It is well known that the optimal control problem is nonconvex. In general, one may not expect a unique solution. However, the local uniqueness of the minimizer for the control functional can be obtained if \(\tau^{\ast}\ll1\).

Suppose that \(U_{1}^{*}(x)\) and \(U_{2}^{*}(x)\) are two given functions which satisfy condition 4.1. Let \(a_{1}(x)\) and \(a_{2}(x)\) be two minimizers of the control problem (2.1) corresponding to \(U_{i}^{*}(x)\) (\(i=1,2\)), respectively, and \(\{U_{i},\varphi_{i}\}\) (\(i=1,2\)) be the solutions of system (4.1)/(4.2) in which \(\bar {a}=a_{i}\) (\(i=1,2\)), respectively.

Set

$$U_{1}-U_{2}=u,\qquad \varphi_{1}-\varphi_{2}= \Phi,\qquad a_{1}-a_{2}=A. $$

Then *U* and Φ satisfy the following equations:

$$\begin{aligned}& \left\{ \textstyle\begin{array}{ll} u_{\tau }-a_{1}(x)(u_{xx}-u_{x})-ru_{x}+ru=A(U_{2xx}-U_{2x}), \\ u(x,0)=0,\\ u(0,\tau)=u(l,\tau)=0. \end{array}\displaystyle \right . \end{aligned}$$

(5.1)

$$\begin{aligned}& \left\{ \textstyle\begin{array}{ll} \frac{1}{a_{2}}\Phi_{\tau}+\Phi_{xx}+\Phi_{x}-r (\frac{\Phi}{a_{2}} )_{x}-\frac{r}{a_{2}}\Phi\\ \quad=(\frac{1}{a_{2}}-\frac{1}{a_{1}})(a_{1}\varphi_{1})_{\tau}-r [(\frac{1}{a_{2}}-\frac{1}{a_{1}})(a_{1}\varphi_{1}) ]_{x} -r(\frac{1}{a_{1}}-\frac{1}{a_{2}})(a_{1}\varphi_{1}),\\ \Phi(x,0)=0,\\ \Phi(0,\tau)=\Phi(l,\tau)=0,\\ \Phi|_{\tau=\tau^{\ast}}=a_{1}U(x,\tau^{\ast})+A(U_{2}(x,\tau^{\ast })-U^{\ast}(x)). \end{array}\displaystyle \right . \end{aligned}$$

(5.2)

### Lemma 5.1

*For problem* (4.2), *we have the following estimate*:

$$\begin{gathered} \max_{\tau\in[0,\tau^{\ast}]} \int_{0}^{l}\varphi^{2}\,dx+ \int_{0}^{\tau ^{\ast}} \int_{0}^{l}\big|\varphi_{x}|^{2} \,dx\,d\tau\leq C \int _{0}^{l}|U\bigl(x,\tau^{\ast} \bigr)-U^{\ast}(x)\big|^{2}\,dx, \\\max_{\tau\in[0,\tau^{\ast}]} \int_{0}^{l}|\varphi_{x}|^{2} \,dx+ \int _{0}^{\tau^{\ast}} \int_{0}^{l}|\varphi_{xx}|^{2} \leq C\big\| U\bigl(x,\tau^{\ast}\bigr)-U^{\ast}(x)\big\| ^{2}_{H^{1}(0,l)}.\end{gathered} $$

### Lemma 5.2

*For any bounded continuous function*
\(g(x)\in C(0, l)\), *we have*

$$\big\| g(x)\big\| _{\infty}\leq\big|g(x_{0})\big|+\sqrt{l}\big\| \nabla g(x) \big\| _{L^{2}(0,l)}, $$

*where*
\(x_{0}\)
*is a fixed point in*
\((0,l)\).

