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Consider the ODE $y^{\prime \prime}(x) = \cos(x) y(x)$ with boundary value conditions $y(0)=1$, $y(1)=2$. Solving it results in a linear combination of Mathieu functions, but what I find more interesting is its graph. So, the questions are:

  1. Is there really a phase transition around 300?

  2. What is the envelope of the graph? That is, are the maxima/minima growing exponentially, or does something else occur?

  3. Is the oscillating part actually periodic (which one would guess from the cos term)?

In general, what qualitative methods are there for answering these questions? I am sure the actual Mathieu function has been studied extensively, but just using its specific properties seems too "rigid".

enter image description here

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By basic Floquet theory, any solution is a linear combination of solutions of the form $u_j(x)e^{\alpha_j x}$, with $u_j$ $2\pi$ periodic and $e^{\alpha_1+\alpha_2}=1$. Doesn't that tell the whole story? – Christian Remling Sep 4 '14 at 18:13
@ChristianRemling: This implies the solution is oscillating with varying amplitude, not periodic. Characteristic values of the Mathieu equation tell when the solution is bona-fide periodic, and it's a tricky business. – Alex R. Sep 4 '14 at 18:21
@AlexR.: It does not imply the solution is oscillating, this follows precisely if you are in a band. However, it completely clarifies the general asymptotics (plane wave or real exponential, modulated by a periodic amplitude). – Christian Remling Sep 4 '14 at 18:24
@ChristianRemling: You're right. I guess I'm taking this question literally as whether the whole solution is periodic or not (clearly from the picture it's not). The solution is of the form: $$c_1 S(0,2,x/2)+c_2C(0,2,x/2)$$ where the two functions are the Mathieu functions. As far as I know, the "characteristic values of the Mathieu equation" determine for which coefficients $a,b$ in $S(a,b,x)$ and $C(a,b,x)$ you have bona-fide periodic solutions. – Alex R. Sep 4 '14 at 18:33
Actually, if you want real solutions in the 'exponential-periodic' form, you have to go to period $4\pi$. There are two independent solutions $y_1(x) = u_1(x)\mu^x$ and $y_2(x) = u_2(x) \mu^{-x}$, where $u_1$ and $u_2$ satisfy $u_i(x+2\pi) = -u_i(x)$ and where $\mu\approx 1.5557$. (Additionally, you can take $u_2(x)=u_1(-x)$.) No nonzero linear combination of these two solutions is actually periodic for any period. – Robert Bryant Sep 5 '14 at 10:19

I would guess that you are just seeing the effects of a dynamical system with a hyperbolic fixed point. Consider the matrix equation $$ A'(x) = \begin{pmatrix} 0 & 1\\ \cos(x) & 0\end{pmatrix} A(x) $$ with initial condition $A(0) = I_2$. Because of the $2\pi$-periodicity of $\cos(x)$, this fundamental solution clearly satisfies $$ A(x+2\pi) = A(x)A(2\pi) $$ where $$ A(2\pi)\approx \begin{pmatrix} -8.065 & -8.273\\ -7.742 & -8.065\end{pmatrix}, $$ a matrix that has eigenvalues $\lambda\approx -16.068$ and $1/\lambda\approx -0.062$. (Note that the ratio of the eigenvalues is about $\lambda^2 \approx 258$.) The action of $A(2\pi)$ on $\mathbb{R}^2$ is hyperbolic, contracting along the $1/\lambda$-eigenspace and expanding along the $\lambda$-eigenspace.

Now, the general solution of your equation satisfies
$$ \begin{pmatrix}y(x)\\y'(x)\end{pmatrix} = A(x)\begin{pmatrix}y(0)\\y'(0)\end{pmatrix}, $$ so $$ \begin{pmatrix}y(x+2k\pi)\\y'(x+2k\pi)\end{pmatrix} = A(x)A(2\pi)^k\begin{pmatrix}y(0)\\y'(0)\end{pmatrix}, $$ and if the initial condition is very near (but does not lie in) the $1/\lambda$-eigenspace of $A(2\pi)$, it will take a few cycles of $2\pi$ for the very small $\lambda$-eigenvector component to grow, but once it does, it will become exponentially dominant. I'm guessing that your boundary values just happen to have hit on such an initial condition.

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Robert, I am being dense, but how do you get the expression for $A(2\pi)?$ – Igor Rivin Sep 5 '14 at 0:47
@IgorRivin: Oh, sorry, I guess I should have mentioned that. I just did numerical integration of the matrix ODE, specifically 4th order Runge-Kutta. (Even Euler's method would have sufficed for this level of accuracy, but I had 4RK handy.) By the way, "Floquet Theory" is a fancy name for "holonomy of connections on $S^1$". – Robert Bryant Sep 5 '14 at 9:44
Robert, thanks, though as for nomenclature, I think some would say that "holonomy of connections on $S^1$" is a fancy name for "Floquet theory." – Igor Rivin Sep 5 '14 at 14:57
@IgorRivin: I agree that what is basic depends on one's point of view, but the content of Floquet's Theorem really is that connections on $S^1$ are classified up to gauge equivalence by their holonomy. In this latter form, not only is the proof trivial, the generalizations to higher dimensional bases and other groups besides $\mathrm{GL}(n,\mathbb{R})$ and the connections with discrete transformation groups, complex monodromy, period mappings, and a host of other phenomena are made more clear, at least in geometry. – Robert Bryant Sep 5 '14 at 16:03
Let $P\to S^1$ be a principal right $G$-bundle with connection $\theta:TP\to\frak{g}$. Pulling $P$ back via the covering map $\pi:\mathbb{R}\to S^1$ yields a (flat) bundle $\tilde P\to\mathbb{R}$ that thus has a global $\tilde\theta$-parallel trivialization $\tilde P\to \mathbb{R}\times G$; i.e., $\tilde\theta$ becomes the canonical left-invariant form on $G$. The deck transformation on $\mathbb{R}\times G$ is $\tau(t,g)=(t{+}2\pi, hg)$ for some $h\in G$ (the holonomy element, unique up to $G$-conjugacy). Conversely, any $h\in G$ defines a bundle with connection over $S^1$ as such a quotient. – Robert Bryant Sep 5 '14 at 16:59

All these questions are basically answered by Floquet theory. The envelope is an exponential, and there is nothing special going on at 300, what you are seeing is just exponential growth, and on the scale on which you are plotting everything less that $10^{49}$ just looks like zero.

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Right, this was what I also conjectured, but it is hard to tell (since the solution blows up very fast). – Igor Rivin Sep 4 '14 at 23:49
In any case, the magic words are "Floquet Theory", one lives and learns... – Igor Rivin Sep 5 '14 at 0:47

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