Wave direction and speed of propagation

[kasnay]

TL;DR

Wave direction and speed propagation

I have an answer to a partial differential equation.
I have the equation coded as followed. I am trying to get this wave to propagate back after it hits a given z value. Can anyone help me figure out the direction in this equation?

upsilon=sqrt(3*((1-nu)/(1+nu))*(B/row));expansion=exp(-z/zeta).*(1-.5.*exp(-upsilon.*step(t)/zeta))-.5.*exp(-abs(z-upsilon.*step(t))/zeta).*sign(z-upsilon.*step(t));eta=(1-R)*((Q*Beta)/(A*zeta*C))*((1+nu)/(1-nu))*expansion;

 

[anuttarasammyak]

However I am not confident to have read or understood the equation you wrote well, only z, not x,y, seems to appear in it. So I suppose the direction which the equation determines would be z.

 

[kasnay]

Sorry please let me clarify. We are ignoring the y and x planes. The wave is traveling in the z direction away from the surface as time increase. What I am trying to do is modify the equation so that the wave starts at infinity far away and moves toward the surface as time increases.
In essence I am trying to change my wave velocity to negative. I have tried inversing my terms with z and t but this creates an infinitely building wave.

 

[anuttarasammyak]

Thanks. I have another preliminary observation that exp(-z/zeta) would diverge at z=$-\infty$ for zeta > 0.

 

[kasnay]

the pde that i worked through (which was also previously done by another) was defined that boundary conditions are from z=0 to z=infinity.

 

[Frabjous]

It is not clear what your function is. What is step? You need to write the equation as a simple function of z and t; i.e. eliminate constants by setting them to convenient values (for example upsilon=1, zeta=1 and eta=1*expansion or are z and t buried in these terms) and use the formatting tools on PF so that it is readable.

 

[kasnay]

once again, i apologize. i have never used a forum for this before.
n(z,t)=(3/2)(1.5/.5)(e^-z(1-.5e^-3t)-.5e^-abs(z-3t)*sgn(z-3t))

 

[Frabjous]

Let’s simplify further
m(z,t) = e^-z(1-.5e^-3t)-.5e^-abs(z-3t)*sgn(z-3t)
= e^-z-.5e^-(z+3t)-.5e^-abs(z-3t)*sgn(z-3t)
= f(z) + g(z+3t) + h(z-3t)
which is a spacially dependent background + a left going wave + a right going wave
Are you sure everything you wrote is transcribed correctly?

 

[anuttarasammyak]

Please find the plot for post 7,8 to investigate.
https://www.wolframalpha.com/input?i=+plot+e^(-z)-0.5e^(-z-3t)-0.5e^(-|z-3t|)*sgn(z-3t)+for+0<z<3,+0<t<3&lang=en
There is a discrepancy along z=3t. For other coefficients you choose, you can change it and show us the link.

 

[kasnay]

So i didnt think to expand the exp(1-.5exp) so that actually made it easy to look at.
Your plot is in essence what i get when I use matlab. I programmed your simplified model and I get that same graph. A crest and trough that moves in the positive direction as time goes toward infinity.
So I also noticed what you are saying about the background left and right wave. Just playing around with it, when I remove the g(z+3t) The wave does not have a noticeable change. However If I remove the h(z-3t) then it just becomes a exponential with no movement in time.

I appreciate your help.

 

[anuttarasammyak]

Just for fun of functions of sinusoidal waves not exponential
https://www.wolframalpha.com/input?i=+plot++Im{e^(-iz)-0.5e^{i(-z-3t)}-0.5e^(-i|z-3t|)*sgn(z-3t)}+for+0<z<10,+0<t<10&lang=en
https://www.wolframalpha.com/input?i=+plot++Re{e^(-iz)-0.5e^{i(-z-3t)}-0.5e^(-i|z-3t|)*sgn(z-3t)}+for+0<z<10,+0<t<10&lang=en

Real wave changes its pattern after arrival of z=3t wave front form z=0.

 

[Frabjous]

Notice at t=0, g=h in the region of interest
For t>0, g quickly runs outside the region leaving its exponentially decaying tail while h moves to the right.

 

[kasnay]

sorry for the late responce, thank you for your help