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

  • $\begingroup$ 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. $\endgroup$ – David Ketcheson Nov 5 '19 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 '19 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. $\endgroup$ – David Ketcheson Nov 6 '19 at 6:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.