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Chapter 5: Q4E (page 267)

In Exercises 3-6, solve the initial value problem \(x'\left( t \right) = Ax\left( t \right)\) for \(t \ge 0\), with \(x\left( 0 \right) = \left( {3,2} \right)\). Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by \(x' = Ax\). Find the directions of greatest attraction and/or repulsion. When the origin is a saddle point, sketch typical trajectories.

4. \(A = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 5}\\1&4\end{aligned}} \right)\)

Short Answer

Expert verified

The required solution is:

\({\rm{x}}\left( t \right) = \frac{{13}}{4}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{3t}} - \frac{5}{4}\left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right){e^{ - t}}\)

The origin is a saddle point.

The direction of greatest repulsion is through the origin, and the eigenvector is\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\).

The direction of greatest attraction is through the origin and eigenvector is \({v_2} = \left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right)\).

Step by step solution

01

System of Differential Equations

The general solutionfor any system of differential equations withthe eigenvalues\({\lambda _1}\)and\({\lambda _2}\)with the respective eigenvectors\({v_1}\)and\({v_2}\)is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here \({c_1}\) and \({c_2}\) are the constants from the initial condition.

02

Find the eigenvalues.

According to the question, we have:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 5}\\1&4\end{aligned}} \right)\)

For eigenvalues:

\(\begin{aligned}{c}\det (A - \lambda I) = 0\\( - 2 - \lambda )(4 - \lambda ) + 5 = 0\\ - 3 - 2\lambda + {\lambda ^2} = 0\\{\lambda _1} = 3,{\lambda _2} = - 1\end{aligned}\)

The obtained eigenvalues are 3 and\( - 1\).

Hence, the origin is a saddle point.

03

Find eigenvectors for both eigenvalues

Now, for\({\lambda _1} = 3\)we have:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 5}&{ - 5}&0\\1&1&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&1&0\\0&0&0\end{aligned}} \right)\)

So,\({x_1} = - {x_2}\)with\({x_2}\)free.

Taking\({x_2} = 1\), we get, eigenvector:

\({v_1} = \left( {\begin{aligned}{ {20}{l}}{ - 1}\\1\end{aligned}} \right)\)

Similarly,for\({\lambda _2} = - 1\)we have:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 1}&{ - 5}&0\\1&5&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{l}}1&5&0\\0&0&0\end{aligned}} \right)\)

So\({x_1} = - 5{x_2}\)with\({x_2}\)free.

Taking\({x_2} = 1\), we get:

\({v_2} = \left( {\begin{aligned}{ {20}{l}}{ - 5}\\1\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\left( {\begin{aligned}{ {20}{l}}{{v_1}}&{{v_2}}&{x(0)}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - 1}&{ - 5}&3\\1&1&2\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&0&{\frac{{13}}{4}}\\0&1&{ - \frac{5}{4}}\end{aligned}} \right)\)

Thus,\({c_1} = \frac{{13}}{4}\)and\({c_2} = - \frac{5}{4}\).

Now, the general solution will be:

\(x(t) = \frac{{13}}{4}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{3t}} - \frac{5}{4}\left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right){e^{ - t}}\)

Hence, this is the required solution.

04

Find the directions of greatest attraction and/or repulsion

Since one eigenvalue is less than 1 and the other is greater, so the origin is a saddle point of the dynamical system described by\(x\prime = Ax\).

The direction of greatest repulsion is through the origin and eigenvector:\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\).

The direction of greatest attraction is through the origin and eigenvector: \({v_2} = \left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \righ

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). For each value of \(a\) in the set \(\left\{ {32,31.9,31.8,32.1,32.2} \right\}\), compute the characteristic polynomial of \(A\) and the eigenvalues. In each case, create a graph of the characteristic polynomial \(p\left( t \right) = \det \left( {A - tI} \right)\) for \(0 \le t \le 3\). If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues of \(a\) changes.

19鈥�23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

If \(p\left( t \right) = {c_0} + {c_1}t + {c_2}{t^2} + ...... + {c_n}{t^n}\), define \(p\left( A \right)\) to be the matrix formed by replacing each power of \(t\) in \(p\left( t \right)\)by the corresponding power of \(A\) (with \({A^0} = I\) ). That is,

\(p\left( t \right) = {c_0} + {c_1}I + {c_2}{I^2} + ...... + {c_n}{I^n}\)

Show that if \(\lambda \) is an eigenvalue of A, then one eigenvalue of \(p\left( A \right)\) is\(p\left( \lambda \right)\).

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

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

7. \(\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\)

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