In this section, we provide the suitable Carleman estimate for the study of the stability of our inverse problem.
Let us consider the operator
$$P=\partial_{t}+a(x)\partial_{xxx}+b(x,t) \partial_{x}+d(x,t) $$
defined on
$$\begin{aligned} \mathcal{V} =& \bigl\{ v\in L^{2} \bigl(0,T;H^{3}(0,L)\bigr)\mid v(0,t)=v_{x}(L,t)=v_{xx}(L,t)=0, t \in(0,T), \\ &{}\textrm{and }Pv\in L^{2}(Q)\bigr\} . \end{aligned}$$
Here \(b\in L^{\infty}(0,T;W^{1,\infty}(0,L))\), \(d\in L^{\infty}(Q)\) and \(a\in\Sigma(a_{0},\alpha)\).
Consider \(\beta\in C^{3}([0,L])\) such that for some \(r>0\) we have
$$0< r\leq\beta(x)\quad\mbox{and}\quad 0< r\leq\beta^{\prime}(x),\quad \forall x \in(0,L). $$
We define, for each \(\lambda>0\), the functions
$$\phi(x,t)=\frac{e^{2\lambda \Vert \beta \Vert _{\infty}}-e^{\lambda\beta (x)}}{t(T-t)},\qquad \theta(x,t)=\frac{e^{\lambda\beta(x)}}{t(T-t)}, $$
for \((x,t)\in Q\). It is not difficult to see that θ satisfies the following properties:
$$\begin{aligned} &{\exists C>0\quad\textrm{such that }\frac{1}{C}\theta\leq\theta_{x} \leq C\theta\quad\mbox{and}} \\ &{\theta^{n}\leq C\theta^{m}\quad\textrm{for~each~positive~integers }n< m.} \end{aligned}$$
Theorem 2.1
Let
ϕ, θ
and
P
be defined as above. There exist
\(s_{0}>0\), \(\lambda_{0}>0\)
and a constant
\(C=C(T,s_{0},\lambda_{0},a_{0},\alpha,L)>0\)
such that, for every
\(s\geq s_{0}\), \(\lambda\geq \lambda_{0}\),
$$\begin{aligned} &{\int_{Q}e^{-2s\phi} \bigl(s^{5} \lambda^{6}\theta ^{5}u^{2}+s^{3} \lambda^{4}\theta^{3}u_{x}^{2}+s \lambda^{2}\theta u_{xx}^{2} \bigr)\,dx\,dt} \\ &{\quad \leq C \int_{Q}e^{-2s\phi} \vert Pu \vert ^{2}\,dx\,dt+C \int_{0}^{T} \bigl(s^{5} \lambda^{5}\theta^{5}e^{-2s\phi}u^{2} \bigr) \bigm| _{x=L}\,dt.} \end{aligned}$$
(2.1)
Proof
Following the method in [21], it is enough to prove (2.1) for \(\widetilde{P}=\partial_{t}+a(x)\partial_{xxx}\). In fact, assume that we have proved (2.1) for P̃, we have
$$\int_{Q}e^{-2s\phi} \vert \widetilde{P}u \vert ^{2}\,dx\,dt\leq C \biggl( \int _{Q}e^{-2s\phi} \vert Pu \vert ^{2}\,dx\,dt+ \int_{Q}e^{-2s\phi }\bigl(u_{x}^{2}+u^{2} \bigr)\,dx\,dt \biggr). $$
By choosing \(s>0\) and \(\lambda>0\) large, it is possible to absorb \(\int_{Q}e^{-2s\phi}(u_{x}^{2}+u^{2})\,dx\,dt\) with the left-hand side of (2.1), concluding that (2.1) also holds for P.
