Skip to main content
Mathematica

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

9
  • $\begingroup$ I tried your approach together with "SpatialDiscretization" -> {"FiniteElement",...} but I get a warning CoefficientArrays::poly: (u^(1,0))[3,t] is not a polynomial. What could be the reason? Thanks! $\endgroup$ Commented Jul 11, 2019 at 8:20
  • 2
    $\begingroup$ @UlrichNeumann "FiniteElement" method cannot handle the b.c. Derivative[1, 0][u][L, t] == 0 (at least now), one needs to rewrite it with NeumannValue. (BTW, some related discussion can be found in the comments under this post. ) (Edit: Oh I forgot it's not necessary to explicitly set zero NeumannValue, one can just remove that b.c., as suggested by user21 below. ) $\endgroup$ Commented Jul 11, 2019 at 8:28
  • 1
    $\begingroup$ @UlrichNeumann, just delete that Derivative[1, 0][u][L, t] == 0 boundary condition. $\endgroup$ Commented Jul 11, 2019 at 8:29
  • 1
    $\begingroup$ @Nasser The default value of Piecewise is 0 i.e. $f(x)=0$ when $x\ge L$ under your definition. $\endgroup$ Commented Jul 11, 2019 at 8:33
  • 1
    $\begingroup$ @xzczd Thanks, FiniteElement seems to be much more robust (concerning the definition of f[x]) $\endgroup$ Commented Jul 11, 2019 at 9:43