Energy Scaling and Asymptotic Properties of One-Dimensional Discrete System with Generalized Lennard-Jones (m, n) Interaction

Abstract

It is well known that elastic effects can cause surface instability. In this paper, we analyze a one-dimensional discrete system which can reveal pattern formation mechanism resembling the “step-bunching” phenomenon for epitaxial growth on vicinal surfaces. The surface steps are subject to long-range pairwise interactions taking the form of a general Lennard-Jones (LJ)-type potential. It is characterized by two exponents m and n describing the singular and decaying behaviors of the interacting potential at small and large distances, and henceforth are called generalized LJ (m, n) potential. We provide a systematic analysis of the asymptotic properties of the step configurations and the value of the minimum energy, in particular their dependence on m and n and an additional parameter $\alpha$ indicating the interaction range. Our results show that there is a phase transition between the bunching and non-bunching regimes. Moreover, some of our statements are applicable for any critical points of the energy, not necessarily minimizers. This work extends the technique and results of Luo et al. (SIAM Multiscale Model Simul 14(2):737–771, 2016) which concentrates on the case of LJ (0,2) potential (originated from the elastic force monopole and dipole interactions between the steps). As a by-product, our result also leads to the well-known fact that the classical LJ (6,12) potential does not demonstrate the step-bunching-type phenomenon.

Proof of Theorem 3: Upper Bounds for \(E_N\)

We first state the following simple lemma without proof.

Lemma 5

(i) If $\phi :[1,+\infty )\rightarrow {\mathbb {R}}$ is nonnegative and monotonically decreasing, then

$$\begin{aligned} \left| \sum _{k=1}^{\lfloor N^\alpha \rfloor } \phi (k)-\int _{1}^{N^\alpha }\phi (x)\,\mathrm {d}x\right| \le \phi (1) \end{aligned}$$

(51)

(ii) If $\phi :[1,+\infty )\rightarrow {\mathbb {R}}$ is nonnegative and monotonically increasing, then

$$\begin{aligned} \left| \sum _{k=1}^{\lfloor N^\alpha \rfloor } \phi (k)-\int _{1}^{N^\alpha }\phi (x)\,\mathrm {d}x\right| \le \phi (N^\alpha ). \end{aligned}$$

(52)

Proof of Theorem 3

We estimate the upper bound of $E[X_N^0]$ where $X_N^0=(x_1^0,\ldots ,x_N^0)^T$ with $x_i^0=(i-1)l_0$ for $i=1,\ldots ,N$ with some appropriate $l_0$. For convenience, we write

$$\begin{aligned} E[X_N^0] = e_m[X_N^0]-e_n[X_N^0] \end{aligned}$$

(53)

where for $-1 < s$ and $s\ne 0$, we have

$$\begin{aligned} e_s[X_N^0]&= \sum \limits _{1\le i<j\le N,\,\,j-i\le \lfloor N^{\alpha }\rfloor }-\frac{1}{s}|x_j-x_i|^{-s} = -\sum _{k=1}^{\lfloor N^\alpha \rfloor }(N-k)\frac{1}{s}k^{-s}l_0^{-s} \nonumber \\&= -\frac{1}{s}l_0^{-s}\left( N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-s}-\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{1-s}\right) , \end{aligned}$$

(54)

while for $s=0$, we have

$$\begin{aligned} e_s[X_N^0]&= \sum \limits _{1\le i<j\le N,\,\,j-i\le \lfloor N^{\alpha }\rfloor }\log |x_j-x_i| = \sum _{k=1}^{\lfloor N^\alpha \rfloor }(N-k)\log (kl_0). \end{aligned}$$

(55)

We will estimate the above summation by applying part (i) of Lemma 5 for $\phi (x)=x^{-s}\,(s>0)$ on $[1,+\infty )$ but part (ii) for $\phi (x)=x^{-s}\,(s<0)$, $\log x$, and $x\log x$ on $[1,+\infty )$.

