A Mechanical Model of Brownian Motion for One Massive Particle Including Slow Light Particles

Abstract

We provide a connection between Brownian motion and a classical mechanical system. Precisely, we consider a system of one massive particle interacting with an ideal gas, evolved according to non-random mechanical principles, via interaction potentials, without any assumption requiring that the initial velocities of the environmental particles should be restricted to be “fast enough”. We prove the convergence of the (position, velocity)-process of the massive particle under a certain scaling limit, such that the mass of the environmental particles converges to 0 while the density and the velocities of them go to infinity, and give the precise expression of the limiting process, a diffusion process.

Appendix

We give the proof of (2.6) and (2.7) in Sect. A.1; give the proof of Lemma 3.33 in Sect. A.2; present several estimates that are used to prove that $I^1 (t)$ and $I^2 (t)$ converge to 0 fast enough in Sect. A.3; finally, in Sect. A.4, we present several necessary estimates with respect to integrals involving $t_1 (v; *)$.

1.1 A.1 Proof of (2.6) and (2.7)

By a simple calculation, we have that $\nabla _i \nabla _j U (x) = \Big ( h''(|x|) - \frac{h'(|x|)}{|x|} \Big ) \frac{x_i}{|x|} \frac{x_j}{|x|} + \delta _{ij} \frac{h'(|x|)}{|x|}$ for any $i, j \in \{ 1, \cdots , d \}$, so $\nabla ^2 U(x) y = \Big ( h''(|x|) - \frac{h'(|x|)}{|x|} \Big ) (y, \frac{x}{|x|}) \frac{x}{|x|} + \frac{h'(|x|)}{|x|} y = h''(|x|) \pi _x y + \frac{h'(|x|)}{|x|} \pi ^{\perp }_x y$. Since $|\frac{h'(|x|)}{|x|}| \le \Vert h'' \Vert _{\infty }$, this implies (2.6).

Also, (2.7) is an easy consequence of (2.6) since $| \nabla U (y_1) - \nabla U (y_2) | = \int _0^1 \nabla ^2 U(y_2 + \theta (y_1-y_2)) (y_1-y_2) d \theta$.

1.2 A.2 Proof of Lemma 3.33

We give the proof of Lemma 3.33 in this section. We first notice the following translation property (Lemma 4.1) and symmetry (Lemma 4.2) of $\varphi$, which are clear heuristically.

Lemma 4.1

For any $(x, v) \in \mathbf{R}^{2d}$, $t \ge 0$ and $X \in \mathbf{R}^d$, we have that

$$\begin{aligned} \varphi ^0 (t, x, v; X) - X = \varphi ^0 (t, x-X, v; \mathbf{0}). \end{aligned}$$

Proof

Just notice that both sides above satisfy the same ODE with the same initial conditions. $\square$

Lemma 4.2

For any $X \in \mathbf{R}^d$ and $(x, v) \in E$, let $\iota (x, v; X) := \Big ( \iota ^0 (x, v; X), \iota ^1 (x, v; X) \Big ) := (2X-x, -v)$. Then we have the following.

$$\begin{aligned} \varphi \big (t, \iota (x, v; X); X\big ) = \iota \big ( \varphi (t, x, v; X); X\big ), \qquad t \ge 0. \end{aligned}$$

Proof

First we notice that $\frac{d}{dt} \varphi ^0 (t, \iota (x, v; X); X) = \varphi ^1 (t, \iota (x, v; X); X)$ and $\frac{d}{dt} \iota ^0 ( \varphi (t, x, v; X); X) = \frac{d}{dt} \Big ( 2X - \varphi ^0 (t, x, v; X) \Big ) = - \varphi ^1 (t, x, v; X) = \iota ^1 ( \varphi (t, x, v; X); X)$. We have by definition that $(\varphi ^0 (0, \iota (x, v; X); X), \varphi ^1 (0, \iota (x, v; X); X) ) = (2X-x, -v) = \iota ( \varphi (0, x, v; X); X).$ So $\varphi (t, \iota (x, v; X); X)$ and $\iota ( \varphi (t, x, v; X); X)$ satisfy the same initial condition. Also,

