Plot dynamics of an ODE

8 vues (au cours des 30 derniers jours)
UserCJ
UserCJ le 21 Sep 2022
Commenté : Cris LaPierre le 22 Sep 2022
I have an ODE and I need to plot with respect to and varying p. Following is the sample plot required.
I managed to draw one plot for and following is the sample code.
[xi,x] = meshgrid(0:0.001:1,0:0.002:1);
p =1
f2 = @(x,a,b) a*b*(x)./(b-2+x));
f1 = @(x,a,b) a*b*(1-x)./(2*b-x);
rs = @(x1,x2,a,b,p) x1.*f1(x,a,b) - p.*(1-x1).*f2(x,a,b);
figure
contour(x1,2,rs(x1,x2,1,2,1),[0 0]);
Next I added a loop for p as follows:
np = 10;
phi = linspace(0,1,np);
for p = 1:np
f2 = @(x,a,b) a*b*(x)./(b-2+x));
f1 = @(x,a,b) a*b*(1-x)./(2*b-x);
rs = @(x1,x2,a,b,p) x1.*f1(x,a,b) - p.*(1-x1).*f2(x,a,b);
R = rs(x1,x2,1,2, p(p));
contour(x1,x2,R);
hold on
end
hold off
But, instead of the required ourput of 10 plots in one graph, I get many plots (>10) in my output. Can someone please help me with this?
  1 commentaire
Cris LaPierre
Cris LaPierre le 21 Sep 2022
Modifié(e) : Cris LaPierre le 21 Sep 2022
You have not copied your code correctly here, as both examples do not run.

Connectez-vous pour commenter.

Réponse acceptée

Cris LaPierre
Cris LaPierre le 21 Sep 2022
I'm assuming your goal is to plot a line for when the function rs is equal to zero for various values of p. You could do that using contour, but I don't think that is the best way. However, you could still use contour to get the (x,y) coordinates of the 0 contour line for each, and then plot those.
np = 10;
[x1,x2] = meshgrid(0:0.001:1,0:0.002:1);
for p = 1:np
f2 = @(x,a,b) a*b*(x)./(b-2+x);
f1 = @(x,a,b) a*b*(1-x)./(2*b-x);
rs = @(xi,x,a,b,p) xi.*f1(x,a,b) - p.*(1-xi).*f2(x,a,b);
R = rs(x1,x2,1,2, p);
figure(1)
M=contour(x1,x2,R,[0 0]);
figure(2)
plot(M(1,2:end),M(2,2:end),'DisplayName',"p="+p)
axis([0 1 0 1])
hold on
end
hold off
legend(Location="northwest")
  4 commentaires
Cris LaPierre
Cris LaPierre le 22 Sep 2022
I would keep the meshgrid approach. That will be faster than looping.
I'm not sure of a best way, but one way to do this without contour is to look for the min(abs(R)) in each row. This won't be the exact zero point, but will be close enough for the visualization.
np = 10;
X = 0.001:0.001:1;
Y = 0.002:0.002:1;
[x1,x2] = meshgrid(X,Y);
for p = 1:np
f2 = @(x,a,b) a*b*(x)./(b-2+x);
f1 = @(x,a,b) a*b*(1-x)./(2*b-x);
rs = @(xi,x,a,b,p) xi.*f1(x,a,b) - p.*(1-xi).*f2(x,a,b);
R = rs(x1,x2,1,2, p);
[xval,xind] = min(abs(R),[],2);
X_z0(:,p) = X(xind);
end
plot(X_z0,Y)
axis([0 1 0 1])
legend(Location="northwest")
Cris LaPierre
Cris LaPierre le 22 Sep 2022

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur 2-D and 3-D Plots dans Help Center et File Exchange

Produits


Version

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by