本文研究一个简化的量子能量输运模型^[1]:

$n_{t}+\mathrm{div}\left[\frac{\varepsilon^{2}}{6}n\nabla\left(\frac{\triangle\sqrt{n}}{\sqrt{n}}\right) -\nabla(nT)+n\nabla V\right]=0,$

(1.1)

$-\mathrm{div}(n\nabla T)=\frac{n}{\tau}(T_{L}(x)-T),$

(1.2)

$\lambda^{2}\triangle V=n-C(x),$

(1.3)

这里电子浓度$n$, 电子温度$T$和电位势$V$为未知函数; 带电粒子杂质$C(x)$和晶格温度$T_{L}(x)$为给定函数; 普朗克常数$\varepsilon>0$, 能量松弛时间$\tau>0$和标度的德拜长度$\lambda>0$为物理参数.模型 (1.1)-(1.3) 可以从量子流体动力学方程组中通过取大时间及小速度极限推导出^[1].文献[1]在周期边界条件下证明了 (1.1)-(1.3) 弱解的整体存在性, 文献[2]得到了其解的半古典极限.

最近文献[3]研究了 (1.1)-(1.3) 一维稳态模型古典解的存在性, 具体来说, 我们考虑了如下边值问题:

$\frac{\varepsilon^{2}}{6}n\left(\frac{(\sqrt{n})_{xx}}{\sqrt{n}}\right)_{x}-(nT)_{x}+nV_{x}=J_{0},$

(1.4)

$-(nT_{x})_{x}=\frac{n}{\tau}(T_{L}(x)-T),$

(1.5)

$\lambda^{2}V_{xx}=n-C(x)\ \ \ \mathrm{in}\ (0,1),$

(1.6)

$n(0)=n(1)=1,\ \ n_{x}(0)=n_{x}(1)=0,\ \ T(0)=T_{0},$

(1.7)

$V(0)=V_{0}=-\frac{\varepsilon^{2}}{6}(\sqrt{n})_{xx}(0)+T_{0},$

(1.8)

其中$J_{0}$表示电流密度, 得到了如下结果:

定理1.1 (见文献[3]) 设$C(x),\ T_{L}(x)\in L^{\infty}(0,1)$, $C(x)>0,\ 0<m_{L}\leq T_{L}(x)\leq M_{L}$, $x\in\ (0,1)$, 则问题 (1.4)-(1.8) 存在古典解$(n,T,V)$使得$n(x)\geq e^{-M}>0$, $x\in(0,1)$, 其中$M$是

$M=\sqrt{\frac{e^{2M}}{\tau m_{L}^{2}}(M_{L}-m_{L})M_{L}+\frac{2(e^{-1}+ \parallel C(x)\log C(x)\parallel_{L^{\infty}(0,1)})}{\lambda^{2}m_{L}}}$

(1.9)

的解.

本文继续证明问题 (1.4)-(1.8) 解的唯一性, 我们的主要结果是

定理1.2 设定理1.1中的条件成立, 再若$m_{L}$充分大且$|J_{0}|$较小, 则问题 (1.4)-(1.8) 的解是唯一的.

注1.1 模型 (1.1)-(1.3) 是一种简化的量子能量输运模型, 对于其他类型的量子能量输运模型, 我们可以参考文献[4-6].

注1.2 当晶格温度$T_{L}(x)$和电子温度$T$都为常数时, 则模型 (1.1)-(1.3) 变成量子漂移-扩散模型, 此类模型的研究结果见文献[7-19].

文献[3]通过指数变换$n=e^{u}$把问题 (1.4)-(1.8) 变成了如下与之等价的问题

$\frac{\varepsilon^{2}}{12}\left(u_{xx}+\frac{u_{x}^{2}}{2}\right)_{xx}-T_{xx}-(u_{x}T)_{x} +\frac{e^{u}-C(x)}{\lambda^{2}}=J_{0}(e^{-u})_{x},$

(2.1)

$-(e^{u}T_{x})_{x}=\frac{e^{u}}{\tau}(T_{L}(x)-T),$

(2.2)

$V(x)=-\frac{\varepsilon^{2}}{12}\left(u_{xx}+\frac{u_{x}^{2}}{2}\right)(x)+T(x)+\int_{0}^{x}u_{x}(s)T(s)ds +J_{0}\int_{0}^{x}e^{-u(s)}ds,$

(2.3)

$u(0)=u(1)=0,\ \ u_{x}(0)=u_{x}(1)=0,\ \ T(0)=T_{0},\ \ T_{x}(0)=T_{x}(1)=0,$

(2.4)

