Bounded non periodic solutions of ${\displaystyle {\dot {x}}=A(t)x}$

asymptoticsdynamical systemslinear algebraordinary differential equationsperiodic functions

I am interested in the behaviour of the solutions of the $2\pi$-periodic differential equation $${\displaystyle {\dot {x}}=A(t)x}$$ with

$$A(\text{t})\text{=}\left(
\begin{array}{cc}
\sin (t) & \sin ^2(t) \cos (t) \\
\sin ^2(t) \cos (t) & \cos (t) \\
\end{array}
\right);$$
I plotted for example the solution $x(t)=(x_1(t),x_2(t))$ with initial condition $x(0)=(x_1(0),x_2(0))=(0,1)$, where $x_1(t)$ is in blue and $x_2(t)$ is in red.

<span class=$x_1(t)$" />
<span class=$x_2(t)$" />

It seems we have boundedness of all solutions on $(-\infty,\infty)$ and these solutions do not seem to be $2\pi$ -periodic. According to the table in Encyclopedia of Mathematics , boundedness of solutions are linked to characteristic exponents being purely imaginary with simple elementary divisors of the indicator matrix. Is there a method to compute the exponents?

So my questions are

1) How to prove boundedness of all solutions on $(-\infty,\infty)$

2) How to prove that these solutions are not periodic.

Best Answer

Consider the fundamental matrix $X(t)$, the solution of $X'(t) = A(t) X(t)$ with $X(0) = \pmatrix{1 & 0\cr 0 & 1\cr}$. Then $X(2\pi) = B$ where $$\det(B) = \exp\left(\int_0^{2\pi} \text{tr}(A(t))\; dt\right) = \exp(0) = 1$$
Numerically (using Maple's numerical DE solver) $$ B = \pmatrix{0.736606947094663 & 0.310166738881922\cr -3.21321753950662 &0.00457171990219575\cr}$$ The characteristic polynomial of $B$ is $$\lambda^2 - \text{tr}(B) \lambda + \det(B)$$ where, as above, $\det(B) = 1$ while numerically $\text{tr}(B) \approx 0.741178666996859$. Since $\text{tr}(B)^2 - 4 \det(B) < 0$ the eigenvalues are complex conjugates with absolute value $1$, say $\exp(\pm c i)$. The fact that their absolute value is $1$ shows that the solutions are bounded.
The solutions are non-periodic if and only if $c$ is not a rational multiple of $\pi$. This is not possible to determine with certainty numerically, but if $c/\pi \approx 0.3791557842561098$ is rational it does not have a small denominator: the continued fraction starts $[0; 2, 1, 1, 1, 3, 7, 2, 2, 1, 1, 10, 3]$.

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