Differential equations – Systems of linear differential equations – Defect matrices 2

Now we consider two different situations where the \(3\times3\) matrix \(A\) has a triple eigenvalue:

1. when the eigenspace is \(1\)-dimensional, then there is only one linearly independent eigenvector and we need two generalized eigenvectors;


2. when the eigenspace is \(2\)-dimensional, then there are two linearly independent eigenvectors and we need only one generalized eigenvector, which however should be chosen carefully.

Examples

1) Consider \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(A=\begin{pmatrix}3&2&-1\\-3&-2&2\\-1&-1&2\end{pmatrix}\). Then we have:

\begin{align*} |A-rI|&=\begin{vmatrix}3-r&2&-1\\-3&-2-r&2\\-1&-1&2-r\end{vmatrix}=\begin{vmatrix}1-r&2&-1\\-1+r&-2-r&2\\0&-1&2-r\end{vmatrix} =\begin{vmatrix}1-r&2&-1\\0&-r&1\\0&-1&2-r\end{vmatrix}\\[2.5mm] &=(1-r)\begin{vmatrix}-r&1\\-1&2-r\end{vmatrix}=(1-r)(r^2-2r+1)=(1-r)^3. \end{align*}

Hence: \(A\) has a triple eigenvalue \(r=1\). Further we have:

\[r=1:\quad\begin{pmatrix}2&2&-1\\-3&-3&2\\-1&-1&1\end{pmatrix}\sim\begin{pmatrix}1&1&0\\0&0&1\\0&0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_1=\text{Span}\left\{\begin{pmatrix}1\\-1\\0\end{pmatrix}\right\}.\]

Hence: \(\mathbf{x}_1(t)=\begin{pmatrix}1\\-1\\0\end{pmatrix}e^t\) is a solution. A second solution has the form \(\mathbf{x}_2(t)=(\mathbf{v}_1t+\mathbf{v}_2)e^{rt}=\mathbf{v}_1te^t+\mathbf{v}_2e^t\). Substitution then gives:

\[\mathbf{v}_1te^t+(\mathbf{v}_1+\mathbf{v}_2)e^t=A\mathbf{v}_1te^t+A\mathbf{v}_2\quad\Longleftrightarrow\quad A\mathbf{v}_1=\mathbf{v}_1\quad\text{and}\quad A\mathbf{v}_2=\mathbf{v}_1+\mathbf{v}_2\quad\Longleftrightarrow\quad (A-I)\mathbf{v}_1=\mathbf{0}\quad\text{and}\quad(A-I)\mathbf{v}_2=\mathbf{v}_1.\]

So the vector \(\mathbf{v}_1\) is an eigenvector of \(A\) corresponding to the eigenvalue \(1\) and \(\mathbf{v}_2\) is a corresponding generalized eigenvector.

Now choose \(\mathbf{v}_1=\begin{pmatrix}1\\-1\\0\end{pmatrix}\), then we have:

\[(A-I)\mathbf{v}_2=\mathbf{v}_1:\quad\left(\left.\begin{matrix}2&2&-1\\-3&-3&2\\-1&-1&1\end{matrix}\,\right|\,\begin{matrix}1\\-1\\0\end{matrix}\right) \sim\left(\left.\begin{matrix}1&1&0\\0&0&1\\0&0&0\end{matrix}\,\right|\,\begin{matrix}1\\1\\0\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{v}_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\quad\text{for instance.}\]

So a second solution is \(\mathbf{x}_2(t)=\mathbf{v}_1te^t+\mathbf{v}_2e^t =\begin{pmatrix}1\\-1\\0\end{pmatrix}te^t+\begin{pmatrix}0\\1\\1\end{pmatrix}e^t=\begin{pmatrix}t\\1-t\\1\end{pmatrix}e^t\).

A third solution has the form \(\mathbf{x}_3(t)=\left(\frac{1}{2}\mathbf{v}_1t^2+\mathbf{v}_2t+\mathbf{v}_3\right)e^{rt} =\frac{1}{2}\mathbf{v}_1t^2e^t+\mathbf{v}_2te^t+\mathbf{v}_3e^t\). Substitution then gives:

\[\tfrac{1}{2}\mathbf{v}_1t^2e^t+(\mathbf{v}_1+\mathbf{v}_2)te^t+(\mathbf{v}_2+\mathbf{v}_3)e^t=\tfrac{1}{2}A\mathbf{v}_1t^2e^t +A\mathbf{v}_2te^t+A\mathbf{v}_3e^t.\]

This implies:

\[A\mathbf{v}_1=\mathbf{v}_1,\quad A\mathbf{v}_2=\mathbf{v}_1+\mathbf{v}_2\quad\text{en}\quad A\mathbf{v}_3=\mathbf{v}_2+\mathbf{v}_3 \quad\Longleftrightarrow\quad(A-I)\mathbf{v}_1=\mathbf{0},\quad(A-I)\mathbf{v}_2=\mathbf{v}_1\quad\text{en}\quad(A-I)\mathbf{v}_3=\mathbf{v}_2.\]

