In this section we state and prove our main result. For this purpose we establish several lemmas.

**Lemma 3.1** *Let* y(\cdot ,\cdot ) *be the solution for system* (7a)-(7d). *Then*

{\int}_{0}^{1}{y}_{x}{y}_{xt}\phantom{\rule{0.2em}{0ex}}dx={y}_{x}(1,t){y}_{t}(1,t)-{\int}_{0}^{1}{y}_{xx}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx,

(11)

{\int}_{0}^{1}x{y}_{x}{y}_{xx}\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{2}{y}_{x}^{2}(1,t)-\frac{1}{2}{\int}_{0}^{1}{y}_{x}^{2}(x,t)\phantom{\rule{0.2em}{0ex}}dx,

(12)

{\int}_{0}^{1}x{y}_{x}{y}_{tt}\phantom{\rule{0.2em}{0ex}}dx={\int}_{0}^{1}({[x{y}_{x}{y}_{t}]}_{t}+\frac{1}{2}{y}_{t}^{2})\phantom{\rule{0.2em}{0ex}}dx-\frac{1}{2}{y}_{t}^{2}(1,t).

(13)

*Proof* See the Appendix. □

Now, we give a property of the energy function *E*.

**Proposition 3.1** *The time*-*derivative of the energy function* *E* *in equation* (10), *along the solution of system* (7a)-(7d) *satisfies*

{E}^{\prime}(t)=u(t){y}_{t}(1,t)\le 0

(14)

*for all* t\ge 0.

*Proof* Differentiating the energy function (10) with respect to *t*, we get

{E}^{\prime}(t)={\int}_{0}^{1}{y}_{t}{y}_{tt}\phantom{\rule{0.2em}{0ex}}dx+[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{\int}_{0}^{1}{y}_{x}{y}_{xt}\phantom{\rule{0.2em}{0ex}}dx.

(15)

According to equation (11) and boundary control (7c), we get

\begin{array}{c}[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{\int}_{0}^{1}{y}_{x}{y}_{xt}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}=u(t){y}_{t}(1,t)-[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{\int}_{0}^{1}{y}_{xx}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx.\hfill \end{array}

(16)

Substituting equation (16) into equation (15) and observing (7a), we obtain

\begin{array}{rl}{E}^{\prime}(t)& =u(t){y}_{t}(1,t)+{\int}_{0}^{1}{y}_{tt}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx-[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{\int}_{0}^{1}{y}_{xx}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx\\ =u(t){y}_{t}(1,t)\end{array}

(17)

for all t\ge 0. We obtain equation (14). □

**Remark 3.1** From Proposition 3.1, we obtain the energy identity for system (7a)-(7d),

E(S)-E(T)={\int}_{T}^{S}u(t){y}_{t}(1,t)\phantom{\rule{0.2em}{0ex}}dt.

Therefore, the energy *E* is a decreasing function of time.

During the subsequent stability analysis, we utilize the following inequality.

**Lemma 3.2** *Let* y(\cdot ,\cdot ) *be the solution for system* (7a)-(7d). *Then*

{\int}_{0}^{1}x{y}_{x}(x,t){y}_{t}(x,t)\phantom{\rule{0.2em}{0ex}}dx\le max\{1,\frac{1}{a}\}E(t)

(18)

*for all* t\ge 0.

*Proof* Applying the Cauchy-Schwarz inequality, we get

\begin{array}{rl}{\int}_{0}^{1}x{y}_{x}(x,t){y}_{t}(x,t)\phantom{\rule{0.2em}{0ex}}dx& \le \frac{1}{2}{\int}_{0}^{1}{x}^{2}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{\int}_{0}^{1}{y}_{t}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \le \frac{1}{2}{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{\int}_{0}^{1}{y}_{t}^{2}\phantom{\rule{0.2em}{0ex}}dx\end{array}

(19)

for all t\ge 0. On the other hand, the definition of energy function (10) implies

\frac{a}{2}{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{\int}_{0}^{1}{y}_{t}^{2}\phantom{\rule{0.2em}{0ex}}dx\le E(t)

for all t\ge 0. It follows from the above inequality that

\frac{1}{2}{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{\int}_{0}^{1}{y}_{t}^{2}\phantom{\rule{0.2em}{0ex}}dx\le \{\begin{array}{ll}E(t),& a\ge 1,\\ \frac{1}{a}E(t),& 0<a<1.\end{array}

(20)

Together with (19) and (20), we get equation (18). Hence we complete the proof of Lemma 3.2. □

Now, we present a Gronwall-type lemma (see Komornik [22], pp.124), which will play an essential role when establishing the stabilization result.

