The eigenvalues and eigenvectors of a matrix A are given.
Consider the corresponding system $x' = Ax$.
(a) Sketch a phase portrait of the system.
(b) Sketch the trajectory passing through the initial point $(2, 3)$.
(c) For the trajectory in part (b) sketch the graphs of $x_1$ versus t and of $x_2$ versus t on the same set of axes.
$$r_1=1,n_1 = \left(
\begin{array}{c}
1\\
2\\
\end{array}
\right)$$
$$r_2=2,n_2 = \left(
\begin{array}{c}
1\\
-2\\
\end{array}
\right)$$
Where $r$ represents the eigenvalues and $n$ represents the eigenvectors.
What i tried
Since both eigenvalues are positive, the plot must be a nodal source with unstable equilibrium. Hence the graph points away from the origin. From what i know, Eigenvalues that are negative will correspond to solutions that will move towards the origin as t increases in a direction that is parallel to its eigenvector. Likewise, eigenvalues that are positive move away from the origin as t increases in a direction that will be parallel to its eigenvector.However im unsure of how to intepret the eigenvectors to plot the graph. Could anyone please explain. Thanks
Best Answer
To visualize the solution, it helps to graph it in the $x_1 x_2$-plane for various $c_1, c_2$. I will map out most of the details and you can add the rest. Start with the first eigenvalue/vector solution and we have in scalar form:
$$x_1 = c_1 e^t, x_2 =2 c_1 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = 2 x_1$. If $c_1 \gt 0$, we are in quadrant $I$ and if $c_1 \lt 0$, we are in quadrant $III$.
Repeating this for the second eigenvalue/vector, we have:
$$x_1 = c_2 e^t, x_2 =-2 c_2 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = -2 x_1$. If $c_2 \gt 0$, we are in quadrant $IV$ and if $c_2 \lt 0$, we are in quadrant $II$.
The solution is a linear combination of these two as:
$$x(t) = c_1 e^{t} \left( \begin{array}{c} 1\\ 2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c} 1\\ -2\\ \end{array}\right)$$
So, we can plot the eigenvectors as (blue is $n_1$ and yellow is $n_2$):
Now, what happens to the solutions as $t \rightarrow + \infty$? Just draw handfuls of those around the lines in each quadrant (hint - they are asymptotic to the line and move away from the origin as this is an unstable node). This will give you a phase portrait that is an unstable node.
For part $b.$, we have the initial point $(2, 3)$. Since we know we are going to be asymptotic to the eigenvector in quadrant $I$, it starts at ($2, 3)$ and goes to this line.
For part $c.$, we have $x(0) = (2, 3)$, with:
$$x(t) = c_1 e^{t} \left( \begin{array}{c} 1\\ 2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c} 1\\ -2\\ \end{array}\right)$$
We end up with $c_1 = \dfrac 74, c_2 = \dfrac 14$, so have:
$$x_1(t) = \frac{7}{4} e^t+\frac{1}{4} e^{2t}, x_2(t) = \frac{7}{2}e^{t}-\frac{1}{2} e^{2t}$$
A plot of this shows:
One last note, it helps to plot this parametrically and we get: