Optimal and non-optimal lattices for non-completely monotone interaction potentials

Abstract

We investigate the minimization of the energy per point $E_f$ among d-dimensional Bravais lattices, depending on the choice of pairwise potential equal to a radially symmetric function $f(|x|^2)$. We formulate criteria for minimality and non-minimality of some lattices for $E_f$ at fixed scale based on the sign of the inverse Laplace transform of f when f is a superposition of exponentials, beyond the class of completely monotone functions. We also construct a family of non-completely monotone functions having the triangular lattice as the unique minimizer of $E_f$ at any scale. For Lennard-Jones type potentials, we reduce the minimization problem among all Bravais lattices to a minimization over the smaller space of unit-density lattices and we establish a link to the maximum kissing problem. New numerical evidence for the optimality of particular lattices for all the exponents are also given. We finally design one-well potentials f such that the square lattice has lower energy $E_f$ than the triangular one. Many open questions are also presented.

Appendix A: Proof of Proposition 6.3

Proof of Proposition 6.3

In the following, we will write $c=\left( \frac{2}{3} \right) \left( \frac{4}{9} \right) ^{p}$ and $I_\lambda =\left[ \frac{4}{9\lambda },\frac{1}{\lambda } \right]$. Let $L\in \mathscr {L}_d$ be a Bravais lattice, then we have, for any $\lambda >0$,

$$\begin{aligned} E_f[\lambda L]=\frac{c}{\lambda ^{p}}\sum _{\begin{array}{c} x\in L \\ |x|<\frac{4}{9\lambda } \end{array}}\frac{1}{|x|^{p}}+\sum _{\begin{array}{c} x\in L \\ |x|\in I_\lambda \end{array}}(2-3\lambda |x|)-\frac{1}{\lambda ^4}\sum _{\begin{array}{c} x\in L\\ |x|>\frac{1}{\lambda } \end{array}} \frac{1}{|x|^4}:=S_1+S_2+S_3. \end{aligned}$$

(A.1)

Step 1. We remark that, for any L with minimal distance 1 (like $\mathbb {Z}^2$ and $\mathsf {A}_2$), therefore if $\lambda >1$, then $S_1=S_2=0$ and we get $E_f[\lambda L]=S_3=-\lambda ^{-4} \zeta _{L}(4)$. Since $\lambda \mapsto \lambda ^{-4}$ is decreasing, it follows that $E_f[\lambda L]$ is increasing in $\lambda$ for $\lambda \in (1,+\infty )$.

Step 2. We now treat the case $L=\mathbb {Z}^2$, for $\lambda \in [4/9,1]$, a range in which $S_1=0$ but $S_2,S_3\ne 0$. Note that the distances to the origin for the square lattice are 1 (achieved 4 times), $\sqrt{2}$ (achieved 4 times), 2 (achieved 4 times) and $\sqrt{5}$ (achieved 8 times). We base our case subdivision on these values.

• (1) Values$4/9\le \lambda \le 1/\sqrt{5}$for$L=\mathbb {Z}^2$. Then

  $$\begin{aligned} E_f[\lambda \mathbb {Z}^2]=40-3(12+4\sqrt{2}+8\sqrt{5})\lambda -\frac{(\zeta _{\mathbb {Z}^2}(4)-5-1/4-8/25)}{\lambda ^4}. \end{aligned}$$

  (A.2)

  Therefore, $\frac{d}{d\lambda }E_f[\lambda \mathbb {Z}^2]\ge 0$ if and only if $\lambda \le \left( \frac{4(\zeta _{\mathbb {Z}^2}(4)-5-1/4-8/25)}{3(12+4\sqrt{2}+8\sqrt{5})} \right) ^{1/5}=:\lambda _1\approx 0.4433$. Thus, $\lambda \mapsto E_f[\lambda \mathbb {Z}^2]$ is decreasing on $[4/9,1/\sqrt{5}]$.

• (2) Values$1/\sqrt{5}< \lambda \le 1/2$for$L=\mathbb {Z}^2$. In this case

  $$\begin{aligned} E_f[\lambda \mathbb {Z}^2]=24-3(12+4\sqrt{2})\lambda -\frac{(\zeta _{\mathbb {Z}^2}(4)-5-1/4)}{\lambda ^4}. \end{aligned}$$

  (A.3)

  The $\tfrac{d}{d\lambda }$-derivative of (A.3) is positive for $\lambda <\left( \frac{4(\zeta _{\mathbb {Z}^2}(4)-5-1/4)}{3(12+4\sqrt{2})} \right) ^{1/5}\approx 0.56$ and thus $\lambda \mapsto E_f[\lambda \mathbb {Z}^2]$ is increasing for $\lambda \in (1/\sqrt{5},1/2]$.

