一类含梯度项的奇异抛物型方程弱解的存在性

张亚茹; 夏莉

doi:10.6054/j.jscnun.2023040

一类含梯度项的奇异抛物型方程弱解的存在性

张亚茹¹,
夏莉^{1, 2, ,}

1.
广东财经大学统计与数学学院, 广州 510320
2.
广东财经大学大数据与教育应用统计实验室, 广州 510320

基金项目:

国家自然科学基金项目 11971200

广东省教育厅科研项目[特色创新项目] 2018KTSCX069

广东省教育厅委托项目 0835-210Z33606691

详细信息

通讯作者:
夏莉, Email: xaleysherry@163.com

中图分类号: O175.26
计量
- 文章访问数: 68
- HTML全文浏览量: 10
- PDF下载量: 29
出版历程
- 收稿日期: 2021-02-12
- 网络出版日期: 2023-08-25
- 刊出日期: 2023-06-24

Existence of Weak Solutions for Some Singular Parabolic Equations with Gradient Terms

ZHANG Yaru¹,
XIA Li^{1, 2, ,}

1.
School of Statistics and Mathematics, Guangdong University of Finance and Economic, Guangzhou 510320, China
2.
Big Data and Educational Statistics Application Laboratory, Guangdong University of Finance and Economic, Guangzhou 510320, China

摘要

摘要: 文章研究一类具有Dirichlet边界条件和初始条件的含梯度项奇异抛物型偏微分方程： $\left\{\begin{array}{l}y_t-y^{\prime \prime}-\frac{\kappa}{r} y^{\prime}+\lambda \frac{\left|y^{\prime}\right|^2}{y^m}=f(r, t) \quad(y \geqslant 0, (r, t) \in(0, 1) \times(0, T]), \\y(0, t)=y(1, t)=0 \ \ \ \ \ \quad(t \in(0, T]), \\y(r, 0)=\varphi(r) \quad(r \in(0, 1)), \end{array}\right.$ 其中, T>0, κ≥0, λ>0, m∈(0, 2)。由于含梯度的奇异抛物型方程中具有奇异项和非线性项, 故先利用抛物正则化方法将方程进行正则化, 再结合上下解方法, 证明了在不同假设条件下的该类方程非负弱解的存在性。最后, 证明了该方程的弱比较原理。
- 存在性 /
- 奇异抛物型方程 /
- 弱解
Abstract: A class of singular parabolic equations with gradient terms is studied in this paper, with Dirichlet boun-dary condition and initial condition, of the from $\left\{\begin{array}{l}y_t-y^{\prime \prime}-\frac{\kappa}{r} y^{\prime}+\lambda \frac{\left|y^{\prime}\right|^2}{y^m}=f(r, t) \quad(y \geqslant 0, (r, t) \in(0, 1) \times(0, T]), \\y(0, t)=y(1, t)=0 \ \ \ \ \ \quad(t \in(0, T]), \\y(r, 0)=\varphi(r) \quad(r \in(0, 1)), \end{array}\right.$ where T>0, κ≥0, λ>0, m∈(0, 2). Since singular parabolic equations containing gradient terms have singular terms and nonlinear terms, the parabolic regularization method is used to regularize the equation, and then combined with the sub-super solutions method, the existence of weak solutions of the equations under different assumptions is proved. Finally, the weak comparison principle of the equations is proved.
- existence /
- singular parabolic equation /
- weak solution

HTML全文

含有梯度项的奇异抛物方程可以描述热化学反应、人口动态学、非牛顿流体、渗流现象和相变理论等领域中的许多实际问题。这类含梯度项的奇异抛物方程可能出现多处奇性项和非线性项, 从而造成理论研究上的本质性困难。而且, 这类奇异方程在一定条件下可以通过某些变换转化为其他类型的方程。因此, 对于带梯度项的奇异抛物型问题的解的研究一直是热点问题。

在过去的几十年里, 学者们获得了许多有关含梯度项奇异抛物方程的理论成果。例如, 假设f(|x|, t)是径向对称函数, A₁为 $\mathbb{R}^N(N \geqslant 2)$ 中的单位球, 学者们研究了

$\left\{\begin{array}{l} y_t-\Delta y+\lambda \frac{|\nabla y|^2}{y}=f(|x|, t) \quad\left((x, t) \in A_1 \backslash\{0\} \times(0, T]\right), \\ y(|x|, t)=0 \quad\left((x, t) \in A_1 \backslash\{0\} \times(0, T]\right), \\ y(|x|, 0)=\varphi(|x|) \quad\left(x \in A_1\right) \end{array}\right.$

(1)

的非径向经典解或弱解的存在性、唯一性和渐近行为^[1-3]; 讨论了与方程(1)相关的一些奇异椭圆方程, 得到了其正则化的结果, 并证明了这些奇异椭圆方程的非负弱解的存在性、唯一性^[4-8]。

其次, 当用p-laplace算子 $\Delta_p y=\operatorname{div}\left(|\nabla y|^{p-2} \nabla y\right)(p>1)$ 代替方程(1)中的算子Δy时, 学者们研究了这类非线性抛物方程或椭圆方程的解的存在性和唯一性^[9-10]；考虑了某些非线性抛物方程边值问题的非负解的存在性^[11]；进一步考虑了某些含超线性奇异梯度项的抛物型方程Cauchy-Dirichlet问题的非负解的存在性^[12]。

此外, 如果定义|x|=r, 则方程(1)转化为方程

$y_t-y^{\prime \prime}-\frac{\kappa}{r} y^{\prime}+\lambda \frac{\left|y^{\prime}\right|^l}{y^m}=f(r, t)$

(2)

的初边值问题, 其中, $y \geqslant 0, (r, t) \in(0, 1) \times(0, T]$ 。在l=2时, ZHOU和LEI^[13]研究了当κ=0, 1≤m < 2时, 问题(2)的经典正解的存在性; ZHOU^[14]研究了在1≤m < 2的情况下, 与问题(2)相关的奇异椭圆边值问题存在正解的充分条件; 夏莉和李敬娜^[15]考虑了问题(2)在m=1的特殊情况下的弱解。在l>0时，XIA和ZHANG^[16]研究了问题(2)的古典解的存在性。

本文主要研究以下含梯度项的奇异抛物型方程

$\left\{\begin{array}{l} y_t-y^{\prime \prime}-\frac{\kappa}{r} y^{\prime}+\lambda \frac{\left|y^{\prime}\right|^2}{y^m}=f(r, t) \\ \quad(y \geqslant 0, (r, t) \in(0, 1) \times(0, T]), \ \ \ \ \ \ \ (3)\\ y(0, t)=y(1, t)=0 \quad(t \in(0, T]), \ \ \ \ \ \ (4)\\ y(r, 0)=\varphi(r) \quad(r \in(0, 1))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \end{array}\right.$

