# Periodic solution to linear differential equations

Let A $$\in L(\mathbb R^2)$$, and $$u(t)$$ be a periodic solution to the following system, with period $$p \gt 0$$. $$\dot x = Ax$$

Show that all solutions will be periodic, with the same period $$p$$

ATTEMPT:

The general solution to the above equation can be expressed as $$x(t) = e^{At}k$$ , where $$k$$ is a constant vector in $$\mathbb R^2$$, and $$e^{At}$$ is the matrix exponential $$e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + ...$$

Now, we are given that $$u(t)$$ is a solution. Hence, for some $$k_0$$, we have $$u(t) = e^{At}k_0$$. From the periodicity property, we have $$u(t+p) = u(t) \forall t$$ $$e^{A(t+p)}k_0 = e^{At}k_0 \forall t$$ $$e^{Ap}k_0 = k_0$$ Hence, we have that for matrix $$e^{Ap}$$, 1 is an eigenvalue, and since $$p>0$$, this means that 0 is an eigenvalue of $$A$$. But for a periodic solution, the eigenvalues of $$A$$ must be complex, to get the cosine and sine terms. Where have I gone wrong? If this approach is wrong, how do I actually solve this?

Hence, we have that for matrix $$e^{Ap}$$, 1 is an eigenvalue, and since $$p>0$$, this means that 0 is an eigenvalue of $$A$$.
This is wrong. The counterexample is $$A=\left(\begin{array}{cc}0&2\pi\\-2\pi &0\end{array}\right),\qquad e^A=I.$$ Here $$A$$ has the eigenvalues $$\pm2\pi i$$.
Let $$u(t)$$ be a nontrivial periodic solution with period $$p$$ and prove that there exists some $$s$$, $$s\in (0,p)$$, such that the vectors $$u(0)$$ and $$u(s)$$ are linearly independent. Suppose the opposite is true, i.e. $$\forall s\in [0,p]\; \exists c(s)\in\mathbb R\, :\; u(s)= c(s)u(0).$$ It means that the periodic solution lays on the straight line determined by the points $$0$$ and $$u(0)$$, which is impossible due to the existence and uniqueness theorem.
Since $$u(t)$$ is a $$p$$-periodic solution, $$e^{Ap}u(0)=u(0)$$ and $$e^{Ap}u(s)=u(s)$$, hence $$e^{Ap}$$ has an eigenbasis $$(u(0),u(s))$$ and the eigenvalues are $$1,1$$. It implies $$e^{Ap}=I$$.
Indeed, it means that $$e^{Ap}$$ is diagonalizable: $$U^{-1}e^{Ap} U= I\;\Rightarrow\; e^{Ap} U=U I\;\Rightarrow\; e^{Ap}=U U^{-1}=I,$$ where $$U$$ is a block matrix of the column vectors $$(u(0),u(s))$$.