$$\begin{aligned} \frac{d^2}{dt^2} \varphi ^0 (t, \iota (x, v; X); X) = - \nabla U ( \varphi ^0 (t, \iota (x, v; X); X) - X ), \end{aligned}$$

and

$$\begin{aligned}&\frac{d^2}{dt^2} \iota ^0 ( \varphi (t, x, v; X); X) = \frac{d^2}{dt^2} \Big ( 2X - \varphi ^0 (t, x, v; X) \Big ) \\&\quad = \nabla U \big ( \varphi ^0 (t, x, v; X) - X \big ) = - \nabla U \big ( (2X - \varphi ^0 (t, x, v; X) ) - X \big ) \\&\quad = - \nabla U \big ( \iota ^0 ( \varphi (t, x, v; X); X) - X \big ). \end{aligned}$$

So $\varphi ^0 (t, \iota (x, v; X); X)$ and $\iota ^0 ( \varphi (t, x, v; X); X)$ satisfy the same ODE, too. Therefore, $\varphi ^0 (t, \iota (x, v; X); X) = \iota ^0 ( \varphi (t, x, v; X); X)$, which implies our assertion. $\square$

Now we are ready to prove the first assertion of Lemma 3.33.

Proof of Lemma 3.33 (1)

We first notice that the left hand side of (3.27) does not depend on $X_0$. Indeed, by definition and Lemma 4.1, we have that

$$\begin{aligned}&\hbox {(LHS) of }(3.27) \\&\quad = \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U \left( - \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v - X_0, v, \mathbf{0})\right) \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v - X_0\right) {\tilde{\nu }} (dx; v). \end{aligned}$$

Writing $x - m^{-1/2} r v - X_0$ as $x - \pi ^{\perp }_v X_0 - m^{-1/2} ( r + m^{1/2} \frac{(X_0, v)}{|v|^2} ) v$, by change of variable $x - \pi ^{\perp }_v X_0 \rightarrow x$, $r + m^{1/2} \frac{(X_0, v)}{|v|^2} \rightarrow r$ for any fix v, this gives us that

$$\begin{aligned}&\hbox {(LHS) of } (3.27) \nonumber \\&\quad = \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U \left( - \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v, v, \mathbf{0})\right) \nonumber \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v\right) {\tilde{\nu }} (dx; v), \end{aligned}$$

(5.1)

the right hand side of which does not depend on $X_0$.

Let J denote the right hand side of (5.1). So it suffices to prove that $J=0$. By Lemma 4.2 with $X=0$, we have that $- \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r, v, \mathbf{0}) = \varphi ^0 (m^{-1/2} s, -x + m^{-1/2} r, -v, \mathbf{0})$. Substituting this into the definition of J, since $\rho (u, - z) = \rho (u, z)$ for any $(u, z) \in [0, \infty ) \times \mathbf{R}^d$ by (A2), we get by change of variable $x \rightarrow -x$ and $v \rightarrow -v$ that

$$\begin{aligned} J= & {} \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U ( \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v, v, \mathbf{0})) \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v\right) {\tilde{\nu }} (dx; v) \\= & {} - J, \end{aligned}$$

hence $J=0$. $\square$

The second assertion of Lemma 3.33 is proved similarly. We first notice the following property of $\psi$.

Lemma 4.3

For any $(x, v) \in E$, $u \in \mathbf{R}$ and $X \in \mathbf{R}^d$,

• 1. $\psi ^0 (u, x, v; X) - X = \psi ^0 (u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0})$.

• 2. $\psi ^0 (u, -x, -v; \mathbf{0}) = - \psi ^0 (u, x, v; \mathbf{0})$.

