Ni<sub>3</sub>S<sub>2</sub> -多孔镍@泡沫镍电极制备及其赝电容性能

赵少飞; 刘鹏; 李婉萍; 曾小红; 余林; 曾华强

doi:10.6054/j.jscnun.2020090

Ni₃S₂ -多孔镍@泡沫镍电极制备及其赝电容性能

赵少飞¹,
刘鹏¹,
李婉萍¹,
曾小红¹,
余林^1, ,,
曾华强^{1, 2, ,}

1.
广东工业大学轻工化工学院，广州 510006
2.
纳米生物实验室，新加坡 138669

基金项目:

国家自然科学基金项目 21306026

国家自然科学基金项目 21576054

国家自然科学基金项目 51678160

详细信息

通讯作者:
余林，教授，Email:gych@gdut.edu.cn

曾华强，教授，Email:hqzeng@nbl.a-star.edu.sg

中图分类号: TQ152;O646
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出版历程
- 收稿日期: 2020-02-29
- 网络出版日期: 2021-01-04
- 刊出日期: 2020-12-24

Preparation of Ni₃S₂-Ni@Ni Foam Electrode and Its Pseudocapacitance Properties

1.
School of Chemical Engineering & Light Industry, Guangdong University of Technology, Guangzhou 510006, China
2.
Nanobio laboratory, Singapore 138669, Singapore

摘要

摘要: 采用氢气鼓泡法预处理泡沫镍集流体，通过循环伏安法原位电沉积Ni₃S₂活性材料，制备了Ni₃S₂-多孔镍@泡沫镍(Ni₃S₂-Ni@NF)电极.通过扫描电子显微镜(SEM)、X射线衍射(XRD)、光电子能谱(XPS)和拉曼光谱(Raman)对物相结构进行了表征，并利用电化学工作站测试了电化学性能.结果表明：制备的Ni₃S₂-Ni@NF材料表现出优异的赝电容性能，在2 mA/cm²的电流密度下，比电容达到4.56 F/cm²，且具有优异的倍率性能(20 mA/cm²的电流密度下，比电容达到4.06 F/cm²)和循环性能(10 mA/cm²的电流密度下，循环1 000次的比电容保持率约73%).
- 氢气鼓泡 /
- 电沉积 /
- 超级电容器 /
- 比电容 /
- 循环性能
Abstract: Ni₃S₂ coated on well-designed hierarchical porous Ni@Ni foam (Ni₃S₂-Ni@NF) was fabricated with the novel method of hydrogen-bubble template electrodeposition and in-situ cyclic voltammetry electrodeposition. The physical phase and morphological properties of the as prepared samples were characterized with scanning electron microscopy (SEM), X-ray diffraction (XRD), X-ray photoelectron spectroscopy (XPS) and Raman spectroscopy. The electrochemical capacitance properties were tested with an electrochemical workstation. The Ni₃S₂-Ni@NF electrode demonstrates a much higher specific capacitance of 4.56 F/cm² at a current density of 2 mA/cm². The Ni₃S₂-Ni@NF also exhibits excellent rate capability (4.06 F/cm² at a current density of 20 mA/cm²) and cycle performance (73% after 1 000 cycles).
- hydrogen-bubble /
- electrodeposition /
- supercapacitance /
- specific capacitance /
- cycle performance

HTML全文

近几十年来，分数阶微分方程及其边值问题受到了许多学者的关注，在很多科学领域中都有着广泛的应用。目前，关于分数阶微分方程边值问题的研究已经有很多成果^[1-13]，但是关于边值条件中带不同分数阶导数的研究相对较少。

薛益民等^[1]运用Guo-Krasnosel'skii's不动点定理，得到如下分数阶微分方程正解的存在性：

$\left\{\begin{array}{l} D^\alpha u(t)+f(t, u(t))=0 \quad(0<t<1), \\ u(0)=D^\beta u(0)=D^\beta u(1)=0, \end{array}\right.$

其中, D^α(2 < α≤3)为Rimann-Liouvile分数阶导数。

张凯斌和陈鹏玉^[3]运用非紧性测度的估计技巧与凝聚映射的不动点指数理论，得到如下分数阶微分方程正解的存在性：

$\left\{\begin{array}{l} -D_{0+}^\alpha u(t)=f(t, u(t)) \quad(0<t<1), \\ u(0)=u^{\prime}(0)=u^{\prime}(1)=\theta, \end{array}\right.$

