Multiple-scale Anlysis. 2

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\operatorname{sgn}}$ $\newcommand{\rank}{\operatorname{rank}}$

### Derivative Expansion Method

The derivative expansion method is probably the most common of the various multiple scale methods. One introduces several time (or length) scales and treats them as independent variables: If $t$ is the original variable and $\varepsilon$ is the small parameter, we introduce the auxiliary time scales $$\tau_1=\varepsilon t,\ \tau_2=\varepsilon ^2t,\ldots,\ \tau_N=\varepsilon^N t \label{eq-8.2.1}$$ and consider $u(t,\varepsilon)=w(t,\tau_1,\ldots,\tau_N)$. Then $$\frac{d \ }{dt}u= \Bigl(\frac{\partial\ }{\partial t}+ \varepsilon \frac{\partial\ }{\partial \tau_1} + \varepsilon^2 \frac{\partial\ }{\partial \tau_2}+\ldots + \varepsilon^N \frac{\partial\ }{\partial \tau_N}\Bigr)w. \label{eq-8.2.2}$$ Let us apply this with $N=2$ to problem (8.1.1)--(8.1.2) \begin{align} &u''+ 2\varepsilon u'+ u=0,\qquad t>0 \label{eq-8.2.3}\\ &u(0)=1,\qquad u'(0)=0. \label{eq-8.2.4} \end{align} Then \begin{align} u' &=\frac{\partial w}{\partial t}+ \varepsilon \frac{\partial w}{\partial \tau_1} + \varepsilon^2 \frac{\partial w}{\partial \tau_2}, \label{eq-8.2.5}\\ u''&=\Bigr(\frac{\partial^2 w}{\partial t^2}+ \varepsilon \frac{\partial w}{\partial \tau_1} + \varepsilon^2 \frac{\partial w}{\partial \tau_2}\Bigr)^2w \label{eq-8.2.6}\\ &=\frac{\partial^2 w }{\partial t^2}+ 2\varepsilon \frac{\partial^2 w }{\partial t\partial \tau_1} + \varepsilon^2 \Bigl(2\frac{\partial^2w }{\partial t\partial \tau_2} + \frac{\partial^2 w }{\partial \tau_1^2}\Bigr) +O(\varepsilon^3). \notag \end{align} and we arrive to \begin{align} &\frac{\partial^2 w}{\partial t^2} + 2\varepsilon \Bigl(\frac{\partial^2 w }{\partial t\partial \tau_1}+ \frac{\partial w}{\partial t}\Bigr)+ \varepsilon^2 \Bigl(2\frac{\partial^w }{\partial t\partial \tau_2} + \frac{\partial^2 w }{\partial \tau_1^2}+ 2\frac{\partial w}{\partial \tau_1} \Bigr) + w =O(\varepsilon^3), \label{eq-8.2.7}\\ &w(0,0,0,\varepsilon)=1, \label{eq-8.2.8}\\ &\frac{\partial w}{\partial t}(0,0,0,\varepsilon)+ \varepsilon \frac{\partial w}{\partial \tau_1}(0,0,0,\varepsilon) + \varepsilon^2 \frac{\partial w}{\partial \tau_2}(0,0,0,\varepsilon)=0. \label{eq-8.2.9} \end{align} We look for a solution to this partial differential equation of the form $$w(t,\tau_1, \tau_2,\varepsilon) = w_0(t,\tau_1, \tau_2)+ \varepsilon w_1(t,\tau_1)+ \varepsilon^2 w_2(t) +O(\varepsilon^3). \label{eq-8.2.10}$$ Equalizing to $0$ coefficients at powers of $\varepsilon$ we find \begin{align} &\varepsilon^0:\ \left\{\begin{aligned} &\frac{\partial ^2w_0}{\partial t^2}+w_0=0,\\ &w_0(0,0,0)=1,\\ &\frac{\partial w_0}{\partial t}(0,0,0)=0; \end{aligned}\right. \label{eq-8.2.11}\\[4pt] &\varepsilon^1:\ \left\{\begin{aligned} &\frac{\partial ^2w_1}{\partial t^2}+w_1= -2\frac{\partial ^2w_0}{\partial t\partial\tau_1} - 2\frac{\partial w_0}{\partial t},\\ &w_1(0,0)=0,\\ &\frac{\partial w_1}{\partial t}(0,0)= - \frac{\partial w_0}{\partial \tau_1}(0,0,0); \end{aligned}\right. \label{eq-8.2.12}\\[4pt] &\varepsilon^2:\ \left\{\begin{aligned} &\frac{\partial ^2w_2}{\partial t^2}+w_2= -2\frac{\partial ^2w_0}{\partial t\partial\tau_2} - \frac{\partial^2 w_0}{\partial \tau_1^2}- 2\frac{\partial w_0}{\partial \tau_1 } - 2\frac{\partial ^2w_1}{\partial t\partial\tau_1}- 2\frac{\partial w_1}{\partial t},\\ &w_2(0)=0,\\ &\frac{\partial w_2}{\partial t}(0)= - \frac{\partial w_0}{\partial \tau_2}(0,0,0)- \frac{\partial w_1}{\partial \tau_1}(0,0). \end{aligned}\right. \label{eq-8.2.13} \end{align} The solution to problem (\ref{eq-8.2.11}) is \begin{align} &w_0(t,\tau_1, \tau_2) = A(\tau_1, \tau_2)\cos(t)+ B(\tau_1, \tau_2)\sin(t),\label{eq-8.2.14}\\ &A(0,0)=1,\quad B(0,0)=0.\label{eq-8.2.15} \end{align} Then equation in (\ref{eq-8.2.12}) is $$\frac{\partial ^2w_1}{\partial t^2}+w_1= 2\Bigl(\frac{\partial A}{\partial\tau_1}+A\Bigr)\sin(t)- 2\Bigl(\frac{\partial B}{\partial\tau_1}+B\Bigr)\cos(t) \label{eq-8.2.16}$$ and to avoid secularities we equalize coefficients here to $0$: $$\frac{\partial A}{\partial\tau_1}+A=0,\qquad \frac{\partial B}{\partial\tau_1}+B=0 \label{eq-8.2.17}$$ and with initial conditions (\ref{eq-8.2.15}) we conclude \begin{align} &A= \alpha(\tau_2) e^{-\tau_1}, && \alpha(0)=1, \label{eq-8.2.18}\\ &B= \beta(\tau_2) e^{-\tau_1}, &&\beta(0)=0. \label{eq-8.2.19} \end{align}

