Introduction

Let H be a real separable Hilbert space and \(K,\,V\) reflexive Banach spaces with \(H'\) and \(V'\) their respective dual spaces. Identifying H with its dual \(H'\) we will assume that

$$\begin{aligned} K \subset V\subset H\cong H'\subset V'\subset K' \end{aligned}$$
(1.1)

where the inclusions are assumed to be dense and continuous and inclusion \(V\subset H\) compact. We will denote by \(\Vert \cdot \Vert _K,\) \(\Vert \cdot \Vert _V,\) \(\Vert \cdot \Vert ,\) \(\Vert \cdot \Vert _{V'}\) the norms in K,  V,  H,  and \(V'\) respectively. The inner product in H and the duality scalar product between V and \(V'\) will be denoted by \((\cdot ,\cdot )\) and \(\langle \cdot ,\cdot \rangle \) respectively. The constant \(c_{HV}\) is such that \(\Vert v\Vert ^2\le c_{HV}\Vert v\Vert ^2_V\) for all \(v\in V,\) the constant \(c_{HK}\) is such that \(\Vert v\Vert ^2\le c_{HK}\Vert v\Vert ^2_K\) for all \(v\in K.\) We are interested in proving the existence of a global solution to the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{rl} \dfrac{d^2u}{dt^2} - A u = B\left( \dfrac{du}{dt}\right) + f(u), t\ge 0,\\ \left( u(0),\dfrac{du}{dt}(0)\right) =(u_0, u_1)\text { in }V\times H, \end{array} \right. \end{aligned}$$
(1.2)

where \(A: V\rightarrow V'\) is a self-adjoint linear operator, \(B: K\rightarrow K'\) is a monotone operator which has the damping role and satisfies a monotonicity condition H2 and \(f: V\rightarrow H\) is a nonlinear map satisfying suitable conditions (see the conditions H1–H10 in the Sects. 2 and 4).

When \(B=0\), there are a great number of results on the non-existence of global weak solutions of (1.2), see Knops–Straughan [5] and references therein. Existence theorems concerning global solutions of (1.2) have been given by some authors under suitable assumptions. Tsutsumi [15] proved that for \(u_0 \in K \), \(u_1 \in H\) and \( f \in L^2(0,T; H)\) there exists at least one function u such that

$$\begin{aligned}&u \in L^{\infty }(0,T;K),\\&u' \in L^{\infty }(0,T;H) \cap L^2(0,T;V),\\&u'' \in L^{2}(0,T;K'). \end{aligned}$$

Yamada [16] showed the existence of a global weak solution satisfying a certain inequality of energy type and especially, he weakens the assumptions of Tsutsumi so that the result can be applied to a wider class of nonlinear partial differential equations. For the asymptotic behavior of solutions, we mention the work of Nakao [6], where the case \(f(u)=0\) and B is the Fréchet derivative of a nonnegative functional \(F_A(u)\) on K was considered and, for each t, B(t) is a bounded operator from V to \(V'\) . Nakao gives the precise rate of decay as \(t \rightarrow \infty \). (see also [18]).

In [3] the additional term \(B(t)\left( \dfrac{du}{dt}\right) \) was considered, where, for each \(0\le t < \infty \), \(B(t)\,:\, V \rightarrow V'\) is a linear operator associated to the symmetric, bilinear form \(b(t,\,\cdot \,, \,\cdot \,)\) in \(V \times V\) with appropriated conditions, that is,

$$\begin{aligned} \dfrac{d^2u}{dt^2} + A u + B(t)\left( \dfrac{du}{dt}\right) + G\left( \dfrac{du}{dt}\right) = f(t), t\ge 0. \end{aligned}$$
(1.3)

Biazutti proved the global solutions and asymptotic behavior, assuming that the operators A and G are not (necessarily) monotone.

Ang and Dinh [1] have studied the existence and uniqueness of solutions of (1.3) in the special case when A and B(t) are like

$$\begin{aligned} Au = - \sum _{i=1}^n \dfrac{\partial }{\partial x_i} \sigma _i (u(x_i)) \,\,\,\,\text{ and }\,\,\,\, B(t)\left( \dfrac{du}{dt}\right) = - \varDelta u' \end{aligned}$$

and G is the monotone operator

$$\begin{aligned} G u' = |u'|^\alpha sgn(u'), \,\, 0 < \alpha \le 1. \end{aligned}$$

Problem (1.2) was studied in [10, 11] considering \(A=\varDelta ,\) \(B=\varDelta _p\) the p-Laplacian for \(p\ge 2,\) \(H=L^2(D),\) \(V=H^1_0(D),\) \(K=W^{1,p}_0(D),\) \(D\subset \mathbb {R}^n\) to be a bounded domain with smooth boundary and f is such that it includes the case when \(f(u)=|u|^{r-1}u\) for \(1\le r\le 5.\) Besides studying the existence of a solution to the problem (1.2) in [10] a decay of the energy was demonstrated. The case when \(p=2\) was studied in [13] which includes the case when \(f:H^1_0(\Omega )\rightarrow L^2(\Omega )\) is globally Lipschitz continuous.

Pereira et al. [8] considered a Kirchhoff plate equations with internal damping and logarithmic nonlinearity where \( -A = \varDelta ^2 \), \(B(s) = s,\) and \(f(u) = u \ln |u|^2\), given by,

$$\begin{aligned}&u_{tt}+\varDelta ^2 u + M(\Vert \nabla u\Vert ^2)(-\varDelta u) + u_{t} = u \ln |u|^2,\text { in }\Omega \times (0,T), \\&u(x,0) = u_0(x), \quad u_{t}(x,0)=u_1(x),\quad x \in \Omega , \\&u(x,t) = \frac{\partial u}{\partial \eta }(x,t)=0, \quad x \in \partial \Omega , \; t\ge 0, \end{aligned}$$

where \(\Omega \) is a bounded domain in \(\mathbb {R}^2 \) with smooth boundary \(\partial \Omega \), \(T>0\) is a fixed but arbitrary real number, M(s) is a continuous function on \([0,+\infty )\) and \(\eta \) is the unit outward normal on \(\partial \Omega \). The existence of weak solutions by Faedo-Galerkin method and exponential stability following the ideas of Nakao [6] was proved.

An extensible beam equation of Kirchhoff type with internal damping and source term was considered in [9], taking into account \( -A = \varDelta ^2 \), \(B(s) = |s|^{p-1}s\) and \(f(u) = |u|^{q-1}u\),

$$\begin{aligned}&u_{tt} + \varDelta ^2 u + M(|\nabla u|^2)(-\varDelta u) + |u_t|^{p-1}u_t = |u|^{q-1}u ~~ \text { in } \Omega \times (0,T), \\&u(x,0)=u_0(x),\ \ u_t(x,0) = u_1(x), ~~ x\in \Omega ,\\&u(x,t) = \dfrac{\partial u}{\partial \eta }(x,t) = 0, ~~ x\in \partial \Omega , t\ge 0, \end{aligned}$$

where \(p \ge 1, \,q > 1\) are real numbers, \(\Omega \) be a bounded domain in \(\mathbb {R}^n\) with smooth boundary \(\partial \Omega \) and M(s) is a continuous function on \([0,+\infty )\). The global well-posedness was proved and the exponential stability of the solution for \(p = 1\) and the polynomial stability for \(p > 1\) were obtained.

The present work intends to provide a more general framework for the existence of a global solution to the Cauchy problem (1.2), focusing on the damping, which must satisfy the monotonicity condition

$$\begin{aligned} \langle B(u)-B(v),u-v \rangle \le \alpha \Vert u-v\Vert ^2 \end{aligned}$$
(*)

for \(\alpha \) constant and u, v in V. Some authors studied the cases \(B(u)=k\varDelta u\) with \(k>0\) constant (see [13]), \(B(u)=\varDelta _p u\) the p-Laplacian for \(p\ge 2,\) (see [10, 11]) and \(B(u)=-a|u|^{m-1}u\) with \(a>0\) and \(m>1\) constants (see [4]). All these cases satisfy \((*)\) with \(\alpha =0.\) The author [2] studied a second order differential equations of hyperbolic type in Banach spaces involving an operator satisfying \((*)\) with \(\alpha =0.\) Thus, the result of the present paper can be applied to cases not taken into consideration in the previous works because \((*)\) is a generalization of the previous works with \(\alpha = 0\), (see Sect. 5).

The present article is organized in the following way: in Sect. 2, we present the basic spaces, the norms, properties, and notations which we are going to work on within the subsequent sections. In Sect. 3, we provide preliminary results such as the existence of global solution on the assumption that the source term is globally Lipschitz and the existence of local solution when the source is locally Lipschitz. We will use these results in the next section. The Sect. 4 is devoted to the main result: The existence of global solution when the source is locally Lipschitz. Finally, in Sect. 5 we illustrate with examples the result obtained in the present paper.