其中, D₀₊^α为Rimann-Liouvile分数阶导数，2 < α≤3。

受文献[1]、[3]的启发，本文考虑如下带有分数阶导数边值条件的分数阶微分方程

$\left\{\begin{array}{l} D_{0+}^v u(t)+h(t) f(t, u(t))=0 \quad(0<t<1, n-1<v \leqslant n), \\ u(0)=u^{\prime}(0)=u^{\prime \prime}(0)=\cdots=u^{(n-2)}(0)=0 \quad(n \geqslant 3), \\ \left(D_{0+}^\alpha u(t)\right)_{t=1}=\sum\limits_{i=1}^{m-2} \beta_i\left(D_{0+}^{\alpha_i} u(t)\right)_{t=\eta_i} \quad\left(1 \leqslant \alpha, \alpha_i \leqslant n-2\right), \end{array}\right.$

(1)

其中, η_i∈(0, 1), 0 < η₁ < η₂ < … < η_m-2 < 1, β_i∈[0, ∞)。需要指出的是，这里的边值条件中带有不同阶数的分数阶导数。

文中首先构建其格林函数，得到相应的相关性质；其次, 运用凸泛函上的不动点指数定理来计算不动点指数，从而得到了方程(1)至少存在一个正解的结论; 最后, 通过一个例子来说明定理的具体应用。

1. 预备知识

首先，给出一些必要的定义和引理，推导出相应的带有分数阶导数边值条件的分数阶微分方程的格林函数，并给出格林函数的一些性质；然后，将方程(1)转化为一个等价的积分方程。

定义1^[6] 函数y: (0, +∞)→ $\mathbb{R}$ 的v>0阶Riemann-Liouville积分定义如下

$I_{0+}^v y(t)=\frac{1}{\varGamma(v)} \int_0^t(t-s)^{v-1} y(s) \mathrm{d} s,$

其中, 等式右边是在(0, +∞)上逐点定义的。

定义2^[6] 函数y: (0, +∞)→ $\mathbb{R}$ 的v>0阶Riemann-Liouville微分定义如下

$D_{0+}^v y(t)=\frac{1}{\varGamma(n-v)}\left(\frac{\mathrm{d}}{\mathrm{d} t}\right)^n \int_0^t \frac{y(s)}{(t-s)^{v-n+1}} \mathrm{~d} s,$

其中，等式右边是在(0, +∞)上逐点定义的，n=[α]+1。

引理1^[6] 假设u∈C(0, 1)∩L[0, 1], 有v>0阶导数D^v₀₊∈C(0, 1)∩L[0, 1], 则

$I_{0+}^v D_{0+}^v u(t)=u(t)+C_1 t^{v-1}+C_2 t^{v-2}+\cdots+C_N t^{v-N},$

其中，C_i∈ $\mathbb{R}$ (i=1, 2, …, N), N是大于或等于v的最小整数。

为下文叙述方便，现给出如下假设条件：

(H₁) $\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_{i=1}^{m-2} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}>0$ 。

(H₂)h: (0, 1)→[0, ∞)连续, h(t)不恒等于0。允许h(t)在t=0, 1处奇异，且

$0<\int_0^1 G(1, t) h(t) \mathrm{d} t<+\infty \text { 。 }$

(2)

(H₃)f: [0, 1]×[0, +∞)→[0, +∞)连续。

引理2 给定y∈C[0, 1], 边值问题

$\left\{\begin{array}{l} D_{0+}^v u(t)+y(t)=0 \quad(0<t<1, n-1<v \leqslant n), \\ u(0)=u^{\prime}(0)=u^{\prime \prime}(0)=\cdots=u^{(n-2)}(0)=0 \quad(n \geqslant 3), \\ \left(D_{0+}^\alpha u(t)\right)_{t=1}=\sum\limits_{i=1}^{m-2} \beta_i\left(D_{0+}^{\alpha_i} u(t)\right)_{t=\eta_i} \\ \left(1 \leqslant \alpha, \alpha_i \leqslant n-2\right) \end{array}\right.$

(3)

有唯一解

$u(t)=\int_0^1 G(t, s) y(s) \mathrm{d} s,$

这里η_i∈(0, 1), 0 < η₁ < η₂ < … < η_m-2 < 1, β_i∈[0, ∞), 其中

$G(t, s)=\left\{\begin{array}{l} \frac{1}{p(0) \varGamma(v)} p(s)(1-s)^{v-\alpha-1} t^{v-1}-(t-s)^{v-1} p(0) \\ (0 \leqslant s \leqslant t \leqslant 1), \\ \frac{1}{p(0) \varGamma(v)}(1-s)^{v-\alpha-1} p(s) t^{v-1} \quad(0 \leqslant t \leqslant s \leqslant 1), \end{array}\right.$

