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A specific method of characteristics problem

  1. Feb 5, 2013 #1
    In my previous PDE class we did nothing with method of chars. Now I am assigned one in a higher class, but all the examples I find online are not helpful for a particular problem I have.
    du/dt+a*du/dx=f(t,x)

    with u(0,x)=0 and f(t,x)=1 if -1<=x<=1, and f(t,x)= 0 otherwise

    Differences between this and typical examples I see in my PDE book and online:
    1. I rarely see non-homogeneous ones.
    2. Initial condition is NEVER equal to 0 in any example I've ever seen.
    3. The solution I'm supposed to "show" is piecewise defined in 5 parts. I've seen a couple examples whose solution is defined into 2 parts, sometimes. But 5?

    Please be gentle.
     
  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    1. If you write out the equation for each regeon, you'll see that f(t,x) is a constant in each case.

    2. shouldn't matter - just plug the zero in to the general solution like you would any other number.

    3. makes no difference - you just have more steps to do. Do it the same way - just divide the DE into the different regions.

    What is the actual problem you have to solve

    You've seen:
    http://en.wikipedia.org/wiki/Method..._of_first-order_partial_differential_equation
     
  4. Feb 5, 2013 #3
    Can you solve the following ODE, subject to the initial condition u = 0 at t = 0?

    [tex]\frac{Du}{Dt}=f(t,x_0+at)[/tex]

    where x0 is a constant.
     
  5. Feb 6, 2013 #4
    Obviously I'm missing something fairly rudimentary.

    Simon:
    I wrote the problem I have to solve.

    du/dt+a*du/dx=f(t,x)

    with u(0,x)=0 and f(t,x)=1 if -1<=x<=1, and f(t,x)= 0 otherwise

    If I write out the equation for each region, I get 2 equations.

    Chestermiller:

    When you use capital D, is that the same as an ordinary derivative?

    I really don't see how to proceed.
     
  6. Feb 6, 2013 #5
    Yes, that is the same as an ordinary derivative. I wanted to use d's, but you had already used d's in your original equation, where you properly should have been using partials. The solution to the equation I gave is u(t,x0+at). So you have an initial value problem, starting out at various locations x0. It is like following the time rate of change of a particle that is moving with velocity a. This is how the method of characteristics works.
     
  7. Feb 6, 2013 #6
    I'm lost. Could you walk me through this?
     
  8. Feb 6, 2013 #7
    Do you know how to make a change of variables from t and x, to t and (x-at)?

    Chet
     
  9. Feb 6, 2013 #8
    Do I 'know how'? I guess not really.
    Ok, let me put my attempt to 'solve' this.

    problem:
    ∂u/∂t+a*∂u/∂x=f(t,x)
    with f(t,x)=1 if -1<=x<=1 and f= 0 otherwise
    u(0,x)=0

    For whatever reason, (I read this a long time ago) I set u(x,t)=U([itex]\tau[/itex], [itex]\xi[/itex]) with [itex]\xi[/itex] = x-at and [itex]\tau[/itex] = t
    which leads to, after partial derivatives, ∂u/∂[itex]\tau[/itex] = f(t,x)

    Now integrate and either get u=[itex]\tau[/itex] for -1<=x<=1 or get 0 for x values other than that.
    So what am I missing?
     
  10. Feb 6, 2013 #9

    Yes. You're very close. So you have ∂u/∂[itex]\tau[/itex] = f(t,x). Now, you substitute [itex]x=\xi+at[/itex] to get

    [tex](\frac{\partial u}{\partial t})_\xi=f(t,\xi+at)[/tex]

    Now you take any value of [itex]\xi[/itex], say [itex]\xi=-5[/itex] at time t = 0 (x = -5), and start integrating with respect to t (at constant [itex]\xi=-5[/itex]), subject to the initial condition u = 0. So, in this example, u(x,t) = 0 until
    x = -5 + at = -1.
    That would be t = 4/a. After that u(t,x) = u(t,-5+at) = (t - 4/a) until x = -5 + at = +1
    That would be at t = 6/a. After that u(t,x)=u(t,-5+at) = 2/a for all subsequent times along the line x = -5 + at.

    So, integrating along the line x = -5 + at, you have:

    [itex]u=0[/itex] for [itex]t\leqslant 4/a[/itex]

    [itex]u=t-4/a[/itex] for [itex]4/a\leqslant t\leqslant 6/a[/itex]

    [itex]u=2/a[/itex] for [itex]6/a\leqslant t[/itex]

    You can do this for different values of [itex]\xi[/itex].
     
  11. Feb 7, 2013 #10
    I have to say I still do not understand several aspects. What exactly is meant by the subscript [itex]\xi[/itex] and why is "u=0 until x=-1"

    Further, this does not appear to be the solution we're supposed to show. :(
     
  12. Feb 7, 2013 #11
    The subscript [itex]\xi[/itex] means "holding [itex]\xi[/itex] constant." So, the derivative with a subscript is the partial of u with respect to t holding [itex]\xi[/itex] constant. The initial condition on u is u=0 for all x.

    You are integrating with respect to t along a line

    x - at=constant.

    The constant is the value of x at t = 0.

    Anywhere you start integrating to the left of x = -1 at time = 0, the integrand f(t,x) is equal to zero. You will find that, throughout the entire region x < -1, the solution for u is going to be u = 0 for all t. But, once your line crosses over into the region -1<x<+1, your integrand f(t,x) suddenly jumps up to 1 (and stays there until your line crosses out of the region again at x = 1).

    I'm afraid I'm not explaining things very well. Hopefully, I said enough to give you the gist of what is happening. If not, all I can say is I did my best.

    Chet
     
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