In this section, we prove the unique continuation for the adjoint system by Carleman estimate.
First, we study the well-posedness of the adjoint system
$$\begin{aligned}& -y_{t}-\bigl(x^{p}y_{x}\bigr)_{x}+ \lambda_{1}y+\lambda_{3}z=f_{1},\quad(x,t)\in(0,1) \times (0,T), \end{aligned}$$
(2.1)
$$\begin{aligned}& -z_{t}-\bigl(x^{p}z_{x}\bigr)_{x}+ \lambda_{2}y+\lambda_{4}z=f_{2},\quad(x,t)\in(0,1) \times (0,T), \end{aligned}$$
(2.2)
$$\begin{aligned}& y(0,t)=0,\qquad y(1,t)=0,\quad t\in(0,T), \end{aligned}$$
(2.3)
$$\begin{aligned}& z(0,t)=0,\qquad z(1,t)=0, \quad t\in(0,T), \end{aligned}$$
(2.4)
$$\begin{aligned}& y(x,T)=y_{T}(x),\qquad z(x,T)=z_{T}(x),\quad x \in(0,1), \end{aligned}$$
(2.5)
where \(f_{1},f_{2}\in L^{2}((0,1)\times(0,T))\), \(y_{T},z_{T}\in L^{2}(0,1)\).
Define \(H_{p}^{1}(0,1)\), \(H_{p}^{2}(0,1)\) are the closures of \(C_{0}^{\infty}(0,1)\) with respect to the norms; see [1],
$$\begin{gathered} \|u\|_{H_{p}^{1}(0,1)}= \biggl( \int_{0}^{1}\bigl(u^{2}+x^{p}u_{x}^{2} \bigr)\,dx \biggr)^{1/2},\quad u\in H_{p}^{1}(0,1), \\ \|u\|_{H_{p}^{2}(0,1)}= \biggl( \int_{0}^{1}\bigl(u^{2}+x^{p}u_{x}^{2}+ \bigl(x^{p}u_{x}\bigr)_{x}^{2}\bigr)\,dx \biggr)^{1/2},\quad u\in H_{p}^{2}(0,1),\end{gathered} $$
respectively. Denote \(\mathbb{B}=L^{\infty}(0,T;L^{2}(0,1))\cap L^{2}(0,T;H_{p}^{1}(0,1))\) and \(\mathbb{D}=L^{2}(0,T;H_{p}^{2}(0,1))\cap H^{1}(0,T;L^{2}(0,1))\) with respect to the norms
$$\begin{gathered} \|u\|_{\mathbb{B}}= \biggl(\sup_{t\in(0,T)} \int_{0}^{1}\bigl(u(x,t)\bigr)^{2}\,dx+ \int _{0}^{T} \int_{0}^{1}\bigl(u^{2}+x^{p}u_{x}^{2} \bigr)\,dx\,dt \biggr)^{1/2}, \quad u\in\mathbb{B}, \\ \|u\|_{\mathbb{D}}= \biggl( \int_{0}^{T} \int_{0}^{1}u_{t}^{2}\,dx+ \int_{0}^{T} \int_{0}^{1}\bigl(u^{2}+x^{p}u_{x}^{2}+ \bigl(x^{p}u_{x}\bigr)_{x}^{2}\bigr)\,dx\,dt \biggr)^{1/2}, \quad u\in \mathbb{D},\end{gathered} $$
respectively.
Definition 2.1
A pair of functions \((y,z)\in\mathbb{B}\times \mathbb{B}\) is called a solution to the system (2.1)-(2.5), if for any \(\varphi,\psi\in\mathbb{B}\) with \(\varphi_{t},\psi_{t}\in L^{2}((0,1)\times(0,T))\) and \(\varphi(x,0)=0\), \(\psi(x,0)=0\), \(x\in(0,1)\), the following integral equalities hold:
$$\begin{gathered} \int_{0}^{T} \int_{0}^{1}\bigl(y\varphi_{t}+x^{p}y_{x} \varphi_{x}+\lambda _{1}y\varphi+\lambda_{3}z \varphi\bigr) \,dx\,dt= \int_{0}^{T} \int _{0}^{1}f_{1}\varphi \,dx\,dt + \int_{0}^{1}y_{T}(x)\varphi(x,T)\,dx, \\ \int_{0}^{T} \int_{0}^{1}\bigl(z\psi_{t}+x^{p}z_{x} \psi_{x}+\lambda_{2}y\psi +\lambda_{4}z\psi \bigr)\,dx\,dt= \int_{0}^{T} \int_{0}^{1}f_{2}\psi \,dx\,dt + \int_{0}^{1}z_{T}(x)\psi(x,T)\,dx.\end{gathered} $$
By energy estimates, one can prove the well-posedness as the case of the single equations.
