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

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)

[Amal Matrafi]

Note: i named Pi here by lambda value.

[Amal Matrafi]

@Bora Eryilmaz I hope this post finds you well,

Can you help me with this problem please.

Answer 1

[Bora Eryilmaz]

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.

[Amal Matrafi]

thank you so much for your response and help.