$p(s)=\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_{s \leqslant \eta_i} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\frac{\eta_i-s}{1-s}\right)^{v-\alpha_i-1}(1-s)^{\alpha-\alpha_i} 。$

证明应用引理1，将微分方程(3)转化为等价的积分方程

$u(t)=C_1 t^{v-1}+C_2 t^{v-2}+\cdots+C_n t^{v-n}-I_{0+}^v y(s) 。$

由u(0)=u′(0)=u″(0)=…=u^(n-2)(0)=0, 可得C₂=C₃=…=C_n=0。又由 $D^\alpha\left[t^{v-1}\right]=\frac{\varGamma(v)}{\varGamma(v-\alpha)} t^{v-\alpha-1}$ , 再代入边值条件 $\left(D_{0+}^\alpha u(t)\right)_{t=1}=\sum\limits_{i=1}^{m-2} \beta_i\left(D_{0+}^{\alpha_i} u(t)\right)_{t=\eta_i}$ , 可得

$\begin{aligned} &C_1 \frac{\varGamma(v)}{\varGamma(v-\alpha)} 1^{v-\alpha-1}-\frac{1}{\varGamma(v-\alpha)} \int_0^1(1-s)^{v-\alpha-1} y(s) \mathrm{d} s=\\ & \sum\limits_{i=1}^{m-2} \beta_i\left[C_1 \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}-\right. \\ & \left.\frac{1}{\varGamma\left(v-\alpha_i\right)} \int_0^{\eta_i}\left(\eta_i-s\right)^{v-\alpha_i-1} y(s) \mathrm{d} s\right], \end{aligned}$

整理得

$\begin{aligned} C_1= & \frac{1}{\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_{i=1}^{m-2} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}} \times \\ & {\left[\frac{1}{\varGamma(v-\alpha)} \int_0^1(1-s)^{v-\alpha-1} y(s) \mathrm{d} s^{-}\right.} \\ & \left.\sum\limits_{i=1}^{m-2} \beta_i \frac{1}{\varGamma\left(v-\alpha_i\right)} \int_0^{\eta_i}\left(\eta_i-s\right)^{v-\alpha_i-1} y(s) \mathrm{d} s\right] 。 \end{aligned}$

于是

$\begin{aligned} u(t)= & \frac{t^{v-1}}{\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_{i=1}^{m-2} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}} \times \\ & {\left[\frac{1}{\varGamma(v-\alpha)} \int_0^1(1-s)^{v-\alpha-1} y(s) \mathrm{d} s-\right.} \\ & \left.\sum\limits_{i=1}^{m-2} \beta_i \frac{1}{\varGamma\left(v-\alpha_i\right)} \int_0^{\eta_i}\left(\eta_i-s\right)^{v-\alpha_i-1} y(s) \mathrm{d} s\right]- \\ & \frac{1}{\varGamma(v)} \int_0^t(t-s)^{v-1} y(s) \mathrm{d} s=\int_0^1\left[\frac{(1-s)^{v-\alpha-1}}{\varGamma(v) p(0)} \times\right. \\ & \left.p(s) t^{v-1} y(s)\right] \mathrm{d} s-\frac{1}{\varGamma(v)} \int_0^t(t-s)^{v-1} y(s) \mathrm{d} s= \\ & \int_0^t \frac{(1-s)^{v-\alpha-1} p(s) t^{v-1}-(t-s)^{v-1} p(0)}{\varGamma(v) p(0)} y(s) \mathrm{d} s+ \\ & \int_t^1 \frac{(1-s)^{v-\alpha-1} p(s) t^{v-1}}{\varGamma(v) p(0)} y(s) \mathrm{d} s=\int_0^1 G(t, s) y(s) \mathrm{d} s 。 \end{aligned}$

