TeachingDifferential equations – Systems of linear differential equations – Fundamental matrices
If \(\{\mathbf{x}_1(t),\ldots,\mathbf{x}_n(t)\}\) is a fundamental set of \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(A\) an \(n\times n\) matrix, that is: \(\mathbf{x}_1(t),\ldots,\mathbf{x}_n(t)\) are solutions and the set \(\{\mathbf{x}_1(t),\ldots,\mathbf{x}_n(t)\}\) is linearly independent, then the matrix
\[\Psi(t):=\Bigg(\mathbf{x}_1(t)\;\ldots\;\mathbf{x}_n(t)\Bigg)\]is called a fundamental matrix of the system \(\mathbf{x}'(t)=A\mathbf{x}(t)\). Hence we have: \(\Psi'(t)=A\Psi(t)\). Further we have that \(\left|\Psi(t)\right|=W(\mathbf{x}_1,\ldots,\mathbf{x}_n)(t)\neq0\). Hence: \(\Psi(t)\) is invertible.
The general solution of \(\mathbf{x}'(t)=A\mathbf{x}(t)\) is \(\mathbf{x}(t)=c_1\mathbf{x}_1(t)+\cdots+c_n\mathbf{x}_n(t)\) with \(c_1,\ldots,c_n\in\mathbb{R}\). This can also be written as
\[\mathbf{x}(t)=\Psi(t)\mathbf{c}\quad\text{with}\quad\mathbf{c}=\begin{pmatrix}c_1\\\vdots\\c_n\end{pmatrix}.\]For the initial-value problem \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(\mathbf{x}(t_0)=\mathbf{x}_0\) we then have:
\[\mathbf{x}(t)=\Psi(t)\mathbf{c}\quad\text{with}\quad\mathbf{c}=\Psi^{-1}(t_0)\mathbf{x}_0\quad\Longrightarrow\quad \mathbf{x}(t)=\Psi(t)\Psi^{-1}(t_0)\mathbf{x}_0.\]Definition: The fundamental matrix \(\Phi(t)\) that satisfies \(\Phi'(t)=A\Phi(t)\) and \(\Phi(0)=I\) is denoted as \(\Phi(t)=\exp(At)=e^{At}\).
Hereby we have:
\[e^{At}=\sum_{n=0}^{\infty}\frac{1}{n!}(At)^n=I+At+\tfrac{1}{2}A^2t^2+\cdots\]and therefore
\[\frac{d}{dt}e^{At}=A+A^2t+\tfrac{1}{2}A^3t^2+\cdots=A(I+At+\tfrac{1}{2}A^2t^2+\cdots)=A\,e^{At}\quad\text{and}\quad e^{A0}=I.\]If \(\Psi(t)\) is an arbitrary fundamental matrix of \(\mathbf{x}'(t)=A\mathbf{x}(t)\), then we have: \(e^{At}=\Psi(t)\Psi^{-1}(0)\).
Examples:
1) Consider \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(A=\begin{pmatrix}2&1\\-5&-2\end{pmatrix}\). Then we have:
\[|A-rI|=\begin{vmatrix}2-r&1\\-5&-2-r\end{vmatrix}=r^2+1\quad\Longrightarrow\quad r=\pm i.\]Further we have:
\[r=i:\quad\begin{pmatrix}2-i&1\\-5&-2-i\end{pmatrix}\quad\Longrightarrow\quad\mathbf{v}=\begin{pmatrix}-1\\2-i\end{pmatrix}\]and then
\[\mathbf{v}e^{it}=\begin{pmatrix}-1\\2-1\end{pmatrix}\left(\cos(t)+i\sin(t)\right)=\begin{pmatrix}-\cos(t)\\2\cos(t)+\sin(t)\end{pmatrix} +i\begin{pmatrix}\sin(t)\\2\sin(t)-\cos(t)\end{pmatrix}.\]Hence: \(\Psi(t)=\begin{pmatrix}-\cos(t)&-\sin(t)\\2\cos(t)+\sin(t)&2\sin(t)-\cos(t)\end{pmatrix}\) is a fundamental matrix. Then we have:
\[\Psi(0)=\begin{pmatrix}-1&0\\2&-1\end{pmatrix}\quad\Longrightarrow\quad\Psi^{-1}(0)=\begin{pmatrix}-1&0\\-2&-1\end{pmatrix}.\]Now we have:
\[e^{At}=\Psi(t)\Psi^{-1}(0)=\begin{pmatrix}-\cos(t)&-\sin(t)\\2\cos(t)+\sin(t)&2\sin(t)-\cos(t)\end{pmatrix} \begin{pmatrix}-1&0\\-2&-1\end{pmatrix}=\begin{pmatrix}\cos(t)+2\sin(t)&\sin(t)\\-5\sin(t)&\cos(t)-2\sin(t)\end{pmatrix}.\]2) Consider \(\mathbf{x}'(t)=A\mathbf{x}(t)\) with \(A=\begin{pmatrix}-1&4\\1&2\end{pmatrix}\). Then we have:
