Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finite difference Method of Wave Equation

  1. Mar 11, 2017 #1
    wave.jpg Hi, Physics forum!
    Just a little push of my doubts I hope somebody could help me with my confusion of one of our home works.
    I know that all boundary conditions are zero. My doubt is how do I interpret (x,y,0)=0.01 source in the figure? Where is it located in the grid. I am hoping someone could help me with this.
     
  2. jcsd
  3. Mar 11, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's in the z direction (which they call the u direction) :smile:. Basically the whole sheet is lifted up 0.01 cm before t=0 and then the edges are being pulled down suddenly to z=0 and stay there. The disturbance propagates inwards over the xy plane.
     
  4. Mar 11, 2017 #3
    So you mean to say at u(0,0,0)=0.01 is the source located?
     
  5. Mar 11, 2017 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. The whole sheet is at an 'amplitude' 0.01 in the u-direction. All x, all y.
     
  6. Mar 11, 2017 #5

    Ibix

    User Avatar
    Science Advisor

    Presumably for ##0<x <45## and ##0 <y <45##. Otherwise you can't satisfy the other boundary conditions.
     
  7. Mar 11, 2017 #6
    Ok, got it. Thank you so much I appreciate it.

    Anyhow, I computed for the CFL=5.0E-5 which is recommended to be at least 0.7071. What can you say about this wave?
     
  8. Mar 11, 2017 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Didn't know about that thing. How do you calculate that ? How about 0.05 ?
     
  9. Mar 11, 2017 #8
    CFL= c Δt/Δx, the Courant–Friedrichs–Lewy (CFL) condition where c- wave propagation constant, Δt & Δx are time step and step size
     
  10. Mar 11, 2017 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So how about 0.05 as the outcome ?

    And: there is an upper bound for C, so 'at least' sounds weird...
     
  11. Mar 11, 2017 #10
    CFL for this one is about 5E-5. I tried in matlab, results seems to diverge. My professor might be deliberately giving off problem like this to develop critical thinking. I don't know might be his mistake probably
     
  12. Mar 11, 2017 #11

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Does that mean your time step is ##10^{-3}## s ?
    A lot of steps to find out what goes on at ##t=5## s ...
     
  13. Mar 11, 2017 #12
    Nope, I think not necessarily. Minimizing time step would actually make CFL→0 which is off 0.7071 value. The book says at those values far from 0.7071 solution tends to diverge. Now, if I tend to maintain CFL of 0.7071 I have a time step of actually, 1400 plus seconds.
    upload_2017-3-12_11-14-52.png
     
  14. Mar 11, 2017 #13
    Here's the program results running at Δt=1500 sec at t=10,000 sec
    upload_2017-3-12_12-3-26.png
     
  15. Mar 12, 2017 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What I meant is that when I calculate ##{c\Delta t\over \Delta x} = {5\times 10^{-4}\times 1\over 0.01 }## I get 0.05 for a time step of 1 s. That is a most reasonable CFL (which of course I Googled before admitting I didn't know about)

    And the screen shot tells us CFL needs to be < 0.7, not 'at least' 0.7

    With a c of 0.0005 cm/s one expects the boundary value to propagate some 0.0025 cm in 5 seconds. Utterly uninteresting. Hence your second post, I assume.
     
  16. Mar 12, 2017 #15

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    My mistake utterly: I used different units for dx and c myself o:) o:)
    It worked for you. I wrongfooted myself ! :-p
     
  17. Mar 12, 2017 #16

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Had some coffee. ##
    {c\Delta t\over \Delta x} = {5\times 10^{-4}{\rm\ cm/s} \times 1 {\rm\, s}\over 1 {\rm\, cm} } = 5\times 10^{-4}##, not ##5\times 10^{-5}## and that divided into 0.7 gives your 1400 s. Hehe...
    Nice tool this is. I would stay on the safe side and work with a cfl of 0.1 cfl max. But the picture tells different: it looks good.

    Still, professor picked a membrane with c = 0.0005 cm/s Really slow stuff ! Could he have meant 0.0005 cm/ms ?

    Looking at your picture the whole central part of the sheet is at -0.05 cm after 10 ks. Would be interesting to see things happening when the waves meet in the center !
     
    Last edited: Mar 12, 2017
  18. Mar 12, 2017 #17
    Yep, probably he had some mistakes on the units, I guess. But, this type of homework is worth to learn than any regular straight forward one. Probably, its deliberately mistaken. I so appreciate your help. Thank you so much.:smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted