基于VCI的2003—2017年广东省干旱时空变化特征分析

李维娇; 王云鹏

doi:10.6054/j.jscnun.2020048

基于VCI的2003—2017年广东省干旱时空变化特征分析

李维娇^{1, 2},
王云鹏^1, ,

1.
中国科学院广州地球化学研究所，广州 510640
2.
中国科学院大学，北京 100049

基金项目:

国家自然科学基金委-广东联合基金重点项目 U1901215

详细信息

通讯作者:
王云鹏，研究员, Email:wangyp@gig.ac.cn

中图分类号: X43
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出版历程
- 收稿日期: 2020-01-20
- 网络出版日期: 2021-03-21
- 刊出日期: 2020-06-24

An Analysis of the Spatial-Temporal Characteristics of Drought in Guangdong Based on Vegetation Condition Index from 2003 to 2017

LI Weijiao^{1, 2},
WANG Yunpeng^1, ,

1.
Guangzhou Institute of Geochemistry, Chinese Academy of Sciences, Guangzhou 510640, China
2.
University of Chinese Academy of Sciences, Beijing 100049, China

摘要

摘要: 为了研究近年来广东省干旱情况的时空变化特征，利用8天250 m分辨率的MODIS地表反射率数据，计算了2003—2017年广东省植被状态指数(Vegetation Condition Index, VCI)，并以VCI为干旱指标，采用趋势分析、频率分析和MK检验等方法，对全省干旱的时空特征和趋势进行了分析.结果表明：广东省2003—2017年间的VCI整体呈现显著的上升趋势，80.3%以上区域的VCI变化趋势倾向率为正值，表示全省大部分地区旱情逐年减轻；全省大部分区域发生干旱的频率在0.2~0.6之间，以中旱为主(发生频率：0.2~0.4)，重旱较少出现(发生频率：0~0.2)；广东省总体干旱面积占比逐年降低，中旱和重旱面积占比也呈现下降的趋势；2017年与2003年相比，总体干旱面积占比减少了39.5%；总体上看，广东省近年来干旱严重程度在逐年减轻，干旱面积也逐年减少.
- 广东省 /
- 中分辨率成像光谱仪 /
- 植被状态指数 /
- 干旱
Abstract: The temporal and spatial changes of drought in Guangdong Province in recent years were studied. The vegetation condition index (VCI) was calculated using MODIS surface reflectance data of 8 days and 250m resolution. The spatial and temporal characteristics and trend of drought in Guangdong were analyzed using trend analysis, frequency analysis and the Mann-Kenddall trend test. The results show that the average VCI of Guangdong has been rising significantly from 2003 to 2017, and the trend rate of VCI change in more than 80.3% of the whole area is positive, indicating that the drought in most regions of the province is alleviated in the study period. The frequency of drought in most regions of the province is between 0.2 and 0.6, the main drought is moderate drought(frequency: 0.2-0.4), and the severe drought is rare (frequency: 0-0.2). The proportion of drought-affected area in Guangdong has also been decreasing gradually, and the proportions of moderate and severe drought-affected areas show the same trend. Compared with 2003, the proportion of drought-affected area in 2017 decreased by 39.5%. In gene-ral, the severity of drought in Guangdong is decreasing year by year.
- Guangdong /
- MODIS /
- vegetation condition index /
- drought

HTML全文

含有m-Laplacian项的拟线性波方程源于非线性Voigt模型, 描述的是粘弹性材料杆的纵向振动, 特别是可描述受外力作用的Ludwick材料的振动^[1-2]. 近年来, 学者们对带有p-Laplacian项的偏微分方程解的性态作了深入研究, 如：PEI等^[3]研究了p-Laplacian型波方程u_tt-Δ_pu-Δu_t=f(u), 其中2 < p < 3且f(u)的指数为超临界情形, 证明了该方程的整体弱解的存在以及当初始能量为负值时其解存在有限时间爆破; NAKAO^[4]在适当假设下, 证明了p-Laplacian项为div{σ(|▽u|²)▽u}的偏微分方程的能量解具有多项式衰减性; RAPOSO等^[5]考虑了带有记忆项和p-Laplacian项但不含非线性源项和强阻尼项的波方程, 主要通过Galerkin方法证明该方程的整体解存在且能量函数也具有多项式衰减性. 此外, 时滞项是系统不稳定的一个重要因素, 含有时滞项的方程受到越来越多学者的关注, 如: MESSAOUDI等^[6]研究了带有强阻尼时滞项的波方程u_tt-Δu-μ₁Δu_t-μ₂Δu_t(t- τ)=0, 当假设|μ₂|≤μ₁时证明方程解的适定性以及能量函数具有指数衰减性; FENG^[7]在一定条件下利用能量扰动法得到方程 ${u_{tt}} - \Delta u + \int_0^t {g\left( {t - s} \right)} \;\Delta u\left( {x, s} \right){\rm{d}}s - {\mu _1}\Delta {u_t} - {\mu _2}\Delta {u_t}\left( {t - \tau } \right) = 0$ 解的指数衰减性; MOHAMMAD和MUHAMMAD^[8]在文献[7]的基础上研究带有非线性源项|u|^p-1u的柯西问题, 证明在初始能量为负时, 解存在有限时间爆破. 更多解的爆破研究可参考文献[9-12].

