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.

17. (M) \(A = \left( {\begin{aligned}{ {20}{c}}{30}&{64}&{23}\\{ - 11}&{ - 23}&{ - 9}\\6&{15}&4\end{aligned}} \right)\)

Short answer

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{23 - 34{\mathop{\rm i}\nolimits} }\\{ - 9 + 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right){e^{\left( {5 + 2i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{23 + 34{\mathop{\rm i}\nolimits} }\\{ - 9 - 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right){e^{\left( {5 - 2i} \right)t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 3}\\1\\1\end{aligned}} \right){e^t}\). The real general solution is of the form

\(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{23\cos 2t + 34\sin 2t}\\{ - 9\cos 2t - 14\sin 2t}\\3\end{aligned}} \right){e^{5t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{23\sin 2t - 34\cos 2t}\\{ - 9\sin 2t + 14\cos 2t}\\3\end{aligned}} \right){e^{5t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 3}\\1\\1\end{aligned}} \right){e^t}\). It is observed that the origin is a repellor since all eigenvalue is positive in their real parts. Every trajectory spirals away from its origin.

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

The given matrix is \(A = \left( {\begin{aligned}{ {20}{c}}{30}&{64}&{23}\\{ - 11}&{ - 23}&{ - 9}\\6&{15}&4\end{aligned}} \right)\).

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

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( {30\,\,\,64\,\,\,23\,;\, - 11\,\,\, - 23\,\,\, - 9;\,6\,\,\,15\,\,\,4} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{5.0000}& + &{2.0000{\mathop{\rm i}\nolimits} }\\{5.0000}& - &{2.0000{\mathop{\rm i}\nolimits} }\\{1.0000}&{}&{}\end{aligned}} \right)\)

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

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

\(\left( {\begin{aligned}{ {20}{c}}{7.6667}& - &{11.3333{\mathop{\rm i}\nolimits} }\\{ - 3.0000}& + &{4.6667{\mathop{\rm i}\nolimits} }\\{1.0000}&{}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{23 - 34{\mathop{\rm i}\nolimits} }\\{ - 9 + 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{ {20}{c}}{7.6667}& + &{11.3333{\mathop{\rm i}\nolimits} }\\{ - 3.0000}& - &{4.6667{\mathop{\rm i}\nolimits} }\\{1.0000}&{}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{23 + 34{\mathop{\rm i}\nolimits} }\\{ - 9 - 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{ {20}{c}}{ - 3.0000}\\{1.0000}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{ {20}{c}}{ - 3}\\1\\1\end{aligned}} \right)\).

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

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{23 - 34{\mathop{\rm i}\nolimits} }\\{ - 9 + 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right){e^{\left( {5 + 2i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{23 + 34{\mathop{\rm i}\nolimits} }\\{ - 9 - 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right){e^{\left( {5 - 2i} \right)t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 3}\\1\\1\end{aligned}} \right){e^t}\), with \({c_1}\) , and \({c_2}\) are arbitrary complex numbers.

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

Rewrite the first eigenfunction of the matrix as shown below:

\(\left( {\begin{aligned}{ {20}{c}}{23 - 34{\mathop{\rm i}\nolimits} }\\{ - 9 + 14{\mathop{\rm i}\nolimits} }\\3\end{aligned}} \right){e^{5t}}\left( {\cos 2t + i\sin 2t} \right) = \left( {\begin{aligned}{ {20}{c}}{23\cos 2t + 34\sin 2t}\\{ - 9\cos 2t - 14\sin 2t}\\3\end{aligned}} \right){e^{5t}} + i\left( {\begin{aligned}{ {20}{c}}{23\sin 2t - 34\cos 2t}\\{ - 9\sin 2t + 14\cos 2t}\\3\end{aligned}} \right){e^{5t}}\)

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

\(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{23\cos 2t + 34\sin 2t}\\{ - 9\cos 2t - 14\sin 2t}\\3\end{aligned}} \right){e^{5t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{23\sin 2t - 34\cos 2t}\\{ - 9\sin 2t + 14\cos 2t}\\3\end{aligned}} \right){e^{5t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 3}\\1\\1\end{aligned}} \right){e^t}\),

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

It is observed that the origin is a repellor since all eigenvalue is positive in their real parts. Every trajectory spirals away from its origin.