Plotting a 3d figure for a function with 4 variables?

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M
M el 13 de Ag. de 2021
Comentada: DGM el 19 de Ag. de 2021
Hi, I have a function with 4 variables (x,y,z,w) which means d(x,y,z,w)=0, and I have to plot a 3D figure for just 3 variable (x,y,z) without specific values. could you please tell me which matlab function is useful?
here is my code:
function [] = critical()
[x,y,z] = meshgrid(-4:0.1:4,-4:0.1:4,-4:0.1:4);
w=1;
%parameter
a=25;K=0.5;K_p=0.4;K_b=0.3; V_s=0.8;K_s=0.2; gamma=6;K_bar=1.5e-4;
%Function
phi_1 = y.^3 ./ ( y.^3 + K ^3 );
phi_2 = w.^2 ./ ( K_p^2 + w.^2);
phi_3= K_p^2 ./ ( K_p^2 + w.^2);
teta = phi_1 .* phi_2 .* z;
aa = phi_3 .* ( 1 - phi_3).*x;
b = teta./(teta+K_b*(teta+aa));
Beta=(1-K_bar*(gamma*(x-y)./y).^2)./(1+(K_s./y).^2);
alpha=a*b*gamma.*(x-(1+(1/gamma).*y))/V_s;
d=alpha-Beta;
fv=isosurface(x,y,z,d,0)
fv = isosurface(d,0)
fvc = isosurface(d,'blue')
pp = patch(isosurface(x,y,z,d,-0))
isonormals(x,y,z,d,pp)
set(pp,'FaceColor','red','EdgeColor','none');
view(3);
axis tight
camlight
end
when I define w as a vector [w]=meshgrid(-4:0.1:4); it has this error:
Error using .*
Matrix dimensions must agree.
Error in criticalaki (line 11)
teta = phi_1 .* phi_2 .* z;
so I put w=1 which is not true and give me the figure below. but I want a soft figure like surface.
  15 comentarios
M
M el 19 de Ag. de 2021
Thanks Adam and DGM for your comments,
DGM, I copied your code and run it, it has an error that "out of memory"...
I did something else, solve "alpha-Beta=0" and found an equation that x is written based on y,z,w and use "surface" or "plot3" to plot it.
DGM
DGM el 19 de Ag. de 2021
If you're running out of memory, with that example, then you'd need to reduce the number of points to something less than 400.
If you managed to get it working with surf, then you probably don't need to bother with my example.

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