Multiple-scale Anlysis. 3

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### Two-variable expansion

Instead of introducing many slow variables one can introduce only one slow variable $\tau \varepsilon t$ and a modified fast variable $$T=(1+\varepsilon^2\nu_2+\varepsilon^3\nu_3+\ldots+\varepsilon^N\nu_N)t \label{eq-8.3.1}$$ with unknown constants $\nu_j$. Then $$\frac{d\ }{dt}= (1+\varepsilon^2\nu_2+\varepsilon^3\nu_3+\ldots+\varepsilon^N\nu_N) \frac{\partial\ }{\partial T}+ \varepsilon \frac{\partial\ }{\partial\tau}. \label{eq-8.3.2}$$ 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-8.3.3}\\ &u(0)=1,\qquad u'(0)=0. \label{eq-8.3.4} \end{align} Again we take $N=2$ and set $\nu=\nu_2$. Then \begin{align} u' &=(1+\varepsilon^2\nu) \frac{\partial w}{\partial T}+ \varepsilon \frac{\partial w}{\partial \tau} , \label{eq-8.3.5}\\ u''&= \Bigl((1+\varepsilon^2\nu)\frac{\partial\ }{\partial T}+ \varepsilon \frac{\partial\ }{\partial\tau}\Bigr)^2w \label{eq-8.3.6}\\ &= \frac{\partial^2w}{\partial T^2}+ 2\varepsilon \frac{\partial^2w}{\partial T\partial \tau }+ \varepsilon^2\Bigl(2\nu \frac{\partial^2w}{\partial T^2} + \frac{\partial^2w}{\partial \tau^2}+ \frac{\partial w}{\partial \tau}\Bigr)=O(\varepsilon^3) \notag \end{align} and we arrive to \begin{align} &\frac{\partial^2w}{\partial T^2}+w + 2\varepsilon \Bigl(\frac{\partial^2w}{\partial T\partial \tau }+ \frac{\partial w}{\partial T }\Bigr) + \varepsilon^2 \Bigl(2\nu \frac{\partial^2w }{\partial T^2} + \frac{\partial^2 w }{\partial \tau ^2}+ 2\frac{\partial w}{\partial \tau} \Bigr) =O(\varepsilon^3), \label{eq-8.3.7}\\ &w(0,0,\varepsilon)=1, \label{eq-8.3.8}\\ &\frac{\partial w}{\partial t}(0,0,\varepsilon)+ \varepsilon \frac{\partial w}{\partial \tau}(0,0,\varepsilon) + \varepsilon^2 \nu \frac{\partial w}{\partial T}(0,0,\varepsilon)=0. \label{eq-8.3.9} \end{align} We look for a solution to this partial differential equation of the form $$w(T, \tau,\varepsilon) = w_0(T, \tau)+ \varepsilon w_1(T,\tau)+ \varepsilon^2 w_2(T,\tau) +O(\varepsilon^3). \label{eq-8.3.10}$$ 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)=1,\\ &\frac{\partial w_0}{\partial T}(0,0)=0; \end{aligned}\right. \label{eq-8.3.11}\\[4pt] &\varepsilon^1:\ \left\{\begin{aligned} &\frac{\partial ^2w_1}{\partial T^2}+w_1= -2\frac{\partial ^2w_0}{\partial T\partial\tau} - 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}(0,0); \end{aligned}\right. \label{eq-8.3.12}\\[4pt] &\varepsilon^2:\ \left\{\begin{aligned} &\frac{\partial ^2w_2}{\partial T^2}+w_2= -2\nu\frac{\partial ^2w_0}{\partial T^2} - \frac{\partial^2 w_0}{\partial \tau ^2}- 2\frac{\partial w_0}{\partial \tau } - 2\frac{\partial ^2w_1}{\partial T\partial\tau}- 2\frac{\partial w_1}{\partial T},\\ &w_2(0)=0,\\ &\frac{\partial w_2}{\partial t}(0)= - \frac{\partial w_0}{\partial \tau}(0,0)- \nu \frac{\partial w_1}{\partial T}(0,0). \end{aligned}\right. \label{eq-8.3.13} \end{align} The solution to problem (\ref{eq-8.3.11}) is \begin{align} &w_0(Т,\tau) = A(\tau)\cos(Т)+ B(\tau)\sin(Т),\label{eq-8.3.14}\\ &A(0)=1,\quad B(0)=0.\label{eq-8.3.15} \end{align} Then equation in (\ref{eq-8.3.12}) is $$\frac{\partial ^2w_1}{\partial Т^2}+w_1= 2(А'+A)\sin(Т)- 2(B'+B)\cos(T) \label{eq-8.3.16}$$ and to avoid secularities we equalize coefficients here to $0$: $$A'+A=0,\qquad B'+B=0 \label{eq-8.3.17}$$ and with initial conditions (\ref{eq-8.3.15}) we conclude $A= e^{-\tau}$, $B= 0$.

