In this section, in order to prove Theorem 2.2, we need the following lemmas.

Define the following weighted energy for (2.1):

$$ E(t)=\frac{1}{2}\int_{0}^{\alpha_{k}(t)} \bigl[\bigl\vert z_{t}(x,t)\bigr\vert ^{2} +\bigl\vert z_{x}(x,t)\bigr\vert ^{2} \bigr]\,dx \quad \mbox{for }t \geq0, $$

where *z* is the solution of (2.1). It follows that

$$ E_{0}\triangleq E(0)=\frac{1}{2}\int_{0}^{1} \bigl[\bigl\vert z^{1}(x)\bigr\vert ^{2}+\bigl\vert z^{0}_{x}(x)\bigr\vert ^{2} \bigr]\,dx. $$

### Lemma 3.1

*For any*
\((z^{0}, z^{1})\in H_{0}^{1}(0, 1)\times L^{2}(0, 1)\)
*and*
\(t\in[0, T]\), *the corresponding solution*
*z*
*of* (2.1) *satisfies*

$$ E(t)-E_{0}=\frac{k(k^{2}-1)}{2}\int_{0}^{t} \bigl\vert z_{x}\bigl(\alpha_{k}(s),s\bigr)\bigr\vert ^{2}\,ds. $$

(3.1)

### Proof

Multiplying the first equation of (2.1) by \(z_{t}\) and integrating on \((0,\alpha_{k}(s))\times(0,t)\), for any \(0< t\leq T\), we get

$$ 0= \int_{0}^{t}\int_{0}^{\alpha_{k}(s)} \bigl[z_{tt}(x,s)z_{t}(x,s) - z_{xx}(x,s)z_{t}(x,s) \bigr]\,dx\,ds \triangleq I_{1}+I_{2}. $$

In the following, we calculate the above two integrals \(I_{i}\) (\(i=1, 2\)). It is easy to check that

$$\begin{aligned} I_{1} =& \int_{0}^{t}\int _{0}^{\alpha_{k}(t)}z_{tt}(x,s)z_{t}(x,s) \,dx\,ds \\ =&\int_{0}^{t}\biggl[\frac{\partial}{\partial s}\int _{0}^{\alpha_{k}(s)} \frac{1}{2}\bigl\vert z_{t}(x,s)\bigr\vert ^{2}\,dx -\frac{1}{2}z_{t}^{2} \bigl(\alpha_{k}(s),s\bigr)\cdot k\biggr]\,ds \\ =&\int_{0}^{\alpha_{k}(t)} \frac{1}{2}\bigl\vert z_{t}(x,t)\bigr\vert ^{2}\,dx-\int_{0}^{1} \frac{1}{2}\bigl\vert z^{1}\bigr\vert ^{2}\,dx- \frac{1}{2}\int_{0}^{t}z_{t}^{2} \bigl(\alpha_{k}(s),s\bigr)\cdot k\,ds. \end{aligned}$$

By the definition of \(\alpha_{k}\) and \(z(\alpha_{k}(s),s)=0\), we have

$$ z_{x}\bigl(\alpha_{k}(s),s\bigr)\cdot k+z_{s}\bigl(\alpha_{k}(s),s\bigr)=0. $$

(3.2)

Therefore, it follows that

$$ I_{1}=\int_{0}^{\alpha_{k}(t)} \frac{1}{2}\bigl\vert z_{t}(x,t)\bigr\vert ^{2} \,dx-\int_{0}^{1} \frac{1}{2}\bigl\vert z^{1}\bigr\vert ^{2}\,dx-\frac{k^{3}}{2}\int _{0}^{t}z_{x}^{2}\bigl( \alpha_{k}(s),s\bigr)\,ds. $$

(3.3)