的弱解, 其中, κ≥0和λ>0都是常数, m∈(0, 2), f(r, t)、φ(r)是分别满足条件(H1)、(H2)的函数:

(H1) f(r, t)∈C(E_T), 且在E_T上f(r, t)>0;

(H2) φ(r)∈C(0, 1)∩H¹(0, 1), 在区间(0, 1)中, φ(r)>0, 且φ(0)=φ(1)=0。

本文使用抛物正则化法和上下解方法, 研究了0<m< 2时, 方程(3)在初边值条件(4)、(5)下的弱解的存在性问题。

1. 预备知识

为了得到问题(3)~(5)的解, 约定:

(D1) E=(0, 1), E_T=(0, 1)×(0, T], E_T=[0, 1]×[0, T];

(D2) C_c^∞(E)是E中具有紧支集的无穷可微函数空间;

(D3) 区间[0, T]中, g(t)为C¹连续函数, 且满足函数值g(t)≥1和一阶导数g′(t)>0;

(D4) 记 $\varrho=2 /(2-m)$ , 由m∈(0, 2)易知 $\varrho>1$ 。如果0 < m≤1, 则约定 $\kappa>L=(\lambda+\varrho-1) / \varrho$ ; 如果1 < m < 2, 则κ>inf_s≥1G(s), 其中 $G:(0, \infty) \rightarrow \mathbb{R}$ 的定义^[2]如下:

$\begin{gathered} G(s)=s^{m-1} \varrho^{-1}\left(\lambda+\varrho^{-1}\right) \cdot\left[\max _{[0, T]} g(t)\right]^{m-1}+ \\ s^{m-2} \varrho^{-2} \cdot \max _{E_T} f \cdot\left[\min _{[0, T]} g(t)\right]^{m-2} 。 \end{gathered}$

其次, 除了f(r, t)、φ(r)(分别简记为f、φ)需要满足的条件(H1)、(H2)外, 还需要给定以下条件:

(H3) 当 $\kappa>0, \lambda>(\kappa+1) / 2$ 时, 有

$\frac{C_1}{2^{\varrho}} g^{\varrho}(0)\left(1-r^2\right)^{\varrho} \leqslant \varphi \leqslant C_2 g(0) \min \left\{r^{\varrho}, (1-r)^{\varrho}\right\} \text {; }$

(H4) 当 $\kappa \in[0, 1], \lambda>0$ 时, 有

$\frac{C_1}{2^{\varrho}} g^{\varrho}(0)\left(1-r^2\right)^{\varrho} \leqslant \varphi \leqslant C_3 g(0) r^{1-\kappa}(1-r) 。$

上述条件中C₁、C₂、C₃均为常数，C₁<1/2, C₂, C₃≥1。

下面对问题(3)~(5)的弱解进行定义。

定义1 如果函数y≥0, 且满足以下条件:

(i) 在E_T上, y几乎处处大于0;

(ii) $y \in V=L^2\left(0, T ; W_0^{1, 2}(E)\right) \cap L^{\infty}\left(E_T\right)$ , 且有 $y_t \in V^* \text { 及 } r^\kappa\left|y^{\prime}\right|^2 \in L^1\left(E_T\right)$ ;

(iii) 对于任意函数 $\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} \in C_c^{\infty}\left(E_T\right)$ , 以下等式成立：

$\int_0^T<y_t, \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}>+\iint_{E_T}\left(y^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}-\frac{\kappa}{r} y^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}+\lambda\left|y^{\prime}\right|^2 \frac{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}}{y^m}-f(r, t) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}\right) \mathrm{d} r \mathrm{~d} t=0\\$

(iv) $\lim _{t \rightarrow 0^{+}} \int_0^1|y(r, t)-\varphi(r)| \mathrm{d} r=0$ ,

则称y为问题(3)~(5)的弱解。

下面将问题(3)~(5)进行正则化, 同时给出正则化问题的古典上下解定义。

显然, 在r=0和y(r, t)=0时, 方程(3)具有奇异性。故需要对问题(3)~(5)进行正则化, 将其转化为以下正则化方程

$\left\{\begin{array}{l} y_{\varepsilon t}-y_{\varepsilon}^{\prime \prime}-\frac{\kappa}{r+\varepsilon^{1 / \varrho}} y_{\varepsilon}^{\prime}+\lambda \frac{\left|y_{\varepsilon}^{\prime}\right|^2}{\left(y_{\varepsilon}+\varepsilon^2\right)^m}=f(r, t) \\ \quad\left(y_{\varepsilon} \geqslant 0, (r, t) \in E_T\right), \\ y_{\varepsilon}(0, t)=y_{\varepsilon}(1, t)=0 \ \ \quad(t \in(0, T]), \\ y_{\varepsilon}(r, 0)=\varphi(r) \quad \ \ (r \in(0, 1)) 。 \end{array}\right.$

(6)

由条件(H1)、(H2)及文献[17], 容易得到问题(6)的古典解y_ε的存在性。为了后续证明过程中表述的简洁，给出以下简记符号：

$F(\xi)=\xi_t-\xi^{\prime \prime},$

$h_{\varepsilon}(r, t, \xi, \eta)=f(r, t)+\frac{\kappa}{r+\varepsilon^{1 / \varrho}} \eta-\lambda \frac{|\eta|^2}{\left(\xi+\varepsilon^2\right)^m} 。$

如果 $0 \leqslant \underline{y} \leqslant C\left(\bar{E}_T\right) \cap C^{2,1}\left(E_T\right)$ , 并满足以下条件

$\begin{gathered} F(y) \leqslant h_{\varepsilon}\left(r, t, \underline{y}, \underline{y}^{\prime}\right) \quad\left((r, t) \in E_T\right), \\ \underline{y}(0, t) \leqslant 0, \underline{y}(1, t) \leqslant 0 \quad(t \in(0, T]), \\ \underline{y}(r, 0) \leqslant \varphi(r) \quad(r \in(0, 1)), \end{gathered}$

则称 $\underline{y}$ 为问题(6)的古典下解。将上述条件中“≤”改为“≥”, 则可得到问题(6)的古典上解y的定义。

令y₀=g(t)(r-r²)/2, 其中g(t)满足条件(D3)。显然, y₀为以下问题的古典解:

$\left\{\begin{array}{l} y_t-y^{\prime \prime}=b(r, t) \quad\left((r, t) \in E_T\right), \\ y(0, t)=y(1, t)=0 \quad(t \in[0, T]), \\ y(r, 0)=y_0 \quad(r \in(0, 1)), \end{array}\right.$

其中, b(r, t)=g′(t)(r-r²)/2+g(t)。显然, 在E_T上有0≤y₀≤rg(t), 并且对∀δ∈(0, 1/2), 在[δ, 1-δ]×[0, T]上有y₀>0。