**Lemma 3.3** *Let* G:{R}_{+}\to {R}_{+} *be a non*-*increasing function*. *Assume that there exists a constant* \omega >0 *such that*

{\int}_{T}^{+\mathrm{\infty}}{G}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt\le \frac{1}{\omega}G(T)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}T\ge 0.

*Then the following estimation is true*, *for all* t\ge 0,

\{\begin{array}{ll}G(t)\le G(0){e}^{1-\omega t}& \mathit{\text{if}}q=0,\\ G(t)\le G(0){(\frac{1+q}{q\omega t})}^{\frac{1}{q}}& \mathit{\text{if}}q0.\end{array}

We give *a priori* estimation for the energy function E(t), which was established in [15]. For the sake of completeness, we give the proof here.

**Lemma 3.4** *The energy function* *E* *in equation* (10), *along the solution of system* (7a)-(7d), *satisfies*

E(t)\le -{\int}_{0}^{1}{[x{y}_{x}{y}_{t}]}_{t}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{y}_{t}^{2}(1,t)+\frac{1}{2a}{u}^{2}({y}_{t}(1,t))

(21)

*for all* t\ge 0.

*Proof* We multiply equation (7a) by x{y}_{x}(x,t) and do integration over [0,1], with respect to *x*. We obtain

\begin{array}{rl}0=& {\int}_{0}^{1}x{y}_{x}({y}_{tt}-[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{y}_{xx})\phantom{\rule{0.2em}{0ex}}dx\\ =& {\int}_{0}^{1}({[x{y}_{x}{y}_{t}]}_{t}+\frac{1}{2}{y}_{t}^{2})\phantom{\rule{0.2em}{0ex}}dx-\frac{1}{2}{y}_{t}^{2}(1,t)\\ -\frac{1}{2}[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]({y}_{x}^{2}(1,t)-{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx),\end{array}

(22)

using equations (13) and (12) in Lemma 3.1. It follows from (10) that

\begin{array}{c}\frac{1}{2}{\int}_{0}^{1}{y}_{t}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}=E(t)+\frac{b}{4}{\left[{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx\right]}^{2}.\hfill \end{array}

(23)

Since a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx\ge a, according to boundary control (7c), we have

[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{y}_{x}^{2}(1,t)\le \frac{1}{a}{u}^{2}({y}_{t}(1,t)).

(24)

Hence, substituting equation (23) into equation (22) and using equation (24), one has

\begin{array}{c}E(t)+\frac{b}{4}{\left[{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx\right]}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-{\int}_{0}^{1}{[x{y}_{x}{y}_{t}]}_{t}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{y}_{t}^{2}(1,t)+\frac{1}{2}[a+b{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]{y}_{x}^{2}(1,t)\hfill \\ \phantom{\rule{1em}{0ex}}\le -{\int}_{0}^{1}{[x{y}_{x}{y}_{t}]}_{t}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{2}{y}_{t}^{2}(1,t)+\frac{1}{2a}{u}^{2}({y}_{t}(1,t)).\hfill \end{array}

Since \frac{b}{4}{[{\int}_{0}^{1}{y}_{x}^{2}\phantom{\rule{0.2em}{0ex}}dx]}^{2}\ge 0, we complete the proof of Lemma 3.4. □

**Lemma 3.5** *For any constant* q\ge 0, *the energy function* *E* *along the solution of system* (7a)-(7d) *satisfies the following estimation*, *for all* S>T\ge 0,

\begin{array}{rl}{\int}_{T}^{S}{E}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt\le & {C}_{1}{E}^{q+1}(T)+\frac{1}{2}{\int}_{T}^{S}{E}^{q}(t){y}_{t}^{2}(1,t)\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2a}{\int}_{T}^{S}{E}^{q}(t){u}^{2}({y}_{t}(1,t))\phantom{\rule{0.2em}{0ex}}dt,\end{array}

(25)

*where* {C}_{1}=\frac{3q+2}{q+1}max\{1,\frac{1}{a}\}.

