\quad\quad本文研究如下的反应扩散系统：\begin{array}{ll} \dfrac{\partial u}{\partial t}=\Delta_{\Omega, 1}^{s} u - v, & (x,t)\in(0, T) \times \Omega, \\[1.5ex] \dfrac{\partial v}{\partial t}=\Delta v - (f-u), & (x,t)\in(0, T) \times \Omega, \\[1.5ex] u(0, x)=f, \ v(0,x)=0, & x \in \Omega, \\[0.8ex] \dfrac{\partial v}{\partial \boldsymbol{n}} = 0, & (x,t)\in(0, T) \times \partial \Omega. \end{array}\tag{1} \quad\quad为了研究这个系统，首先要利用空间 W^{s,p}(\Omega) 的稠密性质，找到一系列 \{f_p\} \subset W^{s,p}(\Omega) 来逼近 f，即 \lim _{p \rightarrow 1^{+}}\left(\left\|f_{p}-f\right\|_{W^{s, p}(\Omega)}+\left\|f_{p}-f\right\|_{L^{2}(\Omega)}\right)=0,其次，要研究扩散项被替换为 1<p<2 时的 p-Laplacian 时的系统 \begin{array}{ll} \dfrac{\partial u}{\partial t}=\Delta_{\Omega, p}^{s} u - v, & (x,t)\in(0, T) \times \Omega, \\[1.5ex] \dfrac{\partial v}{\partial t}=\Delta v - (f_p-u), & (x,t)\in(0, T) \times \Omega, \\[1.5ex] u(0, x)=f_p, \ v(0,x)=0, & x \in \Omega, \\[0.8ex] \dfrac{\partial v}{\partial \boldsymbol{n}} = 0, & (x,t)\in(0, T) \times \partial \Omega. \end{array}\tag{2} \quad\quad现在先给出问题 (2) 的弱解定义. \quad\quad定义 1\quad给定 f_p \in L^{2}(\Omega) \cap W^{s,p}(\Omega)，若存在 (u_p,v_p) 使得 u_{p} \in L^{\infty}\left(0, T ; W^{s, p}(\Omega) \cap L^{2}(\Omega)\right), \dfrac{\partial u_{p}}{\partial t} \in L^{2}\left(Q_{T}\right), \ v_{p} \in L^{\infty}\left(0, T ; H^{1}(\Omega)\right) \cap C\left([0, T] ; L^{2}(\Omega)\right),\ \dfrac{\partial v_{p}}{\partial t} \in L^{2}\left(Q_{T}\right)，对任意的 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,p}(\Omega)) 和 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_T} \frac{\partial u_{p}}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t+\frac{1}{2} \iint_{Q_T^{1,2}} \frac{\left|u_{p}(y)-u_{p}(x)\right|^{p-2}\left(u_{p}(y)-u_{p}(x)\right)}{|x-y|^{N+s p}}\left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_T} v_{p} \varphi_{1} \mathrm{d} x \mathrm{d} t=0, \\ &\iint_{Q_T} \frac{\partial v_{p}}{\partial t} \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \nabla v_{p} \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \left(f_{p}-u_{p}\right) \varphi_{2} \mathrm{d} x \mathrm{d} t=0, \end{aligned}且\ \displaystyle \lim _{t \rightarrow 0^{+}}\left\|u_{p}(x, t)-f_{p}\right\|_{L^{1}(\Omega)}=0, \lim _{t \rightarrow 0^{+}}\left\|v_{p}(x, t)\right\|_{L^{2}(\Omega)}=0，则称 (u_p,v_p) 为问题 (2) 的一个弱解.\quad\quad引理 1\quad(Simon)\quad设 X, \ B, \ Y 为 Banach 空间，B \subset Y 且 X 紧嵌入 B. 若 F 在 L^{\infty}(0,T;X) 中有界且 \dfrac{\partial F}{\partial t} = \left\{\dfrac{\partial f}{\partial t}: \ f \in F\right\} 在 L^{p}(0,T;Y) 中有界 (p>1)，则 F 在 C([0,T];B) 中相对紧.\quad\quad接下来用 Galerkin 方法证明问题 (2) 弱解的存在唯一性，并给出对弱解的估计.\quad\quad定理 1\quad对每一个f_p \in L^{2}(\Omega) \cap W^{s,p}(\Omega)，问题 (2) 存在唯一的弱解 (u_p,v_p)，且存在不依赖于 p 和 s 的常数 M 使得 \begin{aligned} &\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; L^{2}(\Omega)\right)}+\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; W^{s, p}(\Omega)\right)}+\left\|\frac{\partial u_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M,\\ &\left\|v_{p}\right\|_{L^{\infty}\left(0, T ; W^{1, 2}(\Omega)\right)}+\left\|\frac{\partial v_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M. \end{aligned} \quad\quad证明 \quad先来构造逼近解. 记 \omega_1,\ \omega_2,\ \cdots 为 H^{1}(\Omega) \subset W^{s,p}(\Omega) 的正交基，令 u_m(x, t)=\sum_{i=1}^{m} c_{i}^{m}(t) \omega_{i}(x), \quad v_m(x, t)=\sum_{i=1}^{m} d_{i}^{m}(t) \omega_{i}(x),使其满足 \begin{aligned} &\left(\dfrac{\partial u_m}{\partial t},\omega_i\right) = \left(\Delta_{\Omega,p}^s u_m,\omega_i\right)-(v_m,\omega_i),\\ &\left(\dfrac{\partial v_m}{\partial t},\omega_i\right) = \left(\Delta v_m,\omega_i\right)-(f_p-u_m,\omega_i), \end{aligned} \tag{3}即 \begin{aligned} &\int_{\Omega} \dfrac{\partial u_m}{\partial t} \omega_i \mathrm{d} x + \dfrac{1}{2} \int_{\Omega} \int_{\Omega} \frac{\left|u_{m}(y)-u_{m}(x)\right|^{p-2}(u_{m}(y)-u_{m}(x))}{|x-y|^{N+s p}}(\omega_i(y)-\omega_i(x)) \mathrm{d} x \mathrm{d} y \\ &~~~~~~~~~~~~~~~~~~~~~~+ \int_{\Omega} v_m \omega_i \mathrm{d} x=0,\\ &\int_{\Omega} \dfrac{\partial v_m}{\partial t} \omega_i \mathrm{d} x + \int_{\Omega} \nabla v_m \cdot \nabla \omega_i \mathrm{d} x + \int_{\Omega} (f_p-u_m)\omega_i \mathrm{d} x = 0, \end{aligned} \tag{4} 从而得到常微分方程组\begin{aligned} &\dfrac{\mathrm{d}}{\mathrm{d} t} c_i^m(t) = F_i(t,c_1^m(t),c_2^m(t),\cdots,c_m^m(t),d_i^m(t)),\\ &\dfrac{\mathrm{d}}{\mathrm{d} t} d_i^m(t) = G_i(t,d_1^m(t),d_2^m(t),\cdots,d_m^m(t),c_i^m(t)), \end{aligned}其中 i=1,2,\cdots,m，F_i 和 G_i 是某些关于其各位置的连续函数. 此外，初值 c_i^m(0) 可由 u_m(0)=f_p^{m} \in \mathrm{span}\{\omega_1,\omega_2,\cdots,\omega_m\} 按 \omega_{i} 的展开 f_p^{m} = \displaystyle \sum_{i=1}^{m}c_i^m(0) \omega_i(x) 确定，这里 f_p^{m} 在 L^{2}(\Omega) \cap W^{s,p}(\Omega) 中强收敛到 f_p. 而 d_i^m(0)=0，这样就构成了一个常微分方程组的初值问题，由 Peano 存在性定理可知存在 \tau_0>0 使得它在 [0,\tau_0] 上有解. 接下来用 Picard-Lindelöf 定理来证明这个常微分方程组在 [\tau_0,T] 上解的存在唯一性，为此只需验证 F_i 关于其各位置的局部 Lipschitz 连续性. 事实上，只需验证 H(z)=\int_{\Omega} \int_{\Omega} \frac{\left|Kz+B\right|^{p-2}(Kz+B)}{|x-y|^{N+s p}}(\omega_i(y)-\omega_i(x)) \mathrm{d} x \mathrm{d} y 关于 z 的局部 Lipschitz 连续性，其中 z=c_1^m(t), K=\omega_{1}(y)-\omega_{1}(x)，而\ \displaystyle B = \sum_{i=2}^m c_i^m(t)(\omega_i(y)-\omega_i(x)). 由不等式 \left||a|^{p-2}a-|b|^{p-2}b\right| \leqslant 2|a-b|^{p-1}，有 \begin{aligned} & \left|H(z_1) - H(z_2)\right| \\ \leqslant & \int_{\Omega} \int_{\Omega} \frac{\left|\left|Kz_1+B\right|^{p-2}(Kz_1+B)-\left|Kz_2+B\right|^{p-2}(Kz_2+B)\right|} {|x-y|^{N+s p}}(\omega_i(y)-\omega_i(x)) \mathrm{d} x \mathrm{d} y \\ \leqslant & \int_{\Omega} \int_{\Omega} \frac{2 \left|K(z_1-z_2)\right|^{p-1}}{|x-y|^{N+s p}}(\omega_i(y)-\omega_i(x)) \mathrm{d} x \mathrm{d} y, \\ \leqslant & 2|z_1-z_2|^{p-1} \int_{\Omega} \int_{\Omega} \frac{\left|\omega_{1}(t, y)-\omega_{1}(t, x)\right|^{p-1}} {|x-y|^{(N+sp) \tfrac{p-1}{p}}} \frac{\left|\omega_i(y)-\omega_i(x)\right|}{|x-y|^{(N+s p) \tfrac{1}{p}}} \mathrm{d} x \mathrm{d} y \\ \leqslant & 2 [\omega_{1}]_{W^{s,p}(\Omega)}^{p-1} [\omega_{i}]_{W^{s,p}(\Omega)} |z_1-z_2|^{p-1}, \end{aligned}这样一来，就验证了 F_i 局部 Lipschitz 连续性，从而证明了在 [0,T] 上各系数 c_i^m(t) 和 d_i^m(t) 的存在性. 至此我们完成了逼近解的构造.\quad\quad现在对逼近解作先验估计. 将 \omega_{i} 取成 u_m 或 v_m 代入 (4) 得到 \begin{aligned} &\dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} |u_m|^2 \mathrm{d} x + \dfrac{1}{2} \int_{\Omega} \int_{\Omega} \frac{\left|u_{m}(y)-u_{m}(x)\right|^{p}}{|x-y|^{N+s p}} \mathrm{d} x \mathrm{d} y = - \int_{\Omega} u_m v_m \mathrm{d} x,\\ &\dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} |v_m|^2 \mathrm{d} x + \int_{\Omega} |\nabla v_m|^2 \mathrm{d} x = - \int_{\Omega} (f_p-u_m)v_m \mathrm{d} x, \end{aligned} \tag{5}从而由 Cauchy 不等式得到 \dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} \left(|u_m|^2 + |v_m|^2 \right)\mathrm{d} x \leqslant - \int_{\Omega} f_p v_m \mathrm{d} x \leqslant \dfrac{1}{2} \int_{\Omega} |f_p| ^2 \mathrm{d} x + \dfrac{1}{2} \int_{\Omega} \left(|u_m|^2 + |v_m|^2 \right)\mathrm{d} x 再由 Gronwall 不等式就有 \begin{aligned} \int_{\Omega} \left(|u_m|^2 + |v_m|^2 \right)\mathrm{d} x &\leqslant \mathrm{e}^t \left(\int_{\Omega} \left(|u_m(0)|^2 \right)\mathrm{d} x + \int_{0}^t \int_{\Omega} |f_p|^2 \mathrm{d} x \right)\\ &\leqslant \mathrm{e}^t \left(\int_{\Omega} |f_p|^2 \mathrm{d} x + \int_{0}^t \int_{\Omega} |f_p|^2 \mathrm{d} x \right), \end{aligned}故存在只依赖 T 和 \|f_p\|_{L^2(\Omega)} 的常数 C 使得 \sup_{0 \leqslant t \leqslant T} \int_{\Omega} \left(|u_m|^2 + |v_m|^2 \right)\mathrm{d} x \leqslant C, \tag{6}类似地，将 \omega_{i} 取成 \dfrac{\partial u_m}{\partial t} 代入 (4) 的第一个方程，得到 \begin{aligned} &\int_{\Omega} \left(\dfrac{\partial u_m}{\partial t}\right)^2 \mathrm{d} x + \dfrac{1}{2} \int_{\Omega} \int_{\Omega} \frac{\left|u_{m}(y)-u_{m}(x)\right|^{p-2}(u_{m}(y)-u_{m}(x))}{|x-y|^{N+s p}}\left(\dfrac{\partial u_m(y)}{\partial t}-\dfrac{\partial u_m(x)}{\partial