Numerical Methods for the Critical Temperature and Gap Solution of Bogoliubov-Tolmachev-Shirkov Model

Ge Zhihao,, Li Ruihua

 基金资助: 国家自然科学基金.  11971150河南大学一流学科培育项目.  2019YLZDJL08

 Fund supported: the NSFC.  11971150the Cultivation Project of First Class Subject of Henan University.  2019YLZDJL08

Abstract

In the paper, the max-mixed scheme and min-mixed scheme are proposed for the critical temperature and gap solution of Bogoliubov-Tolmachev-Shirkov model in superconductivity theory. For the first time, the numerical solutions of critical temperature and gap solution are given for the above model with the alternating kernel function. The convergence analysis of the numerical method is also given. Also, some numerical examples are presented to verify the results of the paper.

Keywords： Bogoliubov-Tolmachev-Shirkov model ; Critical temperature ; Gap solution ; Convergence analysis

Ge Zhihao, Li Ruihua. Numerical Methods for the Critical Temperature and Gap Solution of Bogoliubov-Tolmachev-Shirkov Model. Acta Mathematica Scientia[J], 2020, 40(6): 1699-1711 doi:

1 引言

$$$\Delta({\bf x}) = \int_{\Omega}{\rm d}{\bf y}K({\bf x}, {\bf y})\varphi_{\beta} ({\bf y}, \Delta({\bf y})),$$$

$$$\Delta(x) = \int_{I'}K(x, y)\frac{\tanh((1/2T)\sqrt{y^{2} +\Delta^{2}(y)})}{\sqrt{y^{2}+\Delta^{2}(y)}}\Delta(y){\rm d}y,$$$

$$$K({\bf x}, {\bf y})>0, K({\bf x}, {\bf y})\leq \sigma({\bf y}), \ \frac{\sigma({\bf x})}{{\bf x}^{2}+1}\in L({{\Bbb R}} ^{3}),$$$

$$$\Delta(x) = \int_{I}K(x, y)\frac{\tanh((1/2T)\sqrt{y^{2} +\Delta^{2}(y)})}{\sqrt{y^{2} +\Delta^{2}(y)}}\Delta(y){\rm d}y,$$$

$$$\begin{array}{l} { } K_{phonon}(x, y)\equiv\frac{K_{1}}{2}>0, |x|, |y|<a, \\ K_{phonon}(x, y) = 0, \quad \mbox{其他};\\ { } K_{Coulomb}(x, y)\equiv-\frac{K_{2}}{2}<0, |x|, |y|<b, \\ K_{Coulomb}(x, y) = 0, \quad \mbox{其他}, \end{array}$$$

$K_1, K_2$是常数, $a>0$是Debye能量, 并且$a = \hbar\omega_{D}$, $b>a$是屏蔽库仑斥力能量范围的上界.

$$$\begin{array}{l} \Delta(x) = \Delta_{1}, |x|<a;\\ \Delta(x) = \Delta_{2}, a<|x|<b;\\ \Delta(x) = 0, \ \ \mbox{其他}. \end{array}$$$

2 BTS模型的数值方法

$f_{\beta}(s) = \frac{\tanh(1/2\beta s)}{s}$, 定义$A_{\beta} $$B_{\beta} 为如下非线性形式: $$B_{\beta}(\Delta) = \Delta\int_{a}^{b}f_{\beta}(\sqrt{\Delta^{2} +x^{2}}){\rm d}x = \Delta\int_{a}^{b} \frac{\tanh(1/2\beta\sqrt{\Delta^{2} +x^{2}})}{\sqrt{\Delta^{2}+x^{2}}}{\rm d}x.$$ 引入变量 u = \Delta_{1}$$ v = -\Delta_{2}$, 由(2.1)和(1.6)式可得

$$$\left\{\begin{array}{ll} u = (K_{1}-K_{2})A_{\beta}(u)+K_{2}B_{\beta}(v), \\ v = K_{2}A_{\beta}(u)-K_{2}B_{\beta}(v). \end{array}\right.$$$

