I was solving a question, in which I was asked to solve the following system of differential equations: $$ \dot x = 3x + 2y, \quad \dot y = -5x - 3y. $$ I got the general solution in $\mathbb R^2$ to be $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} C & D \\ -\frac32 C + \frac12 D & -\frac12 C - \frac 32 D \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}, \quad C,D\in \mathbb R. $$ Now, I want to sketch the phase portrait of this solution (apparently, as we change $C$ and $D$, only the scale of the ellipse changes, not its orientation or ratio between major and minor axes). My question is, how do I determine the orientation of axis lengths of this ellipse (in terms of $C$ and $D$)? Given an arbitrary ellipse of the form $$ \begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}, $$ how would one sketch it?
4 Answers
$$\dot x = 3x + 2y, \quad \dot y = -5x - 3y.$$ $$2\dot x+\dot y = x + y, \quad \dot y+\dot x = -2x - y.$$ Substitute $u=x+y$ and $v=2x+y$: $$\dot v= u ,\quad \dot u=-v$$ $$\implies \dfrac {dv}{du}=-\dfrac u v$$ $$v^2+u^2=C$$ $$(x+y)^2+(2x+y)^2=C$$
late to the party. Given $$ \left( \begin{array}{c} \dot x \\ \dot y \end{array} \right) = \left( \begin{array}{cc} p & q \\ r&s \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right), $$ we can have a constant quadratic form (other than at the origin) only when the trace or the determinant of the coefficient matrix is zero. Zero determinant means a line. More interesting is trace zero.
When $$ \left( \begin{array}{c} \dot x \\ \dot y \end{array} \right) = \left( \begin{array}{cc} p & q \\ r&-p \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) $$ we can readily calculate that $$ \frac{d}{dt} \; \left( rx^2 - 2pxy - q y^2 \right) = 0 $$ The type of conic section we see is determined by the discriminant $$ \Delta = 4p^2 + 4qr ,$$ as well as the sign of the constant $C$ in $ rx^2 - 2pxy - q y^2 = C $
If you set $\vec\alpha=(C,E)$ and $\vec\beta=(D,F)$, then the parametric equation can be written as $\vec p(t)=\vec\alpha\cos t+\vec\beta\sin t$ and vectors $\vec\alpha$, $\vec\beta$ are two conjugate semi-diameters.
But conjugate semidiameters are related to semi-axes $a$, $b$ of the ellipse by Apollonius' formulas: $$ a^2+b^2=|\vec\alpha|^2+|\vec\beta|^2, \quad ab=|\vec\alpha\times\vec\beta| $$ and can thus be computed. To find the directions of the axes one could follow a nice geometric construction, or find the values of $t$ for which the derivative of $|\vec p(t)|^2$ vanishes, obtaining: $$ \tan 2t={2\vec\alpha\cdot\vec\beta \over|\vec\alpha|^2-|\vec\beta|^2}. $$
First write $\cos t$ and $\sin t$ in terms of $x$ and $y$: $$ \begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}\\ \begin{pmatrix} \cos t \\ \sin t \end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}^{-1}\begin{pmatrix} x \\ y\end{pmatrix} $$ The inverse is a $2\times2$ matrix, say $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$. You also know $\cos^2 t+\sin^2t=1$. So $$(\alpha x+\beta y)^2+(\gamma x+\delta y)^2=1$$ You can group the terms together, then use this answer to get the orientation, and something like this answer to find the length of the semiaxes.