t}\right) \mathrm{d} x \mathrm{d} y \\ =& - \int_{\Omega} v_m\dfrac{\partial u_m}{\partial t} \mathrm{d} x \\ \leqslant & \dfrac{1}{2} \int_{\Omega} |v_m|^2 \mathrm{d} x + \dfrac{1}{2} \int_{\Omega} \left(\dfrac{\partial u_m}{\partial t}\right)^2 \mathrm{d} x, \end{aligned} 从而 \int_{\Omega} \left(\dfrac{\partial u_m}{\partial t}\right)^2 \mathrm{d} x + \dfrac{1}{p} \dfrac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} \int_{\Omega} \frac{\left|u_{m}(y,t)-u_{m}(x,t)\right|^{p}}{|x-y|^{N+s p}} \mathrm{d} x \mathrm{d} y \leqslant C,两边在 (0,t) 上积分，得到\iint_{Q_T} \left(\dfrac{\partial u_m}{\partial t}\right)^2 \mathrm{d} x + \dfrac{1}{p} \int_{\Omega} \int_{\Omega} \frac{\left|u_{m}(t,y)-u_{m}(t,x)\right|^{p}}{|x-y|^{N+s p}} \mathrm{d} x \mathrm{d} y \leqslant CT + \dfrac{1}{p} [f_p^m]_{W^{s,p}(\Omega)}^p,\tag{7}将 \omega_{i} 取成 \dfrac{\partial v_m}{\partial t} 代入 (4) 的第二个方程，得到 \begin{aligned} \int_{\Omega} \left(\dfrac{\partial v_m}{\partial t}\right)^2 \mathrm{d} x + \dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d}t} \int_{\Omega} |\nabla v_m|^2\mathrm{d} x &= - \int_{\Omega} (f_p-u_m)\dfrac{\partial v_m}{\partial t} \mathrm{d} x \\ &\leqslant \dfrac{1}{2} \int_{\Omega} \left(\dfrac{\partial v_m}{\partial t}\right)^2 \mathrm{d} x + \int_{\Omega} |u_m|^2 \mathrm{d} x + \int_{\Omega} |f_p|^2 \mathrm{d} x, \end{aligned} 同理有 \iint_{Q_T} \left(\dfrac{\partial v_m}{\partial t}\right)^2 \mathrm{d} x + \int_{\Omega} |\nabla v_m|^2 \mathrm{d} x \leqslant CT + \dfrac{1}{2} \|\nabla v_m(\cdot,0)\|_{L^2(\Omega)}^2.\tag{8} \quad\quad由估计式 (6) 至 (8) 可知存在 \{u_m\} 的子列，仍记作 \{u_m\}，在 L^{\infty}\left(0, T ; W^{s, p}(\Omega) \cap L^{2}(\Omega)\right) 中弱*收敛于 u_p，且 \dfrac{\partial u_{m}}{\partial t} 在 L^{2}\left(Q_{T}\right) 中弱收敛于 \dfrac{\partial u_{p}}{\partial t}；存在 \{v_m\} 的子列，仍记作 \{v_m\}，在 L^{\infty}\left(0, T ; H^{1}(\Omega)\right) 中弱*收敛于 v_p，且 \dfrac{\partial v_{m}}{\partial t} 在 L^{2}\left(Q_{T}\right) 中弱收敛于 \dfrac{\partial v_{p}}{\partial t}，由 H^{1}\left(Q_{T}\right) 紧嵌入 L^{2}\left(Q_{T}\right)，我们还可以要求这样抽出的子列能够使 \{v_m\} 在 L^{2}\left(Q_{T}\right) 中强收敛于 v_p. 另外，由 \sup _{0 \leqslant t \leqslant T}\left\|\frac{\left|u_m(t, y)-u_m(t, x)\right|^{p-2}}{|x-y|^{(N+sp) \frac{p-1}{p}}}\left(u_m(t, y)-u_m(t, x)\right)\right\|_{L^{\tfrac{p}{p-1}}(\Omega \times \Omega)} \leqslant C 可知存在 \xi \in L^{\infty}\left(0,T;L^{\tfrac{p}{p-1}}(\Omega \times \Omega)\right) 和 \{u_m\} 的子列，仍记作 \{u_m\}，在 L^{\infty}\left(0,T;L^{\frac{p}{p-1}}(\Omega \times \Omega)\right) 中 \frac{\left|u_m(t, y)-u_m(t, x)\right|^{p-2}}{|x-y|^{(N+sp) \tfrac{p-1}{p}}}\left(u_m(t, y)-u_m(t, x)\right)\overset{*}{\rightharpoonup} \xi(t,x,y), \quad m \to \infty. \quad\quad接下来对逼近解取极限. 定义算子 A: L^{\infty}\left(0, T ; W^{s, p}(\Omega) \cap L^{2}(\Omega)\right) \to L^{\infty}\left(0,T;L^{\frac{p}{p-1}}(\Omega \times \Omega)\right),u \mapsto\frac{|u(t,y)-u(t,x)|^{p-2}(u(t,y)-u(t,x))}{|x-y|^{(N+p s) \tfrac{p-1}{p}}}, 对 (4) 中的第一个方程在 (0,T) 上积分，并取 \varphi_m (x,t)=\psi_m(x) \kappa(t)，其中 \psi_m \in \mathrm{span}\{\omega_1,\omega_2,\cdots,\omega_m\}, \kappa \in C_0^{\infty}(0,T) 就得到 \begin{aligned} &\iiint_{Q_{T}^{1,2}} Au_m \left(\frac{u_m(t,y)-u_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi_m(t,y)-\varphi_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =& - \iint_{Q_{T}} \frac{\partial u_{m}}{\partial t} (u_m-\varphi_m) \mathrm{d} x \mathrm{d} t - \iint_{Q_{T}} v_{m} (u_m-\varphi_m) \mathrm{d} x \mathrm{d} t, \tag{9}\end{aligned} 由 \{\omega_i\} 在 L^2(\Omega) \cap W^{s,p}(\Omega) 中的稠密性可知存在 \varphi 使得 \varphi_m 在 L^2(Q_T) \cap L^1(0,T;W^{s,p}(\Omega)) 中强收敛到 \varphi. 