Mathematical Engineering - LRT » Maßtheorie und DGL

Beispiel: Lösung einer inhomogenen Differentialgleichung

admin2 — Thu, 11 Mar 2010 19:14:27 +0000

Lösen Sie die Differentialgleichung

Lösung

Da es sich um eine inhomogene Differentialgleichung handelt, müssen wir zuerst die Lösung der homogenen Gleichung

finden. Anschließend suchen wir eine partikuläre Lösung, die die inhomogene DGL erfüllt. Die allgemeine Lösung ist die Summe aus homogener und partikulärer Lösung.

homogene Lösung

Lösungsansatz:

Ableiten und Einsetzen führt auf die charakteristische Gleichung:

Wir lösen die charakteristische Gleichung durch quadratisches Ergänzen:

Dies setzen wir in den Ansatz ein und transformieren schließlich mit der Eulerformel in den reellen Bereich:

Dass diese Funktion die homogene Gleichung erfüllt, sehen wir, wenn wir die Probe durchführen (muss nicht unbedingt gemacht werden):

einsetzen und vereinfachen:

partikuläre Lösung

Als Lösungsansatz verwenden wir einen Ansatz vom “Typ der rechten Seite”. Das bedeutet, wir verwenden als Ansatzfunktion eine Funktion der Klasse der Funktion, die auf der rechten Seite des Gleichheitszeichens steht. In diesem Fall ist das das Produkt aus einer Exponentialfunktion und eines Polynoms zweiten Grades:

Wir bilden die ersten beiden Ableitungen:

Einsetzen in die inhomogene DGL liefert:

vereinfachen:

Da die Exponentialfunktion immer positiv ist, dürfen wir sie kürzen:

Wir führen nun einen Koeffizientenvergleich durch (Vergleich der Vorfaktoren vor und erhalten dadurch die Werte für die Koeffizienten:

Einsetzen in den Lösungsansatz liefert die partikuläre Lösung:

Damit ist die allgemeine Lösung:

Eine mit Maxima durchgeführte Probe bestätigt das Ergebnis.

Aufgabe 7.3 – Differentialgleichung in Matrixform

admin2 — Tue, 08 Dec 2009 16:38:48 +0000

Gegeben sei

A = \left( {\begin{array}{*{20}{c}}
-2 & {-1} & 0 \\
1 & {-2} & {-1} \\
0 & 1 & {-2} \\

\end{array} } \right)
" title="
A = \left( {\begin{array}{*{20}{c}}
-2 & {-1} & 0 \\
1 & {-2} & {-1} \\
0 & 1 & {-2} \\

\end{array} } \right)
" style="vertical-align: -28px; border: none;"/>

Bestimmen Sie alle Lösungen von

x ^{\prime}\left( t \right) = Ax
" title="
x ^{\prime}\left( t \right) = Ax
" style="vertical-align: -4px; border: none;"/>

Warum gehen alle diese Lösungen für t → ∞ gegen 0?

Lösung

A = \left( {\begin{array}{*{20}{c}}
-2 & {-1} & 0 \\
1 & {-2} & {-1} \\
0 & 1 & {-2} \\

\end{array} } \right)
" title="
A = \left( {\begin{array}{*{20}{c}}
-2 & {-1} & 0 \\
1 & {-2} & {-1} \\
0 & 1 & {-2} \\

\end{array} } \right)
" style="vertical-align: -28px; border: none;"/>

x ^{\prime}= Ax
" title="
x ^{\prime}= Ax
" style="vertical-align: 0px; border: none;"/>

Charakteristische Gleichung:

\mathcal{X}_A \left( \lambda \right) = \det \left( {A-\lambda I} \right)
" title="
\mathcal{X}_A \left( \lambda \right) = \det \left( {A-\lambda I} \right)
" style="vertical-align: -4px; border: none;"/>

= \left( {-2-\lambda } \right)^3 +2\left( {-2-\lambda } \right)
" title="
= \left( {-2-\lambda } \right)^3 +2\left( {-2-\lambda } \right)
" style="vertical-align: -4px; border: none;"/>

= \left( {-2-\lambda } \right)\left( {\left( {-2-\lambda } \right)^2 +2} \right)
" title="
= \left( {-2-\lambda } \right)\left( {\left( {-2-\lambda } \right)^2 +2} \right)
" style="vertical-align: -12px; border: none;"/>

= \left( {-2-\lambda } \right)\left( {\lambda ^2 +4\lambda +6} \right) = 0
" title="
= \left( {-2-\lambda } \right)\left( {\lambda ^2 +4\lambda +6} \right) = 0
" style="vertical-align: -6px; border: none;"/>

\Rightarrow \quad \lambda _1 = -2,\quad \lambda _{2,3} = \frac{{-4 \pm \sqrt {16-24} }}
{2} = -2 \pm \sqrt 2 i
" title="
\Rightarrow \quad \lambda _1 = -2,\quad \lambda _{2,3} = \frac{{-4 \pm \sqrt {16-24} }}
{2} = -2 \pm \sqrt 2 i
" style="vertical-align: -6px; border: none;"/>

x_h = x_1 e^{-2t} +c_2 e^{-2t} \cos \left( {\sqrt 2 t} \right)+c_3 e^{-2t} \sin \left( {\sqrt 2 t} \right),\quad \quad c_1 ,c_2 ,c_3 \in \mathbb{R}
" title="
x_h = x_1 e^{-2t} +c_2 e^{-2t} \cos \left( {\sqrt 2 t} \right)+c_3 e^{-2t} \sin \left( {\sqrt 2 t} \right),\quad \quad c_1 ,c_2 ,c_3 \in \mathbb{R}
" style="vertical-align: -6px; border: none;"/>

Wir müssen nun noch die Konstanten c_i bestimmen.

Eigenvektoren:

Av = \lambda v\quad \quad \Rightarrow \quad \quad \left( {A-\lambda I} \right)v = 0
" title="
Av = \lambda v\quad \quad \Rightarrow \quad \quad \left( {A-\lambda I} \right)v = 0
" style="vertical-align: -4px; border: none;"/>

zu λ₁:

\left( {\begin{array}{*{20}{c}}
0 & {-1} & 0 \\
1 & 0 & {-1} \\
0 & 1 & 0 \\

\end{array} } \right)\vec v = 0\quad \quad \Rightarrow \quad \quad \left( {\begin{array}{*{20}{c}}
0 & 0 & 0 \\
1 & 0 & {-1} \\
0 & 1 & 0 \\

\end{array} } \right)\vec v = 0
" title="
\left( {\begin{array}{*{20}{c}}
0 & {-1} & 0 \\
1 & 0 & {-1} \\
0 & 1 & 0 \\

\end{array} } \right)\vec v = 0\quad \quad \Rightarrow \quad \quad \left( {\begin{array}{*{20}{c}}
0 & 0 & 0 \\
1 & 0 & {-1} \\
0 & 1 & 0 \\

\end{array} } \right)\vec v = 0
" style="vertical-align: -28px; border: none;"/>

v_2 = 0
" title="
v_2 = 0
" style="vertical-align: -3px; border: none;"/>

v_1 = v_3 = 1
" title="
v_1 = v_3 = 1
" style="vertical-align: -4px; border: none;"/>

v = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1 \\

\end{array} } \right)
" title="
v = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1 \\

\end{array} } \right)
" style="vertical-align: -28px; border: none;"/>

zu λ₂:

\left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
1 & {-\sqrt 2 i} & {-1} \\
0 & 1 & {-\sqrt 2 i} \\

\end{array} } \right)\vec w = \vec 0
" title="
\left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
1 & {-\sqrt 2 i} & {-1} \\
0 & 1 & {-\sqrt 2 i} \\

\end{array} } \right)\vec w = \vec 0
" style="vertical-align: -28px; border: none;"/>

\left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
0 & 1 & {-\sqrt 2 i} \\
0 & 1 & {-\sqrt 2 i} \\

\end{array} } \right)\vec w = \vec 0\quad \quad \Rightarrow \quad \quad \left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
0 & 1 & {-\sqrt 2 i} \\
0 & 0 & 0 \\

\end{array} } \right)\vec w = \vec 0
" title="
\left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
0 & 1 & {-\sqrt 2 i} \\
0 & 1 & {-\sqrt 2 i} \\

\end{array} } \right)\vec w = \vec 0\quad \quad \Rightarrow \quad \quad \left( {\begin{array}{*{20}{c}}
-\sqrt 2 i & {-1} & 0 \\
0 & 1 & {-\sqrt 2 i} \\
0 & 0 & 0 \\

\end{array} } \right)\vec w = \vec 0
" style="vertical-align: -28px; border: none;"/>

Daraus folgt:

w_2 = \sqrt 2 iw_3
" title="
w_2 = \sqrt 2 iw_3
" style="vertical-align: -3px; border: none;"/>

-\sqrt {2i} w_1 = w_2
" title="
-\sqrt {2i} w_1 = w_2
" style="vertical-align: -4px; border: none;"/>

Wir definieren:

w_2 : = 1
" title="
w_2 : = 1
" style="vertical-align: -3px; border: none;"/>

Der Lösungsvektor ist damit:

