indent 20pt Recently, the quantum drift-diffusion equations attracted many scientists' interest since they are capable to describe quantum confinement and tunneling effects in metal-oxide-semiconductor structures and to simulate ultra-small semiconductor devices [1, 2]. Quantum drift-diffusion models were derived from a Wigner-Boltzmann equation by a moment method [3]. This paper is concerned with the bipolar quantum drift-diffusion model [4]:

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

(1.1)

$p_{t}=\mathrm{div}\left[-\xi\varepsilon^{2}p\nabla\left(\frac{\triangle\sqrt{p}}{\sqrt{p}}\right) +\nabla p+p\nabla V\right],$

(1.2)

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

(1.3)

where the particle density $n$, the hole density $p$ and the electrostatic potential $V$ are unknown variables; the scaled Planck constant $\varepsilon>0$, the scaled Debye length $\lambda>0$ and the ratio of the effective masses of electrons and holes $\xi>0$ are physical parameters; the doping profile $C(x)$ representing the distribution of charged background ions. This type of transient model consists of one or two fourth-order parabolic equations (unipolar or bipolar) coupled to a Poisson equation and has been studied in many works [4--14]. Abdallah and Unterreiter [15] showed the existence of solutions to the stationary model of (1.1)--(1.3) over the multi-dimension bounded domain and carried out the semiclassical limit.

The objective of this paper is to analyze the one-dimensional stationary version of (1.1)--(1.3):

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

(1.4)

$-\varepsilon^{2}p\left(\frac{(\sqrt{p})_{xx}}{\sqrt{p}}\right)_{x}+p_{x}+pV_{x}=J_{1},$

(1.5)

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

(1.6)

where we have let $\lambda=\xi=1$ for convenience as in [4], the electron current density $J_{0}$ and the hole current density $J_{1}$ are two constants. We choose the physically motivated boundary conditions:

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

(1.7)

$p(0)=p(1)=1,\ \ p_{x}(0)=p_{x}(1)=0,$

(1.8)

$V(0)=V_{0}.$

(1.9)

Dividing (1.4) by $n$ and taking the derivative gives

$-\varepsilon^{2}\left(\frac{(\sqrt{n})_{xx}}{\sqrt{n}} \right)_{xx}+\left(\frac{n_{x}}{n}\right)_{x}-(n-p-C(x))=\left(\frac{J_{0}}{n}\right)_{x},$

(1.10)

where we have used the Poisson equation (1.6). Similarly, dividing (1.5) by $p$ and taking the derivative leads to

$-\varepsilon^{2}\left(\frac{(\sqrt{p})_{xx}}{\sqrt{p}} \right)_{xx}+\left(\frac{p_{x}}{p}\right)_{x}+(n-p-C(x))=\left(\frac{J_{1}}{p}\right)_{x}.$

(1.11)

After two exponential transformations $n=e^{u},\ p=e^{v}$, we obtain

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

(1.12)

$-\frac{\varepsilon^{2}}{2}\left(v_{xx}+\frac{v_{x}^{2}}{2}\right)_{xx}+v_{xx} +(e^{u}-e^{v}-C(x))=J_{1}(e^{-v})_{x}$

(1.13)

with the boundary conditions

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

(1.14)

$v(0)=v(1)=0,\ \ \ v_{x}(0)=v_{x}(1)=0.$

(1.15)

As usual, we call $(u,v)\in H_{0}^{2}(0,1)\times H_{0}^{2}(0,1)$ a weak solution of the problem (1.12)--(1.15), if for all $\psi\in H_{0}^{2}(0,1)$ it holds

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}\left(u_{xx}+\frac{u_{x}^{2}}{2}\right)\psi_{xx}dx+\int_{0}^{1} u_{x}\psi_{x}dx+\int_{0}^{1}(e^{u}-e^{v}-C(x))\psi dx =J_{0}\int_{0}^{1}e^{-u}\psi_{x}dx,$

(1.16)

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}\left(v_{xx}+\frac{v_{x}^{2}}{2}\right)\psi_{xx}dx+\int_{0}^{1} v_{x}\psi_{x}dx-\int_{0}^{1}(e^{u}-e^{v}-C(x))\psi dx =J_{1}\int_{0}^{1}e^{-v}\psi_{x}dx.$

(1.17)

Our main results are stated as follows:

Theorem 1.1 (Existence) Let $C(x)\in L^{2}(0,1)$, then there exists a weak solution $(u,v)\in H_{0}^{2}(0,1)\times H_{0}^{2}(0,1)$ of the problem (1.12)--(1.15) for any $J_{0},J_{1}\in\mathbb{R}$.

