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Mathematics LibreTexts

4.2: Linear Systems of Differential Equations

  • Page ID
    17435
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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

    \begin{equation} \label{eq:4.2.1}
    \begin{array}{ccl}
    y'_1&=&a_{11}(t)y_1+a_{12}(t)y_2+\cdots+a_{1n}(t)y_n+f_1(t)\\
    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}
    \end{equation}

    is called a \( \textcolor{blue}{\mbox{linear system}} \). 

    The linear system \eqref{eq:4.2.1} can be written in matrix form as

    \begin{eqnarray*}
    {y'}_n = A_{nn} y_n + f_n,
    \end{eqnarray*}

    or more briefly as

    \begin{equation} \label{eq:4.2.2}
    {\bf y}'=A(t){\bf y}+{\bf f}(t),
    \end{equation}

    where

    \begin{eqnarray*}
    {\bf y} = y_n, \quad A(t) = A_{nn}, \quad \mbox{and} \quad {\bf f}(t) = f_n.
    \end{eqnarray*}

    We call \(A\) the \( \textcolor{blue}{\mbox{coefficient matrix}} \) of \eqref{eq:4.2.2} and \({\bf f}\) the \( \textcolor{blue}{\mbox{forcing function}} \). We'll say that \(A\) and \({\bf f}\) are \( \textcolor{blue}{\mbox{continuous}} \) if their entries are continuous. If \({\bf f}={\bf 0}\), then \eqref{eq:4.2.2} is \( \textcolor{blue}{\mbox{homogeneous}} \); otherwise, \eqref{eq:4.2.2} is  \( \textcolor{blue}{\mbox{nonhomogeneous}} \).

    An initial value problem for \eqref{eq:4.2.2} consists of finding a solution of \eqref{eq:4.2.2} that equals a given constant vector

    \begin{eqnarray*}
    {\bf k} = k_n.
    \end{eqnarray*}

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

    \begin{eqnarray*}
    {\bf y}' = A(t){\bf y} + {\bf f}(t), \quad {\bf y}(t_0) - {\bf k}.
    \end{eqnarray*}

    The next theorem gives sufficient conditions for the existence of solutions of initial value problems for \eqref{eq:4.2.2}. We omit  the proof.

    Theorem \(\PageIndex{1}\)

    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

    \begin{eqnarray*}
    {\bf y}' = A(t) {\bf y} + {\bf f}(t), \quad {\bf y} (t_0) = {\bf k}
    \end{eqnarray*}

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

     

    Proof

    Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

    Example \(\PageIndex{1}\)

    (a) Write the system

    \begin{equation} \label{eq:4.2.3}
    \begin{array}{rcl}
    y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\
    y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t}
    \end{array}
    \end{equation}

    in matrix form and conclude from Theorem \((4.2.1)\) that every initial value problem for \eqref{eq:4.2.3} has a unique solution on \((-\infty,\infty)\).

    (b) Verify that

    \begin{equation} \label{eq:4.2.4}
    {\bf y} = {1\over 5}\left[ \begin{array} \\ 8 \\ 7e^{4t} \end{array} \right] + c_1 \left[ \begin{array} \\ 11 \\ e^{3t} \end{array} \right] + c_2 \left[ \begin{array} \\ 1 \\ {-1} e^{-t} \end{array} \right]
    \end{equation}

    is a solution of \eqref{eq:4.2.3} for all values of the constants \(c_1\) and \(c_2\).

    (c) Find the solution of the initial value problem

    \begin{equation} \label{eq:4.2.5}
    {\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right], \quad {\bf y}(0)= {1 \over 5} \left[ \begin{array} \\ 3 \\ 22 \end{array} \right].
    \end{equation}

     

    Answer

    (a) The system \eqref{eq:4.2.3} can be written in matrix form as 

    \begin{eqnarray*}
    {\bf y}' = \left[\begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right].
    \end{eqnarray*}

    An initial value problem for \eqref{eq:4.2.3} can be written as

    \begin{eqnarray*}
    {\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right], \quad y(t_0) = \left[ \begin{array} \\ {k_1} \\ {k_2} \end{array} \right].
    \end{eqnarray*}

    Since the coefficient matrix and the forcing function are both continuous on \((-\infty,\infty)\), Theorem \((4.2.1)\) implies that this problem has a unique solution on \((-\infty,\infty)\).

     

    (b) If \({\bf y}\) is given by \eqref{eq:4.2.4}, then 

    \begin{eqnarray*}
    A{\bf y}+{\bf f}&=&
    {1\over5} \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\ 8 \\ 7e^{4t} \end{array} \right] +
    c_1 \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\11 \\e^{3t} \end{array} \right] \\
    &&+c_2 \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\ 1 \\ {-1}e^{-t} \end{array} \right]
    + \left[ \begin{array} \\ 2 \\ 1e^{4t} \end{array} \right] \\
    &=&{1\over5} \left[ \begin{array} \\ {22} \\ {23}e^{4t} \end{array} \right] +c_1 \left[ \begin{array} \\ 3 \\ 3e^{3t} \end{array} \right] + c_2 \left[ \begin{array} \\ {-1} \\ 1e^{-t} \end{array} \right]
    + \left[ \begin{array} \\ 2 \\ 1e^{4t} \end{array} \right] \\
    &=&{1\over5} \left[ \begin{array} \\ {32} \\ {28}e^{4t} \end{array} \right] + 3c_1 \left[ \begin{array} \\ 11 \\ e^{3t} \end{array} \right] - c_2 \left[ \begin{array} \\ 1 \\ {-1}e^{-t} \end{array} \right]
    ={\bf y}'.
    \end{eqnarray*}

     

    (c) We must choose \(c_1\) and \(c_2\) in \eqref{eq:4.2.4} so that

    \begin{eqnarray*}
    \displaystyle\frac{1}{5} \left[ \begin{array} \\ 8 \\ 7 \end{array} \right] + c_1 \left[ \begin{array} \\ 1 \\ 1 \end{array} \right] + c_2 \left[ \begin{array} \\ 1 \\ -1 \end{array} \right] = \displaystyle\frac{1}{5} \left[ \begin{array} \\ 3 \\ {22} \end{array} \right],
    \end{eqnarray*}

    which is equivalent to

    \begin{eqnarray*}
    \left[ \begin{array} \\ 1 & 1 \\ 1 & {-1} \end{array} \right] \left[ \begin{array} \\ {c_1} \\ {c_2} \end{array} \right] = \left[ \begin{array} \\ {-1} \\ 3 \end{array} \right].
    \end{eqnarray*}

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

    \begin{eqnarray*}
    {\bf y} = \displaystyle\frac{1}{5} \left[ \begin{array} \\ 8 \\ 7 e^{4t} \end{array} \right] + \left[ \begin{array} \\ 11 \\ e^{3t}-2 \end{array} \right] \left[ \begin{array} \\ 1 \\ {-1} e^{-t} \end{array} \right] 
    \end{eqnarray*}

    is the solution of  \eqref{eq:4.2.5}.

    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} \label{eq:4.2.6}
    P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t),
    \end{equation}

    as developed in Sections 3.1. For example, by rewriting \eqref{eq:4.2.6} as an equivalent linear system it can be shown that Theorem \((4.2.1)\) implies Theorem \((3.1.1)\) (Exercise \((4.2E.12)\).