ode15s

Solving a partial differential equation with ode15s,we know that this ode solver integrates the ode over time direvatives.I need to find the values of these time direvatives,do you have any special command? for example: for i=1:n dudt=uxx(i) end

how can I find these dudt at each time and x(i)? thanks alot

Answer 1

[Andrew Newell]

If you are solving something like

[t,x] = solver(odefun,tspan,y0)

and you want the time derivative at each point t(i), then that is simply

dxdt = 0*x;  % initialize the array
for ii=1:length(t)
  dxdt(ii,:) = odefun(t(ii),x(ii,:));
end

(edited to allow for the possibility that x has more than one component at each time t).

[Bård Skaflestad]

That's more or less what I had in mind. Calling the |odefun| from an |OutputFcn| is just another refinement.

[ahmad]

So if I want to calculate the integral of square of dudt i don't have any special command in my mind,in fact I used trapz but it keeps saying "the length of x should be the same as the first non-singleton variable" or it says that "x should be a vector".

Answer 2

[Bård Skaflestad]

If you're going to solve a PDE by the method of lines (i.e., converting it into a system of ODEs), then you will need to provide a spatial discretisation of the PDE along with suitable initial conditions whence ode15s (or others) will likely assist you in the numerical solution.

Please review your favourite text book on discretisation methods.

[ahmad]

Actually I am solving a PDE by discretization,i did all that and got the matrix of solution which is like u(x,t).but at some other part of my research I need to double integrate over values of dudt,so I need them.I know that ode15s calculates these values and save them probably somewhere,I need to find them and then use them as my integrand.

I hope you understood my problem.

thanks

[Bård Skaflestad]

I'd be *very* surprised if the solver stores all function values throughout the computation. That is an extraordinary amount of data, especially if the spatial resolution is high or if you're solving a multi-dimensional PDE.

However, I'm thoroughly confused about what you're asking. Isn't |dudt| exactly the output of your |odefun|? Can you not calculate these values yourself by simply calling your |odefun| directly at a point (t,u)? This process may even be directed by the |OutputFcn| facility detailed in the |optimset| documentation.

[Bård Skaflestad]

Sorry, I meant |odeset|, not |optimset|.

[ahmad]

I think that you are good at matlab numerical processes,can you please take a look at my program?I need to call the du/dt values solved by ode solver at every point like du/dt(t,x),is there a special command for it?it is true that its a big amount of data.so how can i do the other calculations using these values?

thanks

Answer 3

[Andrew Newell]

You could use deval.

EDIT: Here is a modified version of the example at the link provided:

sol = ode45(@vdp1,[0 20],[2 0]);
t = linspace(0,20,100);
[y,dy] = deval(sol,t,1);
plot(t,dy);

[Bård Skaflestad]

Possibly, but I don't think that's what the OP actually wants. If I understand correctly, he wants to calculate the (temporal) derivative of the solution which |deval| does not do -- or at least not return to the caller. I think he should just call his original |odefun| from an |OutputFcn| and then do something sensible with the result.

I could be mistaken, though, and I'd be happy to stand corrected.

[Andrew Newell]

I think it is the time derivative (as in the example I added), but for a polynomial fitting the solution. I provide a more direct solution in a separate answer.

[Bård Skaflestad]

Indeed it is. The second output of |deval| is the derivative (wrt the independent variable) of the continuous output polynomial fitted to the numerical scheme of the particular solver (ODE45 in the above example).

Yet another detail I'd missed while skimming the documentation of |deval|.