# [SOLVED]Mathieus

#### dwsmith

##### Well-known member
For Mathieu's equation
$$u'' + (\delta + \epsilon\cos 2t)u = 0,$$
use the perturbation method to analytically determine its stability near $\delta = 1$ and $\delta = 4$ when $\epsilon\ll 1$.

How would I do this using a perturbation method?

#### dwsmith

##### Well-known member
For Mathieu's equation
$$u'' + (\delta + \epsilon\cos 2t)u = 0,$$
use the perturbation method to analytically determine its stability near $\delta = 1$ and $\delta = 4$ when $\epsilon\ll 1$.

How would I do this using a perturbation method?
If we take $\delta = 1$, we have $u'' + (1 + \epsilon\cos 2t)u = 0$. I need to find a $u(t) = u_1(t) + u_2(t)$ such that it satisfies $u_1(t) = 1$, $u_1'(t) = 0$, $u_2(t) = 0$, $u_2'(t) = 1$ and $u$ satisfies $u'' + (1 + \epsilon\cos 2t)u = 0$. Since $\epsilon\ll 1$, can we make $\epsilon = 0$? Because if that was the case, then $u = A\cos t\sqrt{\delta} + B\sin t\sqrt{\delta}$.

#### dwsmith

##### Well-known member
Am I on the right track and what does this tell me about its stability?

When $\epsilon = 0$, we have $\ddot{u} + \delta u = 0$.
Given that for these two cases $\delta > 0$, $u(t) = A\cos t\sqrt{\delta} + B\sin t\sqrt{\delta}$ which satisfies $u_1(0) = 1$, $\dot{u}_1(0) = 0$, $u_2(0) = 0$, and $\dot{u}_2(0) = 1$ when $u_1 = A\cos t\sqrt{\delta}$ and $u_2 = B\sin t\sqrt{\delta}$.
The period has to be the same as Mathieu's equation of $\pi$.
So then we have
\begin{alignat*}{3}
\phi & = & \frac{1}{2}\left[u_1(T) + \dot{u}_2(T)\right]\\
& = & \frac{1}{2}\left[\cos \pi\sqrt{\delta} + \cos pi\sqrt{\delta}\right]\\
& = & \cos \pi\sqrt{\delta}
\end{alignat*}
For $\delta = 1$, we have $\phi = -1$.
Then $\left.\phi\right|_{\epsilon = 0} = -1$.
By Floquet, at -1, we need a period of $2T$.
$\epsilon\cos 2t$ admits a $2\pi$ periodic solution.
Taking the Taylor series expansion, we obtain
\begin{alignat}{5}
u(t,\epsilon) & = & u_0(t) + \epsilon u_1(t) + \epsilon^2\frac{u_2(t)}{2} + \mathcal{O}(\epsilon^3) & = & 0\notag\\
& = & (\ddot{u}_0 + \epsilon \ddot{u}_1) + [\delta + \epsilon\cos 2t](u_0 + \epsilon u_1 +\cdots) & = & 0\\
& = & (\ddot{u}_0 + \epsilon \ddot{u}_1) + \delta u_0 + \epsilon\delta u_1 + \epsilon\cos (2t) u_0 + \cdots & = & 0\notag\\
& = & (u_0 + \delta u_0) + \epsilon(\ddot{u}_1 + \delta u_1 + \cos (2t) u_0) + \cdots & = & 0\notag
\end{alignat}
We would need
\begin{alignat*}{3}
\ddot{u}_0 + \delta u_0 & = & 0\\
\ddot{u}_1 + \delta u_1 & = & -\cos (2t)u_0
\end{alignat*}
Now, let's expand $\delta$ by its Taylor series.
$$\delta(\epsilon) = \underbrace{\delta_0}_{= 1} + \epsilon\delta_1 + \mathcal{O}(\epsilon^2)$$
So $\delta$ can be expand in (1).
\begin{alignat*}{3}
(\ddot{u}_0 + \epsilon \ddot{u}_1 + \mathcal{\epsilon^2}) + [(\delta_0 + \epsilon\delta_1 + \mathcal{O}(\epsilon^2)) + \epsilon\cos 2t](u_0 + \epsilon u_1 +\mathcal{O}(\epsilon^2)] & = & 0\\
(\ddot{u}_0 + \delta_0 u_0) + \epsilon(\ddot{u}_1 + \delta_0 u_1 + \cos (2t)u_0 + \delta_1 u_0) + \cdots & = & 0
\end{alignat*}
Since $\delta_0 = 1$, we have
\begin{alignat*}{3}
\ddot{u}_0 + u_0 & = & 0\\
\ddot{u}_1 + u_1 & = & -\cos (2t)u_0 - \delta_1 u_0\\
\vdots & & \vdots
\end{alignat*}
$u_0 = c_1\cos t + c_2\sin t$ which is $2\pi$ periodic and this satisfies $\ddot{u}_0 + u_0 = 0$.
Then $\ddot{u}_1 + u_1 = -\cos (2t)[c_1\cos t + c_2\sin t] - \delta_1[c_1\cos t + c_2\sin t]$.
$$\ddot{u}_1 + u_1 = -\frac{1}{2}c_1[\cos t + \cos 3t] - \frac{1}{2}c_2[\sin 3t - \sin t] - \delta_1 c_1\cos t - \delta_1 c_2\sin t$$
After using $\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha + \beta) + \cos(\alpha - \beta)]$ and $\sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha + \beta) + \sin(\alpha - \beta)],$
we must suppress resonance.
So we have to focus on the $\sin t$ and $\cos t$ terms
$$\ddot{u}_1 + u_1 = \sin t\left[\frac{1}{1}c_2 - \delta_1 c_2\right] + \cos t\left[-\frac{1}{2}c_1 - \delta_1 c_2\right] + \text{the rest}$$
Then we must have
\begin{alignat*}{3}
c_2\left[\frac{1}{2} - \delta_1\right] & = & 0\\
c_1\left[\frac{1}{2} + \delta_1\right] & = & 0
\end{alignat*}
Therefore, $\delta_1 = \pm\frac{1}{2}$.
From the above, we have
$$\delta(\epsilon) = 1 + \frac{1}{2}\epsilon + \mathcal{O}(\epsilon^3)$$
and
$$\delta(\epsilon) = 1 - \frac{1}{2}\epsilon + \mathcal{O}(\epsilon^3).$$

