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Homework Help: First order linear PDE, need help understanding solution/method

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the initial boundary value problem:

    u_t + cu_x = -ku

    u is a function of x,t

    u(x,0) = 0, x > 0
    u(0,t) = g(t), t > 0

    treat the domains x > ct and x < ct differently in this problem. the boundary condition affects the solution in the region x < ct, while the IC affects it in the region x > ct.

    3. The attempt at a solution

    The question previous question walks you through transforming the coordinates to get the general solution:

    A = x - ct B = t

    So by the chain rule:

    u_x = (dA/dx)u_A + (dB/dx)u_B
    u_t = (dA/dt)u_A + (dB/dt)u_B

    dA/dx = 1 dB/dt = 0
    dA/dt = 1 dB/dt = 1


    u_x = u_A
    u_t = -cu_A + u_B

    the new equation is now:

    -cu_A + u_B + cu_A + ku

    = u_B + ku


    u_B = -ku

    --> u = exp(-kB).f(A)
    --> u(x,t) = exp(-kt).f(x-ct)

    the justification given here is: "since the intergration constant is only constant in B, it may depend on A". I don't understand this step, what does that justification mean? I see that all they did was swap from B and A back to x-ct and t though I dont quite understand whats happening.

    Anyway, now we have the general solution, so we can work on the main part of the question:

    u(x,t) = exp(-kt).f(x-ct)

    Since u(x,0) = 0 then f(x) = 0 for x > 0

    Since u(0,t) = g(t) then exp(-kt).f(-ct) = g(t) for t > 0

    Here is where I get confused...

    => f(z) = g(-z/c) exp(-(kz/c)) for z < 0

    then the next step says

    => u(x,t) = 0 for x > ct
    and u(x,t) = exp(-kt).exp(-(k/c)*(x-ct)).g(-(x-ct)/c) for x < ct

    => u(x,t) = exp(-kx/c).g(t-(x/c)) for x < ct

    No idea what happens once they substitute z, and why they are doing the things they've done. Any help is appreciated. Sorry for the difficult notation and thank you in advance!!!
  2. jcsd
  3. Aug 22, 2011 #2


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    Homework Helper

    With four first equation, it's just s trick. The change of variables is perfectly valid, because if you calculate the Jacobian you obtain a non-zero result, that means the co-ordinate system (A,B) is perfectly allowable. Or that I think you mean that if f(x,y)=A(x) + B(y) say, then [itex]\partial f/\partial x=A'(x)[/itex], remember these are partial differentials that you're dealing with, not full ones.

    For the second part of your question, you need to understand that f is a function of a single variable only, for the second part that are introducing a new variable z=-ct, as [itex]t\geqslant 0[/itex], we can say that [itex]z\leqslant 0[/itex], so we have found f(B) in two regions, when [itex]B\leqslant 0[/itex] and when [itex]B\geqslant 0[/itex]. The variable in the equation is x-ct, so for x-ct>0, the solution is zero, i.e. for x>ct and for x-ct<0, the solution is as you mentioned it, .e. for x<ct.
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