本文考虑一类非齐次加权拟线性抛物微分不等式的非平凡非负弱解的非存在性

$ u_{t} \geqslant \Delta_{p} u+a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s}+\omega(x), (x, t) \in S, $

(1)

其中$S: =\mathbf{R}^{N} \times \mathbf{R}^{+}, \mathbf{R}^{+}: =(0, +\infty), N \geqslant 1, p>\frac{2 N}{N+1}$，$\Delta_{p} u: =\operatorname{div}\left(|\nabla u|^{p-2} \nabla u\right), \left\|K^{\frac{1}{r}} u\right\|_{r}^{s}: =\left(\int_{\mathbf{R}^{N}} K(x) \times\right. \left.u^{r}(x, t) \mathrm{d} x\right)^{\frac{s}{r}}, q \geqslant 1, r \geqslant 1, s>0$，非负非平凡反应项$\omega(x) \in L_{\text {loc }}^{1}\left(\mathbf{R}^{N}\right)$，非负初值$u_{0}(x) \in L_{\text {loc }}^{1}\left(\mathbf{R}^{N}\right)$。非负权函数$a(x)$ 和$K(x)$ 为可测且满足如下条件：

$ a(x) \geqslant|x|^{-\alpha}, K(x) \geqslant|x|^{-\beta}, x \in \mathbf{R}^{N} \backslash\{0\}, \alpha, \beta \in \mathbf{R}, $

(2)

其在原点处可能发生奇异或退化现象。注意到，上述不等式(1)称为非局部不等式的原因在于其并非逐点意义下成立且指出，(1)式中含有的加权非局部源项$\left\|K^{\frac{1}{r}} u\right\|_{r}^{s}$ 常出现于种群动力学及人口动力学中^[1-2]。

关于此类问题，Fujita等^[3-5]做出了开创性的研究工作并在全空间上得到了如下具有乘幂型局部源的经典热传导方程

$ u_{t}-\Delta u=u^{q}, \quad q>1, $

非平凡非负经典解的存在性与爆破的临界指标为$q_{F}: =1+2 / N$, 即对任意初值，当$1 < q < q_{F}$时解在有限时刻发生爆破，而当$q>q_{F}$时，对小初值，解整体存在；对大初值，解发生爆破。从此以后，Fujita型结果受到了许多学者的关注并有了各种方向的新的延伸，其中本文中所关心的问题是在全空间上非常有趣的非齐次问题中的延伸。Bandle等^[6]首次在全空间上考虑了非齐次热传导方程$u_{t}-\Delta u=u^{q}+\omega(x)$并结合格林函数法和辅助函数法证明了Fujita型临界指标为

$ q_{I H F}:=1+2 /(N-2)(N \geqslant 3) 。$

显然，它与相应的齐次问题的Fujita临界指标q_F比较更大，是与其齐次稳态不等式$-\Delta u \geqslant u^{q}$的Sobolev临界指标一致。事实上，在全空间上关于含乘幂型局部源的齐次或非齐次方程及不等式的整体解的存在性与爆破的Fujita型临界指标研究已有许多文献和结论^[7-15], 其中，齐次问题的综述性及最新进展见文献[9-11], 非齐次问题的相关研究见文献[12-15]。此外，关于齐次非局部抛物方程及微分不等式解的非存在性研究进展方面，可参考文献[16]的第五节及文献[17-18]。

本文受到Mitideri和Pohozaev^[19-20]及Kartsatos和Kurta^[13]的研究成果启发，在试验函数理论框架内，利用非线性容度法在非局部问题中的扩展技巧，建立一大类非齐次非局部微分不等式(1)非平凡非负弱解的Fujita型非存在性定理。此种研究方法的特点如下：无需利用极大值原理或比较原理，无需假设解在无穷远处的行为或对称性或初始迹。

本文的主要结果陈述如下：

定理1   令$N \geqslant 1, p>\frac{2 N}{1+N}, q \geqslant 1, r \geqslant 1, s>0$, 非负非平凡函数$\omega(x) \in L_{\text {loc }}^{1}\left(\mathbf{R}^{N}\right)$，权函数$a(x)$ 和$K(x)$ 满足公式(2), 指数α, β满足如下关系式：

$ \alpha+\frac{s}{r} \beta<\min \left\{N\left(1+\frac{s}{r}\right)-\frac{N-p}{p-1}, p\left(1+\frac{s}{r}\right)\right\}, $

(3)

其中q、r不同时为1。

(Ⅰ)如果

$ \begin{align*} & \max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\}<q+s< \\ & Q_{I H F}:=(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}, \end{align*} $

(4)

则非齐次非局部微分不等式(1)在集合

$ \begin{gathered} S_{d}:=\left\{u \in W_{\mathrm{loc}}^{1, p}(S): a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s}, \right. \\ \left.|\nabla u|^{p} u^{-d-1} \in L_{\mathrm{loc}}^{1}(S)\right\} \end{gathered} $

上不存在非平凡非负弱解，其中充分小的d满足

$ 0<d<\min \left\{1, p-1, \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}\right\} 。$

(Ⅱ)如果

$ \begin{aligned} & \max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\}<q+s= \\ & Q_{I H F}:=(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}, \end{aligned} $

则非齐次非局部微分不等式(1)在集合

$ \widetilde{S}_{d}:=\left\{\left.u(x, t) \in W_{\mathrm{loc}}^{1, p}(S)| | \nabla u\right|^{p} u^{-1-d} \in L_{\mathrm{loc}}^{1}(S)\right\} $

上不存在非平凡非负弱解，其中充分小的d满足

$ 0<d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)} 。$

注1：定理1中条件(3)保证(4)式为非空区间。

注2：由定理1知，不等式(1)中指标q+s的新的上界为$(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}$。当$s=\alpha=0$时，则完全包含现有的关于含乘幂型局部源项的非齐次扩散方程^{[6, 14-15]}及非齐次局部不等式^[13]给出的关于参数q分类的Fujita型临界指标。同时，值得注意的是，如果取$s=\alpha=0$, 则得到Fujita型临界指标为$p-1+(p-1) \times$ $\frac{p}{N-p}=\frac{N(p-1)}{N-p}$, 其中$p < N$。因此，由$1 < q \leqslant \frac{N(p-1)}{N-p}$易知，它与相应的齐次不等式中得到的Fujita型第一临界指标$p-1+\frac{p}{N}$ (见文献[11])相比更大。此外，当$\frac{s}{r} N-\alpha-\frac{s}{r} \beta \leqslant 0$ 时，它与文献[18]中定理1给出的相应齐次不等式的Fujita型第一临界指标$Q_{F}=p-1+\frac{s}{r}+\frac{p-\alpha-\frac{s}{r} \beta}{N}$之间也可比较大小, 即$Q_{I H F}>Q_{F}$。事实上，

$ Q_{I H F}-Q_{F}=\left(p+\frac{s}{r} N-\alpha-\frac{s}{r} \beta\right) \frac{1}{N}\left(\frac{N(p-1)}{N-p}-1\right) $

由条件

$ \alpha+\frac{s}{r} \beta<\min \left\{N\left(1+\frac{s}{r}\right)-\frac{N-p}{p-1}, p\left(1+\frac{s}{r}\right)\right\} $

和$p < N$知

$ p+\frac{s}{r} N-\alpha-\frac{s}{r} \beta>\left(1+\frac{s}{r}\right) p-\alpha-\frac{s}{r} \beta>0 。$

再由条件$\frac{s}{r} N-\alpha-\frac{s}{r} \beta \leqslant 0$ 和$Q_{I H F}>1$得到

$ \frac{N(p-1)}{N-p}-1=Q_{I H F}-1-(p-1) \frac{\frac{s}{r} N-\alpha-\frac{s}{r} \beta}{N-p}>0 。$

因此，$Q_{I H F}>Q_{F}$。

1 预备知识

本节中介绍一些记号、弱解的定义、试验函数的选取办法并导出预备性引理及精细的非线性容度估计值。

注意到，下文的不同行中出现的正常数C>0可能代表不同常数且其不依赖于弱解u。

首先引入弱解的定义。

定义1   令$\omega(x) \in L_{\text {loc }}^{1}\left(\mathbf{R}^{N}\right)$, 非负函数$u(x, t) \in W_{\text {loc }}^{1, p}(S)$满足

(Ⅰ)$K(x) u^{r}(x, t) \in L_{\text {loc }}^{1}\left(\mathrm{R}^{N}\right)$，

(Ⅱ)$a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \in L_{\text {loc }}^{1}(S)$，

且

$ \begin{gather*} \int_{S} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \varphi \mathrm{~d} x \mathrm{~d} t+\int_{S} \omega(x) \varphi \mathrm{d} x \mathrm{~d} t \leqslant \\ \int_{S} u_{t} \varphi \mathrm{~d} x \mathrm{~d} t+\int_{S}|\nabla u|^{p-2} \nabla u \cdot \nabla \varphi \mathrm{~d} x \mathrm{~d} t, \end{gather*} $

(5)

其中$\forall \varphi \in C_0^{\infty}(S)$为非负，则$u(x, t)$称为非齐次非局部微分不等式(1)的弱解。

接下来，构造带参数的分离变量形式的试验函数。令$B_{R}(0)$是$\mathbf{R}^{N}$中以原点为中心，R>0为半径的单位球。定义函数$\xi_{0}(s) \in C^{1}([0, \infty))$ 使得

$ \begin{gathered} 0 \leqslant \xi_{0}(s) \leqslant 1, \quad \forall s \geqslant 0 ; \xi_{0}(s)=1, 0 \leqslant s \leqslant 1 ; \\ \xi_{0}(s)=0, s \geqslant 2, \end{gathered} $

且存在正常数C>0, 使得

$ \left|\xi_{0}^{\prime}(s)\right| \leqslant C, \forall s \geqslant 0 \text { 。} $

对于空间变量，取

$ \chi(x):=\xi_{0}\left(\frac{|x|}{R}\right), $

因此$\chi(x) \in C_{0}^{1}\left(\mathbf{R}^{N}\right)$且满足

$ \begin{gathered} \chi(x)=1, x \in B_{R}(0) ; \chi(x)=0, x \in \mathbf{R}^{N} \backslash B_{2 R}(0), \\ 0 \leqslant \chi(x) \leqslant 1, \forall x \in \mathbf{R}^{N} ;|\nabla \chi(x)| \leqslant \frac{C}{R}, \forall x \in \mathbf{R}^{N} 。\end{gathered} $

其中C>0是常数。对于时间变量，取

$ \eta(t):=\xi_{0}\left(\frac{t}{R^{\gamma}}\right), $

其中γ>0待定。

现在，对$\forall R>0$, 在S上的非负截断函数定义为

$ \zeta(x, t):=\chi(x) \eta(t), $

则显然$\zeta(x, t) \in C_{0}^{1}(S)$。同时，记支集为

$ \begin{gathered} P(2 R):=\operatorname{supp} \zeta=B_{2 R}(0) \times\left[0, 2 R^{\gamma}\right], \\ P_{1}(R):=\operatorname{supp} \nabla \zeta=\left\{x|R \leqslant|x| \leqslant 2 R\} \times\left[0, 2 R^{\gamma}\right], \right. \\ P_{2}(R):=\operatorname{supp} \zeta_{t}=B_{2 R}(0) \times\left[R^{\gamma}, 2 R^{\gamma}\right] 。\end{gathered} $

下面，引入试验函数

$ \varphi(x, t):=\widetilde{u}_{\varepsilon}^{-d} \zeta^{k}, $

其中

$ \begin{gathered} \widetilde{u}_{\varepsilon}:=\tau+\int_{\mathrm{R}^{N}} \xi_{\varepsilon}(x-y, t) u(y, t) \mathrm{d} y, \\ u_{\tau}(x, t):=\tau+u(x, t), \end{gathered} $

$u(x, t)$ 为非齐次非局部不等式(1)的非负弱解，$\varepsilon>0$，$\tau>0, d>0, k>0$，且$\left(\xi_{\varepsilon}\right)_{\varepsilon>0}$ 表示标准的磨光化算子族。由此可知，对$\forall(x, t) \in S, \widetilde{u}_{\varepsilon}, u_{\tau} \geqslant \tau>0, u_{\tau}(x, t)$按参数τ递增，且$\varphi(x, t) \in C_{0}^{\infty}(S)$。

下面，给出一些预备性引理。

引理1   令$q \geqslant 1, r \geqslant 1, s>0, u(x, t)$ 为非齐次非局部微分不等式(1)的非负弱解，且$u(x, t) \in S_{d}$, 则对$\forall 0 < d < 1, \forall k>0$, 可得到

$ \begin{gather*} \int_{P_{2}(R)} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{1} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t, \end{gather*} $

(6)

其中

$ \kappa_{1}=\frac{q+s-d}{1-d}, \kappa_{1}^{\prime}=\frac{q+s-d}{q+s-1}, \kappa_{2}=\frac{\kappa_{1}}{\kappa_{1}-\left(1+\frac{s}{r}\right)}, $

且$\varepsilon_{1}, C>0$。

证明  通过简单计算可得，在(6)式左端中被积函数等价于

$ \begin{gathered} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1}=\left(a(x)^{\frac{1}{\kappa_{1}}} u^{\frac{q-d}{\kappa_{1}}} \zeta^{\frac{k}{\kappa_{1}}}\right)\left(K(x)^{\frac{s}{\kappa_{1}}} u^{\frac{s}{\kappa_{1}}}\right) \times \\ \left(a(x)^{-\frac{1}{\kappa_{1}}} K(x)^{-\frac{s}{r\kappa_{1}}} u^{1-d-\frac{q+s-d}{\kappa_{1}}}\left|\zeta_{t}\right| \zeta^{\frac{k}{\kappa_{1}^{\prime}}-1}\right) 。\end{gathered} $

为了使上式最后一项中u的指数满足

$ 1-d-\frac{q+s-d}{\kappa_{1}}=0, $

取$\kappa_{1}=\frac{q+s-d}{1-d}$且$\kappa_{2}$ 满足

$ \frac{1}{\kappa_{1}}+\frac{s}{r \kappa_{1}}+\frac{1}{\kappa_{2}}=1 。$

于是，可断言：$\kappa_{1}, \frac{r \kappa_{1}}{s}, \kappa_{2}>1$。事实上，如下几个等式成立：

$ \begin{gathered} \kappa_{1}-1=\frac{q+s-1}{1-d}, \\ \frac{r \kappa_{1}}{s}-1=\frac{r(q+s-d)-s(1-d)}{s(1-d)}= \\ \frac{s d+r(q-1)+s(r-1)+r(1-d)}{s(1-d)} , \\ \kappa_{2}-1=\frac{\left(1+\frac{s}{r}\right)(1-d)}{q+s-d-\left(1+\frac{s}{r}\right)(1-d)}= \\ \frac{\left(1+\frac{s}{r}\right)(1-d)}{\frac{s}{r} d+q-1+\frac{s}{r}(r-1)}。\end{gathered} $

再由假设$q \geqslant 1, r \geqslant 1, s>0$ 及$0 < d < 1$ 可推出，$\kappa_{1}> 1, \frac{r \kappa_{1}}{s}>1$，且$\kappa_{2}>1$。

利用以$\kappa_{1}, \frac{r \kappa_{1}}{s}$ 和$\kappa_{2}$为指数的Hölder不等式推出

$ \begin{gathered} \int_{|x| \leqslant 2 R} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \leqslant \\ \left(\int_{|x| \leqslant 2 R} a(x) u^{q-d} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{1}}}\left(\int_{|x| \leqslant 2 R} K(x) u^{r}(x, t) \mathrm{d} x\right)^{\frac{s}{r\kappa_{1}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{2}}} \leqslant \\ \left(\int_{|x| \leqslant 2 R} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{1}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{2}}}。\end{gathered} $

