Homogeneous solutions in the renormalization-group pertrubation method I am trying to get a handle on the renormalization-group approach to perturbation theory. I understand the overarching approach, but I'm stumbling on the mechanics of early steps when actually doing it. I'm following a simple example in the paper by Bhattacharjee and Ray, which is the underdamped oscillator:
$$
x''(t) + 2\epsilon x'(t) + x(t) = 0 
$$
with initial conditions $x(t_0) = A(t_0)$ and $x'(t_0) = 0$. The RG method starts with a naive perturbation series
$$
x(t) = x_0(t) + \epsilon x_1(t) + \cdots
$$
So that at $O(1)$, we have a simple oscillator, and at each subsequent order, we have forced oscillators
$$\begin{align}
& x_0'' + x_0 = 0 && x_0(t_0)=A(t_0) &x_0'(t_0)=0\\
& x_1'' + x_1 = -2x_0' && x_1(t_0)=0 &x_1'(t_0)=0\\
\end{align}$$
and so on. The $O(1)$ solution that satisfies the initial conditions is
$$
x_0(t) = A(t_0) \cos(t - t_0)
$$
which then produces the $O(\epsilon)$ equation
$$
x_1'' + x_1 = 2A(t_0) \sin(t-t_0)
$$
A particular solution is
$$
X_1(t) = -A(t_0) (t - t_0) \cos(t - t_0)
$$
Bhattacharjee and Ray (and similar papers I've read working the same example) then take this as the entire solution at $O(\epsilon)$ and write down our secular perturbation series as
$$
x(t) = A(t_0) \cos(t - t_0) [ 1 - \epsilon (t - t_0) + O(\epsilon^2)]
$$
I'm terribly confused by this, because there is more to $x_1$ than just the particular solution; we also have a homogeneous solution that ought to satisfy the $O(\epsilon)$ initial conditions, $x_1(t_0) = x_1'(t_0) = 0$. If in include this, I find that
$$
x_1(t) = A(t_0) \sin(t - t_0) - A(t_0) (t - t_0) \cos(t - t_0)
$$
and if we were to continue the procedure to second order, the presence or absence of this $\sin$ term makes a difference to what I find for $x_2$. So my question is, what is the justification for dropping the homogeneous solution, and fixating solely on the particular solutions? I understand that in the RG approach, we are playing all kinds of devious games with the initial condition; is there some argument that the homogeneous solutions beyond $O(1)$ are irrelevant?
 A: I think the presentation of the method by these authors sort of muddles the procedure, conflating two slightly different steps regarding 'initial conditions.' I think the authors are essentially taking a shortcut so that they can neglect one of the two solutions of the equation, which is leading to the confusion. I originally learned this method from this paper by Chen et al., which is worth reading if you haven't seen it before. (Reference 8 in Bhattacharjee and Ray). Chen et al. only apply the initial conditions to the final approximate solution---intermediate initial conditions are really just constants of integration.
First, there are the actual initial or boundary conditions that are to be applied to the true solution $x(t)$. In the referenced paper, the authors apply these to the $O(\epsilon^0)$ solution $x_0(t)$, however these initial conditions really ought to be applied to the full approximation, not just $x_0(t)$. In applying the initial conditions only to $x_0(t)$, the authors eliminate one of the two general solutions of the equation, which is perhaps a shortcut that simplifies some of their calculation, but in general does not seem to be the best way to go about it.
The initial conditions that are supplied to $x_0(t)$ are meant to be arbitrary initial conditions specified at some arbitrary time $t_0$. i.e., you are looking for the general solution. In this case that solution can be written
$$x_0(t) = a(t_0) \cos(t-t_0) + b(t_0) \sin(t-t_0) = A(t_0) \cos(t-t_0 + \Theta(t_0)).$$
You see here though that the two linearly independent solutions appear with two integration constants (amplitudes of the sine and cosine), but using a trig identity we can write this as a single cosine with the two integration constants, the amplitude $A(t_0)$ and the phase $\Theta(t_0)$.
Now, if you move on to $x_1(t)$, the homogeneous solutions are just going to be the same form as the solution for $x_0(t)$, and can just be combined with these terms, so they do not really add anything. Recall that we will allow $A(t_0)$ and $\Theta(t_0)$ (or $a(t_0)$ and $b(t_0)$) to depend on $\epsilon$ in this procedure to cancel out the secular divergences, hence why we can absorb the higher-order homogeneous solutions into the $O(\epsilon^0)$ solution.
Hence, at $O(\epsilon)$ the only solution that matters is the particular solution. If we use the more general solution $x_0(t) = A(t_0) \cos(t-t_0 + \Theta_0)$, we get
$$x_1(t) = -A(t_0)(t-t_0) \cos(t-t_0 + \Theta_0).$$
Thus, our solution to $O(\epsilon)$ is
$$x(t) \approx A(t_0) \cos(t-t_0 + \Theta(t_0)) - \epsilon A(t_0)(t-t_0) \cos(t-t_0 + \Theta_0).$$
Now, the usual way this method is presented (e.g., by Chen et al. cited above) is to introduce an arbitrary intermediate time $\tau$ by writing $t-t_0 = t-\tau + (\tau-t_0)$ and then write $A(t_0) = Z(t_0,\tau) \mathcal A(\tau)$ and $\Theta(t_0) = Z_2(t_0,\tau) + \mathcal \Theta(\tau)$ and expand $Z_1$ and $Z_2$ in a series in powers of $\epsilon$, choosing the coefficients to cancel the divergent terms.