Proof

(1) By definition and Lemma 4.1, we have that

$$\begin{aligned}&\psi ^0 (u, x, v; X) - X = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v, v; X) - X \\&\quad = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v - X, v; \mathbf{0}) \\&\quad = \lim _{s \rightarrow \infty } \varphi ^0 \left( \Big ( u - \frac{(X, v)}{|v|^2} \Big ) + \Big ( s + \frac{(X, v)}{|v|^2} \Big ), x - \pi ^{\perp }_v X - \Big ( s + \frac{(X, v)}{|v|^2} \Big ) v, v; \mathbf{0}\right) \\&\quad = \psi ^0 \left( u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0} \right) . \end{aligned}$$

(2) By definition and Lemma 4.2, we have that

$$\begin{aligned}&\psi ^0 (u, -x, -v; \mathbf{0}) = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, -x + s v, - v; \mathbf{0}) \\&\quad = - \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v, v; \mathbf{0}) = - \psi ^0 (u, x, v; \mathbf{0}). \end{aligned}$$

$\square$

Proof of Lemma 3.33 (2)

By Lemma 4.3 (1) and change of variable $u - \frac{(X, v)}{|v|^2} \rightarrow u$, $x - \pi ^{\perp }_v X \rightarrow x$ for any fixed $v \in \mathbf{R}^d$, we have that

$$\begin{aligned}&\hbox {(LHS) of }(3.28) \\&\quad = \int _{\mathbf{R} \times E} \nabla U \left( - \psi ^0 \Big ( u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0} \Big ) \right) m^{-1} \rho _0 \Big (\frac{1}{2} |v|^2\Big ) du \nu (dx, dv) \\&\quad = \int _{\mathbf{R} \times E} \nabla U \big (- \psi ^0 (u, x, v; \mathbf{0}) \big ) m^{-1} \rho _0 \Big (\frac{1}{2} |v|^2\Big ) du \nu (dx, dv). \end{aligned}$$

Let J denote the right hand side above. Then by Lemma 4.3 (2) and change of variable $x \rightarrow -x$, $v \rightarrow - v$, we get that $J = - J$, hence $J=0$. $\square$

1.3 A.3 Estimates for $I^1 (t)$ and $I^2 (t)$

Lemma 4.4 is used in the estimates with respect to $I^1 (t)$ and $I^2 (t)$ (see Lemmas 3.35 and 3.36).

Lemma 4.4

For any $A>0$, there exists a constant $C>0$ such that for any $B_m \ge 0$ ($m \in (0, 1]$), we have that

• 1.
  $$\begin{aligned} \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|)^2 \lambda _m (dr, dx, dv) \le C_9 m^{-1} B_m, \end{aligned}$$
• 2.
  $$\begin{aligned} E^{P_m} \left[ \Big ( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \mu _{\omega } (dr, dx, dv) \Big )^2 \right] \le C_9 (m^{-2} B_m^2 + 1). \end{aligned}$$

Proof

Let $C := 2 (2 A)^{d-1} \int _{\mathbf{R}^d} (1+|v|^2) \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv$. Then C is finite by assumption. We have that

$$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|^2) \lambda _m (dr, dx, dv) \\&\quad \le \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|^2) m^{-1} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dr {\tilde{\nu }} (dx; v) dv \\&\quad \le C m^{-1} B_m, \end{aligned}$$

which gives us our first assertion.

Therefore, by (3.29), we get that

$$\begin{aligned}&E^{P_m} \left[ \Big ( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \mu _{\omega } (dr, dx, dv) \Big )^2 \right] \\&\le 2 \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|)^2 \lambda _m (dr, dx, dv) \\&\qquad + 2 \left( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \lambda _m (dr, dx, dv) \right) ^2 \\&\le 2 C m^{-1} B_m + 2 (C m^{-1} B_m)^2. \end{aligned}$$

So we get our second assertion by re-choosing C in an appropriate way. $\square$

1.4 A.4 Several Estimates with Respect to Integrals Involving $t_1 (v, *)$

As seen, the valid interaction time of each light particle is $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau$, which is not bounded with respect to (x, v). However, by Lemma 4.5, this does not cause any problem after taking integrals.