定义2 若在E_T中y≥0几乎处处成立, y∈V, y_t∈V^*, 且对∀0≤Ψ∈V, 均有

$\left\{\begin{array}{l} \int_0^T<y_t, \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}>+\iint_{E_T}\left(y^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}-\frac{\kappa}{r+\varepsilon^{1 / \varrho}} y^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}+\lambda\left|y^{\prime}\right|^2 \frac{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}}{\left(y+\varepsilon^2\right)^m}-\right. \\ f(r, t) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}) \mathrm{d} r \mathrm{~d} t \leqslant 0 \quad\left((r, t) \in E_T\right), \\ y(0, t) \leqslant 0, y(1, t) \leqslant 0 \quad(t \in[0, T]), \\ y(r, 0) \leqslant \varphi(r) \quad(r \in[0, 1]), \end{array}\right.$

则称y为问题(6)的弱下解。同理, 将上述符号反号, 可得其弱上解的定义。

2. 主要结论及其证明

首先引入以下引理。

引理1 令 $\underline{y}=C_1 y_0^{\varrho}$ , 其中0≤C₁≤1/2是常数, 则 $\underline{y}$ 是问题(6)的古典下解。

证明由问题(6)可得

$\begin{aligned} & F(\underline{y})-h_{\varepsilon}\left(r, t, \underline{y}, \underline{y}^{\prime}\right)=\underline{y}_t-\underline{y}^{\prime \prime}-f(r, t)-\frac{\kappa}{r+\varepsilon^{1 / \varrho}} \underline{y}^{\prime}+ \\ & \ \ \ \ \lambda \frac{\left|\underline{y}^{\prime}\right|^2}{\left(\underline{\left.y+\varepsilon^2\right)^m}\right.}=C_1 \varrho y_0^{\varrho-1} b(r, t)-C_1 \varrho(\varrho-1) y_0^{\varrho-2}\left|y_0^{\prime}\right|^2- \\ & \ \ \ \ \frac{\kappa C_1 \varrho \varrho_0^{\varrho-1}}{r+\varepsilon^{1 / \varrho}}\left[\frac{1}{2} g(t)(1-2 r)\right]+\lambda \frac{\left|C_1 \varrho y_0^{\varrho-1} y_0^{\prime}\right|^2}{\left(\underline{y}+\varepsilon^2\right)^m}-\\ & f(r, t) \leqslant C_1 \varrho y_0^{\varrho-1} b(r, t)+\kappa C_1 \varrho g(t)\left|y_0^{\prime}\right| y_0^{\varrho-2}+ \\ & \lambda C_1^{2-m} \varrho^2\left|y_0^{\prime}\right|^2-f(r, t) 。 \end{aligned}$

当0 <m≤1时, 先选取C₁<1/2。又因为C₁^2-m<C₁, 故可以选取一个适当的正数C₁=min(1/2, M_{2, m}), 其中 $M_{2, m}=\min _{\bar{E}_T} f \cdot\left[\operatorname { m a x } _ { \overline { E } _ { T } } \left(\varrho y_0^{\varrho-1} b+\kappa \varrho g(t)\left|y_0^{\prime}\right| y_0^{\rho^{-2}}+\right.\right.\left.\left.\lambda \varrho^2\left|y_0^{\prime}\right|^2\right)\right]^{-1}$ , 从而使得 $\underline{y}$ 是问题(6)的古典下解。

当1 <m < 2时, 类似地可选取C₁=min(1/2, M_{2, m}^1/(2-m)), 易证 $\underline{y}$ 是问题(6)的古典下解。证毕。

引理2 假设κ>0, λ>(κ+1)/2, 且条件(H3)成立, 令 $y_{1 \varepsilon}=C_2 g(t)\left(r+\varepsilon^{1 / \varrho}\right)^{\varrho}, y_{2 \varepsilon}=C_2 g(t)\left(1+\varepsilon^{1 / \varrho}-\right.r)^{\varrho}$ , 其中, δ∈(0, δ₀), 0 <δ₀ <1/2, 则y_ε=min(y_1ε, y_2ε)是问题(6)的古典上解。

证明只需证F(y_iε)≥h_ε(r, t, y_iε, y′_iε)(i=1, 2)。因为ε≤y_1ε, 且在[0, T]上g′(t)>0, 因此

$\begin{aligned} & F\left(y_{1 \varepsilon}\right)-h_{\varepsilon}\left(r, t, y_{1 \varepsilon}, y_{1 \varepsilon}^{\prime}\right)=C_2 g^{\prime}(t)\left(r+\varepsilon^{1 / \varrho}\right)^{\varrho}- \\ & \ \ \ \ \ C_2 \varrho\left(\kappa+\varrho^{-1}\right) g(t)\left(r+\varepsilon^{1 / \varrho}\right)^{\varrho^{-2}}+ \\ & \ \ \ \ \ \lambda \varrho^2 C_2^{2-m} g^{2-m}(t) \frac{y_{1 \varepsilon}^m}{\left(y_{1 \varepsilon}+\varepsilon^2\right)^m}-f(r, t) \geqslant \\ & \ \ \ \ \ -C_2 \varrho\left(\kappa^{+} \varrho^{-1}\right) g(t)\left(1+\varepsilon^{1 / \varrho}\right)^{\varrho^{-2}}+ \\ & \ \ \ \ \ \lambda \varrho^2 C_2^{2-m} g^{2-m}(t)(1+\varepsilon)^{-m}-\max _\bar{E_T} f(r, t)= \\ & \ \ \ \ \ -C_2 \varrho\left(\kappa+\varrho^{-1}\right) g(t)+\lambda \varrho^2 C_2^{2-m} g^{2-m}(t)- \\ & \ \ \ \ \ \max _{\bar{E}_T} f(r, t)+S_{\varepsilon}(t), \end{aligned}$

其中,

$\begin{array}{*{20}{l}} {{S_\varepsilon }(t) = {C_2}\varrho \left( {\kappa + {\varrho ^{ - 1}}} \right)g(t)\left[ {1 - {{\left( {1 + {\varepsilon ^{1/\varrho }}} \right)}^{\varrho - 2}}} \right] + } \\ \ \ \ \ \ \ \ {\lambda {\varrho ^2}C_2^{2 - m}{g^{2 - m}}(t)\left[ {{{(1 + \varepsilon )}^{ - m}} - 1} \right]} \end{array}$

又因为g(t)和g^2-m(t)在[0, T]上一致有界, 故对∀t∈[0, T], 当ε→0时, 易证S_ε(t)→0。

受文献[14]的启示, 易知存在正常数C₂≥1, 使得G(C₂) <λ, 其中函数G(s)已由条件(D4)给出。因此, 当0 <m≤1时, 选取以上C₂和某δ₀∈(0, 1/2), 可使得对∀(r, t)∈E_T和∀ε∈(0, δ₀), 有

