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Base Characteristics

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to plot base characteristics using mathematica.
    The characteristic equations are:

    2. Relevant equations
    [tex] \frac{dx}{d\tau}=x u(x,y)[/tex]
    [tex] \frac{dy}{d\tau}=y u(x,y)[/tex]
    [tex] \frac{du}{d\tau}=x^2+y^2[/tex]
    u(x,1)=Sqrt(x^2+1) Initial Data

    3. The attempt at a solution
    How do I solve this with Mathematica. I need to get x(tau,xi),y(tau,xi),u(tau,xi) so I can plot the base characteristics. By the way xi is the other parameter.
     
  2. jcsd
  3. Apr 8, 2009 #2
    i tried for x
    Code (Text):

    ode1 = D[x[\[Tau]], \[Tau]] == x[\[Tau]] u[x[\[Tau]], y[\[Tau]]]
    ode2 = D[y[\[Tau]], \[Tau]] == y[\[Tau]] u[x[\[Tau]], y[\[Tau]]]
    ode3 = D[u[\[Tau]], \[Tau]] == x[\[Tau]]^2 + y[\[Tau]]^2
    DSolve[{ode1, ode2, ode3, x[0] == \[Xi], y[0] == 1,
      u[0] == Sqrt[\[Xi]^2 + 1]}, x[\[Tau]], \[Tau]]

     
    Note: \[Tau]=[tex]\tau[/tex] & \[Xi]=[tex]\xi[/tex]
    Its when I copy pasted the code it turned up like this.
     
  4. Apr 8, 2009 #3
    Any ideas? I think it's how I express the function in mathematica. Should I show x as a function of tau explicitly everywhere?
     
  5. Apr 8, 2009 #4
    What about if i have the intial conditions:
    [tex] x(\xi,0)=\xi[/tex]
    [tex] y(\xi,0)=1 [/tex]
    [tex] u(\xi,0)=\sqrt(\xi^2+1)[/tex]
     
  6. Apr 17, 2009 #5
    What about uncoupling the equations?
     
  7. Apr 17, 2009 #6
    [tex]\hat{}\downarrow\downarrow[/tex][tex]\scshape[/tex]
     
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