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System of PDEs

  1. Nov 15, 2006 #1
    I have a system of two PDEs:

    [tex]y_t+(h_0v)_x=0 \quad (1a)[/tex]
    [tex]v_t+y_x=0 \quad (1b)[/tex],

    where [itex]h_0[/itex] is a constant.

    Then I want to show that (1) has traveling wave solutions of the form

    [tex]y(x,t)=f(x-ut) \quad (2a)[/tex]
    [tex]v(x,t)=g(x-ut) \quad (2b)[/tex],

    where [itex]u[/itex] is the propagation velocity.

    Differentiating (1a) w.r.t. [itex]x[/itex] and (1b) w.r.t. [itex]t[/itex] I conclude that
    [tex]\frac{\partial^2v}{\partial t^2}=h_0\frac{\partial^2v}{\partial x^2}[/tex],
    but I am not sure that this is useful. I have tried a linear change of variables in order to arrive at a solution. Introducing the variables [itex]\alpha=ax+bt[/itex] and [itex]\beta=cx+dt[/itex] I arrive at a solution for [itex]v[/itex], namely either [itex]v=C(x+t)[/itex] or [itex]v=C(x-t)[/itex], depending on how the constants [itex]a,b,c,d[/itex] are chosen. Here [itex]C[/itex] is an arbitrary function. Indeed this looks like (2b), but I would like to put forward a more rigorous argument.

    I would like some thoughts on this problem, and any hint on how to solve it is welcome. And is it possible to say anything about the relationship between [itex]f[/itex] and [itex]g[/itex] in (2)?
     
    Last edited: Nov 15, 2006
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  3. Nov 15, 2006 #2

    HallsofIvy

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    I'm not sure what more you want. Yes, differentiating as you did eliminates y from the equation leaving
    [tex]\frac{\partial^2v}{\partial t^2}=h_0\frac{\partial^2v}{\partial x^2}[/tex]
    which is the wave equation. It is well known that
    [tex]f(x-\sqrt{h_0}t)[/tex]
    and
    [tex]f(x+\sqrt{h_0}t)[/tex]
    are solutions for any twice differentiable function f, and you can show that by substituting them into your equation.
    Of course, if you differentiate the first equation with respect to t and the second with respect to x you can eliminate v, getting the same equation for y.

    If all you want to do is show that
    you don't really need to do all that. Just substitute those functions: yx= f ' and yt= -u f '; vx= g' and vt= -ug'. Substitute those into the equations and you can see that they satisfy both equations as long as u2= h0.
     
  4. Nov 15, 2006 #3
    HallsofIvy, thank you for the reply.

    I guess it suffices to substitute the assumed solutions into the system, and show that they satisfy the differential equations. This is of course the straightforward way to do it. As Einstein said: "Everything should be done as simple as possible, but no simpler."

    What can we say about the relationship between f and g? Obviously, we can read it from the equation, but can we say something in general?
     
  5. Nov 15, 2006 #4

    HallsofIvy

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    Not without initial conditions. For example, if we have some given initial shape, F(x), and are told that the wire or string or whatever is held in that shape and then released (the partial derivative with respect to t is 0 when t=0, a very common situation) then we know that f(x)+ g(x)= F(x) and that cf '(x)- cg'(x)= 0. From the second equation, f '(x)= g'(x) so f(x)= g(x)+ C. From the first then g(x)+ C+ g(x)= F(x) or 2g(x)= F(x)- C so g(x)= (1/2)F(x)- (1/2)C. Then f(x)= g(x)+ C= (1/2)F(x)+ (1/2)C. The solution is (1/2)F(x+ ct)+ (1/2)F(x-ct) (The constants cancel). That is, the initial shape divides into two 1/2 size waves going in opposite directions.

    For any functions, f and g, it is possible to come up with initial conditions that will yield them so nothing can be said in general.
     
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