Now secular terms in (\ref{eq-8.3.16}) vanish and we solve (\ref{eq-8.3.16}) (and use initial conditions) and also we have (\ref{eq-8.3.19}) \begin{align} &w_1(t,\tau )=C(\tau )\cos(T)+D(\tau )\sin(T). \label{eq-8.3.18}\\ &w_0(T,\tau)= e^{-\tau}\cos(T). \label{eq-8.3.19} \end{align} Equation in (\ref{eq-8.3.13}) now becomes \begin{equation*} \frac{\partial ^2w_2}{\partial T^2}+w_2= \bigl[-2(D'+D) +(1+2\nu)e^{-\tau}]\cos(T)+ 2(C'+C)\sin(T) \end{equation*} Again we should remove secular terms: \begin{align*} &C'+C=0,&&C(0)=0,\\ &D'+D=\frac{1}{2}(1+2\nu)e^{-\tau}=0, && D(0)=1. \end{align*} Then $C(\tau)=0$, $D(\tau)=\bigl[ 1+ \frac{1}{2}(1+2\nu)\tau\bigr]e^{-\tau}$. Therefore $$w((T,\tau)\sim w_0 +\varepsilon w_1=e^{-\tau}\Bigl[\cos(T)+\varepsilon \bigl(1+ \frac{1}{2}(1+2\nu)\tau\bigr)\sin(T)\Bigr]. \label{eq-8.3.20}$$ We see, however, that this solution still contains a secular term. Fortunately we still have enough freedom to suppress this secularity: we need only choose $\nu=-\frac{1}{2}$, then $D(\tau)=e^{-\tau}$ and $$w(T,\tau)\sim w_0+\varepsilon w_1=e^{-\tau}\bigl(\cos(T)+\varepsilon\sin(T)\bigr). \label{eq-8.3.21}$$ Finally \begin{multline} u(t,\varepsilon)\sim w\Bigl(\bigl(1-\frac{\varepsilon^2}{2}\bigr)t,\varepsilon t\Bigr)\\ =e^{-\varepsilon t}\Bigl[ \cos \bigl(\bigl(1-\frac{\varepsilon^2}{2}\bigr)t\bigr) + \varepsilon \sin \bigl(\bigl(1-\frac{\varepsilon^2}{2}\bigr)t\bigr)\Bigr] . \label{eq-8.3.22} \end{multline}

Compare three solutions: exact (8.1.3), (8.2.22) and (\ref{eq-8.3.22}) \begin{align*} &e^{-\varepsilon t} \Bigl[ \cos\bigl(\sqrt{1-\varepsilon^2}t\bigr) + \frac{\varepsilon}{\sqrt{1-\varepsilon^2}} \sin\bigl(\sqrt{1-\varepsilon^2}t\bigr) \Bigr],\\ &e^{-\varepsilon t}\Bigl[ \cos\bigl(1-\frac{\varepsilon^2}{2}\bigr)t + \varepsilon \sin(t)\Bigr],\\ &e^{-\varepsilon t} \Bigl[ \cos \bigl(\bigl(1-\frac{\varepsilon^2}{2}\bigr)t\bigr) + \varepsilon \sin \bigl(\bigl(1-\frac{\varepsilon^2}{2}\bigr)t\bigr)\Bigr] \end{align*}