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I am interested in constructing a bifurcation diagram for parameter a in the dynamical system given in the code below. I want to see how parameter changes affect the stability of the system. The answer in

Bifurcation diagrams for multiple equation systems

does not apply for Mathematica 8. The following code include the system of differential equations:

Derivative[1][x][t] == -10^12 x[t] - y[t], 
Derivative[1][y][t] == x[t] - 10^12 y[t] + a z[t], 
Derivative[1][z][t] == -a y[t], 
x[0] == y[0] == 0, z[0] == 1

Thanks

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    $\begingroup$ "bifurcation-diagrams-for-multiple-equation-systems" You mean this question? $\endgroup$
    – user484
    Aug 21, 2013 at 23:21
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    $\begingroup$ Your system is linear, so it will not have very interesting dynamics. You can find the region of stability by rewriting as a matrix and looking at the eigenvalues as a function of a. $\endgroup$
    – bill s
    Aug 21, 2013 at 23:27
  • $\begingroup$ @Rahul Narain. Correct me, if I missed something, but: First: the reference you gave shows a parameters-dependent solution of a system of equations, rather than any bifurcation diagram. In the text of the explanation Mark McClure describes, what is the bifurcation diagram, but does not build any. Second: in this reference even the textual explanation only speaks on the, so-called, local bifurcations. The latter might, of course, be difficult to build in practice. The task of making a program that would automatically draw one might be challenging, at least to me. $\endgroup$
    – Alexei Boulbitch
    Aug 22, 2013 at 7:25
  • $\begingroup$ @Rahul Narain. Continuation. However, it is al least clear, how the problem should be formulated, and it is briefly described in the reference you gave. In contrast, global bifurcations are much more the problem. I remember, when about 20 years ago Viktor Yudovich discovered a new global bifurcation of the Lorenz equation, the resonance was world-wide. Since that time I did not follow the problem, however. Could you please update me, if there are changes. $\endgroup$
    – Alexei Boulbitch
    Aug 22, 2013 at 7:33
  • $\begingroup$ @Rahul Narain (continuation) It is especially interesting, if some automated approaches have been developed, enabling one to numerically build the diagram or, at least, its components. $\endgroup$
    – Alexei Boulbitch
    Aug 22, 2013 at 7:33

1 Answer 1

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Your system is linear, and can be rewritten as dp/dt = m.p where p={x,y,z} and 

m = {{-10^12, -1, 0}, {1, -10^12, a}, {0, -a, 0}};

The stability of the system is given by the eigenvalues of m:

sol = Eigenvalues[m];

Some sample values are: 

sol//.{a->0}
{0, -1000000000000 - I, -1000000000000 + I}

sol//.{a->-0.1}
{-1.*10^12, -1.*10^12, -1.*10^-14}

sol//.{a->0.1}
{-1.*10^12, -1.*10^12, -1.*10^-14}

So -- if you look at the values, you have stability for all a except a=0, where it is marginally stable.

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  • $\begingroup$ Sorry. The original post was faulty. I corrected it. Can you analyze it again? $\endgroup$
    – New Developer
    Aug 22, 2013 at 7:28
  • $\begingroup$ For a linear system such as yours, there is no period doubling, no regions of multiple equilibria. What other information do you want? $\endgroup$
    – bill s
    Aug 22, 2013 at 13:34
  • $\begingroup$ I want to know that how z(t) behave if 0<a<1. Unfortunately, in this case, two of three eigenvalues become zero and mathematica cannot draw the plot. I don't know what should I do to draw the plot. $\endgroup$
    – New Developer
    Aug 22, 2013 at 16:33
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    $\begingroup$ It is marginally stable when the eigenvalues are zero. This means it may diverge or it may oscillate, depending on the initial conditions. $\endgroup$
    – bill s
    Aug 22, 2013 at 16:54

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