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How to use pdeplot with appdesigner?

3 visualizaciones (últimos 30 días)
JClarcq
JClarcq el 5 de Abr. de 2018
Respondida: Ruben Gavín Mulelro el 29 de Mayo de 2021
Hello,
I try to display a gradient contour from a thermal pde solution.
pdeplot(app.data.thermalModelT,'XYData',app.data.temperature(:,end),'Contour','on');
as expected it displays a new figure. Thus I tried
pdeplot(app.UIAxes, app.data.thermalModelT,'XYData',app.data.temperature(:,end),'Contour','on');
like I would do for normal plot. But that is not supported. Is there a solution?
Thanks,
  1 comentario
Mohammad reza Nejati
Mohammad reza Nejati el 4 de Mzo. de 2020
I have the same problem. could anyone help please ?

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Respuestas (2)

Ruben Gavín Mulelro
Ruben Gavín Mulelro el 29 de Mayo de 2021
Same problem.. Have you found any solution?

Angelo Hafner
Angelo Hafner el 14 de Jul. de 2019
Today I was working all day looking for an answer. It folows my code...
nt = 21;
r = linspace(r1,r2,nt);
th = linspace(0,2*pi,nt);
[R,TH] = meshgrid(r,th);
% polar to cartezian (may be not necessary in your case
X = R .* cos(TH);
Y = R .* sin(TH);
The important thing here is to do the meshgrid
querypoints = [X(:),Y(:)]';
uintrp = interpolateSolution(results,querypoints);
and here is reshape
uintrp = reshape(uintrp,size(X));
mesh(X,Y,uintrp)

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