关于t变量，在$\left[R^{\gamma}, 2 R^{\gamma}\right]$ 上积分可得

$ \begin{gathered} \int_{P_{2}(R)} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{1}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{2}}} \mathrm{~d} t 。\end{gathered} $

再利用以$\kappa_{1}$ 和$\kappa_{1}^{\prime}$为指数并带参数ε₁的Young’s不等式可推出

$ \begin{gathered} \int_{P_{2}(R)^{n}} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{1} \int_{P_{2}(R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t \leqslant \\ \varepsilon_{1} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t 。\end{gathered} $

引理1证毕。

引理2   令$p>1, q \geqslant 1, r \geqslant 1, s>0$, 且

$ q+s \geqslant\left(1+\frac{s}{r}\right)(p-1), $

其中$q, r$ 不同时为1。如果$u(x, t)$ 为非齐次非局部微分不等式(1)的非负弱解且$u(x, t) \in S_{d}$，则对$\forall 0 < d < p-1, ~ \forall k>0$，可得

$ \begin{gathered} \int_{P_{1}(R)} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{2} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s\kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{\rho \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} R-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t, \end{gathered} $

(7)

其中

$ \kappa_{3}=\frac{q+s-d}{p-1-d}, \kappa_{3}^{\prime}=\frac{q+s-d}{q+s-(p-1)}, \kappa_{4}=\frac{\kappa_{3}}{\kappa_{3}-\left(1+\frac{s}{r}\right)}, $

且$\varepsilon_{2}, C>0$。

证明  通过直接计算可得，在(7)式左端的被积函数等价于

$ \begin{gathered} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p}= \\ \left(a(x)^{\frac{1}{\kappa_{3}}} u^{\frac{q-d}{\kappa_{3}}} \zeta^{\frac{k}{k_{3}}}\right)\left(K(x)^{\frac{s}{\kappa_{3}}} u^{\frac{s}{\kappa_{3}}}\right) \times \\ \left(a(x)^{-\frac{1}{\kappa_{3}}} K(x)^{-\frac{s}{\kappa_{3}}} u^{p-1-d-\frac{q+s-d}{\kappa_{3}}}|\nabla \zeta|^{p} \zeta^{\frac{k}{\kappa_{3}^{\prime}}-p}\right) 。\end{gathered} $

为了使上式最后一项中$u$ 的指数为

$ p-1-d-\frac{q+s-d}{\kappa_{3}}=0, $

取$\kappa_{3}=\frac{q+s-d}{p-1-d}$ 且$\kappa_{4}$ 满足

$ \frac{1}{\kappa_{3}}+\frac{s}{r \kappa_{3}}+\frac{1}{\kappa_{4}}=1。$

于是，可断言：$\kappa_{3}, \frac{r \kappa_{3}}{s}, \kappa_{4}>1$。事实上，由假设$r \geqslant 1$，$s>0, p>1$ 及$0 < d < p-1$ 易得

$ \begin{gathered} \kappa_{3}-1=\frac{q+s-(p-1)}{p-1-d}>\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{p-1-d} , \\ \frac{r \kappa_{3}}{s}-1=\frac{r(q+s-d)-s(p-1-d)}{s(p-1-d)}> \\ \frac{r(q+s-d)-(r+s)(p-1-d)}{s(p-1-d)}= \\ \frac{s d+r\left[q+s-\left(1+\frac{s}{r}\right)(p-1)\right]}{s(p-1-d)}, \\ \kappa_{4}-1=\frac{\left(1+\frac{s}{r}\right)(p-1-d)}{q+s-(p-1)-\frac{s}{r}(p-1-d)}= \\ \frac{\left(1+\frac{s}{r}\right)(p-1-d)}{\frac{s}{r} d+q+s-\left(1+\frac{s}{r}\right)(p-1)} 。\end{gathered} $

则由假设$q+s \geqslant\left(1+\frac{s}{r}\right)(p-1)$ 可得，$\kappa_{3}>1, \frac{r \kappa_{3}}{s}>1$，$\kappa_{4}>1$。

于是，利用以$\kappa_{3}, \frac{r \kappa_{3}}{s}$，和$\kappa_{4}$ 为指数的Hölder不等式可导出

$ \begin{gathered} \int_{R \leqslant|x| \leqslant 2 R} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \leqslant \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q-d} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{3}}} \times \\ \left.\left(\int_{R \leqslant|x| \leqslant 2 R} K(x) u_{( }^{r} x, t\right) \mathrm{d} x\right)^{\frac{s}{r\kappa_{3}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s \kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{4}}} \leqslant \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{3}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s \kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{4}}} 。\end{gathered} $

关于$t$ 变量，在$\left[0, 2 R^{\gamma}\right]$ 上积分可得

$ \begin{gathered} \int_{P_{1}(R)} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{3}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s \kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{4}}} \mathrm{~d} t。\end{gathered} $

类似地，利用以$\kappa_{3}$ 和$\kappa_{3}^{\prime}$ 为指数且带有参数$\varepsilon_{2}$ 的Young's不等式

$ \begin{gathered} \int_{P_{1}(R)} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{2} \int_{P_{1}(R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s\kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} d x\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t \leqslant \\ \varepsilon_{2} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s\kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t 。\end{gathered} $

引理2证毕。

引理3   令$p>1, q \geqslant 1, r \geqslant 1, s>0$, 且

$ q+s>\left(1+\frac{s}{r}\right)(p-1), $

其中$q$ 和$r$ 不同时为1。如果$u(x, t)$ 是非齐次非局部微分不等式(1)的非负弱解，且$u(x, t) \in S_{d}$(或$\widetilde{S}_{d}$)，则对于

$ \forall 0<d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)} \text { 及 } \forall k>0 , $

可得到

$ \begin{gathered} \int_{{P}_{1}(R)} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \left(\int_{{P}_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} \times \\ \left(\int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s\kappa_{6}}{r\kappa_{5}}}|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{\kappa_{5}^{\prime}}{\kappa_{6}}} \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}^{\prime}}}, \end{gathered} $

(8)

其中

$ \begin{gathered} \kappa_{5}=\frac{q+s}{(p-1)(1+d)}, \kappa_{5}^{\prime}=\frac{q+s}{q+s-(p-1)(1+d)}, \\ \kappa_{6}=\frac{\kappa_{5}}{\kappa_{5}-\left(1+\frac{s}{r}\right)} \text { 。} \end{gathered} $

证明  通过直接计算可得，在(8)式左端的被积函数等价于

$ \begin{gathered} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p}=\left(a(x)^{\frac{1}{\kappa_{5}}} u^{\frac{q}{\kappa_{5}}} \zeta^{\frac{k}{\kappa_{5}}}\right)\left(K(x)^{\frac{s}{r\kappa_{5}}} u^{\frac{s}{\kappa_{5}}}\right) \times \\ \left(a(x)^{-\frac{1}{\kappa_{5}}} K(x)^{-\frac{s}{{r\kappa_{5}}}} u^{(p-1)(1+d)-\frac{q+s}{\kappa_{5}}}|\nabla \zeta|^{p} \zeta^{\frac{k}{\kappa_5^{\prime}}-p}\right) { 。} \end{gathered} $

同样地，为了使上式的最后一项中$u$ 的指数为

$ (p-1)(1+d)-\frac{q+s}{\kappa_{5}}=0, $

取$\kappa_{5}=\frac{q+s}{(p-1)(1+d)}$ 且$\kappa_{6}$ 满足

$ \frac{1}{\kappa_{5}}+\frac{s}{r \kappa_{5}}+\frac{1}{\kappa_{6}}=1。$

于是，可断言：$\kappa_{5}, \frac{r \kappa_{5}}{s}, \kappa_{6}>1$。事实上，由假设$r \geqslant 1$，$s>0$ 及$p>1$ 易得到

