From: Sunipa Som on
I have to solve a equation, which is mixing of ODE and PDE.
(dx/dt)+c(dx/dr)+d(dx/dk)=(dx(k)/dt)
This equation has initial value at t=0.
Right hand side is ODE, and it is in the form dx(k)/dt=c1*x(k), where c1 is constant.
How can I solve this equation?
Thanks for any further help

Sunipa
From: Torsten Hennig on
> I have to solve a equation, which is mixing of ODE
> and PDE.
> (dx/dt)+c(dx/dr)+d(dx/dk)=(dx(k)/dt)
> This equation has initial value at t=0.
> Right hand side is ODE, and it is in the form
> dx(k)/dt=c1*x(k), where c1 is constant.
> How can I solve this equation?
> Thanks for any further help
>
> Sunipa

I do not completely understand your equation.
Are x and x(k) different solution variables ?

And the answer to your question depends on how
you solved the PDE without ODE.

Best wishes
Torsten.
From: Sunipa Som on
Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <1304861797.288066.1267429202355.JavaMail.root(a)gallium.mathforum.org>...
> > I have to solve a equation, which is mixing of ODE
> > and PDE.
> > (dx/dt)+c(dx/dr)+d(dx/dk)=(dx(k)/dt)
> > This equation has initial value at t=0.
> > Right hand side is ODE, and it is in the form
> > dx(k)/dt=c1*x(k), where c1 is constant.
> > How can I solve this equation?
> > Thanks for any further help
> >
> > Sunipa
>
> I do not completely understand your equation.
> Are x and x(k) different solution variables ?
>
> And the answer to your question depends on how
> you solved the PDE without ODE.
>
> Best wishes
> Torsten.

In the left hand side x depends on k,r and t. So by solving left hand side, we will get values of x with respect to k, r and t. We can draw a curve x vs k and r for a particular t.
In the right hand side x depends on k and t. So by solving right hand side, we will get values of x with respect to k and t. I already solved the right hand part by ode function.
But we have to solve whole equation, where we will get values of x with respect to r, k and t. Then we have to draw a curve x vs r and k for a particular t.

Thanks for any further help
Sunipa