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#### dwsmith

##### Well-known member
At the bottom, what do I need $\delta_2$ to be?

When $\delta = 4$, $\phi = 1$.
By Floquet, at 1, we need a period of $T$.
\begin{alignat*}{3}
\ddot{u}_0 + 4u_0 & = & 0\\
\ddot{u}_1 + 4u_1 & = & -\delta_1u_0 - u_0\cos 2t\\
\ddot{u}_2 + 4u_2 & = & -\delta_1u_1 - \delta_2u_0 - u_1\cos 2t
\end{alignat*}
Then $u_0 = a\cos 2t + b\sin 2t$.
Therefore, we have $\ddot{u}_1 + 4u_1 = -\delta_1a\cos 2t - \delta_1b\sin 2t - \frac{a}{2}\cos 4t - \frac{a}{2} - \frac{b}{2}\sin 4t - \frac{b}{2}$.
$$-\delta_1a = 0\quad\text{and}\quad -\delta_1b = 0\quad\Rightarrow\quad\delta_1 = 0$$
So $\ddot{u}_1 + 4u_1 = - \frac{a}{2}\cos 4t - \frac{a}{2} - \frac{b}{2}\sin 4t - \frac{b}{2}$.
Letting
$$u_1 = \sum_{j = 0}^{\infty}c_j\sin it + d_j\cos it.$$
Then $u_1 = \frac{a}{24}\cos 4t - \frac{a}{8} + \frac{b}{24}\sin 4t$.
We have that
\begin{alignat*}{3}
\ddot{u}_2 + 4u_2 & = & -\delta_1\left(\frac{b}{16}\sin3t + \frac{a}{16}\cos 3t\right) - \delta_2a\cos t - \delta_2b\sin t - \frac{b}{32}\sin t - \frac{b}{32}\sin 5t\\
& & - \frac{a}{32}\cos t - \frac{a}{32}\cos 5t\\
& = & - \delta_2a\cos t - \delta_2b\sin t - \frac{b}{32}\sin t - \frac{b}{32}\sin 5t - \frac{a}{32}\cos t - \frac{a}{32}\cos 5t
\end{alignat*}

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