1 Preliminaries

In order to demonstrate that there is a solution to (1.2) we state the following conditions to the maps A and B :  There is a constant \(\delta \in \mathbb {R}\) and positive constants \(\alpha ,\) \(\beta ,\) and k such that the following conditions hold for all v,  \(v_1,\) \(v_2\in V\):

H1::

(Hemicontinuity) The maps \(s\rightarrow \langle A(v_1+sv_2),v\rangle \) and \(s\rightarrow \langle B(v_1+sv_2),v\rangle \) are continuous on \(\mathbb {R}.\)

H2::

(Monotonicity) B is monotone, that is,

$$\begin{aligned} \langle B(v_1)-B(v_2),v_1-v_2\rangle \le \delta \Vert v_1-v_2\Vert ^2. \end{aligned}$$
H3::

(Coercivity) A is coercive, that is,

$$\begin{aligned} \langle Av,v\rangle \le -\beta \Vert v\Vert _V^2. \end{aligned}$$
H4::

(Coercivity) B is coercive, that is,

$$\begin{aligned} \langle B(v),v\rangle \le -\alpha \Vert v\Vert _K^{q},\,\,\text{ for } \text{ a } \text{ constant }\,\,q>2. \end{aligned}$$
H5::

(Growth)

$$\begin{aligned} \Vert Av\Vert _{V'}\le k\Vert v\Vert _V. \end{aligned}$$

With the aim to resolve the initial value problem (1.2), we will endow V with the following inner product \((v,u)_A=-\langle Au ,v\rangle \), taking into account the coercivity, we can demonstrate that the norm \(\Vert u\Vert _A=\sqrt{-\langle Au,u\rangle }\) is equivalent to norm \(\Vert \cdot \Vert _V\) on V.

Remark 1

Observe that

$$\begin{aligned} \left\langle A u, \dfrac{du}{dt}\right\rangle = -\dfrac{1}{2}\frac{d}{dt}\Vert u\Vert _{A}^2. \end{aligned}$$
(2.1)

Now we need to introduce more notations that we will use throughout the present work. We will denote by

$$\begin{aligned} \mathbb {H}=V\times H \end{aligned}$$

the Banach space endowed with the inner product \((\cdot ,\cdot )_{\mathbb {H}}\) defined as follows. Let \(\mathbb {X}_1=(y_1,z_1)\) and \(\mathbb {X}_2=(y_2,z_2)\) in \(\mathbb {H}\)

$$\begin{aligned} (\mathbb {X}_1,\mathbb {X}_2)_{\mathbb {H}}=(y_1,y_2)_{A}+(z_1,z_2). \end{aligned}$$

In the present work, we understand that u is a solution to the problem (1.2) in the following sense.

Definition 1

Let \(T>0\) be a real number. The map u is called a weak solution of (1.2) on [0, T] if \(u\in C([0,T],V)\), \(\dfrac{du}{dt}\in C([0,T],H)\cap L^p(0,T;K),\) \((u(0),\dfrac{du}{dt}(0))=(u_0,u_1)\in H\times V\) and for all \(\phi \in H^1(0,T;H)\cap L^p(0,T;K),\) u satisfies

$$\begin{aligned}&-\int _0^t\!\!\!\left( \dfrac{du(s)}{ds},\dfrac{d\phi (s)}{ds}\right) ds-\int _0^t\!\!\langle Au(s),v(s)\rangle ds +\left. \left\langle \dfrac{du(s)}{ds},\phi (s)\right\rangle \right| ^{t}_{0}\\&\quad \quad \quad \quad = \int ^t_0\!\!(B(u(s)),\phi (s))ds + \int ^t_0(f(u(s)),\phi (s))ds, \end{aligned}$$

for all \(t\in [0,T].\)

2 Global Solutions

To resolve the initial value problem (1.2), we will reformulate it taking \(\dfrac{du}{dt}=v,\) then the equation (1.2) may be rewritten as:

$$\begin{aligned} \begin{array}{rl} \dfrac{d}{dt} \left( \begin{array}{c}u\\ v \end{array}\right) ^t-\,\left( \begin{array}{c}v\\ Au+B(v)+f(u) \end{array}\right) ^t=\left( \begin{array}{c}0\\ 0 \end{array}\right) ^t \end{array}. \end{aligned}$$
(3.1)

So, defining the operator \(\mathcal {A}\) by

$$\begin{aligned} \mathcal {A}\left( \begin{array}{c}u\\ v \end{array}\right) ^t=\left( \begin{array}{c}-v\\ -Au-B(v)-f(u)^t \end{array}\right) ^t \end{aligned}$$

with domain \(D(\mathcal {A})=\{(u,v)\in V\times K:-Au-Bv-f(u)\in H\}\) we can rewrite the problem (1.2) as

$$\begin{aligned} \left\{ \begin{array}{rl} \dfrac{d}{dt} \left( \begin{array}{c}u\\ v \end{array}\right) ^t&{}+\,\mathcal {A}\left( \begin{array}{c}u\\ v \end{array}\right) ^t=\left( \begin{array}{c}0\\ 0 \end{array}\right) ^t, \\ \\ (u(0),v(0))&{}=(u_0,v_0)\in D(\mathcal {A}). \end{array} \right. \end{aligned}$$
(3.2)

2.1 Globally Lipschitz Sources

Consider the following technical lemma.

Lemma 1

Let \(f \,:\, V \rightarrow H \) be a nonlinear map globally Lipschitz with constant L. For \( a \in V \) and \( z \in H\) the operator \(T:K \rightarrow K'\) defined for \(\lambda >0\) by

$$\begin{aligned} T(z) = -\dfrac{1}{\lambda } Az - B(z) - f\left( \dfrac{a+z}{\lambda }\right) + \lambda z \end{aligned}$$
(3.3)

is surjective.

Proof

In order to prove that T is surjective, it suffices to show that T is monotone, maximal and coercive.

T is monotone: Let \(z_1,z_2\in K\). From H2 and H3 we obtain

$$\begin{aligned} \langle T(z_1) - T(z_2), z_1-z_2\rangle&= \lambda \Vert z_1-z_2\Vert ^2 -\dfrac{1}{\lambda } \langle Az_1-Az_2 ,z_1-z_2 \rangle \\&\,\,\,\,\, - \langle B(z_1)-B(z_2) ,z_1-z_2 \rangle \\&\,\,\,\,\, - \left\langle f\left( \dfrac{a+z_1}{\lambda }\right) - f\left( \dfrac{a+z_2}{\lambda }\right) ,z_1-z_2 \right\rangle \\&\ge (\lambda +\delta )\Vert z_1-z_2\Vert ^2 + \dfrac{\beta }{\lambda }\Vert z_1-z_2\Vert ^2_V \\&\,\,\,\,\, - \dfrac{L}{\lambda }\Vert z_1-z_2\Vert _V \Vert z_1-z_2\Vert \\&\ge \left( \lambda +\delta - \dfrac{1}{4\beta \lambda }\right) \Vert z_1-z_2\Vert ^2 \ge 0, \end{aligned}$$

provided \(\lambda +\delta - \dfrac{1}{4\beta \lambda } \ge 0\).

T is maximal: First we prove that T is hemicontinuous. So we need to prove the weak limit

$$\begin{aligned} \text{ w }-\lim \limits _{\mu \rightarrow 0} T(z_1+\mu z_2) = T(z_1) \end{aligned}$$

for every \(z_1,z_2 \in K\).

Since

$$\begin{aligned} \left\langle T(z_1+\mu z_2), \xi \right\rangle&= \dfrac{1}{\lambda }\langle A(z_1 + \mu z_2), \xi \rangle \langle B(z_1 + \mu z_2), \xi \rangle \\&\quad - \left\langle f\left( \dfrac{a+z_1 + \mu z_2}{\lambda }\right) , \xi \right\rangle + \lambda \langle z_1 + \mu z_2, \xi \rangle \end{aligned}$$

for \(\xi \in V,\) taking into account that

$$\begin{aligned}&\left| \left\langle f\left( \dfrac{a+z_1 + \mu z_2}{\lambda }\right) , \xi \right\rangle - \left\langle f\left( \dfrac{a+z_1}{\lambda }\right) , \xi \right\rangle \right| \\&\quad \quad \quad \le \left| \left\langle f\left( \dfrac{a+z_1 + \mu z_2}{\lambda }\right) -f\left( \dfrac{a+z_1}{\lambda }\right) , \xi \right\rangle \right| \\&\quad \quad \quad \le \left\| f\left( \dfrac{a+z_1 + \mu z_2}{\lambda }\right) -f\left( \dfrac{a+z_1 }{\lambda }\right) \right\| \Vert \xi \Vert _V \\&\quad \quad \quad \le \dfrac{|\mu | L}{\lambda }\Vert z_2\Vert _V\Vert \xi \Vert _V, \end{aligned}$$

from H1, we have

$$\begin{aligned} \lim _{\mu \rightarrow 0}\langle T(z_1+\mu z_2), \xi \rangle =\langle T(z_1), \xi \rangle \text { for }\xi \in V. \end{aligned}$$

Now using theorem 1.3 in [2] we conclude that T is maximal.