\vec w = \left( {\begin{array}{*{20}{c}}
-\frac{1}
{{\sqrt 2 i}} \\
1 \\
{\frac{1}
{{\sqrt 2 i}}} \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
\frac{i}
{{\sqrt 2 }} \\
1 \\
{-\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)
" title="
\vec w = \left( {\begin{array}{*{20}{c}}
-\frac{1}
{{\sqrt 2 i}} \\
1 \\
{\frac{1}
{{\sqrt 2 i}}} \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
\frac{i}
{{\sqrt 2 }} \\
1 \\
{-\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)
" style="vertical-align: -34px; border: none;"/>

Das komplex konjugierte:

u = \bar w = \left( {\begin{array}{*{20}{c}}
-\frac{i}
{{\sqrt 2 }} \\
1 \\
{\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)
" title="
u = \bar w = \left( {\begin{array}{*{20}{c}}
-\frac{i}
{{\sqrt 2 }} \\
1 \\
{\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)
" style="vertical-align: -34px; border: none;"/>

Der homogene Lösungsvektor:

\vec x_h \left( t \right) = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1 \\

\end{array} } \right)e^{-2t} +\left( {\begin{array}{*{20}{c}}
\frac{i}
{{\sqrt 2 }} \\
1 \\
{-\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)e^{\left( {-2+\sqrt 2 i} \right)t} +\left( {\begin{array}{*{20}{c}}
-\frac{i}
{{\sqrt 2 }} \\
1 \\
{\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)e^{\left( {-2-\sqrt 2 i} \right)t}
" title="
\vec x_h \left( t \right) = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
1 \\

\end{array} } \right)e^{-2t} +\left( {\begin{array}{*{20}{c}}
\frac{i}
{{\sqrt 2 }} \\
1 \\
{-\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)e^{\left( {-2+\sqrt 2 i} \right)t} +\left( {\begin{array}{*{20}{c}}
-\frac{i}
{{\sqrt 2 }} \\
1 \\
{\frac{i}
{{\sqrt 2 }}} \\

\end{array} } \right)e^{\left( {-2-\sqrt 2 i} \right)t}
" style="vertical-align: -34px; border: none;"/>

Die komplexe Fundamentalmatrix lautet:

\left( {\begin{array}{*{20}{c}}
e^{-2t} & {\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } & {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \\
0 & {e^{\left( {-2+\sqrt 2 i} \right)t} } & {e^{\left( {-2-\sqrt 2 i} \right)t} } \\
{e^{-2t} } & {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } & {\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \\

\end{array} } \right)
" title="
\left( {\begin{array}{*{20}{c}}
e^{-2t} & {\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } & {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \\
0 & {e^{\left( {-2+\sqrt 2 i} \right)t} } & {e^{\left( {-2-\sqrt 2 i} \right)t} } \\
{e^{-2t} } & {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } & {\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \\

\end{array} } \right)
" style="vertical-align: -40px; border: none;"/>

Daraus folgt die reelle Fundamentalmatrix:

\left( {\begin{array}{*{20}{c}}
e^{-2t} & {\operatorname{Re} \left\{ {\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\
0 & {\operatorname{Re} \left\{ {e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\
{e^{-2t} } & {\operatorname{Re} \left\{ {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\

\end{array} } \right)
" title="
\left( {\begin{array}{*{20}{c}}
e^{-2t} & {\operatorname{Re} \left\{ {\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\
0 & {\operatorname{Re} \left\{ {e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\
{e^{-2t} } & {\operatorname{Re} \left\{ {-\frac{i}
{{\sqrt 2 }}e^{\left( {-2+\sqrt 2 i} \right)t} } \right\}} & {\operatorname{Im} \left\{ {\frac{i}
{{\sqrt 2 }}e^{\left( {-2-\sqrt 2 i} \right)t} } \right\}} \\

\end{array} } \right)
" style="vertical-align: -45px; border: none;"/>

[todo berechnen]
[todo zweite frage beantworten]

Aufgabe 4.2 – Messbarkeit und Bildmaß

admin2 — Tue, 08 Dec 2009 16:36:16 +0000

Gegeben sei der Maßraum und der Messraum . Sei weiter

{-1} & {falls} & {\omega \in \left] {0,5} \right]} \\
0 & {falls} & {\omega = 0} \\
1 & {falls} & {sonst} \\

\end{array} } \right." title="f:\mathbb{R} \to \Omega ,\quad \omega \mapsto \left\{ {\begin{array}{*{20}{c}}
{-1} & {falls} & {\omega \in \left] {0,5} \right]} \\
0 & {falls} & {\omega = 0} \\
1 & {falls} & {sonst} \\

\end{array} } \right." style="vertical-align: -29px; border: none;"/>

eine Funktion.

Zeige, dass -messbar ist
Bestimme das Bildmaß

Lösung

f:\mathbb{R} \to \Omega = \left\{ {-1,0,1} \right\}
" title="
f:\mathbb{R} \to \Omega = \left\{ {-1,0,1} \right\}
" style="vertical-align: -5px; border: none;"/>

\omega \mapsto \left\{ {\begin{array}{*{20}{c}}
- 1 & \forall & {\omega \in \left] {0,5} \right]} \\
0 & {falls} & {\omega = 0} \\
1 & {sonst} & {} \\

\end{array} } \right.
" title="
\omega \mapsto \left\{ {\begin{array}{*{20}{c}}
- 1 & \forall & {\omega \in \left] {0,5} \right]} \\
0 & {falls} & {\omega = 0} \\
1 & {sonst} & {} \\

\end{array} } \right.
" style="vertical-align: -29px; border: none;"/>

a )

\mathcal{P}\left( \Omega \right) = \left\{ {\emptyset ,\Omega ,\left\{ {-1} \right\},\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {-1,0} \right\},\left\{ {-1,1} \right\},\left\{ {0,1} \right\}} \right\}
" title="
\mathcal{P}\left( \Omega \right) = \left\{ {\emptyset ,\Omega ,\left\{ {-1} \right\},\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {-1,0} \right\},\left\{ {-1,1} \right\},\left\{ {0,1} \right\}} \right\}
" style="vertical-align: -5px; border: none;"/>

Die Urbilder sind

f^{-1} \left( \emptyset \right) = \emptyset \in \mathcal{B}
" title="
f^{-1} \left( \emptyset \right) = \emptyset \in \mathcal{B}
" style="vertical-align: -4px; border: none;"/>

f^{-1} \left( \Omega \right) = \mathbb{R} \in \mathcal{B}
" title="
f^{-1} \left( \Omega \right) = \mathbb{R} \in \mathcal{B}
" style="vertical-align: -4px; border: none;"/>

f^{-1} \left( {\left\{ {-1} \right\}} \right) = \left] {0,5} \right] \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ {-1} \right\}} \right) = \left] {0,5} \right] \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

f^{-1} \left( {\left\{ 0 \right\}} \right) = \left\{ 0 \right\} \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ 0 \right\}} \right) = \left\{ 0 \right\} \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

f^{-1} \left( {\left\{ 1 \right\}} \right) = \mathbb{R}{{\backslash }}\left[ {0,5} \right] \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ 1 \right\}} \right) = \mathbb{R}{{\backslash }}\left[ {0,5} \right] \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

f^{-1} \left( {\left\{ {-1,0} \right\}} \right) = \left[ {0,5} \right] \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ {-1,0} \right\}} \right) = \left[ {0,5} \right] \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

f^{-1} \left( {\left\{ {-1,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left\{ 0 \right\} \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ {-1,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left\{ 0 \right\} \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

f^{-1} \left( {\left\{ {0,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left] {0,5} \right] \in \mathcal{B}
" title="
f^{-1} \left( {\left\{ {0,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left] {0,5} \right] \in \mathcal{B}
" style="vertical-align: -5px; border: none;"/>

Daraus folgt:

f\,\,ist\,\,\mathcal{B}-\mathcal{P}\left( \Omega \right)-messbar
" title="
f\,\,ist\,\,\mathcal{B}-\mathcal{P}\left( \Omega \right)-messbar
" style="vertical-align: -4px; border: none;"/>

b )

\mu \left( B \right) = \lambda \left( {f^{-1} \left( B \right)} \right)
" title="
\mu \left( B \right) = \lambda \left( {f^{-1} \left( B \right)} \right)
" style="vertical-align: -6px; border: none;"/>

\mu \left( \emptyset \right) = \lambda \left( {f^{-1} \left( \emptyset \right)} \right) = \lambda \left( \emptyset \right) = 0
" title="
\mu \left( \emptyset \right) = \lambda \left( {f^{-1} \left( \emptyset \right)} \right) = \lambda \left( \emptyset \right) = 0
" style="vertical-align: -6px; border: none;"/>

\mu \left( \Omega \right) = \lambda \left( {f^{-1} \left( \Omega \right)} \right) = \lambda \left( \mathbb{R} \right) = \infty
" title="
\mu \left( \Omega \right) = \lambda \left( {f^{-1} \left( \Omega \right)} \right) = \lambda \left( \mathbb{R} \right) = \infty
" style="vertical-align: -6px; border: none;"/>

\mu \left( {\left\{ {-1} \right\}} \right) = \lambda \left( {\left] {0,5} \right]} \right) = 5-0 = 5
" title="
\mu \left( {\left\{ {-1} \right\}} \right) = \lambda \left( {\left] {0,5} \right]} \right) = 5-0 = 5
" style="vertical-align: -5px; border: none;"/>

