Analytic solution to elliptic coupled ODEs

Question

From my numerical solution I see that the solution to $$ \frac{dT}{dt} = a T - b h \\ \frac{dh}{dt} = -c h - d T $$ is an ellipse (where $a, b, c$ and $d$ are constants). Can this be solved analytically?

Answer 1

$$\frac{d}{dt}\begin{pmatrix}T\\h\end{pmatrix} = \begin{pmatrix}a&-b\\-d&-c\end{pmatrix} \begin{pmatrix}T\\h\end{pmatrix} $$ The general solution is $$\begin{pmatrix}T\\h\end{pmatrix}(t) =\exp\left(t\begin{pmatrix}a&-b\\-d&-c\end{pmatrix}\right)C,$$ where $C\in \Bbb R^2$ - an arbitrary constant vector (can be found by using, for example, initial data). Also, you can read about matrix exponent.

The qualitative behaviour of the solution depends on the Jordan normal form of the matrix $\begin{pmatrix}a&-b\\-d&-c\end{pmatrix} $ - i.e. eigenvalues and the structure of eigenvectors and generalised eigenvectors.

If, for example, the eigenvalues are $\pm i \lambda$, $\lambda\in \Bbb R$, then the solution is indeed an ellipse.

Answer 2

HINT

Suppose you extract $h$ from the first equation and you replace everything in the second equation. This should let you with a second order differential equation in $T$ and so, you will arrive to a parametrized definition of $h(t)$ and $T(t)$.

Answer 3

Continuiing TZakrevskiy's answer, the solution of this system is an ellipse iff the eigenvalues of the matrix $A$ are purely imaginary.

This happens iff $$ a-c=0 \quad\text{and}\quad ac+bd<0. $$