Topic 7 Linear systems

Given a linear system of first order differential equation with constant coefficients \[ \vec{y}'=A\vec{y}, \] the general solution is in the form \[ \vec{y}=e^{At}\vec{c}, \] where \[e^{At}=\sum\limits_{n=0}^\infty \frac{A^nt^n}{n!}.\]

It it known that for a matrix \(A\), there is a matrix \(P\) consists of eigenvectors and generalized eigenvectors such that \(PAP^{-1}\) is in the form \[ \begin{pmatrix} J_1 & & \\ &\ddots & \\ & & J_r \end{pmatrix}, \] where \(J_k\) are the Jordan blocks.

For a \(2\times 2\) matrix \(A\), if it has two distinct eigenvalues \(\lambda_1\) and \(\lambda_2\), then \[ e^{At}=\begin{pmatrix} e^{\lambda_1 t}\vec{u} & e^{\lambda_1 t}\vec{v} \end{pmatrix}P^{-1}, \] where \(\vec{u}\) and \(\vec{v}\) are eigenvectors associated to \(\lambda_1\) and \(\lambda_2\) respectively, and \(P=\begin{pmatrix}\vec{u} & \vec{v}\end{pmatrix}\).

If it has a repeated eigenvalue \(\lambda\), let \(\vec{u}\) be a eigenvector and \(\vec{v}\) a generalized eigenvector such that \((A-\lambda I)\vec{v}=\vec{u}\), then \[ e^{At}=\begin{pmatrix} e^{\lambda t}\vec{u} & te^{\lambda t}\vec{u} + e^{\lambda t}\vec{v} \end{pmatrix}P^{-1}, \] where \(P=\begin{pmatrix} \vec{u} & \vec{v} \end{pmatrix}\).

So to solve a linear system of first-order differential equations, we need to find eigenvalues and associated eigenvectors or generalized eigenvectors.

7.1 Eigenvalues and Eigenvectors

In Maple, eigenvalues, eigenvectors can be found using the command Eigenvectors which is supported by the package LinearAlgebra.

To construct a matrix or vector in Maple, one can use the Matrix() or Vector() command, or the shortcut notation <...>.

Example 7.1 Find the eigenvalues and associated generalized eigenvectors of the matrix \[ A:=\begin{pmatrix} -1&3\\-3&5 \end{pmatrix} \]

Solution. Define the matrix using the shortcut notation.

A:=< <-1, -3> | <3, 5> > # or A:=Matrix(2,2, [-1,3,-3,5])

Load the package LinearAlgebra.


Find eigenvalues and eigenvectors of \(A\).


The output is \[ {\color{blue} {\begin{bmatrix} 2\\2 \end{bmatrix},\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} }}. \]

From the output, we see that \(1\) is a repeated eigenvalue. The first colum in the square matrix is an eigenvector of the eigenvalue. Since the eigenvalue is repeated, there will be a generalized eigenvector. The following Maple codes shows how to find it.

IdM:=IdentityMatrix(2): # define a 2-by-2 identity matrix
u:=Column(Eigen[2],1): # extract the eigenvector
v:=LinearSolve(A-2*IdM, u); # solve for a generalized eigenvector

The output shows \[ {\color{blue}{ \begin{bmatrix} -\frac{1}{3}+\_t_{2} \\ \_t_{2} \end{bmatrix} }}, \] where \(\_t_{2}\) is a free variable that can take any value. For example, taking \(\_t_{2}=\frac13\) and then multiply it by 3 yields an eigenvector \[ \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \]

Exercise 7.1 Find the eigenvalues and associated eigenvectors of the matrix \[ A:=\begin{pmatrix} 5&-3\\6&-4 \end{pmatrix} \]

7.2 Solving Linear System of ODE

In Maple, to solve a linear system of first-order differential equations, one can use the command dsolve. Note that in Maple, the matrix multiplication operator is the . symbol instead of the * symbol. The * symbol is to be used for scalar multiplication.

Example 7.2 Solve the linear system \[ \begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} -1&3\\-3&5 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}. \]

Solution. Define the linear system in matrix form.

M := Matrix(2, 2, [-1, 3, -3, 5]):
Y := <y[1](x), y[2](x)>:
LinSys := diff(Y, x) = M . Y;

Solve the linear system

dsolve(LinSys, Y);

The output shows the general solution as

\[ {\color{blue} {\begin{pmatrix} y_1(x \\ y_2(x) \end{pmatrix} = \begin{pmatrix} \frac{e^{2 x}(3 \_C2 x +3 \_C1 -\_C2 )}{3} \\ e^{2 x}(\_C2 x +\_C1 ) \end{pmatrix} }}. \]

Exercise 7.2 Solve the linear system \[ \begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} 5&-3\\4&-2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}. \]