\mu \left( {\left\{ 0 \right\}} \right) = \lambda \left( {\left\{ 0 \right\}} \right) = 0
" title="
\mu \left( {\left\{ 0 \right\}} \right) = \lambda \left( {\left\{ 0 \right\}} \right) = 0
" style="vertical-align: -5px; border: none;"/>

\mu \left( {\left\{ 1 \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left[ {0,5} \right]} \right) = \infty
" title="
\mu \left( {\left\{ 1 \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left[ {0,5} \right]} \right) = \infty
" style="vertical-align: -5px; border: none;"/>

\mu \left( {\left\{ {-1,0} \right\}} \right) = \lambda \left( {\left[ {0,5} \right]} \right) = 5-0 = 5
" title="
\mu \left( {\left\{ {-1,0} \right\}} \right) = \lambda \left( {\left[ {0,5} \right]} \right) = 5-0 = 5
" style="vertical-align: -5px; border: none;"/>

\mu \left( {\left\{ {-1,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left\{ 0 \right\}} \right) = \infty
" title="
\mu \left( {\left\{ {-1,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left\{ 0 \right\}} \right) = \infty
" style="vertical-align: -5px; border: none;"/>

\mu \left( {\left\{ {0,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left] {0,5} \right]} \right) = \infty
" title="
\mu \left( {\left\{ {0,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left] {0,5} \right]} \right) = \infty
" style="vertical-align: -5px; border: none;"/>

Aufgabe 3.5 – Treppenfunktionen

admin2 — Fri, 02 Oct 2009 15:10:27 +0000

Betrachten Sie das Lebesgue-Maß auf . Sei eine Lebesgue-Nullmenge, also .

Existiert dann notwendig ein Punkt mit

Lösung

Bestimmen einer Treppenfunktion f_k:

1. Unterteile den Bildbereich zunächst in zwei Intervalle

\left[ {0,k} \right[,\quad \left[ {k,\infty } \right[
" title="
\left[ {0,k} \right[,\quad \left[ {k,\infty } \right[
" style="vertical-align: -5px; border: none;"/>

Unterteile \left[ {0,k} \right[
" title="
\left[ {0,k} \right[
" style="vertical-align: -5px; border: none;"/> weiter in gleichgroße Teilintervalle:

\left[ {\frac{{i-1}}
{{2^k }},\frac{i}
{{2^k }}} \right[,\quad i = 1,\ldots,k \cdot 2^k
" title="
\left[ {\frac{{i-1}}
{{2^k }},\frac{i}
{{2^k }}} \right[,\quad i = 1,\ldots,k \cdot 2^k
" style="vertical-align: -7px; border: none;"/>

2. Bestimme

f^{-1} \left( {\left[ {\frac{{i-1}}
{{2^k }},\frac{i}
{{2^k }}} \right[} \right),\quad i = 1,\ldots,k \cdot 2^k
" title="
f^{-1} \left( {\left[ {\frac{{i-1}}
{{2^k }},\frac{i}
{{2^k }}} \right[} \right),\quad i = 1,\ldots,k \cdot 2^k
" style="vertical-align: -7px; border: none;"/>

und

f^{-1} \left( {\left[ {k,\infty } \right[} \right)
" title="
f^{-1} \left( {\left[ {k,\infty } \right[} \right)
" style="vertical-align: -5px; border: none;"/>

Bezeichnung für die beiden Urbilder: A_k,i, A_k,∞

f_k \left( x \right) = \sum\limits_{i = 1}^{k \cdot 2^k } {\frac{{i-1}}
{{2^k }} \cdot I_{A_{k,i} } \left( x \right)+k \cdot } I_{A_{k,\infty } } \left( x \right)
" title="
f_k \left( x \right) = \sum\limits_{i = 1}^{k \cdot 2^k } {\frac{{i-1}}
{{2^k }} \cdot I_{A_{k,i} } \left( x \right)+k \cdot } I_{A_{k,\infty } } \left( x \right)
" style="vertical-align: -17px; border: none;"/>

Beispiel

für k = 1:

\left[ {0,1} \right[,\quad \left[ {1,\infty } \right[
" title="
\left[ {0,1} \right[,\quad \left[ {1,\infty } \right[
" style="vertical-align: -5px; border: none;"/>

\left[ {0,\frac{1}
{2}} \right[,\quad \left[ {\frac{1}
{2},1} \right[
" title="
\left[ {0,\frac{1}
{2}} \right[,\quad \left[ {\frac{1}
{2},1} \right[
" style="vertical-align: -6px; border: none;"/>

A_{1,1} = f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right),\quad \quad A_{1,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right),\quad \quad A_{1,\infty } = f^{-1} \left( {\left[ {1,\infty } \right[} \right)
" title="
A_{1,1} = f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right),\quad \quad A_{1,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right),\quad \quad A_{1,\infty } = f^{-1} \left( {\left[ {1,\infty } \right[} \right)
" style="vertical-align: -6px; border: none;"/>

f_1 \left( x \right) = 0 \cdot I_{A_{1,1} } \left( x \right)+\frac{1}
{2} \cdot I_{A_{1,2} } \left( x \right)+1 \cdot I_{A_{1,\infty } } \left( x \right)
" title="
f_1 \left( x \right) = 0 \cdot I_{A_{1,1} } \left( x \right)+\frac{1}
{2} \cdot I_{A_{1,2} } \left( x \right)+1 \cdot I_{A_{1,\infty } } \left( x \right)
" style="vertical-align: -7px; border: none;"/>

Angewendet auf die Aufgabenstellung:

a )

x \mapsto \left| x \right|
" title="
x \mapsto \left| x \right|
" style="vertical-align: -5px; border: none;"/>

k = 1: erster Schritt klar,

zweiter Schritt:

A_{1,1} = f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right) = \left] {-\frac{1}
{2},\frac{1}
{2}} \right[
" title="
A_{1,1} = f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right) = \left] {-\frac{1}
{2},\frac{1}
{2}} \right[
" style="vertical-align: -6px; border: none;"/>

A_{1,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right) = \left] {-1,-\frac{1}
{2}} \right] \cup \left[ {\frac{1}
{2},1} \right[
" title="
A_{1,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right) = \left] {-1,-\frac{1}
{2}} \right] \cup \left[ {\frac{1}
{2},1} \right[
" style="vertical-align: -6px; border: none;"/>

A_{1,\infty } = f^{-1} \left( {\left[ {1,\infty } \right[} \right) = \left] {-\infty ,-1} \right] \cup \left[ {1,\infty } \right[
" title="
A_{1,\infty } = f^{-1} \left( {\left[ {1,\infty } \right[} \right) = \left] {-\infty ,-1} \right] \cup \left[ {1,\infty } \right[
" style="vertical-align: -6px; border: none;"/>

dritter Schritt:

f_1 \left( x \right) = 0 \cdot I_{\left] {-\frac{1}
{2},\frac{1}
{2}} \right[} \left( x \right)+\frac{1}
{2} \cdot I_{\left] {-1,-\frac{1}
{2}} \right] \cup \left[ {\frac{1}
{2},1} \right[} \left( x \right)+1 \cdot I_{\left] {-\infty ,-1} \right] \cup \left[ {1,\infty } \right[}
" title="
f_1 \left( x \right) = 0 \cdot I_{\left] {-\frac{1}
{2},\frac{1}
{2}} \right[} \left( x \right)+\frac{1}
{2} \cdot I_{\left] {-1,-\frac{1}
{2}} \right] \cup \left[ {\frac{1}
{2},1} \right[} \left( x \right)+1 \cdot I_{\left] {-\infty ,-1} \right] \cup \left[ {1,\infty } \right[}
" style="vertical-align: -12px; border: none;"/>

zweite Annäherung: k = 2

erster Schritt klar,

zweiter Schritt:

A_{2,1} = f^{-1} \left( {\left[ {0,\frac{1}
{4}} \right[} \right) = \left] {-\frac{1}
{4},\frac{1}
{4}} \right[
" title="
A_{2,1} = f^{-1} \left( {\left[ {0,\frac{1}
{4}} \right[} \right) = \left] {-\frac{1}
{4},\frac{1}
{4}} \right[
" style="vertical-align: -6px; border: none;"/>

A_{2,2} = \left] {-\frac{1}
{2},-\frac{1}
{4}} \right[ \cup \left] {\frac{1}
{4},\frac{1}
{2}} \right[
" title="
A_{2,2} = \left] {-\frac{1}
{2},-\frac{1}
{4}} \right[ \cup \left] {\frac{1}
{4},\frac{1}
{2}} \right[
" style="vertical-align: -6px; border: none;"/>

A_{2,8} = \left] {-2,-\frac{7}
{4}} \right[ \cup \left] {\frac{7}
{4},2} \right[
" title="
A_{2,8} = \left] {-2,-\frac{7}
{4}} \right[ \cup \left] {\frac{7}
{4},2} \right[
" style="vertical-align: -6px; border: none;"/>

