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.

15. (M) \(A = \left( {\begin{aligned}{ {20}{c}}{ - 8}&{ - 12}&{ - 6}\\2&1&2\\7&{12}&5\end{aligned}} \right)\)

Short answer

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right){e^{ - t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right){e^{ - 2t}}\). It is observed that the origin is a saddle point. A solution of \({c_1} = 0\) attracts the origin but a solution of \({c_2} = {c_3} = 0\) repels.

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

The given matrix is \(A = \left( {\begin{aligned}{ {20}{c}}{ - 8}&{ - 12}&{ - 6}\\2&1&2\\7&{12}&5\end{aligned}} \right)\).

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

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( { - 8\,\,\, - 12\,\,\,\, - 6;\,2\,\,\,\,1\,\,\,\,2;\,7\,\,\,12\,\,\,5} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

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

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

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

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.0000}\\{0.2500}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.2000}\\{0.2000}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.0000}\\{0.0000}\\{ - 1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right)\).

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

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right){e^{ - t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right){e^{ - 2t}}\), with \({c_1}\) , and \({c_2}\) are arbitrary complex numbers.

Describe the shape of the trajectories

It is observed that the origin is a saddle point. A solution of \({c_1} = 0\) attracts the origin but a solution of \({c_2} = {c_3} = 0\) repels the origin.