Infra-red asymptotics

We consider the asymptotics of the integral in Eq. (A1) as $k_1 \to 0$. We employ the independent variables $x$ and $y$, where $k_1 = k x$, $k_2 = k (1 + y)$, $x,y = O(\epsilon)$, $x>0$ and $-x<y<x$. In this limit of $\epsilon \to 0$, $n_1 \gg n, n_2$. In this limit, Eqs. (21a) and (23b), Eqs. (21b) and (23a), and Eqs. (22a) and (22b) correspond to ES, Eqs. (25a, 27b) ID, Eqs. (25b, 27a), PSI, Eqs. (25a, 27b), respectively. Without loss of generality $m$ is set to be positive.

Table I: Asymptotics as $k_1 \to 0$. ES (21a, 23b) gives $\epsilon ^{-a+5}$ owing to the symmetry of $y$. ID (21b, 23a) gives $\epsilon ^{-a-(b-7)/2}$ ( $\epsilon ^{-a+4}$) owing to the second cancellation. PSI (22a, 22b) gives $\epsilon ^{-a-b+5}$. The asymptotics for $b=0$ appear in parentheses.

Assuming the power-law spectrum, $n(\bm{k}, m) = \vert\bm{k}\vert^{-a} \vert m\vert^{-b}$, we make Taylor expansion for the integrand of the kinetic equation (A1) as powers of $\epsilon$, that is $x$ and $y$. Then, we get Table I which shows the leading orders of the each terms according to the asymptotics. The leading order of the collision integral is given by ID when $-3<b<3$. Therefore, we are going to show the procedure to get the leading order of ID (21b) and (23a) below.

As $\epsilon \to 0$, $n_2 \to n$ for ID solutions. Therefore, the leading orders of $f^0_{1,2} \sim n_1 (n_2 - n)$ and $f^2_{0,1} \sim n_1 (n - n_2)$ is $O(\epsilon^{-a-(b-1)/2})$. The order $O(\epsilon^{-a-b/2})$ is canceled as $\epsilon \to 0$. This is called the first cancellation. It must be noted that the leading order when $b=0$ is $1/2$ larger than that when $b \neq 0$ since $\partial n / \partial m = 0$. The leading orders when $b=0$ are written in parentheses in Table I.

The leading order of the integrand in Eq. (A1) is written as

$$T^0_{1,2} - T^1_{2,0} - T^2_{0,1} \propto k^{-2a + 3} m^{-2b+1} \frac{x^{-a - (b+1)/2} y}{\sqrt{(x+y)(x-y)}}$$

$$\times \! \left( -2 a y^2 \! - b \left((1-b) y (x+y) -2x(x-y)\right) + b (b+1) xy \right) .$$

Therefore, the integrand has $O(\epsilon^{-a - (b - 5)/2})$. The term which has $O(\epsilon^{-a -b/2 + 2})$ is canceled since $T^2_{0,1} \to T^0_{1,2}$ (and $T^1_{2,0} \to 0$) as $\epsilon \to 0$. This is the second cancellation.

Finally, we get the leading order of the kinetic equation after integration over $y$ from $-x$ to $x$:

$$\frac{\partial n_{\bm{p}}}{\partial t} \propto - b (1-b) k^{4 - 2 a} m^{1 - 2b} \int_0 x^{-a - (b-5)/2} dx.$$ (10)

The integral has $O(\epsilon^{-a - (b - 7)/2})$. Consequently integral converges if

The integral for the PR spectrum, which gives $O(\epsilon^{-1/4})$, diverges as $k_1 \to 0$. However, the integral for the GM spectrum converges because $b=0$ and the next order is $O(1)$.

It should be noted that the leading order when $b=1$ is $1/2$ larger than that when $b \neq 0,1$ since a balance between first- and second-order derivative is made. The leading orders when $b=1$ are $O(\epsilon^{-a+7/2})$. It is also helpful to note that $T^0_{1,2} - T^1_{2,0} - T^2_{1,0} = O(\epsilon^{-a+2})$ for ES because of no second cancellation. However, the collision integral has $O(\epsilon^{-a+5})$ because of symmetry of $y$. Therefore, the integral which is dominated by ES converges

Similarly, the integral which is dominated by PSI converges