1 Introduction

In a broader sense the term adhesion refers to a physical situation in which two material bodies, during their mechanical evolution, experience a contact interaction which fails in (possibly bounded) regions of space-time and restores after a while. The phenomenon strongly depends on the constitutive properties of the involved materials and a crucial problem relies in understanding the nature of the interaction. The manifestation of such phenomenon occurs at every scale, ranging from DNA molecules to the structural engineering works. Mathematics has looked at these problems, at least in the stationary case, since the seminal paper [1] and in some recent works (see, e.g., [7,8,9,10,11]) one of the authors contributed to the study of the static problem of adhesion of elastic structures by exploiting different constitutive assumptions to the aim of characterizing, in a variational framework, the interplay of the debonding with other constitutive properties.

However the dynamical problem is a different story since at the heart of the question stays the understanding of the attachement-detachement occurrence and how this affects the whole evolution problem. This is a subtle problem since the analytical tools at disposal, such as spatio-temporal estimates in some norms, seem too rough to catch exhaustive quantitative informations, even in short time.

In [3, 4] the simplest mechanical model consisting in elastic string was considered, whereas a discontinuous forcing term was assumed to model the adhesive interaction of the string with a rigid substrate. The resulting mathematical problem is then ruled by an initial boundary value problem for a semilinear second order wave equation and the results in [3, 4] show some tricky peculiarities of the problem. As it is well known in continuum mechanics, the other basic model for one dimensional elastic structures is represented by the so called Bernoulli-Navier beam governing the flexural deformations of a slender material body. In the linear elastic framework the equation expressing the balance of momentum is ruled by a fourth order spatial differential operator. Analogously to [3], we assume a discontinuous forcing term to model the adhesive interaction of the beam with the external environment. One can visualize as a physical situation an elastic beam connected to a rigid substrate through a foundation made of continuous distribution of elastic-breakable springs.

We study the well posedness of the mathematical problem proving existence of global in time solutions in the natural energy space and exploit some features of the solutions to obtain information about the role played by the attachment-detachment occurrence on the dynamical evolution. Indeed, in [2] it was proved that the main effects induced by the nonlinearity at the transition from attached to detached states consist in a loss of regularity of the solution and in a migration of the total energy through the scales. Here we deepen the analysis by using the linearization condition introduced by Gérard in [5] according to which one can conclude that if the semilinear evolution problem satisfies such condition, then the nonlinear forcing term does not induce any new oscillation or energy concentrations [5]. Furthermore we prove an asymptotic result for the long time behavior in the case of bounded solutions, asserting the occurrence of three mutually exclusive states: the trivial one, the totally detached state, the totally attached state.

The paper is organized as follows. In Sect. 2 we formulate the initial boundary value problem. In Sect. 3 we state the main results of the paper consisting in Theorem 3.2 (Existence of solutions), Theorem 3.3 (Characerization of adhesive states), Theorem 3.4 (Long time behavior), Theorem 3.5 (Linearization property). The proofs of these theorems are given respectively in Sects. 4, 5, 6, 7. In Sect. 8 we provide some examples showing non-uniqueness and lack of smoothness of the solutions.

2 Statement of the Problem

Let us consider an elastic beam under Bernoulli–Navier constitutive assumption, occupying in the reference configuration the interval \([0,L]\subset \mathbb {R}\), the balance of linear momentum delivers the semilinear initial boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho \partial _{tt}^2u=-\mu \partial _{xxxx}^4u-\Phi '\left( u\right) &{}\quad t>0,0<x<L,\\ \partial _{xx}^2u(t,0)=\partial _{xx}^2u(t,L)=0&{}\quad t>0,\\ \partial _{xxx}^3u(t,0)=\partial _{xxx}^3u(t,L)=0&{}\quad t>0,\\ u(0,x)=u_0(x)&{}\quad 0<x<L,\\ \partial _tu(0,x)=u_1(x)&{}\quad 0<x<L. \end{array}\right. } \end{aligned}$$
(2.1)

We shall assume that

  1. (H.1)

    \(\Phi \in C(\mathbb {R})\cap C^2(\mathbb {R}{\setminus }\{1,-1\})\), \(\Phi \) is positive, constant in \((-\infty ,-1]\) and in \([1,\infty )\), convex in \([-1,1]\), decreasing in \([-1,0]\), increasing in [0, 1], and \(\Phi (u)\ge \kappa u^2\) in \([-1,1]\) for some constant \(\kappa >0\);

  2. (H.2)

    \(u_0\in H^2(0,L)\), \(u_1\in L^2(0,L)\);

  3. (H.3)

    \(\rho >0\) is the constant material density and \(\mu >0\) is the elastic stiffness.

As a consequence of (H.1), \(\Phi '\) has a jump discontinuity in \(u=\pm 1\) and

$$\begin{aligned} u\in (-\infty ,-1)\cup (1,\infty )&\Rightarrow \Phi '(u)=0,\\ 0<u<1&\Rightarrow 0<\Phi '(u)\le \lim _{u\rightarrow 1^-}\Phi '(u),\\ -1<u<0&\Rightarrow 0>\Phi '(u)\ge \lim _{u\rightarrow -1^+}\Phi '(u). \end{aligned}$$

The set \(\{|u|=1\}\) of these discontinuities separates the attached zone \(\{|u|<1\}\) from the detached one \(\{|u|>1\}\) since the forcing term \(\Phi '(u)\) is thought to model the elastic breakable interaction between the beam and the external environment. Such forcing term confers to the problem a localized nonlinearity which affects the evolution in a significant way.

