How to graph the basic reproduction number R0 against two parameters on a 3D plot with planes?

26 次查看(过去 30 天)
I need help with ploting the basic reproduction number, related to mathematical biology.
In fact, I tried drawing the basic reproduction number as figures in the paper, but I didn't get the same.
The paper containing the basic reproduction number (R0) and respective parameter values has been uploaded here, I hope you will help me in these drawing.
the code:
clear
clc
[X,Y] = meshgrid(0:0.01:0.5, 0:0.01:0.0381);
R0=6.2954e-04;
Z=R0*ones(size(X));
surf(X,Y,Z)
colormap('turbo')
shading interp
xlabel('\mu')
ylabel('\Pi')
zlabel('R_0')
hold on
%Pi = 0.0381;
lambda=X;
d=0.00174;
beta1=0.00174;
beta2=0.00174;
a1=0.104;
a2=0.104;
wm=0.896;
d1=0.747;
d2=0.747;
gma1=0.253;
gma2=0.253;
theta=0.9;
%mu=0.1177;
mu=Y;
R1=lambda.*beta1./mu.*(a1+mu+wm);
R2=lambda.*beta2./mu.*(a2+mu+wm);
R0=max(R1,R2);
surf(X,Y,R0)

采纳的回答

Bora Eryilmaz
Bora Eryilmaz 2023-11-27
The equations and plots in the article look inconsistent. For example, Eq. 3.21 seems to be missing a 1/mu term when they replace S0 = Pi/mu in the expressions.
[omega,mu] = meshgrid(0.1:0.01:0.5, 0.01:0.001:0.03);
beta1 = 0.00174;
beta2 = 0.00174;
a1 = 0.104;
a2 = 0.104;
Pi = 0.0381;
R1 = beta1*Pi./mu./(a1 + mu + omega); % Eqs. 3.21 and 3.22 seem to be missing the 1/mu factor.
R2 = beta2*Pi./mu./(a2 + mu + omega);
R0 = max(R1,R2);
s = surf(omega,mu,R0); % This looks consistent with the bottom right plot in Figure 2, except the vertical scale of R0.
xlabel('omega')
ylabel('mu');
zlabel('R0')
[Pi,mu] = meshgrid(0.1:0.01:0.5, 0.01:0.001:0.03);
R1 = beta1*Pi./mu./(a1 + mu + omega);
R2 = beta2*Pi./mu./(a2 + mu + omega);
R0 = max(R1,R2);
s = surf(Pi,mu,R0); % This looks consistent with the top right plot in Figure 2, except the vertical scale of R0.
xlabel('Pi')
ylabel('mu');
zlabel('R0')
Hope this helps.

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Graphics 的更多信息

产品


版本

R2023a

Community Treasure Hunt

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

Start Hunting!

Translated by