Further, by \(z(0, s)=z(\alpha_{k}(s),s)=0\) on \((0,t)\), we have

$$\begin{aligned} I_{2} =& -\int_{0}^{t} \int_{0}^{\alpha_{k}(s)}z_{xx}(x,s)z_{t}(x,s) \,dx\,ds \\ =&-\int_{0}^{t}\int_{0}^{\alpha_{k}(s)} \biggl\{ \frac{\partial}{\partial x} \bigl[z_{x}(x,s)z_{t}(x,s) \bigr] -z_{x}(x,s)z_{xt}(x,s)\biggr\} \,dx\,ds \\ =&-\int_{0}^{t}z_{x}\bigl( \alpha_{k}(s),s\bigr)z_{t}\bigl(\alpha_{k}(s),s \bigr) \,ds \\ &{}+\int_{0}^{t}\biggl[\frac{\partial}{\partial s}\int _{0}^{\alpha_{k}(s)} \frac{1}{2}\bigl\vert z_{x}(x,s)\bigr\vert ^{2}\,dx -\frac{1}{2}z_{x}^{2} \bigl(\alpha_{k}(s),s\bigr)\cdot k\biggr]\,ds \\ =&k\int_{0}^{t}z_{x}^{2} \bigl(\alpha_{k}(s),s\bigr) \,ds+\int_{0}^{\alpha_{k}(t)} \frac{1}{2}\bigl\vert z_{x}(x,t)\bigr\vert ^{2} \,dx \\ &{}-\int_{0}^{1} \frac{1}{2}\bigl\vert z_{x}^{0}\bigr\vert ^{2}\,dx -\frac{k}{2} \int_{0}^{t}z_{x}^{2}\bigl( \alpha_{k}(s),s\bigr)\,ds \\ =&\int_{0}^{\alpha_{k}(t)} \frac{1}{2}\bigl\vert z_{x}(x,t)\bigr\vert ^{2}\,dx -\int_{0}^{1} \frac{1}{2}\bigl\vert z_{x}^{0}\bigr\vert ^{2}\,dx +\frac{k}{2}\int_{0}^{t}z_{x}^{2} \bigl(\alpha_{k}(s),s\bigr)\,ds. \end{aligned}$$

(3.4)

Therefore, by (3.3) and (3.4) and the definition of \(E(t)\), we obtain

$$ E(t)-E_{0}=\frac{k(k^{2}-1)}{2}\int_{0}^{t} \bigl\vert z_{x}\bigl(\alpha_{k}(s),s\bigr)\bigr\vert ^{2}\,dt. $$

□

### Lemma 3.2

*For any*
\((z^{0}, z^{1})\in H_{0}^{1}(0, 1)\times L^{2}(0, 1)\)
*and*
\(t\in[0, T]\), *the corresponding solution*
*z*
*of* (2.1) *holds*,

$$ \bigl(1-k^{2}\bigr)\int_{0}^{t} \bigl\vert z_{x}\bigl(\alpha_{k}(s),s\bigr)\bigr\vert ^{2}\,ds =2tE(t)-2\int_{0}^{1} xz^{1}z_{x}^{0}\,dx+2\int_{0}^{\alpha_{k}(t)} xz_{t}(t)z_{x}(t)\,dx. $$

(3.5)

### Proof

Multiplying the first equation of (2.1) by \(2xz_{x}\) and integrating on \((0,\alpha_{k}(s))\times(0,t)\), for any \(0< t\leq T\), from (3.2), we have