$ \begin{gathered} \kappa_{5}-1=\frac{q+s-(p-1)(1+d)}{(p-1)(1+d)}=\frac{q+s-(p-1)(1+d)}{(p-1)(1+d)}, \\ \frac{r \kappa_{5}}{s}-1=\frac{q+s-\frac{s}{r}(p-1)(1+d)}{\frac{s}{r}(p-1)(1+d)}> \\ \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)(1+d)}{\frac{s}{r}(p-1)(1+d)}= \\ \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)-d\left(1+\frac{s}{r}\right)(p-1)}{\frac{s}{r}(p-1)(1+d)}, \\ \kappa_{6}-1=\frac{\left(1+\frac{s}{r}\right)(p-1)(1+d)}{q+s-\left(1+\frac{s}{r}\right)(p-1)(1+d)}=\\ \frac{\left(1+\frac{s}{r}\right)(p-1)(1+d)}{q+s-\left(1+\frac{s}{r}\right)(p-1)-d\left(1+\frac{s}{r}\right)(p-1)}。\\ \end{gathered} $

结合条件$0 < d < \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}$易知，$\frac{r \kappa_5}{s}>1$和$\kappa_6>1$。另外，对任意的

$ 0<d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}, $

显然有

$ \begin{gathered} d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}= \\ \frac{q+s-(p-1)}{p-1}-\frac{\frac{s}{r}(q+s)}{\left(1+\frac{s}{r}\right)(p-1)}<\frac{q+s-(p-1)}{p-1} 。\end{gathered} $

因此$\kappa_{5}>1$ 也成立。

利用以$\kappa_{5}, \frac{r \kappa_{5}}{s}$ 和$\kappa_{6}$ 为指数的Hölder不等式可导出

$ \begin{gathered} \int_{R \leqslant|x| \leqslant 2 R} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \leqslant \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{5}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} K(x) u^{r}(x, t) \mathrm{d} x\right)^{\frac{s}{r\kappa_{5}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s \kappa_{6}}{r\kappa_{5}}}|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{6}}} \leqslant \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{5}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s\kappa_{6}}{r\kappa_{5}}}|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{6}}} 。\end{gathered} $

关于t变量，在$\left[0, 2 R^{\gamma}\right]$上积分可得

$ \begin{gathered} \int_{P_{1}(R)} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{5}}} \times \\ \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s\kappa_{6}}{r\kappa_{5}}}|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{6}}} \mathrm{~d} t 。\end{gathered} $

再利用以$\kappa_{5}$ 和$\kappa_{5}^{\prime}$ 为指数的Hölder不等式可导出

$ \begin{gathered} \int_{P_{1}(R)} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} \times \\ {\left[\int _ { 0 } ^ { 2 R ^ { \gamma } } \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} \times\right.\right.} \\ \left.\left.|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{\kappa_{5}^{\prime}}{\kappa_{6}}} \mathrm{~d} t\right]^{\frac{1}{\kappa_{5}^{\prime}}} 。\end{gathered} $

引理3证毕。

引理4   令$q \geqslant 1, r \geqslant 1, s>0, u(x, t)$ 为非齐次非局部微分不等式(1)的非负弱解，$u(x, t) \in S_{d}$(或$\widetilde{S}_{d}$)且$q$ 和$r$ 不同时为1，则

$ \begin{gather*} \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} \times \\ {\left[\int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}} k{-\kappa_{7}}} \mathrm{~d} x\right)^{\frac{\kappa_{7}^{\prime}}{\kappa_{8}}} \mathrm{~d} t\right]^{\frac{1}{\kappa_{7}^{\prime}}}, } \end{gather*} $

(9)

其中

$ \kappa_{7}=q+s, \kappa_{7}^{\prime}=\frac{q+s}{q+s-1}, \kappa_{8}=\frac{\kappa_{7}}{\kappa_{7}-\left(1+\frac{s}{r}\right)}。$

证明  通过直接计算可得，在(9)式左端中被积函数等价于

$ \begin{gathered} u\left|\zeta_{t}\right| \zeta^{k-1}=\left(a(x)^{\frac{1}{\kappa_{7}}} u^{\frac{q}{\kappa_{7}}} \zeta^{\frac{k}{\kappa_{7}}}\right)\left(K(x)^{\frac{s}{r\kappa_{7}}} u^{\frac{s}{\kappa_{7}}}\right) \times \\ \left(a(x)^{-\frac{1}{\kappa_{7}}} K(x)^{-\frac{s}{r\kappa_{7}}} u^{1-\frac{q+s}{\kappa_{7}}}\left|\zeta_{t}\right| \zeta^{\frac{k}{\kappa_{7}^{\prime}}-1}\right) 。\end{gathered} $

为了使上式最后一项中$u$ 的指数为

$ 1-\frac{q+s}{\kappa_{7}}=0, $

取$\kappa_{7}=q+s$ 且恰当的$\kappa_{8}$ 满足

$ \frac{1}{\kappa_{7}}+\frac{s}{r \kappa_{7}}+\frac{1}{\kappa_{8}}=1 。$

由此，可断言：$\kappa_{7}, ~ \frac{r \kappa_{7}}{s}, ~ \kappa_{8}>1$。事实上，由假设$q \geqslant 1$，$r \geqslant 1$ 及$s>0$ 显然有$\kappa_{7}>1$。另外，如下等价关系成立：

$ \begin{gathered} \frac{r \kappa_{7}}{s}-1=\frac{r(q+s)-s}{s}=\frac{s(r-1)+r q}{s}, \\ \kappa_{8}-1=\frac{1+\frac{s}{r}}{q+s-\left(1+\frac{s}{r}\right)}=\frac{1+\frac{s}{r}}{q-1+\frac{s}{r}(r-1)} 。\end{gathered} $

由假设$q \geqslant 1, r \geqslant 1, s>0$，及$q$ 和$r$ 不同时为1可得到，$\frac{r \kappa_{7}}{s}>1$ 和$\kappa_{8}>1$。

利用以$\kappa_{7}, ~ \frac{r \kappa_{7}}{s}$，和$\kappa_{8}$ 为指数的Hölder不等式可推出

$ \begin{gathered} \int_{|x| \leqslant 2 R} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \leqslant \\ \left(\int_{|x| \leqslant 2 R} a(x) u^{q} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{7}}}\left(\int_{|x| \leqslant 2 R} K(x) u^{r}(x, t) \mathrm{d} x\right)^{\frac{s}{r\kappa_{7}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}}{ }^{k-\kappa_{8}}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{8}}} \leqslant \\ \left(\int_{|x| \leqslant 2 R} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{7}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}}{ }^{k-\kappa_{8}}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{8}}} 。\end{gathered} $

关于时间变量$t$，在$\left[R^{\gamma}, 2 R^{\gamma}\right]$ 上积分可得

$ \begin{gathered} \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x\right)^{\frac{1}{\kappa_{7}}} \times \\ \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}} k-\kappa_{8}} \mathrm{~d} x\right)^{\frac{1}{\kappa_{8}}} \mathrm{~d} t 。\end{gathered} $

再利用以$\kappa_{7}$ 和$\kappa_{7}^{\prime}$ 为指数的Hölder不等式可推出

$ \begin{gathered} \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} \times \\ {\left[\int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}} k-\kappa_{8}} \mathrm{~d} x\right)^{\frac{\kappa_{7}^{\prime}}{\kappa_{8}}} \mathrm{~d} t\right]^{\frac{1}{\kappa_{7}^{\prime}}}}。\end{gathered} $

引理4证毕。

利用前面陈述的引理，可导出如下估计。

命题1   令$N \geqslant 1, q \geqslant 1, r \geqslant 1, s>0, p>1$，

$ \begin{gathered} q+s \geqslant\left(1+\frac{s}{r}\right)(p-1), \\ k>\max \left\{\frac{q+s-d}{q+s-1}, \frac{p(q+s-d)}{q+s-(p-1)}\right\}, \end{gathered} $

且$0 < d < \min \{1, p-1\}$。如果$u(x, t)$ 是非齐次非局部微分不等式(1)的非负弱解且$u(x, t) \in S_{d}$，则