由前面所述的收敛关系，有 \begin{aligned} &\iiint_{Q_{T}^{1,2}} \xi \left(\frac{u_p(t,y)-u_p(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi(t,y)-\varphi(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =& - \iint_{Q_{T}} \frac{\partial u_{p}}{\partial t} (u_p-\varphi) \mathrm{d} x \mathrm{d} t - \iint_{Q_{T}} v_{p} (u_p-\varphi) \mathrm{d} x \mathrm{d} t, \tag{10}\end{aligned} 由 L^2 范数的弱下半连续性，有 \begin{aligned} &\underset{m \to \infty}{\underline{\text{lim}}}\iint_{Q_t} \frac{\partial u_{m}}{\partial \tau} (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} \tau \\ =& \underset{m \to \infty}{\underline{\text{lim}}} \dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d} \tau} \iint_{Q_t} |u_m|^2 \mathrm{d} x \mathrm{d} \tau - \lim_{m \to \infty} \iint_{Q_t} \frac{\partial u_{m}}{\partial \tau} \varphi_{m} \mathrm{d} x \mathrm{d} \tau \\ =& \underset{m \to \infty}{\underline{\text{lim}}} \dfrac{1}{2} \int_{\Omega} |u_m|^2 \mathrm{d} x - \lim_{m \to \infty} \dfrac{1}{2} \int_{\Omega} |f_p^m|^2 \mathrm{d} x - \lim_{m \to \infty} \iint_{Q_t} \frac{\partial u_{m}}{\partial \tau} \varphi_{m} \mathrm{d} x \mathrm{d} \tau \\ \geqslant & \dfrac{1}{2} \int_{\Omega} |u_p|^2 \mathrm{d} x - \dfrac{1}{2} \int_{\Omega} |f_p|^2 \mathrm{d} x - \iint_{Q_t} \frac{\partial u}{\partial \tau} \varphi \mathrm{d} x \mathrm{d} \tau \\ =& \iint_{Q_t} \frac{\partial u_{p}}{\partial \tau} (u_{p}-\varphi) \mathrm{d} x \mathrm{d} \tau, \end{aligned}即 \underset{m \to \infty}{\underline{\text{lim}}}\iint_{Q_T} \frac{\partial u_{m}}{\partial t} (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} t \geqslant \iint_{Q_T} \frac{\partial u_{p}}{\partial t} (u_{p}-\varphi) \mathrm{d} x \mathrm{d} t,\tag{11} 由不等式 \left||a|^{p-2}a-|b|^{p-2}b\right| \leqslant 2|a-b|^{p-1}，有 A \varphi_m 在 L^{\infty}(0,T;L^{\frac{p}{p-1}}(\Omega \times \Omega)) 中强收敛到 A \varphi. 利用 p>1 时 t \mapsto t|t|^{p-2} 的单调性，有 \begin{aligned} &\iiint_{Q_{T}^{1,2}} Au_m \left(\frac{u_m(t,y)-u_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi_m(t,y)-\varphi_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ \geqslant & \iiint_{Q_{T}^{1,2}} A \varphi_m \left(\frac{u_m(t,y)-u_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi_m(t,y)-\varphi_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t , \end{aligned}两边取极限并结合 (9) 至 (11) 就得到 \begin{aligned} & \iiint_{Q_{T}^{1,2}} \xi \left(\frac{u_p(t,y)-u_p(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi(t,y)-\varphi(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ = & - \iint_{Q_T} \frac{\partial u_{p}}{\partial t} (u_{p}-\varphi) \mathrm{d} x \mathrm{d} t - \iint_{Q_{T}} v_{p} (u_p-\varphi) \mathrm{d} x \mathrm{d} t \\ \geqslant & - \underset{m \to \infty}{\underline{\text{lim}}}\iint_{Q_T} \frac{\partial u_{m}}{\partial t} (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} t - \lim_{m \to \infty} \iint_{Q_T} v_m (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} t \\ = & \underset{m \to \infty}{\overline{\text{lim}}} \left(-\iint_{Q_T} \frac{\partial u_{m}}{\partial t} (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} t\right) - \lim_{m \to \infty} \iint_{Q_T} v_m (u_{m}-\varphi_{m}) \mathrm{d} x \mathrm{d} t \\ = & \underset{m \to \infty}{\overline{\text{lim}}} \iiint_{Q_{T}^{1,2}} Au_m \left(\frac{u_m(t,y)-u_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi_m(t,y)-\varphi_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ \geqslant & \lim_{m \to \infty} \iiint_{Q_{T}^{1,2}} A \varphi_m \left(\frac{u_m(t,y)-u_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi_m(t,y)-\varphi_m(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ = & \iiint_{Q_{T}^{1,2}} A \varphi \left(\frac{u_p(t,y)-u_p(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi(t,y)-\varphi(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t, \end{aligned}即 \iiint_{Q_{T}^{1,2}} (\xi-A \varphi) \left(\frac{u_p(t,y)-u_p(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} - \frac{\varphi(t,y)-\varphi(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}}\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \geqslant 0, 对任意 \varepsilon>0，在上式中取 \varphi = u_p - \varepsilon \varphi_1，其中 