### Proof

For \(0< x< l\), we have

$$\begin{aligned} \bigl\vert g(x) \bigr\vert &\leq \bigl\vert g(x_{0}) \bigr\vert + \biggl\vert \int_{x_{0}}^{x}g{'}\,dx \biggr\vert \\ &\leq \bigl\vert g(x_{0}) \bigr\vert + \biggl( \int_{0}^{l} 1\,dx \biggr)^{\frac{1}{2}} \biggl( \int_{0}^{l} \vert \nabla g \vert ^{2} \,dx \biggr)^{\frac{1}{2}}.\end{aligned} $$

This completes the proof of Lemma 5.2. □

### Lemma 5.3

*For problem* (5.1), *we have the following estimate*:

$$\begin{aligned}& \max_{\tau\in[0,\tau^{\ast}]} \int_{0}^{l} u^{2}\,dx\leq C\bigl(\max |A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int_{0}^{l} |U_{2xx}-U_{2x}|^{2} \,dx\, d\tau, \end{aligned}$$

(5.3)

$$\begin{aligned}& \int_{0}^{\tau^{\ast}} \int_{0}^{l}|u_{xx}-u_{x}|^{2} \,dx\,d\tau\leq C\bigl(\max|A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{2xx}-U_{2x}|^{2} \, dx\,d\tau, \end{aligned}$$

(5.4)

*where*
*C*
*is a constant*.

### Proof

From equation (5.1), we have, for \(0<\tau\leq\tau^{\ast}\),

$$\begin{aligned} & \int_{0}^{\tau} \int_{0}^{l}a_{1}(u_{xx}-u_{x})^{2} \,dx\,d\tau \\ &\quad= \int_{0}^{\tau} \int_{0}^{l}u_{\tau}(u_{xx}-u_{x}) \,dx\,d\tau- \int _{0}^{\tau} \int_{0}^{l}A(U_{2xx}-U_{2x}) (u_{xx}-u_{x})\,dx\,d\tau \\ &\quad\quad-r \int_{0}^{\tau} \int_{0}^{l}u_{x}(u_{xx}-u_{x}) \,dx\,d\tau+r \int _{0}^{\tau} \int_{0}^{l}u(u_{xx}-u_{x})\,dx \,d\tau \\ &\quad= \int_{0}^{\tau} \int_{0}^{l}(u_{\tau}u_{x})_{x} \,dx\,d\tau- \int _{0}^{\tau} \int_{0}^{l}u_{\tau x}u_{x} \,dx\,d \tau\\ &\qquad- \int_{0}^{\tau} \int_{0}^{l}u_{\tau}u_{x}\,dx\,d \tau+r \int _{0}^{\tau} \int_{0}^{l}u_{x}^{2}\,dx\,d \tau \\ &\qquad+r \int_{0}^{\tau} \int_{0}^{l}u(u_{xx}-u_{x})\,dx \,d\tau- \int_{0}^{\tau } \int_{0}^{l}A(U_{2xx}-U_{2x}) (u_{xx}-u_{x})\,dx\,d\tau.\end{aligned} $$

By noticing the boundary conditions, we have

$$\begin{aligned} & \int_{0}^{\tau} \int_{0}^{l}a_{1}(u_{xx}-u_{x})^{2} \,dx\,d\tau \\ &\quad=- \int_{0}^{\tau} \int_{0}^{l}u_{x}u_{x\tau}\,dx\,d \tau- \int_{0}^{\tau } \int_{0}^{l}u_{\tau}u_{x}\,dx\,d \tau+ r \int_{0}^{\tau} \int_{0}^{l}u_{x}^{2}\,dx\,d \tau \\ &\qquad+r \int_{0}^{\tau} \int_{0}^{l}u(u_{xx}-u_{x})\,dx \,d\tau- \int_{0}^{\tau } \int_{0}^{l}A(U_{2xx}-U_{2x}) (u_{xx}-u_{x})\,dx\,d\tau \\ &\quad= -\frac{1}{2} \int_{0}^{l}u_{x}^{2}\big|_{(x,\tau)} \,dx- \int_{0}^{\tau} \int_{0}^{l}a_{1}u_{x}(u_{xx}-u_{x}) \,dx\,d\tau- \int _{0}^{\tau} \int_{0}^{l}AU_{x}(U_{2xx}-U_{2x}) \,dx\,d\tau \\ &\qquad-r \int_{0}^{\tau} \int_{0}^{l}u(u_{xx}-u_{x})\,dx \,d\tau- \int_{0}^{\tau } \int_{0}^{l}A(U_{2xx}-U_{2x}) (u_{xx}-u_{x})\,dx\,d\tau,\end{aligned} $$

where we have used the deformation of equation (5.1).