证毕。

引理3 函数p(s)在0, 1上单调不减且恒正。

证明因为

$\begin{aligned} & p^{\prime}(s)=\sum\limits_{s \leqslant \eta_i} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(v-\alpha_i-1\right)\left(\eta_i-s\right)^{v-\alpha_i-2} \times \\ &(1-s)^{\alpha_i+1-v}(1-s)^{\alpha-\alpha_i}+\sum\limits_{s \leqslant \eta_i}\left[\beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\alpha_i+1-v\right) \times\right. \\ &\left.\left(\eta_i-s\right)^{v-\alpha_i-1}(1-s)^{\alpha_i-v}(1-s)^{\alpha-\alpha_i}\right]+ \\ & \sum\limits_{s \leqslant \eta_i}\left[\beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\alpha-\alpha_i\right)\left(\eta_i-s\right)^{v-\alpha_i-1}(1-s)^{\alpha_i+1-v} \times\right. \\ &\left.(1-s)^{\alpha-\alpha_i-1}\right]=\sum\limits_{s \leqslant \eta_i}\left[\beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i-s\right)^{v-\alpha_i-2} \times\right. \\ &(1-s)^{\alpha_i-v}(1-s)^{\alpha-\alpha_i}\left(\left(v-\alpha_i-1\right)(1-s)+\right. \\ &\left.\left.\left(\alpha_i+1-v\right)\left(\eta_i-s\right)+\left(\alpha-\alpha_i\right)\left(\eta_i-s\right)\right)\right]=\\ & \sum\limits_{s \leqslant \eta_i}\left[\beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i-s\right)^{v-\alpha_i-2}(1-s)^{\alpha-v} \times\right. \\ & \left(\left(v-\alpha_i-1\right)(1-s)-\left(v-\alpha_i-1\right)\left(\eta_i-s\right)+\right. \\ & \left.\left.\left(\alpha-\alpha_i\right)\left(\eta_i-s\right)\right)\right] \geqslant 0, \end{aligned}$

故p(s)单调不减。

又根据假设H₁知,

$p(0)=\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_{i=1}^{m-2} \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}>0,$

从而知p(s)≥p(0)>0。证毕。

引理4 函数G(t, s)具有如下性质：

(1) ∀t, s∈[0, 1], 有G(t, s)≥0;

(2) ∀t∈[0, 1], 有G(t, s)≤G(1, s);

(3) ∀ $\frac{1}{4}$ ≤t≤34, 有G(t, s)≥( $\frac{1}{4}$ )^v-1G(1, s)。

证明 (1)当0 < s≤t≤1时, 有

$\begin{aligned} G(t, s)= & \frac{1}{p(0) \varGamma(v)}\left[t^{v-1} p(s)(1-s)^{v-\alpha-1}-p(0)(t-s)^{v-1}\right)= \\ & \frac{t^{v-1}}{p(0) \varGamma(v)}\left[p(s)(1-s)^{v-\alpha-1}-p(0)\left(1-\frac{s}{t}\right)^{v-1}\right] \geqslant \\ & \frac{t^{v-1} p(s)}{p(0) \varGamma(v)}\left[(1-s)^{v-\alpha-1}-\left(1-\frac{s}{t}\right)^{v-1}\right] \geqslant \\ & \frac{t^{v-1} p(s)}{p(0) \varGamma(v)}\left[(1-s)^{v-1}-\left(1-\frac{s}{t}\right)^{v-1}\right] \geqslant 0 \text { 。 } \end{aligned}$

当0 < t≤s≤1时, 显然有G(t, s)≥0。

综上可知，∀t, s∈[0, 1], 有G(t, s)≥0。

(2) 因为

$\begin{aligned} & \frac{\partial}{\partial t} G(t, s)= \\ & \left\{\begin{array}{l} \frac{1}{p(0) \varGamma(v)}(v-1) t^{v-2} p(s)(1-s)^{v-\alpha-1}-(v-1)(t-s)^{v-2} p(0) \\ \quad(0 \leqslant s \leqslant t \leqslant 1), \\ \frac{1}{p(0) \varGamma(v)}(v-1)(1-s)^{v-\alpha-1} p(s) t^{v-2} \quad(0 \leqslant t \leqslant s \leqslant 1), \end{array}\right. \end{aligned}$

所以，当0 < s≤t≤1时, 有

$\begin{aligned} \frac{\partial}{\partial t} &G(t, s)=\frac{1}{p(0) \varGamma(v)}\left[(v-1) t^{v-2} p(s)(1-s)^{v-\alpha-1}-\right. \\ & \left.(v-1)(t-s)^{v-2} p(0)\right] \geqslant \frac{(v-1) t^{v-2}}{\varGamma(v)}\left[(1-s)^{v-\alpha-1}-\right. \\ & \left.\left(1-\frac{s}{t}\right)^{v-2}\right] \geqslant 0 \text { 。 } \end{aligned}$

当0 < t≤s≤1时, 显然有 $\frac{\partial}{\partial t}$ G(t, s)≥0。

综上可知，∀t∈[0, 1], 有 $\frac{\partial}{\partial t}$ G(t, s)≥0, 所以G(t, s)关于t单调不减。因此, ∀t∈[0, 1], 有G(t, s)≤G(1, s)。

