ODE45 solver, with changing initial conditions

조회 수: 14 (최근 30일)
Louis De Jager
Louis De Jager 2020년 6월 12일
답변: 强 陈 2024년 4월 7일
I'm trying to numerically find the transition curves for a ODE, my code is supposed to do this by finding the solution to the ode, determining at which point the solution "blows up" and then storing the values for v and epsilon (epp) within an array.
However when running my code I keep on getting the following errors:
Unrecognized function or variable 'ODEvcnt'.
Error in ode2>@(t,y)dtheta(t,y,ODEvcnt,ODEeppcnt) (line 33)
sol = ode45(@(t,y) dtheta(t,y,ODEvcnt,ODEeppcnt),tspan,y0);
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in ode2 (line 33)
sol = ode45(@(t,y) dtheta(t,y,ODEvcnt,ODEeppcnt),tspan,y0);
I have attatched my code bellow:
Any help or advise would be much appreciated, thank you.
  댓글 수: 4
Ameer Hamza
Ameer Hamza 2020년 6월 13일
Where is phi and psi in this equation? What does this graph represent? The ODE is between tau and theta, so how do you get this graph between phi and epsilon.
Louis De Jager
Louis De Jager 2020년 6월 13일
편집: Louis De Jager 2020년 6월 13일
Phi and psi are two different v's with phi being for the region of v=1 and psi being fot the region of v =4 (the two cases v1 and v2)
The ode has various states: a stable state in which it produces a bounded oscillation and an unstable state during which the oscillations (of theta) grows exponentially.
Hence the graph shows the Arnold tongues for the system.
The program tries to identify the specific epsilon value for a given velocity at which the solution becomes unstable (blows up).

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채택된 답변

Star Strider
Star Strider 2020년 6월 13일
This runs without error. You need to determine if it produces the desired result:
syms y x epp t v
%cases
vM = 1; %Max's case
vMM =4; %Miu-Miu's case
%initial condition
initialY = pi/4;
%define expansions
v1 = 0.5:0.001:1.5; %max v
v2 = 3.5:0.001:4.5; % Miu Miu v
tspan = linspace(0,0.01,100); %time scale
epp = 0:0.0001:1; %epsilon possibilities
%storage variables
CritEpp = zeros(1000,2); % critical epsilon array
A = zeros(1,1000); % array to store applitudes
% sol=[t,y]; % 'sol' Is Not Defined At This Point
%definitiion od dtheta
for vcnt = 1:1:1000
for eppcnt = 1:1:10000
y0=[0,initialY,vcnt,eppcnt];
ODEvcnt = vcnt;
ODEeppcnt = eppcnt;
sol = ode45(@(t,y) dtheta(t,y,ODEvcnt,ODEeppcnt),tspan,y0);
A = max([sol.x.' sol.y.'],[],2);
end
if A(vcnt) > epp(eppcnt)
CritEpp(vcnt,:) = [v1(vcnt) epp(eppcnt)];
else
continue;
end
end
function dy = dtheta(t,y,ODEvcnt,ODEeppcnt)
vArr = 0.5:0.001:1.5;
eppArr = 0:0.0001:1;
dy = -(vArr(ODEvcnt))*y - (eppArr(ODEeppcnt))*cos(2*t);
end
These were not defined previously, anywhere in your code:
ODEvcnt = vcnt;
ODEeppcnt = eppcnt;
I took a guess as to what they should be, based on how you use them in your code. Make the appropriate corrections if I guessed wrong.
.
  댓글 수: 4
Louis De Jager
Louis De Jager 2020년 6월 13일
I see, thank you very much for the help.
Enjoy the rest of your day
Star Strider
Star Strider 2020년 6월 13일
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

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추가 답변 (1개)

强 陈
强 陈 2024년 4월 7일
Hello,I am also learning Arnold's tongue recently, can I study your ODEvcnt code?

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