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Homework Help: Difficulties with solution/plotting of a PDE.

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Question attached

    2. Relevant equations



    3. The attempt at a solution

    I'm mostly wondering with c) and also want to check if my solution is correct.

    My solutions for this question are:

    u(x,t)= -1/2 for x <= -1/2*t
    = 1 for -t < x < 1-t
    = 1/2 for x => 1/4*t+1

    lastly, with c), I'm just wondering how to plot 3 space curves on the same axes?(as there are 3 different solutions)

    thanks in advance.
     

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    Last edited: Apr 23, 2010
  2. jcsd
  3. Apr 24, 2010 #2

    LCKurtz

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    It depends on what software package you are using. In Maple it would be something like this. Call your three functions f(x),g(x), h(x)


    plot({f(x),g(x),h(x)},x=a..b);

    where a and b are whatever limits you want.
     
  4. Apr 24, 2010 #3
    Hi LCKurtz, thank you for your help ones again!

    yes, I'm using Maple, as it is the only available computing package to me.

    I tried plotting it the way you said, and it came out looking wrong, but I think it's because I'm doing it wrong, would this sort of PDE be more suitable with a 3D plot?

    Below I've also done a 3D plot, where the xi ranges I worked out by inputting t=0,2,4 respectively to get the lowest and highest possible domain for x for each interval of x. Is there a way to tell Maple to join up the 3 sets of lines so that you can see where there are jump discontinuities?

    Thanks.
     

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  5. Apr 24, 2010 #4

    LCKurtz

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    First, I am not a PDE expert so I'm not sure how much help I can give you. You are to plot your solution for t = 0, 2, and 4. With t fixed you have just a function of x. So why would you use space curve instead of just plot.

    But a more serious question I have is, consider your solution for t = 2. Then

    u(x,2) = -1/2 for x <= -1
    = 1 for -2 < x < -1
    = 1/2 for x >= 3/2

    What if x is between -2 and -1, do you use the first or second part? What about x between -1 and 3/2, what is u there? It doesn't look like your solution is well defined.
     
  6. Apr 24, 2010 #5
    Oh you're quite right, but I have checked my inequalities and followed the example we've been given step by step, is there anyway to rearrange the inequalities so that the solution is defined for all intervals?

    Thanks.
     
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