$ \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant C\left(R^{\delta_{1}}+R^{\delta_{2}}\right), $

(10)

其中

$ \begin{gathered} \delta_{1}=N+\gamma+\frac{\alpha+\frac{s}{r} \beta-\frac{s}{r} N-(q+s) \gamma}{q+s-1}+ \\ d \frac{\frac{s}{r} N-\alpha-\frac{s}{r} \beta+\gamma}{q+s-1}, \\ \delta_{2}=N+\gamma+\frac{(p-1)\left(\alpha+\frac{s}{r} \beta-\frac{s}{r} N\right)-p(q+s)}{q+s-(p-1)}+ \\ d \frac{\frac{s}{r} N-\alpha-\frac{s}{r} \beta+p}{q+s-(p-1)} 。\end{gathered} $

证明  取试验函数为$\varphi(x, t)=\widetilde{u}_{\varepsilon}^{-d} \zeta^{k}$，则显然$\varphi(x, t) \in C_{0}^{1}(S)$。将$\varphi$ 代人到公式(5)中得到

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q} \tilde{u}_{\varepsilon}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)} \omega(x) \tilde{u}_{\varepsilon}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \int_{P(2 R)} u_{t} \tilde{u}_{\varepsilon}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)}\left(|\nabla u|^{p-2} \nabla u\right) \cdot \nabla\left(\widetilde{u}_{\varepsilon}^{-d} \zeta^{k}\right) \mathrm{d} x \mathrm{~d} t。\end{gathered} $

在$L_{\text {loc }}^{1}(S)$中满足$\widetilde{u}_{\varepsilon} \rightarrow u_{\tau}(\varepsilon \rightarrow 0)$, 因此可应用Lebesgue控制收敛定理^[21]，并取极限$\varepsilon \rightarrow 0$可导出

$ \begin{gather*} \int_{P(2 R)} a(x) u^{q} u_{\tau}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)} \omega(x) u_{\tau}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \int_{P(2 R)} u_{t} u_{\tau}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)}\left(|\nabla u|^{p-2} \nabla u\right) \cdot \nabla\left(u_{\tau}^{-d} \zeta^{k}\right) \mathrm{d} x \mathrm{~d} t 。\end{gather*} $

(11)

再由$u_{\tau}(x, t)$ 的定义得

$ \nabla u_{\tau}=\nabla u, \left(u_{\tau}\right)_{t}=u_{t}, $

且在公式(11)右端的被积函数等价于

$ u_{t} u_{\tau}^{-d} \zeta^{k}=\frac{1}{1-d} \frac{\partial u_{\tau}^{1-d}}{\partial t} \zeta^{k} 。$

利用分部积分可推出

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q} u_{\tau}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)} \omega(x) u_{\tau}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ -\frac{k}{1-d} \int_{P(2 R)} u_{\tau}^{1-d} \zeta^{k-1} \zeta, \mathrm{~d} x \mathrm{~d} t- \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{\tau}^{1-d}(x, 0) \zeta^{k}(x, 0) \mathrm{d} x- \\ d \int_{P(2 R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla u\right] u_{\tau}^{-1-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ k \int_{P(2 R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla \zeta\right] u_{\tau}^{-d} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

由于$u_{\tau}>0$ 且$\zeta \geqslant 0$，由上式可推出

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q} u_{\tau}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{\tau}^{1-d}(x, 0) \zeta^{k}(x, 0) \mathrm{d} x+ \\ d \int_{P_{1}(R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla u\right] u_{\tau}^{-1-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)} \omega(x) u_{\tau}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \frac{k}{1-d} \int_{P(2 R)} u_{\tau}^{1-d} \zeta^{k-1}\left|\zeta_{t}\right| \mathrm{d} x \mathrm{~d} t+ \\ k \int_{P_{1}(R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla \zeta\right] u_{\tau}^{-d} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

对于上式右端第二项，可得到

$ \begin{gathered} k \int_{P_{1}(R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla \zeta\right] u_{\tau}^{-d} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ k \int_{P_{1}(R)}|\nabla u|^{p-1}|\nabla \zeta| u_{\tau}^{-d} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t= \\ k \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right)^{\frac{p-1}{p}} u_{\tau}^{-d}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t, \end{gathered} $

再利用以$p, \frac{p}{p-1}$ 为指数且带有参数$\varepsilon_{3}$ 的Young's不等式可推出

$ \begin{gathered} k \int_{P_{1}(R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla \zeta\right] u_{\tau}^{-d} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{3} \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u_{\tau}^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{P_{1}(R)} u_{\tau}^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

于是，可得到

$ \begin{gather*} \int_{P(2 R)} a(x) u^{q} u_{\tau}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{\tau}^{1-d}(x, 0) \zeta^{k}(x, 0) \mathrm{d} x+ \\ \left(d-\varepsilon_{3}\right) \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u_{\tau}^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)} \omega(x) u_{\tau}^{-d} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \frac{k}{1-d} \int_{P_{2}(R)} u_{\tau}^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{P_{1}(R)} u_{\tau}^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t 。\end{gather*} $

(12)

再由公式(12)左端最后一项的非负性可知

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q} u_{\tau}^{-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{\tau}^{1-d}(x, 0) \zeta^{k}(x, 0) \mathrm{d} x+ \\ \left(d-\varepsilon_{3}\right) \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u_{\tau}^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \frac{k}{1-d} \int_{P_{2}(R)} u_{\tau}^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{P_{1}(R)} u_{\tau}^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

由假设$0 < d < \min \{1, p-1\}$ 易知，上式右端中$u_{\tau}$ 的指数均为正数。因此，可利用Beppo－Levi定理^[21] 且取极限$\tau \rightarrow 0$ 可得到

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{0}^{1-d}(x) \zeta^{k}(x, 0) \mathrm{d} x+ \\ \left(d-\varepsilon_{3}\right) \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \frac{k}{1-d} \int_{P_{2}(R)} u^{1-d}\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{P_{1}(R)} u^{p-1-d}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

对上式应用引理1和引理2并整理得

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{0}^{1-d}(x) \zeta^{k}(x, 0) \mathrm{d} x+ \\ \left(d-\varepsilon_{3}\right) \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ \varepsilon_{1} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s\kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t+ \\ \varepsilon_{2} \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s\kappa_{4}}{r\kappa_{3}}}|\nabla \zeta|^{\rho \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-\rho \kappa_{4}} \mathrm{~d} x\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t, \end{gathered} $

利用$u(x, t) \in S_{d}$, 可推出

$ \begin{gathered} \left(1-\varepsilon_{1}-\varepsilon_{2}\right) \int_{P(2 R)} a(x) u^{q-d}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ \frac{1}{1-d} \int_{B_{2 R}(0)} u_{0}^{1-d}(x) \zeta^{k}(x, 0) \mathrm{d} x+ \\ \left(d-\varepsilon_{3}\right) \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s \kappa_{2}}{r \kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t+ \\ C \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s \kappa_{4}}{r \kappa_{3}}}|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}}\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t 。\end{gathered} $

(13)

下面对公式(13)右端两项进行估计。由假设

$ k>\max \left\{\frac{q+s-d}{q+s-1}, \frac{p(q+s-d)}{q+s-(p-1)}\right\} $

可知，上式右端的两个积分式中$\zeta$ 的指数均为正数，且

$ 0 \leqslant \zeta^{\frac{\kappa_{2}}{\kappa_{1}} k-\kappa_{2}}, \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \leqslant 1。$

根据截断函数$\zeta$ 的定义及条件(2)可知

$ \begin{gathered} \left|\zeta_{t}\right| \leqslant C R^{-\gamma}, |\nabla \zeta| \leqslant C R^{-1} ; \\ a(x) \geqslant|x|^{-\alpha}, K(x) \geqslant|x|^{-\beta}, \forall x \in \mathbf{R}^{N} \backslash\{0\} 。\end{gathered} $

