Question

In Exercises 9-18, construct the general solution of \(x' = Ax\) involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories.

14. \(A = \left( {\begin{aligned}{ {20}{c}}{ - 2}&1\\{ - 8}&2\end{aligned}} \right)\)

Short answer

The general complex solution of \(x' = Ax\) is \({c_1}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\4\end{aligned}} \right){e^{\left( {2i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\4\end{aligned}} \right){e^{\left( { - 2i} \right)t}}\). The real general solution is of the form \({c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 2t + \sin 2t}\\{4\cos 2t}\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}{\sin 2t - \cos 2t}\\{4\sin 2t}\end{aligned}} \right)\). The trajectories are ellipse around the origin since the eigenvalues have real parts of zero.

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

The characteristic polynomial is shown below:

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) = \left( { - 2 - \lambda } \right)\left( {2 - \lambda } \right) - \left( { - 8} \right)\left( 1 \right)\\ = - 4 + 2\lambda - 2\lambda + {\lambda ^2} + 8\\ = {\lambda ^2} + 4\end{aligned}\)

Therefore, the eigenvalues of the matrix \(A\) are \(2i\).

Construct the matrix with eigenvalue \(\lambda = 2i\) as shown below:

\(\begin{aligned}{c}A - \left( {2i} \right)I = \left( {\begin{aligned}{ {20}{c}}{ - 2}&1\\{ - 8}&2\end{aligned}} \right) - \left( {\begin{aligned}{ {20}{c}}{2i}&0\\0&{2i}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - 2 - 2i}&1\\{ - 8}&{2 - 2i}\end{aligned}} \right)\end{aligned}\)

Write the matrix into the system of the equation as shown below:

\(\begin{aligned}{c}\left( { - 2 - 2i} \right){x_1} - {x_2} = 0\\ - 8{x_1} - \left( {2 - 2i} \right){x_2} = 0\end{aligned}\)

The equation \(A - \left( {2i} \right)I\) equals to \( - 8{x_1} - \left( {2 - 2i} \right){x_2} = 0\). Therefore, \({x_1} = - \frac{{\left( {2 - 2i} \right)}}{8}\) . The eigenvector corresponds to \(\lambda = 2i\) is \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{1 - i}\\4\end{aligned}} \right)\).

Construct the general solution of \(x' = Ax\)

The complex eigenfunctions \({\mathop{\rm v}\nolimits} {e^{\lambda t}}\) and \(\overline {\mathop{\rm v}\nolimits} {e^{\overline \lambda t}}\) provides a basis for the set of all complex solutions to \(x' = A{\mathop{\rm x}\nolimits} \). The general complex solution of \(x' = Ax\) is \({c_1}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\4\end{aligned}} \right){e^{\left( {2i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\4\end{aligned}} \right){e^{\left( { - 2i} \right)t}}\) with \({c_1}\) and \({c_2}\) are arbitrary complex numbers.

Construct the real general solution and describe the shape of the trajectories

Rewrite \({\mathop{\rm v}\nolimits} {e^{\left( {2i} \right)t}}\) to construct the real general solution as shown below:

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} {e^{\left( {2i} \right)t}} = \left( {\begin{aligned}{ {20}{c}}{1 - i}\\4\end{aligned}} \right)\left( {\cos 2t + i\sin 2t} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{\cos 2t + \sin 2t}\\{4\cos 2t}\end{aligned}} \right) + i\left( {\begin{aligned}{ {20}{c}}{\sin 2t - \cos 2t}\\{4\sin 2t}\end{aligned}} \right)\end{aligned}\)

Therefore, the real general solution is of the form \({c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 2t + \sin 2t}\\{4\cos 2t}\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}{\sin 2t - \cos 2t}\\{4\sin 2t}\end{aligned}} \right)\), where \({c_1}\) and \({c_2}\) are real numbers.

Therefore, the trajectories are ellipse around the origin since the eigenvalues have real parts of zero.