基于文献[3, 6-8]的研究, 本文考虑可描述弹性杆纵波振动的带有强阻尼时滞项的粘弹性方程：

$\left\{\begin{array}{l} u_{t t}-\Delta_{m} u-\Delta u+g * \Delta u-\mu_{1} \Delta u_{t}(x, t)-\mu_{2} \Delta u_{t}(x, t-\tau)= \\ \begin{array}{ll} \ \ \ \ |u|^{p-2} u \ \ \ \ \ \ (x \in \varOmega, t>0), \\ u(x, t)=0 \ \ \ \ (x \in \partial \varOmega, t \geqslant 0), \\ u_{t}(x, t-\tau)=f_{0}(x, t-\tau) \quad(x \in \varOmega, t \in(0, \tau)), \\ u(x, 0)=u_{0}(x), u_{t}(x, 0)=u_{1}(x) \quad(x \in \varOmega), \end{array} \end{array}\right.$

(1)

其中，u(x, t)表示振动的位移, g*Δu表示粘弹性弱阻尼, -Δu_t表示弹性体的内部阻尼, Ω是 $\mathbb{R}$ ⁿ (n≥1)上带有光滑边界∂Ω的有界区域, g(·): $\mathbb{R}$ ⁺→ $\mathbb{R}$ ⁺是正函数, m、μ₁、μ₂、p为常数,

$\Delta_{m} u=\operatorname{div}\left(|\nabla u|^{m-2} \nabla u\right), (g * \Delta u)(t)=\int_{0}^{t} g(t-s) \Delta u(s) \mathrm{d} s.$

本文主要利用凹性方法, 证明当初始能量0 < E(0) < E₁和E(0) < 0时, 方程组(1)的解都存在有限时间的爆破, 并给出爆破时间T^*的上界估计.

1. 预备知识

本文用到Sobolev空间W₀^{1, m}(Ω) (m≥2)和赋范线性空间L^p(Ω)及其范数‖ · ‖_p. 特别地, L²(Ω)的范数表示为‖ · ‖. 正常数K>0为紧嵌入W₀^{1, m}(Ω)↪L^p(Ω)的最佳嵌入常数. 对方程组(1)中涉及到的函数g(·)和正常数m、p假设如下：

(A1) g: $\mathbb{R}$ ⁺→ $\mathbb{R}$ ⁺是非增可微函数, 满足: 1- $\int_{0}^{+\infty }{g\left( s \right)\text{d}s}={{l}_{0}}>0$ , g′(t)≤0, t≥0.

(A2) 当n>m时, m < p≤ $\frac{mn}{n-m}$ ; 当n≤m时, m < p < +∞.

(A3) 常数p和函数g(·) 满足: $\int_{0}^{+\infty }{g\left( s \right)}\text{d}s\le \frac{p\left( p-2 \right)}{{{\left( p-1 \right)}^{2}}}$ .

