Find the axes of an ellipse described by coupled ODEs

Question

I am given the set of equations:

$$ \dot{x} = 5x + 2y, \hspace{2mm} \dot{y} = -17x - 5y $$

so the relevant matrix is $$ A =\begin{pmatrix} 5 & 2 \\ -17 & -5 \end{pmatrix} $$ Now I have found the eigenvalues and eigenvectors to be $$ \lambda_{1,2} = \pm 3i \\ v_1 = \left(1, \hspace{2mm} \frac{1}{2}(3i-5) \right), \hspace{2mm} v_2 = \left(1, \hspace{2mm} -\frac{1}{2}(3i+5)\right) $$

but now I'm required to draw a phase portrait, but I don't know how to determine the axes of the ellipse that this describes.

Answer 1

For $\lambda=3i$ an eigenvector is $$\begin{bmatrix}-2\\5-3i\end{bmatrix}$$. If we let $$P=\begin{bmatrix}-2&0\\5&3\end{bmatrix}$$ and let $$\begin{bmatrix}x\\y\end{bmatrix}=P\begin{bmatrix}u_1\\u_2\end{bmatrix}$$we have $$\begin{bmatrix}\dot{u_1}\\\dot{u_2}\end{bmatrix}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}\begin{bmatrix}u_1\\u_2\end{bmatrix}$$ Thus $u_1=K\sin(3t),u_2=-K\cos(3t)$ so $$u_1^2+u_2^2=K^2$$ $$\begin{bmatrix}u_1\\u_2\end{bmatrix}=\frac{-1}{6}\begin{bmatrix}3&0\\-5&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$ $$(3x)^2+(-5x-2y)^2=C$$ $$34x^2+20xy+4y^2=C$$ $$17x^2+10xy+2y^2=C'$$ Rotate the ellipse into standard position by $$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\cos \theta& -\sin \theta\\\sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}x'\\y'\end{bmatrix}$$ where $\tan (2\theta)=\frac{10}{17-2}=\frac{2}{3}$ and you should have no problem with the axes.

Answer 2

$$\dot{x} = 5x + 2y, \hspace{2mm} \dot{y} = -17x - 5y$$ $$\dfrac {dx}{dy}=\dfrac {5x+2y}{-17x-5y}$$ $$({-17x-5y}) {dx}=( {5x+2y})dy$$ $$-17xdx-5dxy=2ydy$$ After integration it gives: $$ {17}x^2+10xy+2y^2=c$$ $$Ax^2+Bxy+Cy^2=c$$ Rotate the axes of a certain angle. For the angle you have: $$\cot (2 \theta)=\dfrac {A-C}{B}=\dfrac 32$$

Answer 3

At the axes of the ellipse you need $x^T\dot{x}=x^TAx=0$ (where $x$ is the state vector). If you solve it you get the vectors $$\left[\frac{3+\sqrt{13}}{2} ~~~~ 1\right] ~~\text{and}~~ \left[\frac{3-\sqrt{13}}{2} ~~~~ 1\right]$$ which are orthogonal as expected.

Answer 4

If $A(u+iv) = ib (u+iv)$ then $A(u -iv) = -ib (u-iv)$ and with the usual adding and subtracting we get $Au= -bv, Av=bu$ and so with $W=\begin{bmatrix} u & v \end{bmatrix}$ we have $W^{-1} A W = B=\begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix}$.

In particular, with $y=W^{-1} x$, we have $\dot{y} = B y$ and hence we see that the trajectory of $y$ is the circle $\|y\| = 1$, assuming that $\|y(0)\| = 1$.

You want to determine the axes of $\{W y | \|y\| = 1 \} = \{ x | x^T W^{-T} W^{-1} x = 1 \}$.

Using $u = (2,-5)^T, v = (0, 3)^T$ we get $W^{-T} W^{-1} = {1 \over 18} \begin{bmatrix} 17 & 5 \\ 5 & 2 \end{bmatrix}$