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PDE problem, Solve using Method of Characteristics

  1. Jun 20, 2011 #1
    Hi

    This problem occurred on my final and I could not figure it out.

    1. The problem statement, all variables and given/known data

    The problem was a partial differential equation (I forgot the exact equation) but the solution was a hyperbolic function in the form of u(x,y)= f(x+y) + g (x+y), it was part b that gave me the problem.

    Part b asked to solve the initial value problem give u(0,y)=y and u_x_(0,y)=y^2 (I'm not 100% sure what they were, but they were functions of y). I just want to know how I would solve an initial value problem for these equations as we never covered it in class.

    Thanks.
     
  2. jcsd
  3. Jun 30, 2011 #2
    So can anyone help me out?

    Thanks
     
  4. Jul 24, 2011 #3
    So I ordered my exam, it turns out the actual question was:

    Solve u_xx+u_xy-6u_yy = 0. The solution which I solved correctly was a hyperbolic function u (x,y)= f(2x+y)+g(3x-y).

    The IVP is u(0,y) = y^2 and u_x_(0,y)=3y.

    Thanks.
     
  5. Jul 25, 2011 #4

    hunt_mat

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    Homework Helper

    So found the characteristics and now you apply the boundary conditions. What is your problem exactly?

    Mat
     
  6. Jul 26, 2011 #5
    Hi

    I just don't know how to solve the IVP u(0,y) = y^2 and u_x_(0,y)=3y when u (x,y)= f(2x+y)+g(3x-y).

    I found a solved example online (http://www.mathsman.co.uk/DEnotes[2].pdf, the IVP is solved beginning page 48) but I'm very confused on their methodology, so if someone can explain this or do it some different way I would really appreciate it.

    Thanks
     
  7. Jul 27, 2011 #6

    hunt_mat

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    Just differentiate!
    [tex]
    \frac{\partial u}{\partial x}(0,y)=2f'(y)+3g'(-y)=3y
    [/tex]
    Integrating w.r.t. y shows that:
    [tex]
    2f(y)-3g(-y)=\frac{3y^{2}}{2}
    [/tex]
    Likewise setting x=0 in the equations shows that:
    [tex]
    u(0,y)=f(y)+g(-y)=y^{2}
    [/tex]
    To find f and g just add and subtract.
     
  8. Jul 30, 2011 #7
    Hi

    Just one question, when we take derivative of f and g, what is it with respect to? I substituted w = 2x+y and v = 3x-y, so u_x = df/dw*dw/dx + dg/dv*dv/dx and when I integrated w.r.t. y I let dw=dy and dv = -dy which gave me the same equation you had, is this correct?
     
  9. Jul 30, 2011 #8

    hunt_mat

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    Homework Helper

    Sounds good.
     
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