## 52 Reputation

13 years, 359 days

## system...

Indeed, I have obtained the same values following the hint above. Further it would be possibly to look for bifurcations by a parameter for nonlinear and non autonomous systems? for example if I would substitute A by a function   A(t) or at least a function constant on intervals and look again for bifurcations by F1 (here g). Or more, bifurcations by two parameters at once ? Should here be a meaning?

## here is...

Ok. I apologize for the mistakes along my posts. In the attach you can see my real system. Another question: after this computation, how could I do to insert the code into another file (in latex for example, but not exporting the document, I would like with copy/paste) just as it is.  More precisely i want to appear 0.0012 not something times with "e", the Jacobian as a matrix and so on. Thanks a lot!

p.s I have Maple 9.5 and start a sheet running Maple 9.5 not Classic Worksheet 9.5. I note that it runs rather slowly.

## bug?...

Sorry, pars:= {b =1, c=0.2, d=0.5, e=14, f=0.15, g=2.2} (my original system has coefficients like E2, E2..etc) I studied the system using the commands you proposed above; it worked for e, g but as I see not for a.

## I'm back. Trying to look for...

I'm back. 1. Trying to look for bifurcations by a, for the set of values pars:= {F2 =1, E2=0.2, D2=0.5, D1=14, E1=0.15, F1=2.2} Using the command > evalf([allvalues(eval(S,pars))]) I get the the following Error message, (in evala/preproc3) floats not handled yet. What's wrong? 2. Is it posiibly to find two bifurcations points for the same parameter?

## It worked! Some questions 1....

It worked! Some questions 1. Where do appear the initial conditions x(0)=y(0)=0.01? 2. the bifurcation is considered at e=85.4 or at e=85.5? Where should I stop, where are still two equillibria points or at least one or no more one? Well, I tried for my real system (actually I thought is a slight modification but now I think it is significantly modified) So x'=a*b*((x^2+c^2)/(x^2+1))-d*x y'=a*e*((y^3+f^3)/(y^3+1)*(g*x+1))-y (a modification appears in the down term, for the first bracket I have y instead of x ) I followed the above approach with the following changes pars:= { a=7, b = 0.2, c=0.2, d=0.1, f=0.5, g=2.8}; and I got >evalf([allvalues(eval(S,pars))]); [{e = -0.1155507917e-1-.2044789251*I, x = 0.3444817452e-1-.1975123882*I, y = -.5608636000-.9714442513*I}, {e = -.1713064041-.1122464547*I, x = 0.3444817452e-1+.1975123882*I, y = -.5608636000-.9714442513*I}, {e = -0.3143281871e-2-0.5444323904e-2*I, x = 13.93110365, y = -.5608636000-.9714442513*I}, {e = -.1713064041+.1122464547*I, y = -.5608636000+.9714442513*I, x = 0.3444817452e-1-.1975123882*I}, {e = -0.1155507917e-1+.2044789251*I, x = 0.3444817452e-1+.1975123882*I, y = -.5608636000+.9714442513*I}, {x = 13.93110365, y = -.5608636000+.9714442513*I, e = -0.3143281871e-2+0.5444323904e-2*I}, {e= .1828614833+0.9223247044e-1*I, y = 1.121727200, x = 0.3444817452e-1-.1975123882*I}, {e = .1828614833-0.9223247044e-1*I, x = 0.3444817452e-1+.1975123882*I, y = 1.121727200}, {e = 0.006286563743, x = 13.93110365, y = 1.121727200}, {e = -0.1491201246e-1-.2638832880*I, y = -.2228705874-.3860231810*I, x =.3444817452e-1-.1975123882*I}, {x = 0.3444817452e-1+.1975123882*I, e = -.2210736249-.1448558256*I, y = -.2228705874-.3860231810*I}, {x = 13.93110365, y = -.2228705874-.3860231810*I, e = -0.4056454986e-2-0.7025986134e-2*I}, {e = -.2210736249+.1448558256*I, y = -.2228705874+.3860231810*I, x = .3444817452e-1-.1975123882*I}, {x = 0.3444817452e-1+.1975123882*I, e = -0.1491201246e-1+.2638832880*I, y = .2228705874+.3860231810*I}, {x = 13.93110365, y = -.2228705874+.3860231810*I, e = -0.4056454986e-2+0.7025986134e-2*I}, {e = .2359856373+.1190274624*I, y = .4457411749, x = 0.3444817452e-1-.1975123882*I}, {e= .2359856373-.1190274624*I, x = 0.3444817452e-1+.1975123882*I, y = .4457411749}, {e = 0.008112909971, x = 13.93110365, y = .4457411749}] In this case I have two possibilities: {e = 0.006286563743, x = 13.93110365, y = 1.121727200}, {e = 0.008112909971, x = 13.93110365, y = .4457411749}] for e=0.006 I obtain no equilibria points. With display([implicitplot(e006,x=13.92 .. 13.96,y=1.1..1.3,colour=[green,blue]), DEplot(zip(`=`,[D(x)(t),D(y)(t)],eval(D006,{x=x(t),y=y(t)})),[x(t),y(t)],t=1..20,x=13.92..13.96, y=1.1..1.3,arrows=slim)]); I can see the green line; If I take x=13.92..13.95 no line I can see, how do I interprete this fact? for e=0.008 again I do not obtain any equilibria point, but for e=0.0081, I do; what is the conclusion? there is no bifurcation? :(( Where I'm wrong? Thanks for your patience!

## Robert, there were of very...

Robert, there were of very help for me your explanations, but I have some questions. I'm a beginner in Maple, so I'm reading all commands step by step (working with Maple 9.5). I got an error massage after the command > S:= solve([op(F), J[2,2]], [x,y,e]); I understand that you compute J[2,2] (a proper value) at the equilibrium point. OK, I deduce I should try the same algorithm for J[1,1], isn't it? Now remain only two questions: why doesn't work the above command?maybe because working with Maple 9.5 :( and what means _Z? Thanks again you all! carmina
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