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 $p$


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 $(u(0),u(s))$.

  • $\begingroup$ Why does a matrix having both eigenvalues 1 imply it is an identity matrix? $\endgroup$ – Dhanvi Sreenivasan Oct 5 '18 at 6:18
  • $\begingroup$ @Dhanvi Sreenivasan I updated the answer. Matrix having both eigenvalues 1 does not imply that it is an identity matrix. But if it is also diagonalizable (has an eigenbasis), it does. $\endgroup$ – AVK Oct 5 '18 at 6:41

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