I have a time-dependent differential equation:
\dot{x}=-x+b+a*cos(t)
Can I find the fixed point of the system by discretizing the time??
If yes, can anyone tell me how should we do?
If not, why can't we do it?
If the parameter "a" is nonzero, then such system does not have a fixed point. The reason for that is simple:
dx/dt = 0 = const
if and only if
x = b + a*cos(t) = const
which is not possible - the right hand side of the last equation cannot be constantly zero for a nonzero "a".
Have a good day !
P. S. Such system, depending on parameters, may have a stable limit cycle.
The short answer to the question in the text is, No; discretizing time can’t produce equilibria. Discretization can't change the properties of the differential equation.
Regarding the particular equation, it can be solved in closed form, so everything can be calculated.
Indeed, since the exact solution is x(t)=Cexp(-t) +b+(a/2)(cos(t)+sin(t)) the asymptotic form isn't a fixed point; there"s an exponential sensitivity to the initial conditions in the past and an oscillatory form in the future.
Since x(0) = C+a/2 C=x(0)-a/2, the memory of the initial condition fades exponentially fast in the future and, at long times, t>>1, whatever the initial conditon, x(t)=b+(a/2)(cos(t)+sin(t)). What is interesting is that this expression isn't, just the integral of the terms of the RHS, that don't depend on x(t), however.
The choice x(0)=a/2 eliminates the sensitivity in the past.
So considering the generalization, x'(t)= -x + h(t), we find that x(t)=Ce-t+e-tintegral0t(euh(u)), which, once more, shows that the asymptotic behavior in the future is independent of the initial condition and is described by the second term.
There persists an exponential sensitivity to the initial condition in the past, however.
What's interesting now is to remark that, if we expand h(t) in a Fourier series, then it's possible to evaluate the integral in the RHS, since h''(u)=-h(u), if the series is absolutely convergent. If we assume that h(0)=0 (the generalization to h(0)≠0 is easy), then we find that the asymptotic form of the solution, in the future, is given by (1/2)(h(t)-h'(t)). If h(0)≠0, it suffices to add this constant.
In the past, if the source is bounded, the solution is dominated by the first term; in the future it's dominated by (1/2)(h(t)-h'(t)), which shows that the attractor is a limit cycle, only if h(t) is a periodic function.
The general way to study such systems is to write the RHS as -x+b+acos(y), introducing a new variable, y, whose equation is dy/ dt=1 and studying the autonomous system in the plane. It is then clear that the asymptotic form can't be a fixed point, but is determined by the source.
Stam Nicolis : Your answers are contradictory. First, you say that a fixed point exists. Then, you say that the system is equivalant to autononomous system with dy/dt = 1. However, it means that y cannot be constant, so no fixed point exists.
Moreover, your solution of the equation is wrong. General solution of the homogenous part x'=-x is indeed C1*e^(-t). However, particular solution of the non-homogenous part is in the form x(t) = C2*cos(t)+C3*sin(t) + C4. Substituting it into the equation yields:
-C2*sin(t) + C3*cos(t) = -C2*cos(t) - C3*sin(t) - C4 + b + a*cos(t)
which must hold for any time "t". Therefore:
(C3-C2)*sin(t) + (C3+C2-a)*cos(t) + C4 - b = 0
for any time "t", so:
C2 = C3 = a/2
C4 = b
Consequently, the particlular solution of non-homogenous equation is:
a/2*cos(t) + a/2*sin(t) + b
and the general solution of non-homogenous equation is the sum of the particular solution above and the general solution of the homogenous equation, i.e.:
x(t) = C1*e^(-t) + (a/2)*cos(t) + (a/2)*sin(t) + b.
Initial conditions: x(0) = C1 + a/2 + b, so C1 = x(0) - a/2 - b.
My answer can be easily verified numerically. Suppose x(0) = a = b = 1. Then C1 = 1 - 1/2 - 1 = -1/2 and the solution should be:
x(t) = -0.5e^(-t) + 0.5*cos(t) + 0.5*sin(t) + 1.
Plot of this function in the range 0
Stam Nicolis : Please read my answer carefully and review college level maths.
TL;DR. It's easy to focus on trivial errors and let the more substantial ones slip by.
Anyone who thinks they can solve their homework problems by copying from any forum, should learn that isn't how it's done.
What really matters is that, if a system of differential equations isn't autonomous, it's always possible to render it autonomous-and that's one way to understand, whether it can have a fixed point or not, without having to solve the equation. So what matters is that this is the answer to the question posed.
If the equation is non-linear, but non-autonomous, a solution in terms of known functions isn't possible. But the embedding into a system of one more degree of freedom is how it's possible to see that the equilibrium configuration isn't a fixed point.
@Stam Nicolis - brilliant elaboration on deep analysis of a problem, which obviously has no solution from the first glimpse of eye, as I indicated in my first comment, LOL.
You say tl;dr and this is where your problems come from: you post wrong answers before reading the correct one.
Dear Suresh,
I faced the same problem one year ago.
AC and DC equilibrium point analysis approaches can be found in the reference attached. For more, you can read the reference cited in the article also.
Regards
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