(3) 当1/4≤t≤3/4且0≤s≤t时, 有

$\begin{aligned} G(t, s)= & \frac{t^{v-1}}{p(0) \varGamma(v)}\left[p(s)(1-s)^{v-\alpha-1}-p(0) \times\right. \\ & \left.\left(1-\frac{s}{t}\right)^{v-1}\right] \geqslant \frac{t^{v-1}}{p(0) \varGamma(v)}\left[p(s)(1-s)^{v-\alpha-1}-\right. \\ & \left.p(0)(1-s)^{v-1}\right]=t^{v-1} G(1, s) \geqslant\left(\frac{1}{4}\right)^{v-1} G(1, s) 。 \end{aligned}$

当1/4≤t≤3/4且0 < t≤s时, 有

$\begin{aligned} G(t, s)= & \frac{1}{p(0) \varGamma(v)}(1-s)^{v-\alpha-1} p(s) t^{v-1} \geqslant t^{v-1} G(1, s) \geqslant \\ & \left(\frac{1}{4}\right)^{v-1} G(1, s)。 \end{aligned}$

证毕。

在Banach空间C[0, 1]中，定义范数为‖u‖= $\max\limits_{0 \leqslant t \leqslant 1}|u(t)|$ , 令P={u∈C[0, 1]: u(t)≥0, t∈[0, 1]}, 则P是C[0, 1]上的正锥。取P₁={u∈P: $\min\limits_{1 / 4 \leqslant t \leqslant 3 / 4}$ u(t)≥l‖u‖}, 其中l=( $\frac{1}{4}$ )^v-1。

定义如下算子：

$(A u)(t)=\int_0^1 G(t, s) h(s) f(s, u(s)) \mathrm{d} s \quad(t \in[0, 1]) 。$

接下来证明算子A的全连续性。

引理5 设条件(H₁)~(H₃)满足, 则算子A: P₁→P₁全连续。

证明由引理4可知

$\|A u\| \leqslant \int_0^1 G(1, s) h(s) f(s, u(s)) \mathrm{d} s,$

且

$\begin{gathered} \min\limits_{1 / 4 \leqslant t \leqslant 3 / 4}(A u)(t) \geqslant \min\limits_{1 / 4 \leqslant t \leqslant 3 / 4} \int_0^1 t^{v-1} G(1, s) h(s) f(s, u(s)) \mathrm{d} s \geqslant \\ \left(\frac{1}{4}\right)^{v-1}\|A u\|, \end{gathered}$

从而A: P₁→P₁, 且A(P₁)⊂P₁。由Azela-Ascoli定理知, 算子A: P₁→P₁全连续。证毕。

下面介绍凸泛函的2个不动点指数引理。

定义3^[14] 对于锥P上的泛函ρ: P→ $\mathbb{R}$ , 如果∀x, y∈P, t∈[0, 1]，满足

$\rho(t x+(1-t) y) \leqslant t \rho(x)+(1-t) \rho(y),$

则称ρ是锥P上的凸泛函。

引理6^[14] 设P是E中的锥，Ω是E中的有界开集，且θ∈Ω。假设算子A: P∩Ω→P全连续，ρ: P→[0, +∞)是凸泛函，且满足ρ(θ)=0, 并对∀x≠θ, ρ(x)>0。如果ρ(Ax)≤ρ(x), 且当x∈P∩∂Ω时, Ax≠x, 则不动点指数i(A, P∩Ω, P)=1。

引理7^[14] 设P是E中的锥，Ω是E中的有界开集。假设算子A: P∩Ω→P全连续，ρ: P→[0, +∞)是一致连续的凸泛函，且满足ρ(θ)=0, 并对∀x≠θ, ρ(x)>0。如果

(i) $\inf\limits_{x \in P \cap \partial \varOmega} \rho(x)>0$ ;

(ii) ρ(Ax)≥ρ(x)且对∀x∈P∩∂Ω, Ax≠x, 则不动点指数i(A, P∩Ω, P)=0。

2. 主要结论

令

$h_0=\int_0^1 G(1, t) h(t) \mathrm{d} t, h_\tau=\int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t,$

显然有h₀≥h_τ>0。

定理1 假设条件(H₁)~(H₃)成立, s∈[0, 1], 如果存在常数a和b, 使得当a, b>0时, 有

(i) b < a;

(ii) f(s, u(s))≤h₀^-1u (u≤bl^-1h_τ^-1);