This yields

$$\begin{aligned}[b] &\frac{1}{8} \int_{0}^{\tau} \int_{0}^{l}a_{1}(u_{xx}-u_{x})^{2} \,dx\, d\tau+\frac{1}{2} \int_{0}^{l}u_{x}^{2}\big|_{(x,\tau)} \,dx \\ &\quad\leq 4\max_{x\in(0,l)}a_{1}(x) \int_{0}^{\tau} \int_{0}^{l}u_{x}^{2}\, dx\,d \tau+C \int_{0}^{\tau} \int_{0}^{l}U^{2}\,dx\,d\tau \\ &\qquad+C\bigl(\max|A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int _{0}^{l}|U_{2xx}-U_{2x}|^{2} \,dx\,d\tau.\end{aligned} $$

(5.5)

By estimating (5.3) we have

$$\begin{aligned}[b] &\frac{1}{8} \int_{0}^{\tau} \int_{0}^{l}a_{1}(u_{xx}-u_{x})^{2} \,dx\, d\tau+\frac{1}{2} \int_{0}^{l}u_{x}^{2}\big|_{(x,\tau)} \,dx \\ &\quad\leq 4\max_{x\in(0,l)}a_{1}(x) \int_{0}^{\tau} \int_{0}^{l}u_{x}^{2}\, dx\,d \tau+C\bigl(\max|A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int _{0}^{l}|U_{2xx}-U_{2x}|^{2} \,dx\,d\tau.\end{aligned} $$

(5.6)

From Gronwall’s inequality, we have

$$ \int_{0}^{\tau^{\ast}} \int_{0}^{l}(u_{xx}-u_{x})^{2} \,dx\,d\tau+ \int _{0}^{l}u_{x}^{2}\,dx\leq C\bigl(\max|A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int _{0}^{l}(U_{2xx}-U_{2x})^{2} \,dx\,d\tau, $$

(5.7)

which implies the conclusion.

The proof of Lemma 5.3 is thus completed. □

### Lemma 5.4

*For problem* (5.2) *we have the following estimate*:

$$\begin{aligned} &\max_{\tau\in[0,\tau^{\ast}]} \int_{0}^{l}\Phi^{2}\,dx+ \int_{0}^{\tau } \int_{0}^{l}|\Phi_{x}|^{2}\,dx \,d\tau \\ &\quad\leq C\bigl(\max|A|\bigr)^{2} \int_{0}^{\tau^{\ast}} \int_{0}^{l} \bigl(|U_{2x}|^{2}+| \varphi_{1\tau}|^{2} \bigr)\,dx\,d\tau\\ &\qquad+ C\bigl(\max |A|\bigr)^{2} \int_{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast}(x)\big|^{2}\,dx.\end{aligned} $$

### Proof

For problem (5.2), we have

$$\begin{aligned} & \int_{\tau}^{\tau^{\ast}} \int_{0}^{l} \biggl[\frac{1}{a_{2}}\Phi\Phi _{\tau}+\Phi\Phi_{xx}+\Phi\Phi_{x}-r\Phi \biggl( \frac{\Phi}{a_{2}} \biggr)_{x}-\frac{r}{a_{2}}\Phi^{2} \biggr]\,dx\,d\tau \\ &\quad= \int_{\tau}^{\tau^{\ast}} \int_{0}^{l}\Phi \biggl[\biggl(\frac{1}{a_{2}}- \frac {1}{a_{1}}\biggr) (a_{1}\varphi_{1})_{\tau}\\ &\qquad-r \biggl[\biggl(\frac{1}{a_{2}}-\frac {1}{a_{1}}\biggr) (a_{1} \varphi_{1}) \biggr]_{x}-r\biggl(\frac{1}{a_{1}}- \frac{1}{a_{2}}\biggr) (a_{1}\varphi_{1}) \biggr]\,dx\,d \tau.\end{aligned} $$

