Global Existence and Energy Decay of Solutions for a Nondissipative Wave Equation with a Time-Varying Delay Term

Abstract

We consider the energy decay for a nondissipative wave equation in a bounded domain with a time-varying delay term in the internal feedback. We use an approach introduced by Guesmia which leads to decay estimates (known in the dissipative case) when the integral inequalities method due to Haraux-Komornik (Haraux in Nonlinear Partial Differential Equations and Their Applications. Collège de France seminar, Vol. VII (Paris, 1983–1984), pp. 161–179, 1985; Komornik in Exact Controllability and Stabilization: The Multiplier Method, 1994) cannot be applied due to the lack of dissipativity. First, we study the stability of a nonlinear wave equation of the form

in a bounded domain. We consider the general case with a nonlinear function h satisfying a smallness condition and obtain the decay of solutions under a relation between the weight of the delay term in the feedback and the weight of the term without delay. We impose no control on the sign of the derivative of the energy related to the above equation. In the second case we take θ≡const and h(∇u)=−∇Φ⋅∇u. We prove an exponential decay result of the energy without any smallness condition on Φ.

Appendix

We now state some lemmas that we previously used (see proofs in [8, 14]).

Lemma 3

(Sobolev-Poincaré’s inequality)

Let q be a number with 2≤q<+∞ (n=1,2) or 2≤q≤2n/(n−2) (n≥3). Then there is a constant c _∗=c _∗(Ω,q) such that

$$\| u\|_{q}\leq c_{*}\|\nabla u\|_{2} \quad \mbox{\textit{for} } u\in H_{0}^{1}(\varOmega ). $$

Lemma 4

([14])

Let $E: {\mathbb{R}}_{+}\to{ \mathbb{R}}_{+}$ be a non increasing function and $\phi:{\mathbb{R}}_{+}\to{ \mathbb{R}}_{+}$ be an increasing C ¹ function such that

$$\phi(0)=0 \quad\mbox{\textit{and}} \quad\phi(t)\to+\infty\quad\hbox{\textit{as} } t\to+\infty. $$

Assume that there exist σ≥0 and ω>0 such that

$$ \int_S^{+\infty} E^{1+\sigma}(t) \phi'(t) dt \leq\frac{1}{\omega} E^{\sigma}(0)E(S), \quad 0\leq S<+ \infty. $$

(74)

Then

(75)

(76)

In order to state the last lemma, we follow [8, 9] to introduce the function $h:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$. Let r be a non-negative real number, α a strictly positive real number, $\omega:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+*}$ and $\lambda:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ two continuous functions. We set ${\tilde{\lambda}}(t)=\int_{0}^{t}\lambda(\tau) d\tau$ and find

$$ \int_{0}^{+\infty} e^{(r+1){\tilde{\lambda}}(t)} dt=+\infty. $$

(77)

For fixed $s\in \mathbb{R}^{+}$, we define the function $I_{s}:\mathbb{R}^{+}\rightarrow \mathbb{R}$ by

$$I_{s}(t)= \bigl(\omega(s) \bigr)^{r+1}\int _{s}^{t}e^{(r+1){\tilde{\lambda}}(\tau )} d\tau-e^{(r+1){\tilde{\lambda}}(s)} \biggl( \bigl(\alpha\omega (0) \bigr)^{r}+r\int_{0}^{s} \bigl(\omega(\tau) \bigr)^{r+1} d\tau \biggr). $$

We have: $I_{s}\in C^{1}(\mathbb{R}^{+})$, $I'_{s}(t)=(\omega(s))^{r+1}e^{(r+1){\tilde{\lambda}}(t)}>0$,

$$\begin{aligned} I_{s}(0)={}& \bigl(\omega(s) \bigr)^{r+1}\int _{s}^{0}e^{(r+1){\tilde{\lambda}}(\tau )} d\tau\\ &{}-e^{(r+1){\tilde{\lambda}}(s)} \biggl( \bigl(\alpha\omega (0) \bigr)^{r}+r\int_{0}^{s} \bigl(\omega(\tau) \bigr)^{r+1} d\tau \biggr)< 0 \end{aligned} $$

and from (77) lim _t→+∞ I _s (t)=+∞. Therefore I _s has a unique root in $\mathbb{R}^{+*}$ which will be noted g(s) whence we define $g:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+*}$ by

$$ I_{s} \bigl(g(s) \bigr)=0, \quad \forall s\geq0. $$

(78)

