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Chapter 5: Q17E (page 267)

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

Expert verified

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

01

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

02

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.

03

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.

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Most popular questions from this chapter

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

5. \(\left[ {\begin{array}{*{20}{c}}2&1\\-1&4\end{array}} \right]\)

Question: Is \(\lambda = 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)? Why or why not?
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