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,p}(\Omega))，则 \iiint_{Q_{T}^{1,2}} (\xi-A (u_p-\varepsilon \varphi_1)) \frac{\varphi_1(t,y)-\varphi_1(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} \mathrm{d} x \mathrm{d} y \mathrm{d} t \geqslant 0, 令 \varepsilon \to 0^+，则 \iiint_{Q_{T}^{1,2}} (\xi-Au) \frac{\varphi_1(t,y)-\varphi_1(t,x)}{|x-y|^{(N+p s) \tfrac{1}{p}}} \mathrm{d} x \mathrm{d} y \mathrm{d} t \geqslant 0, 这说明 \xi \geqslant Au_p，在上式中用 -\varphi_1 替换 \varphi_1 还有 \xi \leqslant Au_p，因此 \xi = Au_p = \frac{\left|u_p(t, y)-u_p(t, x)\right|^{p-2}}{|x-y|^{(N+sp) \tfrac{p-1}{p}}}\left(u_p(t, y)-u_p(t, x)\right),从而对所有 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,p}(\Omega)) 都有 \begin{aligned}&\iint_{Q_T} \frac{\partial u_{p}}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t+\frac{1}{2} \iint_{Q_T^{1,2}} \frac{\left|u_{p}(y)-u_{p}(x)\right|^{p-2}\left(u_{p}(y)-u_{p}(x)\right)}{|x-y|^{N+s p}}\left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_T} v_{p} \varphi_{1} \mathrm{d} x \mathrm{d} t=0,\end{aligned}同理，由 \{\omega_i\} 在 W^{s,p}(\Omega) 中的稠密性以及上述收敛关系，对 (4) 中第一个方程两边在 (0,T) 上积分可知对所有 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有\iint_{Q_T} \frac{\partial v_{p}}{\partial t} \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \nabla v_{p} \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \left(f_{p}-u_{p}\right) \varphi_{2} \mathrm{d} x \mathrm{d} t=0,由 (6) 至 (8) 结合弱下半连续性可知存在不依赖于 p 和 s 的常数 M 使得 \begin{aligned} &\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; L^{2}(\Omega)\right)}+\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; W^{s, p}(\Omega)\right)}+\left\|\frac{\partial u_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M,\\ &\left\|v_{p}\right\|_{L^{\infty}\left(0, T ; W^{1, 2}(\Omega)\right)}+\left\|\frac{\partial v_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M, \end{aligned}另外由引理 1 和分数阶 Sobolev 空间的紧嵌入定理可知 u_p \in C([0,T];L^1(\Omega)), v_p \in C([0,T];L^2(\Omega)). 至此我们完成了问题 (2) 弱解存在性的证明.\quad\quad最后来证明弱解的唯一性，设 (u_{p1},v_{p1}) 和 (u_{p2},v_{p2}) 是问题 (2) 在相同初边值条件下的两个弱解，记 u_{p12}=u_{p1}-u_{p2}, v_{p12}=v_{p1}-v_{p2}，将 (u_{p1},v_{p1}) 和 (u_{p2},v_{p2}) 代入弱解的定义并作差，则对所有 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,p}(\Omega)) 和 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_t} \frac{\partial u_{p12}}{\partial \tau} \varphi_{1} \mathrm{d} x \mathrm{d} \tau +\frac{1}{2} \iiint_{Q_t^{1,2}} \frac{\left|u_{p1}(y)-u_{p1}(x)\right|^{p-2}\left(u_{p1}(y)-u_{p1}(x)\right)}{|x-y|^{N+s p}}\left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~- \frac{1}{2} \iiint_{Q_t^{1,2}} \frac{\left|u_{p2}(y)-u_{p2}(x)\right|^{p-2}\left(u_{p2}(y)-u_{p2}(x)\right)}{|x-y|^{N+s p}}\left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_t} v_{p12} \varphi_{1} \mathrm{d} x \mathrm{d} \tau=0, \\ &\iint_{Q_t} \frac{\partial v_{p12}}{\partial \tau} \varphi_{2} \mathrm{d} x \mathrm{d} \tau+\iint_{Q_t} \nabla v_{p12} \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} \tau-\iint_{Q_t} u_{p12} \varphi_{2} \mathrm{d} x \mathrm{d} \tau=0, \end{aligned}将以上两式相加并取 \varphi_1=u_{p12}, \varphi_2=v_{p12} 就有 \begin{aligned} &\int_{\Omega}\dfrac{1}{2} \left(|u_{p12}|^2+|w_{p12}|^2\right) \mathrm{d} x +\frac{1}{2} \iiint_{Q_t^{1,2}} \frac{\left|u_{p1}(y)-u_{p1}(x)\right|^{p-2}\left(u_{p1}(y)-u_{p1}(x)\right)}{|x-y|^{N+s p}}\left(u_{p12}(y)-u_{p12}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~- \frac{1}{2} \iiint_{Q_t^{1,2}} \frac{\left|u_{p2}(y)-u_{p2}(x)\right|^{p-2}\left(u_{p2}(y)-u_{p2}(x)\right)}{|x-y|^{N+s p}}\left(u_{p12}(y)-u_{p12}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_t} |\nabla v_{p12}|^2 \mathrm{d} x \mathrm{d} \tau=0, \end{aligned} 再利用 p>1 时 t \mapsto t|t|^{p-2} 的单调性最终有 \int_{\Omega} \left(u_{p1}-u_{p2}\right)^2 \mathrm{d} x + \int_{\Omega} \left(v_{p1}-v_{p2}\right)^2 \mathrm{d} x \leqslant 0,即对几乎所有 t \in [0,T] 都有 u_{p1}=u_{p2}, v_{p1}=v_{p2}，至此已完成问题 (2) 弱解存在唯一性的证明. 定理 1 证毕.\quad\quad现在先给出问题 (1) 的弱解定义. \quad\quad定义 2\quad给定 f \in L^{2}(\Omega) \cap W^{s,1}(\Omega)，若存在 (u,v) 使得