A_{2,\infty } = \left] {-\infty ,-2} \right[ \cup \left] {2,\infty } \right[
" title="
A_{2,\infty } = \left] {-\infty ,-2} \right[ \cup \left] {2,\infty } \right[
" style="vertical-align: -6px; border: none;"/>

b )

x \mapsto \frac{1}
{{x^2 }}
" title="
x \mapsto \frac{1}
{{x^2 }}
" style="vertical-align: -7px; border: none;"/>

k = 1:

erster Schritt ist klar

zweiter Schritt:

f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right) = \left] {-\infty ,-\sqrt 2 } \right[ \cup \left] {\sqrt 2 ,\infty } \right[
" title="
f^{-1} \left( {\left[ {0,\frac{1}
{2}} \right[} \right) = \left] {-\infty ,-\sqrt 2 } \right[ \cup \left] {\sqrt 2 ,\infty } \right[
" style="vertical-align: -6px; border: none;"/>

A_{2,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right)
" title="
A_{2,2} = f^{-1} \left( {\left[ {\frac{1}
{2},1} \right[} \right)
" style="vertical-align: -6px; border: none;"/>

Aufgabe 5.2 – Lösung von separierbaren Differentialgleichungen

admin2 — Mon, 28 Sep 2009 16:07:28 +0000

Bestimme zunächst die allgemeinen Lösungen der separierbaren DGL

y ^{\prime}\left( x \right)+x^2 y\left( x \right)^3 = 0,\quad \quad y\left( 1 \right) = 2
" title="
y ^{\prime}\left( x \right)+x^2 y\left( x \right)^3 = 0,\quad \quad y\left( 1 \right) = 2
" style="vertical-align: -4px; border: none;"/>
y ^{\prime}\left( x \right)\left( {x^2 -6x+5} \right) = xy\left( x \right)+x-3y\left( x \right)-3,\quad \quad y\left( 3 \right) = 1
" title="
y ^{\prime}\left( x \right)\left( {x^2 -6x+5} \right) = xy\left( x \right)+x-3y\left( x \right)-3,\quad \quad y\left( 3 \right) = 1
" style="vertical-align: -6px; border: none;"/>
x ^{\prime}\left( t \right) = -\frac{{t\left( {1-x\left( t \right)^2 } \right)^2 }}
{{x\left( t \right)\left( {1-t^2 } \right)^2 }}
" title="
x ^{\prime}\left( t \right) = -\frac{{t\left( {1-x\left( t \right)^2 } \right)^2 }}
{{x\left( t \right)\left( {1-t^2 } \right)^2 }}
" style="vertical-align: -12px; border: none;"/>

Lösung

a )

y ^{\prime}+x^2 y^3 = 0
" title="
y ^{\prime}+x^2 y^3 = 0
" style="vertical-align: -4px; border: none;"/>

Die konstante Lösung für diese DGL ist:

{\text{y}}\left( x \right) = 0
" title="
{\text{y}}\left( x \right) = 0
" style="vertical-align: -4px; border: none;"/>

Wenn y ≠ 0, können wir dividieren:

\frac{{y ^{\prime}}}
{{y^3 }} = -x^2
" title="
\frac{{y ^{\prime}}}
{{y^3 }} = -x^2
" style="vertical-align: -10px; border: none;"/>

“Ingenieurlösung”:

\frac{{dy}}
{{dx}} \cdot \frac{1}
{{y^3 }} = -x^2
" title="
\frac{{dy}}
{{dx}} \cdot \frac{1}
{{y^3 }} = -x^2
" style="vertical-align: -10px; border: none;"/>

\int_{}^{} {\frac{1}
{{y^3 }}dy} = \int_{}^{} {-x^2 dx}
" title="
\int_{}^{} {\frac{1}
{{y^3 }}dy} = \int_{}^{} {-x^2 dx}
" style="vertical-align: -10px; border: none;"/>

-\frac{1}
{{2y^2 }} = -\frac{{x^3 }}
{3}+c
" title="
-\frac{1}
{{2y^2 }} = -\frac{{x^3 }}
{3}+c
" style="vertical-align: -10px; border: none;"/>

y = y\left( x \right) = \pm \sqrt {\frac{3}
{{-x^3 +3c}} \cdot \frac{{-1}}
{2}} = \pm \sqrt {\frac{3}
{{2x^3 +d}}}
" title="
y = y\left( x \right) = \pm \sqrt {\frac{3}
{{-x^3 +3c}} \cdot \frac{{-1}}
{2}} = \pm \sqrt {\frac{3}
{{2x^3 +d}}}
" style="vertical-align: -12px; border: none;"/>

Anfangswertproblem:

y\left( 1 \right) = 2
" title="
y\left( 1 \right) = 2
" style="vertical-align: -4px; border: none;"/>

Auflösen nach der Konstanten:

2 = \sqrt {\frac{3}
{{2+d}}}
" title="
2 = \sqrt {\frac{3}
{{2+d}}}
" style="vertical-align: -12px; border: none;"/>

4 = \frac{3}
{{2+d}}
" title="
4 = \frac{3}
{{2+d}}
" style="vertical-align: -7px; border: none;"/>

8+4d = 3
" title="
8+4d = 3
" style="vertical-align: -1px; border: none;"/>

d = -\frac{5}
{4}
" title="
d = -\frac{5}
{4}
" style="vertical-align: -6px; border: none;"/>

Daraus folgt die partikuläre Lösung:

y_p \left( x \right) = \sqrt {\frac{3}
{{2x^3 -\frac{5}
{4}}}} = \sqrt {\frac{{12}}
{{8x^3 -5}}}
" title="
y_p \left( x \right) = \sqrt {\frac{3}
{{2x^3 -\frac{5}
{4}}}} = \sqrt {\frac{{12}}
{{8x^3 -5}}}
" style="vertical-align: -14px; border: none;"/>

b )

y ^{\prime}\left( {x^2 -6x+5} \right) = xy+x-3y-3
" title="
y ^{\prime}\left( {x^2 -6x+5} \right) = xy+x-3y-3
" style="vertical-align: -6px; border: none;"/>

Das Ziel ist es, x und y zu trennen. Auf der linken Seite existieren die beiden Variablen schon als getrennte Faktoren eines Produktes. Wir müssen uns also nur um die rechte Seite kümmern. Die Variablen lassen sich wie folgt trennen:

y ^{\prime}\left( {x^2 -6x+5} \right) = x\left( {y+1} \right)-3\left( {y+1} \right)
" title="
y ^{\prime}\left( {x^2 -6x+5} \right) = x\left( {y+1} \right)-3\left( {y+1} \right)
" style="vertical-align: -6px; border: none;"/>

y ^{\prime}\left( {x^2 -6x+5} \right) = \left( {x-3} \right)\left( {y+1} \right)
" title="
y ^{\prime}\left( {x^2 -6x+5} \right) = \left( {x-3} \right)\left( {y+1} \right)
" style="vertical-align: -6px; border: none;"/>

Wir fassen die Variablen nun zusammen und integrieren:

\frac{{y ^{\prime}}}
{{\left( {y+1} \right)}} = \frac{{\left( {x-3} \right)}}
{{\left( {x^2 -6x+5} \right)}}
" title="
\frac{{y ^{\prime}}}
{{\left( {y+1} \right)}} = \frac{{\left( {x-3} \right)}}
{{\left( {x^2 -6x+5} \right)}}
" style="vertical-align: -10px; border: none;"/>

\frac{{y ^{\prime}}}
{{\left( {y+1} \right)}} = \frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}
" title="
\frac{{y ^{\prime}}}
{{\left( {y+1} \right)}} = \frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}
" style="vertical-align: -10px; border: none;"/>

\int_{}^{} {\frac{1}
{{\left( {y+1} \right)}}dy} = \int_{}^{} {\frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}dx}
" title="
\int_{}^{} {\frac{1}
{{\left( {y+1} \right)}}dy} = \int_{}^{} {\frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}dx}
" style="vertical-align: -10px; border: none;"/>

Wir müssen die Annahmen setzen:

y \ne -1
" title="
y \ne -1
" style="vertical-align: -4px; border: none;"/>

x \ne 1
" title="
x \ne 1
" style="vertical-align: -4px; border: none;"/>

x \ne 5
" title="
x \ne 5
" style="vertical-align: -4px; border: none;"/>

für das rechte Integral führen wir eine Partialbruchzerlegung durch:

\frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}} = \frac{A}
{{x-1}}+\frac{B}
{{x-5}}
" title="
\frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}} = \frac{A}
{{x-1}}+\frac{B}
{{x-5}}
" style="vertical-align: -10px; border: none;"/>

\Rightarrow \frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}} = \frac{{A\left( {x-5} \right)+B\left( {x-1} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}
" title="
\Rightarrow \frac{{\left( {x-3} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}} = \frac{{A\left( {x-5} \right)+B\left( {x-1} \right)}}
{{\left( {x-1} \right)\left( {x-5} \right)}}
" style="vertical-align: -10px; border: none;"/>

