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Math Help - Deriving a shock condition

  1. #1
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    Deriving a shock condition

    P(t, x) = f(t - Ax*f(t)) where f(t) = P(t, 0)

    If one has f(t) = sin(t), then a shock occurs in P (infinite slope) for Ax = 1 (A is a constant).

    I am trying to derive this result, i.e. I'm trying to show that when Ax = 1, dP(t, x)/dt has an infinite slope at some t.

    P(t, x) = f(t - Ax*f(t))
    P(t, x) = f(u), u = t - Ax*f(t)


    I'm not sure how to proceed . Thanks for your input.
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  2. #2
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    Re: Deriving a shock condition

    Since you say "I'm trying to show that when Ax = 1, dP(t, x)/dt has an infinite slope at some t" (more correctly, you are trying to show that if dP/dt is infinite for some t, then Ax=1.) the obvious first thing to do is differentiate! What is \frac{\partial P}{\partial t}?
    Thanks from topsquark and algorithm
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  3. #3
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    Re: Deriving a shock condition

    Hi,

    Thanks for your input. That's what I'm trying to do but I'm not getting the result I'm after.

    Take the case where f(t) = sin(t)

    P(t, x) = f(t - Ax*f(t))
     = sin(t - Ax*sin(t))

    I want to show that, for some t, dP/dt is infinite for Ax = 1 (thanks for your clarification). So I set Ax = 1, and differentiate:

    dP/dt = -cos(t - sin(t))cos(t)

    Now, I can't see how the above could ever tend to infinity :S
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  4. #4
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    Re: Deriving a shock condition

    You are right. It can't. Therefore, there is no "shock".

    However, your derivative is incorrect. The derivative of t- Ax sin(t), with respect to t, is 1- Ax cos(t), not just cos(t).
    dP/dt= cos(t- sin(t))(1- Ax cos(t))
    Thanks from topsquark and algorithm
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  5. #5
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    Re: Deriving a shock condition

    Thanks for that.

    But I have a plot of a sine wave shocking - i.e. I know it happens, but I'm struggling to show it analytically.

    The equation which causes a function f(t) (e.g. sin(t)) to shock is P(t, x) as I posted in #1 (called the Poisson solution in my notes).

    If you could offer ideas about how I can show dP/dt -> inf. then I'd highly appreciate it.

    Thanks.
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  6. #6
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    Re: Deriving a shock condition

    Please see attached for the result I'm trying to show.
    Attached Thumbnails Attached Thumbnails Deriving a shock condition-shock-condition.png  
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  7. #7
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    Re: Deriving a shock condition

    That graph is wrong. There is no point of infinity.
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  8. #8
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    Re: Deriving a shock condition

    This is what I'm doing in code to get that plot with an infinite slope (Matlab):

    θ = -pi:pi
    %Below is f(θ)
    f = sin(θ)

    %Set Ax = 1 to get the shockwave
    Ax = 1;

    %Implement the equation in post 1 to get P(θ, x)
    θnew = θ - Ax * f(θ)

    plot(θnew, f) %plots f(θ) against θnew

    This implements the equation in post 1, but you're saying it doesn't?
    Last edited by algorithm; June 22nd 2013 at 02:27 PM.
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