Solving an initial value problem for a PDE
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Having the following initial value problem
with some mathematical computations we reach to an end that an implicit general solution of this pde can have the following form
if we had phi=e^(-x^2) for example,
I have been able to solve a similar problem to this but the genral solution was only a function of x and t, but here we have also u, so how can we possibly do that.
0 Commenti
Risposte (1)
Torsten
il 26 Apr 2022
Modificato: Torsten
il 26 Apr 2022
The method of characteristics gives the equations
dt/ds = 1, t(0) = 0
dx/ds = u, x(0) = x0
du/ds = 0, u(0) = phi(x0)
with solution
x = x0 + phi(x0) * t
Thus to get the solution u(x,t) in (x,t), you will have to solve
x - x0 - phi(x0)*t = 0
for x0.
The solution u(x,t) in (x,t) is then given by u(x,t) = phi(x0).
Vedere anche
Categorie
Scopri di più su Boundary Conditions in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!