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Aufgabe 7.3 – Differentialgleichung in Matrixform

Gegeben sei

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

Bestimmen Sie alle Lösungen von

x ^{\prime}\left( t \right) = Ax

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)

x ^{\prime}= Ax

Charakteristische Gleichung:

\mathcal{X}_A \left( \lambda \right) = \det \left( {A-\lambda I} \right)

= \left( {-2-\lambda } \right)^3 +2\left( {-2-\lambda } \right)

= \left( {-2-\lambda } \right)\left( {\left( {-2-\lambda } \right)^2 +2} \right)

= \left( {-2-\lambda } \right)\left( {\lambda ^2 +4\lambda +6} \right) = 0

\Rightarrow \quad \lambda _1 = -2,\quad \lambda _{2,3} = \frac{{-4 \pm \sqrt {16-24} }} {2} = -2 \pm \sqrt 2 i

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}

Wir müssen nun noch die Konstanten ci bestimmen.

Eigenvektoren:

Av = \lambda v\quad \quad \Rightarrow \quad \quad \left( {A-\lambda I} \right)v = 0

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

v_2 = 0

v_1 = v_3 = 1

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

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

\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

Daraus folgt:

w_2 = \sqrt 2 iw_3

-\sqrt {2i} w_1 = w_2

Wir definieren:

w_2 : = 1

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)

Das komplex konjugierte:

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

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}

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)

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)

Dies könnte man theoretisch noch ausrechnen. Die Werte gehen für t\to\infty alle gegen 0, da das t überall als negativer Exponent der e-Funktion vorkommt.

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