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Answer check transport equation (1st order linear PDE)

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi everyone, I just wanted to double check if I've solved this correctly?

    Given:
    [tex]\left.\frac{du}{dx} + sin(x)\frac{du}{dy} = 0[/tex]

    [tex]\left.-\infty < x < \infty[/tex]

    y > 0

    [tex]\left.u(\frac{\pi}{2} , y ) = y^{2}[/tex]

    Solve the PDE

    2. Relevant equations
    Method of characteristics

    3. The attempt at a solution
    Using method of characteristics, first I said:

    [tex]\left.\frac{dy}{dx}= \frac{sin x}{1}[/tex]

    taking the integral I get
    [tex]\left. y = -cos(x) + C [/tex]

    solving for C:
    C = y + cos(x)

    so now,
    [tex]\left.u(x,y) = f(y+cos(x))[/tex]

    our initial condition gives us the following:
    [tex]\left.u(\frac{\pi}{2},y) = y^{2}[/tex]
    so:
    [tex]\left.u(\frac{\pi}{2},y) = f(y)[/tex]

    which means:
    [tex]\left. f(y) = y^{2}[/tex]

    Therefore:
    [tex]\left. u(x,y) = f(y + cos x) = (y+cos(x))^{2}[/tex]

    I think this should be right but I wanted to double check. Thanks!!!
     
  2. jcsd
  3. Jun 10, 2009 #2

    Mark44

    Staff: Mentor

    Why don't you, then? (Double check, that is.) One thing you should get in the habit of doing is checking your own work. Take the partials of u(x, y) and see if they satisfy your differential equation and initial conditions. If they do, you're golden.
     
  4. Jun 10, 2009 #3
    hm didn't think about that before. thanks
     
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