Unsolvable first order nonlinear differential equation? $\frac{dx(t)}{dt} + \sin(x(t)) = \sin(\omega t)$

I do want to solve the following differential equation analytically: $$\frac{dx(t)}{dt} + \sin(x(t)) = \sin(\omega t)$$ I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?

Thanks for your help!

• No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $\sin x\approx x$. – Yves Daoust Nov 20 '18 at 17:21
• What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable. – R. Caloudis Nov 20 '18 at 17:24
• It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series. – Robert Israel Nov 20 '18 at 17:24
• Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses? – R. Caloudis Nov 20 '18 at 17:26
• The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term. – R. Caloudis Nov 20 '18 at 17:30

Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $$\omega=1$$ You can plot just a fundamental patch $$t\in[0,2\pi]$$, $$x\in[-\pi,\pi]$$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution): Alternative visualization: Edit. Code for the plots, in Mathematica:

sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x == x0},
x, {t, 0, 4 \[Pi]}, {{x0, -3 \[Pi], 3 \[Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 \[Pi], 2.5 \[Pi], .25}], {t, 0, 3.5 \[Pi]}]

per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x == x[2 \[Pi]]},
x, {t, 0, 2 \[Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 \[Pi]}, {x, -\[Pi], \[Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 \[Pi]}, PlotStyle -> Red]]
• Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t. – R. Caloudis Nov 20 '18 at 18:01
• Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions. – Federico Nov 20 '18 at 18:03
• I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system – Federico Nov 20 '18 at 18:05
• A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) \approx 2.604965594$. – Robert Israel Nov 20 '18 at 19:48
• @RobertIsrael Updated to the plot, thanks – Federico Nov 21 '18 at 15:24