$$$u^{0}-(K_{1}-K_{2})A_{h}(u^{0}) \geq K_{2}B_{h}(v), \quad \forall v.$$$

因为函数$A_{h}(u) $$B_{h}(v) 关于参数 \beta 一致有界, 即存在某个常数 C>0 , 使得 \begin{eqnarray} A_{h}(u)\leq C, \ B_{h}(v)\leq C. \end{eqnarray} 由(2.10)式, 可得存在一个常数 \delta_{0}>0 使得 $$\delta_{0}-(K_{1}-K_{2})A_{h}(\delta_{0}) \geq K_{2}B_{h}(v), \quad \forall v.$$ 由(2.11)式和引理2.1可知, 如果 u^{0}\geq \delta_{0} , 则 u^{0} 是一个上解.证毕. 引理2.4 如果 K_1>K_2 , 当 \beta 足够小时, 格式(2.3)只有零解; 当 \beta 足够大时, 格式(2.3)有一个下解 (u_0, v_0) . 从(2.7)式, 容易验证 \begin{eqnarray} A_{h}(u)\leq\frac{1}{2}\beta au, \ B_{h}(v)\leq\frac{1}{2}\beta (b-a)v. \end{eqnarray} 由(2.12)和(2.3)式, 易知当 \beta 足够小时, 格式(2.3)的非负解是平凡的, 即 u = 0, \ v = 0 . 在格式(2.3)中令 v = 0 , 可得 $$u = (K_{1}-K_{2})A_{h}(u).$$ K_1>K_2 的条件下, 由文献[2]可知, 当 \beta 足够大时, (2.13)式有一个正解, 设为 u_0 . v_{0} = 0 , 当 \beta>0 足够大时, 则非负解对 (u_0, v_0) 满足(2.4)式, 即为格式(2.3)的一个下解. 引理2.5 如果 K_1>K_2 , 则最小-混合格式(2.3)存在正解当且仅当它存在非平凡的下解 (u_0, v_0) . 由引理2.3可知, 存在一个绝对常数 u^{0}>0 使得 $$u^{0}-(K_{1}-K_{2})A_{h}(u^{0}) \geq K_{2}B_{h}(v), \quad \forall v.$$ (u_0, v_0) 是格式(2.3)的下解, 如果 v_{1} = v_{0}\geq 0 , 利用(2.3)和(2.14)式, 可得 \begin{eqnarray} u_{2}>0, \ u_{0}\leq u_{2}\leq u^{0}. \end{eqnarray} 不妨假设存在某个正数 l , 使得下式成立 \begin{eqnarray} &&0<u_{0} = u_{1}\leq u_{2}\leq\cdots \leq u_{l}\leq u^{0}, \end{eqnarray} \begin{eqnarray} &&0\leq v_{0} = v_{1}\leq v_{2}\leq\cdots \leq v_{l}, \end{eqnarray} 由(2.17)式可知 $$K_{2}B_{h}(v_{l-1})\leq K_{2}B_{h}(v_{l}).$$ 由(2.3)式, (2.18)式和引理2.1, 可得 \begin{eqnarray} u_{l+1}\geq u_{l}. \end{eqnarray} 由(2.19)式, 可得 $$K_{2}A_{h}(u_{l}) \leq K_{2}A_{h}(u_{l+1}).$$ 由(2.3)式, (2.20)式和引理2.2, 可得 $$v_{l}\leq v_{l+1}.$$ 由(2.14)式, 若 u^{0} 是一个上解, 则 $$u_{l+1}\leq u^{0}.$$ 由数学归纳法原理, (2.16)和(2.17)式普遍成立. 由格式(2.3)的第二个方程可知, 数列 \{v_{n}\} 的有界性可以通过 \{u_{n}\} 的有界性得到.事实上, v_{n}\leq K_{2}A_{h}(u^{0}), \quad n = 1, 2, \cdots .证毕. 引理2.6 \Lambda 是一个区间且 \beta_c'>0 .除此之外, 我们还有 (\beta_{c}', \infty)\subset\Lambda$$ [0, \beta_{c}')\bigcap\Lambda = \phi$.

为了证明结论成立, 只需要说明:如果$\beta\in\Lambda$, 则对任意$\varepsilon>0$, $\beta+\varepsilon\in\Lambda$.事实上, 令$\beta\in\Lambda$, 取$r\in(0, 1)$使得

$$$B_{\beta+\varepsilon}(rv) = B_{\beta}(v).$$$

$\begin{eqnarray} &&u_{n+1}<(K_{1}-K_{2})u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}} f_{\beta+\varepsilon}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|{}\\ &&{\qquad}{\qquad} +K_{2}v_{n}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I'}_{k}} f_{\beta}(\sqrt{v_{n}^{2}+x^{2}})|\overline{I'}_{k}|, {}\\ &&v_{n+1}+K_{2}v_{n+1}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I'}_{k}} f_{\beta}(\sqrt{v_{n+1}^{2}+x^{2}})|\overline{I'}_{k}|{}\\ && <K_{2}u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}} f_{\beta+\varepsilon}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|. \end{eqnarray}$