This yields

$$\begin{aligned}[b] &\frac{1}{2a_{1}} \int_{0}^{l}\Phi^{2}\big|_{(x,\tau)} \,dx+ \int_{\tau }^{\tau^{\ast}} \int_{0}^{l}|\Phi_{x}|^{2}\,dx \,d\tau+ \int_{\tau}^{\tau ^{\ast}} \int_{0}^{l}\frac{r}{a_{2}}\Phi^{2}\,dx \,d\tau \\ &\quad= \int_{0}^{l}\frac{1}{2a_{2}}\big|a_{1}U \bigl(x,\tau^{\ast}\bigr)+A\bigl(U_{2}\bigl(x,\tau ^{\ast} \bigr)-U^{\ast}(x)\bigr)\big|^{2}\,dx \\ &\qquad- \int_{\tau}^{\tau^{\ast}} \int_{0}^{l}\frac{A}{a_{2}}\Phi\varphi _{1\tau}\,dx\,d\tau- \int_{\tau}^{\tau^{\ast}} \int_{0}^{l}(r\varphi _{1x}+r \varphi_{1})\,dx\,d\tau \\ &\quad\leq C \int_{0}^{l}\big|U\bigl(x,\tau^{\ast} \bigr)\big|^{2}+C\bigl(\max|A|\bigr)^{2} \int _{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast}(x)\big|^{2}\,dx \\ &\qquad+ \int_{\tau}^{\tau^{\ast}} \int_{0}^{l}\frac{1}{2a_{2}}\Phi^{2}\,dx \, d\tau +C\bigl(\max|A|\bigr)^{2} \int_{\tau}^{\tau^{\ast}} \int_{0}^{l} |\varphi_{1\tau}|^{2} \,dx\,d\tau \\ &\quad\leq \int_{\tau}^{\tau^{\ast}} \int_{0}^{l}\frac{1}{2a_{2}}\Phi^{2}\,dx \,d\tau +C\bigl(\max|A|\bigr)^{2} \int_{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast}(x)\big|^{2}\, dx \\ &\quad\quad+C\bigl(\max|A|\bigr)^{2} \int_{\tau}^{\tau^{\ast}} \int_{0}^{l} \bigl(|U_{2xx}-U_{2x}|^{2}+ |\varphi_{1\tau}|^{2} \bigr)\,dx\,d\tau.\end{aligned} $$

(5.8)

From (5.8) and Gronwall’s inequality, we have

$$\begin{aligned} & \int_{0}^{l}\Phi^{2}\,dx+ \int_{0}^{\tau^{\ast}} \int_{0}^{l}\Phi _{x}^{2}\,dx \,d\tau \\ &\quad\leq C\bigl(\max|A|\bigr)^{2} \int_{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast }(x)\big|^{2}\,dx \\ &\qquad+C\bigl( \max|A|\bigr)^{2} \int_{\tau}^{\tau^{\ast}} \int_{0}^{l} \bigl(|U_{2x}|^{2}+ |\varphi_{1\tau}|^{2} \bigr)\,dx\,d\tau.\end{aligned} $$

This completes the proof of Lemma 5.4. □

### Theorem 5.1

*Suppose that*
\(a_{1}(x)\), \(a_{2}(x)\)
*are two minimizers of the optimal control problem* (2.1) *corresponding to*
\(U_{1}^{*}(x)\)
*and*
\(U_{2}^{*}(x)\), *respectively*. *If there exists a point*
\(x_{0}\in (0,l)\)
*such that*
\(a_{1}(x_{0})=a_{2}(x_{0})\), *then for*
\(\tau^{\ast}\ll 1\), *we have*

$$a_{1}(x)\equiv a_{2}(x), \quad\textit{for any } x\in(0,l). $$

### Proof

By taking \(h=a_{2}\) when \(\bar{a}=a_{1}\) and taking \(h=a_{1}\) when \(\bar{a}=a_{2}\) in (4.3), we have

$$\begin{aligned}& \begin{aligned}[b] &\int_{0}^{\tau^{\ast}} \int_{0}^{l}\varphi (a_{2}-a_{1}) (U_{1xx}-U_{1x})\,dx\,d\tau \\&\quad+\frac{N}{2} \int_{0}^{l} \frac{\nabla a_{1}}{\sqrt{|\nabla a_{1}|^{2}+\epsilon^{2}}}\cdot \nabla(a_{2}-a_{1})\,dx+\mu \int_{0}^{l} \nabla{a_{1}}\cdot \nabla(a_{2}-a_{1})\,dx\geq0,\end{aligned} \end{aligned}$$