$$\begin{aligned} 0 =& \int_{0}^{t}\int_{0}^{\alpha_{k}(s)} [z_{tt}2xz_{x} - z_{xx}2xz_{x} ]\,dx \,ds \\ =& \int_{0}^{t}\int_{0}^{\alpha_{k}(s)} \biggl\{ \frac{\partial}{\partial t}(2xz_{x}z_{t})- \frac{\partial}{\partial x} \bigl(xz_{t}^{2}+xz_{x}^{2} \bigr)+z_{t}^{2}+z_{x}^{2} \biggr\} \,dx \,ds \\ =& \int_{0}^{t} \biggl[\frac{\partial}{\partial s}\int _{0}^{\alpha_{k}(s)} 2xz_{x}z_{t}\,dx -2 \alpha_{k}(s)z_{x}\bigl(\alpha_{k}(s),s \bigr)z_{t}\bigl(\alpha_{k}(s),s\bigr)\cdot k \biggr]\,ds \\ &{}+\int_{0}^{t}\bigl[\alpha_{k}(s)z_{t}^{2} \bigl(\alpha_{k}(s),s\bigr)+\alpha_{k}(s)z_{x}^{2} \bigl(\alpha _{k}(s),s\bigr)\bigr]\,ds+2\int_{0}^{t}E(s) \,ds \\ =& \int_{0}^{\alpha_{k}(t)} 2xz_{x}(x,t)z_{t}(x,t) \,dx-\int_{0}^{1} 2xz_{x}^{0}z^{1} \,dx \\ &{}-\int_{0}^{t}2\alpha_{k}(s)z_{x} \bigl(\alpha_{k}(s),s\bigr)z_{t}\bigl(\alpha_{k}(s),s \bigr)\cdot k\,ds \\ &{}-\int_{0}^{t}\bigl[\alpha_{k}(s)z_{t}^{2} \bigl(\alpha_{k}(s),s\bigr)+\alpha_{k}(s)z_{x}^{2} \bigl(\alpha _{k}(s),s\bigr)\bigr]\,ds+2\int_{0}^{t}E(s) \,ds \\ =& \int_{0}^{\alpha_{k}(t)} 2xz_{x}(x,t)z_{t}(x,t) \,dx-\int_{0}^{1} 2xz_{x}^{0}z^{1} \,dx+2\int_{0}^{t}E(s)\,ds \\ &{}+\bigl(k^{2}-1\bigr)\int_{0}^{t} \alpha_{k}(s)z_{x}^{2}\bigl(\alpha_{k}(s),s \bigr)\,ds. \end{aligned}$$

From this one concludes that

$$\begin{aligned}& \bigl(1-k^{2}\bigr)\int_{0}^{t} \alpha_{k}(s)z_{x}^{2}\bigl(\alpha_{k}(s),s \bigr)\,ds \\& \quad = \int_{0}^{\alpha_{k}(t)} 2xz_{x}(x,t)z_{t}(x,t) \,dx-\int_{0}^{1} 2xz_{x}^{0}z^{1} \,dx+2\int_{0}^{t}E(s)\,ds. \end{aligned}$$

(3.6)

Similarly, multiplying the first equation of (2.1) by \(2tz_{t}\) and integrating on \((0,\alpha_{k}(s))\times(0,t)\), for any \(0< t\leq T\), we obtain

$$\begin{aligned} 0 =& \int_{0}^{t}\int_{0}^{\alpha_{k}(s)} [z_{tt}2tz_{t} - z_{xx}2tz_{t} ]\,dx \,ds \\ =& \int_{0}^{t}\int_{0}^{\alpha_{k}(s)} \biggl\{ \frac{\partial}{\partial x}(2tz_{x}z_{t})- \frac{\partial}{\partial t} \bigl(tz_{t}^{2}+tz_{x}^{2} \bigr)+z_{t}^{2}+z_{x}^{2} \biggr\} \,dx \,ds. \end{aligned}$$

From this we get

$$ \bigl(1-k^{2}\bigr)\int_{0}^{t}ksz_{x}^{2} \bigl(\alpha_{k}(s),s\bigr)\,ds= 2\int_{0}^{t}E(s) \,ds-2tE(t). $$

(3.7)

Therefore, (3.5) follows by (3.6) and (3.7). □

By the above two lemmas, we obtain the following lemma concerning a growth estimate of the energy function.

### Lemma 3.3

*For any*
\((z^{0}, z^{1})\in H^{1}_{0}(0, 1)\times L^{2}(0, 1)\)
*and*
\(t\in[0, T]\), *from the corresponding solution*
*z*
*of* (2.1) *follows*

$$ \frac{1-k}{(1+k)\alpha_{k}(t)}E_{0}\leq E(t)\leq \frac{1+k}{(1-k)\alpha_{k}(t)}E_{0}. $$

(3.8)

### Proof

From (3.1) and (3.5), it is easy to check that

$$ \frac{2}{k}\bigl(E(t)-E_{0}\bigr)=2tE(t) +\int _{0}^{\alpha_{k}(t)} 2xz_{x}(x,t)z_{t}(x,t) \,dx-\int_{0}^{1} 2xz_{x}^{0}z^{1} \,dx. $$

