I have a linear elliptic PDE (unfortunately not allowed to be shown here) with a constant parameter $\epsilon$ giving the stable solutions qualitatively shown by the functions below.

enter image description here

As we smoothly increase $\epsilon$, the solution changes "qualitatively" from the viewpoint of the number of extrema relating to the solution.

Can I refer to the solution at $\epsilon_{cr}$ as a bifurcation point?

  • $\begingroup$ Why can't you show the PDE? $\endgroup$ Oct 16 '17 at 13:47
  • $\begingroup$ But also, my interpretation if you called it "bifurcation point" would be to ask "what is it that bifurcates"? In your case, it's clearly the zeros of the solution, but is that what's important about the solution? Commonly, one would use the term to indicate a parameter beyond which there are two solutions to the equation. If the "solution" of your problem are the zeros of the function that solves the PDE, then the term may be appropriate. $\endgroup$ Oct 16 '17 at 13:49
  • $\begingroup$ The PDE is a simple elliptic one, without any useful information regarding to the question above. However, the spatially varying coefficients (which introduce epsilon) are deduced from a theory that is currently under consideration for publication. $\endgroup$
    – BalazsToth
    Oct 16 '17 at 14:29
  • 7
    $\begingroup$ Science has arrived at a bad place if you think that you can't even show an equation any more. Surely an equation without any accompanying text does not give anything away... $\endgroup$ Oct 16 '17 at 14:43
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    $\begingroup$ I agree with Wolfgang, but if it helps try comparing your use of the term with how it's used in people.maths.ox.ac.uk/trefethen/ExplODE, Chapter 17. Also, I think since the PDE is not important here, you could use $x-\epsilon \tanh x$ as a model ($\epsilon_{\mathrm{cr}}=1$), I think it would be the same thing. The roots have a bifurcation because the number of roots is a structural property, but the numbers of roots of a function is not a structural property of that function itself. $\endgroup$
    – Kirill
    Oct 16 '17 at 19:13

No -- as long as you keep saying "the solution".


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