To fix ideas, a function satisfying (H.1) is

$$\begin{aligned} \Phi (u)={\left\{ \begin{array}{ll} u^2 &{}\quad \text {if}\quad |u|\le 1,\\ 1 &{}\quad \text {if}\quad |u|> 1. \end{array}\right. } \end{aligned}$$
(2.2)

In particular we have for all \(u\ne \pm 1\)

$$\begin{aligned} \Phi '(u)={\left\{ \begin{array}{ll} 2u &{}\quad \text {if}\quad |u|< 1,\\ 0 &{}\quad \text {if}\quad |u|>1. \end{array}\right. } \end{aligned}$$
(2.3)

The natural energy associated to the Problem (2.1) (i.e. to any solution u to (2.1)), is given at time t by the quantity

$$\begin{aligned} E[u](t)=\int _0^L\left( \frac{\rho (\partial _tu(t,x))^2+\mu (\partial _{xx}^2u(t,x))^2}{2}+\Phi (u(t,x))\right) dx. \end{aligned}$$
(2.4)

In general, the lack of Lipschitz continuity in the nonlinear term \(\Phi '\) suggests we cannot expect the existence of conservative solutions, i.e. solutions that preserve energy. Also the physics underlying the problem, foresees a kind of dissipation when the material is detaching from the substrate and this is accompanied by hysteresis cycles (see e.g. [8, Appendix B]).

3 Main Results

We begin by studying the well-posedness of Problem (2.1) and the regularity of its solutions (see Definition 3.1 below). We prove the existence of Lipshitz continuous dissipative solutions and give examples of distinct solutions to (2.1) which do not depend continuously on the initial conditions.

Definition 3.1

We say that a function \(u:[0,\infty )\times [0,L]\rightarrow \mathbb {R}\) is a dissipative solution of (2.1) if

(i):

\(u\in C([0,\infty )\times [0,L])\);

(ii):

\(\partial _tu,\,\partial _{xx}^2u\in L^\infty (0,\infty ;L^2(0,L))\);

(iii):

u is a weak solution to (2.1), i.e. for every test function \(\varphi \in C^\infty (\mathbb {R}^2)\) with compact support

$$\begin{aligned}&\int _0^\infty \int _0^L \left( \rho u\partial _{tt}^2\varphi +\mu \partial _{xx}^2u\partial _{xx}^2\varphi +h_u\varphi \right) dtdx \nonumber \\&\quad -\int _0^L \rho u_1(x)\varphi (0,x)dx+\int _0^L \rho u_0(x)\partial _t\varphi (0,x)dx=0, \end{aligned}$$
(3.1)

where \(h_u\in \partial \Phi \left( u\right) \), that is the subdifferential of \(\Phi \) at u;

(iv):

u may dissipate energy, i.e. for almost every \(t>0\): \(E[u](t)\le E[u](0)\), namely (see (2.4))

$$\begin{aligned} E[u](t)= & {} \int _0^L\left( \frac{\rho (\partial _tu(t,x))^2+\mu (\partial _{xx}^2u(t,x))^2}{2}+\Phi (u(t,x))\right) dx \nonumber \\\le & {} \int _0^L\left( \frac{\rho (u_1(x))^2+\mu (\partial _{xx}^2u_0(x))^2}{2}+\Phi (u_0(x))\right) dx=E[u](0). \end{aligned}$$
(3.2)

We remind that

$$\begin{aligned} h\in \partial \Phi \left( u\right) \end{aligned}$$

means that \(h:[0,\infty )\times [0,L]\rightarrow \mathbb {R}\) satisfies

$$\begin{aligned} h_u(t,x){\left\{ \begin{array}{ll} =\Phi '(u(t,x)),&{}\text {if} |u(t,x)|<1,\\ =0,&{}\quad \text {if} |u(t,x)|>1,\\ \in [0,\Phi '(1^-)],&{}\quad \text {if} u(t,x)=1,\\ \in [\Phi '(-1^+),0],&{}\quad \text {if} u(t,x)=-1. \end{array}\right. } \end{aligned}$$

Let us state the following theorem asserting the existence of dissipative solutions.

Theorem 3.2

(Existence) If \(\Phi \) satisfies (H.1) and \(u_0\), \(u_1\) satisfy (H.2), then problem (2.1) admits a dissipative solution in the sense of Definition 3.1.

The following result provides a sufficient condition ruling the non-detachment of dissipative solutions in dependence on the initial data.

Theorem 3.3

Let u be a dissipative solution of (2.1). If

$$\begin{aligned} \left\| u_0\right\| _{L^\infty (0,L)}<1\quad \text{ and } \quad E[u](0)<\frac{4\kappa \mu }{3\vee 2\kappa }(\le 2\mu ), \end{aligned}$$
(3.3)

where \(\kappa \) is defined in (H.1), then

$$\begin{aligned} \left\| u\right\| _{L^\infty ((0,\infty )\times (0,L))}<1. \end{aligned}$$
(3.4)

The long time behavior is a very subtle problem for evolutionary partial differential equations, so by restricting the focus on bounded dissipative solutions, we are able to prove the following statement.

Theorem 3.4

(Long time behavior) Let u be a dissipative solution of (2.1) and \(\{t_n\}_{n\in \mathbb {N}}\subset (0,\infty )\) such that \(t_n\rightarrow \infty \). If

$$\begin{aligned} u\in L^\infty ((0,\infty )\times (0,L)), \end{aligned}$$
(3.5)

then there exist a subsequence \(\{t_{n_k}\}_{k\in \mathbb {N}}\) and two constants \(a,\, b\in \mathbb {R}\) such that

$$\begin{aligned} u(t_{n_k},\cdot )\rightharpoonup u_\infty \quad \text {weakly in}\quad H^2(0,L) \,\,\text {as}\,\, k\rightarrow \infty , \end{aligned}$$

where

$$\begin{aligned} u_\infty (x)=ax+b. \end{aligned}$$
(3.6)

Moreover, only one of the following statements can occur

$$\begin{aligned} u_\infty&\equiv 0, \end{aligned}$$
(3.7)
$$\begin{aligned} u_\infty (x)&\ge 1 \text {for every x} \in [0,L], \end{aligned}$$
(3.8)
$$\begin{aligned} u_\infty (x)&\le -1 \text {for every x} \in [0,L]. \end{aligned}$$
(3.9)

The question at the basis of the last subsequent result can be formulated as follows: How the nonlinearity characterizing the forcing term \(\Phi '\) affects the evolution problem? We retain that Gérard’s linearization condition provides a precise mathematical tool to answer to the previous rather vague question. Indeed the absence of further energy concentrations or oscillations due to the nonlinearity, suspected to arise in correspondence of the attachment-detachment process, constitutes an interesting property in itself, also considering the nonuniqueness of the solutions \(\{u_n\}_n\) below.

