linear differential equation problem

Question

Consider the following system of linear differential equations:

$$\begin{split} \frac{dx}{dt}&=−3x+y\\ \frac{dy}{dt}&=x−3y \end{split}$$

1. Find the eigenvalues and eigenvectors associated with the system.
2. Find the equilibrium and determine its stability.
3. Draw the $x$- and $y$-nullclines in the phase plane.
4. Sketch the solution curve for the initial value $x(0)=6$ , $y(0)=2$ in the phase plane.

Answer 1

Outline: Please fill in the details.

Consider the following system of linear differential equations:

$$\begin{split} \frac{dx}{dt} &= −3x+y \\ \frac{dy}{dt} &= x−3y \end{split}$$

$$x(0) = 6, y(0) = 2$$

Our system can be written as:

$$\begin{bmatrix} x'(t) \\ y'(t)\end{bmatrix} = A x(t) = \begin{bmatrix} -3 & 1 \\ 1 & -3\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\end{bmatrix}$$

For the eigenvalues, we find $|A - \lambda I| = 0$, and for the eigenvectors, we solve $[A - \lambda_i I]v_i = 0$. This leads to:

$$\lambda_1 = -4, v_1 = (-1,1), ~ \lambda_2 = -2, v_2 = (1,1)$$

From this, we can write the solution as:

$$\begin{bmatrix} x(t) \\ y(t)\end{bmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$$

We also have the initial condition, $\begin{bmatrix} x(0) \\ y(0)\end{bmatrix} = \begin{bmatrix} 6 \\ 2 \end{bmatrix} $, so we can find the closed-form solution as:

$$\begin{bmatrix} x(t) \\ y(t)\end{bmatrix} = \begin{bmatrix} 2 e^{-4 t} \left(2 e^{2 t}+1\right) \\ 2 e^{-4 t} \left(2 e^{2 t}-1\right)\end{bmatrix}$$

There is a single equilibrium at $(x, y) = (0,0)$. What does the solution or the eigenvalues tell you about its stability (as well as the phase portrait)?

A phase portrait, with nullclines and the two initial points is:

Answer 2

You should try yourself first. Here is a first hint to get you started. Let $$\vec{z} = \pmatrix{x\\y} \quad \text{ and } \quad A = \pmatrix{-3 & 1 \\ 1 & -3}$$ then your system looks like $\vec{z}' = Az$.

Now diagonalize $A$.