Let \(s>0\) and consider the operator \(P_{\phi}\) defined in \(\mathcal {W}_{s}=\{e^{-s\phi}u\mid u\in\mathcal{V}\}\) by
$$P_{\phi}w=e^{-s\phi}\widetilde{P}\bigl(e^{s\phi}w\bigr). $$
We then obtain the decomposition \(P_{\phi}w=P_{1}w+P_{2}w+Rw\), where
$$ \begin{aligned} &{P_{1}w=w_{t}+3as^{2} \phi_{x}^{2}w_{x}+aw_{xxx}+3as^{2} \phi_{x}\phi_{xx}w,} \\ &{P_{2}w= as^{3}\phi_{x}^{3}w+3as \phi_{x}w_{xx}+3s(a\phi_{x})_{x}w_{x},} \\ &{Rw= as\phi_{xxx}w+3as^{2}\phi_{x} \phi_{xx}w+s\phi _{t}w-3sa_{x} \phi_{x}w_{x}-3as^{2}\phi_{x} \phi_{xx}w.} \end{aligned} $$
(2.2)
Thus
$$\begin{aligned} \Vert P_{\phi}w-Rw \Vert _{L^{2}(Q)}^{2}= \Vert P_{1}w \Vert _{L^{2}(Q)}^{2}+2\langle P_{1}w,P_{2}w\rangle_{L^{2}(Q)}+ \Vert P_{2}w \Vert _{L^{2}(Q)}^{2}. \end{aligned}$$
Let us now consider \(\langle P_{1}w,P_{2}w\rangle_{L^{2}(Q)}\).
Claim that
$$\begin{aligned} \langle P_{1}w,P_{2}w\rangle_{L^{2}(Q)}= \int_{Q} \bigl((\cdot )w^{2}+(\cdot)w_{x}^{2}+( \cdot)w_{xx}^{2} \bigr)\,dx\,dt+ \int _{0}^{T}(\cdot)\mid _{x=0}^{x=L}\,dt, \end{aligned}$$
(2.3)
where
$$\begin{aligned} &{(\cdot)w^{2}= \biggl(-\frac{1}{2}s^{3}a \bigl(\phi_{x}^{3}\bigr)_{t}-\frac {3}{2}s^{5} \bigl(a^{2}\phi_{x}^{5}\bigr)_{x} - \frac{1}{2}s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{xxx}+3s^{5}a^{2}\phi _{x}^{4} \phi_{xx} } \\ &{\phantom{(\cdot)w^{2}=} {}+\frac{9}{2}s^{3}\bigl(a^{2}\phi _{x}^{2}\phi_{xx}\bigr)_{xx} - \frac{9}{2}s^{3}\bigl(a\phi_{x}\phi_{xx}(a \phi_{x})_{x}\bigr)_{x} \biggr)w^{2},} \\ &{(\cdot)w_{x}^{2}= \biggl(\frac{3}{2}sa \phi_{xt}-\frac {9}{2}s^{3}\bigl(a^{2} \phi_{x}^{3}\bigr)_{x}+9s^{3}a \phi_{x}^{2}(a\phi_{x})_{x} + \frac{3}{2}s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{x} } \\ &{\phantom{(\cdot)w_{x}^{2}=} {}+\frac{3}{2}s\bigl(a(a\phi _{x})_{x} \bigr)_{xx}-9s^{3}a^{2}\phi_{x}^{2} \phi_{xx} \biggr)w_{x}^{2},} \\ &{(\cdot)w_{xx}^{2}= \biggl(-\frac{3}{2}s \bigl(a^{2}\phi _{x}\bigr)_{x}-3sa(a \phi_{x})_{x} \biggr)w_{xx}^{2},} \\ &{(\cdot)\mid_{x=0}^{x=L}= \biggl(3sa\phi_{x}w_{t}w_{x}+ \frac {3}{2}s^{5}a^{2}\phi_{x}^{5}w^{2} +\frac{9}{2}s^{3}a^{2}\phi_{x}^{3}w_{x}^{2}+s^{3}a^{2} \phi _{x}^{3}ww_{xx} } \\ &{\phantom{(\cdot)\mid_{x=0}^{x=L}=}{} -s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{x}ww_{x} -\frac{1}{2}s^{3}a^{2} \phi_{x}^{3}w_{x}^{2}+ \frac {1}{2}s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{xx}w^{2}+\frac{3}{2}sa^{2}\phi _{x}w_{xx}^{2}} \\ &{\phantom{(\cdot)\mid_{x=0}^{x=L}=}{}-\frac{3}{2}s\bigl(a(a\phi_{x})_{x} \bigr)_{x}w^{2}_{x}+3sa(a\phi _{x})_{x}w_{x}w_{xx}+9s^{3}a^{2} \phi_{x}^{2}\phi_{xx}ww_{x}} \\ &{\phantom{(\cdot)\mid_{x=0}^{x=L}=}{} -\frac{9}{2}s^{3}\bigl(a^{2} \phi_{x}^{2}\phi_{xx}\bigr)_{x}w^{2} + \frac{9}{2}s^{3}a\phi_{x} \phi_{xx}(a\phi _{x})_{x}w^{2} \biggr) \biggm| _{x=0}^{x=L}. } \end{aligned}$$
Indeed, (2.3) follows easily from the following equations:
$$\begin{aligned} &{\int_{Q}as^{3}\phi_{x}^{3}ww_{t}\,dx\,dt=- \int_{Q}\frac {1}{2}s^{3}a\bigl( \phi_{x}^{3}\bigr)_{t}w^{2}\,dx\,dt,} \\ &{\int_{Q}3as\phi_{x}w_{t}w_{xx}\,dx\,dt= \int_{0}^{T} (3sa\phi_{x}w_{t}w_{x} ) \bigm| _{x=0}^{x=L}\,dt} \\ &{\phantom{\int_{Q}3as\phi_{x}w_{t}w_{xx}\,dx\,dt=} {}+ \int_{Q} \biggl(\frac{3}{2}sa\phi _{xt}w_{x}^{2}-3s(a \phi_{x})_{x}w_{t}w_{x} \biggr)\,dx\,dt,} \\ &{\int_{Q}3a^{2}s^{5}\phi_{x}^{5}ww_{x}\,dx\,dt= \int_{0}^{T} \biggl(\frac{3}{2}s^{5}a^{2} \phi_{x}^{5}w^{2} \biggr) \biggm| _{x=0}^{x=L}\,dt - \int_{Q}\frac{3}{2}s^{5} \bigl(a^{2}\phi_{x}^{5}\bigr)_{x}w^{2}\,dx\,dt,} \\ &{\int_{Q}9s^{3}a^{2}\phi_{x}^{3}w_{x}w_{xx}\,dx\,dt = \int_{0}^{T} \biggl(\frac{9}{2}s^{3}a^{2} \phi _{x}^{3}w_{x}^{2}\biggr) \biggm| _{x=0}^{x=L}\,dt - \int_{Q}\frac{9}{2}s^{3}\bigl(a^{2} \phi_{x}^{3}\bigr)_{x}w_{x}^{2}\,dx\,dt,} \\ &{\int_{Q}s^{3}a^{2}\phi_{x}^{3}ww_{xxx}\,dx\,dt= \int_{Q} \biggl(\frac {3}{2}s^{3} \bigl(a^{2}\phi_{x}^{3}\bigr)_{x}w_{x}^{2}- \frac {1}{2}s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{xxx}w^{2} \biggr)\,dx\,dt} \\ &{\phantom{\int_{Q}s^{3}a^{2}\phi_{x}^{3}ww_{xxx}\,dx\,dt=} {}+ \int_{0}^{T} s^{3} \biggl(a^{2}\phi _{x}^{3}ww_{xx}- \bigl(a^{2}\phi_{x}^{3}\bigr)_{x}ww_{x} -\frac{1}{2}a^{2}\phi_{x}^{3}w_{x}^{2}} \\ &{\phantom{\int_{Q}s^{3}a^{2}\phi_{x}^{3}ww_{xxx}\,dx\,dt=} {}+ \frac{1}{2}\bigl(a^{2}\phi _{x}^{3} \bigr)_{xx}w^{2} \biggr) \biggm| _{x=0}^{x=L}\,dt,} \\ &{\int_{Q}3sa^{2}\phi_{x}w_{xx}w_{xxx}\,dx\,dt = \int_{0}^{T} \biggl(\frac{3}{2}sa^{2} \phi_{x}w_{xx}^{2}\biggr) \biggm| _{x=0}^{x=L}\,dt - \int_{Q}\frac{3}{2}s\bigl(a^{2} \phi_{x}\bigr)_{x}w_{xx}^{2}\,dx\,dt,} \\ &{\int_{Q}3sa(a\phi_{x})_{x}w_{x}w_{xxx}\,dx\,dt= \int_{Q} \biggl(\frac {3}{2}s\bigl(a(a \phi_{x})_{x}\bigr)_{xx}w_{x}^{2}-3sa(a \phi _{x})_{x}w_{xx}^{2} \biggr)\,dx\,dt} \\ &{\phantom{\int_{Q}3sa(a\phi_{x})_{x}w_{x}w_{xxx}\,dx\,dt=} {} + \int_{0}^{T} \biggl(-\frac{3}{2}s \bigl(a(a\phi _{x})_{x}\bigr)_{x}w_{x}^{2}+3sa(a \phi_{x})_{x}w_{x}w_{xx} \biggr) \biggm| _{x=0}^{x=L}\,dt,} \\ &{\int_{Q}9s^{3}a^{2}\phi_{x}^{2} \phi_{xx}ww_{xx}\,dx\,dt = \int_{Q} \biggl(\frac{9}{2}s^{3} \bigl(a^{2}\phi_{x}^{2}\phi _{xx} \bigr)_{xx}w^{2}-9s^{3}a^{2} \phi_{x}^{2}\phi_{xx}w_{x}^{2} \biggr)\,dx\,dt} \\ &{\phantom{\int_{Q}9s^{3}a^{2}\phi_{x}^{2} \phi_{xx}ww_{xx}\,dx\,dt =} {}+ \int_{0}^{T} \biggl(9s^{3} a^{2}\phi _{x}\phi_{xx}ww_{x}- \frac{9}{2}s^{3}\bigl(a^{2}\phi_{x}^{2} \phi _{xx}\bigr)_{x}w^{2} \biggr) \biggm| _{x=0}^{x=L}\,dt,} \\ &{\int_{Q}9s^{3}a\phi_{x} \phi_{xx}(a\phi_{x})_{x}ww_{x}\,dx\,dt=- \int _{Q}\frac{9}{2}s^{3}\bigl(a \phi_{x}\phi_{xx}(a\phi_{x})_{x} \bigr)_{x}w^{2}\,dx\,dt} \\ &{\phantom{\int_{Q}9s^{3}a\phi_{x} \phi_{xx}(a\phi_{x})_{x}ww_{x}\,dx\,dt=} {} + \int_{0}^{T} \biggl(\frac{9}{2}s^{3}a \phi _{x}\phi_{xx}(a\phi_{x})_{x}w^{2} \biggr) \biggm| _{x=0}^{x=L}\,dt.} \end{aligned}$$
Similar to [20], we obtain
$$\begin{aligned} &{\int_{Q} \bigl((\cdot)w^{2}+(\cdot)w_{x}^{2}+( \cdot )w_{xx}^{2} \bigr)\,dx\,dt} \\ &{\quad = \int_{Q} \biggl(\frac{9}{2}s^{5} \lambda^{6}a^{2}\beta_{x}^{6}\theta ^{5}w^{2} +9s^{3}\lambda^{4}a^{2} \beta_{x}^{4}\theta^{3}w_{x}^{2}+ \frac {9}{2}s\lambda^{2}a^{2}\beta_{x}^{2} \theta w_{xx}^{2} \biggr)\,dx\,dt+I_{R},} \end{aligned}$$
(2.4)
