I want to plot Isocline i.e want plot (x,y) when system contains step function .

% Parameters
r =0.1; % Define the value of r
k = 50; % Define the value of k
a =0.01; % Define the value of a
e = 0.5; % Define the value of e
m = 0.05; % Define the value of m
F = 1; % Define the value of F
%s =0.1; % Define the value of s
w = 0.1; % Define the value of w
b = 0.001; % Define the value of b
M = 50; % Define the value of M
s =0.1; % Define the value of sx
% Initial conditions
x0 =2; % Define the initial value of x
y0 =1; % Define the initial value of y
initial_conditions = [x0; y0];
% Time span for simulation
tspan = [0, 100]; % Define the time span (e.g., [0, 10])
% Solve the system of differential equations
[t, y] = ode45(@(t, y) kkk(t, y, r, k, a, e, m, F, s, w, b, M, x), tspan, initial_conditions);
Unrecognized function or variable 'x'.

Error in solution>@(t,y)kkk(t,y,r,k,a,e,m,F,s,w,b,M,x) (line 20)
[t, y] = ode45(@(t, y) kkk(t, y, r, k, a, e, m, F, s, w, b, M, x), tspan, initial_conditions);

Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 107)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
% Extract the solution
x_sol = y(:, 1);
y_sol = y(:, 2);
function dydt = kkk(t,x,y, r, k, a, e, m, F, s, w, b, M)
x = y(1);
u = isocline_input(x, F, s, w, b, M);
dxdt = r * x * (1 - x / k) - a * x * y / (1 + q * u * M);
dydt = (e * a * x * y) / (1 + q * u * M) - m * y;
dydt = [dxdt; dydt];
end
function u = isocline_input(x, F,w, b, M, s)
pi0 = (w * F) / (s * (w + b * M));
pi1 = F * (w + b * M) / (s * w);
if x <= pi0
u = 0;
elseif x >= pi1
u = 1;
else
u = (s*x) / (s*x + F) + w * (s*x - F) / (b * (s*x + F) * M);
end
end
I want to plot (x,y) where is my fault please help

7 Comments

[t, y] = ode45(@(t, y) kkk(t, y, r, k, a, e, m, F, s, w, b, M, x), tspan, initial_conditions);
function dydt = kkk(t,x,y, r, k, a, e, m, F, s, w, b, M)
It is confusing that you are passing x as the last parameter in your call, but your function expects to receive x as the second parameter.
sorry x is not parameter here x is prey population and y is predator population
step function split by the value of x
There is few difference and code is not running and please help me
If you have equations that calculate the derivatives of the prey population and the derivatives of the predator population, then they must be combined into a single state vector instead of being separate variables
[t, y] = ode45(@(t, state) kkk(t, state, r, k, a, e, m, F, s, w, b, M), tspan, initial_conditions)
function dxydt = kkk(t, state, r, k, a, e, m, F, s, w, b, M)
x = state(1);
y = state(2);
....
dxydt = zeros(size(state));
dxydt(1) = ...
dxdyt(2) = ...
end
% Parameters
r =0.1; % Define the value of r
k = 50; % Define the value of k
a =0.01; % Define the value of a
e = 0.5; % Define the value of e
m = 0.05; % Define the value of m
F = 1; % Define the value of F
%s =0.1; % Define the value of s
w = 0.1; % Define the value of w
b = 0.001; % Define the value of b
M = 50; % Define the value of M
s =0.1; % Define the value of sx
% Initial conditions
x0 =2; % Define the initial value of x
y0 =1; % Define the initial value of y
initial_conditions = [x0; y0];
% Time span for simulation
tspan = [0, 100]; % Define the time span (e.g., [0, 10])
% Solve the system of differential equations
[t, y] = ode45(@(t, state) kkk(t,state,r, k, a, e, m, F, s, w, b, M), tspan, initial_conditions);
x = state(1); y = state(2);
% [t, y] = ode45(@(t, state) kkk(t, state, r, k, a, e, m, F, s, w, b, M),
% Extract the solution
x_sol = y(:, 1);
y_sol = y(:, 2);
function dydt = kkk(t,x,y, r, k, a, e, m, F, s, w, b, M)
x = y(1);
u = isocline_input(x, F, s, w, b, M);
dxdt = r * x * (1 - x / k) - a * x * y / (1 + q * u * M);
dydt = (e * a * x * y) / (1 + q * u * M) - m * y;
dydt = [dxdt; dydt];
end
function u = isocline_input(x, F,w, b, M, s)
pi0 = (w * F) / (s * (w + b * M));
pi1 = F * (w + b * M) / (s * w);
if x <= pi0
u = 0;
elseif x >= pi1
u = 1;
else
u = (s*x) / (s*x + F) + w * (s*x - F) / (b * (s*x + F) * M);
end
end
sir, please see and help . I have tired
please solve my probelm .I can not understand the error.