(iii) f(s, u(s))≥h_τ^-1u (ah_τ^-1l≤u≤ah_τ^-1l^-1),

则带有分数阶导数边值条件的分数阶微分方程(1)至少存在一个正解。

证明令

$\rho_1(u)=\int_0^1 G(1, t) h(t) u(t) \mathrm{d} t,$

则ρ₁: P₁→[0, +∞)是一致连续的凸泛函，且ρ₁(θ)=0。

∀u∈P₁\{θ}, 有

$\begin{aligned} \rho_1(u) \geqslant & \int_{1 / 4}^{3 / 4} G(1, t) h(t) u(t) \mathrm{d} t \geqslant \\ & l\|u\| \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t>0 。 \end{aligned}$

设Ω₁={u∈C[0, 1]|ρ₁(u) < b}。显然Ω₁是C[0, 1]上的开集，且θ∈Ω₁。

如果u∈P₁∩Ω₁, 则

$\begin{aligned} b \geqslant & \rho_1(u)=\int_0^1 G(1, t) h(t) u(t) \mathrm{d} t \geqslant \\ & l\|u\| \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t=l\|u\| h_\tau 。 \end{aligned}$

因此‖u‖≤bl^-1h_τ^-1, 这意味着P₁∩Ω₁是有界的。

如果u∈P₁∩∂Ω₁, 则ρ₁(u)=b且‖u‖≤bl^-1h_τ^-1，因此

$\begin{aligned} \rho_1(A u)= & \int_0^1\left[G(1, t) h(t) \int_0^1 G(t, s) h(s) f(s, u(s)) \mathrm{d} s\right] \mathrm{d} t \leqslant \\ & \int_0^1\left[G(1, t) h(t) \int_0^1 G(1, s) h(s) f(s, u(s)) \mathrm{d} s\right] \mathrm{d} t \leqslant \\ & \int_0^1 G(1, t) h(t) \mathrm{d} t \int_0^1 G(1, s) h(s) h_0^{-1} u(s) \mathrm{d} s= \\ & \int_0^1 G(1, s) h(s) u(s) \mathrm{d} s=\rho_1(u) 。 \end{aligned}$

假设A在P₁∩∂Ω₁上没有不动点，则由引理6知

$i\left(A, P_1 \cap \varOmega_1, P_1\right)=1。$

令

$\rho_2(u)=\int_{1 / 4}^{3 / 4} G(1, t) h(t) u(t) \mathrm{d} t,$

则ρ₂: P₁→[0, +∞)是一致连续的凸泛函，且ρ₂(θ)=0，ρ₂(u)>0(u∈P₁\θ)。

设Ω₂={u∈C[0, 1]|ρ₂(u), 显然Ω₂是C[0, 1]上的开集。

如果u∈P₁∩Ω₂, 则

$a \geqslant \rho_2(u) \geqslant l\|u\| \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t=l\|u\| h_{\tau }。$

于是‖u‖≤al^-1h_τ^-1, 这意味着P₁∩Ω₂是有界的。

如果u∈P₁∩∂Ω₂, 则ρ₂(u)=a且‖u‖≤al^-1h_τ^-1。由于

$\begin{aligned} a= & \int_{1 / 4}^{3 / 4} G(1, t) h(t) u(t) \mathrm{d} t \leqslant \\ & \|u\| \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t=\|u\| h_\tau, \end{aligned}$

则‖u‖≥ah_τ^-1, 于是

$\min\limits_{1 / 4 \leqslant l \leqslant 3 / 4} u(t) \geqslant l\|u\| \geqslant {lah}_\tau^{-1},$

所以

$\begin{aligned} \rho_2(A u)= & \int_{1 / 4}^{3 / 4}\left[G(1, t) h(t) \int_0^1 G(t, s) h(s) f(s, u(s)) \mathrm{d} s\right] \mathrm{d} t \geqslant \\ & \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t \int_{1 / 4}^{3 / 4} G(1, s) h(s) h_\tau^{-1} u(s) \mathrm{d} s= \\ & \int_{1 / 4}^{3 / 4} G(t, s) h(s) \mathrm{d} s=\rho_2(u) 。 \end{aligned}$

假设A在P₁∩∂Ω₂上没有不动点，由引理7知

$i\left(A, P_1 \cap \varOmega_2, P_1\right)=0 \text { 。 }$

∀u∈P₁∩Ω₁, 有

$\rho_2(u) \leqslant \int_0^1 G(1, t) h(t) u(t) \mathrm{d} t \leqslant \rho_1(u) \leqslant b<a,$

