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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

1 Answer 1

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.

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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
    
mcandril -- ok but what is it that you would like to know about the solution? There are (almost certainly) no solutions in elementary functions. Apart from that one can ask many questions, e.g. whether solutions are bounded, whether 0 is stable, how the solutions depend on the parameters etc. –  algori Sep 22 '11 at 19:43
    
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

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