Solving a system of Non Linear Differential Equations

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kdv0
kdv0 2024-4-4
编辑： Sam Chak 2024-4-4
Hello, I want to solve the following system of non-linear differential equations numerically. Please provide guidance.
  2 个评论
kdv0
kdv0 2024-4-4
编辑：Sam Chak 2024-4-4
Ran in:
Hey I did this. Is it correct? The graphs are straight lines...
x0 = 0;
y0 = 0;
z0 = 10;
Y0 = [x0 y0 z0];
tspan = [0 1000];
[t, Y] = ode45(@rate, tspan, Y0);
Warning: Failure at t=9.804721e-03. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.775558e-17) at time t.
x = Y(:,1);
y = Y(:,2);
z = Y(:,3);
plot(t, x)
function dYdt = rate(~,Y)
r=10;
a=2;
b=2;
g=1;
m=0.5;
c=1;
p=1;
q=4;
s=1;
x = Y(1);
y = Y(2);
z = Y(3);
dxdt = r*x*(1-x/z)-a*m*x*y/(1+g*m*x);
dydt = b*m*x*y/(1+g*m*x)-c*y;
dzdt = p*z^2+q*z+s;
dYdt = [dxdt;
dydt;
dzdt];
end

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回答（2 个）

Sam Chak
Sam Chak 2024-4-4
编辑：Sam Chak 2024-4-4
Ran in:
Hi @kdv0
Apart from the incorrect initial value for z, which should be , the rest of the information in the code is correct. The response for x behaves as expected since it starts from the equilibrium point.
Both time derivatives for x and y are zero at the beginning because the states x and y start from the equilibrium point.
Even though the time derivative for z is decoupled from the influence of the variations in x and y, the rate of change exhibits a quadratic positive behavior when z is greater than zero. This is attributed to the positive values of the coefficients p, q, and s.
In simpler terms, the rate of change increases rapidly as z moves away from zero in the positive direction. Consequently, the response of z becomes unstable, leading to an explosive behavior and integration failure at time .
p = 1;
q = 4;
s = 1;
z = linspace(-14, 10, 2001);
dz = p*z.^2 + q*z + s;
plot(z, dz), grid on, xlabel('z'), ylabel('dz'), title('dz/dt')
xline(0, '-'), yline(0, '-')
KSSV
KSSV 2024-4-4
编辑：KSSV 2024-4-4
Ran in:
This is the code..I am getting hight values...you may check and modify the code.
tspan = [0 1];
y0 = [0 0 100];
sol = ode45(@odefun,tspan,y0) ;
Warning: Failure at t=9.804721e-03. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.775558e-17) at time t.
plot(sol.x,sol.y)
function dydt = odefun(t,y)
r = 10;
a = 2 ;
b = 2 ;
g = 1 ;
m = 0.5 ;
c = 1 ;
p = 1 ;
s = 1 ;
q = 4 ;
dydt = zeros(3,1) ;
dydt(1) = r*y(1)*(1-y(1)/y(3))-a*m*y(1)*y(2)/(1+g*m*y(1)) ;
dydt(2) = b*m*y(1)*y(2)/(1+g*m*y(1))-c*y(2) ;
dydt(3) = p*y(3)^2+q*y(3)+s ;
end

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