于是，对公式(13)右端第一项估计如下：

$ \begin{gathered} \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{2}}{\kappa_{1}}} K(x)^{-\frac{s \kappa_{2}}{r\kappa_{1}}}\left|\zeta_{t}\right|^{\kappa_{2}} \zeta^{\frac{\kappa_{2}}{\kappa_{1}^{\prime}} k-\kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t \leqslant \\ C \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R}|x|^{\frac{\kappa_{2}}{\kappa_{1}} \alpha}|x|^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta} R^{-\gamma \kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t 。\end{gathered} $

利用尺度变换$\xi: =\frac{|x|}{R}, \theta: =\frac{t}{R^{\gamma}}$，可得

$ \begin{aligned} & |x|^{\frac{\kappa_{2}}{\kappa_{1}} \alpha}=(R \xi)^{\frac{\kappa_{2}}{\kappa_{1}} \alpha}=R^{\frac{\kappa_{2}}{\kappa_{1}} \alpha} \xi^{\frac{\kappa_{2}}{\kappa_{1}} \alpha} , \\ & |x|^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta}=(R \xi)^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta}=R^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta} \xi^{\frac{s \kappa_{2}}{\kappa_{1}} \beta}, \end{aligned} $

且

$ \begin{gathered} \int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R}|x|^{\frac{\kappa_{2}}{\kappa_{1}} \alpha}|x|^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta} R^{-\gamma \kappa_{2}} \mathrm{~d} x\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} t \leqslant \\ C R^{\left(\frac{\kappa_{2}}{\kappa_{1}} \alpha+\frac{s \kappa_{2}}{r\kappa_{1}} \beta-\gamma \kappa_{2}+N\right) \frac{\kappa_{1}^{\prime}}{\kappa_{2}}+\gamma} \int_{1}^{2}\left(\int_{\xi \leqslant 2} \xi^{\frac{\kappa_{2}}{\kappa_{1}} \alpha} \xi^{\frac{s \kappa_{2}}{r\kappa_{1}} \beta} \mathrm{~d} \xi\right)^{\frac{\kappa_{1}^{\prime}}{\kappa_{2}}} \mathrm{~d} \theta \leqslant \\ C R^{\left(\frac{\kappa_{2}}{\kappa_{1}} \alpha+\frac{s \kappa_{2}}{r\kappa_{1}} \beta-\gamma \kappa_{2}+N\right) \frac{\kappa_{1}^{\prime}}{\kappa_{2}}+\gamma}:=R^{\delta_{1}} 。\end{gathered} $

类似地，对于公式(13)右端第二项估计如下：

$ \begin{gathered} \int_{0}^{2 R^{\gamma}}\left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{4}}{\kappa_{3}}} K(x)^{-\frac{s \kappa_{4}}{r \kappa_{3}}} \times\right. \\ \left.|\nabla \zeta|^{p \kappa_{4}} \zeta^{\frac{\kappa_{4}}{\kappa_{3}^{\prime}} k-p \kappa_{4}} \mathrm{~d} x\right)^{\frac{\kappa_{3}^{\prime}}{\kappa_{4}}} \mathrm{~d} t \leqslant \\ C R^{\left(\frac{\kappa_{4}}{\kappa_{3}} \alpha+\frac{s \kappa_{4}}{r \kappa_{3}} \beta-p \kappa_{4}+N\right) \frac{\kappa_{3}^{\prime}}{\kappa_{4}}+\gamma}:=R^{\delta_{2}} 。\end{gathered} $

取$\varepsilon_{1} < \frac{1}{2}, \varepsilon_{2} < \frac{1}{2}, \varepsilon_{3} < d$，则由公式(13)左边三项的非负性，命题1得证。

命题2   令$N \geqslant 1, q \geqslant 1, r \geqslant 1, s>0, p>1$，

$ \begin{gathered} q+s >\left(1+\frac{s}{r}\right)(p-1), \\ k >\max \left\{\frac{p(q+s)}{q+s-(p-1)(1+d)}, \frac{q+s}{q+s-1}\right\}, \\ 0<d <\min \left\{1, p-1, \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}\right\}, \end{gathered} $

其中$q$ 和$r$ 不同时为1。如果$u(x, t)$ 是非齐次非局部微分不等式(1)的非负弱解且$u(x, t) \in S_{d}$，则有

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\delta^{\prime}{ }_{1}}+R^{\delta^{\prime}{ }_{2}}+R^{\delta^{\prime}{ }_{3}}+R^{\delta^{\prime}{ }_{4}}\right), $

(14)

其中

$ \begin{gathered} \delta_{1}^{\prime}=\delta_{1}-\gamma=N+ \\ \frac{\alpha+\frac{s}{r} \beta-\frac{s}{r} N-(q+s) \gamma}{q+s-1}+d \frac{\gamma+\frac{s}{r} N-\alpha-\frac{s}{r} \beta}{q+s-1}, \\ \delta_{2}^{\prime}=\delta_{2}-\gamma=N+\frac{(p-1)\left(\alpha+\frac{s}{r} \beta-\frac{s}{r} N\right)-p(q+s)}{q+s-(p-1)}+ \\ d \frac{\frac{s}{r} N-\alpha-\frac{s}{r} \beta+p}{q+s-(p-1)}, \\ \delta_{3}^{\prime}=\delta_{3}-\gamma=\frac{\gamma}{\kappa_{5}^{\prime}}+\frac{N}{\kappa_{6}}+\frac{\alpha}{\kappa_{5}}+\frac{s \beta}{r \kappa_{5}}-p-\gamma, \\ \delta_{4}^{\prime}=\delta_{4}-\gamma=-\frac{\gamma}{\kappa_{7}}+\frac{N}{\kappa_{8}}+\frac{\alpha}{\kappa_{7}}+\frac{s \beta}{r \kappa_{7}}-\gamma, \end{gathered} $

这里$\kappa_{5}, \kappa_{6}, \kappa_{7}, \kappa_{8}$ 与引理3、引理4中一致。

证明  取试验函数为$\varphi(x, t)=\zeta^{k}(x, t)$，则显然，$\varphi(x, t) \in C_{0}^{1}$($S$)。将$\varphi$ 代入到弱解定义(5)中可得到

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ -k \int_{P(2 R)} u \zeta_{t} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)}\left(|\nabla u|^{p-2} \nabla u\right) \cdot \nabla\left(\zeta^{k}\right) \mathrm{d} x \mathrm{~d} t \leqslant \\ k \int_{P(2 R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ k \int_{P(2 R)}\left[|\nabla u|^{p-2} \nabla u \cdot \nabla \zeta\right] \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t \leqslant \\ k \int_{P(2 R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+k \int_{P(2 R)}|\nabla u|^{p-1}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t= \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+k \int_{P_{1}(R)}|\nabla u|^{p-1}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t= \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ k \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \cdot \nabla u \cdot \nabla u\right)^{\frac{p-1}{p}}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

对上式右端最后一项，利用以$p, \frac{p}{p-1}$ 为指数且带有参数$\varepsilon_{4}$ 的Young's不等式可推出

$ \begin{gather*} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ C \varepsilon_{4} \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+ \\ C \int_{P_{1}(R)} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t。\end{gather*} $

(15)

将命题1中公式(10)代入到上式右端第二项，并对右端第一项和第三项分别利用引理4和引理3可导出

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ C\left(R^{\delta_{1}}+R^{\delta_{2}}\right)+C\left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} \times \\ \left(\int _ { 0 } ^ { 2 R ^ { \gamma } } \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s \kappa_{6}}{r\kappa_{5}}} \times\right.\right. \\ \left.\left.|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}} k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{\kappa_{5}^{\prime}}{\kappa_{6}}} \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}^{\prime}}}+ \\ C\left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} \times \\ \left(\int _ { R ^ { \gamma } } ^ { 2 R ^ { \gamma } } \left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s \kappa_{8}}{r\kappa_{7}}} \times\right.\right. \\ \left.\left.\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}} k-\kappa_{8}} \mathrm{~d} x\right)^{\frac{\kappa_{7}^{\prime}}{\kappa_{8}}} \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}^{\prime}}} 。\end{gathered} $