$V(0)=V_{0}=-\frac{\varepsilon^{2}}{12}u_{xx}(0)+T_{0},$

(2.5)

并证明了此问题存在解$(u,T,V)\in H^{4}(0,1)\times H^{2}(0,1)\times H^{2}(0,1)$, 所以为了得到定理1.2, 只需证明问题 (2.1), (2.2), (2.4) 解的唯一性即可.为此有如下定理

定理2.1 设定理1.1中的条件成立, 再若$m_{L}$充分大且$|J_{0}|$较小, 使得

$m_{L}-\frac{\varepsilon M^{2}}{24}\sqrt{6m_{L}}-\frac{|J_{0}|e^{M}}{\sqrt{2}} -\frac{e^{2M}(\sqrt{2\varepsilon}+\sqrt[4]{6m_{L}}\cdot M)(\sqrt{2}e^{2M}+1)(M_{L}-m_{L})}{2\sqrt{2\varepsilon}\cdot\tau}>0,$

(2.6)

则问题 (2.1), (2.2), (2.4) 的解$(u,T)\in H^{4}(0,1)\times H^{2}(0,1)$是唯一的.

注2.1 由 (1.9) 式知, 当$m_{L}$较大时$M$会较小, 从而可以保证当$m_{L}$充分大且$|J_{0}|$较小时 (2.6) 式成立.

在文献[3]的引理2.1中已经得到了如下估计：

$\frac{\varepsilon^{2}}{12}\parallel u_{xx}\parallel_{L^{2}(0,1)}^{2}+\frac{m_{L}}{2}\parallel u_{x}\parallel_{L^{2}(0,1)}^{2}\nonumber\\ \leq \frac{e^{2M}}{2\tau m_{L}}(M_{L}-m_{L})M_{L}+\lambda^{-2}(e^{-1}+\parallel C(x)\log C(x)\parallel_{L^{\infty}(0,1)}),$

(2.7)

$\parallel u\parallel_{L^{\infty}(0,1)}\leq M,$

(2.8)

$0<m_{L}\leq T\leq M_{L},$

(2.9)

其中 (2.9) 式见文献[3]中引理2.1的证明.为了证明定理2.1, 还需要如下估计

引理2.1 设$(u,T)\in H^{4}(0,1)\times H^{2}(0,1)$是问题 (2.1), (2.2), (2.4) 的解, 则

$\parallel u_{x}\parallel_{L^{\infty}(0,1)}\leq\frac{\sqrt[4]{6m_{L}}\cdot M}{\sqrt{\varepsilon}},$

(2.10)

$\parallel T_{x}\parallel_{L^{\infty}(0,1)}\leq\frac{e^{2M}}{\tau}(M_{L}-m_{L}).$

(2.11)

证 由均值不等式及 (2.7) 式, 得

$\frac{\sqrt{m_{L}}}{\sqrt{6}}\varepsilon\parallel u_{xx}\parallel_{L^{2}(0,1)}\parallel u_{x}\parallel_{L^{2}(0,1)}\leq\frac{\varepsilon^{2}}{12}\parallel u_{xx}\parallel_{L^{2}(0,1)}^{2}+\frac{m_{L}}{2}\parallel u_{x}\parallel_{L^{2}(0,1)}^{2}\\ \leq \frac{e^{2M}}{2\tau m_{L}}(M_{L}-m_{L})M_{L}+\lambda^{-2}(e^{-1}+\parallel C(x)\log C(x)\parallel_{L^{\infty}(0,1)}),$

所以再由Hölder不等式及 (1.9) 式, 得

$u_{x}^{2}(x)=2\int_{0}^{x}u_{xx}(s)u_{x}(s)ds\leq2\parallel u_{xx}\parallel_{L^{2}(0,1)}\parallel u_{x}\parallel_{L^{2}(0,1)}\\ \leq \frac{2\sqrt{6}}{\sqrt{m_{L}}\varepsilon}\left[\frac{e^{2M}}{2\tau m_{L}}(M_{L}-m_{L})M_{L}+\lambda^{-2}(e^{-1}+\parallel C(x)\log C(x)\parallel_{L^{\infty}(0,1)})\right]\\ = \frac{2\sqrt{6}}{\sqrt{m_{L}}\varepsilon}\cdot M^{2}\cdot\frac{m_{L}}{2} =\frac{\sqrt{6m_{L}}\cdot M^{2}}{\varepsilon},$

从而 (2.10) 式成立.