类似文献[6], 引进新变量

$z(x, \rho, t)=u_{t}(x, t-\rho \tau)((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)),$

则有

$\tau z(x, \rho, t)+z_{\rho}(x, \rho, t)=0 \quad((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)).$

从而方程组(1)等价于：

$\left\{\begin{array}{l} u_{t t}-\Delta_{m} u-\Delta u+g * \Delta u-\mu_{1} \Delta u_{t}(x, t)-\mu_{2} \Delta z(x, 1, t)= \\ \quad|u|^{p-2} u \quad((x, t) \in \varOmega \times(0, +\infty)), \\ \tau z(x, \rho, t)+z_{\rho}(x, \rho, t)=0 \\ \quad((x, \rho, t) \in \varOmega \times(0, 1) \times(0, +\infty)), \\ u(x, t)=0 \quad((x, t) \in \partial \varOmega \times[0, +\infty)), \\ z(x, \rho, 0)=f_{0}(x, -\rho \tau)\quad((x, \rho) \in \varOmega \times(0, 1)), \\ u(x, 0)=u_{0}(x), u_{t}(x, 0)=u_{1}(x)\quad(x \in \varOmega) . \end{array}\right.$

(2)

定义方程组(2)的修正能量函数为

$\begin{array}{l} E(t):=\frac{1}{2}\left\|u_{t}\right\|^{2}+\frac{1}{m}\|\nabla u\|_{m}^{m}+ \\ \quad \frac{1}{2}\left(1-\int_{0}^{t} g(s) \mathrm{d} s\right)\|\nabla u\|^{2}+\frac{1}{2}(g \circ \nabla u)(t)+ \\ \quad \frac{\xi}{2} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x-\frac{1}{p}\|u\|_{p}^{p} \quad(t \geqslant 0), \end{array}$

(3)

其中

$\begin{array}{c} (g \circ \varphi)(t)=\int_{0}^{t} g(t-s)\|\varphi(t)-\varphi(s)\|^{2} \mathrm{~d} s, \\ \tau\left|\mu_{2}\right|<\xi<\tau\left(2 \mu_{1}-\left|\mu_{2}\right|\right), \mu_{1}>\left|\mu_{2}\right|. \end{array}$

下面给出方程组(1)的弱解定义.

定义 1 任给定初始值(u₀, u₁)∈W₀^{1, m}(Ω)×L²(Ω), 称 $\mathcal{H}$ =(u, u_t)∈C( $\mathbb{R}$ ⁺, W₀^{1, m}(Ω)×L²(Ω))为方程组(1)的弱解, 若满足: $\mathcal{H}$ (0)=(u₀, u₁), 且对任意w∈H₀¹(Ω)，有

$\begin{array}{l} \left(u_{t t}, w\right)+\left(|\nabla u|^{m-2} \nabla u, \nabla w\right)+(\nabla u, \nabla w)- \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(t-s)(\nabla u(s), \nabla w) \mathrm{d} s+\mu_{1}\left(\nabla u_{t}, \nabla w\right)+ \\ \ \ \ \ \ \ \ \ \mu_{2}\left(\nabla u_{t}(t-\tau), \nabla w\right)=\left(|u|^{p-2} u, w\right). \end{array}$

类似文献[5]、[6]、[10]的证明, 可直接得到方程组(2)的局部解的存在唯一性：

定理 1 假设(A1)、(A2)成立, 若初始值满足(u₀, u₁, f₀)∈W₀^{1, m}(Ω)×L²(Ω)×H¹(Ω×(- τ, 0)), 则方程组(2)存在唯一局部解, 且存在足够小的T>0, 有

$\begin{array}{c} u \in C\left([0, T) ; W_{0}^{1, m}(\varOmega)\right), \\ u_{t} \in C\left([0, T) ; H^{1}(\varOmega)\right) \cap L^{2}([0, T) \times \varOmega), \\ z \in C\left([0, T) ; H^{1}(\varOmega \times(0, 1))\right). \end{array}$

引理 1 设u是方程组(2)的解, 则存在正常数C₁>0, 使得

$E^{\prime}(t) \leqslant-C_{1}\left(\left\|\nabla u_{t}\right\|^{2}+\|\nabla z(x, 1, t)\|^{2}\right)<0.$

证明用u_t乘以方程组(2)中的第1个方程, 并在Ω上积分, 得

$\begin{array}{l} \frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2}\left\|u_{t}\right\|^{2}+\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{1}{2}\left(1-\int_{0}^{t} g(s) \mathrm{d} s\right)\|\nabla u\|^{2}+\right. \\ \quad\; \; \left.\frac{1}{2}(g \circ \nabla u)(t)-\frac{1}{p}\|u\|_{p}^{p}\right]=\frac{1}{2}\left(g^{\prime} \circ \nabla u\right)(t)- \\ \quad\; \; \frac{1}{2} g(t)\|\nabla u\|^{2}-\mu_{1}\left\|\nabla u_{t}\right\|^{2}- \\ \quad\; \; \mu_{2} \int_{\varOmega} \nabla z(x, 1, t) \cdot \nabla u_{t}(x, t) \mathrm{d} x. \end{array}$