By this equality, we get

$$ E_{0}-\int_{0}^{1} kxz_{x}^{0}z^{1}\,dx=\alpha_{k}(t)E(t) + \int_{0}^{\alpha_{k}(t)} kxz_{x}(x,t)z_{t}(x,t) \,dx. $$

(3.9)

On the other hand, for each \(t\in(0,T)\) and \(k\in(0,1)\), we obtain the estimate

$$ \biggl\vert \int_{0}^{\alpha_{k}(t)} xz_{x}(x,t)z_{t}(x,t)\,dx\biggr\vert \leq \alpha_{k}(t)E(t). $$

(3.10)

From (3.9) and (3.10), we derive that

$$\begin{aligned}& (1-k)E_{0}\leq(1+k)\alpha_{k}(t)E(t), \\& (1-k)E_{0}\geq(1-k)\alpha_{k}(t)E(t). \end{aligned}$$

By this (3.8) follows. □

### Remark 3.1

Lemma 3.3 implies that

$$ \frac{1-k}{(1+k)\alpha_{k}(T)}E_{0}\leq E(T)\leq \frac{1+k}{(1-k)\alpha_{k}(T)}E_{0}. $$

(3.11)

In the following, we give the proof of Theorem 2.2.

### Proof of Theorem 2.2

*Step* 1. Multiplying the first equation of (2.1) by \([x-\alpha_{k}(t)]z_{x}\) and integrating on \((0,\alpha_{k}(t))\times(0,T)\), it follows that

$$ 0= \int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \bigl\{ z_{tt}\bigl[x-\alpha_{k}(t)\bigr]z_{x} - z_{xx}\bigl[x-\alpha_{k}(t)\bigr]z_{x} \bigr\} \,dx \,dt \triangleq J_{1}+J_{2}. $$

In the following, we calculate the above two integrals \(J_{i}\) (\(i=1, 2\)). By \(z(0, t)=0\) on \((0,T)\), it is easy to check that

$$\begin{aligned} J_{1} =& \int_{0}^{T}\int _{0}^{\alpha_{k}(t)}z_{tt}\bigl[x- \alpha_{k}(t)\bigr]z_{x}\,dx\,dt \\ =&\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \biggl\{ \frac{\partial}{\partial t}\bigl[\bigl(x-\alpha_{k}(t) \bigr)z_{x}z_{t}\bigr] -kz_{x}z_{t}- \bigl(x-\alpha_{k}(t)\bigr)z_{xt}z_{t} \biggr\} \,dx\,dt \\ =&{\bigl.\int_{0}^{\alpha_{k}(t)}\bigl[\bigl(x- \alpha_{k}(t)\bigr)z_{x}z_{t}\bigr] \,dx\bigr|^{T}_{0}} -k\int_{0}^{T} \int_{0}^{\alpha_{k}(t)}z_{x}z_{t}\,dx \,dt \\ &{}-\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \bigl(x-\alpha_{k}(t)\bigr)\frac{\partial }{\partial x}\biggl[ \frac{1}{2}z_{t}^{2}\biggr]\,dx\,dt \\ =&{\bigl.\int_{0}^{\alpha_{k}(t)}\bigl[\bigl(x- \alpha_{k}(t)\bigr)z_{x}z_{t}\bigr] \,dx\bigr|^{T}_{0}} -k\int_{0}^{T} \int_{0}^{\alpha_{k}(t)}z_{x}z_{t}\,dx \,dt \\ &{}-{\biggl.\int_{0}^{T}\bigl(x-\alpha_{k}(t) \bigr)\frac{1}{2}z_{t}^{2}\biggr|^{\alpha_{k}(t)}_{0}} \,dt +\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \frac{1}{2}z_{t}^{2}\,dx\,dt \\ =&{\bigl.\int_{0}^{\alpha_{k}(t)}\bigl[\bigl(x- \alpha_{k}(t)\bigr)z_{x}z_{t}\bigr] \,dx\bigr|^{T}_{0}} -k\int_{0}^{T} \int_{0}^{\alpha_{k}(t)}z_{x}z_{t}\,dx \,dt \\ &{}+\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \frac{1}{2}z_{t}^{2}\,dx\,dt. \end{aligned}$$

