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Homework Help: Help solving 1st order PDE with associated Equation

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve This equation:

    with [itex]\epsilon[/itex] being a very small number from 0 to 1
    and x bounds from neg infinity to pos infinity, t>0, and condition

    2. Relevant equations
    method of associated equation (dx/P=dt/Q=du/R, and so forth)

    3. The attempt at a solution
    So far, I have attempted to apply the method of solving via associated equation and said
    and using the first two equated equations, got that

    Then next was that I used dx/[itex]\epsilon[/itex] = du/1-u and solved to get x/[itex]\epsilon[/itex]+ln(1-u)=C2.
    I then said the general equation is:
    Then I applied the initial condition u(x,0)=sinx to solve for the f(x).
    Attempting to solve this out got me

    1-e^(f(-x[itex]\epsilon[/itex]-x/[itex]\epsilon[/itex]))=sinx, but am lost after that on how to get the exact solution.

    Help will be appreciated! Thanks!
  2. jcsd
  3. Jun 30, 2011 #2


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    Homework Helper

    You mean method of characteristics?
    your equation:
    \epsilon\frac{\partial u}{\partial t}+\epsilon\frac{\partial u}{\partial x}=1-u
    The characteristics are defined by:
    \frac{dt}{ds}=\epsilon ,\quad \frac{dx}{ds}=\epsilon ,\quad\frac{du}{ds}=1-u
    With initial conditions [itex]u(r,0)=\sin r[/itex] with t(0)=0 and x(0)=r. Solving the first two equations shows that [itex]x=\epsilon s+r[/itex]. Then solving the final equation yields [itex]-\log (1-u)=s+C[/itex], to compute C, use the initial condition:
    -\log (1-\sin r)=C
    \log\left(\frac{1-\sin r}{1-u}\right) =s
    You have equations to substitute t and x for r and s.
  4. Jul 20, 2011 #3
    Hello hunt_mat- thanks for the input. I have seen if your method will work, but have trouble understanding the two conditions involving t(0)=0 and x(0)=r- how did u get such boundary conditions?

  5. Jul 20, 2011 #4


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    You initial conditions are called Cauchy data. Basically what happens in 1st order PDEs is that the initial data get propagated and changed along the characteristics. So what you have to do in paramatrise you initial data in one variable (I used r) abd you characteristics with another variable (I called it s).

    So with the initial condition, there was no t involved in you equation to t is zero when s=0 (s=0 is when you're on the initial curve), and so you set t(0)=0. As your initial data is sinx, what you do is set x=r as will be your paramatrisation of your initial condition for x. then you just insert this paramtrisation into the condition for u and that is where you get sin r as the paramatrisation.
  6. Jul 27, 2011 #5
    Thanks for your help hunt_mat.
    Now, the question asks for a second part (let me know if I should move this question to a separate thread) to this same question:

    -For 0<epsilon<<1, obtain zeroth order outer and inner perturbation solutions and construct a uniformly valid composite solution.

    I have tried to use the zeroth order outer and inner methods by first assuming that epsilon goes to 0. However, that gives u=1 for both inner and outer cases, and I am pretty sure that isn't right :p

    Any help will be appreciated. Thanks!
  7. Jul 28, 2011 #6


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    I thought that this would come into it, you need to rescale x and y and look for the dominant balance.
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