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Solve the first order hyperbolic equation

  1. Jul 4, 2011 #1
    Solve the first order hyperbolic equation

    3 du/dx + 2x du/dt =2u

    With initial condition: u(x,0) = 2x+4



    My attempt at a solution

    I usually adopt the method of characteristics:

    dx/a = dt/b = du/c


    So from the above:

    a=3, b=2x and c=2u

    am I on the right track here?
     
  2. jcsd
  3. Jul 4, 2011 #2

    hunt_mat

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    I do this in a slightly different way, my characteristic equations are:
    [tex]
    \dot{x}=3,\quad\dot{t}=2x,\quad \dot{u}=2u
    [/tex]
    Where the dot denotes differentiation to the characteristic variable, [itex]s[/itex]. I write the cauchy data as:
    [tex]
    t(s=0)=0,\quad x(s=0)=r,\quad u(s=0)=2r+4
    [/tex]
    Then just re-arrange to get rid of s and r and that will be your solution. I posted some notes on this in https://www.physicsforums.com/showthread.php?t=467445
     
  4. Jul 4, 2011 #3
    Thank you hunt_mat. I shall read through ur notes. Is the method I was using incorrect for this question then?
     
  5. Jul 4, 2011 #4

    hunt_mat

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    Not an incorrect method, you can get the solution that way, I just think that the solution method that I presented is far easier to implement and you know if you're making a mistake or not.
     
  6. Jul 4, 2011 #5
    So from what u have said, will the quotient of the characteristics be

    dt/dx = 2x/3 du/dx= 2u/3
     
  7. Jul 4, 2011 #6

    hunt_mat

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    I think so, yes. To be perfectly honest, I was never really happy doing things this way.
     
  8. Jul 4, 2011 #7
    Are u referring to the method I have adopted? Sorry for the confusion, I'm just anxious to answer this question.
     
  9. Jul 4, 2011 #8

    hunt_mat

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    So far, I think you're doing fine, but I haven't much experience with the method you're adopting. I have a book with it in though, so I will keep helping you with this method.
     
  10. Jul 4, 2011 #9
    Ok hunt_mat I appreciate your help, now heres where I get stuck:

    Integrating du/dx=2u/3?
     
  11. Jul 4, 2011 #10

    hunt_mat

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    So you have computed the characteristics which are given as [itex]3t-x^{2}=k[/itex] where [itex]k[/itex] is a constant, you know that [itex]u[/itex] is a constant on these characteristics, and you know that characteristics curve passes through the point [itex](t,x)=(0,r)[/itex], what will [itex]u[/itex] be at this point?
     
  12. Jul 4, 2011 #11
    I'm a little confused how u got 3t-x2=k?

    But will u(r,0) =2r+4
     
  13. Jul 4, 2011 #12

    hunt_mat

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    Because I solved the equation:
    [tex]
    \frac{dt}{dx}=\frac{2x}{3}
    [/tex]
    You are right with the second part.
     
  14. Jul 4, 2011 #13
    Yes sorry its because I had it in the form:

    t = x2 /3 +K

    Ok so given u, where do I go next?

    I still need to solve du/dx correct?
     
  15. Jul 4, 2011 #14

    hunt_mat

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    Now you need to compute K, you know that the characteristic passes through the point [itex](t,x)=(0,r)[/itex], so you can use that to compute K. Then you need to integrate your equation for u and use that fact that at the point [itex](t,x)=(0,r)[/itex], [itex]u=2r+4[/itex], so all you have to do is substitute for r.
     
  16. Jul 4, 2011 #15
    So establishing K:

    t = x2 /3 +K

    0= r2 /3 +K

    K=-r2 /3

    Therefore:

    t = x2 /3 - r2 /3

    Correct??
     
  17. Jul 5, 2011 #16

    hunt_mat

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    Good so far. Now do the same thing for [itex]u[/itex] and you should end up with a function with x's and r's in, and now you know an equation for r.
     
  18. Jul 5, 2011 #17
    So the next step is to do:

    du/dx where u = 2r+4

    so u = 2rx +4x

    Is this correct??
     
  19. Jul 5, 2011 #18

    hunt_mat

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    No, you have an equation for u of the form:
    [tex]
    \frac{du}{dx}=\frac{2u}{3}
    [/tex]
    Solve that. You will have a constant to compute, that is when you use the initial condition you were talking about.
     
  20. Jul 5, 2011 #19
    Ok this is the part I can't get past:

    du/dx= 2u/3

    I'm nt sure how to solve the above:
     
  21. Jul 5, 2011 #20

    hunt_mat

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    You integrate it like any other equation:
    [tex]
    \log u-\log u_{0}=\frac{x^{2}}{3}
    [/tex]
    Now we know that at the point [itex](t,x)=(0,r)[/itex], the solution is [itex]u_{0}=2r+4[/itex], so insert this into the equation and you will be left with an equation with lots of x's and r's, well you know an equation for r...
     
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