where \(I_{R}\) gathers the non-dominating terms and satisfies the requirement that, for any \(\varepsilon>0\), there exist \(s_{0}>0\), \(\lambda_{0}>0\) such that if \(s\geq s_{0}\), \(\lambda\geq\lambda _{0}\), we have
$$\begin{aligned} \vert I_{R} \vert \leq\varepsilon \int_{Q} \bigl(s^{5}\lambda^{6} \theta^{5}w^{2} +s^{3}\lambda^{4} \theta^{3}w_{x}^{2}+s\lambda^{2}\theta w_{xx}^{2} \bigr)\,dx\,dt. \end{aligned}$$
(2.5)
Then we consider the boundary terms. Write
$$\begin{aligned} \int_{0}^{T}(\cdot)\mid_{x=0}^{x=L}\,dt =& \int_{0}^{T}\biggl(-\frac {3}{2}s^{5} \lambda^{5}a^{2}\beta_{x}^{5} \theta^{5}w^{2} -8s^{3}\lambda^{3}a^{2} \beta_{x}^{3}\theta^{3}w_{x}^{2} -\frac{3}{2}s\lambda a^{2}\beta_{x}\theta w_{xx}^{2} \\ &{} -3as\lambda\theta\beta_{x}w_{t}w_{x}-a^{2}s^{3} \lambda ^{3}\beta_{x}^{3}\theta_{x}^{3}ww_{xx}+B_{R} \biggr) \biggm| _{x=0}^{x=L}\,dt, \end{aligned}$$
(2.6)
where \(B_{R}\) can be estimated as follows:
$$\begin{aligned} \vert B_{R} \vert =& \biggl\vert -s^{3}\bigl(a^{2}\phi_{x}^{3} \bigr)_{x}ww_{x}+\frac {1}{2}s^{3} \bigl(a^{2}\phi_{x}^{3}\bigr)_{xx}w^{2} \\ &{} -\frac{3}{2}s\bigl(a(a\phi_{x})_{x} \bigr)_{x}w^{2}_{x}+3sa(a\phi _{x})_{x}w_{x}w_{xx} \\ &{} +9s^{3}a^{2}\phi_{x}^{2} \phi_{xx}ww_{x}-\frac {9}{2}s^{3} \bigl(a^{2}\phi_{x}^{2}\phi_{xx} \bigr)_{x}w^{2} +\frac{9}{2}s^{3}a \phi_{x}\phi_{xx}(a\phi_{x})_{x}w^{2} \biggr\vert \\ \leq& C\bigl(s^{3}\lambda^{4}\theta^{3} \vert w \vert \vert w_{x} \vert +s^{3}\lambda^{5} \theta ^{3}w^{2}+s\lambda^{3}\theta w_{x}^{2}+s\lambda^{2}\theta \vert w_{x} \vert \vert w_{xx} \vert \\ &{}+s^{3}\lambda^{4}\theta^{3} \vert w \vert \vert w_{x} \vert +s^{3}\lambda^{4}\theta ^{2}w^{2}+s^{3}\lambda^{5} \theta^{3}w^{2}\bigr). \end{aligned}$$
For any \(\varepsilon>0\), there exist \(s_{0}>0\), \(\lambda_{0}>0\) such that if \(s\geq s_{0}\), \(\lambda\geq\lambda_{0}\), we have
$$\vert B_{R} \vert \leq\varepsilon\bigl(s^{5} \lambda^{5}\theta^{5}w^{2} +s^{3} \lambda^{3}\theta^{3}w_{x}^{2} +s\lambda \theta w_{xx}^{2}\bigr). $$
Now we estimate each term of the right-hand side in (2.6).