(3.12)

Further, it follows that

$$\begin{aligned} J_{2} =& -\int_{0}^{T} \int_{0}^{\alpha_{k}(t)}z_{xx}\bigl[x- \alpha_{k}(t)\bigr]z_{x}\,dx\,dt \\ =&-\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \biggl\{ \bigl(x-\alpha_{k}(t)\bigr)\frac{\partial }{\partial x}\biggl[ \frac{1}{2}z_{x}^{2}\biggr] \biggr\} \,dx\,dt \\ =&-{\biggl.\int_{0}^{T}\bigl(x-\alpha_{k}(t) \bigr)\frac{1}{2}z_{x}^{2}\biggr|^{\alpha_{k}(t)}_{0}} \,dt +\int_{0}^{T}\int_{0}^{\alpha_{k}(t)} \frac{1}{2}z_{x}^{2}\,dx\,dt \\ =&-\int_{0}^{T}\alpha_{k}(t) \frac{1}{2}z_{x}^{2}(0,t)\,dx +\int _{0}^{T}\int_{0}^{\alpha_{k}(t)} \frac{1}{2}z_{x}^{2}\,dx\,dt. \end{aligned}$$

(3.13)

Therefore, by (3.12) and (3.13), we obtain

$$\begin{aligned}& \frac{1}{2}\int_{0}^{T} \alpha_{k}(t)z_{x}^{2}(0,t)\,dx \\& \quad ={\bigl.\int_{0}^{T}E(t)\,dt -k\int _{0}^{T}\int_{0}^{\alpha_{k}(t)}z_{x}z_{t} \,dx\,dt +\int_{0}^{\alpha_{k}(t)}\bigl[\bigl(x- \alpha_{k}(t)\bigr)z_{x}z_{t}\bigr] \,dx\bigr|^{T}_{0}}. \end{aligned}$$

(3.14)

Next, we estimate every term in the right side of (3.14).

For each \(t\in[0, T]\), we have

$$\begin{aligned} \begin{aligned} &\biggl\vert \int_{0}^{\alpha_{k}(t)}\bigl[\bigl(x- \alpha_{k}(t)\bigr)z_{x}z_{t}\bigr]\,dx\biggr\vert \\ &\leq\int_{0}^{\alpha_{k}(t)}\bigl(\alpha_{k}(t)-x \bigr)\frac{1}{2}\bigl(z_{x}^{2}+z_{t}^{2} \bigr)\,dx \\ &\leq\alpha_{k}(t)E(t). \end{aligned} \end{aligned}$$

This inequality implies that

$$ \biggl\vert {\bigl.\int_{0}^{\alpha_{k}(t)}\bigl[ \bigl(x-\alpha_{k}(t)\bigr)z_{x}z_{t}\bigr]\,dx \bigr|^{T}_{0}}\biggr\vert \leq\alpha_{k}(T)E(T)+E_{0}. $$

(3.15)

Further,

$$ \biggl\vert k\int_{0}^{T}\int _{0}^{\alpha_{k}(t)}z_{x}z_{t}\,dx\,dt \biggr\vert \leq k\int_{0}^{T}E(t)\,dt. $$

(3.16)

*Step* 2. In the following, we give the proof of the first inequality in (2.2).

