10.2 Linear Systems of Differential Equations

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

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

is called a linear system.

The linear system (10.2.1) can be written in matrix form as

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

or more briefly as

\begin{equation} {\bf y}'=A(t){\bf y}+{\bf f}(t), \tag{10.2.2}\end{equation}

where

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

We call \(A\) the coefficient matrix of (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 (10.2.2) is homogeneous; otherwise, (10.2.2) is nonhomogeneous.

An initial value problem for (10.2.2) consists of finding a solution of (10.2.2) that equals a given constant vector

\[ {\bf k} =\col kn. \]

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

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

The next theorem gives sufficient conditions for the existence of solutions of initial value problems for (10.2.2). We omit the proof.

Theorem 10.2.1

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

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

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

Example 10.2.1

  1. Write the system

    \begin{equation} \begin{array}{rcl} y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\[3pt] y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t} \end{array} \tag{10.2.3}\end{equation}

    in matrix form and conclude from Theorem 10.2.1 that every initial value problem for (10.2.3) has a unique solution on \((-\infty,\infty)\).

  2. Verify that

    \begin{equation} {\bf y}= {1\over5}\twocol87e^{4t}+c_1\twocol11e^{3t}+c_2\twocol1{-1}e^{-t} \tag{10.2.4}\end{equation}

    is a solution of (10.2.3) for all values of the constants \(c_1\) and \(c_2\).

  3. Find the solution of the initial value problem

    \begin{equation} {\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t},\quad {\bf y}(0)={1\over5}\twocol3{22}. \tag{10.2.5}\end{equation}

Solution (a) The system (10.2.3) can be written in matrix form as

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

An initial value problem for (10.2.3) can be written as

\[ {\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}, \quad y(t_0)=\twocol{k_1}{k_2}. \]

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 (10.2.4), then

\begin{eqnarray*} A{\bf y}+{\bf f}&=& {1\over5}\twobytwo1221\twocol87e^{4t}+ c_1\twobytwo1221\twocol11e^{3t} \\[6pt] &&+c_2\twobytwo1221\twocol1{-1}e^{-t} +\twocol21e^{4t} \\[6pt] &=&{1\over5}\twocol{22}{23}e^{4t}+c_1\twocol33e^{3t}+c_2\twocol{-1}1e^{-t} +\twocol21e^{4t} \\[6pt] &=&{1\over5}\twocol{32}{28}e^{4t}+3c_1\twocol11e^{3t}-c_2\twocol1{-1}e^{-t} ={\bf y}'. \end{eqnarray*}

Solution (c) We must choose \(c_1\) and \(c_2\) in (10.2.4) so that

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

which is equivalent to

\[ \twobytwo111{-1}\twocol{c_1}{c_2}=\twocol{-1}3. \]

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

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

is the solution of (10.2.5).

Remark

The theory of \(n\times n\) linear systems of differential equations is analogous to the theory of the scalar \(n\)-th order equation

\begin{equation} P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t), \tag{10.2.6}\end{equation}

as developed in Sections 9.1. For example, by rewriting (10.2.6) as an equivalent linear system it can be shown that Theorem 10.2.1 implies Theorem 9.1.1 (Exercise 12).

10.2 Exercises

  1. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants \(c_1\) and \(c_2\).

    1. \(\begin{array}{ccl}y'_1&=&2y_1 + 4y_2\\ y_2'&=&4y_1+2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{6t}+c_2\twocol1{-1}e^{-2t}\)

    2. \(\begin{array}{ccl}y'_1&=&-2y_1 - 2y_2\\ y_2'&=&-5y_1 + \phantom{2}y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{-4t}+c_2\twocol{-2}5e^{3t}\)

    3. \(\begin{array}{ccr}y'_1&=&-4y_1 -10y_2\\ y_2'&=&3y_1 + \phantom{1}7y_2;\end{array} \quad {\bf y}=c_1\twocol{-5}3e^{2t}+c_2\twocol2{-1}e^t\)

    4. \(\begin{array}{ccl}y'_1&=&2y_1 +\phantom{2}y_2 \\ y_2'&=&\phantom{2}y_1 + 2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{3t}+c_2\twocol1{-1}e^t\)

    Show answer

    (a) \({\bf y}'=\twobytwo2442{\bf y}\)     (b) \({\bf y}'=\twobytwo{-2}{-2}{-5}1{\bf y}\)

       (c) \({\bf y}'=\twobytwo{-4}{-10}37{\bf y}\)  (d) \({\bf y}'=\twobytwo2112{\bf y}\)

  2. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants \(c_1\), \(c_2\), and \(c_3\).