Theorem 1.2 (Uniqueness) Let $C(x)\in L^{2}(0,1)$. If

$\varepsilon\parallel C(x)\parallel_{L^{2}(0,1)}^{2}+(1+\sqrt{2}|J_{0}|) e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\leq2,$

(1.18)

$\varepsilon\parallel C(x)\parallel_{L^{2}(0,1)}^{2}+(1+\sqrt{2}|J_{1}|) e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\leq2,$

(1.19)

then the problem (1.12)--(1.15) has a unique solution.

Theorem 1.3 (Semiclassical limit) Let $(u_{\varepsilon},v_{\varepsilon})$ be a solution to the problem (1.12)--(1.15) obtained in Theorem 1.1. Then as $\varepsilon\rightarrow0$, maybe for a subsequence,

$u_{\varepsilon}\rightarrow u,\ \ \ v_{\varepsilon}\rightarrow v\ \ \ {\rm weakly\ in}\ H^{1}(0,1)\ {\rm and\ strongly\ in} \ L^{\infty}(0,1)$

(1.20)

and $(u,v)$ is a weak solution of

$u_{xx}-(e^{u}-e^{v}-C(x))=J_{0}(e^{-u})_{x},$

(1.21)

$v_{xx}+(e^{u}-e^{v}-C(x))=J_{1}(e^{-v})_{x}$

(1.22)

subject to the boundary condition

$u(0)=u(1)=0,$

(1.23)

$v(0)=v(1)=0.$

(1.24)

Remark 1.1 Although Abdallah and Unterreiter obtained the stationary bipolar quantum drift-diffusion model in [15], but they did not show the uniqueness. In this paper, we give such a result. In addition, the proof of [15] was based on a Schauder fixed point iteration combined with a minimization procedure, whereas in this paper, we reformulate the model as two coupled fourth-order elliptic equations by using exponential variable transformations and employ the Schauder fixed-point theorem.

This article is organized as follows. In Section 2, we will show the existence of solutions to the problem (1.12)--(1.15) by using the techniques of a priori estimates and Schauder fixed-point theorem. Then we will prove the uniqueness and the semiclassical limit of the solutions in Section 3 and Section 4, respectively.

indent 20pt In order to use the Schauder fixed-point theorem to prove the existence, we need the following lemma:

Lemma 2.1 Let $C(x)\in L^{2}(0,1)$ and let $(u,v)\in H_{0}^{2}(0,1)\times H_{0}^{2}(0,1)$ be a solution of (1.12)--(1.15). Then

$\varepsilon^{2}\parallel u_{xx}\parallel_{L^{2}(0,1)}^{2} +\varepsilon^{2}\parallel v_{xx}\parallel_{L^{2}(0,1)}^{2}+\parallel u_{x}\parallel_{L^{2}(0,1)}^{2}+\parallel v_{x}\parallel_{L^{2}(0,1)}^{2}\leq2\parallel C(x)\parallel_{L^{2}(0,1)}^{2}.$

(2.1)

Proof We use $\psi=u$ as a test function in the weak formulation of (1.12) to obtain

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}\left(u_{xx}^{2}+\frac{1}{2}u_{x}^{2}u_{xx}\right)dx +\int_{0}^{1}u_{x}^{2}dx=-\int_{0}^{1}(e^{u}-e^{v}-C(x))udx +J_{0}\int_{0}^{1}e^{-u}u_{x}dx.$

(2.2)

The boundary condition (1.14) gives

$\int_{0}^{1}u_{x}^{2}u_{xx}dx=\frac{1}{3}[u_{x}^{3}(1)-u_{x}^{3}(0)]=0$

and

$\int_{0}^{1}e^{-u}u_{x}dx=-\int_{0}^{1}(e^{-u})_{x}dx=e^{-u(0)}-e^{-u(1)}=0.$

Consequently, equation (2.2) is equivalent to

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{xx}^{2}dx +\int_{0}^{1}u_{x}^{2}dx=-\int_{0}^{1}(e^{u}-e^{v}-C(x))udx.$