\[|A-rI|=\begin{vmatrix}-1-r&4\\1&2-r\end{vmatrix}=r^2-r-6=(r-3)(r+2)\quad\Longrightarrow\quad r_1=3,\quad r_2=-2.\]Further we have:
\[r_1=3:\quad\begin{pmatrix}-4&4\\1&-1\end{pmatrix}\sim\begin{pmatrix}1&-1\\0&0\end{pmatrix}\quad\Longrightarrow\quad\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}\]and
\[r_2=-2:\quad\begin{pmatrix}1&4\\1&4\end{pmatrix}\sim\begin{pmatrix}1&4\\0&0\end{pmatrix}\quad\Longrightarrow\quad\mathbf{v}_2=\begin{pmatrix}-4\\1\end{pmatrix}.\]Hence: \(\Psi(t)=\begin{pmatrix}e^{3t}&-4e^{-2t}\\e^{3t}&e^{-2t}\end{pmatrix}\) is a fundamental matrix. Then we have:
\[\Psi(0)=\begin{pmatrix}1&-4\\1&1\end{pmatrix}\quad\Longrightarrow\quad\Psi^{-1}(0)=\frac{1}{5}\begin{pmatrix}1&4\\-1&1\end{pmatrix}.\]Then we have:
\[e^{At}=\Psi(t)\Psi^{-1}(0)=\frac{1}{5}\begin{pmatrix}e^{3t}&-4e^{-2t}\\e^{3t}&e^{-2t}\end{pmatrix} \begin{pmatrix}1&4\\-1&1\end{pmatrix}=\frac{1}{5}\begin{pmatrix}e^{3t}+4e^{-2t}&4e^{3t}-4e^{-2t}\\e^{3t}-e^{-2t}&4e^{3t}+e^{-2t}\end{pmatrix}.\]If the \(n\times n\) matrix \(A\) is diagonalizable, then we have: \(A=PDP^{-1}\) for some invertible matrix \(P\) and diagonal matrix \(D\).
If \(P=\Bigg(\mathbf{v}_1\;\ldots\;\mathbf{v}_n\Bigg)\) and \(D=\text{diag}(\lambda_1,\ldots,\lambda_n)\), then we have:
Hence: \(\mathbf{x}(t)=c_1\mathbf{v}e^{\lambda_1t}+\cdots+c_n\mathbf{v}_ne^{\lambda_nt}\) with \(c_1,\ldots,c_n\in\mathbb{R}\).
Now let \(\mathbf{x}(t)=P\mathbf{y}(t)\), then we have:
\[\mathbf{x}'(t)=A\mathbf{x}(t)\quad\Longleftrightarrow\quad P\mathbf{y}(t)=PDP^{-1}P\mathbf{y}(t)\quad\Longleftrightarrow\quad\mathbf{y}'(t)=D\mathbf{y}(t).\]This process is called uncoupling. The system \(\mathbf{x}'(t)=A\mathbf{x}(t)\) is a system of coupled differential equations and the system \(\mathbf{y}'(t)=D\mathbf{y}(t)\) is a system of noncoupled differential equations:
\[\left\{\begin{array}{c}y_1'(t)=\lambda_1y_1(t)\\\vdots\\y_n'(t)=\lambda_ny_n(t)\end{array}\right.\quad\Longrightarrow\quad \left\{\begin{array}{c}y_1(t)=c_1e^{\lambda_1t}\\\vdots\\y_n(t)=c_ne^{\lambda_nt}\end{array}\right.\]and therefore
\[\mathbf{y}(t)=\begin{pmatrix}y_1(t)\\\vdots\\y_n(t)\end{pmatrix}=\begin{pmatrix}c_1e^{\lambda_1t}\\\vdots\\c_ne^{\lambda_nt}\end{pmatrix}\quad\Longrightarrow\quad \mathbf{x}(t)=P\mathbf{y}(t)=\Bigg(\mathbf{v}_1\;\ldots\;\mathbf{v}_n\Bigg)\begin{pmatrix}c_1e^{\lambda_1t}\\\vdots\\c_ne^{\lambda_nt}\end{pmatrix} =c_1\mathbf{v}_1e^{\lambda_1t}+\cdots+c_n\mathbf{v}_ne^{\lambda_nt}.\]Note that: \(e^{Dt}=\text{diag}(e^{\lambda_1t},\ldots,e^{\lambda_nt})\) and \(e^{At}=Pe^{Dt}P^{-1}\).
Example: For \(A=\begin{pmatrix}-1&4\\1&2\end{pmatrix}\) we have seen that \(P=\begin{pmatrix}1&-4\\1&1\end{pmatrix}\) and \(D=\text{diag}(3,-2)\). Then we have that \(P^{-1}=\dfrac{1}{5}\begin{pmatrix}1&4\\-1&1\end{pmatrix}\) and therefore
\[e^{At}=Pe^{Dt}P^{-1}=\frac{1}{5}\begin{pmatrix}1&-4\\1&1\end{pmatrix}\begin{pmatrix}e^{3t}&0\\0&e^{-2t}\end{pmatrix} \begin{pmatrix}1&4\\-1&1\end{pmatrix}=\frac{1}{5}\begin{pmatrix}e^{3t}+4e^{-2t}&4e^{3t}-4e^{-2t}\\e^{3t}-e^{-2t}&4e^{3t}+e^{-2t}\end{pmatrix}.\]Last modified on July 1, 2021