Now secular terms in (\ref{eq-8.2.16}) vanish and we solve (\ref{eq-8.2.16}) (and use initial conditions) and also we have (\ref{eq-8.2.21}) \begin{align} &w_1(t,\tau_1)=C(\tau_1)\cos(t)+D(\tau_1)\sin(t),&& C(0)=0, D(0)=1 \label{eq-8.2.20}\\ &w_0(t,\tau_1,\tau_2)= e^{-\tau_1}\bigl(\alpha(\tau_2)\cos(t)+\beta(\tau_2)\sin(t)\bigr), &&\alpha(0)=1, \beta(0)=0. \label{eq-8.2.21} \end{align} Equation in (\ref{eq-8.2.13}) now becomes \begin{multline*} \frac{\partial ^2w_2}{\partial t^2}+w_2=\\ \bigl[(2\alpha'+\beta)e^{-\tau_1}+2(C'+C)\bigr]\sin(t)+ \bigl[(-2\beta'+\alpha)e^{-\tau_1}-2(D'+D)\bigr]\cos(t) \end{multline*} Again we should remove secular terms: \begin{align*} &\underbracket{(2\alpha'+\beta)}+\underbracket{2e^{\tau_1}(C'+C)},\\ &\underbracket{(-2\beta'+\alpha)}-\underbracket{2e^{\tau_1}(D'+D)}=0. \end{align*} Since $\alpha,\beta$ depend only on $\tau_2$ and $C,D$ only on $\tau_1$ we have separation of variables like in PDE and then each of the marked term must be a constant. For a sake of simplicity we take them $0$: \begin{align*} &2\alpha'+\beta=0,\\ &-2\beta'+\alpha=0,\\ &C'+C=0,\\ &D'+D=0 \end{align*} and using initial conditions $\alpha(0)=1$, $\beta(0)=0$ we see that $\alpha(\tau_2)=\cos(\tau_2/2)$, $\beta(\tau_2)=\sin(\tau_2/2)$ and using initial conditions $C(0)=0$, $D(0)=1$ we see that $C(\tau_1)=0$, $D(\tau_1)=0$. Thus we arrive to \begin{align*} w_0(t,\tau_1,\tau_2)=&e^{-\tau_1}\Bigl[ \cos\bigl(\frac{\tau_2}{2}\bigr)\cos(t)+ \sin\bigl(\frac{\tau_2}{2}\bigr)\sin(t)\Bigr]= e^{-\tau_1}\cos\bigl(t-\frac{\tau_2}{2}\bigr),\\ w_1(t,\tau_1)=&e^{-\tau_1}\sin(t) \end{align*} For $w_2$ we have $w''+w=0$, $w_2(0)=0$, $w_2'(0)=0$ in virtue of (\ref{eq-8.2.13}); then $w_2=0$. Finally $$u(t,\varepsilon)=e^{-\varepsilon t}\Bigl[ \cos\bigl(1-\frac{\varepsilon^2}{2}\bigr)t + \varepsilon \sin(t)\Bigr] \label{eq-8.2.22}$$ which provides an approximation as $\varepsilon^{3}t\ll 1$.