If we now choose \(\mathbf{v}_1=\begin{pmatrix}1\\-1\\0\end{pmatrix}\) and \(\mathbf{v}_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\) as above, then we have:

\[(A-I)\mathbf{v}_3=\mathbf{v}_2:\quad\left(\left.\begin{matrix}2&2&-1\\-3&-3&2\\-1&-1&1\end{matrix}\,\right|\,\begin{matrix}0\\1\\1\end{matrix}\right) \sim\left(\left.\begin{matrix}1&1&0\\0&0&1\\0&0&0\end{matrix}\,\right|\,\begin{matrix}1\\2\\0\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{v}_3=\begin{pmatrix}1\\0\\2\end{pmatrix}\quad\text{for instance.}\]

So a third solution is \(\mathbf{x}_3(t)=\frac{1}{2}\mathbf{v}_1t^2e^t+\mathbf{v}_2te^t+\mathbf{v}_3e^t =\frac{1}{2}\begin{pmatrix}1\\-1\\0\end{pmatrix}t^2e^t+\begin{pmatrix}0\\1\\1\end{pmatrix}te^t+\begin{pmatrix}1\\0\\2\end{pmatrix}e^t =\begin{pmatrix}\frac{1}{2}t^2+1\\-\frac{1}{2}t^2+t\\t+2\end{pmatrix}e^t\).

Now the general solution is \(\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+c_3\mathbf{x}_3(t)\) or equivalently

\[\mathbf{x}(t)=c_1\begin{pmatrix}1\\-1\\0\end{pmatrix}e^t+c_2\begin{pmatrix}t\\1-t\\1\end{pmatrix}e^t+c_3\begin{pmatrix}\frac{1}{2}t^2+1\\-\frac{1}{2}t^2+t\\t+2\end{pmatrix}e^t, \quad c_1,c_2,c_3\in\mathbb{R}.\]

For the matrix \(A\) we now have: \(A=PJP^{-1}\) with \(P=\begin{pmatrix}1&0&1\\-1&1&0\\0&1&2\end{pmatrix}\) and \(J=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}\). Dan geldt: \(P^{-1}=\begin{pmatrix}2&1&-1\\2&2&-1\\-1&-1&1\end{pmatrix}\).

Now let \(\mathbf{x}(t)=P\mathbf{y}(t)\), then we have:

\[\mathbf{x}'(t)=A\mathbf{x}(t)\quad\Longleftrightarrow\quad\mathbf{y}'(t)=J\mathbf{y}(t)\quad\Longleftrightarrow\quad \left\{\begin{array}{l}y_1'(t)=y_1(t)+y_2(t)\\[0.5mm]y_2'(t)=y_2(t)+y_3(t)\\[0.5mm]y_3'(t)=y_3(t)\end{array}\right.\quad\Longleftrightarrow\quad \mathbf{y}(t)=\begin{pmatrix}y_1(t)\\y_2(t)\\y_3(t)\end{pmatrix}=\begin{pmatrix}c_1e^t+c_2te^t+\frac{1}{2}c_3t^2e^t\\c_2e^t+c_3te^t\\c_3e^t\end{pmatrix}.\]

Hence: \(e^{Jt}=\begin{pmatrix}1&t&\frac{1}{2}t^2\\0&1&t\\0&0&1\end{pmatrix}e^t\). Then we have:

\[e^{At}=Pe^{Jt}P^{-1}=\begin{pmatrix}1&0&1\\-1&1&0\\0&1&2\end{pmatrix}\begin{pmatrix}1&t&\frac{1}{2}t^2\\0&1&t\\0&0&1\end{pmatrix} \begin{pmatrix}2&1&-1\\2&2&-1\\-1&-1&1\end{pmatrix}e^t=\begin{pmatrix}1+2t-\frac{1}{2}t^2&2t-\frac{1}{2}t^2&-t+\frac{1}{2}t^2\\ -3t+\frac{1}{2}t^2&1-3t+\frac{1}{2}t^2&2t-\frac{1}{2}t^2\\-t&-t&1+t\end{pmatrix}e^t.\]

2) Consider \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(A=\begin{pmatrix}-1&0&0\\-1&-2&1\\-1&-1&0\end{pmatrix}\). Then we have:

\[|A-rI|=\begin{vmatrix}-1-r&0&0\\-1&-2-r&1\\-1&-1&-r\end{vmatrix}=(-1-r)\begin{vmatrix}-2-r&1\\-1&-r\end{vmatrix} =-(1+r)(r^2+2r+1)=-(1+r)^3.\]

Hence: \(A\) has a triple eigenvalue \(r=-1\). Further we have:

\[r=-1:\quad\begin{pmatrix}0&0&0\\-1&-1&1\\-1&-1&1\end{pmatrix}\sim\begin{pmatrix}1&1&-1\\0&0&0\\0&0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_{-1}=\text{Span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix}\right\}.\]