T is coercive: We just need to show that

$$\begin{aligned} \dfrac{\langle T(z),z\rangle }{\Vert z\Vert _K} \rightarrow \infty , ~ \text{ as } ~ \Vert z\Vert _K\rightarrow \infty . \end{aligned}$$
(3.4)

From H4 we have that \(\langle -B(z),z\rangle \ge \alpha \Vert z\Vert _K^q\) with \(q>2\). Since

$$\begin{aligned} \langle T(z),z\rangle =&\left\langle -\dfrac{1}{\lambda } Az-B(z) - f\left( \dfrac{a+z}{\lambda }\right) + \lambda z, z \right\rangle \\ =&\dfrac{1}{\lambda }\langle -Az,z\rangle + \langle -Bz,z\rangle - \left\langle f\left( \dfrac{a+z}{\lambda }\right) ,z\right\rangle + \lambda \Vert z\Vert ^2 \\ \ge&\dfrac{\beta }{\lambda }\Vert z\Vert _V^2 + \alpha \Vert z\Vert _K^q - \left\langle f\left( \dfrac{a+z}{\lambda }\right) , z\right\rangle + \lambda \Vert z\Vert ^2. \end{aligned}$$

Concerning the third term in the last line we have that

$$\begin{aligned} \left\langle f\left( \dfrac{a+z}{\lambda }\right) ,z\right\rangle \le&\left\| f\left( \dfrac{a+z}{\lambda }\right) \right\| \Vert z\Vert \\ \le&\left[ \left\| f\left( \dfrac{a+z}{\lambda }\right) - f\left( \dfrac{a}{\lambda }\right) \right\| + \left\| f\left( \dfrac{a}{\lambda }\right) \right\| \right] \Vert z\Vert \\ \le&\dfrac{\beta }{2\lambda }\Vert z\Vert ^2_V + \dfrac{L^2}{2\lambda \beta }\Vert z\Vert ^2 + \dfrac{1}{2}\left\| f\left( \dfrac{a}{\lambda }\right) \right\| ^2 + \dfrac{1}{2}\Vert z\Vert ^2. \end{aligned}$$

Thus

$$\begin{aligned} \langle T(z),z\rangle \ge \dfrac{\beta }{2\lambda } \Vert z\Vert _V^2 + \alpha \Vert z\Vert _K^q + \left( \lambda -\dfrac{L^2}{2\lambda \beta } -\dfrac{1}{2}\right) \Vert z\Vert ^2 - \dfrac{1}{2} \left\| f\left( \dfrac{a}{\lambda }\right) \right\| ^2. \end{aligned}$$

Provided \(\lambda >0\) such that \(\lambda -\dfrac{L^2}{2\lambda \beta } -\dfrac{1}{2}>0, \) because \(q>2\) and \(\left\| f\left( \dfrac{a}{\lambda }\right) \right\| ^2\) is constant we have

$$\begin{aligned} \dfrac{1}{\Vert z\Vert _K} \langle T(z),z\rangle \ge c\Vert z\Vert _K^{q-1} -\dfrac{1}{2} \left\| f\left( \dfrac{a}{\lambda }\right) \right\| ^2 \dfrac{1}{\Vert z\Vert _K} \rightarrow \infty ~ \text{ as } ~ \Vert z\Vert _K \rightarrow \infty . \end{aligned}$$

Hence, T is coercive. By corollary 1.3 in [2], \(T\,:\, K \rightarrow K'\) is surjective.

\(\square \)

The following proposition will demonstrate that when there is a solution \((u,v)\in W^{1,\infty }(0,T;\mathbb {H})\) to the problem (3.2) then u is a weak solution to the problem (1.2) when \((u_0,v_0)\in D(\mathcal {A})\). It means that, \(u\in C([0,T],V),\) \(v\in C([0,T],H),\) \(\dfrac{dv}{dt}\in L^{\infty }(0,T;H),\) \(v\in K,\) a.e. \(t\in [0,T],\) and u satisfies

$$\begin{aligned} \left\{ \begin{array}{rl} &{} \dfrac{d^2u}{dt^2} - A u = B\left( \dfrac{du}{dt}\right) + f(u),\ a.e. t\in [0,T], \\ \\ &{} (u(0),v(0)) =(u_0, v_0)\in D(\mathcal {A}). \end{array} \right. \end{aligned}$$
(3.5)

Proposition 1

Suppose that L is the global Lipschitz constant for \(f:V\rightarrow H \). Then, for arbitrary \(T>0\) there is a unique global weak solution \((u,v)\in W^{1,\infty }(0,T;\mathbb {H})\) with \((u(t),v(t))\in D(\mathcal {A})\) a.e. \(t\in [0,T]\) to the problem (3.2).

Proof

We will prove the proposition by using the Kato’s Theorem (p. 180, [12]), thus it is enough to demonstrate that the operator \(\mathcal {A}+\omega I\) is \(m-\)accretive for some \(\omega >0.\)

Step 1: \(\mathcal {A}+\omega I\) is accretive for some \(\omega >0\).

Let \(\mathbb {X}_1, \mathbb {X}_2 \in D(\mathcal {A)}\) with \(\mathbb {X}_i = (y_i,z_i)\), \(i=1,2\). We need to prove that

$$\begin{aligned} \left( -(\mathcal {A}+\omega I)\mathbb {X}_1 - (-(\mathcal {A}+\omega I)\mathbb {X}_2), \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}} \le 0 \end{aligned}$$

or equivalently

$$\begin{aligned} \left( (\mathcal {A}+\omega I)\mathbb {X}_1 - (\mathcal {A}+\omega I)\mathbb {X}_2, \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}} \ge 0. \end{aligned}$$

Observe that the right hand side of the inequality above can be rewritten in the following way

$$\begin{aligned}&\left( (\mathcal {A}+\omega I)\mathbb {X}_1 - (\mathcal {A}+\omega I)\mathbb {X}_2, \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}} \\&\quad \quad =(\mathcal {A}\mathbb {X}_1-\mathcal {A}\mathbb {X}_2,\mathbb {X}_1-\mathbb {X}_2)_{\mathbb {H}} + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2 \\&\quad \quad = (-(z_1 - z_2),- (A(y_1) - A(y_2))\\&\quad \quad \,\,\,\,\,\, - (B(z_1) - B(z_2)) - (f(y_1) - f(y_2)),\mathbb {X}_1-\mathbb {X}_2)_{\mathbb {H}} + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2 \\&\quad \quad = -(z_1 - z_2 , y_1 - y_2)_A - ( Ay_1 - Ay_2),z_1 - z_2) \\&\quad \quad \,\,\,\,\,\, - ( B(z_1) - B(z_2),z_1 - z_2) - ( f(y_1) - f(y_2),z_1 - z_2) + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2. \end{aligned}$$

By virtue of \(f(y_1)-f(y_2), z_1-z_2 \in H,\) applying the Cauchy–Schwarz inequality and using

$$\begin{aligned} \Vert f(u) - f(v)\Vert \le L \Vert u-v\Vert _{V} \end{aligned}$$

we get

$$\begin{aligned} (f(y_1)- f(y_2), z_1-z_2) \le \Vert f(y_1)-f(y_2)\Vert \Vert z_1-z_2\Vert \le L\Vert y_1-y_2\Vert _{V} \Vert z_1-z_2\Vert . \end{aligned}$$

From Young’s inequality we have

$$\begin{aligned}&\left( (\mathcal {A}+\omega I)\mathbb {X}_1 - (\mathcal {A}+\omega I)\mathbb {X}_2, \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}}\\&\quad \quad \ge -(z_1-z_2,y_1-y_2)_A - (A(y_1-y_2),z_1-z_2) \\&\quad \quad \,\,\,\,\,\,\, - (B(z_1) - B(z_2),z_1 - z_2) \\&\quad \quad \,\,\,\,\,\,\, - \dfrac{L}{2}\Vert y_1-y_2\Vert _V^2 -\dfrac{L}{2}\Vert z_1-z_2\Vert ^2 + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2. \end{aligned}$$