(5.9)

$$\begin{aligned}& \begin{aligned}[b]&\int_{0}^{\tau^{\ast}} \int_{0}^{l}\varphi (a_{1}-a_{2}) (U_{2xx}-U_{2x})\,dx\,d\tau\\&\quad +\frac{N}{2} \int_{0}^{l} \frac{\nabla a_{2}}{\sqrt{|\nabla a_{2}|^{2}+\epsilon^{2}}}\cdot \nabla(a_{1}-a_{2})\,dx+\mu \int_{0}^{l} \nabla{a_{2}}\cdot \nabla(a_{1}-a_{2})\,dx\geq0, \end{aligned} \end{aligned}$$

(5.10)

where \(\{U_{i},\varphi_{i}\}\) (\(i=1,2\)) are the solutions of system (4.1)/(4.2) with \(\bar{a}=a_{i}\) (\(i=1,2\)), respectively.

Combining equations (5.9) and (5.10), when \(\epsilon\rightarrow 0\), we have

$$\begin{aligned} & \mu \int_{0}^{l}\big|\nabla(a_{1}-a_{2})\big|^{2} \,dx \\ &\quad\leq \int_{0}^{\tau^{\ast}} \int_{0}^{l}A \bigl(\varphi _{2}(U_{2xx}-U_{2x})- \varphi_{1}(U_{1xx}-U_{1x}) \bigr)\,dx\,d\tau \\ &\quad\leq \int_{0}^{\tau^{\ast}} \int_{0}^{l}\frac {1}{a_{2}}A(U_{2xx}-U_{2x})a_{2} \varphi_{2}\,dx\,d\tau- \int_{0}^{\tau ^{\ast}} \int_{0}^{l}\frac{1}{a_{1}}A(U_{1xx}-U_{1x})a_{1} \varphi_{1}\, dx\,d\tau \\ &\quad\leq \int_{0}^{\tau^{\ast}} \int_{0}^{l} \biggl(\frac {1}{a_{2}}A(U_{2xx}-U_{2x})a_{2} \varphi_{2}-\frac {1}{a_{2}}A(U_{2xx}-U_{2x})a_{1} \varphi_{1} \biggr)\,dx\,d\tau \\ &\qquad+ \int_{0}^{\tau^{\ast}} \int_{0}^{l} \biggl(\frac {1}{a_{2}}A(U_{2xx}-U_{2x})a_{1} \varphi_{1}-\frac {1}{a_{2}}A(U_{1xx}-U_{1x})a_{1} \varphi_{1} \biggr)\,dx\,d\tau \\ &\qquad+ \int_{0}^{\tau^{\ast}} \int_{0}^{l} \biggl(\frac {1}{a_{2}}A(U_{1xx}-U_{1x})a_{1} \varphi_{1}-\frac {1}{a_{1}}A(U_{1xx}-U_{1x})a_{1} \varphi_{1} \biggr)\,dx\,d\tau \\ &\quad\leq \int_{0}^{\tau^{\ast}} \int_{0}^{l}\frac{1}{a_{2}}A^{2} (U_{1xx}-U_{1x})\varphi_{1}\,dx\,d\tau\\ &\qquad- \int_{0}^{\tau^{\ast}} \int _{0}^{l}\frac{1}{a_{2}}A \Phi(U_{2xx}-U_{2x})\,dx\,d\tau \\ &\qquad- \int_{0}^{\tau^{\ast}} \int_{0}^{l}\frac {1}{a_{2}}A(u_{xx}-u_{x})a_{1} \varphi_{1}\,dx\,d\tau.\end{aligned} $$