\quad\quad(i)\quad 对任意 q<\dfrac{N}{N-s}，有 u \in L^{\infty}\left(0, T ; W^{s, 1}(\Omega) \cap L^{2}(\Omega)\right) \cap C([0,T];L^q(\Omega)), \dfrac{\partial u_{p}}{\partial t} \in L^{2}\left(Q_{T}\right), v \in L^{\infty}\left(0, T ; H^{1}(\Omega)\right) \cap C\left([0, T] ; L^{2}(\Omega)\right), \dfrac{\partial v}{\partial t} \in L^{2}\left(Q_{T}\right)；
\quad\quad(ii)\quad存在函数 \eta(t,x,y) \in L^{\infty}(Q_T^{1,2}) 满足 \eta(t,x,y) = -\eta(t,y,x), 在 Q_T^{1,2} 上几乎处处有 \eta(t,x,y) \in \mathrm{sgn}(u(t,y)-u(t,x))，且 \|\eta\|_{L^{\infty}(Q_T^{1,2})} \leqslant 1 使得对任意 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_T} \frac{\partial u}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t+\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}}\eta(t,x,y)\left(\varphi_{1}(t,y)-\varphi_{1}(t,x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_T} v \varphi_{1} \mathrm{d} x \mathrm{d} t=0, \end{aligned}且 \ \displaystyle \lim _{t \rightarrow 0^{+}}\left\|u(x, t)-f\right\|_{L^{1}(\Omega)}=0；\quad\quad(iii)\quad对任意的 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有 \iint_{Q_T} \frac{\partial v}{\partial t} \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \nabla v \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \left(f-u\right) \varphi_{2} \mathrm{d} x \mathrm{d} t=0, 且\ \displaystyle \lim _{t \rightarrow 0^{+}}\left\|v(x, t)\right\|_{L^{2}(\Omega)}=0，则称 (u,v) 为问题 (1) 的一个弱解.\quad\quad接下来用类似 A13 中的逼近方法证明问题 (1) 弱解的存在唯一性. \quad\quad定理 2\quad对每一个 f \in L^{2}(\Omega) \cap W^{s,1}(\Omega)，问题 (1) 存在唯一的弱解.\quad\quad证明\quad对于 1<p<\dfrac{N}{s}，取 s_p = \dfrac{N}{(p^*)'}，其中 p^*=\dfrac{Np}{N-sp} 为 p 的分数阶 Sobolev 共轭指数，那么对于所有 1<p<(N^*)'=\dfrac{N}{N+s-1} 都有 0<s< s_p = \dfrac{N}{(p^*)'} = N +s - \dfrac{N}{p} < N +s - \dfrac{N}{(N^*)'} = N+s-\dfrac{N}{\dfrac{N}{N+s-1}}=1, 那么由定理 1 可知问题\begin{array}{ll} \dfrac{\partial u}{\partial t}=\Delta_{\Omega, p}^{s_p} u - v, & (x,t)\in(0, T) \times \Omega, \\[1.5ex] \dfrac{\partial v}{\partial t}=\Delta v - (f-u), & (x,t)\in(0, T) \times \Omega, \\[1.5ex] u(0, x)=f, \ v(0,x)=0, & x \in \Omega, \\[0.8ex] \dfrac{\partial v}{\partial \boldsymbol{n}} = 0, & (x,t)\in(0, T) \times \partial \Omega. \end{array}存在唯一的弱解 (u_p,v_p)，且存在不依赖于 p 和 s_p 的常数 M 使得 \begin{aligned} &\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; L^{2}(\Omega)\right)}+\left\|u_{p}\right\|_{L^{\infty}\left(0, T ; W^{s_p, p}(\Omega)\right)}+\left\|\frac{\partial u_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M,\\ &\left\|v_{p}\right\|_{L^{\infty}\left(0, T ; W^{1, 2}(\Omega)\right)}+\left\|\frac{\partial v_{p}}{\partial t}\right\|_{L^{2}\left(Q_{T}\right)} \leqslant M, \end{aligned}那么由这两个估计式，引理 1 以及分数阶 Sobolev 空间的紧嵌入定理可知对任意的 q<\dfrac{N}{N-s}，都存在 u \in L^{\infty}\left(0, T ; W^{s, 1}(\Omega) \cap L^{2}(\Omega)\right) \cap C([0,T];L^q(\Omega)), v \in L^{\infty}\left(0, T ; H^{1}(\Omega)\right) \cap C\left([0, T] ; L^{2}(\Omega)\right) 和单调下降趋于 1 的序列 p_n 使得 u_{p_n} 在 L^{\infty}\left(0, T ; L^{2}(\Omega)\right) 中弱*收敛于 u，且 \dfrac{\partial u_{p_n}}{\partial t} 在 L^{2}\left(Q_{T}\right) 中弱收敛于 \dfrac{\partial u}{\partial t}，同时 u_{p_n} 在 C([0,T];L^1(\Omega)) 中强收敛到 u；v_{p_n} 在 L^{\infty}\left(0, T ; H^{1}(\Omega)\right) 中弱*收敛于 v，且 \dfrac{\partial v_{p_n}}{\partial t} 在 L^{2}\left(Q_{T}\right) 中弱收敛于 \dfrac{\partial v}{\partial t}，同时又有 v_{p_n} 在 C([0,T];L^2(\Omega)) 中强收敛到 v. 