Using H2, we obtain

$$\begin{aligned}&\left( (\mathcal {A}+\omega I)\mathbb {X}_1 - (\mathcal {A}+\omega I)\mathbb {X}_2, \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}} \\&\quad \quad \ge -(z_1-z_2,y_1-y_2)_A - (A(y_1-y_2),z_1-z_2) \\&\quad \quad \,\,\,\,\,\,\, - \dfrac{L}{2}\Vert y_1-y_2\Vert _V^2 -\dfrac{L}{2}\Vert z_1-z_2\Vert ^2 \\&\quad \quad \,\,\,\,\,\,\, + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2-\delta \Vert z_1-z_2\Vert ^2. \end{aligned}$$

From the definition of the inner product \((\cdot ,\cdot )_A\) we have

$$\begin{aligned} - (A(y_1-y_2),z_1-z_2) = (z_1-z_2,y_1-y_2)_A, \end{aligned}$$

then we obtain

$$\begin{aligned}&\left( (\mathcal {A}+\omega I)\mathbb {X}_1 - (\mathcal {A}+\omega I)\mathbb {X}_2, \mathbb {X}_1-\mathbb {X}_2 \right) _{\mathbb {H}}\\&\quad \quad \geqq - \dfrac{L+2\delta }{2}\left( \Vert y_1-y_2\Vert _V^2 + \Vert z_1-z_2\Vert ^2\right) + \omega \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2 \\&\quad \quad = \left( -\dfrac{L+2\delta }{2}+ \omega \right) \Vert \mathbb {X}_1-\mathbb {X}_2\Vert _{\mathbb {H}}^2 \ge 0 \end{aligned}$$

whenever \(\omega \ge \dfrac{L+2\delta }{2}\). Thus \(\mathcal {A}+\omega I\) is accretive.

Step 2: \(\mathcal {A}+\omega I\) is m-accretive.

It suffices to show that \(R(A+\omega I +\eta I) = V\times H\) for some \(\eta >0\). Set \(\omega +\eta = \lambda \) and let \((a,b)\in V\times H\) be given. We need to find \((y,z)\in V\times H\) such that

$$\begin{aligned} \left( \mathcal {A}+\lambda I\right) \left( \begin{array}{c} y \\ z \end{array}\right) ^t = \left( \begin{array}{c} a \\ b \end{array}\right) ^t \end{aligned}$$

that is equivalent to

$$\begin{aligned} \left( \begin{array}{c} -z+\lambda y \\ -Ay-B(z)-f(y)+\lambda z \end{array}\right) ^t = \left( \begin{array}{c} a \\ b \end{array}\right) ^t. \end{aligned}$$

We get \(y=\dfrac{a+z}{\lambda }\) and \(-A\left( \dfrac{a+z}{\lambda }\right) - B(z) - f\left( \dfrac{a+z}{\lambda }\right) + \lambda z = b\) and we deduce that

$$\begin{aligned} -\dfrac{1}{\lambda }Az - B(z) - f\left( \dfrac{a+z}{\lambda }\right) + \lambda z = b + \dfrac{1}{\lambda }Aa = b + \dfrac{1}{\lambda }Aa {\mathop {=}\limits ^{\mathrm{def}}} {\hat{b}} \in V'\subset K'.\nonumber \\ \end{aligned}$$
(3.6)

As \( {\hat{b}} \in K'\) we define \(T:K \rightarrow K'\) by

$$\begin{aligned} T(z) = -\dfrac{1}{\lambda } Az - B(z) - f\left( \dfrac{a+z}{\lambda }\right) + \lambda z. \end{aligned}$$
(3.7)

From lemma 1, \(T:K \rightarrow K'\) is surjective, then (3.6) holds for some \(z \in K\). Thus, given \((a,b)\in H\) and therefore \({\hat{b}} = b +\dfrac{1}{\lambda }Aa \in K'\) we find \(z\in K\) such that \(T(z)={\hat{b}}\). Choosing \(y=\dfrac{a+z}{\lambda }\in V\), we obtain

$$\begin{aligned} -\dfrac{1}{\lambda }Az-B(z)-f\left( \dfrac{a+z}{\lambda }\right) + \lambda z= & {} b +\dfrac{1}{\lambda } Aa \Leftrightarrow -A\left( \dfrac{z+a}{\lambda }\right) \\&- B(z) - f(y) + \lambda z = b \end{aligned}$$

implying

$$\begin{aligned} -Ay-B(z) - f(y) = b-\lambda z \in H,\,\,\, (y,z)\in D(\mathcal {A}) \end{aligned}$$

and therefore \(\mathcal {A}+\omega I\) is m-accretive.

Using Kato’s theorem (p. 180, [12]), there is a unique map \(U=(u,v) \in W^{1,\infty }(0,T;\mathbb {H})\), with \(T>0\) arbitrary, which is a solution to

$$\begin{aligned} \left\{ \begin{array}{rl} &{} \dfrac{d}{dt} \left( \begin{array}{c}u\\ v \end{array}\right) ^t+\,(\mathcal {A} + \omega I)\left( \begin{array}{c}u\\ v \end{array}\right) ^t=\omega \left( \begin{array}{c}u\\ v \end{array}\right) ^t \\ \\ &{} (u(0),v(0))= (u_0,v_0)\in D(\mathcal {A}). \end{array} \right. \end{aligned}$$

or equivalently, \(U=(u,v) \in W^{1,\infty }(0,T,\mathbb {H})\) satisfies

$$\begin{aligned} \left\{ \begin{array}{rl} &{} \dfrac{d}{dt} \left( \begin{array}{c}u\\ v \end{array}\right) ^t-\,\left( \begin{array}{c}v\\ Au+Bv+f(u) \end{array}\right) ^t=\left( \begin{array}{c}0\\ 0 \end{array}\right) ^t \\ \\ &{} (u(0),v(0))=(u_0,v_0)\in D(\mathcal {A}). \end{array} \right. \end{aligned}$$

\(\square \)

2.2 Locally Lipschitz Sources

In this section, we allow the source term f to be locally Lipschitz from V into H.

Lemma 2

Suppose that \(f:V\rightarrow H\) is locally Lipschitz and \((u_0,v_0)\in D(\mathcal {A}).\) Then, there is a unique weak solution u to the problem (1.2) such that \(u\in C([0,T],V),\) \(\dfrac{du}{dt}\in C([0,T],V)\cap L^q(0,T;K),\) \(\dfrac{d^2u}{dt^2}\in L^{\infty }(0,T;H),\) for some \(T>0,\) where T depends on \(\Vert (u_0,v_0)\Vert _{\mathbb {H}}\) and f(0). Furthermore, u satisfies the inequality

$$\begin{aligned}&\frac{1}{2}\left\| \dfrac{du(t)}{dt}\right\| ^2+\frac{1}{2}\Vert u(t)\Vert ^2_A+\alpha \!\!\int ^t_0\!\!\!\left\| \dfrac{du(s)}{ds}\right\| ^q_Kds-\int ^t_0\!\!\!\left( f(u(s)),\dfrac{du(s)}{ds}\right) ds\nonumber \\&\quad \le \dfrac{1}{2}\left\| \frac{du}{dt}(0)\right\| ^2+\dfrac{1}{2}\left\| u(0)\right\| ^2_A \end{aligned}$$
(3.8)

for \(t\in [0,T].\)

Proof

To prove the lemma first, we will truncate the source f

$$\begin{aligned} f_M(u):=\left\{ \begin{array}{rl} &{} f(u), \text {if }\Vert u\Vert _A\le M, \\ \\ &{} f\left( \dfrac{Mu}{\Vert u\Vert _A}\right) ,\text {if } \Vert u\Vert _A>M , \end{array} \right. \end{aligned}$$

where \(M^2> 2(\Vert \dfrac{du}{dt}(0)\Vert ^2_H+\Vert u(0)\Vert ^2_A).\)

We will observe that the mapping \(f_M\,:\, V \rightarrow H\) is globally Lipschitz continuous and we will denote its Lipschitz constant by \(L_M\). Consider the truncated problem

$$\begin{aligned} \left\{ \begin{array}{rl} &{} \dfrac{d^2u(t)}{dt^2}= Au+B\left( \dfrac{du(t)}{dt}\right) +f_M(u), \\ \\ &{}(u(0),\dfrac{du}{dt}(0))=(u_0,v_0)\in D(\mathcal {A}). \end{array} \right. \end{aligned}$$
(3.9)