对上式右端第三项和第五项，由假设

$ k>\max \left\{\frac{p(q+s)}{q+s-(p-1)(1+d)}, \frac{q+s}{q+s-1}\right\} $

知，可利用类似于命题1中公式(13)的估计且可推出

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ C\left(R^{\delta_{1}}+R^{\delta_{2}}\right)+C\left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} R^{\delta_{3}}+ \\ C\left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} R^{\delta_{4}} 。\end{gathered} $

由上式左端两项的非负性可知

$ \begin{gather*} \int_{B_{R}(0) \times\left[0, R^{\gamma}\right]} \omega(x) \mathrm{d} x \mathrm{~d} t \leqslant \\ C\left(R^{\delta_{1}}+R^{\delta_{2}}\right)+C\left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} R^{\delta_{3}}+ \\ C\left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} R^{\delta_{4}}。\end{gather*} $

(16)

根据弱解定义中的条件：$a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \in L_{\text {loc }}^{1}(S)$得到

$ \begin{aligned} & \left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{5}}} \leqslant C \\ & \left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} \leqslant C 。\end{aligned} $

又因为$\omega(x)$ 是只关于空间变量$x$ 的函数，故对(16)式两端同时除以$R^{\gamma}$ 可得到

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\delta_{1}-\gamma}+R^{\delta_{2}-\gamma}+R^{\delta_{3}-\gamma}+R^{\delta_{4}-\gamma}\right) 。$

命题2证毕。

命题3   令$N \geqslant 1, q \geqslant 1, r \geqslant 1, s>0, p>1$，

$ \begin{gathered} q+s>\left(1+\frac{s}{r}\right)(p-1), \\ k>\max \left\{\frac{p(q+s)}{q+s-(p-1)(1+d)}, \frac{q+s}{q+s-1}\right\}, \end{gathered} $

且

$ 0<d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}, $

其中$q$ 和$r$ 不同时为1。如果$u(x, t)$ 是非齐次非局部微分不等式(1)的非负弱解且$u(x, t) \in \widetilde{S}_{d}$，则有

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\frac{\delta_{3}}{p}-\gamma}+R^{\delta^{\prime}{ }_{4}}\right), $

(17)

其中

$ \begin{aligned} \delta_{3} & =\frac{\gamma}{\kappa_{5}^{\prime}}+\frac{N}{\kappa_{6}}+\frac{\alpha}{\kappa_{5}}+\frac{s \beta}{r \kappa_{5}}-p, \\ \delta_{4}^{\prime} & =-\frac{\gamma}{\kappa_{7}}+\frac{N}{\kappa_{8}}+\frac{\alpha}{\kappa_{7}}+\frac{s \beta}{r \kappa_{7}}-\gamma, \end{aligned} $

这里$\kappa_{5}, \kappa_{6}, \kappa_{7}, \kappa_{8}$ 与引理3、引理4中给出的一致。

证明  取试验函数$\varphi(x, t)=\zeta^{k}(x, t)$，则显然

$ \varphi(x, t) \in C_{0}^{1}(S)。$

将φ代入到弱解定义(5)中得到

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ -k \int_{P(2 R)} u \zeta_{t} \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ \int_{P(2 R)}\left(|\nabla u|^{p^{-2}} \nabla u\right) \cdot \nabla\left(\zeta^{k}\right) \mathrm{d} x \mathrm{~d} t \leqslant \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+k \int_{P_{1}(R)}|\nabla u|^{p-1}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t= \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ k \int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right)^{\frac{p-1}{p}}|\nabla \zeta| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t 。\end{gathered} $

对上式，利用以$p, \frac{p}{p-1}$ 为指数的Hölder不等式可推出

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ k \int_{P_{2}(R)} u\left|\zeta_{t}\right| \zeta^{k-1} \mathrm{~d} x \mathrm{~d} t+ \\ C\left(\int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{p-1}{p}} \times \\ \left(\int_{P_{1}(R)} u^{(p-1)(1+d)}|\nabla \zeta|^{p} \zeta^{k-p} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{p}} 。\end{gathered} $

对上式右端的第一项和第三项，分别利用引理4和引理3可得

$ \begin{gathered} \int_{P(2 R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|{ }_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t+\int_{P(2 R)} \omega(x) \zeta^{k} \mathrm{~d} x \mathrm{~d} t \leqslant \\ C\left(\int_{P_{1}(R)}\left(|\nabla u|^{p-2} \nabla u \cdot \nabla u\right) u^{-d-1} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{p-1}{p}} \times \\ \left(\int_{P_{1}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{p \kappa_{5}}} \times \\ \left(\int _ { 0 } ^ { 2 R ^ { \gamma } } \left(\int_{R \leqslant|x| \leqslant 2 R} a(x)^{\frac{\kappa_{6}}{\kappa_{5}}} K(x)^{-\frac{s \kappa_{6}}{r \kappa_{5}}} \times\right.\right. \\ \left.\left.|\nabla \zeta|^{p \kappa_{6}} \zeta^{\frac{\kappa_{6}}{\kappa_{5}^{\prime}}k-p \kappa_{6}} \mathrm{~d} x\right)^{\frac{\kappa_{5}^{\prime}}{\kappa_{6}}} \mathrm{~d} t\right)^{\frac{1}{p \kappa_{5}^{\prime}}}+ \\ C\left(\int_{P_{2}(R)} a(x) u^{q}\left\|K^{\frac{1}{r}} u\right\|_{r}^{s} \zeta^{k} \mathrm{~d} x \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}}} \times \\ \left(\int_{R^{\gamma}}^{2 R^{\gamma}}\left(\int_{|x| \leqslant 2 R} a(x)^{-\frac{\kappa_{8}}{\kappa_{7}}} K(x)^{-\frac{s\kappa_{8}}{r\kappa_{7}}}\left|\zeta_{t}\right|^{\kappa_{8}} \zeta^{\frac{\kappa_{8}}{\kappa_{7}^{\prime}} k-\kappa_{8}} \mathrm{~d} x\right)^{\frac{\kappa_{7}^{\prime}}{\kappa_{8}}} \mathrm{~d} t\right)^{\frac{1}{\kappa_{7}^{\prime}}} 。\end{gathered} $

对上式的右端第一项、第三项和第五项，由集合$\widetilde{S}_{d}$的性质及假设

$ k>\max \left\{\frac{p(q+s)}{q+s-(p-1)(1+d)}, \frac{q+s}{q+s-1}\right\} $

知，可利用命题1中对公式(13)的类似估计且可推出

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\frac{\delta_{3}}{p}-\gamma}+R^{\delta^{\prime}{ }_{4}}\right) $

命题3证毕。

2 定理1的证明

本节中给出定理1的详细证明过程。

证明  由假设$N \geqslant 1, q \geqslant 1, r \geqslant 1, s>0, q$ 和$r$ 不同时为1，以及$q+s>\left(1+\frac{s}{r}\right)(p-1)$ 易知，满足命题2的条件。假设$u(x, t)$ 是非局部微分不等式(1)在$S_{d}$ 上的非负弱解，其中$0 < d < \min \left\{1, p-1, \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}\right\}$，则由命题2中公式(14)可知

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\delta^{\prime}{ }_{1}}+R^{\delta^{\prime}{ }_{2}}+R^{\delta^{\prime}{ }_{3}}+R^{\delta^{\prime}{ }_{4}}\right), $

(18)

对于$R$ 的指数$\delta_{1}^{\prime} \sim \delta^{\prime}{ }_{4}$ 整理可得

$ \begin{gathered} \delta_{1}^{\prime}=\frac{N-\gamma}{q+s-1}\left[q+s-q_{1}(\gamma)\right]+d \frac{\gamma+\frac{s}{r} N-\alpha-\frac{s}{r} \beta}{q+s-1}, \\ \delta_{2}^{\prime}=\frac{N-p}{q+s-(p-1)}\left[q+s-q_{2}(\gamma)\right]+d \frac{\frac{s}{r} N+p-\alpha-\frac{s}{r} \beta}{q+s-(p-1)}, \\ \delta_{3}^{\prime}=\frac{N-p}{q+s}\left[q+s-q_{3}(\gamma)\right]-d(p-1) \times \\ \quad \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{q+s}, \\ \quad \delta_{4}^{\prime}=\frac{N-\gamma}{q+s}\left[q+s-q_{4}(\gamma)\right]-d \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{q+s}, \end{gathered} $