$\begin{aligned} & F\left(y_{1 \varepsilon}\right)-h_{\varepsilon}\left(r, t, y_{1 \varepsilon}, y_{1 \varepsilon}^{\prime}\right) \geqslant \\ & \ \ \quad \varrho^2 C_2^{2-m} g^{2-m}(t)\left(\lambda-G\left(C_2\right)\right)+S_{\varepsilon}(t)>0 \text { 。 } \end{aligned}$

当1 <m <2时, 对∀δ∈(0, δ₀)及(r, t)∈E_T, 有

$\begin{aligned} & F\left(y_{1 \varepsilon}\right)-h_{\varepsilon}\left(r, t, y_{1 \varepsilon}, y_{1 \varepsilon}^{\prime}\right) \geqslant \\ & \ \quad \varrho^2 C_2^{2-m} g^{2-m}(t)(\lambda-L)+S_{\varepsilon}(t)>0, \end{aligned}$

其中， $L=(\lambda+\varrho-1) / \varrho$ 。这表明对∀ε∈(0, δ₀), (r, t)∈E_T, F(y_1ε)-h_ε(r, t, y_1ε, y′_1ε)≥0。同理可证F(y_2ε)≥h_ε(r, t, y_2ε, y′_2ε)。证毕。

引理3 假设κ∈[0, 1], λ>0且条件(H4)成立, 令y_ε=C₃g(t)z_ε, 其中 $z_{\varepsilon}=\frac{1}{2-\kappa}\left(r+\varepsilon^{1 / \varrho}\right)^{1-\kappa}(1-r)$ , 则y_ε是问题(6)的古典上解。

证明显然, z_ε是如下问题的古典解:

$\begin{cases}-\left(\left(r+\varepsilon^{1 / \varrho}\right)^\kappa z^{\prime}\right)^{\prime}=1 & \ \ (r \in[0, 1]), \\ z(0)=\frac{1}{2-\kappa} \varepsilon^{(1-\kappa) / \varrho} & (z(1)=0) 。\end{cases}$

(7)

由问题(6)可知 $F(\bar{y})-h_{\varepsilon}\left(r, t, \bar{y}, \bar{y}^{\prime}\right) \geqslant \bar{y}_t-\bar{y}^{\prime \prime}-f(r, t)-\frac{\kappa}{r+\varepsilon{ }^{1 / \varrho}} \bar{y}^{\prime}$ 。在[0, T]上g′(t)>0, 则由式(7)可得

$\begin{aligned} & {\left[F(\bar{y})-h_{\varepsilon}\left(r, t, \bar{y}, \bar{y}^{\prime}\right)\right]\left(r+\varepsilon^{1 / \varrho}\right)^\kappa \geqslant} \\ & \ \ \quad\left[\bar{y}_{\varepsilon t}-\bar{y}^{\prime \prime}-f(r, t)-\frac{\kappa}{r+\varepsilon^{1 / \varrho}} \bar{y}^{\prime}\right]\left(r+\varepsilon^{1 / \varrho}\right)^\kappa= \\ & \ \ \quad C_3 g^{\prime}(t) z_{\varepsilon}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa-C_3 g(t)\left[\left(r+\varepsilon^{1 / \varrho}\right)^\kappa z_{\varepsilon}^{\prime}\right]^{\prime}- \\ & \ \ \quad f(r, t)\left(r+\varepsilon^{1 / \varrho}\right)^\kappa \geqslant C_3 g(t)-f(r, t)\left(r+\varepsilon^{1 / \varrho}\right)^\kappa, \end{aligned}$

选取C₃=max{1, 2^κmax_E_T(f/g)}, 则y是问题(6)的古典上解。证毕。

由引理1至引理3和问题(6)的最大化原理, 可得在E_T中

$\underline{y} \leqslant y_{\varepsilon} \leqslant \bar{y}_{\varepsilon}$

(8)

几乎处处成立。

引理4 对∀ε∈(0, 1), t∈(0, T], (r, t)∈E_t, 有

$\iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left|y_{\varepsilon}^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t+\frac{1}{2} \int_0^1\left(r+\varepsilon^{1 / \varrho}\right)^\kappa y_{\varepsilon}^2 \mathrm{~d} r \leqslant C,$

其中, E_t=(0, 1)×(0, t], C是与ε无关的正常数。

证明由

$\iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(F\left(y_{\varepsilon}\right)-h_{\varepsilon}\left(r, t, y_{\varepsilon}, y_{\varepsilon}^{\prime}\right)\right) y_{\varepsilon} \mathrm{d} r \mathrm{~d} t=0$

可得

$\begin{aligned} & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left|y_{\varepsilon}^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t+\frac{1}{2} \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa \frac{\partial\left|y_{\varepsilon}\right|^2}{\partial t} \mathrm{~d} r \mathrm{~d} t= \\ & -\lambda \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa \frac{\left|y_{\varepsilon}^{\prime}\right|^2 y_{\varepsilon}}{\left(y_{\varepsilon}+\varepsilon^2\right)^m} \mathrm{~d} r \mathrm{~d} t+\iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa f(r, t) y_{\varepsilon} \mathrm{d} r \mathrm{~d} t \leqslant \\ & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa f(r, t) y_{\varepsilon} \mathrm{d} r \mathrm{~d} t_{\circ} \end{aligned}$

由式(8)可知

$\begin{aligned} & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left|y_{\varepsilon}^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t+\frac{1}{2} \int_0^1\left(r+\varepsilon^{1 / \varrho}\right)^\kappa y_{\varepsilon}^2 \mathrm{~d} r \leqslant \\ & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa f(r, t) \bar{y}_{\varepsilon} \mathrm{d} r \mathrm{~d} t+\frac{1}{2} \int_0^1\left(r+\varepsilon^{1 / \varrho}\right)^\kappa \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^2(r) \mathrm{d} r \leqslant C, \end{aligned}$

其中, C是与ε无关的正常数。证毕。

下面给出 $\partial y_{\varepsilon} / \partial t$ 的局部估计。

引理5 对所有的ζ∈C_c^∞(0, 1), ε∈(0, 1), ∀t∈(0, T], 有

$\begin{aligned} & \iint_{E_t}\left[\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|y_{\varepsilon t}\right|^2 \zeta^2\right] \mathrm{d} r \mathrm{~d} t+ \\ & \ \ \quad \int_0^1\left[\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|y_{\varepsilon}^{\prime}\right|^2 \zeta^2\right] \mathrm{d} r \leqslant C_\delta(T), \end{aligned}$

其中, E_t=(0, 1)×(0, t], δ∈(0, 1/2), C_δ是仅与δ相关的正常数。

证明令

$u_{\varepsilon}= \begin{cases}\frac{\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{1-\lambda}}{1-\lambda} & (\lambda \neq 1), \\ \ln \left(y_{\varepsilon}+\varepsilon^{\varrho}\right) & (\lambda=1), \end{cases}$