Thanks to Proposition 1, we can consider \(u^M\) as the global solution such that, \(u^M\in C([0,T],V),\) \(\displaystyle {\dfrac{du^M}{dt}}\in C([0,T],H),\) \(\displaystyle {\dfrac{d^2u^M}{t^2}}\in L^{\infty }(0,T;H),\) and \(\displaystyle {\dfrac{du^M}{dt}(t)}\in K,\) a.e. \(t\in [0,T]\) where \(T>0\) is arbitrarily large. For the sake of convenience, we will denote \(u^M:=u.\) The strong regularity of u allows us to test the equation in (3.9) with \(\dfrac{du}{dt}.\) Thus, using (2.1) and integrating the equation in (3.9) we obtain

$$\begin{aligned} \int _0^t\frac{1}{2}\dfrac{d}{ds}\left\| \frac{du(s)}{ds}\right\| ^2ds&=\int ^t_0-\dfrac{1}{2}\frac{d}{ds}\Vert u\Vert ^2_Ads+\int _0^t\left( B\left( \dfrac{du}{ds}(s),\dfrac{du}{ds}(s)\right) \right) ds\\&\quad \quad +\int ^t_0\left( f_M(u(s)),\dfrac{du}{ds}(s)\right) ds. \end{aligned}$$

Then, by using H4 we have

$$\begin{aligned} \dfrac{1}{2}\left\| \dfrac{du(t)}{dt}\right\| ^2_H+\frac{1}{2}\Vert u(t)\Vert ^2_A+\alpha \int ^t_0\left\| \frac{du(s)}{ds}\right\| ^q_Kds&\le \dfrac{1}{2}\left\| \frac{du}{dt}(0)\right\| ^2+\dfrac{1}{2}\left\| u(0)\right\| ^2_A \nonumber \\&\quad +\int ^t_0\left( f_M(u(s)),\dfrac{du(s)}{ds}\right) ds. \end{aligned}$$
(3.10)

We need to estimate the last two terms of the right hand side of the last inequality. Using the Cauchy–Schwarz and Young’s inequalities and H3, we have

$$\begin{aligned} \left( f_M(u),\dfrac{du}{ds}\right) \le&\Vert f_M(u)\Vert \left\| \dfrac{du}{ds}\right\| \\ \le&c_{HK}\Vert f_M(u)\Vert \left\| \dfrac{du}{ds}\right\| _K \\ \le&c_{HK}c_{\epsilon }\Vert f_M(u)\Vert ^{q'}+\epsilon \left\| \dfrac{du}{ds}\right\| ^q_K \\ \le&c_{HK}c_{\epsilon }\left( \Vert f_M(u)-f_M(0)\Vert +\Vert f_M(0)\Vert \right) ^{q'}+\epsilon \left\| \dfrac{du}{ds}\right\| _K^q \\ \le&c_{HK}c_{\epsilon ,q}L_M^{q'}\Vert u\Vert ^{q'}_V+c_{HK}c_{\epsilon ,q}\Vert f(0))\Vert ^{q'}+\epsilon \left\| \dfrac{du}{ds}\right\| ^q_K \\ \le&\frac{c_{\epsilon ,q}L_M^{q'}c_{HK}}{\beta ^{q'}}\Vert u(t)\Vert ^{q'}_A+c_{HK}c_{\epsilon ,q}\Vert f(0))\Vert _H^{q'}+\epsilon \left\| \dfrac{du}{ds}\right\| ^q_K \end{aligned}$$

where \(\dfrac{1}{q}+\dfrac{1}{q'}=1.\)

Set \(c_M=2c_{\epsilon ,q}L_M^{q'}c_{HK}/\beta ^{q'},\) \(c_f=c_{HK}c_{\epsilon ,q}\Vert f(0))\Vert _H^{q'}\) and

$$\begin{aligned} \xi (t)=\dfrac{1}{2}\left\| \dfrac{du(t)}{dt}(t)\right\| ^2_H+\dfrac{1}{2}\Vert u(t)\Vert ^2_A. \end{aligned}$$

Thus, the last inequality implies that

$$\begin{aligned} \left( f_M(u(t)),\dfrac{du(t)}{dt}\right) \le c_M \xi (t)^{q/2(q-1)}+c_f+\epsilon \left\| \dfrac{du(t)}{dt}\right\| ^q_K. \end{aligned}$$

Since \(q>2,\) we have

$$\begin{aligned} \left( f_M(u(t),\dfrac{du(t)}{dt}\right) \le c_M \xi (t)+c_f+\epsilon \left\| \dfrac{du(t)}{dt}\right\| ^q_K. \end{aligned}$$

Replacing this inequality in (3.10), we have

$$\begin{aligned} \xi (t)+\dfrac{\alpha }{2}\int _0^t\left\| \dfrac{du(s)}{ds}\right\| ^q_Kds\le \xi (0)+C_f T+C_M\int _0^t\xi (s)ds, \end{aligned}$$
(3.11)

for \(\epsilon \) suitable for all \(0\le t\le T.\)

Due to Gronwall’s inequality, we have

$$\begin{aligned} \xi (t)\le (\xi (0)+C_{M,f}T)e^{C_Mt}\le \left( \dfrac{M^2}{4} +C_{M,f}T\right) e^{C_Mt} \end{aligned}$$

for all \(0\le t\le T,\) where \(C_{M,f}\) is a constant that depends on \(c_M\) and \(c_f\).

Taking \(T>0\) such that \(C_{M,f}T\le \dfrac{1}{4}M^2\) the above inequality implies

$$\begin{aligned} \xi (t)\le \dfrac{1}{2}M^2e^{C_Mt}. \end{aligned}$$

Taking \(t\le \dfrac{\ln 2}{C_M},\) we have

$$\begin{aligned} \xi (t)\le \dfrac{M^2}{2}, \end{aligned}$$

consequently, by choosing \(T=\min \left\{ \dfrac{M^2}{4C_{M,f}},\dfrac{\ln 2}{C_M}\right\} ,\) then \(\xi (t)\le \dfrac{M^2}{2}\) for all \(t\in [0,T].\) Therefore, \(f_M(u(t))=f(u(t))\) on the interval [0, T],  implying that u is a solution to the problem (3.9) and as a consequence of the uniqueness of solution for that problem u is a solution to the original problem (1.2) on [0, T] such that \(u\in C([0,T],V),\) \(\dfrac{du}{dt}\in C([0,T],V),\) \(\dfrac{d^2u}{dt}\in L^{\infty }(0,T;H),\) Furthermore, from (3.11) we can state that \(\dfrac{du}{dt}\in L^q(0,T;K).\)

The inequality

$$\begin{aligned}&\dfrac{1}{2}\left\| \dfrac{du(t)}{dt}\right\| ^2+\dfrac{1}{2}\Vert u(t)\Vert ^2_A +\alpha \!\!\int ^t_0\!\!\!\left\| \dfrac{du(s)}{ds}\right\| ^q_Kds-\int ^t_0\!\!\!\left( f(u(s)), \dfrac{du(s)}{ds}\right) ds\\&\quad \le \dfrac{1}{2}\left\| \frac{du}{dt}(0)\right\| ^2+\dfrac{1}{2}\Vert u(0)\Vert ^2_A \end{aligned}$$

for \(t\in [0,T]\) is obtained from (3.10), completing the proof. \(\square \)

3 Global Existence via Potential Well

In this section we use the potential well theory, a powerful tool in the study of the global existence of solution to partial differential equations first developed by Payne and Sattinger [7].

It is well known that the energy of a PDE system is, in some sense, split into kinetic and potential energy. We are able to construct a set of stability, see [17], and prove that there is a valley or a “well” of depth d created in the potential energy. If this height d is strictly positive, we find that, for solutions with initial data in the “good part” of the well, the potential energy of the solution can never escape the well. In general, it is necessary because the energy from the source term can cause the blow-up in finite time. With this approach, we prove the global existence of solutions to (1.2) under different assumptions than that considered in Proposition 1. With this goal we need to introduce the following notation. Consider a nonnegative function \(F:Z\rightarrow \mathbb {R}\) such that \(F(0)=0\), with \(V\subseteq Z\) and the inclusion is assumed to be continuous. We proceed to define the functional \(J: Z \rightarrow \mathbb {R}\) by

$$\begin{aligned} J(u) = \dfrac{1}{2}\Vert u\Vert ^2_{A} -F(u). \end{aligned}$$
(4.1)

For the system (1.2) the full energy is defined by

$$\begin{aligned} E(t) = \dfrac{1}{2}\left\| \dfrac{du(t)}{dt}\right\| ^2 +J(u(t)) \end{aligned}$$
(4.2)

and the positive quadratic energy is given as

$$\begin{aligned} \xi (t) = \dfrac{1}{2}\left\| \frac{du(t)}{dt}\right\| ^2 + \dfrac{1}{2}\Vert u(t)\Vert ^2_A. \end{aligned}$$
(4.3)

We will assume the following additional conditions

H7::

The map F is Gâteaux differentiable and its Gâteaux derivative, \(F'(u,v)\) of F at u in the direction v is given by

$$\begin{aligned} F'(u,v)=(f(u),v), \end{aligned}$$

with f a continuous map.

