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I am using a differential equation method for finding the roots of a function. My function is much more complicated but I can illustrate my problem with a trivial example. The problem occurs when roots coincide and change direction. I use a differential equation method. The example equation is

eq = s^2 + 2 a s + 1;

where a is a parameter (0 < a) and I wish to find the roots as the parameter is varied. To form the differential equation I take two derivatives and put them together as follows

deq = {D[eq, s], D[eq, a]} /. s -> s[a];
de = deq[[1]] s'[a] + deq[[2]] == 0
(* 2 s[a]+(2 a+2 s[a]) (s^\[Prime])[a]\[Equal]0 *)

I set up the differential equation so that I can solve with different initial conditions and over different ranges of the parameter a

ClearAll[sol];
sol[ic_, ae_] := s /. First[NDSolve[{de, s[0] == ic}, s, {a, 0, ae}]]

I have initial conditions from 

ic = s /. Solve[eq == 0 /. a -> 0]
(* {-\[ImaginaryI],\[ImaginaryI]} *)

First case is fine

a1 = 0.999;
rts = sol[#, a1] & /@ ic;
ParametricPlot[Evaluate[{Re[#[a]], Im[#[a]]} & /@ rts], {a, 0, 0.999}]

Mathematica graphics

Second case attempts to go further

a1 = 2;
rts = sol[#, a1] & /@ ic;

NDSolve gets stuck as it attempts to go "round the corner" at -1. Is there a way to get NDSolve to go round the corner? Can I get both solutions going round the corner and choosing opposite directions? (One solution should go towards the origin the other towards infinity.) Thanks for help and suggestions.

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It looks like you're trying to draw a root locus -- there is a special command for this:

RootLocusPlot[s^2 + 2 a s + 1, {a, 0, 2}]

enter image description here

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    $\begingroup$ Thanks.I am familiar with RootLocusPlot[] but my more complicated functions (not always a polynomial) needs a more "personal" touch. It is interesting to note that RootLocusPlot[] does get "round the corner". How does it do that? $\endgroup$
    – Hugh
    Commented Jun 1, 2014 at 17:17
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    $\begingroup$ Also, note that RootLocusPlot[] is the Evans Root Locus and assumes that there is a feedback path. If you look at your solution it does not go through the same points as mine. $\endgroup$
    – Hugh
    Commented Jun 1, 2014 at 17:37
  • $\begingroup$ You can apply the root locus to a simple polynomial rather than one with a feedback loop (there is an example in the docs). But in any case, it can only handle polynomials -- if you stick a Log or Exp or some nonlinearity, you would need to do it yourself. $\endgroup$
    – bill s
    Commented Jun 1, 2014 at 17:53

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