Question: Find the leading-order approximation for times of order $\epsilon^{-1}$ to $$\ddot{x} + x = y,~~\dot{y} = \epsilon(xy – 2y^2),~~x(0) = 1,~~\dot{x}(0) = 0,~~y(0)=1.$$
My approach: Let the fast time scale of oscillation be $\tau = t$ and the slow time scale of amplitude drift be $T = \epsilon t$. Further, $$x(t;\epsilon) = X(\tau, T; \epsilon) = X_0(\tau, T) + \epsilon X_1(\tau, T) + O(\epsilon^2),$$ $$y(t;\epsilon) = Y(\tau, T; \epsilon) = Y_0(\tau, T) + \epsilon Y_1(\tau, T) + O(\epsilon^2).$$
At $O(1)$, the set of PDEs is:
$$(X_0)_{\tau\tau} + X_0 = Y_0,~\frac{\partial Y_0}{\partial \tau} = 0,$$ subject to the initial conditions $X_0(0,0) = 1$, $\frac{\partial X_0}{\partial \tau}(0,0) = 0,~Y_0(0,0) = 1$. Solving the above system leads us to $$Y_0(\tau, T) = f(T),~X_0(\tau, T) = A(T)e^{i\tau} + A^*(T)e^{-i\tau} + f(T) = \left[Ae^{i\tau} + cc\right] + f(T)$$ with $f(0) = 1$ and $A(0) = 0$ (which is weird and leads to problems).
At $O(\epsilon)$, the set of PDEs is:
$$(Y_1)_{\tau} = -(Y_0)_T + X_0Y_0 – 2Y_0^2,$$ $$\frac{\partial^2 X_1}{\partial \tau^2} + X_1 = -2\frac{\partial^2 X_0}{\partial\tau \partial T} + Y_1,$$
subject to the initial condition $Y_1(0,0) = 0$ (note that we do not require the initial conditions on $X_1$ to solve the problem). Solving the first PDE leads us to $$Y_1(\tau, T) = -f_T\tau + fA\left[-ie^{i\tau} + cc\right] – f^2\tau + g(T),$$ where $g(0) = 0$. Therefore, $$\frac{\partial^2 X_1}{\partial \tau^2} + X_1 = \left[e^{i\tau}(-2iA_T-fAi) + cc\right] + \tau\left[-f_{T} -f^2\right] + (\text{non-secular terms}).$$
We need to set the first two terms on the right-hand side of the last PDE to $0$ to avoid secular behavior. Thus, we must have $$\frac{df}{dT} + f^2 = 0,$$ subject to $f(0) =1$, which implies $$f(T) = \frac{1}{T+1}.$$ Further, assuming $A(T) = R(T)e^{i\theta(T)}$, we must also have (amplitude equation):
$$A_T + \frac{1}{2}fA = R_T + iR\theta_T + \frac{1}{2}fR = 0.$$ Equating the real and imaginary parts, we arrive at the following ODEs:
$$\frac{dR}{dT}+\frac{1}{2}\frac{R}{T+1} = 0,~R\frac{d\theta}{dT} = 0,$$
subject to $R(0) = 0$ and $\theta(0)$ can be any angle (problem 1).
Solving the ODE on $R$ gives us $R(T) \equiv 0$ (problem 2). This implies, in this case, we would not be able to solve the ODE on $\theta$ (problem 3).
I might be wrong, but I feel that either $x(0) = 0$ or $y(0) = 0$; both cannot be equal to $1$. It would be great if someone could confirm. Thanks a lot in advance!
Best Answer
Trying out how this looks like numerically, integrate with $ϵ=0.05$ as example.
This looks like $x=y+ϵu$, $u(0)=0$, that is, in your expansion the $X_0$ component does not oscillate, the oscillation happens in $X_1$, so what you computed is actually correct, you just need to advance to the next step.
Alternatively transform first the system without series expansion. Eliminate $x$ using $y-x = -ϵu$ and $x-2y = -(y-ϵu)$, and in the derivative $$ ϵ v \overset{Df.} = \dot x =\underbrace{-ϵy(y-ϵu)}_{=\dot y} + ϵ\dot u \\ \implies \dot u = v + y(y-ϵu). $$ Now insert into the first equation $$ \dot v = \frac{1}{ϵ}\ddot x = \frac{1}{ϵ}(y-x)=-u ,~~v(0)=0 $$ and as already used, the second equation transforms to $$ \dot y = - ϵy(y-ϵu) $$ Plotting this system shows all components at the same scale
One could now switch to a systematic perturbation expansion, but let's go just one step further in the manual transformation. Set $w=v+y^2$, $w(0)=1$, then \begin{align} \dot u &= w - ϵyu\\ \dot w &= -u - 2ϵy^2(y-ϵu)\\ \dot y &= - ϵy(y-ϵu) \end{align} Thus with $z=u+iw$ $$ \dot z=-iz-ϵy(Re(z)+i2y(y-ϵRe(z))),\\ z(0)=i\implies z=ie^{-it}+O(ϵ) $$
At level 0 this properly decouples as $u=\sin t$, $w=\cos t$, $y=1$. For a two-time-scales approach you would next introduce the slow variable $T=ϵt$ and equip the solution with slowly changing constants $$ z_0(t,T)=A(T)e^{-it}+B(T)e^{it}, y_0(t,T)=C(T) \\ A(0)=i,~~B(0)=0,~~ C(0)=1. $$ On the next level, with $z=z_0+ϵz_1+...$, etc. \begin{alignat}{1} \dot z_1 &= -iz_1 - C(T)Re(A(T)e^{-it}+B(T)e^{it})-i2C(T)^3&-A'(T)e^{-it}-B'(T)e^{-it}\\ \dot y_1 &= - C(T)^2 -C'(T) \end{alignat}
The secular terms in the first equation are those that contain $e^{-it}$. To avoid resonances, set them to zero $$ 0=\frac{C(T)}2(A(T)+\overline{B(T)}+A'(T) $$ The remaining equation gives $$ \dot z_1+iz_1 = + \overline{A'(T)}e^{it} - 2iC(T)^3 - B'(T)e^{it} $$ It appears natural to demand $\overline{A'(T)} - B'(T) = 0$, giving $A(T)-\overline{B(T)}=i$ and using $C=-C'/C$ leading to $$ A(T)=\frac i2(1+C(T)), \\ B(T)=\frac i2(1-C(T)). $$ It remains to solve $$ \dot z_1+iz_1=-2iC^3, ~~ z_1(0)=0,\\ z_1 = -2C^3 + 2C_1(T) e^{-it}, ~~~C_1(0)=1. $$
So in reverse the steps are
giving the plot
This looks qualitatively correct, quantitatively only in the short term.