    1. \(\begin{array}{ccr}y'_1&=&- y_1+2y_2 + 3y_3 \\ y_2'&=&y_2 + 6y_3\\y_3'&=&- 2y_3;\end{array}\)

      \({\bf y}=c_1\threecol110e^t+c_2\threecol100e^{-t}+c_3\threecol1{-2}1e^{-2t}\)

    2. \(\begin{array}{ccc}y'_1&=&\phantom{2y_1+}2y_2 + 2y_3 \\ y_2'&=&2y_1\phantom{+2y_2} + 2y_3\\y_3'&=&2y_1 + 2y_2;\phantom{+2y_3}\end{array}\)

      \({\bf y}=c_1\threecol{-1}01e^{-2t}+c_2\threecol0{-1}1e^{-2t}+c_3\threecol111e^{4t}\)

    3. \(\begin{array}{ccr}y'_1&=&-y_1 +2y_2 + 2y_3\\ y_2'&=&2y_1 -\phantom{2}y_2 +2y_3\\y_3'&=&2y_1 + 2y_2 -\phantom{2}y_3;\end{array}\)

      \({\bf y}=c_1\threecol{-1}01e^{-3t}+c_2\threecol0{-1}1e^{-3t}+c_3\threecol111e^{3t}\)

    4. \(\begin{array}{ccr}y'_1&=&3y_1 - \phantom{2}y_2 -\phantom{2}y_3 \\ y_2'&=&-2y_1 + 3y_2 + 2y_3\\y_3'&=&\phantom{-}4y_1 -\phantom{3}y_2 - 2y_3;\end{array}\)

      \({\bf y}=c_1\threecol101e^{2t}+c_2\threecol1{-1}1e^{3t}+c_3\threecol1{-3}7e^{-t}\)

    Show answer

    (a) \({\bf y}'=\threebythree{-1}2301600{-2}{\bf y}\)   (b) \({\bf y}'=\threebythree022202220{\bf y}\)

       (c) \({\bf y}'=\threebythree{-1}222{-1}222{-1}{\bf y}\)   (d) \({\bf y}'=\threebythree3{-1}{-1}{-2}324{-1}{-2}{\bf y}\)

  3. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution.

    1. \(\begin{array}{ccl}y'_1 &=&\phantom{-2}y_1+\phantom{4}y_2\\ y_2'&=&-2y_1 + 4y_2,\end{array} \begin{array}{ccr}y_1(0)&=&1\\y_2(0)&=&0;\end{array}\)   \({\bf y}=2\twocol11e^{2t}-\twocol12e^{3t}\)

    2. \(\begin{array}{ccl}y'_1 &=&5y_1 + 3y_2 \\ y_2'&=&- y_1 + y_2,\end{array} \begin{array}{ccr}y_1(0)&=&12\\y_2(0)&=&-6;\end{array}\)   \({\bf y}=3\twocol1{-1}e^{2t}+3\twocol3{-1}e^{4t}\)

    Show answer

    (a) \({\bf y}'=\twobytwo11{-2}4{\bf y},\; {\bf y}(0)=\twocol10\)   (b) \({\bf y}'=\twobytwo53{-1}1{\bf y},\; {\bf y}(0)=\twocol9{-5}\)

  4. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution.