Theorem 2.1
There exists a unique solution
\((y,z)\in\mathbb{B}\times\mathbb{B}\)
to the problem (2.1)-(2.5) satisfying
$$\begin{gathered} \|y\|_{\mathbb{B}}+\|z\|_{\mathbb{B}}+\big\| x^{p}y_{x}(0,t) \big\| _{L^{2}(T_{1},T_{2})}+\big\| x^{p}y_{x}(0,t)\big\| _{L^{2}(T_{1},T_{2})}+ \big\| x^{p}z_{x}(0,t)\big\| _{L^{2}(T_{1},T_{2})} \\ \quad\le C_{1} \bigl(\|f_{1}\|_{L^{2}((0,1)\times(0,T))}+\|f_{2}\| _{L^{2}((0,1)\times(0,T))}+\|y_{T}\|_{L^{2}(0,1)}+\|z_{T} \|_{L^{2}(0,1)} \bigr),\end{gathered} $$
where
\(C_{1}\)
is depending only on
T, \(T_{1}\), \(\|\lambda_{i}\|_{L^{\infty}((0,1)\times(0,T))}\), \(i=1,2,3,4\). Further, if
\((y_{T},z_{T})\in H_{p}^{1}(0,1)\times H_{p}^{1}(0,1)\), then there exists a constant
\(C_{2}\)
depending only on
T, \(T_{1}\), \(\|\lambda_{i}\|_{L^{\infty}((0,1)\times (0,T))}\), \(i=1,2,3,4\), such that
$$\begin{aligned} \|y\|_{\mathbb{D}}+\|z\|_{\mathbb{D}} \le C_{2} \bigl(\|f_{1} \|_{L^{2}((0,1)\times(0,T))}+\|f_{2}\|_{L^{2}((0,1)\times (0,T))}+\|y_{T} \|_{H_{p}^{1}(0,1)}+\|z_{T}\|_{H_{p}^{1}(0,1)} \bigr). \end{aligned}$$
The proof is similar to Proposition 2.1 in [11] and Proposition 2.1 in [12].
Next, we prove the unique continuation. Consider the problem
$$\begin{aligned}& w_{t}+\bigl(x^{p}w_{x}\bigr)_{x}=F,\quad(x,t) \in(0,1)\times(0,T), \end{aligned}$$
(2.6)
$$\begin{aligned}& w(0,t)=w(1,t)=0,\quad t\in(0,T), \end{aligned}$$
(2.7)
where \(F\in L^{2}((0,1)\times(0,T))\). Then we have the following two lemmas.
Lemma 2.1
(Theorem 2.3 [10])
Let
\(w\in\mathbb{D}\)
be the solution to the problem (2.6) and (2.7) and satisfying
$$\bigl(x^{p}w_{x}\bigr) (0,t)=\bigl(x^{p}w_{x} \bigr) (1,t)=0. $$
Then, for fixed
\(q\in(1-p,1-p/2)\), there exist two positive constants
C
and
\(s_{0}\)
such that, for all
\(s\ge s_{0}\),
$$\begin{gathered} \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}w^{2}e^{-2s x^{q}l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}w_{x}^{2} e^{-2s x^{q} l(t)}\,dx\,dt \\ \quad\le C \int_{0}^{T} \int_{0}^{1}F^{2}e^{-2s x^{q} l(t)}\,dx\,dt,\end{gathered} $$
where
\(l(t)=\frac{1}{t(T-t)}\).
From Lemma 2.1, we can prove the Carleman estimate for the system (2.1)-(2.5).
