0
$\begingroup$

I am interested in problems of the form $$ u_t = F(u) + S(u) $$ where $F(u) = - div(f(u))$ and $S(u)$ is a stiff source term. I am looking for any existing works which develop Lax-Wendroff type schemes for such stiff problems using an IMEX approach to deal with stiff source. So far my search has yielded nothing more than Crank-Nicholson for the source [1] which is not good for stiff case. Is there any fundamental difficulty in constructing Lax-Wendroff schemes which are L-stable ?

The LW schemes I am looking at are of the form as used in [2]. Perhaps LW is not a good name, they seem to be called two-derivative schemes. At second order, they are basically an LW scheme.

[1] Yuangao Zhang and Behrouz Tabarrok, Modifications to the Lax–Wendroff scheme for hyperbolic systems with source terms, https://doi.org/10.1002/(SICI)1097-0207(19990110)44:1%3C27::AID-NME485%3E3.0.CO;2-0

[2] Christlieb et al., Explicit Strong Stability Preserving Multistage Two-Derivative Time-Stepping Schemes, DOI: 10.1007/s10915-016-0164-2

$\endgroup$
3
  • 1
    $\begingroup$ Lax-Wendroff is a full (time+space) discretization; it doesn't fit in the method of lines framework. Meanwhile, "L-stable" refers to a time discretization and only makes sense within the method of lines. So I can't really make sense of your question. $\endgroup$ Nov 5, 2019 at 7:24
  • $\begingroup$ @DavidKetcheson Is there a different concept to develop IMEX Lax-Wendroff methods to deal with stiff source terms ? I was thinking of L-stable along these lines. If LW scheme is like $u^{n+1}=G(F,S,\Delta t)u^n$ then if I apply it to the case, $F(u)=0$, $S(u)=\lambda u$, $Re(\lambda)<0$, then I want $|G|\to 0$ as $Re(\lambda)\Delta t \to -\infty$. $\endgroup$
    – cfdlab
    Nov 5, 2019 at 7:37
  • $\begingroup$ The schemes in [2] are not Lax-Wendroff type (at least by any meaning of the term that I understand). For starters, they are "multistage" whereas perhaps the biggest selling point of Lax-Wendroff is that it is not multistage. $\endgroup$ Nov 6, 2019 at 6:19

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.