Theorem 3.5

(Linearization property) Let \(\{u_{0,n}\}_n\subset H^2(0,L),\,\{u_{1,n}\}_n\subset L^2(0,L)\), \(\{u_n\}_n\) be a sequence of dissipative solutions of (2.1) in correspondence of such initial data, and \(v_n\) be the dissipative solution of the linearized problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho \partial _{tt}^2v_n=-\mu \partial _{xxxx}^4v_n&{}\quad t>0,0<x<L,\\ \partial _{xx}^2v_n(t,0)=\partial _{xx}^2v_n(t,L)=0&{}\quad t>0,\\ \partial _{xxx}^3v_n(t,0)=\partial _{xxx}^3v_n(t,L)=0&{}\quad t>0,\\ v_n(0,x)=u_{0,n}(x)&{}\quad 0<x<L,\\ \partial _tv_n(0,x)=u_{1,n}(x)&{}\quad 0<x<L. \end{array}\right. } \end{aligned}$$
(3.10)

If

$$\begin{aligned}&u_{0,n}\rightharpoonup 0\,\, \text {weakly in}\,\, H^2(0,L), \end{aligned}$$
(3.11)
$$\begin{aligned}&u_{1,n}\rightharpoonup 0\,\, \text {weakly in}\,\, L^2(0,L), \end{aligned}$$
(3.12)
$$\begin{aligned}&\limsup _n\int _0^L\left( \frac{\rho u_{1,n}^2+\mu (\partial _{xx}^2u_{0,n})^2}{2}\right) dx<\frac{4\kappa \mu }{3\vee 2\kappa }(\le 2\mu ), \end{aligned}$$
(3.13)

where \(\kappa \) is defined in (H.1), then the following linearization condition holds true

$$\begin{aligned} \left\| \partial _t(u_n- v_n)\right\| _{L^\infty (0,T;L^2(0,L))}+\left\| \partial _{xx}^2(u_n- v_n)\right\| _{L^\infty (0,T;L^2(0,L))}\rightarrow 0 \end{aligned}$$
(3.14)

for every \(T\ge 0\) as \(n\rightarrow \infty \).

4 Existence of Dissipative Solutions

This section is dedicated to the proof of Theorem 3.2.

Our argument is based on the approximation of the Neumann problem (2.1) with a sequence of Neumann problems (4.1) characterized by smooth source terms and smooth initial data. More precisely, let \(\{u_{0,n}\}_{n\in \mathbb {N}},\,\{u_{1,n}\}_{n\in \mathbb {N}}\subset C^\infty ([0,L]),\{\Phi _n\}_{n\in \mathbb {N}}\subset C^\infty (\mathbb {R})\), for every \(n\in \mathbb {N}\) consider the approximating problems

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho \partial _{tt}^2u_n=-\mu \partial _{xxxx}^4u_n-\Phi _n'(u_n) &{}\quad t>0,0<x<L,\\ \partial _{xx}^2u_n(t,0)=\partial _{xx}^2u_n(t,L)=0 &{}\quad t>0,\\ \partial _{xxx}^3u_n(t,0)=\partial _{xxx}^3u_n(t,L)=0 &{}\quad t>0,\\ u_n(0,x)=u_{0,n}(x)&{}\quad 0<x<L,\\ \partial _tu_n(0,x)=u_{1,n}(x)&{}\quad 0<x<L, \end{array}\right. } \end{aligned}$$
(4.1)

where \(\{u_{0,n}\}_{n\in \mathbb {N}},\,\{u_{1,n}\}_{n\in \mathbb {N}},\) \(\{\Phi _n\}_{n\in \mathbb {N}}\) are sequences of smooth approximations of \(u_0,\,u_1\), and \(\Phi \) respectively, i.e. they satisfy the following requirements

$$\begin{aligned} \begin{aligned}&u_{0,n}\rightarrow u_0\quad \text {in}\,\, H^1(0,L),\quad u_{1,n}\rightarrow u_1\quad \text {in}\,\, L^2(0,L),\quad \Phi _n\rightarrow \Phi \quad \text {uniformly in}\,\, \mathbb {R},\\&\Phi _n'\rightarrow \Phi ' \quad \text {pointwise and uniformly in}\,\, \mathbb {R}{\setminus }\left( (-1-\varepsilon ,-1+\varepsilon )\cup (1-\varepsilon ,1+\varepsilon )\right) \text {for every} \varepsilon ,\\&|u|\ge 1+\varepsilon \Rightarrow \Phi _n'(u)=0,\quad \varepsilon>0,\> n\in \mathbb {N},\\&\left\| u_{0,n}\right\| _{H^1(0,L)}\le C,\quad \left\| u_{1,n}\right\| _{L^2(0,L)}\le C, \quad 0\le \Phi _n,\,\Phi _n'\le C,\quad n\in \mathbb {N},\\&u_{0,n}^{\prime \prime }(0)=u_{0,n}^{\prime \prime }(L)=u_{1,n}(0)=u_{1,n}(L)=0,\quad n\in \mathbb {N}, \end{aligned} \end{aligned}$$
(4.2)

where C is a positive constant which does not depend on n.