It is easy to deduce that
$$\begin{aligned} &{w_{x} =-s\phi_{x}e^{-s\phi}u+e^{-s\phi}u_{x},} \\ &{w_{t}=-s \phi _{t}e^{-s\phi}u+e^{-s\phi}u_{t},} \\ &{w_{xx} =-s\phi_{xx}e^{-s\phi}u+s^{2} \phi_{x}^{2}e^{-s\phi}u-2s\phi _{x}e^{-s\phi}u_{x}+e^{-s\phi}u_{xx}.} \end{aligned}$$
Noting that \(u(0,t)=u_{t}(0,t)=u_{x}(L,t)=u_{xx}(L,t)=0\), \(t\in(0,T)\), we have
$$ \begin{aligned} &{w(0,t)=0,\qquad w(L,t)=e^{-s\phi(L,t)}u(L,t),} \\ &{w_{t}(0,t)=0,\qquad w_{t}(L,t)=e^{-s\phi(L,t)}u_{t}(L,t)-s \phi _{t}(L,t)e^{-s\phi(L,t)}u(L,t),} \\ &{w_{x}(0,t)=e^{-s\phi(0,t)}u_{x}(0,t),} \\ &{w_{x}(L,t)=s\lambda\beta_{x}(L)\theta(L,t)e^{-s\phi(L,t)}u(L,t),} \\ &{w_{xx}(0,t)=2s\lambda\beta_{x}(0)\theta(0,t)e^{-s\phi (0,t)}u_{x}(0,t)+e^{-s\phi(0,t)}u _{xx}(0,t),} \\ &{w_{xx}(L,t)=s\lambda \bigl(\beta_{xx}(L)+\lambda\beta _{x}^{2}(L)+s\lambda\beta_{x}^{2}(L) \theta(L,t) \bigr)\theta (L,t)e^{-s\phi(L,t)}u (L,t).} \end{aligned} $$
(2.7)
Since \(\beta_{x}\geq r>0\) and \(\theta>0\), we can obtain from (2.7)
$$\begin{aligned} &{ \frac{3}{2}s^{5}\lambda^{5}a^{2} \beta_{x}^{5}\theta^{5}w^{2} +8s^{3}\lambda^{3}a^{2}\beta_{x}^{3} \theta^{3}w_{x}^{2} +\frac{3}{2}s\lambda a^{2}\beta_{x}\theta w_{xx}^{2} \biggm| _{x=0}\geq0,} \\ &{\biggl\vert \int_{0}^{T} (-3as\lambda\beta_{x}\theta w_{t}w_{x} ) \mid _{x=0} \,dt \biggr\vert =0,} \\ &{\biggl\vert \int_{0}^{T} \bigl(-a^{2}s^{3} \lambda^{3}\beta^{3}_{x}\theta ^{3} ww_{xx} \bigr) \bigm| _{x=0} \,dt \biggr\vert =0.} \end{aligned}$$
Taking (2.7) into account and choosing s and λ large enough, we obtain
$$\begin{aligned} &{\int_{0}^{T} \biggl(-\frac{3}{2}s^{5} \lambda^{5}a^{2}\beta _{x}^{5} \theta^{5}w^{2} -8s^{3}\lambda^{3}a^{2} \beta_{x}^{3}\theta^{3}w_{x}^{2} -\frac{3}{2}s\lambda a^{2}\beta_{x}\theta w_{xx}^{2} \biggr) \biggm| _{x=0}^{x=L}\,dt} \\ &{\quad \geq-C \int_{0}^{T}\bigl(s^{5} \lambda^{5}\theta^{5}w^{2} +s^{3} \lambda^{3}\theta^{3}w_{x}^{2} +s\lambda \theta w_{xx}^{2}\bigr)\bigm| _{x=L}\,dt} \\ &{\quad \geq-C \int_{0}^{T}(s^{5}\lambda^{5} \theta^{5}e^{-2s\phi}u^{2})\mid _{x=L}\,dt,} \\ &{\biggl\vert \int_{0}^{T} (-3as\lambda\beta_{x}\theta w_{t}w_{x} ) \mid _{x=0}^{x=L} \,dt \biggr\vert } \\ &{\quad = \biggl\vert 3as\lambda \int_{0}^{T} (\beta_{x}\theta w_{t}w_{x}) \mid _{x=L}\,dt \biggr\vert } \\ &{\quad = \biggl\vert \int_{0}^{T} \bigl(3as^{2} \lambda^{2}\beta ^{2}_{x}\theta^{2}e^{-2s\phi}u_{t}u -3as^{3}\lambda^{2}\beta^{2}_{x} \theta^{2}\phi_{t}e^{-2s\phi }u^{2} \bigr) \bigm| _{x=L}\,dt \biggr\vert } \\ &{\quad = \biggl\vert \int_{0}^{T} \biggl(-\frac{3}{2}as^{2} \lambda ^{2}\bigl(\beta^{2}_{x}\theta^{2} \bigr)_{t}e^{-2s\phi}u^{2} -3as^{3} \lambda^{2}\beta^{2}_{x}\theta^{2} \phi_{t}e^{-2s\phi }u^{2} \biggr) \biggm| _{x=L}\,dt \biggr\vert } \\ &{\quad \leq C \int_{0}^{T} \bigl(s^{2} \lambda^{2}\theta^{3}e^{-2s\phi }u^{2}+s^{3} \lambda^{2}\theta^{4}e^{-2s\phi}u^{2} \bigr) \bigm| _{x=L}\,dt\quad \bigl(\mbox{since } \vert \phi_{t} \vert + \vert \theta_{t} \vert \leq C\theta^{2}\bigr)} \\ &{\quad \leq C \int_{0}^{T} \bigl(s^{5} \lambda^{5}\theta^{5}e^{-2s\phi }u^{2} \bigr) \bigm| _{x=L}\,dt,} \\ &{\biggl\vert \int_{0}^{T} \bigl(-a^{2}s^{3} \lambda^{3}\beta^{3}_{x}\theta ^{3} ww_{xx} \bigr) \bigm| _{x=0}^{x=L} \,dt \biggr\vert } \\ &{\quad = \biggl\vert a^{2}s^{3}\lambda^{3} \int_{0}^{T} \bigl(\beta^{3}_{x} \theta ^{3} ww_{xx}\bigr) \bigm| _{x=L}\,dt \biggr\vert } \\ &{\quad = \biggl\vert a^{2}s^{3}\lambda^{3} \int_{0}^{T} \bigl(\beta^{3}_{x} \theta^{3} \bigl(s\lambda\beta_{xx}\theta+s \lambda^{2}\beta^{2}_{x}\theta +s^{2} \lambda^{2}\beta^{2}_{x}\theta^{2} \bigr)e^{-2s\phi}u^{2}\bigr) \bigm| _{x=L}\,dt \biggr\vert } \\ &{\quad \leq C \int_{0}^{T} \bigl(s^{5} \lambda^{5}\theta^{5}e^{-2s\phi }u^{2} \bigr) \mid _{x=L}\,dt.} \end{aligned}$$
Therefore
$$\int_{0}^{T}(\cdot)\mid _{x=0}^{x=L}\,dt \geq-C \int_{0}^{T} \bigl(s^{5} \lambda^{5}\theta^{5}e^{-2s\phi}u^{2} \bigr) \bigm| _{x=L}\,dt. $$
Combining (2.3)-(2.5), consequently,
$$\begin{aligned} &{\int_{Q} \bigl(s^{5}\lambda^{6} \theta^{5}w^{2}+s^{3}\lambda ^{4} \theta^{3}w_{x}^{2}+s\lambda^{2}\theta w_{xx}^{2} \bigr)\,dx\,dt+ \int_{Q}\bigl( \vert P_{1}w \vert ^{2}+ \vert P_{2}w \vert ^{2}\bigr)\,dx\,dt} \\ &{\quad\leq C \int_{Q} \vert P_{\phi}w \vert ^{2}\,dx\,dt+C \int_{0}^{T} \bigl(s^{5} \lambda^{5}\theta^{5}e^{-2s\phi}u^{2} \bigr) \bigm| _{x=L}\,dt.} \end{aligned}$$
(2.8)
Returning w to \(e^{-s\phi}u\), we can obtain (2.1). □