H8::

There is a real number \(\gamma >2\) such that \((f(u),u)^{1/\gamma }\) is a norm on Z and

$$\begin{aligned} (f(u),u)^{1/\gamma }\le C_3\Vert u\Vert _V \end{aligned}$$
(4.4)

where \(C_3>0\) is a constant. We will denote \(\Vert u\Vert _{Z}:=(f(u),u)^{1/\gamma }.\)

H9::

There is a positive constant \(C_4<\dfrac{1}{2}\) such that

$$\begin{aligned} F(u)\le C_4\Vert u\Vert _Z^{\gamma }. \end{aligned}$$

Associated with the J we have the well known Nehari Manifold given by

$$\begin{aligned} \mathcal {N} := \{u\in V{\setminus } \{0\}; \langle J'(u),u\rangle = 0 \} \end{aligned}$$

and, equivalently,

$$\begin{aligned} \mathcal {N}= \{u\in V{\setminus }\{0\}; \Vert u\Vert ^2_A= (f(u),u)\}. \end{aligned}$$

We define, as in the Mountain Pass Theorem the depth of the well

$$\begin{aligned} d := \inf \limits _{u\in V{\setminus }\{0\}} \sup \limits _{\lambda >0} J(\lambda u). \end{aligned}$$

We introduce the following assumption

H10::

For d above defined we suppose that

$$\begin{aligned} d = \inf _{u \in \mathcal {N}} J(u) > 0. \end{aligned}$$

We now define the potential well \(\mathcal {W}\)

$$\begin{aligned} \mathcal {W} := \{u\in V; J(u)<d \} \end{aligned}$$

and partition it into two sets

$$\begin{aligned} \begin{array}{rl} \mathcal {W}_1 := &{} \{u\in \mathcal {W}; \Vert u\Vert ^2_A > (f(u),u)\} \cup \{0\}, \\ \mathcal {W}_2 := &{} \{u\in \mathcal {W}; \Vert u\Vert ^2_V < (f(u),u)\}. \end{array} \end{aligned}$$

We refer to \(\mathcal {W}_1\) as the “good” part of the well.

Lemma 3

The inequality (3.8) can be rewritten in this way

$$\begin{aligned} E(t)+ \alpha \int ^t_0\left\| \dfrac{du(s)}{ds}\right\| ^q_Kds\le E(0). \end{aligned}$$
(4.5)

Proof

From (3.8), H7 and H9 we get

$$\begin{aligned}&\dfrac{1}{2}\left\| \dfrac{du(t)}{dt}\right\| ^2+\dfrac{1}{2}\Vert u(t)\Vert ^2_A+\alpha \int ^t_0\left\| \dfrac{du(s)}{ds}\right\| ^q_K ds-\int ^t_0\dfrac{d}{ds}(F(u(s)))ds\\&\quad \le \dfrac{1}{2}\left\| \dfrac{du}{dt}(0)\right\| ^2+\dfrac{1}{2}\Vert u(0)\Vert ^2_A, \end{aligned}$$

this inequality implies (4.5). \(\square \)

Theorem 1

Under the assumptions H1–H10, given \(u_0\in \mathcal {W}_1\), \(E(0)<d\). The weak solution u obtained in the Lemma 2 is a global solution and T can be considered arbitrarily large.

Proof

As in [10], we will do our argumentation in three steps.

Step 1: (\(\mathcal {W}_1\) is invariant with respect to (1.2)). From (4.5) we obtain

$$\begin{aligned} E(t)\le E(0)<d, \ {\text {for all}}\, t\in [0,T) \end{aligned}$$
(4.6)

then \(J(u(t))<d\) for all \(t\in [0,T).\) Since \(J(u(t))\le E(t)\) we get that \(u(t)\in \mathcal {W}\) for all \(t\in [0,T).\) Now, we will demonstrate that \(u(t)\in \mathcal {W}_1\) for all \(t\in [0,T).\) Suppose, contrary to our claim, that there is \(t_0\in [0,T)\) such that \(u(t_0)\notin \mathcal {W}_1,\) therefore \(u(t_0)\in \mathcal {W}_2\) and thus \(\Vert u(t_0)\Vert _A^2<(f(u(t_0),u(t_0)).\)

Since \(u\in C([0,T],V)\) and \(V\hookrightarrow Z,\) we have \(\Vert u(t)\Vert _A^2-(f(u(t)),u(t))\) is continuous. As \(u_0\in \mathcal {W}_1\) then

$$\begin{aligned} \Vert u(0)\Vert _A^2-(f(u(0),u(0))>0 \end{aligned}$$

and since

$$\begin{aligned} \Vert u(t_0)\Vert _A^2-(f(u(t_0),u(t_0))<0, \end{aligned}$$

it follows that there exists \(s\in (0,t_0)\) such that

$$\begin{aligned} \Vert u(s)\Vert _A^2=(f(u(s),u(s)). \end{aligned}$$

Therefore we can take

$$\begin{aligned} t^{\star }=\sup \{s\in (0,t_0):\, \Vert u(s)\Vert _A^2=(f(u(s),u(s))\}. \end{aligned}$$
(4.7)

As a consequence \(\Vert u(t^{\star })\Vert ^2_A=(f(u(t^{\star })),u(t^{\star }))\) and \(u(t)\in \mathcal {W}_2\) for all \(t^{\star }<t\le t_0.\) We will take into consideration two possibilities:

Possibility 1: Suppose that \(\Vert u(t^{\star })\Vert _A^2\ne 0.\) Then \(u(t^{\star })\in \mathcal {N}.\) Using our assumption H10 on d we get \(J(u(t^{\star }))\ge d.\) It implies that

$$\begin{aligned} E(t^{\star })=\dfrac{1}{2}\left\| \dfrac{du(t^{\star })}{dt}\right\| _A^2+J(u(t^{\star }))\ge d \end{aligned}$$

which contradicts (4.6).

Possibility 2: Suppose that \(\Vert u(t^{\star })\Vert _A^2=0.\) Since \(u(t)\in \mathcal {W}_2\) for all \(t^{\star }<t\le t_0,\)

$$\begin{aligned} \Vert u(t)\Vert _A^2<(f(u(t),u(t)),\text { for all }t^{\star }<t\le t_0. \end{aligned}$$
(4.8)

The regularity of u implies that

$$\begin{aligned} \lim _{t\rightarrow t^{\star \,+}}\Vert u(t)\Vert _A^2=0. \end{aligned}$$
(4.9)

From H8 and (4.8) we have

$$\begin{aligned} \Vert u(t)\Vert ^2_A<(f(u(t)),u(t))\le C\Vert u(t)\Vert _A^{\gamma }\text { for all }t^{\star }<t\le t_0, \end{aligned}$$

where C is constant depending on \(\gamma .\)

Hence,

$$\begin{aligned} \Vert u(t)\Vert ^2_A(1-C\Vert u(t)\Vert _A^{\gamma -2})<0,\text { for all }t^{\star }<t\le t_0, \end{aligned}$$

leads to

$$\begin{aligned} \Vert u(t)\Vert _A>\left( \dfrac{1}{C}\right) ^{1/(\gamma -2)} \end{aligned}$$

which is a contradiction to (4.9). Thus, \(u(t)\in \mathcal {W}_1\) for all \(t\in [0,T)\) implying that \(\mathcal {W}_1\) is invariant with respect to (1.2).

