I am solving a physics problem and done almost everything to a couple of last steps. I currently have two differential equations which I need to solve for functions$x(t)$ and $y(t)$:

$$ \ddot x =\dot y ω\cos(ax) $$ and $$ \ddot y =-\dot x ω\cos(ax) $$

where a,ω are real and positive constants

I tried rewriting both equations in this form:

$$\frac{d}{dt} \dot x =\dot y ω\cos(ax) $$ and $$\frac{d}{dt} \dot y =-\dot x ω\cos(ax) $$

And then insert equation 1 to equation 4 and equation 2 into equation 3.

This yielded me the $y(t)$ functon, which I currently know, however, the new differential equation that I need to solve for $x(t)$ looks like that: $$ \dddot{x} + ω^2\cos^2(ax)\dot{x}=0 $$

I'm not exactly sure whether my method of solving was correct in the first place, but if it was, could anyone give me hints on how to solve the last equation?

  • $\begingroup$ should the arguments to the trig functions $ax$ or $at?$ also you don't need all $*$ to indicate multiplication. $\endgroup$ – abel Feb 21 '15 at 16:08
  • $\begingroup$ Yeah, that's the point. It's the x in there and not t. I'll take out the '*' in a sec, sorry about that :( $\endgroup$ – Henrikas Feb 21 '15 at 16:11

The given coupled equations can be decoupled and still remain in second order.

$$ \ddot x =\dot y\, ω\,cos(at) ...(1*) $$ $$ \ddot y =-\dot x \,ω\,cos(at) ...(2*) $$

Divide (1*) by (2*) and re-arrange

$$ \dot x \ddot x + \dot y \ddot y = 0 $$

Integrating, $$ \dot x ^2 + \dot y^2 = c^2 ...(3*) $$ where c is an arbitrary constant.

This is satisfied by taking $$ \dot x = \pm \, c \cos \phi , \dot y = \pm \, c \sin \phi ...(4*) $$

where $\phi$ is any new arbitrary variable of time t.

We can here take advantage of the geometrical differential arc relations

$$ \frac{dx}{ds} = \pm \cos \phi, \frac{dy}{ds} = \pm \sin \phi,...(5*) $$

Without loss of generality we can now put $ t=s $ = arc length with $ c = 1.$ ...(6*)

Denoting derivative with dash for new arc variable instead of dot for time variable,

$$ x^{''} = \pm \sin \phi \cdot \phi^{'}, y^{''} = \pm \cos \phi \cdot \phi^{'}...(7*)$$

Plugging (7*) into (1*)

$$ \frac{d \phi}{ds} = \pm \omega \cos (a s) ...(8*) $$

Physically it means that its curvature $ \frac{d \phi}{ds} $ is a $ \sin/\cos $ oscillation

It can be noted that due to its oscillatory nature either sign yields qualitativly the same solution.The order of DE can still be raised.

As it was an interesting case, getting curious to see its integral, plotted (x-y) with boundary conditions: $ s =0 , \phi = 0 $ for the case $ a =1 ...(9*) $


Finally,integrating (8*) twice we obtain

$$ x(s)=\int\cos( \omega/a\,(\sin a s)) ds; y(s)=\int \sin( \omega/a\, (\sin a s)) ds ...(10*) $$

where independent variables $ t, s $ are same.

enter image description here

You can see the curvature sinusoidal behavior as required from (8*).


i will give it a try. let us define $$u = \frac{dx}{dt}, y=\frac{dy}{dt}$$ and rewrite the equations as $$\frac{du}{dt} = kv\cos ax, \, \frac{dv}{dt} = -ku\cos ax, \tag 1$$

from $(1),$ we get $$\frac{dv}{du} = - \frac uv$$ which has the integral $$u^2 + v^2 = A^2, u = A\cos s, v = A\sin s \text{ where $A$ is a constant and $s$ is a function of $t$. }$$ putting back in $(1),$ gives $$-A\sin s \frac{ds}{dt} = kA\sin s\cos ax$$ so we have $$\frac{ds}{dt} = -k\cos ax,\, \frac{dx}{dt} = A\cos s \tag2 $$

this gives rise to $$\frac{dx}{ds} = -\frac{A\cos s}{k\cos ax} $$ which can be integrated to yield $$ \sin s = -\frac k{aA}( \sin ax +C) \tag3$$

finally, you need to solve the following first order equation for $x:$

$$\frac{dx}{dt} = A\sqrt{1 - \sin^2 s} \text{ where $\sin s$ is given by } (3) $$

  • $\begingroup$ Thnks! But I still have a question: in the last equation we still have A, which is the square root of u^2+v^2 and both of those are functions dependant on t. If I insert (3) into the last equation, the A goes under the square root. How do I solve the last equation when I have $\dot{x}$ and $\dot{y}$ under the root? $\endgroup$ – Henrikas Feb 21 '15 at 22:13
  • $\begingroup$ @Henrikas, $A$ is a constant. all the time dependence is in $s.$ look at the differential equation $(1).$ that makes $u^2 + v^2$ constant. $\endgroup$ – abel Feb 22 '15 at 0:49

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