Empirical source terms with PDEPE

6 vues (au cours des 30 derniers jours)
Matthew Hunt
Matthew Hunt le 7 Mai 2020
Suppose I'm solving an equation:
With initial condition and boundary conditions and
Now I want to import the source term, as data, so I will have a Nx2 array [t Q] with data. Can I just treat that as I would a constant?
The code I have is:
function sol=temp_pde(t,R,X,source)
m=1; %Sets the geometry to cylindrical
global theta kappa h Q;
theta=X(1); kappa=X(2); h=X(3);
r=linspace(0,R,800);
t=source(:,1);
Q=source(:,2);
sol=pdepe(m,pdefun,icfun,bcfun,r,t);
end
function [c,f,s] = pdefun(r,t,u,dudx)
global Q theta kappa;
c = theta;
f = kappa*dudx;
s = Q;
end
function u0 = icfun(r)
u0 = 0;
end
function [pl,ql,pr,qr] = bcfun(xl,ul,xr,ur,t)
global h kappa;
pl = 0;
ql = 1;
pr = h*ur;
qr = kappa;
end
Where I obtain my t parameter from my data in source. Would this work?
  2 commentaires
Bjorn Gustavsson
Bjorn Gustavsson le 7 Mai 2020
Is a time-variable heating-rate that is the same over the entire region?
It is better to wrap the heating-rate into a function to make it independent of time, something like this:
Qfcn = @(t,r,u) ones(size(r))*interp1(source(:,1),source(:,2),'pchip');
Now will be a function of time that pdepe can evaluate at any time - it will use some time-steps (adaptive or fixed) that will not necessarily line up with the time-steps of your source.
(Check the documentation for how the right-hand-side function should be defined in terms of input variables and the like.)
HTH
Matthew Hunt
Matthew Hunt le 7 Mai 2020
\dot{Q} is purely a function of time, so is homogeneous.

Connectez-vous pour commenter.

Réponses (1)

Bjorn Gustavsson
Bjorn Gustavsson le 7 Mai 2020
To the best of my understanding you should be able to define your pde-function something like this:
function [c,f,s] = pde_interpS(x,t,u,dudx,tQ,Q,theta,kappa)
c = theta;
f = kappa*dudx;
s = interp1(tQ,Q,t,'pchip');
end
Then you can convert that function to a function of t, x, u and dudx with the standard @()-trick. My preference is to avoid globals as much as possible, but you can use that too.
HTH
  6 commentaires
Matthew Hunt
Matthew Hunt le 11 Mai 2020
I spotted my error, I should have "@" signs, now I have a new problem, I get the following error:
246 [c,f,s] = feval(pde,xi(1),t(1),U,Ux,varargin{:});
247 if any([size(c,1),size(f,1),size(s,1)]~=npde)
248 error(message('MATLAB:pdepe:UnexpectedOutputPDEFUN',sprintf('%d',npde)))
Error using pdepe (line 248)
Unexpected output of PDEFUN. For this problem PDEFUN must return three column vectors of length 1.
Bjorn Gustavsson
Bjorn Gustavsson le 11 Mai 2020
Seems like your pde-function that should return your c, f and s either doesn't return all three of those variables, or some of them are not set to column vectors - either one or more of them are empty or of the wrong size - perhaps one of them becomes a row-vector. If you set a debug-stop in the pde-function you can check the sizes of the outputs there.
HTH

Connectez-vous pour commenter.

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by