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.