For any \(n\in \mathbb {N}\), (4.1) admits a classical solution for short time thanks to the Cauchy–Kowaleskaya Theorem (see [14]). Furthermore, for such a problem, solutions are indeed global in time thanks to the following results. Let \(u_n\) be the unique classical solution to (4.1).

Lemma 4.1

(Energy conservation). Classical solution to (4.1) preserves energy.

Proof

Set \(E_n:=E[u_n]\) the energy corresponding to \(u_n\). We have to prove that the function \(t\mapsto E_n(t)\) is constant (with constant value \(E_n(0)\)). Indeed, we have that

$$\begin{aligned} E_n'(t)&=\frac{d}{dt}\int _0^L\left( \frac{\rho (\partial _tu_n)^2+\mu (\partial _{xx}^2u_n)^2}{2}+\Phi _n(u_n)\right) dx\\&=\int _0^L\left( \rho \partial _tu_n\partial _{tt}^2u_n+\mu \partial _{xx}^2u_n\partial _{txx}^3u_n+\Phi _n'(u_n)\partial _tu_n\right) dx\\&=\int _0^L\partial _tu_n\underbrace{\left( \rho \partial _{tt}^2u_n+\mu \partial _{xxxx}^4u_n+\Phi _n'(u_n)\right) }_{=0}dx=0. \end{aligned}$$

\(\square \)

As a consequence of energy conservation, since the functions \(\Phi _n\) are positive, we have the following boundedness result.

Corollary 4.2

The sequences \(\{\partial _tu_n\}_{n\in \mathbb {N}}\) and \(\{\partial _{xx}^2u_n\}_{n\in \mathbb {N}}\) are bounded in \(L^\infty (0,\infty ;L^2(0,L))\).

Lemma 4.3

( \(L^2\) estimate). The sequence \(\{u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty (0,T;L^2(0,L))\), for every \(T>0\).

Proof

Using the Hölder inequality

$$\begin{aligned} \int _0^L u_n^2(t,x)dx&=\int _0^L\left( u_{0,n}(x)+\int _0^t \partial _su_n(s,x)ds\right) ^2dx\\&\le 2\int _0^L u_{0,n}^2(x)dx+2 \int _0^L\left( \int _0^t |\partial _su_n(s,x)|ds\right) ^2dx\\&\le 2\int _0^L u_{0,n}^2(x)dx+2t \int _0^t\int _0^L (\partial _su_n(s,x))^2dxds\\&\le 2\int _0^L u_{0,n}^2(x)dx+2t^2 \sup _{s\ge 0}\int _0^L (\partial _su_n(s,x))^2dx , \end{aligned}$$

the claim follows from Corollary 4.2. \(\square \)

The following result follows from Corollary 4.2 by a straightforward application of Gagliardo-Nirenberg Interpolation Inequality (see for instance [12, Theorem at page 125]).

Lemma 4.4

(\(H^1\) estimate) The sequence \(\{\partial _x u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty (0,T;L^2(0,L))\), for every \(T>0\).

Lemma 4.5

(\(L^\infty \) estimate) The sequence \(\{u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty ((0,T)\times (0,L))\), for every \(T>0\).

Proof

Fix \(0<t<T\). Lemmas 4.1 and 4.3 and Corollary 4.2 imply that \(\{u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty (0,T;H^1(0,L))\). Since \(H^1(0,L)\subset L^\infty (0,L)\) we have

$$\begin{aligned} |u_n(t,x)|&\le \left\| u_n(t,\cdot )\right\| _{L^\infty (0,L)}\le c\left\| u_n(t,\cdot )\right\| _{H^1(0,L)}\\&\le c\left\| u_n\right\| _{L^\infty (0,T;H^1(0,L))},\quad (t,x)\in (0,T)\times (0,L), \end{aligned}$$

for some constant \(c>0\) depending only on L. Therefore

$$\begin{aligned} \left\| u_n\right\| _{L^\infty ((0,T)\times (0,L))}\le c\left\| u_n\right\| _{L^\infty (0,T;H^1(0,L))}, \end{aligned}$$

that gives the claim. \(\square \)

Lemma 4.6

(Space Lipschitz estimate). The sequence \(\{\partial _x u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty ((0,T)\times (0,L))\), for every \(T>0\).

Proof

Fix \(0<t<T\) and \(0<x<L\). Lemmas 4.1, 4.3, and 4.4 imply that \(\{u_n\}_{n\in \mathbb {N}}\) is bounded in \(L^\infty (0,T;H^2(0,L))\). Since \(H^1(0,L)\subset L^\infty (0,L)\) we have, for every \((t,x)\in (0,T)\times (0,L)\),

$$\begin{aligned} |\partial _x u_n(t,x)|&\le \left\| \partial _x u_n(t,\cdot )\right\| _{L^\infty (0,L)}\\&\le c\left\| \partial _x u_n(t,\cdot )\right\| _{H^1(0,L)}\le c\left\| u_n(t,\cdot )\right\| _{H^2(0,L)}\\&\le c\left\| u_n\right\| _{L^\infty (0,T;H^2(0,L))}, \end{aligned}$$

for some constant \(c>0\) depending only on L. Therefore

$$\begin{aligned} \left\| \partial _x u_n\right\| _{L^\infty ((0,T)\times (0,L))}\le c\left\| u_n\right\| _{L^\infty (0,T;H^2(0,L))}, \end{aligned}$$

that gives the claim. \(\square \)

Proof of Theorem 3.2

Thanks to Lemmas 4.1, 4.3 and [13, Theorem 5] there exists a function u satisfying items (i) and (ii) in Definition 3.1 and a function \(h_u\in L^\infty ((0,T)\times (0,L)),\, h_u\in \partial \Phi (u),\) such that, passing to a subsequence,

$$\begin{aligned} \begin{aligned} u_n\rightharpoonup u&\quad \text {in} \,\,H^1((0,T)\times (0,L))\,\, \text {and in}\,\, L^2(0,T;H^2(0,L)), \text {for each}\,\, T\ge 0,\\ u_n \rightarrow u&\quad \text {in}\,\, L^\infty ((0,T)\times (0,L)),\,\, \text {for each}\,\, T\ge 0,\\ \Phi _n'(u_n) \rightharpoonup h_u&\quad \text {in}\,\, L^p((0,T)\times (0,L)),\,\, \text {for each}\,\, T\ge 0 \,\,\text {and}\,\, 1\le p<\infty . \end{aligned} \end{aligned}$$
(4.3)

