How to solve this equation for Laplace transform with matlab?
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10 Comments
darova
on 21 Feb 2020
What do you mean by 'solving'? Is it a surface? What are those equations?
soe min aung
on 25 Feb 2020
darova
on 25 Feb 2020
Did you try to build those surface/equations? Can you show your attempts?
Where is the problem?
soe min aung
on 26 Feb 2020
Walter Roberson
on 26 Feb 2020
Since you are using the symbolic toolbox anyhow, use piecewise() to define your equation. Then when you have it in piecewise form, ude rewrite() specifying the 'Heaviside' option. With the equation in heaviside form, the symbolic toolbox laplace transform functions can create appropriate expressions.
soe min aung
on 26 Feb 2020
Walter Roberson
on 26 Feb 2020
Edited: Walter Roberson
on 26 Feb 2020
syms eta0 L v t x y k k1 k2 s h w
T1 = (eta0*v*t)/(2*L)*(1-cos(pi/50*x)).*(1-cos(pi/100*(y+150))) ;
T2 = (eta0*v*t)/L*(1-cos(pi/50*x)) ;
T3 = (eta0*v*t)/(2*L)*(1-cos(pi/50*x)).*(1-cos(pi/100*(y-150))) ;
R = @(V,L,H,X)heaviside(V-L)*heaviside(H-V)*X
T = simplify( R(x,0,100,R(y,-150,-50,T1) + R(y,-50,50,T2) + R(y,50,150,T3)) )
Now you can do laplace() on T. However, when you do so, you need to specify which variable you are transforming with respect to. Likewise when you fourier, make sure you do so with respect to the correct variable.
Also, you need to carefully examine what happens at the exact boundaries.
soe min aung
on 27 Feb 2020
Walter Roberson
on 28 Feb 2020
Your zeta is a symbolic expression in k and y according to symvar(), but it also contains k2 inside fourier() calls.
Perhaps you should do
kvals = 0:100;
yvals = linspace(-200,200,101);
[K,Y] = meshgrid(kvals, yvals);
Zsym = subs(zeta, {k, y}, {K, Y});
Z = double(Zsym);
surf(K,Y,Z)
However in practice this does not work: after the substitution, double() is not able to resolve some of the values to numeric. I discovered that some of them can be converted to a better form by using simplify(Zsym), but that is slow.
soe min aung
on 2 Mar 2020
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