Are the following phase portraits drawn (well, at least sketched) correctly, in terms of their behaviour?

Question

I have a linear system: $\frac{dx}{dt}=ax+y; \frac{dy}{dt}=-x+ay$.

I'm trying to investigate pictorially what happens for $a$ negative, $0$ and positive. Now, I know that the eigenvalues I eventually get are: $\lambda=a\pm i$ for $a \neq0$; $\lambda=\pm i$ for $a=0$.

Here are my sketches. My question is: are the following sketches accurate (in terms of the nature of the respective critical points) and is the behaviour of the system indicated correctly?

Answer 1

Given the system:

$$\dfrac{dx}{dt}=ax+y \\ \dfrac{dy}{dt}=-x+ay$$

We can find the Jacobian matrix as:

$$A = \begin{bmatrix}a & 1\\-1 & a\end{bmatrix}$$

The eigenvalues of the this matrix are given by:

$$|A - \lambda I | = 0 \implies \lambda_{1,2} = a ~ \pm ~ i$$

• For $a \lt 0$, we get a stable spiral.
• For $a = 0$, we get a semi-stable center.
• For $a \gt 0$, we get an unstable spiral.

From the above results, you can draw the solution curves.

We can also explicitly solve for the system and arrive at:

$$x(t) = e^{a t}(c_1 \cos t + c_2 \sin t) \\ ~~~y(t) = e^{at}(-c_1 \sin t + c_2 \cos t)$$

Of course, using that closed-form solution, you can do a parametric plot of the solutions.

Here is an animated phase portrait for varying $a \in [-11, 11]$ in steps of $1$: