Wave direction and speed of propagation

  • #1
10
0
TL;DR Summary
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;
 

Answers and Replies

  • #2
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.
 
  • #3
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.
 
  • #4
Thanks. I have another preliminary observation that exp(-z/zeta) would diverge at z=##-\infty## for zeta > 0.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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))
 
  • #8
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?
 
Last edited:
  • #10
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.
 
  • #12
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.
 
  • #13
sorry for the late responce, thank you for your help
 

Suggested for: Wave direction and speed of propagation

Replies
6
Views
721
Replies
7
Views
788
Replies
3
Views
1K
Replies
4
Views
2K
Replies
11
Views
1K
Replies
3
Views
905
Back
Top