(4)

用 $-\frac{\xi }{\tau }\Delta z$ 乘以方程组(2)的第2个方程, 并在Ω×(0, 1)上积分，得

$\begin{array}{l} \frac{\xi}{2} \frac{\mathrm{d}}{\mathrm{d} t} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x+ \\ \quad\; \;\frac{\xi}{2 \tau}\left(\|\nabla z(x, 1, t)\|^{2}-\left\|\nabla u_{t}\right\|^{2}\right)=0. \end{array}$

(5)

由Young不等式^[13], 可得

$\begin{array}{l} -\mu_{2} \int_{\varOmega} \nabla z(x, 1, t) \cdot \nabla u_{t}(x, t) \mathrm{d} x \leqslant \\ \quad\; \;\frac{\left|\mu_{2}\right|}{2}\|\nabla z(x, 1, t)\|^{2}+\frac{\left|\mu_{2}\right|}{2}\left\|\nabla u_{t}\right\|^{2}. \end{array}$

(6)

由式(4)~(6), 可得

$\begin{array}{l} E^{\prime}(t) \leqslant \frac{1}{2}\left(g^{\prime} \circ \nabla u\right)(t)-\frac{1}{2} g(t)\|\nabla u\|^{2}+\\ \quad\; \;\left(\frac{\xi}{2 \tau}+\frac{\left|\mu_{2}\right|}{2}-\mu_{1}\right)\left\|\nabla u_{t}\right\|^{2}+\left(\frac{\left|\mu_{2}\right|}{2}-\frac{\xi}{2 \tau}\right)\|\nabla z(x, 1, t)\|^{2}. \end{array}$

(7)

因此, 令 $0<{{C}_{1}}\le \min \left\{ {{\mu }_{1}}-\frac{\left| {{\mu }_{2}} \right|}{2}-\frac{\xi }{2\tau }, \frac{\xi }{2\tau }-\frac{\left| {{\mu }_{2}} \right|}{2} \right\}$ , 由式(7)和假设(A1)可知结论成立.

2. 解的爆破

下面考虑方程组(2)的初始能量为正值和负值时解的爆破情况. 首先, 引入3个正常数：

$K_{1}=\max \{1, K\}, \zeta_{1}=\frac{K_{1}^{p m /(m-p)}}{m}, E_{1}=\left(1-\frac{m}{p}\right) \zeta_{1}.$

引理 2 若假设(A1)、(A2)成立, 且满足

$E(0)<E_{1}, \zeta_{1}<\frac{1}{m}\left\|\nabla u_{0}\right\|_{m}^{m}+\frac{l}{2}\left\|\nabla u_{0}\right\|^{2} \leqslant \frac{K_{1}^{-m}}{m},$

(8)

则存在正常数ζ₂>ζ₁, 使得

$\frac{1}{m}\|\nabla u(t)\|_{m}^{m}+\frac{l}{2}\|\nabla u(t)\|^{2} \geqslant \zeta_{2} \quad(t \geqslant 0),$

(9)

$\|u(t)\|_{p}^{p} \geqslant\left(m^{\frac{1}{m}} K_{1}\right)^{p} \zeta_{2}^{p / m} \quad(t \geqslant 0).$

(10)

证明由式(3)和假设(A1)、(A2), 可得

$\begin{array}{l} E(t) \geqslant \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-\frac{1}{p}\|u\|_{p}^{p} \geqslant \\ \quad\quad\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-\frac{K_{1}^{p}}{p}\|\nabla u\|_{m}^{p} \geqslant \\ \quad\quad\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)- \\ \quad\quad\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{p}\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)^{\frac{p}{m}}= \\ \quad\quad\zeta(t)-\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{p} \zeta^{\frac{p}{m}}(t):=f(\zeta(t)). \end{array}$

(11)

其中ζ(t)= $\frac{1}{m}\left\| \nabla u \right\|_{m}^{m}+\frac{l}{2}{{\left\| \nabla u \right\|}^{2}}$ . 易知f(ζ(t))在ζ₁>0处有最大值E₁, 且有