From the assumption of Theorem 5.1, there exists a point \(x_{0}\in(0,l)\) such that

$$ A(x_{0})=a_{1}(x_{0})-a_{2}(x_{0})=0. $$

(5.11)

From Lemma 5.2 we have

$$ \max_{x\in[0,l]}| A|\leq C \biggl( \int_{0}^{l} |\nabla A| ^{2}\,dx \biggr)^{1/2}. $$

(5.12)

From (5.3), (5.11) and Hölder’s inequality, we have

$$\begin{aligned} &\mu \int_{0}^{l}|\nabla A|^{2}\,dx \\ &\quad\leq C\max|A|\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}\Phi^{2}\,dx\,d\tau }\cdot \sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{2xx}-U_{2x}|^{2} \,dx\, d\tau} \\ &\qquad+C\max|A|^{2}\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|\varphi _{1}|^{2} \,dx\,d\tau} \cdot\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{1xx}-U_{1x}|^{2} \,dx\, d\tau} \\ &\qquad+C\max|A|\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|\varphi_{1}|^{2}\, dx\,d\tau} \cdot\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|u_{xx}-u_{x}|^{2} \,dx\, d\tau} \\ &\quad\leq C\max|A|^{2}\sqrt{\tau^{\ast} \biggl( \int_{0}^{\tau^{\ast}} \int _{0}^{l} \bigl(|U_{2x}|^{2}+| \varphi_{1\tau}|^{2} \bigr)\,dx\,d\tau+ \int _{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast}(x)\big|^{2}\,dx \biggr)} \\ &\qquad\cdot\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{2xx}-U_{2x}|^{2} \, dx\,d\tau} \\ &\qquad+C\max|A|^{2}\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|\varphi _{1}|^{2} \,dx\,d\tau} \cdot\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{1xx}-U_{1x}|^{2} \,dx\, d\tau} \\ &\qquad+C\max|A|^{2}\sqrt{\tau^{\ast} \int_{0}^{\tau^{\ast}} \int _{0}^{l}|\varphi_{1}|^{2} \,dx\,d\tau} \cdot\sqrt{ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{2xx}-U_{2x}|^{2} \,dx\, d\tau}.\end{aligned} $$

From Lemmas 5.1, 5.3 and 5.4, we have

$$\begin{gathered} \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{1xx}-U_{1x}|^{2} \,dx\,d\tau\leq C,\qquad \int_{0}^{\tau^{\ast}} \int_{0}^{l}|U_{2xx}-U_{2x}|^{2} \,dx\,d\tau \leq C, \\ \int_{0}^{\tau^{\ast}} \int_{0}^{l}|\varphi_{1}|^{2} \,dx\,d\tau\leq C\tau^{*},\qquad \int_{0}^{\tau^{\ast}} \int_{0}^{l}\varphi_{1\tau}^{2}\,dx \, d\tau\leq C,\\ \int_{0}^{l}\big|U_{2}\bigl(x, \tau^{\ast}\bigr)-U^{\ast}(x)\big|^{2}\,dx\leq C.\end{gathered} $$

Based on the above analysis, we have

$$ \mu \int_{0}^{l}|\nabla A|^{2}\,dx \leq C \max|A|^{2}\bigl(\sqrt{\tau^{\ast}}+\tau^{\ast}\bigr)\leq C\bigl(\sqrt{\tau^{\ast}}+\tau^{\ast}\bigr) \int_{0}^{l}|\nabla A|^{2}\,dx. $$

(5.13)

Choose \(\tau^{\ast}\leq1\) such that

$$ \frac{C}{\mu}\bigl(\sqrt{\tau^{\ast}}+\tau^{\ast}\bigr)= \frac{1}{4}. $$

(5.14)

Combining (5.13) and (5.14), we can easily obtain

$$ \int_{0}^{l} |\nabla A|^{2}\,dx \leq0. $$

(5.15)

This yields

$$A(x)=a_{1}(x)-a_{2}(x)=a_{1}(x_{0})-a_{2}(x_{0}) \equiv0, \quad x\in(0,l). $$

This completes the proof of Theorem 5.1. □