I'm trying to surf a function f that takes 2 values: 1 if (x,y) is in A and 2 if (x,y) is in B. How can I do that? I attach also a drawing for x,y.

 Réponse acceptée

Torsten
Torsten le 21 Mai 2023
Define the function as
f = @(x,y) (y>=2*x) + 2*(y<2*x);
and make a surface plot.

17 commentaires

Alexandra Roxana
Alexandra Roxana le 21 Mai 2023
Modifié(e) : Alexandra Roxana le 21 Mai 2023
But the function doesn't depend on x and y. It is a constant function. And I want to go further with more triangles dividing the domain. And to make it have a different value within each triangle.
Torsten
Torsten le 21 Mai 2023
Modifié(e) : Torsten le 21 Mai 2023
The function has exactly the values you want:
f = @(x,y) (y>=2*x) + 2*(y<2*x);
f(0,1)
ans = 1
f(1,0)
ans = 2
But I admit: Defining a function with prescribed values on each triangle of a triangulation will soon become a hard job. And I really don't understand what it should be good for ...
Alexandra Roxana
Alexandra Roxana le 21 Mai 2023
I want the value to be 1 in every x and y, not just x=0 and y=1.
Alexandra Roxana
Alexandra Roxana le 21 Mai 2023
I've been thinking of this problem for weeks and can't seem to come to some results.
Alexandra Roxana
Alexandra Roxana le 21 Mai 2023
Modifié(e) : Alexandra Roxana le 21 Mai 2023
Long story short: I want to calculate the pressure on each triangle from the red domain. The left-hand side of the formula is the pressure on each triangle; the right-hand side is the sum of values of each node from the triangle. The thing is the pressure is constant on each triangle and it has to look like that in the surf. Before it looked like a continuous function which isn't right. And I wanted at first to begin with a simple example.
f = @(x,y) (y>=2*x) + 2*(y<2*x);
x = linspace(0,1,100);
y = linspace(0,2,100);
[X,Y] = ndgrid(x,y);
surf(X,Y,f(X,Y))
shading interp
colorbar
Alexandra Roxana
Alexandra Roxana le 22 Mai 2023
It looks really good but there's a slight problem: what can I modify so I can see only the two triangles in the figure? It must look like a jump from 1 to 2.
Star Strider
Star Strider le 22 Mai 2023
Use the view function.
Or:
f = @(x,y) (y>=2*x) + 2*(y<2*x);
x = linspace(0,1,100);
y = linspace(0,2,100);
[X,Y] = ndgrid(x,y);
contourf(X,Y,f(X,Y))
colorbar
Alexandra Roxana
Alexandra Roxana le 22 Mai 2023
No, it has to look like the first one without the middle part, in 3D also.
Alexandra Roxana
Alexandra Roxana le 22 Mai 2023
Probably the function has to be modified, I momentarily don't see how.
Alexandra Roxana
Alexandra Roxana le 22 Mai 2023
The marked part must not appear
You might want to play with this representation:
x1 = [0,0,1];
y1 = [0,2,2];
z1 = [1,1,1];
x2 = [0,1,1];
y2 = [0,0,2];
z2 = [2,2,2];
hold on
s1 = fill3(x1,y1,z1,'r');
s2 = fill3(x2,y2,z2,'b');
hold off
view(25,45)
zlim ([0 3])
rotate(s1,[0 0 1],180)
rotate(s2,[0 0 1],180)
Alexandra Roxana
Alexandra Roxana le 23 Mai 2023
Modifié(e) : Alexandra Roxana le 23 Mai 2023
Unfortunately that's not going to help me very much because I want to go further and do the same thing but with more triangles like in this case and in each triangle the function to have a different value:
Alexandra Roxana
Alexandra Roxana le 23 Mai 2023
Modifié(e) : Alexandra Roxana le 23 Mai 2023
Thank you for all your help!
Alexandra Roxana
Alexandra Roxana le 27 Mai 2023
The method from the link seems to work; however I'm a bit confused with the values of n1, n2 and n3. I don't know what values I should put: if I put [1,1,1] the first coordinates are different, the rest is the same as should be. So I'm guessing the choice of n1, n2 and n3 is wrong.

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