$f^{\prime}(\zeta)=1-\frac{\left(m^{\frac{1}{m}} K_{1}\right)^{p}}{m} \zeta^{\frac{p-m}{m}}.$

(12)

当ζ ∈(0, ζ₁) 时, f′(ζ)>0, 所以f(ζ)在(0, ζ₁)上严格单调递增; 当ζ∈(ζ₁, +∞)时, f′(ζ) < 0, 所以f(ζ)在(ζ₁, +∞)上严格单调递减. 又由于E(0) < E₁=f(ζ₁), 则存在正常数ζ₂ ∈(ζ₁, +∞), 使得f(ζ₂)=E(0). 由式(11)可知f(ζ(0))≤E(0)=f(ζ₂), 所以ζ(0)≥ζ₂.

假设存在t₁>0, 有 $\frac{1}{m}\left\| \nabla u\left( {{t}_{1}} \right) \right\|_{m}^{m}+\frac{l}{2}{{\left\| \nabla u\left( {{t}_{1}} \right) \right\|}^{2}}<{{\zeta }_{2}}$ . 进一步地, 由式(11)和f(ζ)的单调性可得E(t₂)≥ f(ζ(t₂))>f(ζ₂)=E(0), 在t ∈(0, T)上与E(t) < E(0)矛盾, 故式(9)成立.

再一次利用式(11), 可得

$\begin{aligned} &\frac{1}{p}\|u\|_{p}^{p} \geqslant \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-E(t) \geqslant\\ &\ \ \ \ \frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}-E(0) \geqslant \zeta_{2}-E(0)=\\ &\ \ \ \ \frac{\left(m^{\frac{1}{m}} K\right)^{p}}{p} \zeta_{2}^{p / m}. \end{aligned}$

(13)

因此, 式(10)成立.

定理 2 在引理2的假设下, 进一步假设(A3)成立, 则方程组(2)的解存在有限时间爆破.

证明设函数

$L_{1}(t)=G_{1}^{1-\alpha}(t)+\varepsilon \int_{\varOmega} u u_{t} \mathrm{~d} x+\frac{\varepsilon \mu_{1}}{2}\|\nabla u\|^{2} \quad(t \in[0, T)),$

(14)

其中, ε>0为足够小的正实数, 且

$0<\alpha \leqslant \min \left\{\frac{m-2}{p}, \frac{m-2}{2 m}\right\},$

(15)

$G_{1}(t):=E_{2}-E(t), E_{2} \in\left(E(0), E_{1}\right).$

(16)

下面将证明存在C>0, 使L₁(t)满足微分不等式

$L_{1}^{\prime}(t) \geqslant C L_{1}^{q}(t) \quad(q>1).$

首先, 由式(3)、(16)和引理1, 得

$\begin{array}{l} 0<G_{1}(0)<G_{1}(t) \leqslant\\ \ \ \ \ E_{2}-\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)+\frac{1}{p}\|u\|_{p}^{p} \quad(t \geqslant 0). \end{array}$

由引理2, 有

$E_{2}-\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right) \leqslant E_{2}-\zeta_{2} \leqslant E_{1}-\zeta_{1}=-\frac{m}{p} \zeta_{1}.$

因此

$0<G_{1}(0)<G_{1}(t) \leqslant \frac{1}{p}\|u\|_{p}^{p}.$

(17)

对L₁(t)关于时间t求导, 有

$\begin{array}{l} L_{1}^{\prime}(t)=(1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon\left\|u_{t}\right\|^{2}+ \\ \ \ \ \ \ \ \ \ \varepsilon \int_{\varOmega} u u_{t t} \mathrm{~d} x+\varepsilon \mu_{1} \int_{\varOmega} \nabla u \cdot \nabla u_{t} \mathrm{~d} x= \\ \ \ \ \ \ \ \ \ (1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon\left\|u_{t}\right\|^{2}-\varepsilon\|\nabla u\|_{m}^{m}- \\ \ \ \ \ \ \ \ \ \varepsilon\|\nabla u\|^{2}+\varepsilon \int_{\varOmega} \nabla u(t) \int_{0}^{t} g(t-s) \nabla u(s) \mathrm{d} s \mathrm{~d} x- \\ \ \ \ \ \ \ \ \ \varepsilon \mu_{2} \int_{\varOmega} \nabla u(x, t) \cdot \nabla z(x, 1, t) \mathrm{d} x+\varepsilon\|u\|_{p}^{p}. \end{array}$

