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.

18. (M) \(A = \left( {\begin{aligned}{ {20}{c}}{53}&{ - 30}&{ - 2}\\{90}&{ - 52}&{ - 3}\\{20}&{ - 10}&2\end{aligned}} \right)\)

Short answer

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}1\\2\\0\end{aligned}} \right){e^{ - 7t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{6 + 2i}\\{9 + 3i}\\{10}\end{aligned}} \right){e^{\left( {5 + i} \right)t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{6 - 2i}\\{9 - 3i}\\{10}\end{aligned}} \right){e^{\left( {5 - i} \right)t}}\). The real general solution is of the form

\(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}1\\2\\0\end{aligned}} \right){e^{ - 7t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{6\cos t - 2\sin t}\\{9\cos t - 3\sin t}\\{10\cos t}\end{aligned}} \right){e^{5t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{6\sin t + 2\cos t}\\{9\sin t + 3\cos t}\\{10\cos t}\end{aligned}} \right){e^{5t}}\). If \({c_2} = {c_3} = 0\), then trajectories tend toward the origin, whereas in the other case, the trajectories spiral away from the origin.

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

The given matrix is \(A = \left( {\begin{aligned}{ {20}{c}}{53}&{ - 30}&{ - 2}\\{90}&{ - 52}&{ - 3}\\{20}&{ - 10}&2\end{aligned}} \right)\).

Use the MATLAB code to compute the eigenvalues of the matrix \(A\) as shown below:

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( {53\,\,\, - 30\,\,\, - 2;\,90\,\,\, - 52\,\,\, - 3;\,20\,\,\, - 10\,\,\,2} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{ - 7.0000}&{}&{}\\{5.0000}& + &{1.0000{\mathop{\rm i}\nolimits} }\\{5.0000}& - &{1.0000{\mathop{\rm i}\nolimits} }\end{aligned}} \right)\)

Use the MATLAB code to compute the eigenvector of the matrix A as shown below:

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 1 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{0.5000}\\{1.0000}\\{0.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}1\\2\\0\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 2 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{0.6000}& + &{0.2000{\mathop{\rm i}\nolimits} }\\{0.9000}& + &{0.3000{\mathop{\rm i}\nolimits} }\\{1.0000}&{}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{6 + 2{\mathop{\rm i}\nolimits} }\\{9 + 3{\mathop{\rm i}\nolimits} }\\{10}\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 3 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{0.6000}& - &{0.2000{\mathop{\rm i}\nolimits} }\\{0.9000}& - &{0.3000{\mathop{\rm i}\nolimits} }\\{1.0000}&{}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{ {20}{c}}{6 - 2{\mathop{\rm i}\nolimits} }\\{9 - 3{\mathop{\rm i}\nolimits} }\\{10}\end{aligned}} \right)\).

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

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}1\\2\\0\end{aligned}} \right){e^{ - 7t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{6 + 2i}\\{9 + 3i}\\{10}\end{aligned}} \right){e^{\left( {5 + i} \right)t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{6 - 2i}\\{9 - 3i}\\{10}\end{aligned}} \right){e^{\left( {5 - i} \right)t}}\), with \({c_1}\) , and \({c_2}\) are arbitrary complex numbers.

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

Rewrite the second eigenfunction of the matrix as shown below:

\(\left( {\begin{aligned}{ {20}{c}}{6 + 2i}\\{9 + 3i}\\{10}\end{aligned}} \right){e^{5t}}\left( {\cos t + i\sin t} \right) = \left( {\begin{aligned}{ {20}{c}}{6\cos t - 2\sin t}\\{9\cos t - 3\sin t}\\{10\cos t}\end{aligned}} \right){e^{5t}} + i\left( {\begin{aligned}{ {20}{c}}{6\sin t + 2\cos t}\\{9\sin t - 3\cos t}\\{10\sin t}\end{aligned}} \right){e^{5t}}\)

Therefore, the real general solution is of the form as shown below:

\(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}1\\2\\0\end{aligned}} \right){e^{ - 7t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{6\cos t - 2\sin t}\\{9\cos t - 3\sin t}\\{10\cos t}\end{aligned}} \right){e^{5t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{6\sin t + 2\cos t}\\{9\sin t + 3\cos t}\\{10\cos t}\end{aligned}} \right){e^{5t}}\),

Where \({c_1},{c_2},\) and \({c_2}\) are real numbers.

It is observed that the if \({c_2} = {c_3} = 0\), then trajectories tend toward the origin, while in some cases, the trajectories spiral away from the origin.