则P₁∩Ω₁⊂P₁∩Ω₂, 从而有

$i\left(A, P_1 \cap\left(\varOmega_2 \ \bar{\varOmega}_1\right), P_1\right)=-1,$

说明算子A在P₁∩(Ω₂\Ω₁)上至少有一个不动点, 即微分方程(1)至少存在一个正解。

定理2 假设条件(H₁)~(H₃)成立, s∈[0, 1], 如果存在常数a和b, 使得当0 < b时，有

(i) b < al²h_τ²h₀^-1;

(ii) f(s, u(s))≥h_τ^-1u (blh_τ^-1≤u≤bl^-1h_τ^-1);

(iii) f(s, u(s))≤ah₀^-1 (u≤al^-1),

则带有分数阶导数边值条件的分数阶微分方程(1)至少存在一个正解。

证明根据(i)有

$b l^{-1} h_\tau^{-1}<a l^2 h_\tau^2 h_0^{-1} l^{-1} h_\tau^{-1}=a l h_\tau h_0^{-1}<a l^{-1} \text { 。 }$

∀u≤bl^-1h_τ^-1, 有

$h_\tau^{-1} u \leqslant h_\tau^{-1} b l^{-1} h_\tau^{-1}<\left(h_\tau^{-1}\right)^2 a l^2 h_\tau^2 h_0^{-1} l^{-1}=a l h_0^{-1}<a h_0^{-1} 。$

设

$\rho_1(u)=\max\limits_{1 / 4 \leqslant t \leqslant 3 / 4} u(t), \rho_2(u)=\int_{1 / 4}^{3 / 4} G(1, t) h(t) u(t) \mathrm{d} t,$

显然有ρ_i: P₁→[0, +∞)是一致连续凸泛函，且ρ_i(θ)=0(i=1, 2)。

∀u∈P₁\{θ}, 有

$\rho_1(u) \geqslant l\|u\|>0, \rho_2(u) \geqslant \int_{1 / 4}^{3 / 4} G(1, t) h(t) l\|u\| \mathrm{d} t>0 \text { 。 }$

令Ω₁={u∈C[0, 1]|ρ₂(u) < b}, Ω₂={u∈C[0, 1]|ρ₁(u)。显然Ω₁和Ω₂是C[0, 1]上的开集，且θ∈Ω₁。

如果u∈P₁∩Ω₁, 则

$b \geqslant \rho_2(u) \geqslant l\|u\| \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t=l\|u\| h_\tau 。$

因此，‖u‖≤bl^-1h_τ^-1, 这意味着P₁∩Ω₁是有界的。进一步有

$\rho_1(u) \leqslant\|u\| \leqslant b l^{-1} h_\tau^{-1}<a l^2 h_\tau^2 h_0^{-1} l^{-1} h_\tau^{-1}=a l h_\tau h_0^{-1}<a l<a,$

所以P₁∩Ω₁⊂P₁∩Ω₂。

如果u∈P₁∩Ω₂, 则

$a \geqslant \rho_1(u) \geqslant l\|u\|,$

于是‖u‖≤al^-1, 这意味着P₁∩Ω₂是有界的。

假设A在P₁∩∂Ω₁和P₁∩∂Ω₂上没有不动点。如果u∈P₁∩∂Ω₁, 则b=ρ₂(u)≤‖u‖h_τ, 且

$\min\limits_{1 / 4 \leqslant l \leqslant 3 / 4} u(t) \geqslant l\|u\| \geqslant l b h_\tau^{-1},$

则

$\begin{array}{r} \rho_2(A u)=\int_{1 / 4}^{3 / 4} G(1, t) h(t)\left[\int_0^1 G(t, s) h(s) f(s, u(s)) \mathrm{d} s\right] \mathrm{d} t \geqslant \\ \int_{1 / 4}^{3 / 4} G(1, t) h(t) \mathrm{d} t \int_{1 / 4}^{3 / 4} G(1, s) h(s) h_\tau^{-1} u(s) \mathrm{d} s=\rho_2(u)。 \end{array}$

所以, 由引理7知

$i\left(A, P_1 \cap \varOmega_1, P_1\right)=0_{\text {。 }}$

如果u∈P₁∩∂Ω₂, 则

$\begin{gathered} \rho_1(A u)=\max\limits_{1 / 4 \leqslant t \leqslant 3 / 4} \int_0^1 G(t, s) h(s) f(s, u(s)) \mathrm{d} s \leqslant \\ \int_0^1 G(1, s) h(s) a h_0^{-1} \mathrm{~d} s=a=\rho_1(u) 。 \end{gathered}$