(2.2) 式两边在$(0,x)$上积分, 得

$T_{x}=-\frac{e^{-u}}{\tau}\int_{0}^{x}e^{u}(T_{L}(x)-T)dx,$

所以由 (2.8) 及 (2.9) 式知 (2.11) 式成立.

定理2.1的证明 设$(u_{1},T_{1}),\ (u_{2},T_{2})\in H^{4}(0,1)\times H^{2}(0,1)$为问题 (2.1), (2.2), (2.4) 的两个解.用$T_{1}-T_{2}$分别作为

$-(e^{u_{1}}T_{1x})_{x}=\frac{e^{u_{1}}}{\tau}(T_{L}(x)-T_{1})$

和

$-(e^{u_{2}}T_{2x})_{x}=\frac{e^{u_{2}}}{\tau}(T_{L}(x)-T_{2})$

的试验函数并两式相减, 得

$\int_{0}^{1}e^{u_{1}}(T_{1}-T_{2})_{x}^{2}dx=-\int_{0}^{1}T_{2x}(e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})_{x}dx -\frac{1}{\tau}\int_{0}^{1}e^{u_{1}}(T_{1}-T_{2})^{2}dx\nonumber\\ +\frac{1}{\tau}\int_{0}^{1}(T_{L}(x)-T_{2}) (e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})dx\nonumber\\ \leq -\int_{0}^{1}T_{2x}(e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})_{x}dx\nonumber\\ +\frac{1}{\tau}\int_{0}^{1}(T_{L}(x)-T_{2})(e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})dx.$

(2.12)

由 (2.8) 式知

$\int_{0}^{1}e^{u_{1}}(T_{1}-T_{2})_{x}^{2}dx\geq e^{-M}\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx.$

(2.13)

由拉格朗日中值定理及 (2.8) 式知

$|e^{u_{1}}-e^{u_{2}}|\leq e^{M}|u_{1}-u_{2}|,$

所以再由 (2.11) 式, Hölder不等式及Poincaré不等式, 得

$-\int_{0}^{1}T_{2x}(e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})_{x}dx\leq\frac{e^{2M}}{\tau}(M_{L}-m_{L}) \int_{0}^{1}e^{M}|u_{1}-u_{2}|\cdot|(T_{1}-T_{2})_{x}|dx\nonumber\\ \leq \frac{e^{3M}}{\tau}(M_{L}-m_{L})\left[\int_{0}^{1}(u_{1}-u_{2})^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}}\nonumber\\ \leq \frac{e^{3M}}{\sqrt{2}\tau}(M_{L}-m_{L})\left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}}.$

(2.14)

利用 (2.9) 式, 类似上式估计, 可以得到

$\frac{1}{\tau}\int_{0}^{1}(T_{L}(x)-T_{2})(e^{u_{1}}-e^{u_{2}})(T_{1}-T_{2})dx\nonumber\\ \leq \frac{e^{M}}{2\tau}(M_{L}-m_{L})\left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}}.$

(2.15)

由 (2.12)-(2.15) 式可得

$\left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}}\leq\frac{e^{2M}(\sqrt{2}e^{2M}+1)(M_{L}-m_{L})} {2\tau}\left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}}.$

(2.16)

用$u_{1}-u_{2}$分别作为

$\frac{\varepsilon^{2}}{12}\left(u_{1xx}+\frac{u_{1x}^{2}}{2}\right)_{xx}-T_{1xx}-(u_{1x}T_{1})_{x} +\frac{e^{u_{1}}-C(x)}{\lambda^{2}}=J_{0}(e^{-u_{1}})_{x}$

和

$\frac{\varepsilon^{2}}{12}\left(u_{2xx}+\frac{u_{2x}^{2}}{2}\right)_{xx}-T_{2xx}-(u_{2x}T_{2})_{x} +\frac{e^{u_{2}}-C(x)}{\lambda^{2}}=J_{0}(e^{-u_{2}})_{x}$

的试验函数并两式相减, 得

$\frac{\varepsilon^{2}}{12}\int_{0}^{1}(u_{1}-u_{2})_{xx}^{2}dx +\frac{\varepsilon^{2}}{24}\int_{0}^{1}(u_{1x}^{2}-u_{2x}^{2})(u_{1}-u_{2})_{xx}dx +\int_{0}^{1}(T_{1}-T_{2})_{x}(u_{1}-u_{2})_{x}dx\nonumber\\ +\int_{0}^{1}T_{1}(u_{1}-u_{2})_{x}^{2}dx+\int_{0}^{1}u_{2x}(T_{1}-T_{2})(u_{1}-u_{2})_{x}dx +\frac{1}{\lambda^{2}}\int_{0}^{1}(e^{u_{1}}-e^{u_{2}})(u_{1}-u_{2})dx\nonumber\\ = -J_{0}\int_{0}^{1}(e^{-u_{1}}-e^{-u_{2}})(u_{1}-u_{2})_{x}dx.$