(18)

由Young不等式, 有

$\begin{array}{l} \int_{\varOmega} \nabla u(t) \int_{0}^{t} g(t-s) \nabla u(s) \mathrm{d} s \mathrm{~d} x= \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(t-s) \int_{\varOmega} \nabla u(t) \cdot(\nabla u(s)-\nabla u(t)) \mathrm{d} x \mathrm{~d} s+ \\ \ \ \ \ \ \ \ \ \int_{0}^{t} g(s) \mathrm{d} s\|\nabla u\|^{2} \geqslant-\delta_{1}(g \circ \nabla u)(t)+\\ \ \ \ \ \ \ \ \ \left(1-\frac{1}{4 \delta_{1}}\right) \int_{0}^{t} g(s) \mathrm{d} s\|\nabla u\|^{2}, \end{array}$

(19)

$\begin{array}{l} -\mu_{2} \int_{\varOmega} \nabla u(x, t) \cdot \nabla z(x, 1, t) \mathrm{d} x \geqslant\\ \quad-\frac{\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}-\left|\mu_{2}\right| \delta_{2}\|\nabla z(x, 1, t)\|^{2}. \end{array}$

(20)

取0<a<1, δ₁=p(1-a)/2, 将式(19)、(20)代入式(18), 可得

$\begin{array}{l} L_{1}^{\prime}(t) \geqslant(1-\alpha) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\varepsilon p(1-a) G_{1}(t)+\\ \quad\varepsilon\left(1+\frac{p(1-a)}{2}\right)\left\|u_{t}\right\|^{2}+\varepsilon\left[\frac{p(1-a)}{2}-1+\right.\\ \quad\left.\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s\right]\|\nabla u\|^{2}+\\ \quad\varepsilon a\|u\|_{p}^{p}+\frac{\varepsilon p(1-a) \xi}{2} \int_{\varOmega} \int_{0}^{1}|\nabla z(x, \rho, t)|^{2} \mathrm{~d} \rho \mathrm{d} x+\\ \quad\varepsilon\left(\frac{p(1-a)}{m}-1\right)\|\nabla u\|_{m}^{m}-\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}-\\ \quad\varepsilon\left|\mu_{2}\right| \delta_{2}\|\nabla z(x, 1, t)\|^{2}-\varepsilon p(1-a) E_{2}. \end{array}$

(21)

下面对式(21)中一些项进行估计. 令δ₂=MG₁^-α(t) (M>0), 由引理1可得

$\begin{array}{l} -\varepsilon \mid \mu_{2} \mid \delta_{2}\|\nabla z(x, 1, t)\|^{2}=\\ \quad -\varepsilon M\left|\mu_{2}\right| G_{1}^{-\alpha}(t)\|\nabla z(x, 1, t)\|^{2} \geqslant \\ \quad \frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}} E^{\prime}(t) G_{1}^{-\alpha}(t)=-\frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}} G_{1}^{-\alpha}(t) G_{1}^{\prime}(t). \end{array}$

(22)

利用Hölder不等式^[14]和嵌入W₀^{1, m}(Ω)↪L^p(Ω), 结合式(17), 有

$\begin{array}{l} -\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2}=-\frac{\varepsilon\left|\mu_{2}\right|}{4 M} G_{1}^{\alpha}(t)\|\nabla u\|^{2} \geqslant \\ \quad -\frac{\varepsilon\left|\mu_{2}\right||\varOmega|^{\frac{m-2}{m}}}{4 M p^{\alpha}}\|u\|_{p}^{p \alpha}\|\nabla u\|_{m}^{2} \geqslant \\ \quad -\frac{\varepsilon\left|\mu_{2}\right||\varOmega|^{\frac{m-2}{m}} K^{p \alpha}}{4 M p^{\alpha}}\|\nabla u\|_{m}^{p \alpha+2} . \end{array}$

(23)

进一步地, 由 ${{z}^{\nu }}\le z+1\le \left( 1+\frac{1}{b} \right)\left( z+b \right)$ (z≥0, 0 < ν≤1, b≥0)和式(23), 可得

$-\frac{\varepsilon\left|\mu_{2}\right|}{4 \delta_{2}}\|\nabla u\|^{2} \geqslant-\frac{\varepsilon C_{2}}{M}\left(\|\nabla u\|_{m}^{m}+G_{1}(t)\right),$

(24)

其中 ${C_2}: = \frac{{\left| {{\mu _2}} \right|{{\left| \mathit{\Omega } \right|}^{\frac{{m - 2}}{m}}}{K^{p\alpha }}\left( {1 + {G_1}\left( 0 \right)} \right)}}{{4{p^\alpha }{G_1}\left( 0 \right)}}$ .