Without loss of generality, we assume $N\ge 2$. We first give some useful upper bounds for $e_m[X^0_N]$ and $-e_n[X^0_N]$ with $m,n\ne 0$:

• (i) for $-1<m<0$,

  $$\begin{aligned} e_m[X^0_N]&=-\frac{1}{m}l_0^{-m}\left( N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-m}-\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{1-m}\right) \nonumber \\&\le \frac{1}{|m|}l_0^{-m}\left( N\int _1^{N^\alpha }x^{-m}\,\mathrm {d}x+N^{1-\alpha m}\right) \nonumber \\&\le \frac{1}{|m|}l_0^{-m}\left( \frac{1}{1-m}N^{1+\alpha (1-m)}-\frac{1}{1-m}N+N^{1-\alpha m}\right) \nonumber \\&\le \frac{1}{|m|}l_0^{-m}\left( \frac{1}{1-m}N^{1+\alpha (1-m)}+N^{1-\alpha m}\right) . \end{aligned}$$

  (56)

• (ii) for $0<m<1$,

  $$\begin{aligned} e_m[X^0_N]&= \frac{1}{m}l_0^{-m}\left( \sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{1-m}-N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-m}\right) \nonumber \\&\le \frac{1}{m}l_0^{-m}\left( \int _1^{N^\alpha }x^{1-m}\,\mathrm {d}x+N^{\alpha (1-m)}-N\int _1^{N^\alpha }x^{-m}\,\mathrm {d}x+N\right) \nonumber \\&\le \frac{1}{m}l_0^{-m}\left( \frac{1}{2-m}N^{\alpha (2-m)}-\frac{1}{1-m}N^{1+\alpha (1-m)}+\frac{1}{1-m}N+N+N^{\alpha (1-m)}\right) \nonumber \\&\le \frac{1}{m}\left( \frac{1}{2-m}-\frac{1}{1-m}\right) l_0^{-m}N^{1+\alpha (1-m)}+\frac{1}{m}\left( 2+\frac{1}{1-m}\right) l_0^{-m}N\nonumber \\&\le -\frac{1}{m(1-m)(2-m)}l_0^{-m}N^{1+\alpha (1-m)}+\frac{3}{m(1-m)}l_0^{-m}N. \end{aligned}$$

  (57)

• (iii) for $1<m$,

  $$\begin{aligned} e_m[X^0_N]&= -\frac{1}{m}l_0^{-m}\left( N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-m}-\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{1-m}\right) \nonumber \\&\le -\frac{1}{m}l_0^{-m}\left( N\sum _{k=1}^{2}k^{-m}-\sum _{k=1}^{\lfloor N^\alpha \rfloor }1\right) \nonumber \\&\le -\frac{1}{m}l_0^{-m}(N+2^{-m}N-N^\alpha )\nonumber \\&\le -\frac{1}{m}(2l_0)^{-m}N. \end{aligned}$$

  (58)

• (iv) for $0<n<1$,

  $$\begin{aligned} -e_n[X^0_N] \le \frac{1}{n}l_0^{-n}N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-n}&\le \frac{1}{n}l_0^{-n}N\left( \int _{1}^{N^\alpha } x^{-n}\,\mathrm {d}x+1\right) \nonumber \\&\le \frac{1}{n(1-n)}l_0^{-n}N^{1+\alpha (1-n)}. \end{aligned}$$

  (59)

• (v) for $1<n$,

  $$\begin{aligned} -e_n[X^0_N] \le \frac{1}{n}l_0^{-n}N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-n}&\le \frac{1}{n}l_0^{-n}N\left( \int _{1}^{N^\alpha } x^{-n}\,\mathrm {d}x+1\right) \nonumber \\&= \frac{1}{n}l_0^{-n}N\left( 1+\frac{1}{n-1}-\frac{1}{n-1}N^{-\alpha (n-1)}\right) \nonumber \\&\le \frac{1}{n-1}l_0^{-n}N. \end{aligned}$$

  (60)

Now we proceed to prove the theorem. We remark that the classification and computation in the following cases are quite tedious. As the goal is to obtain upper bounds, we certainly would like the bounds to be “as low as” possible and preferably with negative prefactor. Hence, in some cases, we will be very careful in choosing the constants—see the sub-cases A1, A2, and C4 in the following.

Case (A): $-1< m< n <1$, $0 < \alpha \le 1$. Overall, we will choose $l_0 \sim N^{-\alpha }$ and the bounds obtained are of the type $E_N \lesssim N$. The first two cases cover the regime $mn > 0$ while the rest cover the regime $mn\le 0$.