(2.3)

Similarly, using $\psi=v$ as a test function in the weak formulation of (1.13) and using the boundary condition (1.15) we get

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}v_{xx}^{2}dx +\int_{0}^{1}v_{x}^{2}dx=-\int_{0}^{1}(e^{u}-e^{v}-C(x))vdx.$

(2.4)

Summing up (2.3) and (2.4), we have

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{xx}^{2}dx+\frac{\varepsilon^{2}}{2}\int_{0}^{1}v_{xx}^{2}dx +\int_{0}^{1}u_{x}^{2}dx+\int_{0}^{1}v_{x}^{2}dx=-\int_{0}^{1}(e^{u}-e^{v}-C(x))(u-v)dx.$

(2.5)

The monotonicity of $x\mapsto e^{x}$ implies

$-\int_{0}^{1}(e^{u}-e^{v})(u-v)dx\leq0.$

From the Young inequality and the Poincaré inequality,

$\begin{eqnarray*} \int_{0}^{1}C(x)(u-v)dx&\leq&\frac{1}{2}\int_{0}^{1}u^{2}dx+\frac{1}{2}\int_{0}^{1}v^{2}dx+\int_{0}^{1}C(x)^{2}dx\\ &\leq&\frac{1}{2}\int_{0}^{1}u_{x}^{2}dx+\frac{1}{2}\int_{0}^{1}v_{x}^{2}dx+\int_{0}^{1}C(x)^{2}dx.\end{eqnarray*}$

So (2.5) can be estimated as

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{xx}^{2}dx+\frac{\varepsilon^{2}}{2}\int_{0}^{1}v_{xx}^{2}dx +\frac{1}{2}\int_{0}^{1}u_{x}^{2}dx+\frac{1}{2}\int_{0}^{1}v_{x}^{2}dx\leq\int_{0}^{1}C(x)^{2}dx.$

This proves the lemma.

Proof of Theorem 1.1 Consider the following linear problems for given $(\rho,\eta)\in W_{0}^{1,4}(0,1)\times W_{0}^{1,4}(0,1)$ with test functions $\psi\in H_{0}^{2}(0,1)$:

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{xx}\psi_{xx}dx +\frac{\sigma\varepsilon^{2}}{4}\int_{0}^{1}\rho_{x}^{2}\psi_{xx}dx +\int_{0}^{1}u_{x}\psi_{x}dx+\sigma\int_{0}^{1}(e^{\rho}-e^{\eta}-C(x))\psi dx=\sigma J_{0}\int_{0}^{1}e^{-\rho}\psi_{x}dx,$

(2.6)

$\frac{\varepsilon^{2}}{2}\int_{0}^{1}v_{xx}\psi_{xx}dx +\frac{\sigma\varepsilon^{2}}{4}\int_{0}^{1}\eta_{x}^{2}\psi_{xx}dx +\int_{0}^{1}v_{x}\psi_{x}dx-\sigma\int_{0}^{1}(e^{\rho}-e^{\eta}-C(x))\psi dx=\sigma J_{1}\int_{0}^{1}e^{-\eta}\psi_{x}dx,$

(2.7)

where $\sigma\in[0,1]$. We define the bilinear form

$a(u,\psi)=\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{xx}\psi_{xx}dx+\int_{0}^{1}u_{x}\psi_{x}dx,$

(2.8)

and the linear functional

$F(\psi)=-\frac{\sigma\varepsilon^{2}}{4}\int_{0}^{1}\rho_{x}^{2}\psi_{xx}dx -\sigma\int_{0}^{1}(e^{\rho}-e^{\eta}-C(x))\psi dx+\sigma J_{0}\int_{0}^{1}e^{-\rho}\psi_{x}dx.$

(2.9)

Since the bilinear form $a(u,\psi)$ is continuous and coercive on $H_{0}^{2}(0,1)\times H_{0}^{2}(0,1)$ and the linear functional $F(\psi)$ is continuous on $H_{0}^{2}(0,1)$, we can apply the Lax-Milgram theorem to obtain the existence of a solution $u\in H_{0}^{2}(0,1)$ of (2.6). Similarly there exists a solution $v\in H_{0}^{2}(0,1)$ to (2.7). Thus, the operator

