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 ?
where
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.
I'm not sure how to proceed . Thanks for your input.
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 ?
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
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:
Now, I can't see how the above could ever tend to infinity :S
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.
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?