I'm trying to generate a naive perturbation series solution (with all associated secular terms included) to the Rayleigh equation:

\begin{equation} \frac{d^2y}{dt^2} + y = \epsilon \bigg(\frac{dy}{dt} - \frac{1}{3}\bigg(\frac{dy}{dt}\bigg)^3\bigg). \end{equation}

If I plug in the perturbation series

\begin{equation} y = y_0 + \epsilon y_1 + \mathcal{O}(\epsilon^2), \end{equation}

and solve for the $y_n$ order by order, I get $y_0 = Ae^{it}$ + complex conjugate, and, courtesy of Mathematica,

\begin{equation} y_1 = \alpha e^{it} + \beta t e^{it} + \gamma e^{3it} + \text{complex conjugate}, \end{equation} where \begin{equation} \alpha = \frac{iA}{4}\big(1 - A\bar{A}) + C, \end{equation} where $C$ is the (complex) constant of integration specific to $y_1$. The other two constants are \begin{equation} \beta = \frac{A}{2}\big(1 - A\bar{A}\big), \end{equation} and \begin{equation} \gamma = -\frac{iA^3}{24}. \end{equation}

PROBLEM: I'm having difficulty matching boundary conditions. Say my initial conditions are $y(0) = y0$ and $dy/dt(0) = \text{v}0$. I believe that in order for my series solution to be valid for all $\epsilon$, I must have $y_0(0) = y0$, $dy_0/dt(0) = \text{v}0$, and $y_n(0) = 0$ and $dy_n/dt(0) = 0$ for all $n \ge 1$.

QUESTION 1: Is that true in general for perturbation series? i.e. that $y_n(0) = 0$ and $dy_n/dt(0) = 0$ for all $n \ge 1$?

QUESTION 2: When I go to match $y_1(0) = 0$, I seem to require that $\alpha + \gamma = 0$, so $C$ should be whatever makes $\alpha = iA^3/24$. But this seems to mean that I have used my first boundary condition ($y_1(0) = 0$) to determine both the real and imaginary parts of $C$. In particular, I have not used the condition $dy_1/dt(0) = 0$ to help set C, and as one might expect, the resulting solution \begin{equation} y_1 = \frac{iA^3}{24}e^{it} + \frac{A}{2}\big(1 - A\bar{A}\big)te^{it} - \frac{iA^3}{24}e^{3it} + \text{complex conjugate} \end{equation} yields $dy_1/dt(0) \ne 0$. Where's the catch? I have some reason to believe this is the right expression for $y_1$ since a published research article uses this form, but I don't understand how it's possible for $dy_1/dt(0)$ to not equal 0. Any insight would be greatly appreciated.

  • $\begingroup$ I realized that my mistake lay in taking $\alpha + \gamma$ = 0, when in fact the condition is $\alpha + \gamma + \bar{\alpha} + \bar{\gamma} = 0$. This provides leeway for terms of the form $B - \bar{B}$ in $\alpha$ that are automatically canceled out by the complex conjugate when evaluated at t=0, but don't cancel when you take the derivative (since the complex conjugate terms have an extra minus sign). So, as one would expect, you can match both boundary conditions after all. $\endgroup$ – Dan Dec 11 '13 at 16:27

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