*Proof* According to inequality (21) in Lemma 3.4, we have, for all 0\le T<S,

\begin{array}{rl}{\int}_{T}^{S}{E}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt\le & -{\int}_{T}^{S}{E}^{q}(t){\int}_{0}^{1}{[x{y}_{x}{y}_{t}]}_{t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2}{\int}_{T}^{S}{E}^{q}(t){y}_{t}^{2}(1,t)\phantom{\rule{0.2em}{0ex}}dt+\frac{1}{2a}{\int}_{T}^{S}{E}^{q}(t){u}^{2}({y}_{t}(1,t))\phantom{\rule{0.2em}{0ex}}dt.\end{array}

(26)

Moreover, using integration by parts, we get

\begin{array}{rl}{\int}_{T}^{S}{E}^{q}(t){\int}_{0}^{1}{[x{y}_{x}{y}_{t}]}_{t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt=& {\int}_{T}^{S}{[{E}^{q}(t){\int}_{0}^{1}x{y}_{x}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx]}_{t}\phantom{\rule{0.2em}{0ex}}dt\\ -{\int}_{T}^{S}q{E}^{\prime}(t){E}^{q-1}(t){\int}_{0}^{1}(x{y}_{x}{y}_{t})\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.\end{array}

Hence, inequality (26) becomes

\begin{array}{rcl}{\int}_{T}^{S}{E}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt& \le & {A}_{1}+{A}_{2}+\frac{1}{2}{\int}_{T}^{S}{E}^{q}(t){y}_{t}^{2}(1,t)\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2a}{\int}_{T}^{S}{E}^{q}(t){u}^{2}({y}_{t}(1,t))\phantom{\rule{0.2em}{0ex}}dt,\end{array}

(27)

where

\begin{array}{c}{A}_{1}:=-{\int}_{T}^{S}{[{E}^{q}(t){\int}_{0}^{1}x{y}_{x}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx]}_{t}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ {A}_{2}:={\int}_{T}^{S}q{E}^{\prime}(t){E}^{q-1}(t){\int}_{0}^{1}(x{y}_{x}{y}_{t})\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}

Firstly, we estimate {A}_{1} and {A}_{2},

\begin{array}{rcl}{A}_{1}& =& -{\int}_{T}^{S}{[{E}^{q}(t){\int}_{0}^{1}x{y}_{x}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx]}_{t}\phantom{\rule{0.2em}{0ex}}dt\\ \le & |{E}^{q}(T){\int}_{0}^{1}x{y}_{x}(x,T){y}_{t}(x,T)\phantom{\rule{0.2em}{0ex}}dx|+|{E}^{q}(S){\int}_{0}^{1}x{y}_{x}(x,S){y}_{t}(x,S)\phantom{\rule{0.2em}{0ex}}dx|\phantom{\rule{1em}{0ex}}(\text{by (18)})\\ \le & max\{1,\frac{1}{a}\}{E}^{q+1}(T)+max\{1,\frac{1}{a}\}{E}^{q+1}(S)\\ \le & 2max\{1,\frac{1}{a}\}{E}^{q+1}(T),\end{array}

(28)

where the last inequality follows from the fact E(t) is a decreasing function. On the other hand,

\begin{array}{rcl}{A}_{2}& =& {\int}_{T}^{S}q{E}^{\prime}(t){E}^{q-1}(t){\int}_{0}^{1}x{y}_{x}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int}_{T}^{S}|q{E}^{\prime}(t){E}^{q-1}(t)|\left|{\int}_{0}^{1}x{y}_{x}{y}_{t}\phantom{\rule{0.2em}{0ex}}dx\right|\phantom{\rule{0.2em}{0ex}}dt\\ \le & max\{1,\frac{1}{a}\}|{\int}_{T}^{S}q{E}^{\prime}(t){E}^{q}(t)\phantom{\rule{0.2em}{0ex}}dt|\phantom{\rule{1em}{0ex}}(\text{by (18)})\\ =& \frac{q}{q+1}max\{1,\frac{1}{a}\}|{E}^{q+1}(S)-{E}^{q+1}(T)|\\ \le & \frac{q}{q+1}max\{1,\frac{1}{a}\}{E}^{q+1}(T).\end{array}

(29)

Finally, inserting the two inequalities, (28) and (29), in (27), we get inequality (25). This completes the proof of Lemma 3.5. □

We now state the main stabilization result for system (7a)-(7d).

**Theorem 3.1** *Assume that assumption* (H) *holds*. *Then there exist three constants* {k}_{1},{k}_{2},\sigma >0 *such that*, *for all* t\ge 0,

\{\begin{array}{ll}E(t)\le {k}_{1}{e}^{-\sigma t}& \mathit{\text{if}}r=1,\\ E(t)\le {k}_{2}{t}^{-\frac{2}{r-1}}& \mathit{\text{if}}r1.\end{array}

(30)

**Remark 3.2** It is worth to mention that Theorem 2.4 in [13] can be viewed as a special cases of Theorem 3.1. Indeed, in the linear control case (6), the exponential stability in Theorem 3.1 coincides with the result in [13].