Hence: \(\mathbf{x}_1(t)=\begin{pmatrix}1\\0\\1\end{pmatrix}e^{-t}\) and \(\mathbf{x}_2(t)=\begin{pmatrix}0\\1\\1\end{pmatrix}e^{-t}\) are solutions. A third solution has the form \(\mathbf{x}_3(t)=(\mathbf{u}t+\mathbf{v})e^{-t}\). Substitution gives:

\[-\mathbf{u}te^{-t}+(\mathbf{u}-\mathbf{v})e^{-t}=A\mathbf{u}te^{-t}+A\mathbf{v}e^{-t}\quad\Longleftrightarrow\quad A\mathbf{u}=-\mathbf{u}\quad\text{and}\quad A\mathbf{v}=\mathbf{u}-\mathbf{v}\quad\Longleftrightarrow\quad (A+I)\mathbf{u}=\mathbf{0}\quad\text{and}\quad(A+I)\mathbf{v}=\mathbf{u}.\]

Hence: \(\mathbf{u}\) is an eigenvector of \(A\) corresponding to the eigenvalue \(-1\) and \(\mathbf{v}\) is a corresponding generalized eigenvector. Now we need to choose the eigenvector \(\mathbf{u}=\alpha\begin{pmatrix}1\\0\\1\end{pmatrix}+\beta\begin{pmatrix}0\\1\\1\end{pmatrix} =\begin{pmatrix}\alpha\\\beta\\\alpha+\beta\end{pmatrix}\) such that \((A+I)\mathbf{v}=\mathbf{u}\) is consistent:

\[(A+I)\mathbf{v}=\mathbf{u}:\;\;\left(\left.\begin{matrix}0&0&0\\-1&-1&1\\-1&-1&1\end{matrix}\,\right|\,\begin{matrix}\alpha\\\beta\\\alpha+\beta\end{matrix}\right) \;\;\Longrightarrow\;\;\alpha=0\;\;\text{en}\;\;\beta=1\;\;\text{bijvoorbeeld}\;\;\Longrightarrow\;\;\mathbf{u}=\begin{pmatrix}0\\1\\1\end{pmatrix} \;\;\text{and}\;\;\mathbf{v}=\begin{pmatrix}0\\0\\1\end{pmatrix}\;\;\text{for instance.}\]

Hence: \(\mathbf{x}_3(t)=(\mathbf{u}t+\mathbf{v})e^{-t}=\begin{pmatrix}0\\1\\1\end{pmatrix}te^{-t}+\begin{pmatrix}0\\0\\1\end{pmatrix}e^{-t} =\begin{pmatrix}0\\t\\t+1\end{pmatrix}e^{-t}\) is a third solution.

Now the general solution is \(\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+c_3\mathbf{x}_3(t)\) or equivalently

\[\mathbf{x}(t)=c_1\begin{pmatrix}1\\0\\1\end{pmatrix}e^{-t}+c_2\begin{pmatrix}0\\1\\1\end{pmatrix}e^{-t}+c_3\begin{pmatrix}0\\t\\t+1\end{pmatrix}e^{-t}, \quad c_1,c_2,c_3\in\mathbb{R}.\]

For the matrix \(A\) we now have: \(A=PJP^{-1}\) with \(P=\begin{pmatrix}1&0&0\\0&1&0\\1&1&1\end{pmatrix}\) and \(J=\begin{pmatrix}-1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}\). Then we have: \(P^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\-1&-1&1\end{pmatrix}\).

Now let \(\mathbf{x}(t)=P\mathbf{y}(t)\), then we have:

\[\mathbf{x}'(t)=A\mathbf{x}(t)\quad\Longleftrightarrow\quad\mathbf{y}'(t)=J\mathbf{y}(t)\quad\Longleftrightarrow\quad \left\{\begin{array}{l}y_1'(t)=-y_1(t)\\[0.5mm]y_2'(t)=-y_2(t)+y_3(t)\\[0.5mm]y_3'(t)=-y_3(t)\end{array}\right.\quad\Longleftrightarrow\quad \mathbf{y}(t)=\begin{pmatrix}y_1(t)\\y_2(t)\\y_3(t)\end{pmatrix}=\begin{pmatrix}c_1e^{-t}\\c_2e^{-t}+c_3te^{-t}\\c_3e^{-t}\end{pmatrix}.\]

Hence: \(e^{Jt}=\begin{pmatrix}1&0&0\\0&1&t\\0&0&1\end{pmatrix}e^{-t}\). Then we have:

\[e^{At}=Pe^{Jt}P^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\1&1&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&t\\0&0&1\end{pmatrix} \begin{pmatrix}1&0&0\\0&1&0\\-1&-1&1\end{pmatrix}e^{-t}=\begin{pmatrix}1&0&0\\1-2t&1-t&t\\-t&-t&1+t\end{pmatrix}e^{-t}.\]

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Last modified on July 1, 2021