10.2: Linear Systems of Differential Equations

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A first order system of differential equations that can be written in the form

\[\label{eq:10.2.1} \begin{array}{ccl} y'_1&=&a_{11}(t)y_1+a_{12}(t)y_2+\cdots+a_{1n}(t)y_n+f_1(t)\\[4pt] y'_2&=&a_{21}(t)y_1+a_{22}(t)y_2+\cdots+a_{2n}(t)y_n+f_2(t)\\[4pt] &\vdots\\[4pt] y'_n& =&a_{n1}(t)y_1+a_{n2}(t)y_2+\cdots+a_{nn}(t)y_n+f_n(t)\end{array} \]

is called a linear system.

The linear system Equation \ref{eq:10.2.1} can be written in matrix form as

\[\col{y'}n=\matfunc ann\col yn+\colfunc fn, \nonumber \]

or more briefly as

\[\label{eq:10.2.2} {\bf y}'=A(t){\bf y}+{\bf f}(t), \]

where

\[\bf y=\col yn,\quad A(t)=\matfunc ann,\quad \text{and} \quad{\bf f}(t)=\colfunc fn. \nonumber \]

We call \(A\) the coefficient matrix of Equation \ref{eq:10.2.2} and \({\bf f}\) the forcing function. We’ll say that \(A\) and \({\bf f}\) are continuous if their entries are continuous. If \(\bf f={\bf 0}\), then Equation \ref{eq:10.2.2} is homogeneous; otherwise, Equation \ref{eq:10.2.2} is nonhomogeneous.

An initial value problem for Equation \ref{eq:10.2.2} consists of finding a solution of Equation \ref{eq:10.2.2} that equals a given constant vector

\[\bf k =\col kn. \nonumber \]

at some initial point \(t_0\). We write this initial value problem as

\[\bf y'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)={\bf k}.\nonumber \]

The next theorem gives sufficient conditions for the existence of solutions of initial value problems for Equation \ref{eq:10.2.2}. We omit the proof.

Theorem 10.2.1 : Existence

Suppose the coefficient matrix \(A\) and the forcing function \({\bf f}\) are continuous on \((a,b)\), let \(t_0\) be in \((a,b)\), and let \({\bf k}\) be an arbitrary constant \(n\)-vector. Then the initial value problem

\[\bf y'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)= \bf k \nonumber \]

has a unique solution on \((a,b)\).

Example 10.2.1

1. Write the system \[\label{eq:10.2.3} \begin{array}{rcl} y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\[4pt] y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t} \end{array} \] in matrix form and conclude from Theorem 10.2.1 that every initial value problem for Equation \ref{eq:10.2.3} has a unique solution on \((-\infty,\infty)\).
2. Verify that \[\label{eq:10.2.4} {\bf y}= {1\over5}\twocol87e^{4t}+c_1\twocol11e^{3t}+c_2\twocol1{-1}e^{-t} \] is a solution of Equation \ref{eq:10.2.3} for all values of the constants \(c_1\) and \(c_2\).
3. Find the solution of the initial value problem \[\label{eq:10.2.5} {\bf y}'=\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right] {\bf y}+\twocol21e^{4t},\quad {\bf y}(0)={1\over5}\twocol3{22}. \]

Solution a

The system Equation \ref{eq:10.2.3} can be written in matrix form as

\[{\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}.\nonumber \]

An initial value problem for Equation \ref{eq:10.2.3} can be written as

\[{\bf y}'=\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right] {\bf y}+\twocol21e^{4t}, \quad y(t_0)=\twocol{k_1}{k_2}. \nonumber \]

Since the coefficient matrix and the forcing function are both continuous on \((-\infty,\infty)\), Theorem 10.2.1 implies that this problem has a unique solution on \((-\infty,\infty)\).

Solution b

If \({\bf y}\) is given by Equation \ref{eq:10.2.4}, then

\[\begin{align*} A{\bf y}+{\bf f}&= {1\over5}\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right]\twocol87e^{4t}+ c_1\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right]\twocol11e^{3t} +c_2\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right]\twocol1{-1}e^{-t} +\twocol21e^{4t}\\[4pt] &= {1\over5}\twocol{22}{23}e^{4t}+c_1\twocol33e^{3t}+c_2\twocol{-1}1e^{-t} +\twocol21e^{4t}\\[4pt] &= {1\over5}\twocol{32}{28}e^{4t}+3c_1\twocol11e^{3t}-c_2\twocol1{-1}e^{-t} \\[4pt] &={\bf y}'.\end{align*} \]

Solution c

We must choose \(c_1\) and \(c_2\) in Equation \ref{eq:10.2.4} so that

\[{1\over5}\twocol87+c_1\twocol11+c_2\twocol1{-1}={1\over5}\twocol3{22},\nonumber \]

which is equivalent to

\[ \left[\begin{array}{cc}{1}&{1}\\[4pt]{1}&{-1}\end{array} \right] \twocol{c_1}{c_2}=\twocol{-1}3.\nonumber \]

Solving this system yields \(c_1=1\), \(c_2=-2\), so

\[{\bf y}={1\over5}\twocol87e^{4t}+\twocol11e^{3t}-2\twocol1{-1}e^{-t}\nonumber \]

is the solution of Equation \ref{eq:10.2.5}.

Note

The theory of \(n \times n\) linear systems of differential equations is analogous to the theory of the scalar n-th order equation \[\label{eq:10.2.6} P_{0}(t)y^{(n)}+P_{1}(t)y^{(n-1)}+\cdots +P_{n}(t)y=F(t) \] as developed in Sections 9.1. For example by rewriting Equation \ref{eq:10.2.6} as an equivalent linear system it can be shown that Theorem 10.2.1 implies Theorem 9.1.1.