则 $u_{e t}=\left(y_{\varepsilon}+\varepsilon^Q\right)^{-\lambda} y_{\varepsilon t}, u_{\varepsilon}^{\prime}=\left(y_{\varepsilon}+\varepsilon^Q\right)^{-\lambda} y_{\varepsilon}^{\prime}$ , 故式(6)可以写为以下形式：

$\left(r+\varepsilon^{1 / \varrho}\right)^\kappa u_{\varepsilon t}-\left(\left(r+\varepsilon^{1 / \varrho}\right)^\kappa u_{\varepsilon}^{\prime}\right)^{\prime}=\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-\lambda} f(r, t)_{\circ}$

首先, 考虑等式

$\begin{array}{*{20}{l}} {\iint_{{E_t}} {\left[ {{{\left( {r + {\varepsilon ^{1/\varrho }}} \right)}^\kappa }\left[ {{{\left| {{u_{\varepsilon t}}} \right|}^2} - {{\left( {{y_\varepsilon } + {\varepsilon ^\varrho }} \right)}^{ - \lambda }}f(r,t){u_{\varepsilon t}}} \right] - } \right.}} \\ \ \ \ \ \ \ \ \ \ {\left. {{{\left( {{{\left( {r + {\varepsilon ^{1/\varrho }}} \right)}^\kappa }u_\varepsilon ^\prime } \right)}^\prime }{u_{\varepsilon t}}} \right]{\zeta ^2}{\text{d}}r{\text{d}}t = 0}。 \end{array}$

(9)

注意到式(9)中

$\begin{aligned} & -\iint_{E_t}\left(\left(r+\varepsilon^{1 / \varrho}\right)^\kappa u_{\varepsilon}^{\prime}\right)^{\prime} u_{\varepsilon t} \zeta^2 \mathrm{~d} r \mathrm{~d} t= \\ & \ \ \ \ \ \ \ \ \ \quad \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left[\frac{1}{2}\left|u_{\varepsilon t}^{\prime}\right|^2 \zeta^2+2 u_{z t} u_{\varepsilon}^{\prime} \zeta \zeta^{\prime}\right] \mathrm{d} r \mathrm{~d} t_{\circ} \end{aligned}$

(10)

因为ζ∈C_c^∞(0, 1), $\exists \delta \in(0,1 / 2)$ , 使得在(0, 1)\[δ, 1-δ]上, ζ≡0, ζ′≡0。再采用与参考文献[18]相似的方法, 由式(9)、式(10)、HÖlder不等式和Young不等式可得

$\begin{aligned} & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|y_{e t}\right|^2 \zeta^2 \mathrm{~d} r \mathrm{~d} t+ \\ & \ \ \ \ \ \ \ \ \int_0^1\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|y_{\varepsilon}^{\prime}\right|^2 \zeta^2 \mathrm{~d} r \leqslant \\ & \ \ \ \ \ \ \ \ C \int_\delta^{1-\delta} \int_0^T\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left[f^2 \zeta^2+\left|y_{\varepsilon}\right|^2\left|\zeta^{\prime}\right|^2\right] \mathrm{d} r \mathrm{~d} t+ \\ & \ \ \ \ \ \ \ \ \int_\delta^{1-\delta}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(r)+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}\right|^2 \zeta^2 \mathrm{~d} r_{\circ} \end{aligned}$

在 $[\delta, 1-\delta] \times[0, T] \text { 上, } 0<\underline{y} \leqslant y_{\varepsilon}, \left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda} \leqslant\left(\min _{[\delta, 1-\delta] \times[0, T]} y\right)^{-2 \lambda}<\infty$ ，则由引理4可知

$\begin{aligned} C & \int_\delta^{1-\delta} \int_0^T\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\varrho}\right)^{-2 \lambda}\left[f^2 \zeta^2+\left|y_{\varepsilon}\right|^2\left|\zeta^{\prime}\right|^2\right] \mathrm{d} r \mathrm{~d} t+ \\ & \int_\delta^{1-\delta}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(r)+\varepsilon^{\varrho}\right)^{-2 \lambda}\left|\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}\right|^2 \zeta^2 \mathrm{~d} r \leqslant C_\delta(T), \end{aligned}$

因此

$\begin{aligned} & \iint_{E_t}\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\prime}\right)^{-2 \lambda}\left|y_{\varepsilon t}\right|^2 \zeta^2 \mathrm{~d} r \mathrm{~d} t+ \\ & \ \ \ \ \ \ \ \ \quad \int_0^1\left(r+\varepsilon^{1 / \varrho}\right)^\kappa\left(y_{\varepsilon}+\varepsilon^{\prime}\right)^{-2 \lambda}\left|y_{\varepsilon}^{\prime}\right|^2 \zeta^2 \mathrm{~d} r \leqslant C_\delta(T) 。 \end{aligned}$

证毕。

引理6 对∀ε∈(0, 1), ∀r₁, r₂∈[δ, 1-δ]和∀t₁, t₂∈[0, T], 有

$\left|y_{\varepsilon}\left(r_2, t_2\right)-y_{\varepsilon}\left(r_1, t_1\right)\right| \leqslant C_{1 \delta}\left|r_2-r_1\right|^{V_e}+C_{2 \delta}\left|t_2-t_1\right|^{1 / \varrho^2},$

其中, δ∈(0, 1/2), C_1δ、C_2δ仅与δ有关, 且C_1δ为非负常数, C_2δ是正常数。

证明现假设r₁ <r₂, 则 $\exists \bar{r} \in[\delta, 1-\delta]$ , 有

$\begin{gathered} \left|y_{\varepsilon}\left(r_2, t_2\right)-y_{\varepsilon}\left(r_1, t_1\right)\right| \leqslant\left|y_{\varepsilon}\left(r_2, t_2\right)-y_{\varepsilon}\left(\bar{r}, t_2\right)\right|+ \\ \quad\left|y_{\varepsilon}\left(\bar{r}, t_2\right)-y_{\varepsilon}\left(\bar{r}, t_1\right)\right|+\left|y_{\varepsilon}\left(\bar{r}, t_1\right)-y_{\varepsilon}\left(r_1, t_1\right)\right|_{\circ} \end{gathered}$

(11)

由引理5, 可得

$\begin{aligned} & \left|y_{\varepsilon}\left(r_2, t_2\right)-y_{\varepsilon}\left(\bar{r}, t_2\right)\right|=\left|\int_{\bar{r}}^{r_2} y_{\varepsilon}^{\prime}\left(t_2\right) \mathrm{d} r\right| \leqslant \\ & \quad\left(\int_{\bar{r}}^{r_2}\left|y_{\varepsilon}^{\prime}\left(t_2\right)\right|^{\varrho /\left(\varrho^{-1)}\right.} \mathrm{d} r\right)^{1-1 / \varrho}\left|r_2-\bar{r}\right|^{1 / \varrho} \leqslant \\ & \quad\left(\int_\delta^{1-\delta}\left|y_{\varepsilon}^{\prime}\left(t_2\right)\right|^{\varrho / \varrho-1)} \mathrm{d} r\right)^{1-1 / \varrho}\left|r_2-\bar{r}\right|^{1 / \varrho} \leqslant \\ & \quad C_\delta\left|r_2-\bar{r}\right|^{1 / \varrho} 。 \end{aligned}$

