Plot 4D function with respect to two variables and a parameter

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UserCJ on 30 Apr 2022
Commented: Torsten on 1 May 2022
There is a differential equation . I would like to change the value of parameter value of A from to and plot C against and Aas shown in the picture below.
I tried it in MATLAB as follows:
DY3 = {};
PARAMA = linspace(-0.067,0.067,20);%parameter A
for i=1:numel(PARAMA)
param.A = PARAMA(i);
otherparams;%m file that contains other parameters
Tend = 200;
Nt = 100;
RHS = @(t,x) func(t,x,param,1);
x0 = [0.001,0.004,0.006]; %Initial condition
t = linspace(0,Tend,Nt); %TSPAN
options = odeset('RelTol',1e-4,'AbsTol',1e-4);
sol = ode45(RHS, t, x0, options);
[y,yp] = deval(sol,t);
x1 = y(1,:);
x2 = y(2,:);
dy3 = yp(3,:);%evaluate derivative
DY3{i} = dy3;
M = cell2mat(DY3);
[X1,X2,PARAMA] = meshgrid(x1,x2,PARAMA);
[faces,verts] = isosurface(x1,x2,PARAMA,M,0);
Whe I run this code I get the following error message.
Error using isosurface (line 79)
V must be a 3D array.
Can someone please help me to resolve this? Thank you very much!

Accepted Answer

Torsten on 30 Apr 2022
I don't think that you really want to solve the differential equation.
To get volumetric data for DY3, you will have to do
  1. Specify ranges for x1, x2 and parama using the linspace command.
  2. Use [X1,X2,PARAMA] = meshgrid(x1,x2,parama) to get volumetric input data.
  3. Evaluate your function F as DY3 = F(PARAMA,B,C,X1,X2)
  4. Plot the isosurface to the value 1e-4, e.g.
Torsten on 1 May 2022
Isosurfaces can be drawn from 3d-data (i.e. data that are given on a volume).
dy3 = f(x1,x2,Parama) can be 3d data if x1, x2 and Parama are independently prescribed and f is then used to evaluate them to get dy3.
In an ODE solution, x1 and x2 don't span a surface (2d), but only give a curve (1d). This means that DY3 is not 3d as necessary to get isosurfaces, but only 2d - thus itself a surface.

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