# Showing the existence of an unbounded solution to a periodic system of ODE

We consider the system $$\begin{pmatrix} x \\ y\\ z\end{pmatrix}' = \begin{pmatrix} \cos^4(t) && 0 && -\sin(2t) \\ \sin(4t) && \sin(t) && -4 \\ -\sin(5t) && 0 && -\cos(t) \end{pmatrix} \begin{pmatrix} x \\ y\\ z\end{pmatrix}.$$ I'd like to show it has at least one unbounded solution. Since it's $2\pi$-periodic, I can use Floquet theory. I know what I have to show is that it has a characteristic exponent $\lambda$ with positive real part, since then $e^{\lambda t}p(t)$ will be a solution to the ODE where $p(t)$ is a non-vanishing $2\pi$-periodic function. This solution will go to infinity as $t\to \infty$, hence will be unbounded.

The main issue is that I'm not sure how to compute the monodromy matrix. The matrix in the ODE doesn't look too nice, so I don't think I could compute a fundamental matrix directly. Any suggestions? Also, are there any standard tricks on how to find a monodromy matrix if a fundamental matrix is too hard to compute?

Liouville's Formula; I think this problem can be solved using Liouville's Formula, like this:

For the sake of brevity in $$\LaTeX$$, let us agree to denote the coefficient matrix in this problem by $$A(t)$$, thus:

$$A(t) = \begin{bmatrix} \cos^4(t) && 0 && -\sin(2t) \\ \sin(4t) && \sin(t) && -4 \\ -\sin(5t) && 0 && -\cos(t) \end{bmatrix}; \tag{1}$$

then if we set

$$\mathbf r(t) = \begin{pmatrix} x(t)\\ y(t)\\ z(t) \end{pmatrix}, \tag{2}$$

the differential equation in question becomes

$$\dot {\mathbf r} (t) = A(t) \mathbf r (t); \tag{3}$$

we consider in the usual manner a fundamental solution matrix $$X(t, t_0)$$ for (3); that is, $$X(t, t_0)$$ is a $$3 \times 3$$ matrix function of $$t$$ satisfying

$$\dot X(t, t_0) = A(t) X(t, t_0) \tag{4}$$

with

$$X(t_0, t_0) = I, \tag{5}$$

the $$3 \times 3$$ identity matrix. It will be noted that the columns of $$X(t, t_0)$$ are themselves solutions of (3), and that if

$$\mathbf r_0 = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}, \tag{6}$$

then $$X(t, t_0)\mathbf r_0$$ is the unique solution to (3) with

$$\mathbf r(t_0) = \mathbf r_0, \tag{7}$$

since

$$\dfrac{d}{dt} (X(t, t_0)\mathbf r_0) = \dot X(t, t_0) \mathbf r_0 = (A(t)X(t, t_0)) \mathbf r_0 = A(t)(X(t, t_0) \mathbf r_0); \tag{8}$$

it follows that the matrix $$X(t, t_0)$$ encodes all essential information about the solution space of (3).

We consider the matrix $$X(t + 2\pi, t_0)$$; we have

$$\dot X(t + 2\pi, t_0) = A(t + 2\pi)X(t + 2\pi, t_0) = A(t)X(t + 2\pi, t_0), \tag{9}$$

since $$A(t)$$ is periodic of period $$2\pi$$: $$A(t + 2\pi) = A(t)$$; thus $$X(t + 2\pi, t_0)$$ satisfies the same differential equation as does $$X(t, t_0)$$, but with initial condition $$X(t_0 + 2\pi, t_0)$$. Since

$$X(t_0 + 2\pi, t_0) = IX(t_0 + 2\pi) = X(t_0, t_0)X(t_0 + 2\pi, t_0), \tag{10}$$

it follows from the linearity of (3), and the uniqueness of solutions that for any $$t \in \Bbb R$$ we have

$$X(t + 2\pi, t_0) = X(t, t_0)X(t_0 + 2\pi, t_0) ; \tag{11}$$

we see from (11) that further

$$X(t + 4\pi, t_0) = X(t, t_0)X(t_0 + 2\pi, t_0)X(t + 2\pi, t_0) = X(t, t_0)X^2(t + 2\pi, t_0) , \tag{12}$$

and from here a very simple induction, the completion of which is left to the reader, establishes

$$X(t + 2n\pi, t_0) = X(t, t_0)X^n(t_0 + 2\pi, t_0) , \tag{13}$$

for any positive $$n \in \Bbb Z$$. (13) indicates that the long-term growth of the solution matrix $$X(t, t_0)$$ is intimately tied in with the expansive/contractive properties of the matrix $$X(t_0 + 2\pi, t_0)$$. In particular, if $$X(t_0 + 2\pi, t_0)$$ has an eigenvalue $$\lambda$$ with $$\vert \lambda \vert > 1$$, and corresponding eigenvector $$\mathbf v$$, that is

