Finite difference Method of Wave Equation

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1. Mar 11, 2017

Riverbirdy

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. Mar 11, 2017

BvU

It's in the z direction (which they call the u direction) . 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.

3. Mar 11, 2017

Riverbirdy

So you mean to say at u(0,0,0)=0.01 is the source located?

4. Mar 11, 2017

BvU

No. The whole sheet is at an 'amplitude' 0.01 in the u-direction. All x, all y.

5. Mar 11, 2017

Ibix

Presumably for $0<x <45$ and $0 <y <45$. Otherwise you can't satisfy the other boundary conditions.

6. Mar 11, 2017

Riverbirdy

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?

7. Mar 11, 2017

BvU

Didn't know about that thing. How do you calculate that ? How about 0.05 ?

8. Mar 11, 2017

Riverbirdy

CFL= c Δt/Δx, the Courant–Friedrichs–Lewy (CFL) condition where c- wave propagation constant, Δt & Δx are time step and step size

9. Mar 11, 2017

BvU

So how about 0.05 as the outcome ?

And: there is an upper bound for C, so 'at least' sounds weird...

10. Mar 11, 2017

Riverbirdy

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

11. Mar 11, 2017

BvU

Does that mean your time step is $10^{-3}$ s ?
A lot of steps to find out what goes on at $t=5$ s ...

12. Mar 11, 2017

Riverbirdy

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.

13. Mar 11, 2017

Riverbirdy

Here's the program results running at Δt=1500 sec at t=10,000 sec

14. Mar 12, 2017

BvU

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.

15. Mar 12, 2017

BvU

My mistake utterly: I used different units for dx and c myself
It worked for you. I wrongfooted myself !

16. Mar 12, 2017

BvU

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
17. Mar 12, 2017

Riverbirdy

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.