\Rightarrow \left( {x-3} \right) = A\left( {x-5} \right)+B\left( {x-1} \right)
" title="
\Rightarrow \left( {x-3} \right) = A\left( {x-5} \right)+B\left( {x-1} \right)
" style="vertical-align: -4px; border: none;"/>

x = 5 \to 2 = 4B \Rightarrow B = \frac{1}
{2}
" title="
x = 5 \to 2 = 4B \Rightarrow B = \frac{1}
{2}
" style="vertical-align: -6px; border: none;"/>

x = 1 \to -2 = -4A \Rightarrow A = \frac{1}
{2}
" title="
x = 1 \to -2 = -4A \Rightarrow A = \frac{1}
{2}
" style="vertical-align: -6px; border: none;"/>

Auflösung des Integrals:

\ln \left( {y+1} \right) = \frac{1}
{2}\ln \left( {\left( {x-1} \right)\left( {x-5} \right)} \right)+c
" title="
\ln \left( {y+1} \right) = \frac{1}
{2}\ln \left( {\left( {x-1} \right)\left( {x-5} \right)} \right)+c
" style="vertical-align: -6px; border: none;"/>

c )

x ^{\prime} \left( t \right) = -\frac{{t\left( {1-x\left( t \right)^2 } \right)^2 }}
{{x\left( t \right)\left( {1-t^2 } \right)^2 }}
" title="
x ^{\prime} \left( t \right) = -\frac{{t\left( {1-x\left( t \right)^2 } \right)^2 }}
{{x\left( t \right)\left( {1-t^2 } \right)^2 }}
" style="vertical-align: -12px; border: none;"/>

\int_{}^{} {\frac{{xdx}}
{{\left( {1-x^2 } \right)^2 }} = -\int_{}^{} {\frac{{tdt}}
{{\left( {1-t^2 } \right)^2 }}} }
" title="
\int_{}^{} {\frac{{xdx}}
{{\left( {1-x^2 } \right)^2 }} = -\int_{}^{} {\frac{{tdt}}
{{\left( {1-t^2 } \right)^2 }}} }
" style="vertical-align: -12px; border: none;"/>

-\frac{1}
{{2\left( {1-x^2 } \right)}} = \frac{1}
{{2\left( {1-t^2 } \right)}}+c,\quad \quad c \in \mathbb{R}
" title="
-\frac{1}
{{2\left( {1-x^2 } \right)}} = \frac{1}
{{2\left( {1-t^2 } \right)}}+c,\quad \quad c \in \mathbb{R}
" style="vertical-align: -10px; border: none;"/>

Die implizite Lösung lässt sich nicht eindeutig nach x = x(t) bzw. t auflösen.

konstante Lösung:

x\left( t \right) = \pm 1
" title="
x\left( t \right) = \pm 1
" style="vertical-align: -4px; border: none;"/>

Aufgabe 7.1 – Differentialgleichungssysteme aus DGL höherer Ordnung

admin2 — Mon, 28 Sep 2009 15:59:46 +0000

Überführen Sie die folgenden DGL bzw. DGL-Systeme in DGL-Systeme erster Ordnung.

y^{^{\prime\prime}} \left( x \right)+a_1 \left( x \right)y ^{\prime}\left( x \right)+a_0 \left( x \right)y\left( x \right) = b\left( x \right)
" title="
y^{^{\prime\prime}} \left( x \right)+a_1 \left( x \right)y ^{\prime}\left( x \right)+a_0 \left( x \right)y\left( x \right) = b\left( x \right)
" style="vertical-align: -4px; border: none;"/>
Ri ^{\prime}\left( t \right)+Li^{^{\prime\prime}} \left( t \right)+\frac{1}
{C}i\left( t \right) = -U_0 \omega \cdot \sin \left( {\omega t} \right),\quad \quad R,L,C,U_0 ,\omega > 0
" title="
Ri ^{\prime}\left( t \right)+Li^{^{\prime\prime}} \left( t \right)+\frac{1}
{C}i\left( t \right) = -U_0 \omega \cdot \sin \left( {\omega t} \right),\quad \quad R,L,C,U_0 ,\omega > 0
" style="vertical-align: -6px; border: none;"/>
Das System zweiter Ordnung:

x^{^{\prime\prime}} \left( t \right) = f_1 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" title="
x^{^{\prime\prime}} \left( t \right) = f_1 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" style="vertical-align: -4px; border: none;"/>

y^{^{\prime\prime}} \left( t \right) = f_2 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" title="
y^{^{\prime\prime}} \left( t \right) = f_2 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" style="vertical-align: -4px; border: none;"/>

z^{^{\prime\prime}} \left( t \right) = f_3 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" title="
z^{^{\prime\prime}} \left( t \right) = f_3 \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)
" style="vertical-align: -4px; border: none;"/>
F\left( {x,y\left( x \right),y ^{\prime} \left( x \right),y^{^{\prime\prime}} \left( x \right)} \right) = 0
" title="
F\left( {x,y\left( x \right),y ^{\prime} \left( x \right),y^{^{\prime\prime}} \left( x \right)} \right) = 0
" style="vertical-align: -12px; border: none;"/>

Lösung

a )

y^{^{\prime\prime}} = -a_1 y ^{\prime} -a_0 y+b
" title="
y^{^{\prime\prime}} = -a_1 y ^{\prime} -a_0 y+b
" style="vertical-align: -4px; border: none;"/>

\vec z = \left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)
" title="
\vec z = \left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)
" style="vertical-align: -17px; border: none;"/>

z_1 = y
" title="
z_1 = y
" style="vertical-align: -4px; border: none;"/>

z_2 = y ^{\prime} = z_1 ^{\prime}
" title="
z_2 = y ^{\prime} = z_1 ^{\prime}
" style="vertical-align: -6px; border: none;"/>

z_2 ^{\prime} = y^{^{\prime\prime}} = -a_1 z_2 -a_0 z_1 +b
" title="
z_2 ^{\prime} = y^{^{\prime\prime}} = -a_1 z_2 -a_0 z_1 +b
" style="vertical-align: -5px; border: none;"/>

\vec z ^{\prime} = \left( {\begin{array}{*{20}{c}}
z_1 ^{\prime} \\
{z_2 ^{\prime} } \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
0 & 1 \\
{-a_0 } & {-a_1 } \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
0 \\
b \\

\end{array} } \right)
" title="
\vec z ^{\prime} = \left( {\begin{array}{*{20}{c}}
z_1 ^{\prime} \\
{z_2 ^{\prime} } \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
0 & 1 \\
{-a_0 } & {-a_1 } \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
0 \\
b \\

\end{array} } \right)
" style="vertical-align: -17px; border: none;"/>

b )

i^{^{\prime\prime}} = -\frac{R}
{L}i ^{\prime} -\frac{1}
{{LC}}i-\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)
" title="
i^{^{\prime\prime}} = -\frac{R}
{L}i ^{\prime} -\frac{1}
{{LC}}i-\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)
" style="vertical-align: -6px; border: none;"/>

\vec z = \left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)
" title="
\vec z = \left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)
" style="vertical-align: -17px; border: none;"/>

z_1 = i
" title="
z_1 = i
" style="vertical-align: -4px; border: none;"/>

z_2 = i ^{\prime} = z_1 ^{\prime}
" title="
z_2 = i ^{\prime} = z_1 ^{\prime}
" style="vertical-align: -6px; border: none;"/>

z_2 ^{\prime} = i^{^{\prime\prime}} = -\frac{R}
{L}z_2 -\frac{1}
{{LC}}z_1 -\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)
" title="
z_2 ^{\prime} = i^{^{\prime\prime}} = -\frac{R}
{L}z_2 -\frac{1}
{{LC}}z_1 -\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)
" style="vertical-align: -6px; border: none;"/>

\vec z ^{\prime} = \left( {\begin{array}{*{20}{c}}
z_1 ^{\prime} \\
{z_2 ^{\prime} } \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
0 & 1 \\
{\frac{1}
{{LC}}} & {-\frac{R}
{L}} \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
0 \\
{-\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)} \\

\end{array} } \right)
" title="
\vec z ^{\prime} = \left( {\begin{array}{*{20}{c}}
z_1 ^{\prime} \\
{z_2 ^{\prime} } \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
0 & 1 \\
{\frac{1}
{{LC}}} & {-\frac{R}
{L}} \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
z_1 \\
{z_2 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
0 \\
{-\frac{{U_0 }}
{L}\omega \sin \left( {\omega t} \right)} \\

\end{array} } \right)
" style="vertical-align: -18px; border: none;"/>

c )

w_1 = x
" title="
w_1 = x
" style="vertical-align: -4px; border: none;"/>

w_2 = x ^{\prime} = w_1 ^{\prime}
" title="
w_2 = x ^{\prime} = w_1 ^{\prime}
" style="vertical-align: -6px; border: none;"/>

w_3 = y
" title="
w_3 = y
" style="vertical-align: -4px; border: none;"/>

w_4 = y ^{\prime} = w_3 ^{\prime}
" title="
w_4 = y ^{\prime} = w_3 ^{\prime}
" style="vertical-align: -5px; border: none;"/>

w_5 = z
" title="
w_5 = z
" style="vertical-align: -3px; border: none;"/>

w_6 = z ^{\prime} = w_4 ^{\prime}
" title="
w_6 = z ^{\prime} = w_4 ^{\prime}
" style="vertical-align: -5px; border: none;"/>

