0
$\begingroup$

Consider the differential equation $x'=x+cos(t) $

a) find the general solution of this equation

b) prove there is a unique periodic solution for this equation

c) compute the Poincaré map $p:{t=0} \to {t=2\pi}$ for this equation and use this to verify again that there is a unique periodic solution.

I solved part a) and found $x(t)=\frac{1}{2}sin(t)-\frac{1}{2}cos(t)+ce^t$ but I'm unsure how to prove the solution is periodic or how to do the map.

$\endgroup$
2
$\begingroup$

b) Intuitively it is clear that the solution is periodic if and only if $c=0$, that is when $$x(t) = \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t).$$ However, one can still provide a somewhat more rigorous argument in support to this claim.

A periodic solution is a solution of the equation with the property that there exists a positive real number $T>0$ such that $x(t+T) = x(t)$ for all $t \in \mathbb{R}$. Let's write this condition explicitly: \begin{align} x(t+T) &= \frac{1}{2}\sin(t+T) - \frac{1}{2}\cos(t+T) + c\, e^{t+T}\\ &= \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t) + c\, e^{t} = x(t). \end{align} In other words, for any $t \in \mathbb{R}$ we apply some trigonometric formulas and group together the terms: \begin{align} 0 &= x(t+T) - x(t) \\ &= \frac{1}{2}\sin(t+T) - \frac{1}{2}\cos(t+T) + c\, e^{t+T} - \frac{1}{2}\sin(t) + \frac{1}{2}\cos(t) + c\, e^{t} \\ &= \frac{1}{2}\cos(T)\sin(t) + \frac{1}{2}\sin(T)\cos(t) \\ &- \frac{1}{2}\cos(T)\cos(t) + \frac{1}{2}\sin(T)\sin(t) \\ &+ c\, e^T\, e^{t} - \frac{1}{2}\sin(t) + \frac{1}{2}\cos(t) + c e^{t} \\ &= \left(\frac{1}{2}\cos(T) + \frac{1}{2}\sin(T) - \frac{1}{2}\right)\sin(t) + \left(\frac{1}{2}\sin(T) - \frac{1}{2}\cos(T) + \frac{1}{2}\right)\sin(t) + \left(c\,e^T - c\right)\, e^t \end{align} The the three functions $\,\,\sin(t), \,\,\cos(t)$ and $e^t$ are lineraly independent which means that \begin{align} \left(\frac{1}{2}\cos(T) + \frac{1}{2}\sin(T) - \frac{1}{2}\right)\sin(t) + \left(\frac{1}{2}\sin(T) - \frac{1}{2}\cos(T) + \frac{1}{2}\right)\sin(t) + \left(c\,e^T - c\right)\, e^t = 0 \end{align} for all $t \in \mathbb{R}$ the coefficients in front of each of them are zero, i.e. \begin{align} \frac{1}{2}\cos(T) + \frac{1}{2}\sin(T) - \frac{1}{2}&=0\\ \frac{1}{2}\sin(T) - \frac{1}{2}\cos(T) + \frac{1}{2} &= 0\\ c\,e^T - c& = 0 \end{align} which is the same as \begin{align} \cos(T) + \sin(T) &= 1\\ \sin(T) - \cos(T) &= - 1\\ c\,e^T &= c \end{align} To simplify, add the first two equations and then subtract them to obtain the equivalent system \begin{align} \sin(T) &= 0\\ \cos(T) &= 1\\ c\,e^T &= c \end{align} If $c=0$, then the system has solutions in the form of $T_k = 2\pi k$ for $k \in \mathbb{Z}$. The period $T$ should be the smallest positive number, so $T = 2\pi$. If $c \neq 0$ then the third equation becomes $e^T=1$ which holds if and only if $T=0$. In this case, $T$ is not a period as it is not positive. Therefore the only possible case is $c=0$ which means $$x(t) = \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t).$$

c) Let $x \in \mathbb{R}$. The Poincare map $P : \mathbb{R} \to \mathbb{R}$ is defined as follows: take a solution $x(t)$ of your equation such that $x(0)=x$. Then $P(x) = x(2\pi)$, which is the period of the equation.

Now, $$x = x(0) = \frac{1}{2}\sin(0) - \frac{1}{2}\cos(0) + c\, e^{0}= -\frac{1}{2} + c$$, so $c = x + \frac{1}{2}$. Therefore the solution that satisfies $x(0)=x$ is $$ x(t) = \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t) + \left(\frac{1}{2} + x\right)\, e^{t}.$$ Therefore $$P(x) = x(2\pi) = \frac{1}{2}\sin(2\pi) - \frac{1}{2}\cos(2\pi) + \left(\frac{1}{2} + x\right)\, e^{2\pi}$$ which yields $$P(x) = e^{2\pi}\, x + \frac{1}{2}e^{2\pi} - \frac{1}{2}.$$ This is a linear function in $x$ so a periodic solution of period $2\pi$ corresponds to a fixed point of $P$, i.t. $P(x) = x$. Solve the latter equation and you get that $x_0 = -\frac{1}{2}$ is the only fixed point.

Therefore the solutions $x(t)$ with the property that $x(0) = -\frac{1}{2}$ is the only periodic solution with that period which is exactly $$x(t) = \frac{1}{2}\sin(t) - \frac{1}{2}\cos(t).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.