由假设(A3)和p>m, 可得

$\begin{array}{l} \frac{p(1-a)}{2}-1+\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s>0, \\ \frac{p(1-a)}{m}-1 \leqslant p(1-a)-2+\left(2-\frac{1}{p(1-a)}-p(1-a)\right)(1-l). \end{array}$

因此, 令a>0充分小, 取适当的E₂ ∈(E(0), E₁), 并结合引理2可得

$\begin{array}{l} \varepsilon\left[\frac{p(1-a)}{2}-1+\left(1-\frac{1}{2 p(1-a)}-\frac{p(1-a)}{2}\right) \int_{0}^{t} g(s) \mathrm{d} s\right] \times \\ \quad \|\nabla u\|^{2}+\varepsilon\left(\frac{p(1-a)}{m}-1\right)\|\nabla u\|_{m}^{m}-\varepsilon p(1-a) E_{2} \geqslant \\ \quad \varepsilon(p(1-a)-m)\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right)- \\ \quad \varepsilon p(1-a) E_{2} \geqslant \varepsilon C_{3}\left(\frac{1}{m}\|\nabla u\|_{m}^{m}+\frac{l}{2}\|\nabla u\|^{2}\right), \end{array}$

(25)

其中C₃=p(1-a)-m- $\frac{{p\left( {1 - a} \right){E_2}}}{{{\zeta _2}}}$ ≥0.

将式(22)~(25)代入式(21), 可得

$\begin{aligned} L_{1}^{\prime}(t) & \geqslant\left(1-\alpha-\frac{\varepsilon M\left|\mu_{2}\right|}{C_{1}}\right) G_{1}^{-\alpha}(t) G_{1}^{\prime}(t)+\\ & \varepsilon\left(p(1-a)-\frac{C_{2}}{M}\right) G_{1}(t)+\varepsilon\left(1+\frac{p(1-a)}{2}\right)\left\|u_{t}\right\|^{2}+\\ & \varepsilon\left(\frac{C_{3}}{m}-\frac{C_{2}}{M}\right)\|\nabla u\|_{m}^{m}+\frac{\varepsilon l C_{3}}{2}\|\nabla u\|^{2}+\varepsilon a\|u\|_{p}^{p} . \end{aligned}$

(26)

在式(26)中, 当a>0充分小、M>0足够大时, 可使p(1-a)- $\frac{{{C_2}}}{M}$ ≥0, 且 $\frac{{{C_3}}}{m} - \frac{{{C_2}}}{M}$ ≥0. 进一步地, 令ε>0充分小, 可使1-α- $\frac{{\varepsilon M\left| {{\mu _2}} \right|}}{{{C_1}}}$ ≥0. 故存在正常数C₄>0, 使得

$\begin{array}{l} L_{1}^{\prime}(t) \geqslant C_{4}\left(\left\|u_{t}\right\|^{2}+\|\nabla u\|_{m}^{m}+\|\nabla u\|^{2}+\|u\|_{p}^{p}+G_{1}(t)\right) \\ \quad\quad(t \geqslant 0) . \end{array}$

(27)

另一方面, 由式(14)可得

$L_{1}^{1 /(1-\alpha)}(t) \geqslant C_{5}\left(G_{1}(t)+\left(\int_{\varOmega} u u_{t} \mathrm{~d} x\right)^{\frac{1}{1-\alpha}}+\|\nabla u\|^{\frac{2}{1-\alpha}}\right).$

(28)

利用式(15)、(17), 结合Young不等式、Poincaré不等式^[13], 得

$\left(\int_{\varOmega} u u_{t} \mathrm{~d} x\right)^{\frac{1}{1-\alpha}} \leqslant C_{6}\left(G_{1}(t)+\|\nabla u\|_{m}^{m}+\left\|u_{t}\right\|^{2}\right) .$

(29)