    1. \(\begin{array}{ccr}y'_1&=&6y_1 + 4y_2 + 4y_3 \\ y_2'&=&-7y_1 -2y_2 - y_3,\\y_3'&=&7y_1 + 4y_2 + 3y_3\end{array},\; \begin{array}{ccr}y_1(0)&=&3\\ y_2(0)&=&-6\\ y_3(0)&=&4\end{array}\)

      \({\bf y}=\threecol1{-1}1e^{6t}+2\threecol1{-2}1e^{2t}+\threecol0{-1}1e^{-t}\)

    2. \(\begin{array}{ccr}y'_1&=& \phantom{-}8y_1 + 7y_2 +\phantom{1}7y_3 \\ y_2'&=&-5y_1 -6y_2 -\phantom{1}9y_3,\\y_3'&=& \phantom{-}5y_1 + 7y_2 +10y_3,\end{array}\ \begin{array}{ccr}y_1(0)&=&2\\ y_2(0)&=&-4\\ y_3(0)&=&3\end{array}\)

      \({\bf y}=\threecol1{-1}1e^{8t}+\threecol0{-1}1e^{3t}+\threecol1{-2}1e^t\)

    Show answer

    (a) \({\bf y}'=\threebythree644{-7}{-2}{-1}743{\bf y},\; {\bf y}(0) =\threecol3{-6}4\)

       (b) \({\bf y}'=\threebythree877{-5}{-6}{-9}57{10}{\bf y},\; {\bf y}(0) =\threecol2{-4}3\)

  5. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants \(c_1\) and \(c_2\).

    1. \(\begin{array}{ccc}y'_1&=&-3y_1+2y_2+3-2t \\ y_2'&=&-5y_1+3y_2+6-3t\end{array}\)

      \({\bf y}=c_1\left[\begin{array}{c}2\cos t\\3\cos t-\sin t\end{array}\right]+c_2\left[\begin{array}{c}2\sin t\\3\sin t+\cos t \end{array}\right]+\twocol1t\)

    2. \(\begin{array}{ccc}y'_1&=&3y_1+y_2-5e^t \\ y_2'&=&-y_1+y_2+e^t\end{array}\)

      \({\bf y}=c_1\twocol{-1}1e^{2t}+c_2\left[\begin{array}{c}1+t\\-t\end{array} \right]e^{2t}+\twocol13e^t\)

    3. \(\begin{array}{ccl}y'_1&=&-y_1-4y_2+4e^t+8te^t \\ y_2'&=&-y_1-\phantom{4}y_2+e^{3t}+(4t+2)e^t\end{array}\)

      \({\bf y}=c_1\twocol21e^{-3t}+c_2\twocol{-2}1e^t+\left[\begin{array}{c} e^{3t}\\2te^t\end{array}\right]\)

    4. \(\begin{array}{ccc}y'_1&=&-6y_1-3y_2+14e^{2t}+12e^t \\ y_2'&=&\phantom{6}y_1-2y_2+7e^{2t}-12e^t\end{array}\)

      \({\bf y}=c_1\twocol{-3}1e^{-5t}+c_2\twocol{-1}1e^{-3t}+ \left[\begin{array}{c}e^{2t}+3e^t\\2e^{2t}-3e^t\end{array}\right]\)

    Show answer

    (a) \({\bf y}'=\twobytwo{-3}2{-5}3{\bf }+\twocol{3-2t}{6-3t}\)   (b) \({\bf y}'=\twobytwo31{-1}1{\bf y}+\left[\begin{array}{c}-5e^t\\e^t\end{array}\right]\)

  6. Convert the linear scalar equation

    \[ P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y(t)=F(t) \tag*{\rm (A)} \]

    into an equivalent \(n\times n\) system

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

    and show that \(A\) and \({\bf f}\) are continuous on an interval \((a,b)\) if and only if (A) is normal on \((a,b)\).

  7. A matrix function

    \[ Q(t)=\matfunc qrs \]

    is said to be differentiable if its entries \(\{q_{ij}\}\) are differentiable. Then the derivative \(Q'\) is defined by

    \[ Q'(t)=\matfunc {q'}rs. \]

    1. Prove: If \(P\) and \(Q\) are differentiable matrices such that \(P+Q\) is defined and if \(c_1\) and \(c_2\) are constants, then

      \[ (c_1P+c_2Q)'=c_1P'+c_2Q'. \]

    2. Prove: If \(P\) and \(Q\) are differentiable matrices such that \(PQ\) is defined, then

      \[ (PQ)'=P'Q+PQ'. \]

  8. Verify that \(Y' = AY\).

    1. \(\dst{Y = \twobytwo {e^{6t}}{e^{-2t}} {e^{6t}}{-e^{-2t}}, \quad A = \twobytwo 2 4 4 2}\)

