Given a System of Differential Equations, Show All Solutions in Phase Portrait are Elliptical

Question

I found the eigenvalue, $\lambda =i$

Then I did: det(A-$\lambda$)•(V) = $0$ where V = (x, y) is the eigenvector.

I Let x = 10, then solved for y:

y = 3 + i, therefore V $= (10, 3 + i)$

After separating the complex and real part of the solution the solution to the system is:

Y(t) $= k_1\left[10cos\left(t\right),\:3cos\left(t\right)-sin\left(t\right)\right]+k_2\left[10sin\left(t\right),\:cos\left(t\right)+3sin\left(t\right)\right]$

I know that up until this point, my answer is correct, and I understand it up to this point.

I do not understand how to show that all solutions on the phase portrait for this system are elliptical. How do I take the solution and form it in a way that proves this?

Help would be greatly appreciated. TIA!

Answer 1

As you already obtained

$$ x_1(t) = C_1\cos(t)+(10C_2-3C_1)\sin(t)\\ x_2(t) = C_2\cos(t)+(3C_2-C_1)\sin(t) $$

or

$$ x_1(t) = a\cos(t)+b\sin(t)\\ x_2(t) = c\cos(t)+d\sin(t) $$

solving for $\sin(t),\cos(t)$

$$ \sin(t) = \frac{c x_1-a x_2}{bc-ad}\\ \cos(t) = \frac{d x_1-b x_2}{a d-b c} $$

and then

$$ \sin^2(t) + \cos^2(t) = 1 = \left(\frac{c x_1-a x_2}{bc-ad}\right)^2+\left(\frac{d x_1-b x_2}{a d-b c}\right)^2 $$

or

$$ (bc-ad)^2 = \left(c x_1-a x_2\right)^2+\left(d x_1-b x_2\right)^2 $$

which represent a general slanted ellipse.

NOTE

$$ b c-a d = C_1^2-6C_1C_2+10C_2^2 > 0 \ \ \mbox{for} \ \ C_1\ne 0,\ \ \mbox{and}\ \ C_2 \ne 0 $$

Attached the phase plane with an orbit associated to the initial conditions $x_1(0) = x_2(0) = 0.1$

Answer 2

The coefficient matrix is similar to $$J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ You can find a suitable change of basis matrix $B$ by taking real and imaginary parts of one of the eigenvectors. In particular, we can take $$B=\begin{bmatrix}10&0\\3&-1\end{bmatrix}$$ to get the decomposition $BJB^{-1}$ for this system’s coefficient matrix.

The solutions curves of the original system are therefore affine images of the curves $$\exp(tJ)\mathbf C = \begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix} \begin{bmatrix}c_1\\c_2\end{bmatrix}.$$ Eliminating $t$ from the resulting parametric equations yields $x^2+y^2=c_1^2+c_2^2$, which is the equation of a circle, so the solution curves to the original system are ellipses.