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Section 27.4 Examples

Let's do a larger example than those discussed in Discovery guide 27.1 or Section 27.3 as an initial value problem.

Example 27.4.1. Three coupled differential equations.

Consider the following system, with the provided initial values:

\begin{align*} \amp \left\{\begin{array}{rcrcrcr} y_1' \amp = \amp -9 y_1 \amp - \amp 14 y_2 \amp + \amp 14 y_3 \text{,} \\ y_2' \amp = \amp 9 y_1 \amp + \amp 16 y_2 \amp - \amp 13 y_3 \text{,} \\ y_3' \amp = \amp 2 y_1 \amp + \amp 4 y_2 \amp - \amp 1 y_3 \text{,} \end{array}\right. \amp \amp \left\{\begin{array}{rcr} y_1(0) \amp = \amp 6 \text{,} \\ y_2(0) \amp = \amp -4 \text{,} \\ y_3(0) \amp = \amp 4 \text{.} \end{array}\right. \end{align*}

In matrix form, we express this as

\begin{equation*} \ddt \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \left[\begin{array}{rrr} -9 \amp -14 \amp 14 \\ 9 \amp 16 \amp -13 \\ 2 \amp 4 \amp -1 \end{array}\right] \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}\text{.} \end{equation*}

The characteristic polynomial of the coefficient matrix is

\begin{equation*} \lambda^3 - 6 \lambda^2 - \lambda + 30 = (\lambda + 2)(\lambda - 3)(\lambda - 5)\text{,} \end{equation*}

so the eigenvalues are \(\lambda_1 = -2\text{,}\) \(\lambda_2 = 3\text{,}\) \(\lambda_3 = 5\text{.}\) Computing eigenspaces by row reducing \(\lambda I - A\) for the various eigenvalues as usual, we obtain

\begin{gather*} E_{\lambda_1}(A) = \Span\left\{ \left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array}\right] \right\} \text{,} \qquad E_{\lambda_2}(A) = \Span\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \right\} \text{,}\\ E_{\lambda_3}(A) = \Span\left\{ \left[\begin{array}{r} -1 \\ 2 \\ 1 \end{array}\right] \right\} \text{.} \end{gather*}

So we can use the transition matrix

\begin{equation*} P = \left[\begin{array}{rrr} -2 \amp 0 \amp -1 \\ 1 \amp 1 \amp 2 \\ 0 \amp 1 \amp 1 \end{array}\right] \end{equation*}

and its inverse to convert between the original system and a decoupled one involving transformed functions

\begin{equation*} \uvec{w}(t) = \inv{P} \uvec{y}(t) \text{.} \end{equation*}

With our ordering of the columns of \(P\text{,}\) the general solution of the decoupled system is

\begin{equation*} \uvec{w}(t) = \begin{bmatrix} c_1 e^{-2 t} \\ c_2 e^{3 t} \\ c_3 e^{5 t} \end{bmatrix} \text{,} \end{equation*}

where \(c_1,c_2,c_3\) are arbitrary parameters. The general solution to the coupled system is then obtained by converting back via \(\uvec{y}(t) = P \uvec{w}(t)\text{,}\) yielding

\begin{equation*} \left\{\begin{array}{rcrcrcr} y_1(t) \amp = \amp -2 c_1 e^{-2 t} \amp \amp \amp - \amp c_3 e^{5 t} \text{,} \\ y_2(t) \amp = \amp c_1 e^{-2 t} \amp + \amp c_2 e^{3 t} \amp + \amp 2 c_3 e^{5 t} \text{,} \\ y_3(t) \amp = \amp \amp \amp c_2 e^{3 t} \amp + \amp c_3 e^{5 t} \text{.} \end{array}\right. \end{equation*}

It's easier to work with the decoupled system to determine the parameter values that correspond to the initial values. Convert the vector of initial values \(\uvec{y}(0)\) to a vector of initial values for the decoupled system:

\begin{equation*} \uvec{w}(0) = \inv{P} \uvec{y}(0) = \left[\begin{array}{r} 2 \\ 14 \\ -10 \end{array}\right] \text{.} \end{equation*}

These three transformed initial values are precisely the values of the parameters \(c_1,c_2,c_3\) we need to create a particular solution that meets the initial values for \(y_1(t),y_2(t),y_3(t)\) provided at the beginning of this example.

Inserting these parameter values, we obtain particular solution

\begin{equation*} \left\{\begin{array}{rcrcrcr} y_1(t) \amp = \amp -4 e^{-2 t} \amp \amp \amp + \amp 10 e^{5 t} \text{,} \\ y_2(t) \amp = \amp 2 e^{-2 t} \amp + \amp 14 e^{3 t} \amp - \amp 20 e^{5 t} \text{,} \\ y_3(t) \amp = \amp \amp \amp 14 e^{3 t} \amp - \amp 10 e^{5 t} \text{.} \end{array}\right. \end{equation*}