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Diff Eqs

  1. Sep 21, 2005 #1
    A rod of uniform elastic material of length 1/2 lies along the X axis with its left end fixed at x=0. At time t=0, an identical rod hits the riht end of the first rod with a speed of v. The second rod is thereafter kept alongside the first rod, and neither end is fixed. If the Young's Modulus and the density of the rods are such taht c=1, and if the displacement u is a generalized solution of the wave equation, find u(x,t) for t>0. Sketch the x-t diagram

    im guessing the conditions look like
    g(x) = [tex] \frac{\partial^2 u}{\partial t^2} - c^2 \frac{\partial^2 u}{\partial x^2} = 0[/tex]
    [tex] u(0,t) = 0, for t>0 [/tex]
    [tex] \frac{\partial u}{\partial t} (1,t) = 0 [/tex]
    [tex] \frac{\partial u}{\partial t} (\frac{1}{2},t) = v [/tex]
    well f(x) =0 because it is not stated otherwise

    so g(x) = -v for 0<x<1/2
    and g(x) = v for 1/2<x<1
    the left hand side rod would go downward after its right hand side was hit. Also the right hand rod would go upward because of the impulse of the left side rod.

    First of all am i right? Also is there any derivation of sorts that i need to put down... because this question is actually part 2 of a question of similar kind - however of the first part involved the right side's right end to be fixed. There the part between 0 and 1/2 was zero and the 1/2 to 1 interval was -v.
    Also would the x-t graph be a wave that looks like a negative sine wave??

    Any kind of help would be appreciated! Thank you!
  2. jcsd
  3. Sep 22, 2005 #2


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    Hey Stunner, it's crucial to set up the problem like a work of art. Really. This is what I think your problem is:

    [tex]\text{DE:}\quad u_{xx}=u_{tt} \quad 0\leq x \leq 1/2 [/tex]

    [tex]\text{BC:}\quad u(0,t)=0\quad u_x(1/2,t)=0[/tex]

    [tex]\text{IC:}\quad u(x,0)=0\quad u_t(x,0)=g(x)=-2x[/tex]

    (where v=2 in the downward direction)

    If it's not then correct it. Note according to the boundary conditions, the left end is fixed and the right end is free to move up and down.

    Note the initial conditions: Starts just flat with a downward velocity equal to -2x.

    Well in this case, to use D'Alembert's formula, we need to periodically odd-extend g(x) on the left side and periodically even-extend it on the right side because of the derivative. The first plot is g(x). Second plot is g(x) plus the odd-extension to -1/2, third plot adds the even extension to x=1. I'm pretty sure this is right but it's been a while for me. So you'll need to square it with yourself ok (or someone here correct it)?

    Tell you what, suppose this is the correct problem. Could you calculate:


    Attached Files:

    Last edited: Sep 22, 2005
  4. Sep 22, 2005 #3
    how did you figure out that g(x) = -2x? Thus, why is the v=2?

    how does the deriavtive tell you to extend the function even or odd? Is that because of how this function is located in the coordinate system??

    Also [tex]u(1/2,3/4)[/tex] extends from -1/4 to +5/4. THus the function from -1/4 to 1/2 is -2x, and the function from 1/2 to 5/4 is 2(x-0.5) -1. AM i right?

    So [tex] u(\frac{1}{2},\frac{5}{4}) = \frac{1}{2} \int_{-1/4}^{1/2} -2x dx + \int_{1/2}^{5/4} 2x-2 dx [/tex] is this correct?

    and that yields -3/16.
  5. Sep 22, 2005 #4


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    I'm just guessing the velocity is -2x ok? I'm not really sure what your problem is. That extenstion stuf, well it's to model the wave equation on an infinite domain so we extend the initial conditions accordingly. Why the even extension for derivatives? Hum, well I guess I'd need to review that one. The even-extension you calculated is correct.

    Anyway, I get u(1/2,3/4)=-0.1875 and this agrees with numerical methods via Mathematica. That code is below as well as a plot of the results. Note how the left side stays stationary but the right side, which is free to move, cycles up and down.

    Make sure and bracket the integrals too in the solution::

    [tex]u(\frac{1}{2},\frac{3}{4}) = \frac{1}{2}\left( \int_{-1/4}^{1/2} -2x dx + \int_{1/2}^{5/4} (2x-2) dx\right) [/tex]

    Here's the Mathematica code. You have access to Mathematica?

    Code (Text):
    NDSolve[{D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}], u[x, 0] == 0,
        Derivative[0, 1][u][x, 0] == -2 x, u[0, t] == 0,
        Derivative[1, 0][u][1/2, t] == 0}, u, {x, 0, 1/2}, {t, 0, 2},
      AccuracyGoal -> 25]

    Plot3D[Evaluate[u[x, t] /. First[%]], {x, 0, 1/2}, {t, 0, 2},
      ViewPoint -> {2.492, -1.959, 1.184}]
    f[x_, t_] := Evaluate[u[x, t] /. sol1];
    f[1/2, 3/4]

    Attached Files:

    Last edited: Sep 22, 2005
  6. Sep 22, 2005 #5
    ok we figured out u(1/2,5/4) but that gives the position of the rod at that point in time for that part of the rod. But doesnt the question ask for u(x,t) in general? Or is this satisfactory?

    THank you for your help so far i have really understood what to do in these situations! I guess ill have to learn this all on my own since we didnt even come close to this in class.
  7. Sep 22, 2005 #6


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    I don't think so. To calculate u(x,t) from t in (0,2) and x in (0,1/2) you'd need to use the extension shown below. That is from x-t to x+t in that range, the extended g(x) would range from -2 to 5/2. And the problems of course, is that the integrals change depending on where the values are.

    I suppose if one had to plot it via D'Alembert's formula, would need to write a C++ or Mathematica program to calculate each value of u(x,t) as a set of integrals depending on the values of x and t.

    Also I'm pretty sure all this is correct but I'll assume you'll check everything with what you're doing in class Ok. :smile:

    Attached Files:

  8. Sep 22, 2005 #7
    well it is not a plot i am interested in, though. SO are you saying that the g(x) changes for all intervals? But i am quite certain they want the way the g(x) behaves on the interval from 0 to 1/2 and from 1/2 to 1, nothing more. That is how the book answered a question similar to this one.

    Also can you have a look at this question - quite similar to the one you helped me answer earlier (with the extension of functions and such)
  9. Sep 22, 2005 #8


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    Ok, then g(x) on the interval (0,1) is just the first triangle wave in that plot above. As far as g(x) changing, well the odd and even extensions change and depending on what those extensions are, the integrals would change depending on what the extension is in the interval of integration.
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