Lemma 4.5

Fix any $T>0$ and $n \in \mathbf{N}$. Then there exists a constant C such that

• 1. for any $\beta > 0$ such that $2 \varepsilon _1^{-1/2} \beta < d - 1$, $u_0 \in \mathbf{R}$, $k \in \{ 0, 1, 2 \}$, $s \in [0, T \wedge \sigma _n]$ and any $u (r, s) \in [0, T \wedge \sigma _n]$,

  $$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{ \{ |x| \le R_0 \} } 1_{ \left\{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|)^k \\&\qquad \exp \Big ( \beta t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \Big ) (\lambda + {\overline{\lambda }}) (dr, dx, dv) \le C m^{-1/2}, \end{aligned}$$

• 2. for any $-\infty< c_1< c_2 < \infty$ (which may depend on m),

  $$\begin{aligned} \int _{\mathbf{R} \times E} 1_{ \left\{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le m^{\alpha } \right\} } 1_{ \left\{ r \in [c_1, c_2] \right\} } (|v|+1)^2 (\lambda + {\overline{\lambda }}) (dr, dx, dv) \le C m^{\alpha (d-1) - 1} (c_2-c_1), \end{aligned}$$

• 3. for any $s \in [0, T \wedge \sigma _n]$ and $u (r, s) \in [0, T \wedge \sigma _n]$,

  $$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{ \left\{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 m^{\alpha } \right\} } 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } (\lambda + {\overline{\lambda }}) (dr, dx, dv) \\&\qquad \times t_1 \big (v, |x-\pi ^{\perp }_v X(u (r, s))|\big )^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \ell \right) \\&\qquad \le C m^{\alpha \Big ( (d-1 - \ell (1+C_1) \epsilon _1^{-1/2} \Big ) - \frac{1}{2}} \Big ( \log m^{-1} \Big )^{k+1}, \quad k, \ell \in \{ 0, 1 \}, \end{aligned}$$
• 4.
  $$\begin{aligned} \int _E t_1 (v, |x-\pi ^{\perp }_v X|)^k 1_{ \left\{ |x-\pi ^{\perp }_v X| \le R_U + 1 \right\} } \rho \left( \frac{1}{2} |v|^2\right) \nu (dx, dv) \le C, \quad k \in \{ 0, 1, \cdots , 7 \}. \end{aligned}$$

Proof

The proofs of these assertions are similar – first apply the change of variable $y = x - \pi ^{\perp }_v X (\cdot )$ for any fixed v and r, then apply Fubini’s Theorem to switch the order of integrations with respect to y and r if necessary, and finally estimate the integral with respect to x. We present the proofs of (1) and (3) in the following. The proofs of (2) and (4) are similar but simpler, since we do not need Fubini’s Theorem in these proofs, and we omit them here.

(1) By definition and (A3), we get that the left hand side of (1) is dominated by

$$\begin{aligned}&m^{-1} \int _{\mathbf{R}^d} 2 \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dx; v) \\&\qquad \times 1_{ \left\{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 R_0 \right\} } 1_{ \left\{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } \\&\qquad \times t_1 \left( v, |x-\pi ^{\perp }_v X(u (r, s))|\right) ^k \exp \left( \beta t_1 \big (v, |x-\pi ^{\perp }_v X(u (r, s))|\big ) \right) . \end{aligned}$$

Applying the change of variable $y = x-\pi ^{\perp }_v X(u (r, s))$ for any fixed v and r, this is equal to

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad 1_{ \{ |y| \le 2 R_0 \} } 1_{ \{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \Big ( \beta t_1 (v, |y|) \Big ). \end{aligned}$$