    2. \(\dst{Y = \twobytwo {e^{-4t}} {-2e^{3t}} {e^{-4t}} {5e^{3t}}, \quad A = \twobytwo {-2} {-2} {-5} {1}}\)

    3. \(\dst{Y = \twobytwo {-5e^{2t}} {2e^t} {3e^{2t}} {-e^t}, \quad A = \twobytwo {-4} {-10} 3 7}\)

    4. \(\dst{Y = \twobytwo {e^{3t}} {e^t} {e^{3t}} {-e^t}, \quad A = \twobytwo 2 1 1 2}\)

    5. \(Y = \left[\begin{array}{crr} e^t&e^{-t}& e^{-2t}\\ e^t&0&-2e^{-2t}\\ 0&0&e^{-2t}\end{array}\right], \quad A = \threebythree {-1} 2 {3} {0} 1 6 0 0 {-2}\)

    6. \(\dst{Y = \cthreebythree {-e^{-2t}} {-e^{-2t}} {e^{4t}} 0 {\phantom{-} e^{-2t}} {e^{4t}} {e^{-2t}} 0 {e^{4t}}, \quad A = \threebythree 0 2 2 2 0 2 2 2 0}\)

    7. \(\dst{Y = \cthreebythree {e^{3t}} {e^{-3t}} 0 {e^{3t}} 0 {-e^{-3t}} {e^{3t}} {e^{-3t}} {\phantom{-}e^{-3t}}, \quad A = \threebythree {-9}66{-6}36{-6}63}\)

    8. \(Y = \left[\begin{array}{crr} e^{2t}&e^{3t}& e^{-t}\\ 0&-e^{3t}&-3e^{-t}\\ e^{2t}&e^{3t}&7e^{-t}\end{array}\right] , \quad A = \threebythree 3 {-1} {-1}{-2} 3 2 4 {-1} {-2}\)

  9. Suppose

    \[ {\bf y}_1=\twocol{y_{11}}{y_{21}}\mbox{\quad and \quad}{\bf y}_2=\twocol{y_{12}}{y_{22}}\]

    are solutions of the homogeneous system

    \[ {\bf y}'=A(t){\bf y}, \tag*{\rm (A)} \]

    and define

    \[Y= \twobytwo{y_{11}}{y_{12}}{y_{21}}{y_{22}}. \]

    1. Show that \(Y'=AY\).

    2. Show that if \({\bf c}\) is a constant vector then \({\bf y}= Y{\bf c}\) is a solution of (A).

    3. State generalizations of (a) and (b) for \(n\times n\) systems.

  10. Suppose \(Y\) is a differentiable square matrix.

    1. Find a formula for the derivative of \(Y^2\).

    2. Find a formula for the derivative of \(Y^n\), where \(n\) is any positive integer.

    3. State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.

    Show answer

    (a)  \(\dst{d\over dt}Y^2=Y'Y+YY'\)

         (b)  \(\dst{d\over dt}Y^{n}= Y'Y^{n-1}+YY'Y^{n-2}+Y^2Y'Y^{n-3}+\cdots+Y^{n-1}Y'= \sum_{r=0}^{n-1} Y^rY'Y^{n-r-1}\)

  11. It can be shown that if \(Y\) is a differentiable and invertible square matrix function, then \(Y^{-1}\) is differentiable.

    1. Show that (\(Y^{-1})'= -Y^{-1}Y'Y^{-1}\). (Hint: Differentiate the identity \(Y^{-1}Y=I\).)

    2. Find the derivative of \(Y^{-n}=\left(Y^{-1}\right)^n\), where \(n\) is a positive integer.

    3. State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.

  12. Show that Theorem 10.2.1 implies Theorem 9.1.1.

    Hint

    Write the scalar equation

    \[ P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x) \]

    as an \(n\times n\) system of linear equations.

  13. Suppose \({\bf y}\) is a solution of the \(n\times n\) system \({\bf y}'=A(t){\bf y}\) on \((a,b)\), and that the \(n\times n\) matrix \(P\) is invertible and differentiable on \((a,b)\). Find a matrix \(B\) such that the function \({\bf x}=P{\bf y}\) is a solution of \({\bf x}'=B{\bf x}\) on \((a,b)\).

    Show answer

    \(B=(P'+PA)P^{-1}\).