$$X(t_0 + 2\pi, t_0)\mathbf v = \lambda \mathbf v \tag{14}$$

we find

$$X(t + 2n\pi, t_0) \mathbf v = X(t, t_0) X^n(t_0 + 2\pi, t_0)\mathbf v = X(t, t_0)\lambda^n \mathbf v = \lambda^n X(t, t_0) \mathbf v, \tag{15}$$

whence, since $$\vert \lambda \vert > 1$$, the solution $$X(t, t_0)\mathbf v$$ grows without bound as $$t \to \infty$$. We further observe that $$X(t, t_0)$$ is nonsingular for all $$t$$, and $$[t_0, t_0 + 2\pi]$$ is compact, so $$\Vert X(t, t_0) \mathbf v \Vert$$ is bounded below away from $$0$$ on $$[t_0, t_0 + 2\pi]$$ by some real $$\mu > 0$$:

$$\Vert X(t, t_0) \mathbf v \Vert > \mu > 0, \; t_0 \le t \le t_0 + 2\pi; \tag{16}$$

thus we affirm that $$\Vert X(t + 2n\pi, t_0)\mathbf v \Vert > \vert \lambda^n \vert \mu$$, which shows that $$\Vert X(t + 2n\pi, t_0)\mathbf v \Vert \to \infty$$ as $$n \to \infty$$ independently of $$t_0$$ or $$t$$; $$X(t, t_0)$$ is nonsingular since $$X(t_0, t_0) = I$$, and the columns of $$I$$ are linearly independent, and linear independence or dependence of solutions is preserved under the flow of a first-order linear ordinary differential equation.

The preceding discussion indicates that computation of the eigenvalues of $$X(t_0 + 2\pi, t_0)$$ may be decisive in determining the stability or instability of solutions to (3); however, it is in general a difficult task to explicitly solve (4) for $$X(t, t_0)$$, and hence equally challenging to find its eigenvalues. But in some cases, such as the present one, progress may be made via a less than direct route; such is it situation here.

If it were possible to evaluate $$\det X(t_0 + 2\pi, t_0)$$ and to show that $$\det X(t_0 + 2\pi, t_0) > 1$$, then we could affim that $$\vert \lambda \vert > 1$$ for at least one eigenvalue of $$X(t_0 + 2 \pi, t_0)$$ and hence conclude that the system (3) is unstable. Fortunately, for the present problem this is the case, thanks to Liouville's Formula.

Liouville's Formula asserts that, given a system such as (4), $$\det X(t_0 + 2\pi, t_0)$$ evolves according to the scalar differential equation

$$\dfrac{d \det X(t, t_0)}{dt} = \operatorname{Tr}(A) \det X(t, t_0), \tag{17}$$

which has an immediate solution, given that $$X(t_0, t_0) = I$$,

$$\det X(t, t_0) = \exp(\displaystyle \int_{t_0}^t \operatorname{Tr}(A(s))ds) \det X(t_0, t_0) = \exp(\displaystyle \int_{t_0}^t \operatorname{Tr}(A(s))ds); \tag{18}$$

we have

$$\operatorname{Tr}(A(t)) = \cos^4 t + \sin t - \cos t, \tag{19}$$

whence

$$\displaystyle \int_{t_0}^{t_0 + 2\pi} \operatorname{Tr}(A(s))ds = \int_{t_0}^{t_0 + 2\pi}\cos^4 s ds + \int_{t_0}^{t_0 + 2\pi}\sin s ds - \int_{t_0}^{t_0 + 2\pi}\cos s ds; \tag{20}$$

the last two integrals on the right of (20) vanish and we are left with

$$\displaystyle \int_{t_0}^{t_0 + 2\pi}\operatorname{Tr}(A(s))ds = \int_{t_0}^{t_0 + 2\pi}\cos^4 s ds; \tag{21}$$

we now refer to this MSE post where it is shown that

$$\cos^4 t = \dfrac{3 + 4 \cos(2t) + \cos(4t)}{8}, \tag{22}$$

whence

$$\displaystyle \int_{t_0}^{t_0 + 2\pi} \operatorname{Tr}(A(s))ds = \int_{t_0}^{t_0 + 2\pi} \dfrac{3}{8} ds + \dfrac{1}{2}\int_{t_0}^{t_0 + 2\pi}\cos(2s)ds + \dfrac{1}{2}\int_{t_0}^{t_0 + 2\pi}\cos(4s)ds$$ $$= \displaystyle \int_{t_0}^{t_0 + 2\pi} \dfrac{3}{8} ds = \dfrac{3\pi}{4}; \tag{23}$$

then by (18) we find

$$\det X(t_0 + 2\pi, t_0) = e^{3\pi / 4} > 1; \tag{24}$$

it now follows from (24) that $$X(t_0 + 2\pi, t_0)$$ has an eigenvalue $$\lambda$$ with $$\lambda > 1$$; hence the system (3) is in fact unstable, i.e., has at least one unbounded solution.

• Thank you! I didn't think to utilize the determinant to get a characteristic multiplier with magnitude larger than 1, definitely remembering that one. – Curious Jul 27 '17 at 7:44
• @Curious: no problemo! – Robert Lewis Jul 27 '17 at 15:00