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System of two differential equations with trigonometric functions

  1. Feb 23, 2012 #1
    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.
     
    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 24, 2012 #2
    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.
     
  4. Feb 24, 2012 #3
    ...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 ?
     
    Last edited: Feb 24, 2012
  5. Feb 24, 2012 #4
    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.
     
  6. Feb 25, 2012 #5
    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]
     
  7. Feb 25, 2012 #6
    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.
     
  8. Feb 26, 2012 #7
    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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