由 H^{1}\left(Q_{T}\right) 紧嵌入 L^{2}\left(Q_{T}\right)，我们还可以要求这样抽出的子列能够使 \{v_m\} 在 L^{2}\left(Q_{T}\right) 中强收敛于 v_p. \quad\quad由问题 (2) 的弱解定义可知，对任意的 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s_{p_n},p_n}(\Omega)) 和 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_T} \frac{\partial u_{p_n}}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t+\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{\left|u_{p_n}(y)-u_{p_n}(x)\right|^{p_n-2}\left(u_{p_n}(y)-u_{p_n}(x)\right)}{|x-y|^{N+s_{p_n} p_n}}\left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_T} v_{p_n} \varphi_{1} \mathrm{d} x \mathrm{d} t=0, \\ &\iint_{Q_T} \frac{\partial v_{p_n}}{\partial t} \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \nabla v_{p_n} \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \left(f-u_{p_n}\right) \varphi_{2} \mathrm{d} x \mathrm{d} t=0, \end{aligned} \tag{12}结合以上的收敛性不难得到 \iint_{Q_T} \frac{\partial v}{\partial t} \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \nabla v \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} t+\iint_{Q_T} \left(f-u\right) \varphi_{2} \mathrm{d} x \mathrm{d} t=0, 此即问题 (1) 弱解定义中的 (iii). 接下来要验证的是问题 (1) 弱解定义中的 (ii). 利用 Hölder 不等式并注意到 N+s_p p =(N+s)p，对于任意 1<p<\dfrac{N}{N+s-1} 和 1<\sigma<\dfrac{p}{p-1}，都有\begin{aligned} &\left\| \left| \frac{u_{p}(t,y)-u_{p}(t,x)}{|x-y|^{N+s}}\right|^{p-2} \frac{u_{p}(t,y)-u_{p}(t,x)}{|x-y|^{N+s}} \right\|_{L^{\sigma}(Q_T^{1,2})}^{\sigma} \\ \leqslant & \left\| \left| \frac{u_{p}(t,y)-u_{p}(t,x)}{|x-y|^{N+s}}\right|^{p-2} \frac{u_{p}(t,y)-u_{p}(t,x)}{|x-y|^{N+s}} \right\|_{L^{\frac{p}{p-1}}(Q_T^{1,2})}^{\sigma} \left|Q_T^{1,2}\right|^{1-\frac{p-1}{p} \sigma} \\ = & \left(\iiint_{Q_T^{1,2}} \frac{\left|u_{p}(t,y)-u_{p}(t,x)\right|^p}{|x-y|^{N+s_p p}} \mathrm{d} x \mathrm{d} y \mathrm{d} t \right)^{\frac{p-1}{p} \sigma} \left|Q_T^{1,2}\right|^{1-\frac{p-1}{p} \sigma} \\ \leqslant & \| u_p \|_{L^p(0,T;W^{s_p,p}(\Omega))}^{\frac{p-1}{p} \sigma} \left|Q_T^{1,2}\right|^{1-\frac{p-1}{p} \sigma} \\ \leqslant & C, \end{aligned}因此存在 \{u_{p_n}\} 的子列，仍记作 \{u_{p_n}\}，使得在 Q_T^{1,2} 中 \left| \frac{u_{p_{n}}(t,y)-u_{p_{n}}(t,x)}{|x-y|^{N+s}}\right|^{p-2} \frac{u_{p_{n}}(t,y)-u_{p_{n}}(t,x)}{|x-y|^{N+s}} \rightharpoonup \eta(t,x,y), \quad n \to \infty,其中 \eta \in L^{\sigma}(Q_T^{1,2})，\eta(t,x,y)=-\eta(t,y,x). 由 L^{\sigma} 范数的弱下半连续性，\| \eta \|_{L^{\sigma}(Q_T)}^{\sigma} \leqslant \underset{n \to \infty}{\underline{\text{lim}}} \left\| \left| \frac{u_{p_{n}}(t,y)-u_{p_{n}}(t,x)}{|x-y|^{N+s}}\right|^{p_n-2} \frac{u_{p_{n}}(t,y)-u_{p_{n}}(t,x)}{|x-y|^{N+s}} \right\|_{L^{\sigma}(Q_T)}^{\sigma} \leqslant C, 在 \| \eta \|_{L^{\sigma}(Q_T)} \leqslant C^{\frac{1}{\sigma}} 中令 \sigma \to +\infty 就得到 \| \eta \|_{L^{\infty}(Q_T)} \leqslant 1. 令 \varphi_1(x,t)=\psi(x)\kappa(t)，其中 \psi \in C^{\infty}(\overline{\Omega}), \kappa \in C_0^{\infty}(0,T)，则\begin{aligned} &\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}} \eta(t,x,y) (\varphi_1(t,y)-\varphi_1(t,x))\mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =&\lim_{n \to \infty} \frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}} \left| \frac{u_{p_n}(t,y)-u_{p_n}(t,x)}{|x-y|^{N+s}}\right|^{p_n-2} \frac{u_{p_n}(t,y)-u_{p_n}(t,x)}{|x-y|^{N+s}} (\varphi_1(t,y)-\varphi_1(t,x)) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =&\lim_{n \to \infty} \frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s_{p_n} p_n}} \left|u_{p_{n}}(t,y)-u_{p_{n}}(t,x)\right|^{p_n-2} (u_{p_{n}}(t,y)-u_{p_{n}}(t,x)) (\varphi_1(t,y)-\varphi_1(t,x)) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =&-\lim_{n \to \infty} \iint_{Q_T} \frac{\partial u_{p_n}}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t-\lim_{n \to \infty} \iint_{Q_T} v_{p_n} \varphi_{1} \mathrm{d} x \mathrm{d} t \\ =&-\iint_{Q_T} \frac{\partial u}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t- \iint_{Q_T} v \varphi_{1} \mathrm{d} x \mathrm{d} t, \end{aligned}利用 C^{\infty}(\overline{\Omega}) 在 W^{s, 1}(\Omega) 中稠密可知其对任意 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_T} \frac{\partial u}{\partial t} \varphi_{1} \mathrm{d} x \mathrm{d} t+\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}}\eta(t,x,y)\left(\varphi_{1}(t,y)-\varphi_{1}(t,x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_T} v \varphi_{1} \mathrm{d} x \mathrm{d} t=0. \end{aligned} \quad\quad接下来证明在 Q_T^{1,2} 上几乎处处有 \eta(t,x,y) \in \mathrm{sgn}(u(t,y)-u(t,x)). 