\vec w ^{\prime} = \left( {\begin{array}{*{20}{c}}
w_1 ^{\prime} \\
{w_2 ^{\prime} } \\
{w_3 ^{\prime} } \\
{w_4 ^{\prime} } \\
{w_5 ^{\prime} } \\
{w_6 ^{\prime} } \\

\end{array} } \right) = \left( {\begin{array}{*{20}{c}}
& 1 & {} & {} & {} & {} \\
{} & {} & {} & {} & {} & {} \\
{} & {} & {} & 1 & {} & {} \\
{} & {} & {} & {} & {} & {} \\
{} & {} & {} & {} & {} & 1 \\
{} & {} & {} & {} & {} & {} \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
w_1 \\
{w_2 } \\
{w_3 } \\
{w_4 } \\
{w_5 } \\
{w_6 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
\\
{f_1 \left( {w_1 ,w_3 ,w_5 } \right)} \\
{} \\
{f_2 \left( {w_1 ,w_3 ,w_5 } \right)} \\
{} \\
{f_3 \left( {w_1 ,w_3 ,w_5 } \right)} \\

\end{array} } \right)
" title="
\vec w ^{\prime} = \left( {\begin{array}{*{20}{c}}
w_1 ^{\prime} \\
{w_2 ^{\prime} } \\
{w_3 ^{\prime} } \\
{w_4 ^{\prime} } \\
{w_5 ^{\prime} } \\
{w_6 ^{\prime} } \\

\end{array} } \right)\left( {\begin{array}{*{20}{c}}
w_1 \\
{w_2 } \\
{w_3 } \\
{w_4 } \\
{w_5 } \\
{w_6 } \\

\end{array} } \right)+\left( {\begin{array}{*{20}{c}}
\\
{f_1 \left( {w_1 ,w_3 ,w_5 } \right)} \\
{} \\
{f_2 \left( {w_1 ,w_3 ,w_5 } \right)} \\
{} \\
{f_3 \left( {w_1 ,w_3 ,w_5 } \right)} \\

\end{array} } \right)
" style="vertical-align: -61px; border: none;"/>

d )

w_1 = y
" title="
w_1 = y
" style="vertical-align: -4px; border: none;"/>

w_2 = y ^{\prime} = w_1 ^{\prime}
" title="
w_2 = y ^{\prime} = w_1 ^{\prime}
" style="vertical-align: -6px; border: none;"/>

F\left( {x,w_1 ,w_2 ,w_2 ^{\prime} } \right) = 0
" title="
F\left( {x,w_1 ,w_2 ,w_2 ^{\prime} } \right) = 0
" style="vertical-align: -5px; border: none;"/>

G\underbrace {\left( {w_1 ^{\prime} ,w_2 } \right)}_{x,w_1 ,w_1 ^{\prime} ,w_2 ,w_2 ^{\prime} } = w_1 ^{\prime} -w_2 = 0
" title="
G\underbrace {\left( {w_1 ^{\prime} ,w_2 } \right)}_{x,w_1 ,w_1 ^{\prime} ,w_2 ,w_2 ^{\prime} } = w_1 ^{\prime} -w_2 = 0
" style="vertical-align: -37px; border: none;"/>

Aufgabe 7.2 – allgemeine Lösung von Differentialgleichungen

Mon, 28 Sep 2009 15:39:12 +0000

Bestimmen Sie die allgemeinen Lösungen der Differentialgleichungen

y^{\prime\prime} \left( x \right)-2y^{\prime} \left( x \right)+1y \left( x \right) = 0
" title="
y^{\prime\prime} \left( x \right)-2y^{\prime} \left( x \right)+1y \left( x \right) = 0
" style="vertical-align: -4px; border: none;"/>
y^{\prime\prime} \left( x \right) +2y ^{\prime} \left( x \right) -3y \left( x \right) = e^x +\sin \left( x \right)
" title="
y^{\prime\prime} \left( x \right) +2y ^{\prime} \left( x \right) -3y \left( x \right) = e^x +\sin \left( x \right)
" style="vertical-align: -4px; border: none;"/>

Lösung

a )

y^{\prime\prime} -2y ^{\prime}+y = 0
" title="
y^{\prime\prime} -2y ^{\prime}+y = 0
" style="vertical-align: -4px; border: none;"/>

charakteristische Gleichung:

\lambda ^2 -2\lambda +1 = \left( {\lambda -1} \right)^2 = 0
" title="
\lambda ^2 -2\lambda +1 = \left( {\lambda -1} \right)^2 = 0
" style="vertical-align: -4px; border: none;"/>

\Rightarrow \lambda _{1,2} = 1
" title="
\Rightarrow \lambda _{1,2} = 1
" style="vertical-align: -6px; border: none;"/>

Da es eine doppelte Nullstelle gibt, muss die Konstante variiert werden. Das Ergebnis lautet:

y\left( x \right) = y_h \left( x \right) = c_1 e^x +c_2 xe^x ,\quad \quad \quad c_1 ,c_2 \in \mathbb{R}
" title="
y\left( x \right) = y_h \left( x \right) = c_1 e^x +c_2 xe^x ,\quad \quad \quad c_1 ,c_2 \in \mathbb{R}
" style="vertical-align: -4px; border: none;"/>

b )

y^{\prime\prime} +2y ^{\prime}-3y = e^x +\sin x
" title="
y^{\prime\prime} +2y ^{\prime}-3y = e^x +\sin x
" style="vertical-align: -4px; border: none;"/>

charakteristische Gleichung:

\lambda ^2 +2\lambda -3 = 0
" title="
\lambda ^2 +2\lambda -3 = 0
" style="vertical-align: -1px; border: none;"/>

\Rightarrow \left( {\lambda +1} \right)^2 = 4\quad \quad \Rightarrow \quad \quad \lambda _{1,2} = 1,-3
" title="
\Rightarrow \left( {\lambda +1} \right)^2 = 4\quad \quad \Rightarrow \quad \quad \lambda _{1,2} = 1,-3
" style="vertical-align: -6px; border: none;"/>

y_h \left( x \right) = c_1 e^{-3x} +c_2 e^x ,\quad \quad \quad c_1 ,c_2 \in \mathbb{R}
" title="
y_h \left( x \right) = c_1 e^{-3x} +c_2 e^x ,\quad \quad \quad c_1 ,c_2 \in \mathbb{R}
" style="vertical-align: -4px; border: none;"/>

Wir suchen nun eine partikuläre Lösung durch Variation der Konstanten. Dazu nutzen wir die Fundamentalmatrix X und schreiben

y_p = Xc
" title="
y_p = Xc
" style="vertical-align: -6px; border: none;"/>

In die Differentialgleichung eingesetzt ergibt sich daraus:

y ^{\prime}= Ay+b = X ^{\prime}c+Xc ^{\prime}= AXc+b
" title="
y ^{\prime}= Ay+b = X ^{\prime}c+Xc ^{\prime}= AXc+b
" style="vertical-align: -4px; border: none;"/>

X = \left( {\begin{array}{*{20}{c}}
y_1 & {y_2 } \\
{y_1 ^{\prime}} & {y_2 ^{\prime}} \\

\end{array} } \right)
" title="
X = \left( {\begin{array}{*{20}{c}}
y_1 & {y_2 } \\
{y_1 ^{\prime}} & {y_2 ^{\prime}} \\

\end{array} } \right)
" style="vertical-align: -17px; border: none;"/>

b = \left( {\begin{array}{*{20}{c}}
0 \\
f \\

\end{array} } \right)
" title="
b = \left( {\begin{array}{*{20}{c}}
0 \\
f \\

\end{array} } \right)
" style="vertical-align: -17px; border: none;"/>

f\left( x \right) = e^x +\sin x
" title="
f\left( x \right) = e^x +\sin x
" style="vertical-align: -4px; border: none;"/>

y_1 c_1 ^{\prime}+y_2 c_2 ^{\prime}= 0\quad \quad \wedge \quad \quad y_1 ^{\prime}c_1 ^{\prime}+y_2 ^{\prime}c_2 ^{\prime}= f
" title="
y_1 c_1 ^{\prime}+y_2 c_2 ^{\prime}= 0\quad \quad \wedge \quad \quad y_1 ^{\prime}c_1 ^{\prime}+y_2 ^{\prime}c_2 ^{\prime}= f
" style="vertical-align: -6px; border: none;"/>

\left( I \right):\quad \quad \underbrace {c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = 0}_{-c_1 e^{-3x} = c_2 ^{\prime}e^x }\quad \quad \wedge \quad \quad -3c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = e^x
" title="
\left( I \right):\quad \quad \underbrace {c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = 0}_{-c_1 e^{-3x} = c_2 ^{\prime}e^x }\quad \quad \wedge \quad \quad -3c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = e^x
" style="vertical-align: -34px; border: none;"/>

