Multiple-scale Anlysis. 4

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Rayleigh Oscillator

For $a > 0$ consider the nonlinear problem \begin{align} &u''+u=\varepsilon (u'-\frac{1}{3}u'^3), \label{eq-8.4.1}\\ &u(0,\varepsilon)=0, \qquad u'(0,\varepsilon)=2a. \label{eq-8.4.2} \end{align} To remove secularities at leading-order, it is suffcient to introduce a single slow variable, in which case the two-variable expansion and the derivative expansion methods (with $N = 1$) are equivalent. So, $\tau\varepsilon t$, $$u(t,\varepsilon)=w(t,\tau,\varepsilon)=w_0(t,\tau)+\varepsilon (t,\tau)+O(\varepsilon^2). \label{eq-8.4.3}$$ Then \begin{align} &\varepsilon^0:\ \left\{\begin{aligned} &\frac{\partial ^2w_0}{\partial t^2}+w_0=0,\\ &w_0(0,0)=0,\\ &\frac{\partial w_0}{\partial t}(0,0)=2a; \end{aligned}\right. \label{eq-8.4.4}\\[4pt] &\varepsilon^1:\ \left\{\begin{aligned} &\frac{\partial ^2w_1}{\partial t^2}+w_1= -2\frac{\partial ^2w_0}{\partial t\partial\tau} - -2\frac{\partial w_0}{\partial t}- \frac{1}{3}\bigl(\frac{\partial w_0}{\partial t}\bigr)^3,\\ &w_1(0,0)=0,\\ &\frac{\partial w_1}{\partial T}(0,0)=0. \end{aligned}\right. \label{eq-8.4.5} \end{align} The solution to problem (\ref{eq-8.4.4}) could be rewritten as \begin{align} &w_0(t,\tau) = A(\tau)\sin\bigl(t+\theta(\tau)\bigr),\label{eq-8.4.6}\\ &A(0)=2a>0,\quad\theta(0)=0.\label{eq-8.4.7} \end{align} Then equation in (\ref{eq-8.4.5}) is \begin{multline} \frac{\partial ^2w_1}{\partial ΠΆ^2}+w_1=\\ \bigl(A-2A'-\frac{1}{4}A^3\Bigr) \cos (t+\theta) +2A\theta' \sin(t+\theta) -\frac{1}{12} A^3\cos (3(t+\theta)) \label{eq-8.4.8} \end{multline} where we have expressed the right-hand side directly in terms of Fourier harmonics using the relation $\cos^3(t)=\frac{1}{4}\cos(3t)+\frac{3}{4}\cos(t)$.

To avoid secularities we equalize coefficients here to $0$: \begin{align} &A-2A'-\frac{1}{4}A^3=0,\label{eq-8.4.9}\\ &2A\theta'=0.\label{eq-8.4.10} \end{align} We can integrate (\ref{eq-8.4.9}): $8A'=a(2-A)(2+A)$ and' \begin{multline*} \int d\tau= \int \frac{-8\,dA}{A(A-2)(A+2)}= \int \Bigl(\frac{2}{A}-\frac{1}{A-2}-\frac{1}{A+2}\Bigr)\,d\tau=\\ \ln \Bigl(\frac{A^2}{A^2-4}\Bigr)+\ln \alpha. \end{multline*} Then $(A^2-4)/A^2=\alpha e^{-\tau}$ with $\alpha=(a^2-1)/a^2$ because $A(0)=2a$. Therefore \begin{equation*} A(\tau)=\frac{2a}{\sqrt{a^2-(a^2-1)e^{-\tau}}}>0 \end{equation*} and $$u(t,\varepsilon)=\frac{2a\sin(t)}{\sqrt{a^2-(a^2-1)e^{-\varepsilon t)}}} \label{eq-8.4.11}$$ because $\theta(\tau)$ is constant, equal $\theta(0)=0$.

Since $A(\tau)\to 2$ as $\tau\to +\infty$ we see that the solution approaches a limit cycle as $t\to +\infty$. Indeed: \begin{align*} &u(t,\varepsilon)\sim\frac{2a\sin(t)}{\sqrt{a^2-(a^2-1)e^{-\varepsilon t)}}}\\ &u'(t,\varepsilon)\sim\frac{2a\cos(t)}{\sqrt{a^2-(a^2-1)e^{-\varepsilon t)}}} \end{align*}