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A naive question regarding MATLABs definition of parabolic/elliptical PDEs (pdepe)

2 vues (au cours des 30 derniers jours)
This is a shortly put question, and possibly very obvious, but MATLAB says in the documentation of pdepe (here) that is solves 1D parabolic and elliptical PDEs. However, any definition I've come across for PDEs' conic classification holds that the coefficients in the PDE expansion should be constant, i.e.
where
For pdepe these coefficients are functions however. How does MATLAB classify them as parabolic or elliptical if the coefficients are variable? Is it that it must follow the inequality below?

Réponse acceptée

Torsten
Torsten le 28 Avr 2024
Modifié(e) : Torsten le 28 Avr 2024
Second-order linear partial differential equations (PDEs) are classified as either elliptic, hyperbolic, or parabolic. Any second-order linear PDE in two variables can be written in the form ... where A, B, C, D, E, F, and G are functions of x and y.
Loosely speaking, elliptic for "pdepe" means the equation has the form D*d^2u/dx^2 - v*du/dx + s = 0 and parabolic means it has the form du/dt = D*d^2u/dx^2 + v*du/dx + s.
You shouldn't care so much about dependency of the coefficients. The important thing is that the equation doesn't have the form du/dt =- v*du/dx + s or - v*du/dx + s = 0 (hyperbolic type).
  2 commentaires
David Gillcrist
David Gillcrist le 28 Avr 2024
Is it safe to say then that so long as the second order term for x is there than it can be solved (at least in principle) with pdepe?
Torsten
Torsten le 28 Avr 2024
Modifié(e) : Torsten le 28 Avr 2024
The presence of the second-derivative term is crucial because "pdepe" expects 2 boundary conditions for each PDE that you implement. If you have no second-order term, no (if no spatial derivatives are present) or only one (if only a first-order spatial derivative is present) boundary condition is mathematically necessary ( and sufficient ) to define a well-posed problem. The remaining boundary conditions that you would have to impose to make "pdepe" work would be artificial ones.
Another aspect why a second-order term is necessary is the discretization scheme used for the first-order spatial derivatives. It usually is not stable if no stabilizing second-order term is present.

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