-3c_1 ^{\prime}e^{-3x} -c_1 ^{\prime}e^{-3x} = -4c_1 ^{\prime}e^{-3x} = e^x
" title="
-3c_1 ^{\prime}e^{-3x} -c_1 ^{\prime}e^{-3x} = -4c_1 ^{\prime}e^{-3x} = e^x
" style="vertical-align: -6px; border: none;"/>

c_1 ^{\prime}= -\frac{1}
{4}e^{4x} \quad \quad \Rightarrow \quad \quad c_1 = -\frac{1}
{{16}}e^{4x}
" title="
c_1 ^{\prime}= -\frac{1}
{4}e^{4x} \quad \quad \Rightarrow \quad \quad c_1 = -\frac{1}
{{16}}e^{4x}
" style="vertical-align: -7px; border: none;"/>

c_2 ^{\prime}= \frac{1}
{4}e^{4x} e^{-3x} e^{-x} = \frac{1}
{4}\quad \quad \Rightarrow \quad \quad c_2 = \frac{1}
{4}x
" title="
c_2 ^{\prime}= \frac{1}
{4}e^{4x} e^{-3x} e^{-x} = \frac{1}
{4}\quad \quad \Rightarrow \quad \quad c_2 = \frac{1}
{4}x
" style="vertical-align: -6px; border: none;"/>

y_{1p} \left( x \right) = -\frac{1}
{{16}}e^{4x} e^{-3x} +\frac{1}
{4}xe^x = -\frac{1}
{{16}}e^x +\frac{1}
{4}xe^x
" title="
y_{1p} \left( x \right) = -\frac{1}
{{16}}e^{4x} e^{-3x} +\frac{1}
{4}xe^x = -\frac{1}
{{16}}e^x +\frac{1}
{4}xe^x
" style="vertical-align: -7px; border: none;"/>

Zweiter Schritt:

\left( {II} \right):\quad \quad y_1 c_1 ^{\prime}+y_2 c_2 ^{\prime}= 0\quad \quad \wedge \quad \quad y_1 ^{\prime}c_1 ^{\prime}+y_2 ^{\prime}c_2 ^{\prime}= \sin x
" title="
\left( {II} \right):\quad \quad y_1 c_1 ^{\prime}+y_2 c_2 ^{\prime}= 0\quad \quad \wedge \quad \quad y_1 ^{\prime}c_1 ^{\prime}+y_2 ^{\prime}c_2 ^{\prime}= \sin x
" style="vertical-align: -6px; border: none;"/>

-c_1 ^{\prime}e^{-3x} = c_2 ^{\prime}e^x \quad \quad \wedge \quad \quad -3c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = \sin x
" title="
-c_1 ^{\prime}e^{-3x} = c_2 ^{\prime}e^x \quad \quad \wedge \quad \quad -3c_1 ^{\prime}e^{-3x} +c_2 ^{\prime}e^x = \sin x
" style="vertical-align: -6px; border: none;"/>

-3c_1 ^{\prime}e^{-3x} -c_1 ^{\prime}e^{-3x} = -3c_1 ^{\prime}e^{-3x} = \sin x
" title="
-3c_1 ^{\prime}e^{-3x} -c_1 ^{\prime}e^{-3x} = -3c_1 ^{\prime}e^{-3x} = \sin x
" style="vertical-align: -6px; border: none;"/>

c_1 ^{\prime}= -\frac{1}
{4}e^{3x} \sin x
" title="
c_1 ^{\prime}= -\frac{1}
{4}e^{3x} \sin x
" style="vertical-align: -6px; border: none;"/>

c_1 = -\frac{1}
{4}\int_{}^{} {e^{3x} \sin xdx} = -\frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{\left( {3+i} \right)x} dx} } \right\}
" title="
c_1 = -\frac{1}
{4}\int_{}^{} {e^{3x} \sin xdx} = -\frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{\left( {3+i} \right)x} dx} } \right\}
" style="vertical-align: -6px; border: none;"/>

= -\frac{1}
{4}\operatorname{Im} \left\{ {\frac{1}
{{3+i}}e^{\left( {3+i} \right)x} } \right\} = -\frac{1}
{4}\operatorname{Im} \left\{ {\frac{{3-i}}
{{10}}e^{\left( {3+i} \right)x} } \right\}
" title="
= -\frac{1}
{4}\operatorname{Im} \left\{ {\frac{1}
{{3+i}}e^{\left( {3+i} \right)x} } \right\} = -\frac{1}
{4}\operatorname{Im} \left\{ {\frac{{3-i}}
{{10}}e^{\left( {3+i} \right)x} } \right\}
" style="vertical-align: -12px; border: none;"/>

= -\frac{1}
{4}\left( {e^{3x} \operatorname{Im} \left\{ {\frac{{3-i}}
{{10}}\left( {\cos x+i\sin x} \right)} \right\}} \right)
" title="
= -\frac{1}
{4}\left( {e^{3x} \operatorname{Im} \left\{ {\frac{{3-i}}
{{10}}\left( {\cos x+i\sin x} \right)} \right\}} \right)
" style="vertical-align: -7px; border: none;"/>

= -\frac{1}
{4}e^{3x} \left( {\frac{3}
{{10}}\sin x-\frac{1}
{{10}}\cos x} \right)
" title="
= -\frac{1}
{4}e^{3x} \left( {\frac{3}
{{10}}\sin x-\frac{1}
{{10}}\cos x} \right)
" style="vertical-align: -7px; border: none;"/>

Nun suchen wir die Funktion c₂:

c_2 ^{\prime}= \frac{1}
{4}e^{3x} \sin xe^{-3x} e^{-x} = \frac{1}
{4}e^{-x} \sin x
" title="
c_2 ^{\prime}= \frac{1}
{4}e^{3x} \sin xe^{-3x} e^{-x} = \frac{1}
{4}e^{-x} \sin x
" style="vertical-align: -6px; border: none;"/>

c_2 = \frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{-x} \sin xdx} } \right\}
" title="
c_2 = \frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{-x} \sin xdx} } \right\}
" style="vertical-align: -6px; border: none;"/>

= \frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{\left( {-1+i} \right)x} dx} } \right\}
" title="
= \frac{1}
{4}\operatorname{Im} \left\{ {\int_{}^{} {e^{\left( {-1+i} \right)x} dx} } \right\}
" style="vertical-align: -6px; border: none;"/>

= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{1}
{{-1+i}}e^{\left( {-1+i} \right)x} } \right\}
" title="
= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{1}
{{-1+i}}e^{\left( {-1+i} \right)x} } \right\}
" style="vertical-align: -12px; border: none;"/>

= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{{-1-i}}
{2}e^{\left( {-1+i} \right)x} } \right\}
" title="
= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{{-1-i}}
{2}e^{\left( {-1+i} \right)x} } \right\}
" style="vertical-align: -6px; border: none;"/>

= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{{-1-i}}
{2}e^{-x} \left( {\cos x+i\sin x} \right)} \right\}
" title="
= \frac{1}
{4}\operatorname{Im} \left\{ {\frac{{-1-i}}
{2}e^{-x} \left( {\cos x+i\sin x} \right)} \right\}
" style="vertical-align: -6px; border: none;"/>

= \frac{1}
{4}e^{-x} \left( {-\frac{1}
{2}\sin x-\frac{1}
{2}\cos x} \right)
" title="
= \frac{1}
{4}e^{-x} \left( {-\frac{1}
{2}\sin x-\frac{1}
{2}\cos x} \right)
" style="vertical-align: -6px; border: none;"/>

Wir erhalten die zweite Partikulärlösung:

y_{2p} \left( x \right) = e^{3x} \left( {-\frac{1}
{{40}}\cos x-\frac{3}
{{40}}\sin x} \right)e^{-3x} +e^{-x} \left( {-\frac{1}
{8}\sin x-\frac{3}
{{40}}\cos x} \right)e^x
" title="
y_{2p} \left( x \right) = e^{3x} \left( {-\frac{1}
{{40}}\cos x-\frac{3}
{{40}}\sin x} \right)e^{-3x} +e^{-x} \left( {-\frac{1}
{8}\sin x-\frac{3}
{{40}}\cos x} \right)e^x
" style="vertical-align: -6px; border: none;"/>

= -\frac{1}
{{10}}\cos x-\frac{1}
{5}\sin x
" title="
= -\frac{1}
{{10}}\cos x-\frac{1}
{5}\sin x
" style="vertical-align: -7px; border: none;"/>

Die Gesamtlösung ist:

y\left( x \right) = y_h +y_{1p} +y_{2p}
" title="
y\left( x \right) = y_h +y_{1p} +y_{2p}
" style="vertical-align: -6px; border: none;"/>

Aufgabe 7.4 – Eulersches Polygonzugverfahren

admin2 — Wed, 24 Jun 2009 20:19:14 +0000

Bestimmen Sie mit der Eulerschen Polygonzugmethode und der Schrittweite h = 0,1 eine Näherungslösung zum Anfangswertproblem

y^{\prime} \left( x \right) = y \left( x \right)+1,\quad \quad y\left( 0 \right) = 0,\quad \quad h = 0,1 \quad \quad \quad \quad x \in \left[ 0, 0.2 \right]
" title="
y^{\prime} \left( x \right) = y \left( x \right)+1,\quad \quad y\left( 0 \right) = 0,\quad \quad h = 0,1 \quad \quad \quad \quad x \in \left[ 0, 0.2 \right]
" style="vertical-align: -5px; border: none;"/>

Lösung

y^{\prime} = y+1,\quad \quad y\left( 0 \right) = 0,\quad \quad h = 0,1
" title="
y^{\prime} = y+1,\quad \quad y\left( 0 \right) = 0,\quad \quad h = 0,1
" style="vertical-align: -4px; border: none;"/>