$S:\ W_{0}^{1,4}(0,1)\times W_{0}^{1,4}(0,1)\times[0,1]\rightarrow W_{0}^{1,4}(0,1)\times W_{0}^{1,4}(0,1), \ \ \ (\rho,\eta,\sigma)\mapsto (u,v)$

is well defined. Moreover, it is continuous and compact since the embedding $H_{0}^{2}(0,1)\hookrightarrow W_{0}^{1,4}(0,1)$ is compact. Furthermore, $S(\rho,\eta,0)=(0,0)$. Following the steps of the proof of Lemma 2.1, we can show that $\parallel u\parallel_{H_{0}^{2}(0,1)}+\parallel v\parallel_{H_{0}^{2}(0,1)}\leq \mathrm{const}.$ for all $(u,v,\sigma)\in W_{0}^{1,4}(0,1)\times W_{0}^{1,4}(0,1)\times[0,1]$ satisfying $S(u,v,\sigma)=(u,v)$. Therefore, the existence of a fixed point $(u,v)$ with $S(u,v,1)=(u,v)$ follows from the Schauder fixed-point theorem. This fixed point is a solution of (1.12)--(1.15).

To prove the uniqueness, we need the following lemma:

Lemma 3.1 Let $(u,v)$ be a solution of (1.12)--(1.15) obtained in Theorem 1.1. Then

$\parallel u\parallel_{L^{\infty}(0,1)},\ \parallel v\parallel_{L^{\infty}(0,1)}\leq\sqrt{2}allel C(x)\parallel_{L^{2}(0,1)},$

(3.1)

$\parallel u_{x}\parallel_{L^{\infty}(0,1)},\ \parallel v_{x}\parallel_{L^{\infty}(0,1)} \leq\frac{2\parallel C(x)\parallel_{L^{2}(0,1)}}{\sqrt{\varepsilon}}.$

(3.2)

Proof For simplicity, we only treat with the case of $u$. (3.1) can be concluded directly from (2.1) and the Poincaré-Sobolev inequality:

$\parallel u\parallel_{L^{\infty}(0,1)}\leq\parallel u_{x}\parallel_{L^{2}(0,1)}\leq \sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}.$

We observe that, due to the boundary conditions for $u_{x}$,

$u_{x}(x)^{2}=2\int_{0}^{x}u_{x}(s)u_{xx}(s)ds\leq2\parallel u_{x}\parallel_{L^{2}(0,1)}allel u_{xx}\parallel_{L^{2}(0,1)}$

and thus by the Young inequality and (2.1)

$\begin{eqnarray*}&& \parallel u_{x}\parallel_{L^{\infty}(0,1)}\leq\sqrt{2}\sqrt{\parallel u_{x}\parallel_{L^{2}(0,1)}allel u_{xx}\parallel_{L^{2}(0,1)}}\\ & \leq& \frac{\sqrt{2}}{2\sqrt{\varepsilon}}\parallel u_{x}\parallel_{L^{2}(0,1)} +\frac{\sqrt{2\varepsilon}}{2}\parallel u_{xx}\parallel_{L^{2}(0,1)}\\ & \leq& \frac{2\parallel C(x)\parallel_{L^{2}(0,1)}}{\sqrt{\varepsilon}}.\end{eqnarray*}$

Proof of Theorem 1.2 Let $(u_{1},v_{1}),\ (u_{2},v_{2})\in H_{0}^{2}(0,1)\times H_{0}^{2}(0,1)$ be two weak solutions of (1.12)--(1.15). The weak formulations of the difference of the equations satisfied by $(u_{1},v_{1})$ and $(u_{2},v_{2})$, with the test functions $u_{1}-u_{2}$ and $v_{1}-v_{2}$, respectively, read as follows:

$\begin{eqnarray}&&\frac{\varepsilon^{2}}{2}\int_{0}^{1}(u_{1}-u_{2})_{xx}^{2}dx+\frac{\varepsilon^{2}}{4}\int_{0}^{1} (u_{1x}^{2}-u_{2x}^{2})(u_{1}-u_{2})_{xx}dx+\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\nonumber\\ & =& -\int_{0}^{1}(e^{u_{1}}-e^{v_{1}}-e^{u_{2}}+e^{v_{2}})(u_{1}-u_{2})dx -J_{0}\int_{0}^{1}(e^{-u_{1}}-e^{-u_{2}})(u_{1}-u_{2})_{x}dx,\end{eqnarray}$

