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System of two differential equations with trigonometric functions

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mnb96
#1
Feb23-12, 09:45 AM
P: 625
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)?

[tex]\frac{\partial x}{\partial t} = x - \frac{1}{2}\sin(2x)[/tex]
[tex]\frac{\partial y}{\partial t} = y \; \sin^2(x)[/tex]

(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.
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bigfooted
#2
Feb24-12, 02:32 AM
P: 285
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.
mnb96
#3
Feb24-12, 03:34 AM
P: 625
...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:

[tex]y(x)=\pm \sqrt{\frac{-2x}{\tan(x)}+c}[/tex]

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 ?

hawaiifiver
#4
Feb24-12, 02:36 PM
P: 57
System of two differential equations with trigonometric functions

Quote Quote by mnb96 View Post
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)?

[tex]\frac{\partial x}{\partial t} = x - \frac{1}{2}\sin(2x)[/tex]
[tex]\frac{\partial y}{\partial t} = y \; \sin^2(x)[/tex]

(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.
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.
mnb96
#5
Feb25-12, 11:52 AM
P: 625
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
[tex]\frac{\partial x}{\partial t} = x(s,t)[/tex]
is not tx(t,s). It is instead [itex]x(s,t)=e^t c(s)[/itex]
hawaiifiver
#6
Feb25-12, 03:59 PM
P: 57
Quote Quote by mnb96 View Post
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
[tex]\frac{\partial x}{\partial t} = x(s,t)[/tex]
is not tx(t,s). It is instead [itex]x(s,t)=e^t c(s)[/itex]
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.
alan2
#7
Feb26-12, 11:19 AM
P: 199
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}.


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