(2.17)

由 (2.10) 式及Young不等式, 得

$\frac{\varepsilon^{2}}{24}\int_{0}^{1}(u_{1x}^{2}-u_{2x}^{2})(u_{1}-u_{2})_{xx}dx =\frac{\varepsilon^{2}}{24}\int_{0}^{1}(u_{1x}+u_{2x})(u_{1}-u_{2})_{x}(u_{1}-u_{2})_{xx}dx\nonumber\\ \geq-\frac{\varepsilon^{\frac{3}{2}}\cdot\sqrt[4]{6m_{L}}\cdot M}{12}\int_{0}^{1}|(u_{1}-u_{2})_{x}| \cdot|(u_{1}-u_{2})_{xx}|dx\nonumber\\ \geq -\frac{\varepsilon^{2}}{24}\int_{0}^{1}(u_{1}-u_{2})_{xx}^{2}dx -\frac{\varepsilon M^{2}}{24}\sqrt{6m_{L}}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.$

(2.18)

由Hölder不等式及 (2.16) 式, 得

$\int_{0}^{1}(T_{1}-T_{2})_{x}(u_{1}-u_{2})_{x}dx\geq-\left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}}\nonumber\\ \geq-\frac{e^{2M}(\sqrt{2}e^{2M}+1)(M_{L}-m_{L})}{2\tau}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.$

(2.19)

由 (2.9) 式知

$\int_{0}^{1}T_{1}(u_{1}-u_{2})_{x}^{2}dx\geq m_{L}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.$

(2.20)

由 (2.10) 式, Hölder不等式, Poincaré不等式及 (2.16) 式, 得

$\int_{0}^{1}u_{2x}(T_{1}-T_{2})(u_{1}-u_{2})_{x}dx\geq-\frac{\sqrt[4]{6m_{L}}\cdot M}{\sqrt{\varepsilon}} \int_{0}^{1}|T_{1}-T_{2}|\cdot|(u_{1}-u_{2})_{x}|dx\nonumber\\ \geq-\frac{\sqrt[4]{6m_{L}}\cdot M}{\sqrt{2\varepsilon}} \left[\int_{0}^{1}(T_{1}-T_{2})_{x}^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}}\nonumber\\ \geq-\frac{\sqrt[4]{6m_{L}}\cdot M\cdot e^{2M}(\sqrt{2}e^{2M}+1)(M_{L}-m_{L})}{2\sqrt{2\varepsilon}\cdot\tau} \int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.$

(2.21)

由函数$e^{x}$的单调递增性可知

$\frac{1}{\lambda^{2}}\int_{0}^{1}(e^{u_{1}}-e^{u_{2}})(u_{1}-u_{2})dx\geq0.$

(2.22)

由格朗日中值定理, Hölder不等式及Poincaré不等式, 得

$-J_{0}\int_{0}^{1}(e^{-u_{1}}-e^{-u_{2}})(u_{1}-u_{2})_{x}dx \leq|J_{0}|e^{M}\int_{0}^{1}|u_{1}-u_{2}|\cdot|(u_{1}-u_{2})_{x}|dx\nonumber\\ \leq\frac{|J_{0}|e^{M}}{\sqrt{2}}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.$

(2.23)

由 (2.17)-(2.23) 式可得

$\frac{\varepsilon^{2}}{24}\int_{0}^{1}(u_{1}-u_{2})_{xx}^{2}dx +C_{0}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\leq0,$

(2.24)

其中

$C_{0}=m_{L}-\frac{\varepsilon M^{2}}{24}\sqrt{6m_{L}}-\frac{|J_{0}|e^{M}}{\sqrt{2}} -\frac{e^{2M}(\sqrt{2\varepsilon}+\sqrt[4]{6m_{L}}\cdot M)(\sqrt{2}e^{2M}+1)(M_{L}-m_{L})}{2\sqrt{2\varepsilon}\cdot\tau}.$

由 (2.24) 式及条件 (2.6) 式可知$u_{1}=u_{2}$, 再由 (2.16) 式得$T_{1}=T_{2}$.