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Feb 3, 2018 at 5:11 history tweeted twitter.com/StackMma/status/959655681870753792
Feb 2, 2018 at 20:46 history edited Nasser CC BY-SA 3.0
corrected latex
Feb 2, 2018 at 19:22 comment added anderstood @MichaelE2 The \underline{Summary} makes me believe the details were already written before :)
Feb 2, 2018 at 18:57 comment added Michael E2 A simple reminder that for an eigenvalue you have to have a nonzero eigenvector would have been enough. Just a brain fart. :)
Feb 2, 2018 at 17:37 history edited Nasser CC BY-SA 3.0
typo
Feb 2, 2018 at 17:31 history edited Nasser CC BY-SA 3.0
added solution back to answer comment
Feb 2, 2018 at 12:13 history edited Nasser CC BY-SA 3.0
removed my solution as not needed
Feb 2, 2018 at 11:56 comment added xzczd I agree NeumannValue is somewhat hard to use. (You may want to read this post. ) Hope DEigenvalues etc. will be more flexible and powerful in future versions.
Feb 2, 2018 at 11:46 comment added Nasser @xzczd Ok, thanks. I am used to using DSolve as above to find the eigenvalues since it is easier to use than the syntax of DEigenvalues, but I did not realize that the trivial solution will be also given by DSolve. This was the confusing part for me. So I should switch to DEigenvalues from now on to check my solution.
Feb 2, 2018 at 11:43 vote accept Nasser
Feb 2, 2018 at 11:39 comment added xzczd "but it also claims zero is an eigenvalue" No, it doesn't. It just claims when $\lambda=0$, the BVP has a solution. If one wants to find the eigenvalue, DEigenvalues in principle can be used, but something like DEigenvalues[{-y''[x] + NeumannValue[-y[x], x == 1] + NeumannValue[y[x], x == 0]}, y[x], {x, 0, 1}, 5] doesn't work, NDEigenvalues works, though…
Feb 2, 2018 at 11:34 answer added Sjoerd Smit timeline score: 9
Feb 2, 2018 at 10:34 history edited Nasser CC BY-SA 3.0
typo
Feb 2, 2018 at 10:22 history asked Nasser CC BY-SA 3.0