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Differential eq (implicit sol)

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data
    i have a differential equation.

    ∂h/∂t + [g sin a h²/v]∂h/∂x = 0. where h = h(x,t).
    i need to show by substitution that the (implicit) general solution for h is h = f(x - (g sin a h²t/v)) where f is an arbitrary differential function of a single variable.

    3. The attempt at a solution
    ∂h/∂x = ∂f/∂x
    ∂h/∂t = ∂f/∂t * (-g sin a h²/v)
    subbing in gives
    ∂f/∂t*(-g sin a h²/v) + (g sin a h²/v)*∂f/∂x = 0
    cancelling g sin a h²/v
    -∂f/∂t + ∂f/∂x = 0
    so now i have to show... ∂f/∂t = ∂f/∂x ???
    seems like i'm going in circles...

    any help will be appreciated
  2. jcsd
  3. Sep 20, 2008 #2

  4. Sep 20, 2008 #3
    I'm not sure where the brackets should be, so let me just confirm that this is the problem:

    [tex]\frac{\partial h}{\partial t} + \frac{gsin(a)h^2}{v}\frac{\partial h}{\partial x} = 0[/tex]

    [tex] h=f(x-\frac{gsin(a)h^{2}t}{v})[/tex]
  5. Sep 20, 2008 #4
    Okay I worked through it, that must be the correct form. Your problem is that you're interpreting f as a function of two variables. As xaos said, write h=f(y(x,t)). Then:

    [tex] \frac{\partial h}{\partial t} = f'(y)\frac{\partial}{\partial t}(y) [/tex]
  6. Sep 20, 2008 #5


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    f is a function of a single variable so "[itex]\partial f/\partial x[/itex]" and "[itex]\partial f/\partial t[/itex]" are meaningless. Having that "h" inside f makes it complicated: More correctly
    [tex]\partial h/\partial x= f'(x- (g sin(ah^2t/v))(1- 2aght/v)(-g cos(ah^2t/v))(2aht/v)\partial h/\partial x[/tex]
    [tex]\partial h/\partial t= f'(x- (g sin(ah^2t/v)(-2aght/v)(-gcos(ah^2t/v))((2aht/v)\partial h/\partial t+ ah^2/v)[/tex]

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