Analyze Stability of Mathieu's Equation w/ Perturbation Method

[Dustinsfl]

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?

 

[Dustinsfl]

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 $u = A\cos t\sqrt{\delta} + B\sin t\sqrt{\delta}$.

 

[Dustinsfl]

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
$$
\delta(\epsilon) = 1 - \frac{1}{2}\epsilon + \mathcal{O}(\epsilon^3).
$$

 

[Dustinsfl]

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*}

 

[blue_raver22]

Mathieu's equation is a well-known differential equation in mathematics and physics that arises in many areas of science, including mechanics, quantum mechanics, and astrophysics. It is a second-order ordinary differential equation of the form:

$$
u'' + (\delta + \epsilon\cos 2t)u = 0,
$$

where $\delta$ and $\epsilon$ are constants. The stability of this equation is of great interest, as it determines the behavior of solutions to the equation and has important implications for the physical system it represents.

To analyze the stability of Mathieu's equation using the perturbation method, we first need to define a small parameter $\epsilon$ that represents the perturbation in the equation. In this case, we can set $\epsilon\ll 1$ to represent a small perturbation.

Next, we need to expand the solution $u$ in a power series in terms of $\epsilon$:

$$
u = u_0 + \epsilon u_1 + \epsilon^2 u_2 + \cdots
$$

where $u_0$, $u_1$, $u_2$, etc. are functions of $t$. Substituting this expansion into the original equation and equating coefficients of $\epsilon$, we get a series of equations:

$$
\begin{align}
\mathcal{O}(\epsilon^0): \quad & u_0'' + \delta u_0 = 0 \\
\mathcal{O}(\epsilon^1): \quad & u_1'' + \delta u_1 = -\cos 2t u_0 \\
\mathcal{O}(\epsilon^2): \quad & u_2'' + \delta u_2 = -\cos 2t u_1 \\
& \vdots
\end{align}
$$

Solving the first equation, we get $u_0 = A\cos(\sqrt{\delta}t) + B\sin(\sqrt{\delta}t)$, where $A$ and $B$ are constants determined by initial conditions. This solution is valid for all values of $\delta$, including $\delta = 1$ and $\delta = 4$.

For the second equation, we can use the method of undetermined coefficients to find a particular solution $u_{1p}$ that satisfies $u_{1p}'' + \delta u_{1p}