由引理6知，

$i\left(A, P_1 \cap \varOmega_2, P_1\right)=1。$

综上可得

$i\left(A, P_1 \cap\left(\varOmega_2 \backslash \bar{\varOmega}_1\right), P_1\right)=1,$

说明算子A在P₁∩(Ω₂\Ω₁)上至少有一个不动点, 即微分方程(1)至少存在一个正解。

3. 数值例子

为了说明定理的应用性, 下面给出一个具体的实例。

例1 考虑如下的分数阶微分方程

$\left\{\begin{array}{l} D_{0+}^{\frac{9}{2}} u(t)+\frac{0.6 \mathrm{e}^{0.2021 u}}{1+t}=0 \quad(0<t<1), \\ u(0)=u^{\prime}(0)=u^{\prime \prime}(0)=u^{\prime \prime \prime}(0)=0, \\ \left(D_{0+}^{\frac{5}{2}} u(t)\right)_{t=1}=\frac{1}{8}\left(D_{0+}^{\frac{3}{2}} u(t)\right)_{t=\frac{1}{9}}+\frac{1}{6}\left(D_{0+}^{\frac{1}{2}} u(t)\right)_{t=\frac{1}{3}}, \end{array}\right.$

(5)

其中 $v=\frac{9}{2}, \alpha=\frac{5}{2}, n=5, \alpha_1=\frac{3}{2}, \alpha_2=\frac{1}{2}, \eta_1=\frac{1}{9}, \eta_2=\frac{1}{3}$ , $\beta_1=\frac{1}{8}, \beta_2=\frac{1}{6}, l=\left(\frac{1}{4}\right)^{\frac{7}{2}}=\frac{1}{128}, h(t)=\frac{1}{1+t}$ , f(t, u(t))=0.6e^{0.202 1u}。取a≈6.18, b≈0.001。由

$\begin{gathered} p(0)=\frac{\varGamma(v)}{\varGamma(v-\alpha)}-\sum\limits_i \beta_i \frac{\varGamma(v)}{\varGamma\left(v-\alpha_i\right)}\left(\eta_i\right)^{v-\alpha_i-1}= \\ \varGamma\left(\frac{9}{2}\right)\left(1-\frac{1}{1296}-\frac{1}{972}\right)>0, \end{gathered}$

且

$\begin{gathered} G(1, t) \leqslant \frac{1}{p(0)} \approx 0.0096 \quad(0<t<1), \\ G(1, t) \geqslant \frac{0.2421875}{\varGamma(9 / 2)} \approx 0.0023 \quad\left(\frac{1}{4} \leqslant t \leqslant \frac{3}{4}\right) 。 \end{gathered}$

取

$h_0=\int_0^1 G(1, t) \frac{1}{1+t} \mathrm{~d} t, h_\tau=\int_{1 / 4}^{3 / 4} G(1, t) \frac{1}{1+t},$

有

$h_0=\int_0^1 G(1, t) \frac{1}{1+t} \mathrm{~d} t, h_\tau=\int_{1 / 4}^{3 / 4} G(1, t) \frac{1}{1+t},$

经过推导，当u≤39.752 6≤bl^-1h_τ^-1时，f(t, u(t))≤150.757 4u≤h₀^-1u; 当al^-1h_τ^-1≥u≥62.024 6≥alh_τ^-1时，f(t, u(t))≥1 284.652 9u≥h_τ^-1u。

综上可知，微分方程(5)满足定理1的3个条件，则该方程至少存在一个正解。

图 1 不同材料的SEM图

Figure 1. The SEM images of different materials

下载: 全尺寸图片幻灯片

图 2 不同材料的XPS谱

Figure 2. The XPS spectra of different materials

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图 3 Ni₃S₂-Ni@NF的XPS谱

Figure 3. The XPS spectra of Ni₃S₂-Ni@NF

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图 4 不同材料在5 mV/s扫速下的CV曲线

Figure 4. The CV curves of different samples at a scan rate of 5 mV/s

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图 5 Ni₃S₂-Ni@NF在不同扫速下的CV曲线及峰电流与扫速的关系

Figure 5. The CV curves of the Ni₃S₂-Ni@NF at various scan rates and the relationship between the peak current and the scan rate

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图 6 Ni₃S₂-Ni@NF在不同电流密度下的放电曲线及比电容

Figure 6. The discharge curves of Ni₃S₂-Ni@NF and the speci-fic capacitance of different samples under various current densities

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图 7 不同材料的循环曲线

Figure 7. The cycling performance of different samples

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