Mathematica Asked on November 15, 2021
currently I’m working on a problem in which I need to solve an equation of the following form:
$$mathcal{L}left[frac{partial f(x,t) }{partial x} + frac{partial f(x,t)}{partial t}right] = frac{1}{y^{3/4}}mathcal{L}left[frac{partial^{2}f(x,t)}{partial t^{2}}right] $$
Where $mathcal{L}$ stands for the Laplace transformed and $ y $ is corresponding variable to $x$ after performing the Laplace transform.
I have the following doubts:
Although the derivative in the RHS is respect t. Do I have to apply the transform over the two variables $x, y$ ?
I’ve been trying to solve the equation using the following code in mathematical but it seems that it’s not working:
A = 10(*Some number*)
RHS = (1/y^(3/4))LaplaceTransform[D[f(x,t),{x,2}],x,s]
LHS = LaplaceTranform[D[f(x,t),x] + D[f(x,t),t],x,s]
ICOND = {f[0,t] == 0, f[1000,t] == 1000 - A, f[x,1] == Max[x-A,0]}
Sol = NSolve[{LHS == RHS,ICOND},c,{x,0.1,1000},{t,0,1}]
What would be a plausible way for solving the first equation I wrote? Are there any books or PDF’s which talk about solving this kind of equations?
I’m new to Mathematica, I’ve looked around an I haven’t found any that resembles my doubt.
Thank you for reading.
Update 1.
Another doubt. Does someone have any hints on how would this Laplace transform is to be applyed?
$$ mathcal{L}left[tfrac{partial^{2}f(x,t)}{partial t^{2}}right] = ?$$
Considering instead
$$mathcal{L}[frac{partial f(x,y,t) }{partial x} + frac{partial f(x,y,t)}{partial t}] = frac{1}{y^{3/4}}mathcal{L}[frac{partial^{2}f(x,y,t)}{partial t^{2}}] $$
and calling
$ g_y(x,t) = f(x,y,t) $
we can write as
$$ mathcal{L}[frac{partial g_y(x,t) }{partial x} + frac{partial g_y(x,t)}{partial t}] = frac{1}{y^{3/4}}mathcal{L}[frac{partial^{2}g_y(x,t)}{partial t^{2}}] $$
with Laplace transform
$$ frac{partial}{partial x}G_y(x,s) + s G_y(x,s)-frac{s^2}{y^{3/4}}G_y(x,s) = g_y(x,0)-frac{1}{y^{3/4}}left(s g_y(x,0)+g_y'(x,0)right) $$
or
$$ frac{partial}{partial x}G_y(x,s) + left(s-frac{s^2}{y^{3/4}}right)G_y(x,s) = g_y(x,0)-frac{1}{y^{3/4}}left(s g_y(x,0)+g_y'(x,0)right) $$
now assuming $g_y(x,0) = g'_y(x,0) = 0$ we can proceed with
Gy[x, s] /. DSolve[D[Gy[x, s], x] + (s - s^2/y^(3/4)) Gy[x, s] == 0, Gy, {x, s}]
Answered by Cesareo on November 15, 2021
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