# Simple system of ODEs with periodic coefficients

I am stuck with a little problem that I cannot solve mith the standard methods I learn at university. I have a system of coupled ODEs:

$f'(t) = P \cos(k t + \Phi_1) g(t)$

$g'(t) = Q \cos(k t + \Phi_2) f(t)$

From what I read, the is no teaching book-way to solve such a system. I found question 66172 to be similar, but my values are $P,Q$ up to $\approx 10^3-10^4$ and $k$,$\Phi_{1,2}$ basically arbitrary ($k$ ranges from $-10^5$ to $10^5$ explicitly including 0). Only $t$ may be restricted to the order of magnitude around $10^{-2}$. Boundary conditions may be $f(0)=1$, $g(0)=0$ and nice to have would be $f(0) \approx g(0)$

I have done numerical calculations and for my numbers, the solutions do not look to crazy. For $k=0$ the system is solvable and basically gives oscillations between the two functions. For $k\neq0$ I basically geo Oszillations withs higher frequency and lower amplitude (e.g., $f(t)$ oszillating between 1 and 0.8 and $g(t)$ between 0 and 0.2). For $k\gg10^5$, the system becomes mostly stationary.

I would really appreciate suggestions for an analytical approach, at least as an approximation.

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If they are NOT differential operators, then notice that by substituting the second equation into the first you get either $f(t) = 0, or$PQ\cos(kt + \Phi_1) \cos(k t + \Phi_2) = 1,$which means that$k=0.$Voting to close. – Igor Rivin Sep 20 '11 at 14:42 Some reaction of the OP would be appreciated... One other possibility is that$f'(t)=..." was meant... – András Bátkai Sep 20 '11 at 17:08
@Willie: Damn, sorry. Of course, the equations should be in normal form and I forgot the ′. Just corrected it. – mcandril Sep 21 '11 at 7:21
@abatkai Yes, that's it. Suppose it's the weather, could sleep all day right now :) – mcandril Sep 21 '11 at 7:22
What does "I found question 66172 to be similar" mean? – David Roberts Sep 21 '11 at 7:37

If $t\to A(t)$ is a $C^1$-smooth matrix valued function on $\mathbb{R}$, then the solution of the ODE system $x'=A(t)x, x(0)=x_0$ can be written as $x(t)=Pe^{\int_0^t A(s) ds} x_0$ where $$Pe^{\int_0^t A(s) ds}=\lim_{n\to\infty}\left( (1+t\frac{A(t_0)}{n})\cdot (1+t\frac{A(t_1)}{n})\cdot\cdots\cdot (1+t\frac{A(t_n)}{n})\right), t_j=(1-\frac{j}{n})t.$$
This expression is called the path ordered exponential. Note that if $A$ is constant, we get $Pe^{\int_0^t A(s) ds}=e^{At}$ as expected. Note also that when all $A(t)$'s are simultaneously diagonalizable the path ordered exponential can be written using ordinary (1-variable) exponentials. In general $Pe^{\int_0^t A(s) ds}$ can't be expressed using "elementary functions", but it does have some nice properties, e.g. $Pe^{\int_t^r A(s) ds}Pe^{\int_0^t A(s) ds}=Pe^{\int_0^r A(s) ds}$ etc.
I don't see how these nice properties help me, though. :) What I am trying to solve is the $k$ dependent coupling between two quanteties for a given value of $t$. – mcandril Sep 22 '11 at 18:45
algori -- Mainly how the solution depends on the parameters. The system describes a physical experiment, where k is modified to find information about $P$,$Q$ and the $\Phi$s. – mcandril Sep 23 '11 at 10:10
Mcandril, algori's solution tells you a lot. For example, if $k=0$, your matrix $A(t)$ is constant in time; the solution has simple features. If $k\not=0$, there are three regimes to consider: $|k|\approx 0$, $|k|\rightarrow +\infty$, and the intermediate regimes. The solution in the asymptotic regimes can, and should, be analysed asymptotically (see, eg. Bender and Orszag for inspiration). Also, a word of caution re numerical simulations for large $k$: are you sure you don't have a very stiff ODE system (large separation of time scales)? Did you use a DAE solver? – Nilima Nigam Oct 20 '11 at 18:47