From (3.8), (3.11), and (3.14)-(3.16), it follows that

$$\begin{aligned}& \frac{1}{2}\int_{0}^{T} \alpha_{k}(t)z_{x}^{2}(0,t)\,dx \\& \quad \geq(1-k)\int_{0}^{T}E(t)\,dt- \alpha_{k}(T)E(T)-E_{0} \\& \quad \geq(1-k)\int_{0}^{T}\frac{1-k}{(1+k)\alpha_{k}(t)}E_{0} \,dt-\frac{1+k}{1-k}E_{0}-E_{0} \\& \quad = \biggl[\frac{(1-k)^{2}}{(1+k)k}\ln(1+kT)-\frac{1+k}{1-k}-1 \biggr]E_{0}. \end{aligned}$$

If \(T>T^{*}_{k}\), we have \(\frac{(1-k)^{2}}{(1+k)k}\ln(1+kT)-\frac {1+k}{1-k}-1>0\). Also,

$$ \int_{0}^{T}\alpha_{k}(t)z_{x}^{2}(0,t) \,dx \geq C \biggl[\frac{(1-k)^{2}}{(1+k)k}\ln(1+kT)-\frac{1+k}{1-k}-1 \biggr] \bigl( \bigl\vert z^{0}\bigr\vert ^{2}_{H^{1}_{0}(0, 1)}+\bigl\vert z^{1}\bigr\vert ^{2}_{L^{2}(0, 1)}\bigr). $$

*Step* 3. In the following, we prove the second inequality in (2.2).

From (3.8), (3.11), and (3.14)-(3.16), one concludes that

$$\begin{aligned}& \frac{1}{2}\int_{0}^{T} \alpha_{k}(t)z_{x}^{2}(0,t)\,dx \\& \quad \leq(1+k)\int_{0}^{T}E(t)\,dt+ \alpha_{k}(T)E(T)+E_{0} \\& \quad \leq(1+k)\int_{0}^{T}\frac{1+k}{(1-k)\alpha_{k}(t)}E_{0} \,dt+\frac{1+k}{1-k}E_{0}+E_{0} \\& \quad = \biggl[\frac{(1+k)^{2}}{(1-k)k}\ln(1+kT)+\frac{1+k}{1-k}+1 \biggr]E_{0} \\& \quad \leq C \biggl[\frac{(1+k)^{2}}{(1-k)k}\ln(1+kT)+\frac{1+k}{1-k}+1 \biggr] \bigl(\bigl\vert z^{0}\bigr\vert ^{2}_{H^{1}_{0}(0, 1)}+\bigl\vert z^{1}\bigr\vert ^{2}_{L^{2}(0, 1)}\bigr). \end{aligned}$$

□

### Remark 3.2

Theorem 2.2 implies that, for any \((z_{0}, z_{1})\in H^{1}_{0}(0, 1)\times L^{2}(0, 1)\), the corresponding solution *z* of (2.1) satisfies \(z_{x}(0, \cdot)\in L^{2}(0, T)\).

### Remark 3.3

It is easy to check that

$$T_{0}^{*}\triangleq\lim_{k\rightarrow0}T^{*}_{k}= \lim_{k\rightarrow0} \frac{e^{\frac{2k(1+k)}{(1-k)^{3}}}-1}{k} =\lim_{k\rightarrow0} \frac{\frac{2k(1+k)}{(1-k)^{3}}}{k} =2. $$

It is well known that the wave equation (1.2) in the cylindrical domain is null controllable at any time \(T>T_{0}^{*}\).

### Remark 3.4

By a similar method, we obtain the following.

Let \(T>\frac{e^{\frac{2k(1+k)}{(1-k)^{2}}}-1}{k}\). For any \((z^{0}, z^{1})\in H_{0}^{1}(0, 1)\times L^{2}(0, 1)\), there exists a constant \(C>0\) such that the corresponding solution *z* of (2.1) satisfies

$$ C \bigl(\bigl\vert z^{0}\bigr\vert ^{2}_{H^{1}_{0}(0, 1)}+ \bigl\vert z^{1}\bigr\vert ^{2}_{L^{2}(0, 1)} \bigr)\leq \int_{0}^{T} \alpha_{k}(t)\bigl\vert z_{x}\bigl(\alpha_{k}(t),t\bigr)\bigr\vert ^{2} \,dt \leq C \bigl(\bigl\vert z^{0}\bigr\vert ^{2}_{H^{1}_{0}(0, 1)}+ \bigl\vert z^{1}\bigr\vert ^{2}_{L^{2}(0, 1)} \bigr). $$

Therefore, we get the controllability of (1.2) when the control is put on the moving endpoint. But the controllability time is larger than that of [7].