其中

$ \begin{gathered} q_{1}(\gamma):=\frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-\gamma}, \\ q_{2}(\gamma):=(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}, \\ q_{3}(\gamma):=(p-1) \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}, \\ q_{4}(\gamma):=\frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-\gamma} 。\end{gathered} $

现在，由条件$q+s>\max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\}$，$p < N$ 以及$\gamma < N$ 易知，$\delta_{1}^{\prime} \sim \delta_{4}^{\prime}$ 中$q+s-q_{i}(\gamma), i=1$，$2, 3, 4$ 的系数全为正，即

$ \begin{gathered} \frac{N-\gamma}{q+s-1}>0, \frac{N-p}{q+s}>0 , \\ \frac{N-p}{q+s-(p-1)}>0, \frac{N-\gamma}{q+s}>0 。\end{gathered} $

显然，$q_{2}(\gamma)$ 是一个与$\gamma$ 无关的常数，且对于任意的$0 < \gamma < N, p < N$，恒有$q_{3}(\gamma)>q_{2}(\gamma), q_{4}(\gamma)>q_{1}(\gamma)$。

下面，考虑$q_{1}(\gamma)$ 和$q_{2}(\gamma)$ 的交点，为此需求解方程$q_{1}(\gamma)=q_{2}(\gamma)$，即

$ \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-\gamma}=(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p} 。$

由条件$\alpha+\frac{s}{r} \beta < N\left(1+\frac{s}{r}\right)-\frac{N-p}{p-1}$ 易知$\alpha+\frac{s}{r} \beta < N+\frac{s}{r} N$ 成立。因此，上述方程$q_{1}(\gamma)=q_{2}(\gamma)$ 存在唯一解

$ \gamma=N-\frac{N-p}{p-1}, $

且由条件$p < N$ 以及$p>\max \left\{1, \frac{2 N}{1+N}\right\}$ 知$0 < \gamma < N$，并直接计算得到

$ q_{1}(\gamma)=q_{2}(\gamma)=(p-1) \frac{N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p}:=Q_{I H F} 。$

由此断言：当$\alpha+\frac{s}{r} \beta < \min \left\{N\left(1+\frac{s}{r}\right)-\frac{N-p}{p-1}, p(1+\right. \left.\left.\frac{s}{r}\right)\right\}$ 时，

$ Q_{I H F}>\max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\} 。$

事实上

$ Q_{I H F}-1=\frac{(p-1)\left[N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta\right]-(N-p)}{N-p} 。$

由条件$\alpha+\frac{s}{r} \beta < N\left(1+\frac{s}{r}\right)-\frac{N-p}{p-1}$ 易知

$ (p-1)\left[N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta\right]-(N-p)>0, $

且$Q_{I H F}>1$。

类似地，直接计算可得到

$ Q_{I H F}-\left(1+\frac{s}{r}\right)(p-1)=(p-1) \frac{p\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-p} 。$

由条件$\alpha+\frac{s}{r} \beta < p\left(1+\frac{s}{r}\right)$ 知，$p\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta>$ 0且

$ Q_{I H F}>\left(1+\frac{s}{r}\right)(p-1) \text { 。} $

总之，$Q_{I H F}>\max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\}$。

下面，分两种情形讨论(18)式右端$R$ 的指数。

情形1   当$\max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\} < q+s < Q_{I H F}$ 且$q$ 和$r$ 不同时为1时，$\exists \gamma>0$ 使得

$ q+s<\min \left\{q_{1}(\gamma), q_{2}(\gamma), q_{3}(\gamma), q_{4}(\gamma)\right\} \text { 。} $

取$d$ 充分小，则$R$ 的指数$\delta_{1}^{\prime} \sim \delta_{4}^{\prime}$ 均为负数并在(18)式中取极限$R \rightarrow+\infty$ 得到

$ \int_{\mathbf{R}^{N}} \omega(x) \mathrm{d} x=0 。$

由此可得，对于a.e.$x \in \mathbf{R}^{N}$，有$\omega(x) \equiv 0$ 且与$\omega(x)$ 的非平凡性导致矛盾，即微分不等式(1)在集合$S_{d}$ 上不存在非平凡非负弱解，其中$d$ 充分小且满足

$ 0<d<\min \left\{1, p-1, \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}\right\} 。$

情形2   当$\max \left\{1, \left(1+\frac{s}{r}\right)(p-1)\right\} < q+s=Q_{I H F}$ 且$q$ 和$r$ 不同时为1时，根据$Q_{I H F}>\max \left\{1, \left(1+\frac{s}{r}\right) \times\right. (p-1)\}$ 不难验证，命题3中条件全部成立。

现在，假设$u(x, t)$ 是微分不等式(1)在$\widetilde{S}_{d}$ 上的非负弱解，其中$0 < d < \frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)}$，则可由命题3中公式(17)得到

$ \int_{B_{R}(0)} \omega(x) \mathrm{d} x \leqslant C\left(R^{\frac{\delta_{3}}{p}-\gamma}+R^{\delta^{\prime}{ }_{4}}\right), $

(19)

其中

$ \begin{gathered} \delta_{3}=N+\gamma-p-(p-1) \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{q+s}- \\ d(p-1) \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{q+s}, \\ \delta_{4}^{\prime}=\frac{N-\gamma}{q+s}\left[q+s-q_{4}(\gamma)\right], \\ q_{4}(\gamma)=\frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{N-\gamma} 。\end{gathered} $

对于$R$ 的指数$\frac{\delta_{3}}{p}-\gamma$ 直接计算可得

$ \begin{gathered} \frac{\delta_{3}}{p}-\gamma= \\ \frac{(q+s)(N+\gamma-p)-(p-1)\left[\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta\right]}{p(q+s)}- \\ \gamma-d(p-1) \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{p(q+s)}= \\ \frac{(q+s)(N-p)-(p-1)\left[N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta\right]}{p(q+s)}+ \\ \frac{(1-p)(q+s+1) \gamma}{p(q+s)}-d(p-1) \frac{\gamma+N\left(1+\frac{s}{r}\right)-\alpha-\frac{s}{r} \beta}{p(q+s)} 。\end{gathered} $

(20)

取d充分小并将条件$q+s=Q_{I H F}$代入到(20)式可得

$ \frac{\delta_{3}}{p}-\gamma=-\frac{(p-1)(q+s+1) \gamma}{p(q+s)}<0 。$

此外，当$q+s=Q_{I H F}$ 时，利用$q+s-q_{1}(\gamma)=0$ 以及$q_{4}(\gamma)>q_{1}(\gamma)$ 易得，$q+s-q_{4}(\gamma) < 0$ 且$\delta_{4}^{\prime} < 0$。

在公式(19)中，取极限$R \rightarrow+\infty$可得

$ \int_{\mathbf{R}^{N}} \omega(x) \mathrm{d} x=0 。$

显然，与$\omega(x)$ 的非平凡性导致矛盾，即非齐次非局部微分不等式(1)在集合$\widetilde{S}_{d}$ 上无非平凡非负弱解，其中

$ 0<d<\frac{q+s-\left(1+\frac{s}{r}\right)(p-1)}{\left(1+\frac{s}{r}\right)(p-1)} 。$

定理1证毕。

3 结语

本文给出了当$p < N$时微分不等式(1)柯西问题非平凡非负弱解的非存在性结论，而当$p \geqslant N$ 时能否建立非存在性结果及解的存在性问题尚未探讨。此外，在不等式(1)中能否把p-Laplacian算子换成更一般算子来研究是下一步的研究工作目标。