We have yet to verify that u is a weak solution of (2.1) i.e. Definition 3.1-item (iii). Let \(\varphi \in C^\infty (\mathbb {R}^2)\) be a test function with compact support, since \(u_n\) is a solution to (4.1), we have that for every n

$$\begin{aligned} \int _0^\infty \int _0^L&\left( \rho u_n\partial _{tt}^2\varphi +\mu \partial _{xx}^2u_n\partial _{xx}^2\varphi +\Phi _n'(u_n)\varphi \right) dxdt\\&-\int _0^L \rho u_{1,n}(x)\varphi (0,x)dx+\int _0^L \rho u_{0,n}(x)\partial _t\varphi (0,x)dx=0. \end{aligned}$$

Then, by taking the limit as \(n\rightarrow \infty \), (3.1) follows by using (4.2) and (4.3).

Finally, due to (4.2) and (4.3) we have

$$\begin{aligned} \partial _tu_n\rightharpoonup \partial _tu&\quad \text {in}\,\, L^p(0,T;L^2(0,L)), \,\,\text {for each}\,\, T\ge 0\,\, \text {and}\,\, 1\le p<\infty ,\\ \partial _{xx}^2u_n\rightharpoonup \partial _{xx}^2u&\quad \text {in}\,\, L^p(0,T;L^2(0,L)),\,\, \text {for each}\,\, T\ge 0 \,\,\text {and}\,\, 1\le p<\infty ,\\ \Phi _n(u_n) \rightarrow \Phi (u)&\quad \text {in}\,\, L^\infty ((0,T)\times (0,L)),\,\, \text {for each}\,\, T\ge 0. \end{aligned}$$

Therefore, Definition 3.1-item (iv) follows by the lower semicontinuity of the \(L^2\) norm with respect to the weak convergence by taking into account Lemma 4.1. \(\square \)

5 Adhesive States

This section is dedicated to the proof of Theorem 3.3.

Proof of Theorem 3.3

Since u is continuous by (3.3) there exists \(\tau >0\) such that in the short time interval \([0,\tau ]\) we have

$$\begin{aligned} |u(t,x)|<1,\quad (t,x)\in [0,\tau ]\times [0,L]. \end{aligned}$$

So we can define \(\tau ^*\) as follows

$$\begin{aligned} \tau ^*=\sup \left\{ \tau >0 \;|\; |u(t,x)|<1 \,\, \text{ for } \text{ all } \,\,(t,x)\in [0,\tau ]\times [0,L]\right\} . \end{aligned}$$

We claim that

$$\begin{aligned} \tau ^*=\infty . \end{aligned}$$
(5.1)

Observe that, due to (H.1),

$$\begin{aligned} \Phi (u(t,x))\ge \kappa u^2(t,x),\quad (t,x)\in [0,\tau ^*)\times [0,L]. \end{aligned}$$

Therefore

$$\begin{aligned} E[u](t)\ge \int _0^L\left( \frac{\rho (\partial _tu(t,x))^2+\mu (\partial _{xx}^2u(t,x))^2}{2}+\kappa (u(t,x))^2\right) dx,\quad t\in [0,\tau ^*), \end{aligned}$$

and, in particular,

$$\begin{aligned} E[u](t)\ge \frac{\mu \left\| \partial _{xx}^2u(t,\cdot )\right\| _{L^2(0,L)}^2}{2}+\kappa \left\| u(t,\cdot )\right\| _{L^2(0,L)}^2,\quad t\in [0,\tau ^*). \end{aligned}$$

Since u is dissipative, using the Sobolev embedding \(H^1(0,L)\subset L^\infty (0,L)\) (see [6, Theorem 8.5]) and Lemma 4.4, we have from every \(t\in [0,\tau ^*)\)

$$\begin{aligned} \left\| u(t,\cdot )\right\| _{L^\infty (0,L)}^2&\le \frac{\left\| u(t,\cdot )\right\| _{H^1(0,L)}^2}{2}=\frac{\left\| u(t,\cdot )\right\| _{L^2(0,L)}^2+\left\| \partial _x u(t,\cdot )\right\| _{L^2(0,L)}^2}{2}\\&\le \frac{3}{4}\left\| u(t,\cdot )\right\| _{L^2(0,L)}^2+\frac{\left\| \partial _{xx}^2u(t,\cdot )\right\| _{L^2(0,L)}^2}{4}\\&\le \left( \frac{3}{4\kappa }\vee \frac{1}{2}\right) \left( \frac{\left\| \partial _{xx}^2u(t,\cdot )\right\| _{L^2(0,L)}^2}{2}+\kappa \left\| u(t,\cdot )\right\| _{L^2(0,L)}^2\right) \\&\le \frac{1}{\mu }\left( \frac{3}{4\kappa }\vee \frac{1}{2}\right) E[u](t)\le \frac{1}{\mu }\left( \frac{3}{4\kappa }\vee \frac{1}{2}\right) E[u](0)<1, \end{aligned}$$

(where the last inequality holds thanks to (3.3)) that proves (5.1). \(\square \)

6 Long Time Behavior

This section is dedicated to the proof of Theorem 3.4.

Let u be a dissipative solution of (2.1) satisfying (3.5).

STEP 1. We begin by deducing the effective asymptotic problem.

