# Stability of Mathieu equation: $x''(t)+\cos t \,x(t)=0$

The equation $$x''(t)+\cos t \,x(t)=0 \quad (1)$$ can be transformed to the system: $$\vec{x}'= \begin{pmatrix} 0 & 1\\ -\cos t & 0 \end{pmatrix} \vec{x}=A(t) \cdot x(t)$$ with minimum period $$T=2\pi$$. Let $$\mu_1,\mu_2$$ be its characteristic values. A theorem gives: $$\mu_1\mu_2=\exp\Bigg\{\int_0^{2\pi} tr(A(t))dt\Bigg\}=1 \quad (2)$$ Therefore, the Wronskian of any two linearly independent solutions satisfies: $$W(t+2\pi)=W(t) \quad (3)$$

Does $$(3)$$ imply that all solutions are bounded and thus we have asymptotic stability? If not, in what way could we use $$(2)$$ to determine $$(1)$$'s stability?

• Determining the numerical solution of $\Phi'(t)=A(t)\Phi(t)$, $\Phi(0)=I_2$ gives $$\Phi(2\pi)=\pmatrix{-8.06518375& 14.50923829\\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace. – Lutz Lehmann Dec 1 '18 at 11:25

Hint :

If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $$\mu_j$$ for your given system, it is $$|\mu_j| <1$$, then the system is unstable. But, that's true, if you can prove that :

$$\mu_1\mu_2 = 1 \Rightarrow |\mu_1\mu_2| = 1 \Leftrightarrow |\mu_1| = \frac{1}{|\mu_2|} <1, \; \text{if} \; \mu_1, \mu_2 \neq 1$$

In order to conclude that, use the case of the characteristic matrix being periodic, thus $$\Phi(t+T) = \Phi(t) \Rightarrow \Phi(2 \pi) = \Phi(0)E \Rightarrow E= \Phi(0)^{-1}\Phi(2\pi)$$.

Now, the characteristic values will be the eigenvalues of the matrix $$E$$. You can calculate them (or approximate them) and conclude if $$\mu_1,\mu_2 \neq 1$$.

You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.

Graphs :

(A phase portrait for a certain $$t$$) For a simple case of time $$t$$ such that $$\cos t = -1$$, the system has the image of the phase portrait :

$$\qquad \qquad \qquad \quad$$

which is a saddle, thus unstable.

Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :

$$\qquad \qquad \qquad \qquad$$

• You cannot use the linearization to infer the stability of a time dependent system. – Artem Nov 30 '18 at 21:25
• Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1. – Artem Nov 30 '18 at 21:28
• @Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying). – Rebellos Nov 30 '18 at 21:45
• @Artem Any issues now for that -1 ? – Rebellos Nov 30 '18 at 21:54
• 1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect. – Artem Nov 30 '18 at 22:19