Step 2: (\(\Vert u(t)\Vert _A\) is controlled by the depth of the well.) Since \(E(t)<d\) and \(u(t)\in \mathcal {W}_1\) for all \(t\in \left[ 0,T\right) ,\)

$$\begin{aligned} d>J(u(t))&=\dfrac{1}{2}\Vert u(t)\Vert _A^2-F(u(t)) \nonumber \\&>\dfrac{1}{2}(f(u(t)),u(t))-F(u(t)) \nonumber \\&=\dfrac{1}{2}\Vert u(t)\Vert _Z^{\gamma }-F(u(t)). \end{aligned}$$
(4.10)

From H9 we get

$$\begin{aligned} d>\dfrac{1}{2}\Vert u(t)\Vert _Z^{\gamma }-C_4\Vert u(t)\Vert _Z^{\gamma }=\left( \dfrac{1}{2}-C_4\right) \Vert u(t)\Vert _Z^{\gamma },\text { for all }t\in \left[ 0,T\right) , \end{aligned}$$

that implies

$$\begin{aligned} F(u(t))\le C_4\Vert u(t)\Vert _Z^{\gamma }<\dfrac{2dC_4}{1-2C_4}\text { for all }t\in \left[ 0,T\right) . \end{aligned}$$
(4.11)

Therefore, since \(E(t)<d\) we obtain

$$\begin{aligned} \dfrac{1}{2}\Vert u(t)\Vert _A^2<d+F(u(t))< d+\dfrac{2dC_4}{1-2C_4}\text { for all }t\in \left[ 0,T\right) . \end{aligned}$$

Step 3: (The solution is a global solution.) Rewriting the inequality (4.5) we get

$$\begin{aligned} \xi (t)+\alpha \int ^t_0\left\| \dfrac{du}{dt}(s)\right\| ^q_Kds\le E(0)+F(u(t))\text { for all }t\in \left[ 0,T\right) . \end{aligned}$$

From (4.11) we obtain

$$\begin{aligned} \xi (t)+\alpha \int ^t_0\left\| \dfrac{du}{dt}(s)\right\| ^q_Kds\le E(0)+F(u(t))<d+\dfrac{2dC_4}{1-2C_4}\text { for all }t\in \left[ 0,T\right) . \end{aligned}$$

Thus

$$\begin{aligned} \xi (t)<d+\dfrac{2dC_4}{1-2C_4}\text { for all }t\in \left[ 0,T\right) . \end{aligned}$$

By standard continuation argument and the Lemma 2, we conclude the proof. \(\square \)

As a consequence of the last theorem we obtain the dissipative property of the system (1.2).

Corollary 1

Let u be a solution of (1.2) with initial data \(u_0\in \mathcal {W}_1\) e \(u_1\in H\). Then

$$\begin{aligned} \dfrac{{\text {d}}}{{\text {d}}\!t} E(t) \le -\alpha \left\| \dfrac{du}{dt}\right\| ^q_{K}. \end{aligned}$$
(4.12)

for a constant \(q>2.\)

Proof

Using the duality we get for \(\dfrac{du(t)}{dt}\in K\)

$$\begin{aligned} \left\langle \dfrac{d^2u}{dt^2} , \dfrac{du}{dt}\right\rangle - \left\langle Au , \dfrac{du}{dt}\right\rangle = \left\langle B\left( \dfrac{du}{dt}\right) , \dfrac{du}{dt} \right\rangle + \left\langle f(u) ,\dfrac{du}{dt} \right\rangle . \end{aligned}$$
(4.13)

Taking into account (1.1) and assumptions H4, (2.1) and H7 it is easy to see that

$$\begin{aligned} \langle \dfrac{d^2u}{dt^2}, \dfrac{du}{dt}\rangle =&\dfrac{1}{2} \dfrac{{\text {d}}}{{\text {d}}\!t} \Vert \dfrac{du}{dt}\Vert ^2, \\ \left\langle Au ,\dfrac{du}{dt} \right\rangle =&-\dfrac{1}{2} \dfrac{{\text {d}}}{{\text {d}}\!t} \Vert u\Vert ^2_A, \\ \left\langle B\left( \dfrac{du}{dt}\right) ,\dfrac{du}{dt} \right\rangle \le&-\alpha \left\| \dfrac{du}{dt}\right\| ^q_{K}, ~~ \text { for } q>2 \\ \left\langle f(u) ,\dfrac{du}{dt} \right\rangle =&\dfrac{{\text {d}}}{{\text {d}}\!t} F(u(t). \end{aligned}$$

Placing these estimates into (4.13) and using the definition of full energy E(t) we get (4.12). \(\square \)

Remark 2

Note that in the previous corollary we have \(E(t)>0\), because we take the initial data in \(\mathcal {W}_1,\) the good part of the well.

4 Application

This section is devoted to providing some examples to illustrate our result.

Example 1

Let \(T>0\) a fixed number. Our first example has a close relation with the works [10, 11]. Specifically we will take \(f\in C^1(\mathbb {R})\) given by

$$\begin{aligned} f(u)=|u|^{r-1} u^{2i+1} \end{aligned}$$
(5.1)

where \(1<r<3,\) \(2<r+2i+1<6\) and \(i=0,1.\) There are two reasons to take this form for f; the first is that this map must satisfy H10. To get it we need to use the Theorem 4.2 from [14], the conditions to use that theorem oblige that f have this appearance; the second reason is that the work [10] considers the case \(f(u)=|u|^{r-1}u\) where \(1<r\le 5\) as the prototype to apply their results (see Pag. 4365), so we need to exhibit a different map to apply our results.

We consider \(H=L^2(D),\) \(V=H^1_0(D),\) \(K=W^{1,p}_0\) where \(D\subset \mathbb {R}^3\) is a bounded domain with sufficiently smooth boundary \(\Gamma \) and \(2<p\) a real number. Moreover, for the coefficients of the equation in the problem (1.2) we will take \(A=a(\cdot )\varDelta ,\) where \(0<m\le a(t)\le M\) for constants m and M and \(0\le t\le T\) and the damping B is given by:

$$\begin{aligned} \begin{array}{rl} B&{}:W^{1,p}_0(D)\rightarrow W^{-1,p'}(D),\\ \\ \langle Bu,\theta \rangle &{}=-a_1(\Vert u\Vert _{1,p})\displaystyle {\int _D}|\nabla u|^{p-2}\nabla u\nabla \theta dx=a(\Vert u\Vert _{1,p})\langle \varDelta _pu,\theta \rangle , \end{array} \end{aligned}$$

where \(\langle \cdot ,\cdot \rangle \) denotes the duality between \(W^{-1,p'}(D)\) and \(W^{1,p}_0(D)\) with \(\dfrac{1}{p}+\dfrac{1}{p'}=1,\) \(\Vert u\Vert _{1,p}\) the norm on \(W^{1,p}_0(D)\) and \(a:\mathbb {R}\rightarrow \mathbb {R}\) is a non-decreasing continuous function such that \(0< m_1\le a(x)\le M_1\) for all \(x \in \mathbb {R}\) and \(m_1\) and \(M_1\) real numbers. \(\varDelta ,\) \(\varDelta _p\) denote the Laplacian and the p-Laplacian, respectively.

Thus, we have the following problem

$$\begin{aligned} \left\{ \begin{array}{rl} &{} \dfrac{d^2u}{dt^2} - A u = B\left( \dfrac{du}{dt}\right) + |u|^{r-1} u^{2i+1},\ \text { in }D\times (0,T), \\ &{} \left( u(0),\dfrac{du(0)}{dt}\right) =(u_0, u_1)\in H^1_0(D)\times L^2(D), \\ &{} u=0 \text { on } \Gamma \times (0,T). \end{array} \right. \end{aligned}$$
(5.2)

It is not hard to demonstrate that A satisfies H1 and also H3 and H5 for suitable constants.

On another hand, the map B satisfies

$$\begin{aligned} \langle B(v_1)-B(v_2),v_1-v_2 \rangle&=-a(\Vert v_1\Vert _{1,p})\Vert v_1\Vert ^p_{1,p}-a(\Vert v_1\Vert _{1,p})\langle \varDelta _pv_1,v_2\rangle \\&\quad -a(\Vert v_2\Vert _{1,p})\langle \varDelta _pv_2,v_1\rangle -a(\Vert v_2\Vert _{1,p})\Vert v_2\Vert ^p_{1,p} \end{aligned}$$

since

$$\begin{aligned}&\displaystyle {\int _D}|\nabla v_1|^{p-2}\nabla v_1\nabla v_2 dx\le \Vert v_1\Vert ^{p-1}_{1,p}\Vert v_2\Vert _{1,p}\text { and }\displaystyle {\int _D}|\nabla v_2|^{p-2}\nabla v_2\nabla v_1 dx\\&\quad \le \Vert v_2\Vert ^{p-1}_{1,p}\Vert v_1\Vert _{1,p} \end{aligned}$$

and using the fact that the function a is non-negative and non-decreasing

$$\begin{aligned} \langle B(v_1)-B(v_2),v_1-v_2 \rangle&\le -a(\Vert v_1\Vert _{1,p})\Vert v_1\Vert ^p_{1,p}+a(\Vert v_1\Vert _{1,p}) \Vert v_1\Vert ^{p-1}_{1,p}\Vert v_2\Vert _{1,p} \\&\quad + a(\Vert v_2\Vert _{1,p}) \Vert v_2\Vert ^{p-1}_{1,p}\Vert v_1\Vert _{1,p}-a(\Vert v_2\Vert _{1,p})\Vert v_2\Vert ^p_{1,p} \\&\le -(a(\Vert v_1\Vert _{1,p})\Vert v_1\Vert ^{p-1}_{1,p} \\&\quad - a(\Vert v_2\Vert _{1,p})\Vert v_2\Vert ^{p-1}_{1,p})(\Vert v_1\Vert _{1,p}-\Vert v_2\Vert _{1,p})\le 0. \end{aligned}$$

Thus, H2 is satisfied with \(\delta =0\). Moreover, the map B satisfies the assumption H4 with \(\alpha =m_1\) and \(q=p\).