Consider the functions

$$\begin{aligned} u_\tau (t,x)=u(\tau t, x),\quad \tau>0,\>t\ge 0,\>x\in [0,L]. \end{aligned}$$

\(u_\tau \) is a dissipative solution of the initial boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \frac{\rho \partial _{tt}^2u_\tau }{\tau ^2}=-\mu \partial _{xxxx}^4u_\tau -\Phi '\left( u_\tau \right) &{}\quad t>0,0<x<L,\\ \partial _{xx}^2u_\tau (t,0)=\partial _{xx}^2u_\tau (t,L)=0&{}\quad t>0,\\ \partial _{xxx}^3u_\tau (t,0)=\partial _{xxx}^3u_\tau (t,L)=0&{}\quad t>0,\\ u_\tau (0,x)=u_0(x)&{}\quad 0<x<L,\\ \partial _tu_\tau (0,x)=\tau u_1(x)&{}\quad 0<x<L, \end{array}\right. } \end{aligned}$$
(6.1)

in the sense of Definition 3.1, namely

(iii):

for every test function \(\varphi \in C^\infty (\mathbb {R}^2)\) with compact support

$$\begin{aligned} \begin{aligned} \int _0^\infty \int _0^L&\left( -\frac{\rho \partial _tu_\tau }{\tau ^2}\partial _t\varphi + \mu \partial _{xx}^2u_\tau \partial _{xx}^2\varphi +h_{\tau }\varphi \right) dtdx-\int _0^L \frac{\rho u_1(x)}{\tau }\varphi (0,x)dx=0, \end{aligned} \end{aligned}$$
(6.2)

where \(h_{\tau }\in \partial \Phi \left( u_\tau \right) \), that is the subdifferential of \(\Phi \) at \(u_\tau \);

(iv):

\(u_\tau \) may dissipate energy, i.e. for almost every \(t>0\):

$$\begin{aligned} \begin{aligned} \int _0^L&\left( \frac{\rho (\partial _tu_\tau (t,x))^2}{2\tau ^2}+\frac{\mu (\partial _{xx}^2u_\tau (t,x))^2}{2}+\Phi (u_\tau (t,x))\right) dx\\&\quad \le \int _0^L\left( \frac{\rho (u_1(x))^2}{2}+\frac{\mu (\partial _{xx}^2u_0(x))^2}{2}+\Phi (u_0(x))\right) dx. \end{aligned} \end{aligned}$$
(6.3)

Thanks to (H.1), (3.5), and (6.3),

$$\begin{aligned}&\{ u_\tau \}_{\tau>0} \,\,\text {is bounded in}\,\, L^\infty (0,\infty ;H^2(0,L)),\\&\{h_{\tau }\}_{\tau >0} \,\,\text {is bounded in}\,\, L^\infty ((0,\infty )\times (0,L)), \end{aligned}$$

there exists two functions \(U\in L^\infty (0,\infty ;H^2(0,L)),\, H\in L^\infty ((0,\infty )\times (0,L))\) such that, passing to a subsequence,

$$\begin{aligned} u_\tau \overset{\star }{\rightharpoonup }U&\quad \text {weakly}-\star \,\, \text {in}\,\, L^\infty _{loc}((0,\infty )\times (0,L)) \,\,\text {as}\,\, \tau \rightarrow \infty ,\\ h_\tau \overset{\star }{\rightharpoonup }H&\quad \text {weakly}-\star \,\, \text {in}\,\, L^\infty _{loc}((0,\infty )\times (0,L)) \,\,\text {as}\,\, \tau \rightarrow \infty . \end{aligned}$$

Using (6.3)

$$\begin{aligned}&\{\partial _tu_\tau /\tau \}_{n\in \mathbb {N}} \,\,\text {is bounded in}\,\, L^\infty (0,\infty ;L^2(0,L)), \end{aligned}$$

therefore as \(\tau \rightarrow \infty \) in (6.2) we get

$$\begin{aligned} \int _0^\infty \int _0^L \left( \mu \partial _{xx}^2U\partial _{xx}^2\varphi +H\varphi \right) dtdx=0, \end{aligned}$$
(6.4)

namely \(U=U(x)\), \(H=H(x)\) and the effective asymptotic problem is

$$\begin{aligned} {\left\{ \begin{array}{ll} -\mu \partial _{xxxx}^4U=H,&{}\quad 0<x<L,\\ \partial _{xx}^2U(0)= \partial _{xx}^2U(L)=0,&{}{}\\ \partial _{xxx}^3U(0)=\partial _{xxx}^3U(L)=0.&{}{} \end{array}\right. } \end{aligned}$$
(6.5)

STEP 2. We exploit more subtle characterizations of the limit functions U and H. To this aim we fix a sequence \(\{t_n\}_{n\in \mathbb {N}}\subset (0,\infty )\) such that \(t_n\rightarrow \infty \) and study the convergence of the sequence

$$\begin{aligned} \{u(t_n,\cdot )\}_{n\in \mathbb {N}}. \end{aligned}$$

Since we have the dissipation inequality (3.2) and the assumption (3.5), we gain

$$\begin{aligned}&\{u(t_n,\cdot )\}_{n\in \mathbb {N}} \,\,\text {is bounded in}\,\, H^2(0,L),\\&\{h_u(t_n,\cdot )\}_{n\in \mathbb {N}} \,\,\text {is bounded in}\,\, L^\infty (0,L). \end{aligned}$$

Therefore there exist two functions \(u_\infty \in H^2(0,L),\, h_\infty \in L^\infty (0,L)\) such that passing to a subsequence

$$\begin{aligned} \begin{aligned}&u(t_n,\cdot )\rightharpoonup u_\infty \quad \text {weakly in}\,\, H^2(0,L)\,\, \text {as}\,\, n\rightarrow \infty ,\\&u(t_n,\cdot )\rightarrow u_\infty \quad \text {a.e. in}\,\, (0,L) \,\,\text {as}\,\, n\rightarrow \infty ,\\&h_u(t_n,\cdot )\overset{\star }{\rightharpoonup }h_\infty \quad \text {weakly}-\star \,\, \text {in}\,\, L^\infty (0,L)\,\, \text {as}\,\, n\rightarrow \infty . \end{aligned} \end{aligned}$$
(6.6)

