Clear Filters
Clear Filters

The Tsunami model has need to fixed this problem

31 views (last 30 days)
aoming
aoming on 7 Apr 2024 at 13:16
Commented: aoming on 8 Apr 2024 at 1:11
hi,
I found that the "https://ww2.mathworks.cn/help/symbolic/solve-a-partial-differential-equation-tsunami-simulation.html#d126e37561", it has the problem that soliton = @(x,t) A.*sech(sqrt(3/4*g*A/H)*(x/c2+t)).^2; ,but it not Conservation of Dimensions . I think that it should be soliton = @(x,t) A.*sech(sqrt(3/4*g*A/H^2)*(x/c2+t)).^2. And I want to confirm that. Thank you!
  2 Comments
Torsten
Torsten on 7 Apr 2024 at 14:10
The unit of
sqrt(3/4*g*A/H^2)
is sqrt(m)/s. I doubt this can be correct.
aoming
aoming on 8 Apr 2024 at 1:09
Thanks! thank you.But I actually have an another problem,
the code :S = fft(soliton(-0.8*TimeScale*c2, linspace(0,TimeScale,2*(Nt/2)-1)))';
S = repmat(S,1,Nx);
I would like to ask where the initial position of the waves is? In the -0.8*TimeScale*c2? and why it is minus value?

Sign in to comment.

Answers (1)

the cyclist
the cyclist on 7 Apr 2024 at 15:46
Disclaimer: I am not an expert in this!
I was not able to track down the exact (Goring & Reichlan) reference from the documentation, and therefore not the exact formula. But, I did find some related materials.
I'm guessing that you might be being fooled by having seen the formula where kappa is inside sech(), rather than g. Looking over some other reference, it seems to me that
kappa ~ g / H
which is where the "missing" dimension of H is hiding.
Another place that may have fooled you is that in the documentation, A is a dimensionless (divided by H) amplitude.
I hope that helps.
  1 Comment
aoming
aoming on 8 Apr 2024 at 1:11
Thanks,cyclist!
But I actually have an another problem
the code :S = fft(soliton(-0.8*TimeScale*c2, linspace(0,TimeScale,2*(Nt/2)-1)))';
S = repmat(S,1,Nx);
I would like to ask where the initial position of the waves is? In the -0.8*TimeScale*c2? and why it is minus value?or the initial position is in the x_max = L2 + 12;?

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!