• (3) Values$1/2<\lambda \le 1/\sqrt{2}$for$L=\mathbb {Z}^2$. In this case

  $$\begin{aligned} E_f[\lambda \mathbb {Z}^2]=16-3(4+4\sqrt{2})\lambda -\frac{(\zeta _{\mathbb {Z}^2}(4)-5)}{\lambda ^4}. \end{aligned}$$

  (A.4)

  Now for the critical value $\left( \frac{4(\zeta _{\mathbb {Z}^2}(4)-5)}{3(4+4\sqrt{2})} \right) ^{1/5}=:\lambda _2\approx 0.6765$ we find that $\lambda \mapsto E_f[\lambda \mathbb {Z}^2]$ is increasing on $(1/2,\lambda _2]$ and decreasing on $[\lambda _2,1/\sqrt{2}]$.

• (4) Values$1/\sqrt{2}<\lambda \le 1$for$L=\mathbb {Z}^2$. In this case

  $$\begin{aligned} E_f[\lambda \mathbb {Z}^2]=8-12\lambda -\frac{(\zeta _{\mathbb {Z}^2}(4)-4)}{\lambda ^4}. \end{aligned}$$

  (A.5)

  Therefore, defining $\left( \frac{4(\zeta _{\mathbb {Z}^2}(4)-4)}{12} \right) ^{1/5}=:\lambda _3\approx 0.9245$ the map $\lambda \mapsto E_f[\lambda \mathbb {Z}^2]$ is increasing on $[1/\sqrt{2},\lambda _3]$ and decreasing on $[\lambda _3,1]$.

Now, comparing the values of $E_f[\lambda L]$ for $L=\mathbb {Z}^2$ for $\lambda \in \{1/\sqrt{5},1/\sqrt{2},1\}$, we find that, based on the above discussion and on Step 1,

$$\begin{aligned} \displaystyle \min _{\lambda \ge 4/9}E_f[\lambda \mathbb {Z}^2]=E_f\left[ \frac{1}{\sqrt{5}}\mathbb {Z}^2 \right] \approx -19.108745 \end{aligned}$$

(A.6)

Step 3. By performing a similar discussion as in Steps 1 and 2, based on the distances to the origin for points in $L=\mathsf {A}_2$ lower than 9 / 4, that are $1,\sqrt{3}$ and 2 (all achieved 6 times), we obtain

$$\begin{aligned} \displaystyle \min _{\lambda \ge 4/9}E_f[\lambda \mathsf {A}_2]=E_f\left[ \frac{4}{9}\mathsf {A}_2 \right] \approx -19.013358. \end{aligned}$$

(A.7)

Step 4. We now assume that $\lambda <\frac{4}{9}$ and we compute a lower bound for the energy (A.1). We bound $S_1$ by the first term in the sum and $S_3$ by the sum over the whole lattice without the constraint $|x|>\tfrac{1}{\lambda }$, and we obtain

$$\begin{aligned} S_1> \frac{4\left( \frac{2}{3} \right) \left( \frac{4}{9} \right) ^p}{\lambda ^p},\quad S_3>-\frac{\zeta _{\mathbb {Z}^2}(4)}{\lambda ^4}. \end{aligned}$$

(A.8)

For $S_2$, we use the fact that $\#\{ x\in \mathbb {Z}^2; |x|\le r \}=\pi r^2 + R(r)$ where $|R(r)|\le 2\sqrt{2}\pi r$. We therefore get

$$\begin{aligned} S_2=\sum _{\begin{array}{c} x\in \mathbb {Z}^2 \\ |x|\in I_\lambda \end{array}}(2-3\lambda |x|)&> \left( 2-\frac{4}{3\lambda }\right) \#\{x\in \mathbb {Z}^2; |x|\in I_\lambda \} \end{aligned}$$

(A.9)

$$\begin{aligned}&> \left( 2-\frac{4}{3\lambda }\right) \left( \frac{\pi }{\lambda ^2} -\frac{16\pi }{81\lambda ^2}+\frac{2\sqrt{2}\pi }{\lambda } +\frac{8\sqrt{2}\pi }{9\lambda } \right) \end{aligned}$$