On the other hand, g is continuous due to the continuity of ω. Also we have

$$I_{s}(s)=-e^{(r+1){\tilde{\lambda}}(s)} \biggl( \bigl(\alpha\omega(0) \bigr)^{r}+r\int_{0}^{s} \bigl(\omega( \tau) \bigr)^{r+1} d\tau \biggr)<0, $$

hence g(s)>s, and lim _s→+∞ g(s)=+∞. Therefore, g is surjective from $\mathbb{R}^{+}$ to [g(0),+∞[. Now let t∈ ]g(0),+∞[ be fixed. We define the function $J_{t}:[0, t]\rightarrow \mathbb{R}^{+}$ by

The function J _t is positive and differentiable on [0,t] and we have:

Since $J'_{t}(s)$ has the same sign as I _s (t), then $J'_{t}> 0$ holds on the right of 0 (because t>g(0)) and on the left of 0 (because g(s)>s). Then J _t has a maximum on [0,t] at least in one point s ₀∈ ]0,t[ satisfying $I_{s_{0}}(t)=0$, hence s ₀∈g ⁻¹({t}).

Now, we define the function $h: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ by:

$$ h(t)= \left \{ \begin{array}{l@{ \quad}l} 0 & \mbox{if } t\in[0, g(0)], \\ \max g^{-1}(\{t\}) & \mbox{if } t\in\,]g(0), +\infty]. \end{array} \right . $$

(79)

We have, for all t>g(0):h(t)∈g ⁻¹({t}) and I _h(t)(t)=0.

If ω is a constant, then g is an increasing function (it suffices to derive the equality (78)) and in this case

$$h(t)= \everymath{\displaystyle } \left \{ \begin{array}{l@{ \quad}l} 0 & \mbox{if } t\in\biggl[0, D^{-1} \biggl(\frac{\alpha^{r}}{\omega} \biggr) \biggr], \\[10pt] g^{-1}\bigl(\{t\}\bigr)=K^{-1}\bigl(D(t)\bigr) & \mbox{if } t\in \biggl]D^{-1} \biggl(\frac{\alpha^{r}}{\omega} \biggr), +\infty \biggr], \end{array} \right . $$

where K and D are two functions defined on $\mathbb{R}^{+}$ by

$$K(t)=D(t)+e^{(r+1){\tilde{\lambda}}(t)} \biggl(rt+ \frac{\alpha^{r}}{ \omega} \biggr), \quad D(t)= \int _{0}^{t} e^{(r+1){\tilde{\lambda}}(\tau)} d\tau. $$

Lemma 5

([8])

Let $E: {\mathbb{R}}_{+}\to{ \mathbb{R}}_{+}$ be a differentiable function, $\lambda\in \mathbb{R}^{+}$, $a_{3}\in \mathbb{R}^{+}$, $a_{1}, a_{2}:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+*}=(0, +\infty)$ and $\lambda: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ three continuous functions. Assume that there exist r,p≥0 such that

$$a_{3}(r+1) \sup_{t\geq0} \bigl\{\lambda(t) \bigr\}< 1 $$

and for all 0≤s≤T<+∞

$$ \begin{cases} \int_s^{T} E^{r+1}(t) dt\leq a_{1} E (s)+ a_{2} E^{p+1}(s)+ a_{3}E^{r+1}(T), & \forall0\leq s\leq T, \cr E'(t)\leq\lambda(t) E(t), & \forall t \geq0 . \end{cases} $$

(80)

Then E satisfies the following estimate:

(81)

(82)

where ${\tilde{\lambda}}(t)=\int_{0}^{t}\lambda(\tau) d\tau$, h is defined by (79) with $\alpha=\frac{1}{E(0)}$ and $\omega=\frac{1}{a}$ such that

$$ a(s)=\frac{a_{1}(s)+a_{2}(s) (d(s) )^{p}+a_{3}(s) (d(s) )^{r}}{ 1-a_{3}(r+1)\sup_{t\geq 0} \{\lambda(t) \}} $$

(83)

with

$$\begin{aligned} &d(s)=\min \biggl\{E(0) e^{{\tilde{\lambda}}(s)}, \biggl(\frac{b(0)E(0)}{ f_{0}(s)} \biggr)^{{1}/{(r+1)}} \biggr\} ,\\ & \quad f_{0}(s)=e^{-(r+1){\tilde{\lambda}}(s)} \int _{0}^{s}e^{(r+1){\tilde{\lambda}}(\tau)} d\tau \end{aligned} $$

and

$$b(s)=\frac{a_{1}(s)+a_{2}(s)E^{p}(s)+a_{3}(s)E^{r}(s)}{ 1-a_{3}(r+1)\sup_{t\geq 0} \{\lambda(t) \}}, \quad \forall s \geq0. $$