Due to the result in STEP 1, we know that the functions \(u_\infty \) and \(h_\infty \) must satisfy the effective problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\mu \partial _{xxxx}^4u_\infty =h_\infty ,&{}\quad 0<x<L,\\ \partial _{xx}^2u_\infty (0)= \partial _{xx}^2u_\infty (L)=0,&{}{}\\ \partial _{xxx}^3u_\infty (0)=\partial _{xxx}^3u_\infty (L)=0.&{}{} \end{array}\right. } \end{aligned}$$
(6.7)

Moreover, by (6.6) we have also that

$$\begin{aligned} h_\infty \in \partial \Phi '(u_\infty ). \end{aligned}$$
(6.8)

By multiplying (6.7) by \(u_\infty \), integrating over (0, L), and recalling (6.8), we get

$$\begin{aligned} \int _0^L \mu (\partial _{xx}^2u_\infty )^2dx=-\int _0^L h_\infty u_\infty dx\le 0, \end{aligned}$$

so

$$\begin{aligned} \partial _{xx}^2u_\infty \equiv 0. \end{aligned}$$

Therefore, we can conclude that (3.6) holds.

Using (3.6) in (6.7) we have also that

$$\begin{aligned} h_\infty \equiv 0, \end{aligned}$$

hence, due to (6.8), only one within (3.7), (3.8), (3.9) can occur.

7 Linearization Property

Proof of Theorem 3.5

Thanks to (3.11)

$$\begin{aligned} u_{0,n}\rightarrow 0\quad \text {uniformly in} [0,L]. \end{aligned}$$

Therefore, using also (3.13), the assumptions of Theorem 3.3 are fulfilled, hence

$$\begin{aligned} \left\| u_n\right\| _{L^\infty ((0,\infty )\times (0,L))}<1. \end{aligned}$$
(7.1)

In addition, the function

$$\begin{aligned} w_n=u_n-v_n \end{aligned}$$

is a dissipative solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho \partial _{tt}^2w_n=-\mu \partial _{xxxx}^4w_n-\Phi '(u_n)&{}\quad t>0,0<x<L,\\ \partial _{xx}^2w_n(t,0)=\partial _{xx}^2w_n(t,L)=0&{}\quad t>0,\\ \partial _{xxx}^3w_n(t,0)=\partial _{xxx}^3w_n(t,L)=0&{}\quad t>0,\\ w_n(0,x)=\partial _tw_n(0,x)=0&{}\quad 0<x<L. \end{array}\right. } \end{aligned}$$
(7.2)

Multiplying (7.2) by \(\partial _tw_n\) we gain

$$\begin{aligned} \frac{d}{dt}\int _0^L\frac{\rho (\partial _tw_n)^2+\mu (\partial _{xx}^2w_n)^2}{2}dx&=\int _0^L\Phi '(u_n)\partial _tw_n dx\\&\le \frac{1}{2\rho }\int _0^L\Phi '(u_n)^2dx+\int _0^L\frac{\rho (\partial _tw_n)^2}{2}dx\\&\le \frac{1}{2\rho }\int _0^L\Phi '(u_n)^2dx+\int _0^L\frac{\rho (\partial _tw_n)^2+\mu (\partial _{xx}^2w_n)^2}{2}dx \end{aligned}$$

and applying the Gronwall Lemma

$$\begin{aligned} \int _0^L\frac{\rho (\partial _tw_n(t,x))^2+\mu (\partial _{xx}^2w_n(t,x))^2}{2}dx\le \frac{1}{2\rho }\int _0^t\int _0^L e^{t-s}\Phi '(u_n(s,x))^2dsdx. \end{aligned}$$
(7.3)

Thanks to Theorem 3.2 and [13, Theorem 5], and (7.1) there exists a function u satisfying items (i) and (ii) in Definition 3.1 such that, passing to a subsequence,

$$\begin{aligned} \begin{aligned}&\left\| u\right\| _{L^\infty ((0,\infty )\times (0,L))}\le 1,\\&u_n\rightharpoonup u \quad \text {in}\,\, H^1((0,T)\times (0,L)) \,\,\text {and in}\,\, L^2(0,T;H^2(0,L)), \,\,\text {for each}\,\, T\ge 0,\\&u_n \rightarrow u \quad \text {in}\,\, L^\infty ((0,T)\times (0,L)),\,\, \text {for each}\,\, T\ge 0,\\&\Phi '(u_n) \rightharpoonup \Phi '(u) \quad \text {in}\,\, L^p((0,T)\times (0,L)), \,\,\text {for each} \,\,T\ge 0 \,\,\text {and}\,\, 1\le p<\infty . \end{aligned} \end{aligned}$$
(7.4)

Therefore u is a distributional solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho \partial _{tt}^2u=-\mu \partial _{xxxx}^4u-\Phi '(u)&{}\quad t>0,\,\,0<x<L,\\ \partial _{xx}^2u(t,0)=\partial _{xx}^2u(t,L)=0&{}\quad t>0,\\ \partial _{xxx}^3u(t,0)=\partial _{xxx}^3u(t,L)=0&{}\quad t>0,\\ u(0,x)=\partial _tu(0,x)=0&{}\quad 0<x<L. \end{array}\right. } \end{aligned}$$
(7.5)

Since u takes values in \([-1,1]\) and \(\Phi \) is \(C^2\) therein we can differentiate (7.5) and get

$$\begin{aligned} \rho \partial _{ttt}^3u=-\mu \partial _{txxxx}^4u-\Phi ''(u)\partial _tu. \end{aligned}$$

