}{1-2\cdot {\mathfrak {Re}}(z_{a})\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}}}\\\gamma _{_{2}}(t)&=&{\frac {\overbrace {z_{a,2}} ^{\in \mathbb {R} }}{1-2\cdot {\mathfrak {Re}}(z_{a})\cdot (t-a)+|z_{a}|^{2}\cdot (t-a)^{2}}}\\\end{array}}}
Plotten Sie nun den Weg durch Festlegung des Punktes
(
γ
1
(
t
)
,
γ
2
(
t
)
)
∈
R
2
{\displaystyle (\gamma _{1}(t),\gamma _{2}(t))\in \mathbb {R} ^{2}}
t
∈
[
a
,
b
]
{\displaystyle t\in [a,b]}
Animation - differential equation - solution plotted as curve in OpenSource Geogebra Bestimmen Sie Weg für eine Vektorfeld
f
(
z
)
=
z
n
{\displaystyle f(z)=z^{n}}
f
:
G
→
R
{\displaystyle f:G\to \mathbb {R} }
[
a
,
b
]
⊂
G
{\displaystyle [a,b]\subset G}
Bestimmen Sie Weg allgemein für eine Vektorfeld, das durch eine meromorphe Funktion
f
:
G
→
R
{\displaystyle f:G\to \mathbb {R} }
γ
{\displaystyle \gamma }
[
a
,
b
]
{\displaystyle [a,b]}
γ
{\displaystyle \gamma }
[
a
,
b
]
{\displaystyle [a,b]}
![{\displaystyle \gamma :[a,b]\to \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ba8e67ece48ed58462781818669d29170ae4809)
![{\displaystyle t\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7f3050ace6dc0dd95250c418528da28eb477ffe)
![{\displaystyle [a,b]\subset U}](https://wikimedia.org/api/rest_v1/media/math/render/svg/823faf478e9265672a07a5c4d9e233abfe27ccdf)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle t_{o}\notin [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee8e6ec008c2888752ff3ea921f069c2e2a432a8)
![{\displaystyle t_{o}={\tfrac {1+a}{z_{a}}}\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f670f79f1f59c7f751a740a9d08aa86e1f552dc2)
![{\displaystyle [a,b]:=[1,3]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10794bb17593d9c6007070cc96665783219b0049)
![{\displaystyle [a,t]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b7e4d6424116f83d1d8eadea305b1ccc9c8744e)
![{\displaystyle \gamma _{1}:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8260c4752c91d786e7df75406c95c4a452b37b91)
![{\displaystyle \gamma _{2}:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0041edc05a5a5beadc3a38654de6592205751fc6)
![{\displaystyle [a,b]\subset G}](https://wikimedia.org/api/rest_v1/media/math/render/svg/307cf3c74f5457581437ee6bd69118b45dc2f39f)