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 Feb21 comment What boundary is added when NDSolve::bcart pops up? You mixed up -1 and 1 in your code, right? Feb16 comment Dramatic speed difference of code on Matlab and Mathematica @zjx1805 I believe you'll manage to improve the complete code if you fully understand the answers you got. BTW, See my edit for another 2X speedup. Feb15 comment Dramatic speed difference of code on Matlab and Mathematica @zjx1805 Then what's the timing for my code? Feb15 comment Dramatic speed difference of code on Matlab and Mathematica Seems that this hasn't been done: Hello, welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! Feb14 comment FindFit runs extremely slow for my Function,how do I optimize this? Add Method -> {Automatic, "SymbolicProcessing" -> False} in NIntegrate may help. Feb13 comment How can I invoke the solution of NDSolve to determine a parameter in my equation just inside NDSlove? @lxy No, they're the same, when being a definite solution problems. The solutions of your 2 examples are different, but it's only true for general solution. When you set proper b.c.s to get a unique solution, they'll be the same, try DSolve[{y'[x] + y[x] == a f[t], y[0] == 0}, y[x], x] and DSolve[{D[y'[x] + y[x] == a f[t], x], y[0] == 0, y'[0] == a f[t]}, y[x], x]. (y'[0] == a f[t] can be easily deduced from the first equation set.) And this is exactly your case. Of course if Derivative[2, 0][h][0, t] == Derivative[2, 0][h][L, t] is mistakenly given, then it's another story. Feb13 comment Making table and plotting results of NDSolve problem There are some simple mistakes in your question, but I think the most troublesome part is that under h = 1 it seems to be impossible to find a proper initial guess for shooting method. Using the approach in this post, the best initial guess I can find is sol[h_, L_, i_] := sol[h, L, i] = {i, NDSolve[{f[x]^3 # & /@ (f''[x] == f[x] (h/f[x]^2)^2 - f[x] (1 - f[x]^2)), f[0] == f[L] == 1}, {f}, {x, 0, L}, Method -> {"Shooting", "StartingInitialConditions" -> {f[L] == 1, f'[L] == i}}]}; sol[1, 8, 9000000003/10000000000] Feb13 comment VectorScale Explanation What do you mean by saying "My version (10.0.0) does not support sfun."? (You did set it in your second sample.) Anyway, +1, I didn't know StreamPoints etc. can be used inside VectorPlot! Feb13 comment VectorScale Explanation 1. Is "With unitlen = 1 (middle)" a typo? 2. So you think aratio is short for arrowhead ratio rather than aspect ratio? Feb13 comment Graphics3D: Opacity limitations I'm afraid this isn't what OP wants, he wants "the lines further from the eye to be "grayer" than the lines closer to the eye, as if there were a little fog inside the sphere", while this only differs "front" and "behind". Feb12 comment Obtaining the number of iterations used in ReplaceRepeated But you're asking for "a way to obtain the actual number of replacement operation", not a option of ReplaceRepeated that demonstrate the actual number of replacement operation :) Feb12 comment Obtaining the number of iterations used in ReplaceRepeated possible duplicate of How to visualize pattern matching process? Feb12 comment Graphics3D: Opacity limitations @celtschk I got the same disappointing result as OP in v8.0.4 and v9.0.1, Vista 32bit. Can this be a platform or even hardware related issue? Feb12 comment NDSolve grid refinement for PDEs I think it'll be better if you can add the specific equations to your question, maybe someone can find a combination of options that'll circumvent your problem. Feb11 comment replacement rules from a pattern and a matching expression Er……how can I make "Failed!" appear in the result? Feb11 comment Alternative to Series @Hawk Yes. Seems that it can be arbitrary small. Just tried x[1][0] == 10^-160. Feb11 comment Alternative to Series I fixed it for you. NDSolve outputs an InterpolatingFunction, which is already a differentiable approximation. Well, I can post an answer if you like, the graph under your i.c.s isn't interesting though. Feb11 comment Alternative to Series Your code still contains typos, you'd better fix them. The only troublesome part is the x[1][0] == 0 condition, if an approximate nonzero condition like x[1][0] == 10^-10 is acceptable, then the method in the post linked above will handle your equation set. Feb11 comment Alternative to Series If a numeric series solution is desired, then I think this is a possible duplicate of Series expansion of InterpolatingFunction obtained from NDSolve Feb11 comment How to visualize pattern matching process? I tried to improve the expressions of your question. (Wish I didn't make it worse… ) Well, To be honest, personally I feel your Edit a little verbose…