Deriving a shock condition

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

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.

$\displaystyle P(t, x) = f(t - Ax*f(t))$

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

I'm not sure how to proceed :) . Thanks for your input.

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 $\displaystyle \frac{\partial P}{\partial t}$?

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 $\displaystyle f(t) = sin(t)$

$\displaystyle P(t, x) = f(t - Ax*f(t))$

$\displaystyle = sin(t - Ax*sin(t))$

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:

$\displaystyle dP/dt = -cos(t - sin(t))cos(t)$

Now, I can't see how the above could ever tend to infinity :S

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))

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.

1 Attachment(s)

Re: Deriving a shock condition

Please see attached for the result I'm trying to show.

Re: Deriving a shock condition

That graph is wrong. There is no point of infinity.

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?