Math 119 A, Midterm


ODEs and Dynamical Systems

Solve the 2x2 ODE systems $\dot x = Ax$ by finding the eigenvalues and eigenvectors and exponentiating the matrix P^-1AP,

1.

$$A = {\left(\begin{array}{cc} -4 &5\\ -1 &2 \end{array}\right)}.$$

Solution:

$$det {\left(\begin{array}{cc} \lambda+4 &-5\\ 1 &\lambda-2 \end{array}\right)}$$

$$= (\lambda-1)(\lambda+3)=0.$$

The eigenvalues are $\lambda = 1, \: \: \lambda = -3.$The eigenvectors are found by solving the system

$${\left(\begin{array}{cc} \lambda+4 &-5\\ 1 &\lambda-2 \end{a... ...ay}\right)}= {\left(\begin{array}{c}0\\ 0\end{array}\right)}.$$

The eigenvector corresponding to $\lambda = 1$ is

$${\left(\begin{array}{c}1\\ 1\end{array}\right)},$$

and the eigenvector corresponding to $\lambda = -3$ is

$${\left(\begin{array}{c}5\\ 1\end{array}\right)}.$$

The solution in y-space is

$$y(t) = {\left(\begin{array}{cc} e^{-3t} &0\\ 0 &e^{t} \end{array}\right)y_o}.$$

The solution in x-space is

$$x(t) = {\left(\begin{array}{cc} \frac{5}{4}e^{-3t}-\frac{1}{4... ...}&-\frac{1}{4}e^{-3t}+\frac{5}{4}e^{t} \end{array}\right)x_o}.$$

The computations for the next two matricies are similar,

2.

$$A = {\left(\begin{array}{cc} 1 &-4\\ 1 &1 \end{array}\right)}.$$

The eigenvalues are $\lambda = 1 \pm 2i$ and the eigenvectors are

$${\left(\begin{array}{c}\pm 2i\\ 1\end{array}\right)}.$$

The solution in y-space is

$$y(t) = {e^{t}\left(\begin{array}{cc} cos(2t) & -sin(2t)\\ sin(2t) &cos(2t) \end{array}\right)y_o}.$$

The solution in x-space is

$$x(t) = {\left(\begin{array}{cc} cos(2t) & -2sin(2t)\\ \frac{1}{2}sin(2t) &cos(2t) \end{array}\right)x_o}.$$

3.

$$A = {\left(\begin{array}{cc} -1 &1\\ 0 &-1 \end{array}\right)}.$$

The eigenvalue is $\lambda = -1$, with mulitplicity two, and the eigenvector is

$${\left(\begin{array}{c}1\\ 0\end{array}\right)},$$

whereas

$${\left(\begin{array}{c}0\\ 1\end{array}\right)},$$

is a generalized eigenvector. The matrix is already in Jordan normal form. The solution in x-space is

$$x(t) = {\left(\begin{array}{cc} e^{-t} &te^{-t}\\ 0 &e^{-t} \end{array}\right)x_o}.$$

Draw the phase portraits for the three ODEs above and classify the flow as sinks, sources, centers, etc.

4.

$$A = {\left(\begin{array}{cc} -4 &5\\ -1 &2 \end{array}\right)},$$

  

Figure: A saddel.

see Figure .

5.

  

Figure: A source.

$$A = {\left(\begin{array}{cc} 1 &-4\\ 1 &1 \end{array}\right)},$$

see Figure .

6.

  

Figure: A sink.

$$A = {\left(\begin{array}{cc} -1 &1\\ 0 &-1 \end{array}\right)},$$

see Figure .

7.

Solve the IVP,

$$\dot x = Ax, \: \: x(0) = x_o.$$

with

$$A = {\left(\begin{array}{cccc} -1 &-2 &1 &0 \\ 2 &-1 &0 &1 \\ 0 &0 &-1 &-2\\ 0 &0 &2 &-1 \end{array}\right)}.$$

What happens to the solutions x(t) as $t \to \infty$?

The matrix is in Jordan normal form. The solution is,

$$x(t)= {e^{-t}\left(\begin{array}{cccc} cos(2t) &-sin(2t) &tco... ... & -sin(2t)\\ 0 &0 &sin(2t) &cos(2t) \end{array}\right)x_o},$$

as $t \to \infty$ the solutions $x(t) \to 0$ because of the exponential decay.