Applying Fubini’s Theorem to switch the order of integrations with respect to x and r, since $t_1 (v, *) \ge \tau$, this is dominated by

$$\begin{aligned} 4 m^{-1/2} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _{E_v} 1_{ \{ |y| \le 2 R_0 \} } t_1 (v, |y|)^{k+1} \exp \Big ( \beta t_1 (v, |y|) \Big ) {\tilde{\nu }} (dy; v). \end{aligned}$$

So in order to prove (1), it suffices to prove that

$$\begin{aligned} \sup _{v \in \mathbf{R}^d} \int _{E_v} 1_{ \{ |y| \le 2 R_0 \} } t_1 (v, |y|)^{k+1} \exp \Big ( \beta t_1 (v, |y|) \Big ) {\tilde{\nu }} (dy; v) < \infty . \end{aligned}$$

(5.2)

By the definition of $t_1 (v, |y|)$ and the help of polar coordinates, it in turn suffice to prove that

$$\begin{aligned} \int _{\left[ 0, (2 R_0) \wedge \frac{R_U}{2} \wedge (2 \varepsilon _3)\right] } r^{- \beta 2 \varepsilon _1^{-1/2} + d - 2} \Big ( \log \frac{1}{r} \Big )^{k+1} dr < \infty . \end{aligned}$$

(5.3)

On the other hand, by (2.12), we have that (5.3) holds as long as $- \beta 2 \varepsilon _1^{-1/2} + d - 2 > -1$. This completes our proof of (1).

(3) By definition and (A3), we get that the left hand side of (3) is dominated by

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dx; v) \\&\qquad \times 1_{ \{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \} } \\&\qquad \times t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \ell \right) . \end{aligned}$$

Applying the change of variable $y = x-\pi ^{\perp }_v X(u (r, s))$ for any fixed v and r, this is equal to

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) \\&\quad = 2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _{E_v} {\tilde{\nu }} (dy; v) \int _\mathbf{R} dr \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) \\&\le 4 m^{-\frac{1}{2}} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } t_1 (v, |y|)^{k+1} \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) . \end{aligned}$$

It suffices to consider the case where $m>0$ is small enough such that $2 m^{\alpha } \le \frac{R_U}{2} \wedge (2 \varepsilon _3)$. So by the definition of $t_1 (v, |y|)$, it suffices to prove the following:

$$\begin{aligned}&\sup _{v \in \mathbf{R}^d} \int _{E_v} {\tilde{\nu }} (dy; v) 1_{ \{ |y| \le 2 m^{\alpha } \} } (\log |y|^{-1})^{k+1} \exp \left( \frac{1}{2} \ell (1+C_1) 2 \epsilon _1^{-1/2} \log \frac{2 \varepsilon _3}{|y|} \right) \nonumber \\&\le C m^{\alpha \Big ( (d-1 - \ell (1+C_1) \epsilon _1^{-1/2} \Big )} \Big ( \log m^{-1} \Big )^{k+1}. \end{aligned}$$

(5.4)

On the other hand, by changing variables to polar coordinates, the left hand side of (5.4) is equal to

$$\begin{aligned} C \int _0^{2 m^{\alpha }} r^{d-2} r^{- \ell (1+C_1) \epsilon _1^{-1/2}} \Big ( \log r^{-1} \Big )^{k+1} dr. \end{aligned}$$

For any fixed $\ell \in \{ 0, 1 \}$, we have that $d - 1 - \ell (1+C_1) \epsilon _1^{-1/2} > 0$. Now we get our assertion by the well-known result that for any $p \ne -1$,

$$\begin{aligned}&\int r^p \log r d r = \frac{r^{p+1}}{p+1} \Big ( \log r - \frac{1}{p+1} \Big ), \\&\int r^p \Big ( \log r \Big )^2 d r = \frac{r^{p+1}}{p+1} \left( \Big ( \log r \Big )^2 - \frac{2}{p+1} \log r + \frac{2}{(p+1)^2} \right) . \end{aligned}$$

$\square$