A slightly easier version of this approach (see, e.g., this paper) that avoids introducing this expansion and picking the coefficients by hand is to simply use $t_0$ as the arbitrary time at the outset, since in my presentation I am using $t_0$ as an arbitrary initial time; I will later take $t=0$ to be the actual initial time. We can proceed by differentiating $x(t)$ with respect to $t_0$, demanding that $x(t)$ does not actually depend on this arbitrary time $t_0$:
$$\frac{\partial x(t)}{\partial t_0} = \frac{\partial A(t_0)}{\partial t_0} \cos(t-t_0 + \Theta(t_0))  - A(t_0) \sin(t-t_0 + \Theta(t_0))\left(-1 + \frac{\partial \Theta(t_0)}{\partial t_0}\right) - \epsilon \left\{ \frac{\partial A(t_0)}{\partial t_0} (t-t_0) \cos(t-t_0 + \Theta(t_0)) - A(t_0) \cos(t-t_0 + \Theta(t_0)) - A(t_0) \sin(t-t_0+\Theta(t_0))\left(-1 + \frac{\partial \Theta(t_0)}{\partial t_0}\right) \right\}.$$
Now, we set $t = t_0$ and demand that $\partial x(t)/\partial t_0\Big|_{t = t_0} = 0$. Before we set $t \rightarrow t_0$, though, note that we have several factors of $\sin(t-t_0+\Theta(t_0))$ and $\cos(t-t_0+\Theta(t_0))$ ---independent terms, so we can group the terms multiplied by each factor and demand they are separately zero (since the RG equation must hold for arbitrary $t$). After we set $t = t_0$, the secular terms will vanish, giving the system of differential equations
$$\frac{\partial A(t_0)}{\partial t_0} = - \epsilon A(t_0)$$
and
$$\frac{\partial \Theta(t_0)}{\partial t_0} = 1.$$
The solutions are $A(t_0) = A e^{-\epsilon t}$ and $\Theta(t_0) = t_0 + \phi$, for integration constants $A$ and $\phi$ that do not depend on $t_0$. We may insert these into our approximation for $x(t)$, and then finally set $t_0 = t$---it is an arbitrary time that cannot affect our solution by construction, so we may choose any value for it now, and $t$ is a convenient choice. We thus have our uniformly valid solution
$$x(t) = A e^{-\epsilon t} \cos(t + \phi).$$
Now, we can finally apply the actual initial conditions of the problem. If we want $x(0) = B$ and $x'(0) = 0$, then we get the system of equations $A \cos\phi = B$ and $\sin \phi = -\epsilon \cos\phi$. This gives $\phi = -\tan^{-1}(\epsilon)$ and $A = B\sqrt{1+\epsilon^2}$. Thus, our solution is
$$x(t) \approx B\sqrt{1+\epsilon^2} e^{-\epsilon t} \cos\left(t - \tan^{-1}(\epsilon)\right).$$
Because we only worked to $O(\epsilon)$, we should not necessarily trust the $\epsilon^2$ terms appearing instead the functions here---or, perhaps more accurately, there are $O(\epsilon^2)$ terms that are missing here. If we compare to the exact solution, which for these initial conditions is
$$x(t) = \frac{B}{\sqrt{1-\epsilon^2}} \cos\left(t\sqrt{1-\epsilon^2} - \tan^{-1}\left(\frac{\epsilon}{\sqrt{1-\epsilon^2}} \right) \right),$$
we see where the $\epsilon^2$ terms in our approximate solution come from: $B/\sqrt{1-\epsilon^2} \approx B\sqrt{1+\epsilon^2}$ (using $1/(1-\epsilon^2) \approx 1 + \epsilon^2$) and $\tan^{-1}(\epsilon/\sqrt{1-\epsilon^2}) \approx \tan^{-1}(\epsilon)$, but the solution misses a couple of other terms that depend on $\epsilon^2$.
Edit: A follow-up note on determining the independent set of RG equations, because I was looking through Chen et al. again and the procedure is a bit unclear in my opinion, and my earlier claim as to how to get the independent equations was also a little bit unclear/off. I edited the above text to be clearer, but essentially there are two potential issues in determining the independent RG equations: one is that you need to identify the independent functions of time $t$ before setting $t = t_0$ to eliminate the secular terms. The other is that, depending on the problem, there may apparently be several independent functions of time in problems with nonlinearities. For example, in the Rayleigh equation problem in Chen et al., there are functions $\sin(t)$, $\cos(t)$, $\cos(3t)$, and $\sin(3t)$ that all show up in the derivatives of the $O(\epsilon)$ calculation, and are independent functions, which at first glance looks like there are more independent functions than renormalized parameters. However, the equations for $\cos(3t)$ and $\sin(3t)$ turn out to be satisfied to $O(\epsilon^2)$. This complication of having too many independent functions does not show up in the example addressed here because the original ODE is linear, but shows up in nonlinear problems, so I thought I would add this note.
