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These are replies submitted by J4James

@Preben Alsholm You have read my mind. Thanks.

How to equate the terms in the exponents from the output? 


to get

2*alpha[2]-alpha[3] = 2*alpha[1]-alpha[3]


@Carl Love 


I think, the issue is with replacing infinity with 10000.

@Preben Alsholm Your are right, I'm looking for a stable limit cylce and now need to follow whatever you have suggested.


@Preben Alsholm 

with ,obsrange=false the output look more complicated. 


arrows=medium,axes=boxed,scene= [x(t),y(t)],obsrange=false)


@Carl Love 

The desired one if possible, should look like this,

@Carl Love Thanks 

I'm getting empty plot when I do this


color=blue,linecolor=magenta,arrows=medium,axes=boxed,scene= [x(t),y(t)]);


@Preben Alsholm 

Here is the PDF.



@Preben Alsholm 

Sorry for the confusion. 

How does the procedure (when using procedural input) have to look for a delay differential equation if it is at all available to the user? 



@Preben Alsholm 

I'm wondering maybe Maple can still handle DDEs with a Proc.


@Preben Alsholm 

Probably, the bugged results matches with the published one. 

 With my Maple 18, I get the error.

The delays are constant and positive. In this case, tau1 and tau2 (delays) are assumed to be 10.

The history is constant for this problem

x(t)=1, y(t)=1, v(t)=5, and z(t)=1. t<=0.


I tried your example with Maple 18 but get the same error. 

@Preben Alsholm

But i get this 


Eq1 := diff(x(t), t) = 1-.1*x(t)-0.5e-3*x(t)*v(t)/(1+0.1e-5*v(t));

Eq2 := diff(y(t), t) = 0.5e-3*x(t-10)*v(t-10)/(1+0.1e-5*v(t-10))-.3*y(t)-y(t)*z(t);

Eq3 := diff(v(t), t) = 200*y(t-10)-8*v(t);

Eq4 := diff(z(t), t) = 2*y(t)*z(t)-.15*z(t);

ics := x(0) = 1, y(0) = 1, v(0) = 5, z(0) = 1;


Error, (in dsolve/numeric/process_input) input system must be an ODE system, got independent variables {t, t-10}

Any comment?

@Kitonum  FIrst of all, thanks for your interest.

I think, you are just trying to replicate the Figure?



@Markiyan Hirnyk Thanks dear for a very nice and helpfull explanation. If there is something else, I have nowhere else to go?

I still don't understand, why the authors need to present inaccurate plot (Fig 2.).


@Markiyan Hirnyk 

How you manange to consider 0 < t and t < theta[m], 1/2-(1/2)*phi+t*phi/theta[m], 1)  from the assumptions.?

0<t part is clear but  not t < theta[m].


@Carl Love Is it possible to use maple to solve Eq 1


to get the solution

x = phi*(theta[m]-1/x)/theta[m]+(1-phi)*(1/2); ?


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