*Proof of Theorem 3.1* We distinguish two cases related to the parameter *r* to establish the energy decay rate.

Case (I): r=1;

Case (II): r>1.

In Case (I), we choose q=0. According to hypothesis (H), we know that

{s}^{2}\le -\frac{1}{{L}_{1}}u(s)s,\phantom{\rule{2em}{0ex}}{u}^{2}(s)\le -{L}_{2}u(s)s\phantom{\rule{1em}{0ex}}\text{for all}s\in R.

Hence, from inequality (25) and equation (14), we deduce that, for all S>T\ge 0,

\begin{array}{rl}{\int}_{T}^{S}E(t)\phantom{\rule{0.2em}{0ex}}dt& \le {C}_{1}E(T)+\frac{1}{2}{\int}_{T}^{S}[{y}_{t}^{2}(1,t)+\frac{1}{a}{u}^{2}({y}_{t}(1,t))]\phantom{\rule{0.2em}{0ex}}dt\\ \le {C}_{1}E(T)-\frac{a+{L}_{1}{L}_{2}}{2a{L}_{1}}{\int}_{T}^{S}u(t){y}_{t}(1,t)\phantom{\rule{0.2em}{0ex}}dt\\ ={C}_{1}E(T)+\frac{a+{L}_{1}{L}_{2}}{2a{L}_{1}}{\int}_{T}^{S}-{E}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\\ \le {C}_{2}E(T),\end{array}

(31)

where {C}_{2}={C}_{1}+\frac{1}{2}(\frac{1}{{L}_{1}}+\frac{{L}_{2}}{a}) and {C}_{1} is given in Lemma 3.5.

Now we deal with Case (II). In this case, we choose q=\frac{r-1}{2}>0. We first admit the following fact (the proof is given in the Appendix).

**Claim 1** *For any* \delta >0, *we have the following estimates*, *for all* S>T\ge 0,

\begin{array}{r}{\int}_{T}^{S}{E}^{q}(t){y}_{t}^{2}(1,t)\phantom{\rule{0.2em}{0ex}}dt\le \frac{r-1}{r+1}{\delta}^{\frac{r+1}{r-1}}{\int}_{T}^{S}{E}^{q\frac{r+1}{r-1}}(t)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{{\int}_{T}^{S}{E}^{q}(t){y}_{t}^{2}(1,t)\phantom{\rule{0.2em}{0ex}}dt\le}+\frac{2{\delta}^{-\frac{r+1}{2}}}{(r+1){L}_{1}}E(T)+\frac{1}{(q+1){L}_{1}}{E}^{q+1}(T),\end{array}

(32)

\begin{array}{r}{\int}_{T}^{S}{E}^{q}(t){u}^{2}({y}_{t}(1,t))\phantom{\rule{0.2em}{0ex}}dt\le \frac{r-1}{r+1}{\delta}^{\frac{r+1}{r-1}}{\int}_{T}^{S}{E}^{q\frac{r+1}{r-1}}(t)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{{\int}_{T}^{S}{E}^{q}(t){u}^{2}({y}_{t}(1,t))\phantom{\rule{0.2em}{0ex}}dt\le}+\frac{2{L}_{2}^{r}{\delta}^{-\frac{r+1}{2}}}{(r+1)}E(T)+\frac{{L}_{2}}{(q+1)}{E}^{q+1}(T).\end{array}

(33)

Now, inserting inequalities (32) and (33) into (25), we obtain, for all S>T\ge 0,

\begin{array}{rl}{\int}_{T}^{S}{E}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt\le & (\frac{1}{2}+\frac{1}{2a})\frac{r-1}{r+1}{\delta}^{\frac{r+1}{r-1}}{\int}_{T}^{S}{E}^{q\frac{r+1}{r-1}}(t)\phantom{\rule{0.2em}{0ex}}dt\\ +{C}_{3}{E}^{q+1}(T)+{C}_{4}E(T),\end{array}

(34)

where {C}_{3}={C}_{1}+\frac{a+{L}_{1}{L}_{2}}{2a(q+1){L}_{1}}, {C}_{4}=\frac{1}{(r+1){L}_{1}}{\delta}^{-\frac{r+1}{2}}+\frac{{L}_{2}^{r}}{(r+1)a}{\delta}^{-\frac{r+1}{2}} and {C}_{1} is given in Lemma 3.5. Now we choose \delta ={[(1+\frac{1}{a})\frac{r-1}{r+1}]}^{-\frac{r-1}{r+1}}. Then it is obvious that (1+\frac{1}{a})\frac{r-1}{r+1}{\delta}^{\frac{r+1}{r-1}}=1. Hence, inequality (34) becomes

{\int}_{T}^{S}{E}^{q+1}(t)\phantom{\rule{0.2em}{0ex}}dt\le \frac{1}{2}{\int}_{T}^{S}{E}^{q\frac{r+1}{r-1}}(t)\phantom{\rule{0.2em}{0ex}}dt+{C}_{3}{E}^{q+1}(T)+{C}_{4}E(T).