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Answers (1)

Your code needs q but you did not define any q, so I had to pick SOME value for q in order to debug.
You need to change the definition of q to something appropriate for your situation.
% Parameters
r =0.1; % Define the value of r
k = 50; % Define the value of k
a =0.01; % Define the value of a
e = 0.5; % Define the value of e
m = 0.05; % Define the value of m
F = 1; % Define the value of F
%s =0.1; % Define the value of s
w = 0.1; % Define the value of w
b = 0.001; % Define the value of b
M = 50; % Define the value of M
s =0.1; % Define the value of sx
%user's code does not define any q, but we need a value of it to debug
rng(655321)
q = rand() * 10 %define the value of q
q = 3.1402
% Initial conditions
x0 =2; % Define the initial value of x
y0 =1; % Define the initial value of y
initial_conditions = [x0; y0];
% Time span for simulation
tspan = [0, 100]; % Define the time span (e.g., [0, 10])
% Solve the system of differential equations
[t, y] = ode45(@(t, state) kkk(t, state, r, k, a, e, m, F, s, w, b, M, q), tspan, initial_conditions);
% Extract the solution
x_sol = y(:, 1);
y_sol = y(:, 2);
subplot(2,1,1); plot(t, x_sol); title('x');
subplot(2,1,2); plot(t, y_sol); title('y');
function dydt = kkk(t, state, r, k, a, e, m, F, s, w, b, M, q)
x = state(1);
y = state(2);
u = isocline_input(x, F, s, w, b, M);
dxdt = r * x * (1 - x / k) - a * x * y / (1 + q * u * M);
dydt = (e * a * x * y) / (1 + q * u * M) - m * y;
dydt = [dxdt; dydt];
end
function u = isocline_input(x, F, s, w, b, M)
pi0 = (w * F) / (s * (w + b * M));
pi1 = F * (w + b * M) / (s * w);
if x <= pi0
u = 0;
elseif x >= pi1
u = 1;
else
u = (s*x) / (s*x + F) + w * (s*x - F) / (b * (s*x + F) * M);
end
end

2 Comments

why are you maintion 'u' as isclinefun
Here u is an adaptive step function .Which is contained in our system. My paper says that
1.when q=0.02,F=1
x=0:50;y=0:20,The system has an equilibrium point.
2.when q=0.03,F=1.714
x=0:50;y=0:20,The system has three equilibrium points
3.when q=0.02,F=1.8
x=0:50;y=0:20,The system has an equilibrium point.
4.when q=0.02,F=1.37
x=0:50;y=0:20,The system has an equilibrium point
5.
when q=0.03,F=1.613
x=0:50;y=0:20,The system have two equilibrium points
Give me suggetions how i can draw this plot(x,y)
I actually find the equilibrium points for choice of q and alternative resource F
What exactly do you mean by "u is an Adaptive Step Function"? I tried searching the keywords, but they seem unrelated. A closer look shows that depends on the parameter that is defined through 3 static (non-dynamic) If-Else statements, making it looks like the scheduling approach.

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Asked:

mks
on 1 Aug 2023

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on 1 Aug 2023

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