(A.10)

$$\begin{aligned}&=-\frac{260\pi }{243\lambda ^3}-\left( \frac{104\sqrt{2}\pi }{27\lambda ^2} -\frac{130\pi }{81\lambda ^2} \right) +\frac{52\sqrt{2}\pi }{9\lambda }. \end{aligned}$$

(A.11)

Thus, we have obtained

$$\begin{aligned} E_f[\lambda \mathbb {Z}^2]>\frac{4\left( \frac{2}{3} \right) \left( \frac{4}{9} \right) ^p}{\lambda ^p}-\frac{\zeta _{\mathbb {Z}^2}(4)}{\lambda ^4}-\frac{260\pi }{243\lambda ^3} -\left( \frac{104\sqrt{2}\pi }{27\lambda ^2}-\frac{130\pi }{81\lambda ^2} \right) +\frac{52\sqrt{2}\pi }{9\lambda } .\nonumber \\ \end{aligned}$$

(A.12)

We now want to determine a value $\lambda <4/9$ such that $E_f[\lambda \mathbb {Z}^2]>E[\frac{1}{\sqrt{5}}\mathbb {Z}^2]$. A sufficient condition is, setting $X=\lambda ^{-1}$, to know all the $X>\frac{9}{4}$ satisfy

$$\begin{aligned}&4\left( \frac{2}{3} \right) \left( \frac{4}{9} \right) ^p X^p+\frac{52\sqrt{2}\pi }{9}X \nonumber \\&\quad \ge \zeta _{\mathbb {Z}^2}(4) X^4 +\frac{260\pi }{243}X^3 +\left( \frac{104\sqrt{2}\pi }{27}+\frac{130\pi }{81} \right) X^2 +E\left[ \frac{1}{\sqrt{5}}\mathbb {Z}^2\right] .\nonumber \\ \end{aligned}$$

(A.13)

Defining the following coefficient

$$\begin{aligned} \alpha _4:=\zeta _{\mathbb {Z}^2}(4),\quad \alpha _3:=\frac{260\pi }{243}, \quad \alpha _2:=\left( \frac{104\sqrt{2}\pi }{27}-\frac{130\pi }{81} \right) , \end{aligned}$$

it follows by a direct estimate that (A.13) holds if

$$\begin{aligned} X\ge U_p:=\max \left\{ \left( \frac{9}{8}\left( \frac{9}{4} \right) ^p \alpha _4 \right) ^{\frac{1}{p-4}}, \left( \frac{9}{8}\left( \frac{9}{4} \right) ^p \alpha _3 \right) ^{\frac{1}{p-3}}, \left( \frac{9}{8} \left( \frac{9}{4} \right) ^p \alpha _2 \right) ^{\frac{1}{p-2}}\right\} . \end{aligned}$$

We observe that $U_p\rightarrow \frac{9}{4}$ as $p\rightarrow +\infty$ and is $U_p$ decreasing in p for large p. Therefore, by continuity of $\lambda \mapsto E_f[\lambda \mathbb {Z}^2]$, for any $\varepsilon >0$, there exists $p_0$ such that

$$\begin{aligned} \left| \min _{\lambda<4/9} E_f[\lambda \mathbb {Z}^2] -E_f\left[ \frac{4}{9}\mathbb {Z}^2\right] \right| <\varepsilon . \end{aligned}$$

Since $E_f[\frac{4}{9}\mathbb {Z}^2]>E_f[\frac{1}{\sqrt{5}}\mathbb {Z}^2]$, it follows that, for enough large p,

$$\begin{aligned} \min _{\lambda >0}E_f[\lambda \mathbb {Z}^2]=E_f\left[ \frac{1}{\sqrt{5}}\mathbb {Z}^2\right] \approx -19.108745. \end{aligned}$$

The same argument can be repeated for $L=\mathsf {A}_2$, obtaining that for any $\varepsilon >0$, there exists $p_0$ such that for any $p>p_0$ there holds

$$\begin{aligned} \left| \min _{\lambda<4/9} E_f[\lambda \mathsf {A}_2] -E_f\left[ \frac{4}{9}\mathsf {A}_2\right] \right| <\varepsilon . \end{aligned}$$

Therefore, by (A.7), $\min _{\lambda>0} E_f[\lambda \mathsf {A}_2]> E_f[\frac{4}{9} \mathsf {A}_2]-\varepsilon >E_f[\frac{1}{\sqrt{5}}\mathbb {Z}^2]$ for $\varepsilon >0$ sufficiently small, which in turn is achievable for $p_0$ sufficiently large. These choices allow to complete the proof. $\square$