(A1) $0<m<n<1$. Collecting (59) and (57) together with $l_0=C_0 N^{-\alpha }$ and $C_0=\left( \frac{2m(1-m)(2-m)}{n(1-n)}\right) ^{\frac{1}{n-m}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \left( \frac{1}{n(1-n)}C_0^{-n}-\frac{1}{m(1-m)(2-m)}C_0^{-m}\right) N^{1+\alpha }\nonumber \\&\quad +\frac{3}{m(1-m)}C_0^{-m}N^{1+\alpha m}\nonumber \\&\le -\frac{1}{2m(1-m)(2-m)}C_0^{-m}N^{1+\alpha }+\frac{3}{m(1-m)}C_0^{-m}N^{1+\alpha m}, \end{aligned}$$

(61)

where we have used the fact that

$$\begin{aligned} \frac{1}{n(1-n)}C_0^{-n}-\frac{1}{m(1-m)(2-m)}C_0^{-m}=-\frac{1}{2m(1-m)(2-m)}C_0^{-m}. \end{aligned}$$

(A2) $-1<m<n<0$. For $-1<n<0$, we have

$$\begin{aligned} -e_n[X^0_N]&= \frac{1}{|n|}l_0^{-n}\left( \sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{1-n}-N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-n}\right) \nonumber \\&\le \frac{1}{|n|}l_0^{-n}\left( \int _1^{N^\alpha }x^{1-n}\,\mathrm {d}x+N^{\alpha (1-n)}-N\int _1^{N^\alpha }x^{-n}\,\mathrm {d}x+N^{1-\alpha n}\right) \nonumber \\&\le \frac{1}{|n|}l_0^{-n}\left( \frac{1}{2-n}N^{\alpha (2-n)}-\frac{1}{1-n}N^{1+\alpha (1-n)}+\frac{1}{1-n}N+2N^{1-\alpha n}\right) \nonumber \\&\le \frac{1}{|n|}\left( \frac{1}{2-n}-\frac{1}{1-n}\right) l_0^{-n}N^{1+\alpha (1-n)}+\frac{1}{|n|}\left( 2+\frac{1}{1-n}\right) l_0^{-n}N^{1-\alpha n}\nonumber \\&\le -\frac{1}{|n|(1-n)(2-n)}l_0^{-n}N^{1+\alpha (1-n)}+\frac{3}{|n|}l_0^{-n}N^{1-\alpha n}. \end{aligned}$$

(62)

Collecting (62) and (56) together with $l_0=C_0 N^{-\alpha }$ and $C_0=\left( \frac{|m|(1-m)}{2|n|(1-n)(2-n)}\right) ^{\frac{1}{n-m}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \left( \frac{1}{|m|(1-m)}C_0^{-m}-\frac{1}{|n|(1-n)(2-n)}C_0^{-n}\right) N^{1+\alpha }\nonumber \\&\quad +\frac{1}{|m|}C_0^{-m}N+\frac{3}{|n|}C_0^{-n}N\nonumber \\&\le -\frac{1}{2|n|(1-n)(2-n)}C_0^{-n}N^{1+\alpha }+\left( \frac{1}{|m|}C_0^{-m}+\frac{3}{|n|}C_0^{-n}\right) N, \end{aligned}$$

(63)

where we have used the fact that

$$\begin{aligned} \frac{1}{|m|(1-m)}C_0^{-m}-\frac{1}{|n|(1-n)(2-n)}C_0^{-n}=-\frac{1}{2|n|(1-n)(2-n)}C_0^{-n}. \end{aligned}$$

(A3) $0=m<n<1$ and $0<\alpha <1$. We set $l_0=N^{-\alpha }\le 1$. Then,

$$\begin{aligned} e_m[X^0_N]&= -\sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log (k l_0)+N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log (k l_0)\nonumber \\&= -\sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log k+N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log k+(\log l_0)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le N\int _1^{N^\alpha }\log x\,\mathrm {d}x+N\log N^\alpha -\alpha (\log N)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le N^{1+\alpha }\log N^\alpha +N\log N^\alpha -\alpha N^{1+\alpha }\log N+\alpha N\log N+\alpha N^{2\alpha }\log N\nonumber \\&= 2N\log N^\alpha +N^{2\alpha }\log N^\alpha . \end{aligned}$$

(64)

Collecting (59) and (64) together with $l_0=N^{-\alpha }$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{n(1-n)}N^{1+\alpha }+2N\log N^\alpha +N^{2\alpha }\log N^\alpha \nonumber \\&\le \frac{1}{n(1-n)}N^{1+\alpha }+\frac{3}{n}N^{1\wedge 2\alpha }\log N^\alpha . \end{aligned}$$