To demonstrate H7, we can take

$$\begin{aligned} Z=L^{r+1+2i}(D),\quad F(u)=\dfrac{1}{r+1+2i}\Vert u\Vert _{L^{r+1+2i}(D)}^{r+1+2i}, \end{aligned}$$

it is possible to demonstrate that

$$\begin{aligned} F'(u,v)=(|u|^{r-1}u^{2i+1},v)\text { for } u,v\in L^{r+1}(D). \end{aligned}$$

To verify H8, we observe that

$$\begin{aligned} (f(u),u)=\Vert u\Vert _{L^{r+1+2i}(D)}^{r+1+2i} \end{aligned}$$

and since \(H^1_0(D)\hookrightarrow L^{r+1+2i}(D),\) H8 is satisfied to \(\gamma =r+1+2i\) and an adequate constant \(C_3\).

With respect to the assumption H9, it is satisfied with \(C_4=\dfrac{1}{r+1+2i}.\)

In order to obtain a global solution to the problem 5.2, using the Theorem 1 we need to describe the correspondent objects with 4.1, 4.2 and 4.3.

The full energy E(t) of a system is split into kinetic and potential energy. For the system (5.2) the full energy is given by

$$\begin{aligned} E(t) = \dfrac{1}{2}\left\| \dfrac{du}{dt}\right\| ^2 + \dfrac{1}{2}\Vert u\Vert _A^2 - \dfrac{1}{r+1+2i}\Vert u\Vert ^{r+1+2i}_{r+1+2i}. \end{aligned}$$

We define the functional \(J: V \rightarrow \mathbb {R}\)

$$\begin{aligned} J(u) = \dfrac{1}{2}\Vert u\Vert _A^2 - \dfrac{1}{r+1+2i}\Vert u\Vert ^{r+1+2i}_{r+1+2i}. \end{aligned}$$
(5.3)

Then, we write

$$\begin{aligned} E(t) = \dfrac{1}{2}\left\| \dfrac{du}{dt}\right\| ^2 + J(u(t)) \end{aligned}$$
(5.4)

and the positive quadratic energy as

$$\begin{aligned} \xi (t) = \dfrac{1}{2}\left\| \dfrac{du}{dt}\right\| ^2 + \dfrac{1}{2}\Vert u\Vert _A^2. \end{aligned}$$
(5.5)

The Gâteaux derivative, \(J'(u,v)\) of J at u in the direction v is given by

$$\begin{aligned} J'(u,u) = -(v,u)_A - \left( |u|^{r-1}u^{2i+1},v\right) . \end{aligned}$$

Thus, the Nehari Manifold

$$\begin{aligned} \mathcal {N}= \{u\in V{\setminus } \{0\}; \langle J'(u),u\rangle = 0 \} \end{aligned}$$

where \(J'(u)\) is the Frechét derivate at u.

Equivalently,

$$\begin{aligned} \mathcal {N}= \{u\in V{\setminus }\{0\}; \Vert u\Vert ^2_A = \Vert u\Vert ^{r+1+2i}_{r+1+2i}\}. \end{aligned}$$

Analogously,

$$\begin{aligned} d = \inf \limits _{u\in V{\setminus }\{0\}} \sup \limits _{\lambda >0} J(\lambda u). \end{aligned}$$

Moreover, since \(1<r+1+2i< 6\) from [14] (Theorem 4.2) we have that H10 is satisfied. So, the potential well \(\mathcal {W}\) is defined by

$$\begin{aligned} \mathcal {W}= \{u\in V; J(u)<d \} \end{aligned}$$

and partition it into the two sets

$$\begin{aligned} \begin{array}{rl} \mathcal {W}_1= &{} \{u\in \mathcal {W}; \Vert u\Vert ^2_A > \Vert u\Vert ^{r+1+2i}_{r+1+2i}\} \cup \{0\}, \\ \mathcal {W}_2= &{} \{u\in \mathcal {W}; \Vert u\Vert ^2_A < \Vert u\Vert ^{r+1+2i}_{r+1+2i}\}. \end{array} \end{aligned}$$

To apply the Theorem 1 to get a global solution to the system (5.2) there only remains to prove that \(f:H^1_0(D)\rightarrow L^2(D)\) is locally Lipschitz. In fact, it is a consequence of the inequality

$$\begin{aligned} \Vert |u|^{r-1}u^{1+2i}-|v|^{r-1}v^{1+2i}\Vert ^2_{L^2(D)}\le&c_{r}\Vert (|u|^{r-1+2i}+|v|^{r-1+2i})(u-v)\Vert _{L^2(D)}\\ \le&c_{r}\Vert |u|^{r-1+2i}+|v|^{r-1+2i}\Vert ^2_{L^{\left( \frac{(r+1+2i)2}{r-1+2i}\right) }(D)}\\&\Vert u-v\Vert _{L^{r+1+2i}(D)}^2\\ \le&c_{r}(\Vert u\Vert ^{2(r-1+2i)}_{L^{2(r+1+2i)}(D)}+\Vert v\Vert ^{2(r-1+2i)}_{L^{2(r+1+2i)}(D)})\\&\Vert u-v\Vert _{H^1_0(D)}^2, \end{aligned}$$

where \(c_r\) is a constant that depends on r.

Then, under the conditions of Theorem 1 we have a global solution to the system (5.2).

Example 2

For the second example, we will take \(f\in C^1(\mathbb {R})\) and A as the example 1. Furthermore, we consider \(H=L^2(D),\) \(V=H^1_0(D),\) \(K=W^{1,p}_0\) where \(D\subset \mathbb {R}^3\) is a bounded domain with sufficiently smooth boundary \(\Gamma \) and \(2<p\) a real number. Moreover, for coefficients of the equation in the problem (1.2) we will take B given by

$$\begin{aligned} \begin{array}{rl} B\,\,:\,\,W^{1,p}_0(D)\rightarrow W^{-1,p'}(D),\\ \\ \langle B(u),\phi \rangle =\langle \varDelta _pu,\phi \rangle -\displaystyle {\int _D}u\phi dx,\text { for }\phi \in W^{1,p}_0(D). \end{array} \end{aligned}$$
(5.6)

Thus, we have the following problem

$$\begin{aligned} \left\{ \begin{array}{rl} \dfrac{d^2u}{dt^2} - A u =B\left( \dfrac{du}{dt}\right) + f(u),\ \text { in }D\times (0,T),\\ \left( u(0),\dfrac{du}{dt}(0)\right) =(u_0, u_1)\in H^1_0(D)\times L^2(D),\\ u=0 \text { on }\Gamma \times (0,T). \end{array} \right. \end{aligned}$$
(5.7)

We claim that the problem (5.7) has a global solution. In fact, it is not hard to demonstrate that the map B satisfies H1, due to the fact that \(\varDelta _p\) satisfies H2 with \(\delta =0\) (see [11]) the map B satisfies H2 with \(\delta =-1.\) In fact,

$$\begin{aligned} \langle B(v_1)-B(v_2),v_1-v_2\rangle =&\langle \varDelta _p(v_1)-\varDelta _p(v_2),v_1-v_2\rangle -\int _D|v_1-v_2|^2dx\\ \le&-\Vert v_1-v_2\Vert ^2_{L^2(D)}. \end{aligned}$$

About H4, since

$$\begin{aligned} \langle B(v),v\rangle =&\langle \varDelta _p(v),v\rangle -\int _D|v|^2dx\\ \le&-\Vert v\Vert _{W^{1,p}_0(D)}-\Vert v\Vert ^2_{L^2(D)} \\ \le&-\Vert v\Vert _{W^{1,p}_0(D)}, \end{aligned}$$

then the map B satisfies H4 with \(\alpha =1\) and \(q=p.\) As the map A satisfies the assumptions H1, H3 and H5–H9, along with H10, then we demonstrate the claim.

5 Final Comments

The present manuscript contributes to increment the literature in partial differential equations theory. The objective is to study, in a general framework, a Cauchy problem involving a wave equation with damping operator on Banach spaces with the presence of a nonlinear source term. The main focus is on the damping, which was considered to satisfy a monotonicity condition. This condition is a generalization of some cases studied before by some authors and allow us to provide examples not considered before. Although the focus is on the damping, our examples try to consider a nonlinear source term different than other authors.