再由式(15)、(17), 有

$\begin{array}{l}\|\nabla u\|^{\frac{2}{1-\alpha}}=\left(\|\nabla u\|^{m}\right)^{\frac{2}{m(1-\alpha)}} \leqslant\\ \quad\left(1+\frac{1}{G_{1}(0)}\right)\left(G_{1}(t)+\|\nabla u\|_{m}^{m}\right) \leqslant \\ \quad\left(1+\frac{1}{G_{1}(0)}\right)\left(G_{1}(t)+|\varOmega|^{\frac{m-2}{2}}\|\nabla u\|_{m}^{m}\right). \end{array}$

(30)

可得

$L_{1}^{1 /(1-\alpha)}(t) \leqslant C_{6}\left(G_{1}(t)+\|\nabla u\|_{m}^{m}+\left\|u_{t}\right\|^{2}\right) .$

(31)

最后, 综合式(27)、(31)可知：存在C>0, 使得

$L_{1}^{\prime}(t) \geqslant C L_{1}^{1 /(1-\alpha)}(t) \quad(t \geqslant 0).$

(32)

对式(32)在(0, t)上积分, 有

$L_{1}^{1 /(1-\alpha)}(t) \geqslant \frac{1}{L_{1}^{-\alpha /(1-\alpha)}(0)-\frac{C \alpha}{1-\alpha} t},$

所以, 存在T^*≤ $\frac{{1 - \alpha }}{{C\alpha L_1^{\alpha /\left( {1 - \alpha } \right)}\left( 0 \right)}}$ , 当t趋于T^*时, L₁(t)趋于无穷, 即方程组(2)的解爆破.

定理 3 若假设(A1)~(A3)成立, 且初始能量E(0) < 0时, 则方程组(2)的解同样存在有限时间爆破.

证明设函数

$L_{2}(t)=G_{2}^{1-\beta}(t)+\varepsilon \int_{\varOmega} u u_{t} \mathrm{~d} x+\frac{\varepsilon \mu_{1}}{2}\|\nabla u\|^{2} \quad(t \in[0, T)),$

(33)

其中, ε>0为足够小的正实数, 0 < β≤min $\left\{ {\frac{{m - 2}}{p}, \frac{{m - 2}}{{2m}}} \right\}$ , 且G₂(t)=-E(t).

由式(3)、(16)和引理1, 可得0 < G₂(0) < G₂(t)≤ $\frac{1}{p}\left\| u \right\|_p^p$ (t≥0). 参照定理2的证明方法(只需令E₂=0), 定理得证.

图 1 研究区的数字地形图

Figure 1. The digital elevation model of the study area

下载: 全尺寸图片幻灯片

图 2 2003—2017年广东省的VCI及其距平指数变化趋势

Figure 2. The trends of VCI and AVCI in Guangdong from 2003 to 2017

下载: 全尺寸图片幻灯片

图 3 2003—2017年广东省VCI的MK检验结果

Figure 3. The Mann-Kendall mutation analysis results of VCI in Guangdong from 2003 to 2017

下载: 全尺寸图片幻灯片

图 4 2003—2017年广东省VCI变化趋势的空间分布

Figure 4. The spatial distribution of the VCI trend from 2003 to 2017 in Guangdong

下载: 全尺寸图片幻灯片

图 5 广东省干旱发生频率空间分布

Figure 5. The spatial distribution of drought frequency in Guangdong

下载: 全尺寸图片幻灯片

图 6 2003—2017年广东省干旱面积占比的距平指数变化

Figure 6. The AVCI of proportion of drought area in Guangdong from 2003 to 2017

下载: 全尺寸图片幻灯片

表 1 广东省2003—2017年干旱面积占比

Table 1 The proportion of drought-affected area in Guangdong from 2003 to 2017 %

年份	干旱占比	中旱占比	重旱占比
2003	46.1	34.2	11.9
2004	48.8	36.0	12.8
2005	49.3	34.6	14.7
2006	46.2	33.6	12.6
2007	45.5	33.4	12.1
2008	43.8	32.5	11.3
2009	42.4	32.1	10.3
2010	39.2	28.7	10.5
2011	43.2	32.8	10.4
2012	37.8	27.4	10.4
2013	36.5	25.8	10.7
2014	38.8	28.4	10.4
2015	33.6	23.5	10.1
2016	30.4	20.6	9.8
2017	27.9	20.2	7.7

下载: 导出CSV

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