# System of two differential equations with trigonometric functions

1. Feb 23, 2012

### mnb96

Hello,
do you have any strategy to suggest in order to solve the following system of partial differential equations in x(s,t) and y(s,t)?

$$\frac{\partial x}{\partial t} = x - \frac{1}{2}\sin(2x)$$
$$\frac{\partial y}{\partial t} = y \; \sin^2(x)$$

(note that the partial differentiation is always with respect to t).
In case it might be useful, I can safely assume that the codomain of x(s,t) and y(s,t) is [-1,1].

I already tried with Maple and Mathematica but they only give me numerical solutions.
An approximation would be ok for me, as long as I get a closed form for x and y.

I was also wondering if you think there might exist another system of coordinates in which this system is easier to solve

Thanks.

Last edited: Feb 23, 2012
2. Feb 24, 2012

### bigfooted

You might try dividing one equation by the other (left hand side by left hand side and right hand side by right hand side) and get a single ode for dy/dx. The result is of the form y'=f(x)*y.
I am surprised that mathematica is not able to solve this.

3. Feb 24, 2012

### mnb96

...it is also possible that I don't know how to use properly those packages, as I am a total beginner. For Mathematica I used the following syntax:
------ DSolve[{D[x[t], t] == x[t] - 1/2*Sin[2*x[t]], D[y[t], t] == y[t]*Sin[x[t]]^2}, {x[t], y[t]}, t]

while for Maple:
------ dsolve({diff(x(t), t) = x(t)-(1/2)*sin(2*x(t)), diff(y(t), t) = y(t)*sin(x(t))^2}, {x(t), y(t)})

With both packages I obtain a numerical solution involving integrals, which I guess indicates a failure in finding a closed form solution.

***********
*** EDIT: ***
***********
I tried to divide the first equation by the second, as you suggested and obtained:

$$y(x)=\pm \sqrt{\frac{-2x}{\tan(x)}+c}$$

but now we only have y(x), and I guess we still need to solve the first differential equation in order to obtain x(t), am I right?
Am I also free to set c=0 ?

Last edited: Feb 24, 2012
4. Feb 24, 2012

### hawaiifiver

Are you looking for x(s, t) and y(s, t). Then you just integrate with respect to t surely.

x(s, t) = xt - t/2 sin(2x) + f(s)

y(s,t) = y t sin^2 (2x) + f(s)

where f(s) is a function of s.

5. Feb 25, 2012

### mnb96

Hi hawaiifiver...I am afraid something went wrong with your solution, because you treated x(s,t) as a constant, and did not take into account that x is a function of t.
For instance the solution of the simper differential equation
$$\frac{\partial x}{\partial t} = x(s,t)$$
is not tx(t,s). It is instead $x(s,t)=e^t c(s)$

6. Feb 25, 2012

### hawaiifiver

Perhaps you could elaborate on how you found x(s,t). I thought of x as a function of two variables, i.e. s and t.

7. Feb 26, 2012

### alan2

The first equation, in x, is separable. Once you have x, including an arbitrary function of s, substitute that into the equation for y which become separable. I don't see how to do the first integration of 1/{x-sin(2x)/2}.