I have the following system of ordinary differential equations with $\gamma(t),M(t),$ and $\beta(t)$ strictly positive and smooth.

$\dot{g_1}(t) = \gamma(t)M(t){g_2}(t)$
$\dot{g_2}(t) = \beta(t)\frac{1}{M(t)}{g_1}(t)$

with initial conditions


$0\le t \le 1$

How would I go about finding a non-trivial solution?

Editing to add that $\gamma(t), M(t),$ and $\beta(t)$ are finite on $[0,1]$

2nd edit: Corrected typo in domain of $t$

  • $\begingroup$ I'm having a little trouble figuring out how "$0 < t < 1$" fits into your question. Is it the range of $t$ over which you seek a solution? $\endgroup$ – Robert Lewis Nov 3 '14 at 23:16
  • $\begingroup$ Yes, that is correct. It is the domain on which the functions are defined and should be solved. $\endgroup$ – ELC Nov 3 '14 at 23:17
  • $\begingroup$ If $M(t)$ is bounded away from zero, why is the problem considered in these terms, rather than absorbing $M(t), 1/M(t)$ into the factors $\gamma(t),\beta(t)$ respectively? $\endgroup$ – hardmath Nov 14 '14 at 12:19
  • $\begingroup$ The problem originated from a biological system in which $\gamma(t), M(t),$ and $\beta(t)$ have specific meaning, which is why I prefer to keep the system in its current form instead of absorbing $M(t)$. $\endgroup$ – ELC Nov 15 '14 at 19:15

EEDDIITT: given a system $$ f_1' = g f_2, \; \; \; f_2' = h f_2, $$ you get separate equations $$ f_1'' = \left( \frac{g'}{g} \right) f_1' + g h f_1 $$ $$ f_2'' = \left( \frac{h'}{h} \right) f_2' + g h f_2. $$

ORIGINAL: Currently see no hope of a closed form solution in general. Bounds are available, and numerical solution always is, assuming you have your coefficient functions in an explicit manner.

There is one circumstance with a solution, a fairly restrictive one. Let $$ A(t) = \int_0^t \gamma(s) M(s) ds, $$ $$ B(t) = \int_0^t \frac{\beta(s)}{ M(s)} ds. $$ The restrictive condition is $$ \frac{A}{B} $$ constant. In that case, the fundamental solution matrix is $$ \left( \begin{array}{cc} \cosh \sqrt {AB} & \frac{\sqrt A \sinh \sqrt {AB}}{\sqrt B} \\ \frac{\sqrt B \sinh \sqrt {AB}}{\sqrt A} & \cosh \sqrt {AB} \end{array} \right). $$ With your initial condition, $$ g_1(t) = \cosh \sqrt {AB} + \frac{\sqrt A \sinh \sqrt {AB}}{\sqrt B}, $$ $$ g_2(t) = \cosh \sqrt {AB} + \frac{\sqrt B \sinh \sqrt {AB}}{\sqrt A}. $$

The theorem used here is that, for a square matrix $H(t),$ if $H$ and $\dot{H}$ commute, then $$ \frac{d}{dt} e^H = e^H \dot{H } = \dot{H} e^H. $$


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