同理可得式(11)右端第3项 $\left|y_{\varepsilon}\left(\bar{r}, t_1\right)-y_{\varepsilon}\left(r_1, t_1\right)\right| \leqslant C_\delta\left|r_1-\bar{r}\right|^{1 / \varrho}$ 。

取Δt=|t₂-t₁|, 使得 $r_1+\Delta t^{1 / \varrho} \in[\delta, 1-\delta]$ 。由引理5、Hölder不等式和Young不等式, 可得

$\begin{aligned} & \int_{r_1}^{r_1+\Delta t^{1 / e}}\left|y_e\left(r, t_2\right)-y_{\varepsilon}\left(r, t_1\right)\right| \mathrm{d} r \leqslant \\ & \int_{t_1}^{t_2} \int_{r_1+\Delta t^{1 / e}}^{r_1}\left|y_{e t}\right| \mathrm{d} r \mathrm{~d} t \leqslant\left(\int_{t_1}^{t_2} \int_{r_1}^{r_1+\Delta t / / e}\left|y_{e t}\right|^{k(\varrho)} \mathrm{d} r \mathrm{~d} t\right)^{\frac{1}{k(\varrho)}} \Delta t^{1 / \varrho^2} \leqslant \\ & \left(\int_0^T \int_\delta^{1-\delta}\left|y_{e t}\right|^{k(\varrho)} \mathrm{d} r \mathrm{~d} t\right)^{\frac{1}{k(e)}} \Delta t^{1 / \varrho^2} \leqslant C_\delta \Delta t^{1 / \varrho^2}, \end{aligned}$

(12)

其中， $k(\varrho)=\left(\varrho^2+\varrho\right) /\left(\varrho^2+\varrho-1\right)$ 。

另外, $\exists \bar{r} \in\left[r_1, r_1+\Delta t^{1 / \varrho}\right]$ , 满足

$\int_{r_1}^{r_1+\Delta t^{1 / e}}\left(y_{\varepsilon}\left(r, t_2\right)-y_{\varepsilon}\left(r, t_1\right)\right) \mathrm{d} r=\left(y_{\varepsilon}\left(\bar{r}, t_2\right)-y_{\varepsilon}\left(\bar{r}, t_1\right)\right) \Delta t^{1 / \varrho},$

再结合式(12)可得

$\left|y_{\varepsilon}\left(r, t_2\right)-y_{\varepsilon}\left(r, t_1\right)\right| \leqslant C_{2 \delta} \Delta t^{1 / \varrho^2} 。$

若不存在这样的, 则由 $\bar{r} \in\left[r_1, r_1+\Delta t^{1 / \varrho}\right]$ , 可知 $r_1<r_2<\bar{r} \leqslant r_1+\Delta t^{1 / \varrho}$ 。因此

$0<\bar{r}-r_1 \leqslant \Delta t^{1 / \varrho}, 0<\bar{r}-r_2 \leqslant \Delta t^{1 / \varrho} \text { 。 }$

最后, 由式(11)可得

$\left|y_{\varepsilon}\left(r_2, t_2\right)-y_{\varepsilon}\left(r_1, t_1\right)\right| \leqslant C_{1 \delta}\left|r_2-r_1\right|^{1 / \varrho}+C_{2 \delta}\left|t_2-t_1\right|^{1 / \varrho^2} \text {, }$

证毕。

由引理6, 可知 $\left\|y_{\varepsilon}\right\|_{C^{1 / \varrho}, 1 / \varrho^2((0, 1) \times[0, \tau])}$ 一致有界。利用Arzelá -Ascoli定理和取对角线序列方法, 由引理4、引理5和式(8)可知: 当ε→0时, 在C((0, 1)×(0, T])中, 存在子序列{y_ε}一致收敛到某个函数y∈C((0, 1)×(0, T])∩V, 即有

$y_{\varepsilon} \rightarrow y, (r, t) \in E_T ;$

(13)

在L²(E_T)中, 有

$y_{\varepsilon}^{\prime} \rightharpoonup y^{\prime} \text {; }$

(14)

在L²(0, T; L_loc²(0, 1))和V^*中, 均有

$\frac{\partial y_{\varepsilon}}{\partial t} \rightharpoonup \frac{\partial y}{\partial t}$

(15)

式(14)、(15)中“⇀”表示弱收敛。

从引理4和Lukin定理可知

$\iint_{E_t} r^\kappa\left|y^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t \leqslant \underline{\lim } \iint_{E_s} r^\kappa\left|y_{\varepsilon}^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t \leqslant C(T) 。$

(16)

引理7 对∀ζ∈C_c^∞(0, 1), 当ε→0时, 有

$\left.\iint_{E_t} r^{\kappa+1} \zeta^2|| y_{\varepsilon}^{\prime}\right|^2-\left|y^{\prime}\right|^2 \mid \mathrm{d} r \mathrm{~d} t \rightarrow 0 \text { 。 }$

证明

$\iint_{E_t} r^{\kappa+1} \zeta^2\left(y_{\varepsilon}-y\right)\left(F\left(y_{\varepsilon}\right)-h_{\varepsilon}\left(r, t, y_{\varepsilon}, y_{\varepsilon}^{\prime}\right)\right) \mathrm{d} r \mathrm{~d} t=0,$

将其展开后可得

$\begin{gathered} \begin{array}{*{20}{l}} {\iint_{{E_t}} {{r^{\kappa + 1}}}{\zeta ^2}y_\varepsilon ^\prime \left( {{y_\varepsilon } - {y^\prime }} \right){\text{d}}r{\text{d}}t + \iint_{{E_t}} {{r^{\kappa + 1}}}{\zeta ^2}\left( {{y_\varepsilon } - y} \right){y_{\varepsilon t}}{\text{d}}r{\text{d}}t + } \\ \ \ \ \ \ \ {2\iint_{{E_t}} {{r^{\kappa + 1}}}y_\varepsilon ^\prime \zeta {\zeta ^\prime }\left( {{y_\varepsilon } - y} \right){\text{d}}r{\text{d}}t + \iint_{{E_t}} {\left[ {\left( {\kappa + 1 - \frac{{\kappa r}}{{r + {\varepsilon ^{1/\varrho }}}}} \right){r^\kappa }{\zeta ^2} \times } \right.}} \\ \ \ \ \ \ \ {\left. {y_\varepsilon ^\prime \left( {{y_\varepsilon } - y} \right)} \right]{\text{d}}r{\text{d}}t + \lambda \iint_{{E_t}} {{r^{\kappa + 1}}}{\zeta ^2}{{\left( {{y_\varepsilon } + {\varepsilon ^\varrho }} \right)}^{ - 1}}\left( {{y_\varepsilon } - y} \right){{\left| {y_\varepsilon ^\prime } \right|}^2}{\text{d}}r{\text{d}}t - } \\ \ \ \ \ \ \ {\iint_{{E_t}} {{r^{\kappa + 1}}}{\zeta ^2}\left( {{y_\varepsilon } - y} \right)f(r,t){\text{d}}r{\text{d}}t = 0} 。 \end{array} \end{gathered}$