Recalling q=\frac{r-1}{2}, we get q+1=q\frac{r+1}{r-1}=\frac{r+1}{2}. Hence, the above inequality is rewritten as

\begin{array}{rcl}{\int}_{T}^{S}{E}^{\frac{r+1}{2}}(t)\phantom{\rule{0.2em}{0ex}}dt& \le & 2{C}_{3}{E}^{\frac{r+1}{2}}(T)+2{C}_{4}E(T)\\ \le & 2({C}_{3}{E}^{\frac{r-1}{2}}(0)+{C}_{4})E(T),\end{array}

(35)

where the last inequality follows from Remark 3.1. Finally, by letting S\to +\mathrm{\infty} in (31), (35) and using Lemma 3.3 with G(t)=E(t), we complete the proof of Theorem 3.1. □

**Remark 3.3** According to the proof of Theorem 3.1, it is easy to see that the constants *σ*, {k}_{1} and {k}_{2} in Theorem 3.1 can be chosen as, respectively, {\sigma}^{-1}=2max\{1,\frac{1}{a}\}+\frac{1}{2}(\frac{1}{{L}_{1}}+\frac{{L}_{2}}{a}), {k}_{1}=eE(0), and {k}_{2}=E(0){[2\frac{r+1}{r-1}({C}_{3}{E}^{\frac{r-1}{2}}(0)+{C}_{4})]}^{\frac{2}{r-1}} with {C}_{3}=\frac{3r-1}{r+1}max\{1,\frac{1}{a}\}+\frac{a+{L}_{1}{L}_{2}}{a(r+1){L}_{1}}, {C}_{4}=\frac{1}{r+1}(\frac{1}{{L}_{1}}+\frac{{L}_{2}^{r}}{a}){[(1+\frac{1}{a})\frac{r-1}{r+1}]}^{\frac{r-1}{2}}. This means that the coefficients of the exponential or polynomial decay rate are exactly determined only by the initial tension *a*, the initial energy E(0) and the feedback control *u*. However, in the polynomial decay case, the order of decay rate is determined only by the feedback control *u*.

Finally, it is shown that the boundary control *u* stabilizes the nonlinear Kirchhoff string.

**Theorem 3.2** *Assume that assumption* (H) *holds*. *Then there exist two constants* {k}_{1}^{\prime},{k}_{2}^{\prime}>0 *such that for all* x\in (0,1) *and* t\ge 0,

\{\begin{array}{ll}|y(x,t)|\le {k}_{1}^{\prime}{e}^{-\frac{\sigma t}{2}}& \mathit{\text{if}}r=1,\\ |y(x,t)|\le {k}_{2}^{\prime}{t}^{-\frac{1}{r-1}}& \mathit{\text{if}}r1,\end{array}

(36)

*where* {k}_{1}^{\prime}=\sqrt{\frac{2{k}_{1}}{a}}, {k}_{2}^{\prime}=\sqrt{\frac{2{k}_{2}}{a}} *and* {k}_{1}, {k}_{2}, *σ* *are given in Theorem * 3.1.

*Proof* According to the fact that {y}_{x}(0,t)=0, for all t\ge 0, we get

\begin{array}{rl}|y(x,t)|& =|{\int}_{0}^{x}{y}_{x}(z,t)\phantom{\rule{0.2em}{0ex}}dz|\le {\int}_{0}^{1}|{y}_{x}(z,t)|\phantom{\rule{0.2em}{0ex}}dz\\ \le {({\int}_{0}^{1}{y}_{x}^{2}(x,t)\phantom{\rule{0.2em}{0ex}}dx)}^{\frac{1}{2}}\phantom{\rule{1em}{0ex}}(\text{by the Cauchy inequality})\\ \le \sqrt{\frac{2E(t)}{a}}\phantom{\rule{1em}{0ex}}(\text{by equation (14)})\end{array}

(37)

for all x\in (0,1) and t\ge 0. By combining (37) with (30) in Theorem 3.1, we complete the proof of Theorem 3.2. □