Theorem 2.2
Let
\((y,z)\in\mathbb{D}\times\mathbb{D}\)
be the solution to the system (2.1)-(2.4) and suppose that, for a.e. \(t\in(0,T)\),
$$\bigl(x^{p}y_{x}\bigr) (0,t)=\bigl(x^{p}y_{x} \bigr) (1,t)=\bigl(x^{p}z_{x}\bigr) (0,t)= \bigl(x^{p}z_{x}\bigr) (1,t)=0. $$
Then, for fixed
\(q\in(1-p,1-p/2)\), there exist positive constants
\(C_{1}\)
and
\(s_{1}\)
such that, for all
\(s\ge s_{1}\),
$$\begin{gathered} \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}y^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}y_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \qquad{}+ \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}z^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}z_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \quad\le C_{1} \biggl( \int_{0}^{T} \int_{0}^{1}|f_{1}|^{2}e^{-2s x^{q}l(t)}\,dx\,dt+ \int _{0}^{T} \int_{0}^{1}|f_{2}|^{2}e^{-2s x^{q} l(t)}\,dx\,dt \biggr). \end{gathered}$$
Proof
It follows from Lemma 2.1 that there exist C and \(s_{0}\) such that, for all \(s\ge s_{0}\),
$$\begin{gathered} \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}y^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}y_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \qquad{}+ \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}z^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}z_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \quad\le C \biggl( \int_{0}^{T} \int_{0}^{1}|f_{1}|^{2}e^{-2s x^{q} l(t)}\,dx\,dt+ \int _{0}^{T} \int_{0}^{1}|f_{2}|^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \qquad{}+\bigl(\|\lambda_{1}\|_{L^{\infty}(0,1)\times(0,T)}^{2}+\| \lambda_{2}\| _{L^{\infty}(0,1)\times(0,T)}^{2}\bigr) \int_{0}^{T} \int_{0}^{1}|y|^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \qquad{}+\bigl(\|\lambda_{3}\|_{L^{\infty}(0,1)\times(0,T)}^{2}+\| \lambda_{4}\| _{L^{\infty}(0,1)\times(0,T)}^{2}\bigr) \int_{0}^{T} \int_{0}^{1}|z|^{2}e^{-2s x^{q} l(t)}\,dx\,dt \biggr).\end{gathered} $$
Note that \(2p+3q-4<0\) due to \(q\in(1-p,1-p/2)\). Take
$$s_{1}=\max\Biggl\{ s_{0}, 2^{-5/3}T^{2}C^{1/3} \Biggl(\sum_{i=1}^{4}\|\lambda_{i} \| _{L^{\infty}(0,1)\times(0,T)}^{2}\Biggr)^{1/3}\Biggr\} . $$
Then, for \(s>s_{1}\), we have
$$\begin{aligned} \begin{gathered} \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}y^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}y_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \qquad{}+ \int_{0}^{T} \int_{0}^{1}s^{3}l^{3}(t)x^{2p+3q-4}z^{2}e^{-2s x^{q} l(t)}\,dx\,dt + \int_{0}^{T} \int_{0}^{1}s l(t)x^{2p+q-2}z_{x}^{2}e^{-2s x^{q} l(t)}\,dx\,dt \\ \quad\le 2C \biggl( \int_{0}^{T} \int_{0}^{1}|f_{1}|^{2}e^{-2s x^{q} l(t)}\,dx\,dt+ \int _{0}^{T} \int_{0}^{1}|f_{2}|^{2}e^{-2s x^{q} l(t)}\,dx\,dt\biggr).\end{gathered} \end{aligned}$$
The proof is complete. □
Similar to the proof of Theorem 3.1 [10] and Proposition 4.2 [12], one can prove the following unique continuation properties.
Theorem 2.3
Let
\((y,z)\in\mathbb{D}\times\mathbb{D}\)
be the solution to the system (2.1)-(2.4) and suppose that, for almost every
\(t\in(0,T)\),
$$\bigl(x^{p}y_{x}\bigr) (0,t)=\bigl(x^{p}z_{x} \bigr) (0,t)=0. $$
If
\(f_{1}(x,t)=f_{2}(x,t)=0\), then
\(y(x,t)=0\), \(z(x,t)=0\), where
\((x,t)\in (0,1)\times(0,T)\).
Theorem 2.4
Let
\((y,z)\in\mathbb{B}\times\mathbb{B}\)
be the solution to the system (2.1)-(2.5) and suppose that, for almost every
\(t\in(0,T)\),
$$\bigl(x^{p}y_{x}\bigr) (0,t)=\bigl(x^{p}z_{x} \bigr) (0,t)=0. $$
If
\(f_{1}(x,t)=f_{2}(x,t)=0\), then
\(y(x,t)=0\), \(z(x,t)=0\), where
\((x,t)\in (0,1)\times(0,T)\).