(3.3)

$\begin{eqnarray}&&\frac{\varepsilon^{2}}{2}\int_{0}^{1}(v_{1}-v_{2})_{xx}^{2}dx+\frac{\varepsilon^{2}}{4}\int_{0}^{1} (v_{1x}^{2}-v_{2x}^{2})(v_{1}-v_{2})_{xx}dx+\int_{0}^{1}(v_{1}-v_{2})_{x}^{2}dx\nonumber\\ &=&\int_{0}^{1}(e^{u_{1}}-e^{v_{1}}-e^{u_{2}}+e^{v_{2}})(v_{1}-v_{2})dx -J_{1}\int_{0}^{1}(e^{-v_{1}}-e^{-v_{2}})(v_{1}-v_{2})_{x}dx.\end{eqnarray}$

(3.4)

Using (3.2) and the Young inequality, we can estimate the second integral on the left-hand side of (3.3) as

$\begin{eqnarray}&& \frac{\varepsilon^{2}}{4}\int_{0}^{1} (u_{1x}^{2}-u_{2x}^{2})(u_{1}-u_{2})_{xx}dx\nonumber\\ &=& \frac{\varepsilon^{2}}{4}\int_{0}^{1} (u_{1x}+u_{2x})(u_{1}-u_{2})_{x}(u_{1}-u_{2})_{xx}dx\nonumber\\ &\geq& -\varepsilon^{\frac{3}{2}}\parallel C(x)\parallel_{L^{2}(0,1)}\int_{0}^{1} |(u_{1}-u_{2})_{xx}|\cdot|(u_{1}-u_{2})_{x}|dx\nonumber\\ & \geq& -\frac{\varepsilon^{2}}{2}\int_{0}^{1} (u_{1}-u_{2})_{xx}^{2}dx-\frac{\varepsilon}{2}\parallel C(x)\parallel_{L^{2}(0,1)}^{2}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.\end{eqnarray}$

(3.5)

The mean value theorem and estimate (3.1) for $v$ yields

$\begin{equation}|e^{v_{1}}-e^{v_{2}}|\leq e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}|v_{1}-v_{2}|.\end{equation}$

(3.6)

The monotonicity of $x\mapsto e^{x}$, inequality (3.6), the Young inequality and the Poincaré inequality leads to

$\begin{eqnarray}&& -\int_{0}^{1}(e^{u_{1}}-e^{v_{1}}-e^{u_{2}}+e^{v_{2}})(u_{1}-u_{2})dx \leq \int_{0}^{1}(e^{v_{1}}-e^{v_{2}})(u_{1}-u_{2})dx\nonumber\\ &\leq& e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\int_{0}^{1}|v_{1}-v_{2}|\cdot|u_{1}-u_{2}|dx\nonumber\\ &\leq& \frac{1}{2}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\left[\int_{0}^{1}(u_{1}-u_{2})^{2}dx+\int_{0}^{1}(v_{1}-v_{2})^{2}dx \right]\nonumber\\ & \leq& \frac{1}{4}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx+\int_{0}^{1}(v_{1}-v_{2})_{x}^{2}dx \right].\end{eqnarray}$

(3.7)

For the estimate of the second integral on the right-hand side of (3.3), we obtain similarly as above

$\begin{eqnarray}&& -J_{0}\int_{0}^{1}(e^{-u_{1}}-e^{-u_{2}})(u_{1}-u_{2})_{x}dx\nonumber\\ &\leq& |J_{0}|e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\int_{0}^{1}|u_{1}-u_{2}|\cdot|(u_{1}-u_{2})_{x}|dx\nonumber\\ & \leq& |J_{0}|e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\left[\int_{0}^{1}(u_{1}-u_{2})^{2}dx\right]^{\frac{1}{2}} \left[\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\right]^{\frac{1}{2}}\nonumber\\ &\leq&\frac{|J_{0}|}{\sqrt{2}}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx,\end{eqnarray}$

(3.8)

where we have used the Hölder inequality in the second inequality of (3.8). By (3.3), (3.5), (3.7) and (3.8), we get

