Clear Filters
Clear Filters

How to use pdeplot with appdesigner?

2 views (last 30 days)
JClarcq
JClarcq on 5 Apr 2018
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,

Answers (2)

Ruben Gavín Mulelro
Ruben Gavín Mulelro on 29 May 2021
Same problem.. Have you found any solution?

Angelo Hafner
Angelo Hafner on 14 Jul 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)

Community Treasure Hunt

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

Start Hunting!