Using ODE45 to solve Rossler equations

Question

jacob Mitch 2019-11-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/490052-using-ode45-to-solve-rossler-equations

回答： Tanggo Tang 2021-3-29

The Rossler equations are defined as

x'(t)=-x(t)-y(t)
y'(t)=ax(t)+y(t)
z'(t)=b+z(t)(x(t)-c) 

and I am trying to use ode45 to solve them

I have so far created a ross function file

function dx = ross(t,x)
%ross: Computes the derivatives involved in solving the
%ross equations.
a=-0.1;
b=2;
c=5.7;
%Right hand sides
dx=zeros(3,1);
dx(1)=-1*x(1)-1*x(2);
dx(2)=a*x(1)+x(2);
dx(3)=b+(x(3)*(x(1)-c));

and I am trying the below code but it doesnt stop running and seems to be stuck in a loop

clear all
      x0=[-8 8 27];
      tspan=[0,20];
     [t,x]=ode45(@ross,tspan,x0)
     %or run
      [t,x]=ode45(@lorenz,tspan,x0)
      %BUT NOT BOTH IN THE SAME CODE (1 OR THE OTHER LORENZ OR ROSS)
     plot3(x(:,1),x(:,2),x(:,3),'b','linewidth',1.5)
     

however if I have instead of @ross but have @lorenz

[t,x]=ode45(@lorenz,tspan,x0)

and try

     function dx = lorenz(t,x)
     % Parameters
     sigma=10;
     beta=8/3;
     rho=28;
     % Differential Equations
     dx=zeros(3,1);
     dx(1)=sigma*(x(2)-x(1));
     dx(2)=rho*x(1)-x(2)-x(1)*x(3);
     dx(3)=x(1)*x(2)-beta*x(3);

it seems to work fine, my question is where am I going wrong with my ross equations

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darova 2019-11-8

Try

tspan = [0 2];

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Answer 1

Jeremy 2019-11-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/490052-using-ode45-to-solve-rossler-equations#answer_400581

You are correct ode45 appears to be stuck. But if you try

ode15s

it appears to work

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Jeremy 2019-11-8

One more thing - are you plotting the equation solutions vs themselves instead of vs. time intentionally?

figure, hold on
plot(t,x(:,1),t,x(:,2),t,x(:,3),'LineWidth',2)

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Answer 2

Tanggo Tang 2021-3-29

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/490052-using-ode45-to-solve-rossler-equations#answer_661251

Hi you have err in ross attraction

function  dx = ross(t,x)
a = 0.2;
b = 0.2;
c = 5.7 ;
dx=zeros(3,1);
dx(1)=-x(2)-x(3);   %here
dx(2)=x(1)+a*x(2);  %and here
dx(3)=b+(x(3)*(x(1)-c));
end