We are given the system
\begin{align}
\begin{cases}
\dot{x} &= -y\\
\dot{y} &= x + ay
\end{cases} \ , \ a >0
\end{align}
which, by the way, is called Rossler System. More info on this here. I am asked to "solve" them, ie. find the equations that fulfill it and then draw the phase plane.
For the first part we have a lot of methods to go about this. The usual one is by way of elimination. Deriving the first equation and substituting in the second we have the system
\begin{align}
\ddot{x}-a \dot{x}+x=0.
\end{align}
The characteristic equation is
\begin{align}
\lambda^2-a\lambda+1=0 &\implies \Delta=a^2-4.
\end{align}
Here's the first problem I am having: am I supposed to take every option into account?
Ie.
- If $\boxed{a=2}$ (negative root is not accepted since $a>0$), then we have
\begin{align}
\ddot{x} – 2\dot{x}+x=0 &\implies x(t)=c_1e^t+c_2te^t
\end{align}
and we must continue the system to find $y(t)$. Through a lot of calculations we can do
\begin{align}
y(t) = -c_1 t e^t – c_2 t^2 e^t.
\end{align}
In that specific case, what does the system "tell" us?
At any rate the phase portrait can be sketched by the eigenvalues of the matrix
\begin{align}
\begin{pmatrix}
0 & -1 \\ 1 & 2
\end{pmatrix}
\end{align}
which we have already (through the system) calculated to be $\lambda = 1$. Now we know that if we put $\lambda = 1$ in the characteristic polynomial, we get
\begin{align}
\begin{pmatrix}
-1 & -1\\
1 & 1
\end{pmatrix} \implies \begin{pmatrix} -1 \\ 1 \end{pmatrix} \ \text{null space.}
\end{align}
How can we draw this?
-Am I supposed to do the same now for $a \in (0,2)$ and $a \in (2, + \infty)$?
Best Answer
I shall provide an intuitive method to sketch the phase portraits which I have learnt back at university years ago. In simpler terms, it shows the trajectory of your gradient field. It is relatively straightforward given a first-order $2 \times 2$ matrix linear ODE system. We shall leverage on the info you have provided above and some Linear Algebra to analyze the dynamics of the system.
We can rewrite our system of linear ODE in matrix form.
$$ \begin{bmatrix} \dot x \\ \dot y \\ \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & a \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} $$ $$ \implies A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & a \\ \end{bmatrix} $$ For a $2 \times 2$ system, the characteristic polynomial can be given by: $$ p(\lambda) = \lambda^2-\text{tr}(A)\lambda + \text{det}(A) $$ The roots or the eigenvalues to the characteristic polynomial is given by setting $p(\lambda)=0$. Then: $$ \lambda_{1,2}=\frac{\text{tr}(A)}{2} \pm \sqrt{\Big(\frac{\text{tr}(A)}{2}\Big)^2-\text{det}(A)} $$ If $A$ is diagonalized, ie. $A=Q\Lambda Q^{-1}$, where $Q$ consists of eigenvectors $q_i$ and eigenvalues in $\Lambda$, we can find our eigenvectors by computing: $$q_i = \begin{bmatrix} -a_{12} \\ a_{11} - \lambda_i \\ \end{bmatrix} $$ The eigenvectors will show the trajectory of the phase portrait.
From the discriminant $\Delta$ and $a > 0$, we will have to analyze the system for $a \in (0,2),$ $a \in (2,\infty),$ and $a =2$. I will skip the computations and present the concept directly. I leave it to you to DIY.
Case $1$: $a \in (0,2)$. This implies $\Delta<0$. The eigenvalues are complex. For simplicity, let us assume $a=1$. $$\lambda_{1,2} = \frac{1}{2} \pm i \frac{\sqrt3}{2} $$ For this case, we shall just consider one of the eigenvectors, for the corresponding eigenvalue has a positive square-root. Its real and imaginary parts of the (complex) eigenvector will determine the trajectory of the field. $$q_1 = \begin{bmatrix} -(-1) \\ 0 - \big( \frac{1}{2} + i\frac{\sqrt3}{2} \big) \\ \end{bmatrix} = \begin{bmatrix} 1 \\ - \frac{1}{2} - i\frac{\sqrt3}{2} \\ \end{bmatrix} $$ $$\implies q_{\text{re}} = \begin{bmatrix} 1 \\ - \frac{1}{2}\\ \end{bmatrix}, -q_{\text{im}} \begin{bmatrix} 0 \\ \frac{\sqrt 3}{2}\\ \end{bmatrix} $$ As the eigenvalue is complex with $\text{Re}\{\lambda_{1,2}\} > 0$, it is an unstable focus. We shall sketch $q_{\text{re}}$ and $-q_{\text{im}}$ in the phase plane as follows.
Case $2$: $a \in (2,\infty)$. This implies $\Delta > 0$. The eigenvalues are real and not the same. For simplicity, let us assume $a=3$. $$ \lambda_1 = \frac{3}{2} + \frac{\sqrt{5}}{2} = \frac{1}{2}(3 + \sqrt5) \implies q_1 = \begin{bmatrix} -(-1) \\ 0 - \frac{1}{2}(3 + \sqrt5)\\ \end{bmatrix} = \begin{bmatrix} 1 \\ - \frac{1}{2}(3 + \sqrt5)\\ \end{bmatrix} $$ $$ \lambda_2 = \frac{3}{2} - \frac{\sqrt{5}}{2} = \frac{1}{2}(3 - \sqrt5) \implies q_2 = \begin{bmatrix} -(-1) \\ 0 - \frac{1}{2}(3 - \sqrt5)\\ \end{bmatrix} = \begin{bmatrix} 1 \\ - \frac{1}{2}(3 - \sqrt5)\\ \end{bmatrix} $$ $$\lambda_i: \lambda_1 > \lambda_2 > 0 $$ The eigenvalue with the highest magnitude, i.e. $|\lambda_i|$ is known as the 'fast' eigenvalue. The trajectory will take the direction of the eigenvector corresponding to the 'fast' eigenvalue as it converges to (or diverges from) the origin. Since $\lambda_1 > \lambda_2 > 0$, it is an unstable node.
Case $3$: $a = 2$. This implies $\Delta = 0$. The eigenvalues are real and the same (repeated). This shows a degenerate system. This also means $A$ is not diagonalizable. The trajectory take straight lines through the origin. We can deduce the trajectory by multiplying $A$ with the standard basis vectors as follows: $$ \lambda_1 = \lambda_2 = \frac{2}{2} = 1 $$ $$ q_1 = \begin{bmatrix} 0 & -1 \\ 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} $$ $$ q_2 = \begin{bmatrix} 0 & -1 \\ 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ \end{bmatrix} $$ Since $\lambda_1 = \lambda_2 > 0$, it shows an unstable degenerate node.
Alternatively, you can check out this video on how to sketch phase portraits: https://www.youtube.com/watch?v=dpbRUQ-5YWc
Hope it helps. Cheers!