(65)

(A4) $0=m<n<1$ and $\alpha =1$. We set $l_0=N^{-1}\le 1$. Then,

$$\begin{aligned} e_m[X^0_N]&= -\sum _{k=1}^{N}k\log (k l_0)+N\sum _{k=1}^{N}\log (k l_0)\nonumber \\&= -\sum _{k=1}^{N}k\log k+N\sum _{k=1}^{N}\log k+(\log l_0)\left( N^2-\sum _{k=1}^{N} k \right) \nonumber \\&\le -\int _1^N x\log x\,\mathrm {d}x+N\log N+N\int _1^{N}\log x\,\mathrm {d}x+N\log N-\frac{1}{2}N(N-1)\log N\nonumber \\&\le -\frac{1}{2}N^2\log N+\frac{1}{4}N^2+N^2\log N+2N\log N-\frac{1}{2}N^2\log N+\frac{1}{2}N\log N\nonumber \\&\le 3N\log N+\frac{1}{4}N^{2}. \end{aligned}$$

(66)

Collecting (59) and (66) together with $l_0=N^{-1}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{n(1-n)}N^{2}+3N\log N+\frac{1}{4}N^{2}\nonumber \\&\le \left( \frac{1}{n(1-n)}+\frac{1}{4}\right) N^{2}+\frac{3}{n}N\log N. \end{aligned}$$

(67)

(A5) $-1<m<0=n$ and $0<\alpha <1$. We set $l_0=N^{-\alpha }\le 1$. Then,

$$\begin{aligned} -e_n[X^0_N]&= \sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log (k l_0)-N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log (k l_0)\nonumber \\&= \sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log k-N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log k-(\log l_0)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le \int _1^{N^\alpha }x\log x\,\mathrm {d}x+N^\alpha \log N^\alpha -N\int _1^{N^\alpha }\log x\,\mathrm {d}x+N\log N^\alpha \nonumber \\&\quad +\alpha (\log N)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le \frac{1}{2}N^{2\alpha }\log N^\alpha +N^\alpha \log N^\alpha -N^{1+\alpha }\log N^\alpha +N^{1+\alpha }\nonumber \\&\quad -N+N\log N^\alpha +\alpha N^{1+\alpha }\log N\nonumber \\&\le N^{1+\alpha }+ \left( \frac{1}{2}N^{2\alpha }+N^{\alpha }+N\right) \log N^\alpha . \end{aligned}$$

(68)

Collecting (56) and (68) together with $l_0=N^{-\alpha }$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{|m|(1-m)}N^{1+\alpha }+\frac{2}{|m|}N+N^{1+\alpha }+\left( \frac{1}{2}N^{2\alpha }+N^{\alpha }+N\right) \log N^\alpha \nonumber \\&\le \left( \frac{1}{|m|(1-m)}+1\right) N^{1+\alpha }+\frac{5}{|m|}N^{1\wedge 2\alpha }\log N^\alpha . \end{aligned}$$

(69)

(A6) $-1<m<0=n$ and $\alpha =1$. We set $l_0=N^{-1}\le 1$. Then,

$$\begin{aligned} -e_n[X^0_N]&= \sum _{k=1}^{N}k\log (k l_0)-N\sum _{k=1}^{N}\log (k l_0)\nonumber \\&= \sum _{k=1}^{N}k\log k-N\sum _{k=1}^{N}\log k-(\log l_0)\left( N^2-\sum _{k=1}^{N} k \right) \nonumber \\&\le \int _1^{N}x\log x\,\mathrm {d}x+N\log N-N\int _1^{N}\log x\,\mathrm {d}x+N\log N\nonumber \\&\quad +(\log N)\frac{1}{2}N(N-1)\nonumber \\&\le \frac{1}{2}N^{2}\log N+N\log N-N^{2}\log N+N^2-N+N\log N+\frac{1}{2}N^{2}\log N\nonumber \\&\le N^2+2N\log N. \end{aligned}$$

(70)

Collecting (56) and (70) together with $l_0=N^{-\alpha }$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{|m|(1-m)}N^{2}+\frac{2}{|m|}N+N^2+2N\log N\nonumber \\&\le \left( \frac{1}{|m|(1-m)}+1\right) N^{2}+\frac{4}{|m|}N\log N. \end{aligned}$$

