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I have 2 functions of time $f(t),g(t)$ and a condition for the time-derivative of a third function $h(t)$, say $$\dot{h}(t)=\dot{g}(t)\cos{f(t)},$$ so $h$ is defined provided a value for $h(0)$ (as $h(t)=h(0)+\int_0^t\dot{h}(t)$). We can choose $h(0)$ as we like while maintaining this condition.

I have certain functions, say $A(t), B(t)$ that depend on $f$, $g$ and $h$ in convoluted ways, i.e. $$A(t) = \cos(f(t))\sin(g(t))\sin(h(t))-\sin(f(t))\cos(g(t)),$$ $$B(t)=-\sin(f(t))\sin(g(t))\sin(h(t))+\cos(f(t))\cos(g(t)),$$

but I know for a fact that there is a value of $h(0)$ that solves $\dot{A}(t)=0, \dot{B}(t)=0$ (in this case multiple values, since the solution is $2\pi$ periodic).

I want to find the value of $h(0)$ from this set of conditions, and any other relations between $f,\dot{f},g,\dot{g}$ that my arise from it, should the conditions be strict enough.

What mathematical tools do I need to find it?

Reasons why I'm stuck with the problem:

  • I cannot analytically integrate the condition for $\dot{h}(t)$, so I cannot get an expression for $h(t)$ that does not involve an integral and I want to avoid integro-differential equations.
  • Therefore, whenever I derive $A(t)$ and $B(t)$, I get dependences on $h(t)$ that I wish I could swap to dependences on $\dot{h}(t)$ and $h(0)$ to apply the condition, but I cannot.
  • Without such simplifications, I don't know how to proceed with my expressions for $\dot{A}(t),\dot{B}(t)$.

Any wild idea, topic that I can read/explore... will be much appreciated!

Thank you!

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  • $\begingroup$ are $f$ and $g$ given? what are they? you do want some information to make progress. $\endgroup$ Commented Nov 21 at 16:15
  • $\begingroup$ They are general expressions that come from a physical problem. Analyzing symmetries in the physical problem, you can argue that $f$ and $g$ obey some condition such that it is guaranteed that $\dot{A}(t)=0, \dot{B}(t)=0$ for a valid choice of $h$. Solving the system of equations $\dot{A}(t)=0, \dot{B}(t)=0$ will clarify the choice of $h$ and eventually show an explicit form of the relation between $f$, $g$ and their derivatives, which is not apparent at first sight. $\endgroup$ Commented Nov 21 at 16:53

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