$\begin{eqnarray} &&u_{n+1}\leq(K_{1}-K_{2})u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}} f_{\beta+\varepsilon}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|{}\\ &&{\qquad}{\qquad}+K_{2}rv_{n}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I'}_{k}} f_{\beta+\varepsilon}(\sqrt{(rv_{n})^{2}+x^{2}})|\overline{I'}_{k}|, {}\\ &&rv_{n+1}+K_{2}rv_{n+1}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I'}_{k}} f_{\beta+\varepsilon}(\sqrt{(rv_{n+1})^{2}+x^{2}})|\overline{I'}_{k}|{}\\ &&\leq K_{2}u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}} f_{\beta+\varepsilon}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|. \end{eqnarray}$

图 4

$\begin{eqnarray} \beta'_{c}\geq\beta_{c}\geq\beta''_{c}. \end{eqnarray}$

由(2.2), (2.3)和(2.6)式, 容易验证(2.26)式成立.

$$$\beta'_{c}\geq\beta_{c}.$$$

$$$\beta_{c}\geq\beta''_{c}.$$$

$$$\begin{array}{l} u+(K_{2}-K_{1})A_{\beta}(u) = K_{2}B_{\beta}(v), \\ v+K_{2}B_{\beta}(v) = K_{2}A_{\beta}(u). \end{array}$$$

$\begin{eqnarray} &&u_{n+1}+(K_{2}-K_{1})u_{n+1}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I}_{k}}f_{\beta}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}| {}\\ && = K_{2}v_{n}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I'}_{k}}f_{\beta}(\sqrt{v_{n}^{2}+x^{2}})|\overline{I'}_{k}|, {}\\ &&v_{n+1}+K_{2}v_{n+1}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I'}_{k}}f_{\beta}(\sqrt{v_{n+1}^{2}+x^{2}})|\overline{I'}_{k}|{}\\ && = K_{2}u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}}f_{\beta}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|, \ n = 1, 2, \cdots ;\ v = v_{0}, \end{eqnarray}$

$\begin{eqnarray} &&u_{0}+(K_{2}-K_{1})u_{0}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I}_{k}} f_{\beta}(\sqrt{u_{0}^{2} +x^{2}})|\overline{I}_{k}| \leq K_{2}v_{0}\sum\limits_{k}^{N} \min\limits_{x\in\overline{I'}_{k}} f_{\beta}(\sqrt{v_{0}^{2} +x^{2}})|\overline{I'}_{k}|, {}\\ &&v_{0}+K_{2}v_{0} \sum\limits_{k}^{N}\max\limits_{x\in\overline{I'}_{k}} f_{\beta}(\sqrt{v_{0}^{2} +x^{2}})|\overline{I'}_{k}| \leq K_{2}u\sum\limits_{k}^{N} \min\limits_{x\in\overline{I}_{k}} f_{\beta}(\sqrt{u_{0}^{2} +x^{2}})|\overline{I}_{k}|. \end{eqnarray}$

$\begin{eqnarray} &&u_{n+1}+(K_{2}-K_{1})u_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}}f_{\beta}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|{}\\ && = K_{2}v_{n}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I'}_{k}}f_{\beta}(\sqrt{v_{n}^{2}+x^{2}})|\overline{I'}_{k}|, {}\\ &&v_{n+1}+K_{2}v_{n+1}\sum\limits_{k}^{N}\min\limits_{x\in\overline{I'}_{k}}f_{\beta}(\sqrt{v_{n+1}^{2}+x^{2}})|\overline{I'}_{k}|{}\\ && = K_{2}u_{n+1}\sum\limits_{k}^{N}\max\limits_{x\in\overline{I}_{k}}f_{\beta}(\sqrt{u_{n+1}^{2}+x^{2}})|\overline{I}_{k}|, \ n = 1, 2, \cdots ;\ v = v_{0}. \end{eqnarray}$

$\begin{eqnarray} A'_{h}(u)& = &u\sum\limits_{k}^{N}\min\limits_{x\in\overline{I}_{k}} f_{\beta}(\sqrt{u^{2} +x^{2}})|\overline{I}_{k}|, \end{eqnarray}$

$\begin{eqnarray} P_{h}(u)& = &u+(K_{2}-K_{1})u\sum\limits_{k}^{N} \max\limits_{x\in\overline{I}_{k}}f_{\beta} (\sqrt{u^{2}+x^{2}})|\overline{I}_{k}|. \end{eqnarray}$

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