(17)

当ε→0, 采用类似文献[18]的方法, 由引理4、引理5、Hölder不等式和Lebesgue控制收敛定理, 可得

$\begin{gathered} I_1=\iint_{E_t} r^{\kappa+1} \zeta^2\left(y_{\varepsilon}-y\right) y_{\varepsilon t} \mathrm{~d} r \mathrm{~d} t \rightarrow 0, \\ I_2=\iint_{E_t} r^{\kappa+1} y_{\varepsilon}^{\prime} \zeta \zeta^{\prime}\left(y_{\varepsilon}-y\right) \mathrm{d} r \mathrm{~d} t \rightarrow 0, \\ I_3=\iint_{E_t}\left(\kappa+1-\frac{\kappa r}{r+\varepsilon^{1 / \varrho}}\right) r^\kappa \zeta^2 y_{\varepsilon}^{\prime}\left(y_{\varepsilon}-y\right) \mathrm{d} r \mathrm{~d} t \rightarrow 0 \end{gathered}$

已知ζ∈C_c^∞(0, 1), 且在C((0, 1)×[0, T])上, y_ε一致收敛到y，则由式(8)、引理4和Lebesgue控制收敛定理, 当ε→0时, 可得

$\begin{gathered} \begin{array}{*{20}{l}} {{I_4} = \lambda \iint_{{E_t}} {{r^{\kappa + 1}}}{\zeta ^2}{{\left( {{y_\varepsilon } + {\varepsilon ^\varrho }} \right)}^{ - 1}}\left( {{y_\varepsilon } - y} \right){{\left| {y_\varepsilon ^\prime } \right|}^2}{\text{d}}r{\text{d}}t} \\ \ \ {\quad C\lambda {{\max }_{{E_T}}}\left| {\zeta \left( {{y_\varepsilon } - y} \right)} \right| \to 0} \end{array} \end{gathered}$

(18)

由式(18)和式(16), 当ε→0时, 有

$\iint_{E_t} r^{n+1} \zeta^2 y_{\varepsilon}^{\prime}\left(y_{\varepsilon}-y\right) \mathrm{d} r \mathrm{~d} t \rightarrow 0 \text { 。 }$

由式(13), 当ε→0时, 有

$\iint_{E_t} r^{\kappa+1} \zeta^2 y_{\varepsilon}^{\prime}\left(y_z-y\right)^{\prime} \mathrm{d} r \mathrm{~d} t \rightarrow 0 \text { 。 }$

因此

$\iint_{E_t} r^{\kappa+1} \zeta^2\left|y_z^{\prime}-y^{\prime}\right| \mathrm{d} r \mathrm{~d} t \rightarrow 0 \text { 。 }$

最后, 在V^*中, 由式(15)和引理4, 可得

$\begin{aligned} & \iint_{E_t} r^{\kappa+1} \zeta^2 \|\left. y_{\varepsilon}^{\prime}\right|^2-\left|y^{\prime}\right|^2 \mid \mathrm{d} r \mathrm{~d} t \leqslant \\ & \ \ \ \ C\left(\iint_{E_t} r^{\kappa+1} \zeta^2\left(\left|y_{\varepsilon}^{\prime}\right|^2+\left|y^{\prime}\right|^2\right) \mathrm{d} r \mathrm{~d} t\right)^{\frac{1}{2}} \times \\ & \quad\left(\iint_{E_t} r^{\kappa+1} \zeta^2\left|y_{\varepsilon}^{\prime}-y^{\prime}\right|^2 \mathrm{~d} r \mathrm{~d} t\right)^{\frac{1}{2}} \rightarrow 0, \end{aligned}$

证毕。

注1 在证明引理7的过程中, 可以进一步证得问题(3)~(5)的弱解y∈C((0, 1)×[0, T])。

本文主要结论如下:

定理1 令κ>0, λ>(κ+1)/2, 假设条件(H1)、(H2)成立, 且函数φ满足条件(H3), 则问题(3)~(5)至少存在一个弱解。更进一步, ∀t∈0, T, 有y′(0, t)=y′(1, t)=0。

证明由引理7和式(13)~(15), 易知y满足定义1中的条件(i)~(iii)。由式(8), 当ε→0时, 有 $\lim _{r \rightarrow 0^{+}} y(r, t)=\lim _{r \rightarrow 0^{-}} y(r, t)=0$ 。则可令y(0, t)=y(1, t)=0(t∈(0, T]), 从而可知y(r, t)满足边值条件(4)。

接下来, 仍采用文献[18]类似的方法, 易证y满足定义1中的条件(iv)。

此外, 由式(8)得

$\begin{aligned} & C_1 g^e(t)\left(r-r^2\right)^{\varrho} \leqslant y(r, t) \leqslant \\ & \quad C_2 g(t) \cdot \min \left\{r^{\varrho}, (1-r)^{\varrho}\right\} \quad\left((r, t) \in E_T\right) 。 \end{aligned}$

从而

$\begin{gathered} 0 \leqslant y^{\prime}(0, t)=\lim _{r \rightarrow 0^{+}} \frac{y(r, t)}{r} \leqslant \lim _{r \rightarrow 0^{+}} C_2 g(t) \frac{r^{\varrho}}{r}=0, \\ 0=\lim _{r \rightarrow 1^{-}} C_2 g(t) \frac{(1-r)^{\varrho}}{r-1} \leqslant y^{\prime}(1, t)=\lim _{r \rightarrow 1^{-}} \frac{y(r, t)}{r-1} \leqslant 0 。 \end{gathered}$

因此, y′(0, t)、y′(1, t)存在且y′(0, t)=y′(1, t)=0, t∈(0, T]。证毕。

定理2 令0≤κ <1, λ>0, 假设条件(H1)、(H2)成立, 并且函数φ满足条件(H4), 则问题(3)~(5)至少存在1个弱解。

定理2的证明过程类似定理1，在此略。

最后, 给出问题(6)的弱比较原理：

定理3 令κ>0, λ>0, 假设y₁、y₂分别为问题(6)的弱下解和弱上解, 如果在(0, 1)内y₂(r, 0)≥y₁(r, 0)几乎处处成立, 且在(0, T)内y₂(0, t)≥y₁(0, t), y₂(1, t)≥y₁(1, t)几乎处处成立, 则在E_T上y₂≥y₁几乎处处成立。