Euler:

y_{i+1} = y_i +h \cdot f\left( {x_i ,y_i } \right),\quad \quad x_{i+1} = x_i +h
" title="
y_{i+1} = y_i +h \cdot f\left( {x_i ,y_i } \right),\quad \quad x_{i+1} = x_i +h
" style="vertical-align: -4px; border: none;"/>

y_1 = 0+0,1\left( {0+1} \right) = 0,1
" title="
y_1 = 0+0,1\left( {0+1} \right) = 0,1
" style="vertical-align: -4px; border: none;"/>

y_2 = 0,1+0,1\left( {0,1+1} \right) = 0,21
" title="
y_2 = 0,1+0,1\left( {0,1+1} \right) = 0,21
" style="vertical-align: -4px; border: none;"/>

\begin{array}{*{20}{c}}
i & 0 & 1 & 2 \\
{x_i } & 0 & {0,1} & {0,2} \\
{y_i } & 0 & {0,1} & {0,21} \\

\end{array}
" title="
\begin{array}{*{20}{c}}
i & 0 & 1 & 2 \\
{x_i } & 0 & {0,1} & {0,2} \\
{y_i } & 0 & {0,1} & {0,21} \\

\end{array}
" style="vertical-align: -27px; border: none;"/>

Exakte Werte:

y₁ = 0,105
y₂ = 0,221

Lösungsfunktion:

y\left( x \right) = e^x -1
" title="
y\left( x \right) = e^x -1
" style="vertical-align: -4px; border: none;"/>

Aufgabe 4.3 – Riemann und Lebesgue

admin2 — Mon, 15 Jun 2009 19:46:59 +0000

Berechne

\int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)}
" title="
\int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)}
" style="vertical-align: -10px; border: none;"/>

Lösung

x \mapsto x^4
" title="
x \mapsto x^4
" style="vertical-align: 0px; border: none;"/> ist messbar und Riemann-integrierbar, es existiert also ein Lebesgue-Integral.

\int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)} = \int_0^1 {x^4 dx} = \frac{1}
{5}
" title="
\int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)} = \int_0^1 {x^4 dx} = \frac{1}
{5}
" style="vertical-align: -10px; border: none;"/>

Beispiel, bei dem das Integral nicht Riemann-integrierbar ist:

Treppenfunktion

\frac{{j-1}}
{{2^k }} \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right)^{\frac{1}
{4}} ,\left( {\frac{j}
{{2^k }}} \right)^{\frac{1}
{4}} } \right[}
" title="
\frac{{j-1}}
{{2^k }} \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right)^{\frac{1}
{4}} ,\left( {\frac{j}
{{2^k }}} \right)^{\frac{1}
{4}} } \right[}
" style="vertical-align: -38px; border: none;"/>

Wir schreiben statt dessen

\left( {\frac{{j-1}}
{{2^k }}} \right)^4 \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right)^{{\frac{1}
{4}} ^4} ,\left( {\frac{j}
{{2^k }}} \right)^{{\frac{1}
{4}} ^4 }} \right[} = \left( {\frac{{j-1}}
{{2^k }}} \right)^4 \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right),\left( {\frac{j}
{{2^k }}} \right)} \right[}
" title="
\left( {\frac{{j-1}}
{{2^k }}} \right)^4 \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right)^{{\frac{1}
{4}} ^4} ,\left( {\frac{j}
{{2^k }}} \right)^{{\frac{1}
{4}} ^4 }} \right[} = \left( {\frac{{j-1}}
{{2^k }}} \right)^4 \cdot I_{\left[ {\left( {\frac{{j-1}}
{{2^k }}} \right),\left( {\frac{j}
{{2^k }}} \right)} \right[}
" style="vertical-align: -39px; border: none;"/>

Integral über die Treppenfunktion (Berechnung mit Lebesgue):

\sum\limits_{j = 1}^{2^k } {\left( {\frac{{j-1}}
{{2^k }}} \right)^4 \left( {\frac{j}
{{2^k }}-\frac{{j-1}}
{{2^k }}} \right)} = \frac{1}
{{2^{5k} }}\sum\limits_{j = 0}^{2^k -1} {\left( {j-1} \right)^4 } = \frac{1}
{{2^{5k} }}\sum\limits_{j = 1}^{2^k } {j^4 }
" title="
\sum\limits_{j = 1}^{2^k } {\left( {\frac{{j-1}}
{{2^k }}} \right)^4 \left( {\frac{j}
{{2^k }}-\frac{{j-1}}
{{2^k }}} \right)} = \frac{1}
{{2^{5k} }}\sum\limits_{j = 0}^{2^k -1} {\left( {j-1} \right)^4 } = \frac{1}
{{2^{5k} }}\sum\limits_{j = 1}^{2^k } {j^4 }
" style="vertical-align: -19px; border: none;"/>

= \frac{1}
{{2^{5k} }} \cdot \frac{1}
{{30}} \cdot 2^k \left( {2^k -1} \right)\left( {2^{k+1} -1} \right)\left( {3 \cdot 2^{k2} -3 \cdot 2^k -1} \right)
" title="
= \frac{1}
{{2^{5k} }} \cdot \frac{1}
{{30}} \cdot 2^k \left( {2^k -1} \right)\left( {2^{k+1} -1} \right)\left( {3 \cdot 2^{k2} -3 \cdot 2^k -1} \right)
" style="vertical-align: -7px; border: none;"/>

= \frac{1}
{{30}} \cdot \frac{{2^k }}
{{2^k }} \cdot \frac{{2^k -1}}
{{2^k }} \cdot \frac{{2^{k+1} -1}}
{{2^k }} \cdot \frac{{3 \cdot 2^{k2} -3 \cdot 2^k -1}}
{{2^{2k} }}
" title="
= \frac{1}
{{30}} \cdot \frac{{2^k }}
{{2^k }} \cdot \frac{{2^k -1}}
{{2^k }} \cdot \frac{{2^{k+1} -1}}
{{2^k }} \cdot \frac{{3 \cdot 2^{k2} -3 \cdot 2^k -1}}
{{2^{2k} }}
" style="vertical-align: -7px; border: none;"/>

Grenzübergang:

k \to \infty \quad \quad \quad \Rightarrow \quad \quad \frac{1}
{{30}} \cdot 1 \cdot 1 \cdot 2 \cdot 3 = \frac{1}
{5} = \int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)}
" title="
k \to \infty \quad \quad \quad \Rightarrow \quad \quad \frac{1}
{{30}} \cdot 1 \cdot 1 \cdot 2 \cdot 3 = \frac{1}
{5} = \int_{\left[ {0,1} \right]}^{} {x^4 d\lambda \left( x \right)}
" style="vertical-align: -10px; border: none;"/>

Aufgabe 4.5 – Satz von Fubini

admin2 — Mon, 15 Jun 2009 19:43:29 +0000

Berechne mit Hilfe des Satzes von Fubini das Integral

\int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\left( {\frac{x}
{y}-\frac{y}
{x}} \right)d\lambda ^2 \left( {x,y} \right)}
" title="
\int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\left( {\frac{x}
{y}-\frac{y}
{x}} \right)d\lambda ^2 \left( {x,y} \right)}
" style="vertical-align: -12px; border: none;"/>

Lösung

Aus

folgt wegen der Linearität

= \int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\frac{x}
{y}d\lambda ^2 \left( {x,y} \right)} -\int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\frac{y}
{x}d\lambda ^2 \left( {x,y} \right)}
" title="
= \int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\frac{x}
{y}d\lambda ^2 \left( {x,y} \right)} -\int_{\mathbb{R}^2 }^{} {I_{\left[ {1,2} \right[ \times \left[ {1,2} \right[} \left( {x,y} \right)\frac{y}
{x}d\lambda ^2 \left( {x,y} \right)}
" style="vertical-align: -9px; border: none;"/>

Mit Hilfe des Satzes von Fubini wandeln wir dies in ein Riemann-Integral um:

= \int_1^2 {\int_1^2 {\frac{x}
{y}dxdy} } -\int_1^2 {\int_1^2 {\frac{y}
{x}dydx} }
" title="
= \int_1^2 {\int_1^2 {\frac{x}
{y}dxdy} } -\int_1^2 {\int_1^2 {\frac{y}
{x}dydx} }
" style="vertical-align: -9px; border: none;"/>

Dabei wurde die Integrationsreihenfolge im zweiten Integral vertauscht. Die Berechnung ergibt:

= \int_1^2 {\frac{1}
{y}\left[ {\frac{{x^2 }}
{2}} \right]_1^2 dy} -\int_1^2 {\frac{1}
{x}\left[ {\frac{{y^2 }}
{2}} \right]_1^2 dx} = \frac{3}
{2}\ln 2-\frac{3}
{2}\ln 2 = 0
" title="
= \int_1^2 {\frac{1}
{y}\left[ {\frac{{x^2 }}
{2}} \right]_1^2 dy} -\int_1^2 {\frac{1}
{x}\left[ {\frac{{y^2 }}
{2}} \right]_1^2 dx} = \frac{3}
{2}\ln 2-\frac{3}
{2}\ln 2 = 0
" style="vertical-align: -14px; border: none;"/>