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Topic: Integration with variable limits
Replies: 17   Last Post: Aug 20, 2013 10:20 AM

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Sanaa

Posts: 154
Registered: 3/20/12
Re: Integration with variable limits
Posted: Aug 16, 2013 1:07 PM
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"Torsten" wrote in message <kulcpv$jj9$1@newscl01ah.mathworks.com>...
> "Sanaa" wrote in message <kulb7b$qrm$1@newscl01ah.mathworks.com>...
> > "Torsten" wrote in message <kul93r$t5q$1@newscl01ah.mathworks.com>...
> > > "Sanaa" wrote in message <kul7kv$a10$1@newscl01ah.mathworks.com>...
> > > >
> > > > > > y_(n+1)=x_n;
> > > > > > x_(n+1)= x_n + integration (f(y_(n+1)(s))) ds.

> > > > >
> > > > > So your final aim is to solve the delay differential equation
> > > > > dx(t)/dt = rho*x(t-r)*(1-x(t-r))
> > > > > with x given on an interval of length r at the beginning, r and rho constant over time ?
> > > > >
> > > > > Best wishes
> > > > > Torsten.

> > > >
> > > > No. I don't want to solve the original delay differential equations, I wish to solve the discrete system after a certain discretization method applied to it.
> > > > Thanks a lot in advance.

> > >
> > > But if the discretrization method is given, you are also given how the above integral is approximated.
> > > Or are you _searching_ for an adequate discretization method for the delay differential equation ?
> > >
> > > Best wishes
> > > Torsten.

> >
> > Thanks once again for your kind reply. The discretization method here is very simple, the step methods, I am not searching for the method, i just don't know how to approximate the integral. I have read about trapz and cumtrapz but my only problem now is the variable limits of integration. Using trapz or cumptrapz don't allow me to specify the variable limits; I am really confused.

>
> First take care that your step length divides r - otherwise you will have trouble.
> Say you have the delay differential equation from above
> dx/dt = rho*x(t-r)*(1-x(t-r))
> with x given on the interval [-r,0].
> Now take a step dt such that dt divides r (say n*dt=r)
> x(t+dt)-x(t) = integral_{t'=t}^{t'=t+dt} (rho*x(t'-r)*(1-x(t'-r)) dt')
> The integral can be approximated by the trapezoidal rule:
> integral_{t'=t}^{t'=t+dt} (rho*x(t'-r)*(1-x(t'-r)) dt') ~
> dt*(rho*x(t+dt-r)*(1-x(t+dt-r))+rho*x(t-r)*(1-x(t-r)))/2.
> Thus your recursion reads
> x(t+dt)=x(t)+dt/2*rho*(x(t+dt-r)*(1-x(t+dt-r))+x(t-r)*(1-x(t-r)))
> Inserting n*dt=r gives
> x(t+dt)=x(t)+dt/2*rho*(x(t-(n-1)*dt)*(1-x(t-(n-1)*dt))+x(t-n*dt)*(1-x(t-n*dt)))
> or as discrete recursion
> x_(k+1)=x_k+dt/2*rho*(x_(k-(n-1))*(1-x_(k-(n-1)))+x_(k-n)*(1-x_(k-n)))
> for k=0,1,2,...
> The values x(-(n-1)),x(-(n-2)),...,x(0) are given by the initial condition on the interval
> [-r,0].
> Does this help ?
>
> Best wishes
> Torsten.


I am sorry for any inconvenience. I want the symbolic integration of the function (whatever the original problem is and whatever the dis. method is); that is,
y_(n+1)=x_n;
x_(n+1)= x_n + integration (f(y_(n+1)(s))) ds.
where f(y(t)) = rho*y(t)*(1-y(t)).
From the discussion above, I think this task is very difficult.
Anyway, thank you very much for your help.



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