(71)

(A7) $-1<m<0$, $0<n<1$. Collecting (59) and (56) together with $l_0=N^{-\alpha }$, we obtain

$$\begin{aligned} E[X^0_N]&\le \left( \frac{1}{n(1-n)}+\frac{1}{|m|(1-m)}\right) N^{1+\alpha }+\frac{1}{|m|}N. \end{aligned}$$

(72)

Case (B): $-1< m < n =1$. In this case, we set $l_0=N^{-\alpha }$. Then,

$$\begin{aligned} -e_n[X^0_N]&= l_0^{-1}\left( N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-1}-\lfloor N^\alpha \rfloor \right) \le N^{1+\alpha }\left( \int _1^{N^\alpha }x^{-1}\,\mathrm {d}x+1\right) \nonumber \\&=N^{1+\alpha }\log N^\alpha +N^{1+\alpha }. \end{aligned}$$

(73)

Collecting (56), (57), (64), and (66), we obtain

$$\begin{aligned} e_m[X_N^0]&= \left\{ \begin{array}{ll} \frac{1}{|m|(1-m)}N^{1+\alpha }+\frac{1}{|m|}N, &{} -1<m<0,\\ -\frac{1}{m(1-m)(2-m)}N^{1+\alpha }+\frac{3}{m(1-m)}N^{1+\alpha m}, &{} 0<m<1,\\ 2N\log N^\alpha +N^{2\alpha }\log N^\alpha , &{} m=0, 0<\alpha <1,\\ 3N\log N+\frac{1}{4}N^{2}, &{} m=0, \alpha =1, \end{array} \right. \end{aligned}$$

(74)

which can be summarized as

$$\begin{aligned} E[X^0_N]&\le CN^{1+\alpha }\log N. \end{aligned}$$

(75)

Case (C): $-1< m< 1 < n$. Overall, we will take $l_0\sim N^{-\frac{\alpha (1-m)}{n-m}}$ but in Case C4, it is crucial that we obtain a negative prefactor. Hence, the choice of the constant in $l_0$ is important.

(C1) $-1<m<0$. Collecting (60) and (56) together with $l_0=N^{-\frac{\alpha (1-m)}{n-m}}$, we obtain

$$\begin{aligned} E[X^0_N]\le \left( \frac{1}{n-1}+ \frac{1}{|m|(1-m)}\right) N^{1+\frac{n(1-m)\alpha }{n-m}}+\frac{1}{|m|}N^{1-\alpha +\frac{n(1-m)\alpha }{n-m}}. \end{aligned}$$

(76)

(C2) For $m=0$ and $0<\alpha <1$, we set $l_0=N^{-\frac{\alpha }{n}}\le 1$. Thus,

$$\begin{aligned} e_m[X^0_N]&= -\sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log (k l_0)+N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log (k l_0)\nonumber \\&= -\sum _{k=1}^{\lfloor N^\alpha \rfloor }k\log k+N\sum _{k=1}^{\lfloor N^\alpha \rfloor }\log k+(\log l_0)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le N\int _1^{N^\alpha }\log x\,\mathrm {d}x+N\log N^\alpha -\frac{\alpha }{n}(\log N)\left( N\lfloor N^\alpha \rfloor -\sum _{k=1}^{\lfloor N^\alpha \rfloor } k \right) \nonumber \\&\le N^{1+\alpha }\log N^\alpha +N\log N^\alpha -\frac{\alpha }{n}N^{1+\alpha }\log N+\frac{\alpha }{n}N\log N+\frac{\alpha }{n}N^{2\alpha }\log N\nonumber \\&= \frac{(n-1)\alpha }{n}N^{1+\alpha }\log N+2(N+N^{2\alpha })\log N. \end{aligned}$$

(77)

Collecting (60) and (77) together with $l_0=N^{-\frac{\alpha }{n}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{n-1}N^{1+\alpha }+\frac{(n-1)\alpha }{n}N^{1+\alpha }\log N+2(N+N^{2\alpha })\log N\nonumber \\&\le \frac{(n-1)\alpha }{n}N^{1+\alpha }\log N+CN^{1+\alpha }. \end{aligned}$$

(78)