在上式中令 \varphi_1 = u，在 (12) 的第一个式子中令 \varphi_1 = u_{p_n} 就有 \begin{aligned} &\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{|u_{p_n}(t,y)-u_{p_n}(t,x)|^{p_n}}{|x-y|^{(N+s)p_n}} \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ =& -\iint_{Q_T} \frac{\partial u_{p_n}}{\partial t} u_{p_n} \mathrm{d} x \mathrm{d} t- \iint_{Q_T} v_{p_n} u_{p_n} \mathrm{d} x \mathrm{d} t \\ =& -\iint_{Q_T} \left(\frac{\partial u_{p_n}}{\partial t} u_{p_n} - \frac{\partial u}{\partial t} u\right) \mathrm{d} x \mathrm{d} t- \iint_{Q_T} \left(v_{p_n} u_{p_n} - vu\right)\mathrm{d} x \mathrm{d} t \\ &~ +\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}}\eta(t,x,y)\left(\varphi_{1}(t,y)-\varphi_{1}(t,x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} t, \end{aligned}令 n \to \infty 就有 \underset{n \to \infty}{\overline{\text{lim}}} \frac{1}{2}\iiint_{Q_T^{1,2}} \frac{\left|u_{p_{n}}(t,y)-u_{p_{n}}(t,x)\right|^{p_{n}}}{|x-y|^{(N+s) p_{n}}} \mathrm{d} x \mathrm{d} y \mathrm{d} t \leqslant \frac{1}{2}\iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}} \eta(t,x,y) (u(y,t) - u(x,t)) \mathrm{d} x \mathrm{d} y \mathrm{d} t,再利用 [\cdot]_{W^{s,1}(\Omega)} 在 L^{1}(\Omega) 的下半连续性和 Hölder 不等式就有\begin{aligned} &\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}}|u(t,y)-u(t,x)| \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ \leqslant& \underset{n \to \infty}{\underline{\text{lim}}} \frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}}\left|u_{p_{n}}(t,y)-u_{p_{n}}(t,x)\right| \mathrm{d} x \mathrm{d} y \mathrm{d} t \\ \leqslant&\underset{n \to \infty}{\underline{\text{lim}}} \frac{1}{2}\left(\iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{(N+s) p_{n}}}\left|u_{p_{n}}(t,y)-u_{p_{n}}(t,x)\right|^{p_{n}} \mathrm{d} x \mathrm{d} y \mathrm{d} t\right)^{\frac{1}{p_{n}}}\left|Q_T^{1,2}\right|^{\frac{p_{n}-1}{p_{n}}} \\ \leqslant&\frac{1}{2} \iiint_{Q_T^{1,2}} \frac{1}{|x-y|^{N+s}} \eta(t,x, y)(u(t,y)-u(t,x)) \mathrm{d} x \mathrm{d} y \mathrm{d} t, \end{aligned}结合 \| \eta \|_{L^{\infty}(Q_T^{1,2})} \leqslant 1 就得到 \eta(t,x,y) \in \mathrm{sgn}(u(t,y)-u(t,x)), \mathrm{a.e.}\ (t,x,y) \in Q_T^{1,2} . 至此我们完成了问题 (1) 弱解存在性的证明.\quad\quad最后来证明弱解的唯一性. 设 (u_{1},v_{1}) 和 (u_{2},v_{2}) 是问题 (1) 在相同初边值条件下的两个弱解，记 u_{12}=u_{1}-u_{2}, v_{12}=v_{1}-v_{2}，那么由弱解定义可知，存在 \eta_1 \in \mathrm{sgn}(u_1(t,y)-u_1(t,x)), \eta_2 \in \mathrm{sgn}(u_2(t,y)-u_2(t,x))，对所有 \varphi_1 \in L^2(Q_T) \cap L^1(0,T;W^{s,1}(\Omega)) 和 \varphi_2 \in L^2(Q_T) \cap L^1(0,T;H^{1}(\Omega)) 都有 \begin{aligned} &\iint_{Q_t} \frac{\partial u_{12}}{\partial \tau} \varphi_{1} \mathrm{d} x \mathrm{d} \tau +\frac{1}{2} \iiint_{Q_t^{1,2}} \frac{\eta_1(t,x,y)-\eta_2(t,x,y)}{|x-y|^{N+s p}} \left(\varphi_{1}(y)-\varphi_{1}(x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + \iint_{Q_t} w_{12} \varphi_{1} \mathrm{d} x \mathrm{d} \tau =0, \\ &\iint_{Q_t} \frac{\partial v_{12}}{\partial \tau} \varphi_{2} \mathrm{d} x \mathrm{d} \tau +\iint_{Q_t} \nabla v_{12} \cdot \nabla \varphi_{2} \mathrm{d} x \mathrm{d} \tau-\iint_{Q_t} u_{12} \varphi_{2} \mathrm{d} x \mathrm{d} \tau =0, \end{aligned}将以上两式相加并取 \varphi_1=u_{12}, \varphi_2=v_{12} 就有 \begin{aligned} &\dfrac{1}{2} \int_{\Omega} \left(|u_{12}|^2+|w_{12}|^2\right) \mathrm{d} x +\dfrac{1}{2} \iiint_{Q_t^{1,2}} \frac{\eta_1(t,x,y)-\eta_2(t,x,y)}{|x-y|^{N+s p}} \left(u_{12}(t,y)-u_{12}(t,x)\right) \mathrm{d} x \mathrm{d} y \mathrm{d} \tau \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ \iint_{Q_t} |\nabla v_{12}|^2 \mathrm{d} x \mathrm{d} \tau=0, \end{aligned} 再利用 r \mapsto \mathrm{sgn}(r) 的单调性最终有 \int_{\Omega} \left(u_{1}-u_{2}\right)^2 \mathrm{d} x + \int_{\Omega} \left(v_{1}-v_{2}\right)^2 \mathrm{d} x \leqslant 0,即对几乎所有 t \in [0,T] 都有 u_{1}=u_{2}, v_{1}=v_{2}，至此已完成问题 (1) 弱解存在唯一性的证明. 定理 2 证毕.