$\begin{eqnarray}&& \left[1-\frac{\varepsilon}{2}\parallel C(x)\parallel_{L^{2}(0,1)}^{2}-\frac{1+2\sqrt{2}|J_{0}|}{4}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\right]\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\nonumber\\ & \leq& \frac{1}{4}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\int_{0}^{1}(v_{1}-v_{2})_{x}^{2}dx.\end{eqnarray}$

(3.9)

Employing the same techniques as above, we can estimate (3.4) as

$\begin{eqnarray}&& \left[1-\frac{\varepsilon}{2}\parallel C(x)\parallel_{L^{2}(0,1)}^{2}-\frac{1+2\sqrt{2}|J_{1}|}{4}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\right]\int_{0}^{1}(v_{1}-v_{2})_{x}^{2}dx\nonumber\\ & \leq& \frac{1}{4}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx.\end{eqnarray}$

(3.10)

It follows from (3.9) and (3.10) that

$\begin{eqnarray}&& \left[1-\frac{\varepsilon}{2}\parallel C(x)\parallel_{L^{2}(0,1)}^{2}-\frac{1+\sqrt{2}|J_{0}|}{2}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\right]\int_{0}^{1}(u_{1}-u_{2})_{x}^{2}dx\nonumber\\ && +\left[1-\frac{\varepsilon}{2}\parallel C(x)\parallel_{L^{2}(0,1)}^{2}-\frac{1+\sqrt{2}|J_{1}|}{2}e^{\sqrt{2}\parallel C(x)\parallel_{L^{2}(0,1)}}\right]\int_{0}^{1}(v_{1}-v_{2})_{x}^{2}dx\nonumber\\ & \leq& 0.\end{eqnarray}$

(3.11)

This inequality and (1.18), (1.19) implies $u_{1}=u_{2},\ v_{1}=v_{2}$ in $(0,1)$.

Proof of Theorem 1.3 From Lemma 2.1 and the Poincaré inequality we obtain a uniform $H^{1}(0,1)$ bound for $u_{\varepsilon}$ and $v_{\varepsilon}$. Then there exists a subsequence of $(u_{\varepsilon},v_{\varepsilon})$ (not relabeled) such that (1.20) holds. The weak formulations of (1.12) and (1.13) read, for any $\psi\in C_{0}^{\infty}(0,1)$, after integration by parts,

$\begin{eqnarray}&&-\frac{\varepsilon^{2}}{2}\int_{0}^{1}u_{\varepsilon}\psi_{xxxx}dx -\frac{\varepsilon^{2}}{4}\int_{0}^{1}u_{\varepsilon,x}^{2}\psi_{xx}dx\nonumber\\ &=& \int_{0}^{1} u_{\varepsilon,x}\psi_{x}dx+\int_{0}^{1}(e^{u_{\varepsilon}}-e^{v_{\varepsilon}}-C(x))\psi dx-J_{0}\int_{0}^{1}e^{-u_{\varepsilon}}\psi_{x}dx,\end{eqnarray}$

(4.1)

$\begin{eqnarray}&&-\frac{\varepsilon^{2}}{2}\int_{0}^{1}v_{\varepsilon}\psi_{xxxx}dx -\frac{\varepsilon^{2}}{4}\int_{0}^{1}v_{\varepsilon,x}^{2}\psi_{xx}dx\nonumber\\ &=& \int_{0}^{1} v_{\varepsilon,x}\psi_{x}dx-\int_{0}^{1}(e^{u_{\varepsilon}}-e^{v_{\varepsilon}}-C(x))\psi dx-J_{1}\int_{0}^{1}e^{-v_{\varepsilon}}\psi_{x}dx.\end{eqnarray}$

(4.2)

Convergences (1.20) allow us to pass to the limit $\varepsilon\rightarrow0$ in the above equations, observing that the left-hand sides of (4.1) and (4.2) vanish in the limit:

$0=\int_{0}^{1} u_{x}\psi_{x}dx+\int_{0}^{1}(e^{u}-e^{v}-C(x))\psi dx-J_{0}\int_{0}^{1}e^{-u}\psi_{x}dx,$

(4.3)

$0=\int_{0}^{1} v_{x}\psi_{x}dx-\int_{0}^{1}(e^{u}-e^{v}-C(x))\psi dx-J_{1}\int_{0}^{1}e^{-v}\psi_{x}dx.$

(4.4)

This shows the weak forms of (1.21) and (1.22) hold.