(C3) For $m=0$ and $\alpha =1$, we set $l_0=N^{-\frac{1}{n}}\le 1$. Thus,

$$\begin{aligned} e_m[X^0_N]&= -\sum _{k=1}^{N}k\log (k l_0)+N\sum _{k=1}^{N}\log (k l_0)\nonumber \\&= -\sum _{k=1}^{N}k\log k+N\sum _{k=1}^{N}\log k+(\log l_0)\left( N^2-\sum _{k=1}^{N} k \right) \nonumber \\&\le -\int _1^N x\log x\,\mathrm {d}x+N\log N+N\int _1^{N}\log x\,\mathrm {d}x+N\log N-\frac{1}{2n}N(N-1)\log N\nonumber \\&\le -\frac{1}{2}N^2\log N+\frac{1}{4}N^2+N^2\log N+2N\log N-\frac{1}{2n}N^2\log N+\frac{1}{2n}N\log N\nonumber \\&\le \frac{n-1}{2n}N^2\log N+3N\log N+\frac{1}{4}N^{2}. \end{aligned}$$

(79)

Collecting (60) and (79) together with $l_0=N^{-\frac{1}{n}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \frac{1}{n-1}N^{2}+\frac{n-1}{2n}N^2\log N+3N\log N+\frac{1}{4}N^{2}\nonumber \\&\le \frac{n-1}{2n}N^{2}\log N+CN^2. \end{aligned}$$

(80)

(C4) $0<m<1$. Collecting (60) and (57) together with $l_0=C_0 N^{-\frac{\alpha (1-m)}{n-m}}$ and $C_0=\left( \frac{2m(1-m)(2-m)}{n-1}\right) ^{\frac{1}{n-m}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \left( \frac{1}{n-1}C_0^{-n}-\frac{1}{m(1-m)(2-m)}C_0^{-m}\right) N^{1+\frac{n(1-m)\alpha }{n-m}}\nonumber \\&\quad +\frac{3}{m(1-m)}C_0^{-m} N^{1+\frac{m(1-m)\alpha }{n-m}}\nonumber \\&\le -\frac{C_0^{-n}}{n-1}N^{1+\frac{n(1-m)\alpha }{n-m}} +\frac{3C_0^{-m}}{m(1-m)} N^{1+\frac{m(1-m)\alpha }{n-m}}, \end{aligned}$$

(81)

where we have used the fact that $\frac{1}{n-1}C_0^{-n}-\frac{1}{m(1-m)(2-m)}C_0^{-m}=-\frac{1}{n-1}C_0^{-n}$.

Case (D): $1=m < n$. We set $l_0 = 1$. Then,

$$\begin{aligned} e_m[X^0_N]&= l_0^{-1}\left( -N\sum _{k=1}^{\lfloor N^\alpha \rfloor }k^{-1}+\lfloor N^\alpha \rfloor \right) \nonumber \\&\le -N\int _1^{N^\alpha }x^{-1}\, \mathrm {d}x+2N =-N\log N^\alpha +2N. \end{aligned}$$

(82)

By (60) with $l_0=1$, we have

$$\begin{aligned} -e_n[X^0_N] \le \frac{1}{n-1}N. \end{aligned}$$

(83)

Therefore,

$$\begin{aligned} E[X^0_N]&\le -N\log N^\alpha +\left( 2+\frac{1}{n-1}\right) N. \end{aligned}$$

(84)

Case (E). There are two cases depending on whether $\alpha >0$ or not. But in either cases, $l_0 \sim 1$.

(E1) $1<m<n$ and $0<\alpha \le 1$. Collecting (60) and (58) together with $l_0=(\frac{2^{m+1}m}{n-1})^{\frac{1}{n-m}}$, we obtain

$$\begin{aligned} E[X^0_N]&\le \left( \frac{1}{n-1}l_0^{-n}-\frac{1}{m}(2l_0)^{-m}\right) N\nonumber \\&=-\frac{1}{n-1}l_0^{-n}N, \end{aligned}$$

(85)

where we have used the fact that $\frac{1}{n-1}l_0^{-n}-\frac{2^{-m}}{m}l_0^{-m}=-\frac{1}{n-1}l_0^{-n}$.

(E2) $-1< m < n$ and $\alpha =0$. Let $l_0=1$. Then, $E[X^0_N]=(N-1)\min V\le CN.$

All the cases are thus considered. $\square$