证明由弱下解和弱上解的定义可得：对∀Ψ∈L²(0, T; W₀^{1, 2}(0, 1)∩L^∞(E_T)), 有

$\begin{aligned} & \int_0^T\left\langle\frac{\partial y_2}{\partial t}, \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}\right\rangle+\iint_{E_t}\left[y_2^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}-\frac{\kappa}{r+\varepsilon^{1 / \underline{\underline{o}}}} y_2^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}+\lambda\left|y_2^{\prime}\right|^2 \frac{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}}{\left(y_2+\varepsilon^2\right)^m}-\right. \\ & \ \ \ \ \quad f(r, t) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}] \mathrm{d} r \mathrm{~d} t \geqslant 0, \end{aligned}$

(19)

$\begin{aligned} & \int_0^T\left\langle\frac{\partial y_1}{\partial t}, \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}\right\rangle+\iint_{E_t}\left[y_1^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\prime}-\frac{\kappa}{r+\varepsilon^{1 / \underline{e}}} y_1^{\prime} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}+\lambda\left|y_1^{\prime}\right|^2 \frac{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}}{\left(y_1+\varepsilon^2\right)^m}-\right. \\ & \ \ \ \ \quad f(r, t) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}] \mathrm{d} r \mathrm{~d} t \leqslant 0, \end{aligned}$

(20)

其中, E_t=(0, 1)×(0, t], t∈[0, T]。

定义函数 $\rho(s):(0, \infty) \rightarrow \mathbb{R}$ 为

$\rho(s)=\left\{\begin{array}{l} \int_{a_m}^s \mathrm{e}^{-\frac{\lambda}{1-m} y^{1-m}} \mathrm{~d} y \quad(m \neq 1), \\ \ln s \quad(m=1, \lambda=1), \\ \frac{s^{1-\lambda}}{1-\lambda} \quad(m=1, \lambda \neq 1), \end{array}\right.$

其中, 若m∈(0, 1), 则令a_m=0; 若m≥1, $\exists c_1, c_2>0$ , 使得y_i≥c_i(i=1, 2)成立, 则可取a_m=min(c₁, c₂)。

注意到

$\rho^{\prime}(s)=\left\{\begin{array}{l} \mathrm{e}^{-\frac{\lambda}{1-m^s} s^{1-m}} \quad(m \neq 1), \\ s^{-\lambda} \quad(m=1) \end{array}\right.$

且

$\rho^{\prime \prime}(s)=\left\{\begin{array}{l} \left(-\lambda s^{-m}\right) \mathrm{e}^{-\frac{\lambda}{1-m^{\prime}} s^{1-m}} \quad(m \neq 1), \\ -\lambda s^{-\lambda-1} \quad(m=1), \end{array}\right.$

可得

$\frac{\rho^{\prime \prime}(s)}{\rho^{\prime}(s)}=-\lambda s^{-m} \text { 。 }$

通过以上变换, 式(3)转换为

$[\rho(y)]_t=[\rho(y)]_{r r}+\left[\rho^{\prime}(y)\right] f(r, t)+\frac{\kappa}{r} y^{\prime},$

故不等式(19)、(20)分别可变为

$\begin{array}{*{20}{l}} {\iint_{{E_t}} {\left\{ {{{\left[ {\rho \left( {{y_2}} \right)} \right]}_t}\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} + \left[ {\rho \left( {{y_2}} \right)} \right],{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} ^\prime } - \left[ {{\rho ^\prime }\left( {{y_2}} \right)} \right]f(r,t)\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} - } \right.}} \\ \ \ \ \ \ \ \ {\left. {\frac{\kappa }{r}y_2^\prime \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} } \right\}{\text{d}}r{\text{d}}t0} \end{array}$

(21)

$\begin{array}{*{20}{l}} {\iint_{{E_t}} {\left\{ {{{\left[ {\rho \left( {{y_1}} \right)} \right]}_t}\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} + \left[ {\rho \left( {{y_1}} \right)} \right],{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} ^\prime } - \left[ {{\rho ^\prime }\left( {{y_1}} \right)} \right]f(r,t)\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} - } \right.}} \\ \ \ \ \ \ \ \ {\left. {\frac{\kappa }{r}y_1^\prime \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} } \right\}{\text{d}}r{\text{d}}t0} \end{array}$

(22)

记w=ρ(y₁)-ρ(y₂)=(ρ₁-ρ₂), 再由式(22)减去式(21), 可得

$\begin{array}{*{20}{l}} {\iint_{{E_t}} {\left\{ {{w_t}\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} + {w_r}{\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} ^\prime } + \left[ {{w^\prime }\left( {{y_2}} \right) - {w^\prime }\left( {{y_1}} \right)} \right]f(r,t)\mathit{\boldsymbol{ \boldsymbol{\varPsi} }} - } \right.}} \\ \ \ \ \ \ \ \ \left.\frac{\kappa}{r}\left(y_1^{\prime}-y_2^{\prime}\right) \Psi\right\} \mathrm{d} r \mathrm{d} t \leqslant 0 。 \end{array}$

同时, 注意到\$, 其中(w)₊=max{0, w}, 则有

$\begin{aligned} & \frac{1}{2} \int_0^1\left(w_{+}\right)^2 \mathrm{~d} t+\iint_{E_t}\left(w_{+}\right)_r^2 \mathrm{~d} r \mathrm{~d} t+\iint_{E_t}\left[w^{\prime}\left(y_2\right)-\right. \\ & \ \ \ \ \ \left.w^{\prime}\left(y_1\right)\right] f(r, t) w_{+} \mathrm{d} r \mathrm{~d} t-\iint_{E_t} \frac{\kappa}{r}\left(y_1^{\prime}-y_2^{\prime}\right) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} \mathrm{d} r \mathrm{~d} t \leqslant 0 。 \end{aligned}$

在[a_m, ∞)中, w′≥0且w″≤0, 所以, 在E_T中

$\left[w^{\prime}\left(y_2\right)-w^{\prime}\left(y_1\right)\right] f(r, t) w_{+} \geqslant 0$

几乎处处成立。

又因为y′₁≤y′₂, 故

$-\iint_{E_t} \frac{\kappa}{r}\left(y_1^{\prime}-y_2^{\prime}\right) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} \mathrm{d} r \mathrm{~d} t \geqslant 0 \text { 。 }$

综上可知

$\frac{1}{2} \int_0^1\left(w_{+}\right)^2 \mathrm{~d} t+\iint_{E_t}\left(w_{+}\right)_r^2 \mathrm{~d} r \mathrm{~d} t=0 \text { 。 }$

由此可得, 在E_T上w₊=0几乎处处成立, 故在E_T上y₂≥y₁几乎处处成立。证毕。

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