The differential equation problem with variable solution by using ode45
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abhishek singh
am 31 Okt. 2019
Kommentiert: Walter Roberson
am 1 Nov. 2019
I have four coupled diffrential equation shown bellow :-
In which a ,b ,c ,d ,e,f are constant and x[t] we will get from the solution of this second order diffrential equation .how to write code for it in matlab.plz help
4 Kommentare
Akzeptierte Antwort
darova
am 31 Okt. 2019
Here is an idea
[t1,x1] = ode45(@F1,ts,x0); % solve second order
[t2,s] = ode45(@(t,s)F2(t,s,t1,x1(:,1)),ts,s0); % solve system
function ds = F2(t,s,t1,x1)
x = interp1(t1,x1,t); % extract x
ds(1,1) = a*x-b ...
ds(2,1) = c*x*s(1) ...
end
3 Kommentare
darova
am 31 Okt. 2019
Bearbeitet: darova
am 31 Okt. 2019
Here is how your code should look like
function main
%% your constants
[t1,x1] = ode45(@noscillator,[0 10],[0 1]);
[t2,s1] = ode45(@(t,s) xotss(t,s,t1,x1(:,1)), [0 10], [1 0 0 0]);
plot(t2,s1(:,1))
function xdot=noscillator(t,x)
xdot(1) = x(2);
xdot(2) = -(omega^2)*x(1)-3*(gamma/m)*x(1)^2 - 4*(beta/m)*x(1)^3 + (V/m)*cos(w *t);
xdot=xdot';
end
function dxdt = xotss(t,s,t1,x1)
x = interp1(t1,x1,t);
dxdt(1) = (a*x-b)*s(1) + c*x*s(2);
% ...
dxdt = -1i*dxdt'
end
end
Weitere Antworten (6)
abhishek singh
am 31 Okt. 2019
1 Kommentar
darova
am 31 Okt. 2019
It means
सदस्यता सूचकांकों को वास्तविक धनात्मक पूर्णांक या तार्किक होना चाहिए।
In your language. Any ideas what the problem it might be?
Walter Roberson
am 31 Okt. 2019
[t1,x1] = ode45(@noscillator,[0:100],[0 1]);
[t2,s1] = ode45(@(t,s) xotss(t,s,t1,x1(:,1)), [0:100], [1 0 0 0]);
for ti = 0:1:100
rho11(ti+1)=s1(ti+1,1).*s1(ti+1,1)'-s1(ti+1,3).*s1(ti+1,3)';
rho12(ti+1)=s1(ti+1,1).*s1(ti+1,2)'+s1(ti+1,3).*s1(ti+1,4)';
rho21(ti+1)=s1(ti+1,2).*s1(ti+1,1)'+s1(ti+1,4).*s1(ti+1,3)';
rho22(ti+1)=s1(ti+1,2).*s1(ti+1,2)'-s1(ti+1,4).*s1(ti+1,4)';
end
3 Kommentare
Walter Roberson
am 1 Nov. 2019
Note that s1(ti+1,1)' means the conjugate complex transpose of s1(ti+1,1) . It is, however, a scalar, so transpose does not make any change. The You are also expecting real-valued results, so the conjugate is probably not makeing any changes. I suspect you are doing the equivalent of squaring the value.
I worry that you might have that that s1(ti+1,1)' is the derivative of s1(ti+1,1) .
abhishek singh
am 1 Nov. 2019
1 Kommentar
Rik
am 1 Nov. 2019
Please do not post your comments as answer. Their order can change, which makes it confusing.
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