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Let $\frac{d^2}{dt^2}x=\begin{pmatrix}1 &1 \\ 0 &a\end{pmatrix}x$. For which $a \in {\mathbb{R}}$ there exist periodic solutions?


I think only for $a<0$ there can be periodic solutions because then we get non-real eigenvalues. So we have

$$\frac{d^2}{dt^2}x_2=ax_2. \tag{1}\label{eq1}$$

$P(\lambda)=\lambda^2-a=0$ so $\lambda_1=-\sqrt{-a}i$, $\lambda_2=\sqrt{-a}i$.

Each function $e^{-\sqrt{-a}ti}$, $e^{\sqrt{-a}ti}$ is a solution to (\ref{eq1}) so their linear combination is also a solution to (\ref{eq1}).

So we have $$x_2(t)=E_1\cos(\sqrt{-a}t)+E_2\sin(\sqrt{-a}t).$$

We put $x_2$ to $$\frac{d^2}{dt^2}x_1=x_1+x_2, \tag{2}\label{eq2}$$

and we have $\frac{d^2}{dt^2}x_1-x_1=E_1\cos(\sqrt{-a}t)+E_2\sin(\sqrt{-a}t)$.

We look for a solution in a form: $C\cos(\sqrt{-a}t)+D\sin(\sqrt{-a}t)$. By putting it to the (\ref{eq2}) we get that $C = E_1/(a-1)$, $D = E_2/(a-1)$.

And the solution of homogeneous equation of (\ref{eq2}) is $D_1e^t+D_2e^{-t}$.

So we have

$$x_1(t)=\frac{E_1}{a-1}\cos(\sqrt{-a}t)+\frac{E_2}{a-1}\sin(\sqrt{-a}t) + D_1e^t+D_2e^{-t}.$$

So the periodic solutions exist when $a<0$. and I think whenever $D_1$ or $D_2$ $=0$ there are such solutions.


Is it a good approach to this problem?

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  • $\begingroup$ I assumed $a<0$. It is wrong? $\endgroup$
    – frugo
    May 14, 2021 at 16:11

1 Answer 1

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I presume the differential equation should be $$\dfrac{d^2}{dt^2} x = \pmatrix{1 & 1\cr 0 & a\cr} x$$

In general for $$ \dfrac{d^2}{dt^2} x = A x$$ $x = e^{\lambda t} v$ is a solution if $v$ is an eigenvector of $A$ for eigenvalue $\lambda^2$. This is periodic if $\lambda^2 < 0$. The eigenvalues of your $A = \pmatrix{1 & 1\cr 0 & a\cr}$ are $1$ and $a$. So yes, you need $a < 0$.

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