Multiplying by \(\partial _{tt}^2u\), using (H.1) and (ii) in Definition 3.1 through a regularization argument we get

$$\begin{aligned} \frac{d}{dt}\int _0^L\frac{\rho (\partial _{tt}^2u)^2+\mu (\partial _{txx}^3u)^2}{2}dx&=\int _0^L\Phi ''(u)\partial _tu \partial _{tt}^2udx\\&\le \frac{1}{2\rho }\int _0^L\Phi ''(u)^2(\partial _tu)^2dx+\int _0^L\frac{\rho (\partial _{tt}^2u)^2}{2}dx\\&\le c+\int _0^L\frac{\rho (\partial _{tt}^2u)^2+\mu (\partial _{txx}^3u)^2}{2}dx. \end{aligned}$$

Thanks to the Gronwall Lemma

$$\begin{aligned} \int _0^L\frac{\rho (\partial _{tt}^2u(t,x))^2+\mu (\partial _{txx}^3u(t,x))^2}{2}dx\le c(e^t-1),\quad t\ge 0. \end{aligned}$$

As a consequence u is an energy preserving solution of (7.5) and then it must be the trivial one. Eventually, (7.3) concludes the proof. \(\square \)

8 Non-uniqueness and Lack of Smoothness

This section is devoted to exploit some qualitative properties of (2.1) through explicit analytical examples evidencing the lack of uniqueness and smoothness of solutions. A key mechanism ruling these phenomena relies in the transition between the two configurations induced by the discontinuity affecting the forcing term \(\Phi '\). In particular, the first two examples show the lack of uniqueness while the last one and the numerical experiments enlighten the occurrences of lack of smoothness.

Example 8.1

Let \(\varepsilon >0\) and set \(\rho =\mu =1\). Consider the function

$$\begin{aligned} \Phi _\varepsilon (u)={\left\{ \begin{array}{ll} \frac{2-\varepsilon }{2}u^2,&{}\quad \text {if}\,\, |u|\le 1,\\ \frac{2-\varepsilon }{\varepsilon }\left( (1+\varepsilon )\left( u-\frac{1}{2}\right) -\frac{u^2}{2}\right) ,&{}\quad \text {if} \,\,1\le u\le 1+\varepsilon ,\\ \frac{\varepsilon -2}{\varepsilon }\left( (1+\varepsilon )\left( u+\frac{1}{2}\right) +\frac{u^2}{2}\right) ,&{}\quad \text {if}\,\, -1-\varepsilon \le u\le -1,\\ \frac{(2-\varepsilon )(1+\varepsilon )}{2},&{}\quad \text {if}\,\, |u|\ge 1+\varepsilon . \end{array}\right. } \end{aligned}$$

We have

$$\begin{aligned} \Phi _\varepsilon '(u)={\left\{ \begin{array}{ll} (2-\varepsilon )u,&{}\quad \text {if}\,\, |u|\le 1,\\ \frac{2-\varepsilon }{\varepsilon }(1+\varepsilon -u),&{}\quad \text {if}\,\, 1\le u\le 1+\varepsilon ,\\ \frac{\varepsilon -2}{\varepsilon }(1+\varepsilon +u),&{}\quad \text {if}\,\, -1-\varepsilon \le u\le -1,\\ 0,&{}\quad \text {if} |u|\ge 1+\varepsilon . \end{array}\right. } \end{aligned}$$

The functions

$$\begin{aligned} u_\varepsilon (t,x)=(1-\varepsilon )\cos \left( \sqrt{2-\varepsilon }\,t\right) ,\quad v_\varepsilon (t,x)=1+\varepsilon \end{aligned}$$

solve

$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _{tt}^2u_\varepsilon =-\partial _{xxxx}^4u_\varepsilon -\Phi _\varepsilon '(u_\varepsilon ),&{}\quad t>0,\,\,0<x<L,\\ \partial _{xx}^2u_\varepsilon (t,0)=\partial _{xx}^2u_\varepsilon (t,L)=0,&{}\quad t>0,\\ \partial _{xxx}^3u_\varepsilon (t,0)=\partial _{xxx}^3u_\varepsilon (t,L)=0,&{}\quad t>0,\\ u_\varepsilon (0,x)=1-\varepsilon ,&{}\quad 0<x<L,\\ \partial _tu_\varepsilon (0,x)=0,&{}\quad 0<x<L, \end{array}\right. } \end{aligned}$$
(8.1)
$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _{tt}^2v_\varepsilon =-\partial _{xxxx}^4v_\varepsilon -\Phi _\varepsilon '(v_\varepsilon ),&{}\quad t>0,\,\,0<x<L,\\ \partial _{xx}^2v_\varepsilon (t,0)=\partial _{xx}^2v_\varepsilon (t,L)=0,&{}\quad t>0,\\ \partial _{xxx}^3v_\varepsilon (t,0)=\partial _{xxx}^3v_\varepsilon (t,L)=0,&{}\quad t>0,\\ v_\varepsilon (0,x)=1+\varepsilon ,&{}\quad 0<x<L,\\ \partial _tv_\varepsilon (0,x)=0,&{}\quad 0<x<L. \end{array}\right. } \end{aligned}$$
(8.2)

As \(\varepsilon \rightarrow 0\) we have

$$\begin{aligned} u_\varepsilon (t,x)\rightarrow u(t,x)=\cos \left( \sqrt{2}\,t\right) ,\quad v_\varepsilon (t,x)\rightarrow v(t,x)=1, \end{aligned}$$

and u and v provides two different solutions of (2.1) in correspondence of the initial data

$$\begin{aligned} u_0(x)=1,\quad u_1(x)=0. \end{aligned}$$

The energies associated to (8.1) and (8.2) are

$$\begin{aligned} E_\varepsilon [u_\varepsilon ](t)&=\int _0^L\left( \frac{(\partial _tu_\varepsilon (t,x))^2+(\partial _{xx}^2u_\varepsilon (t,x))^2}{2}+ \Phi _\varepsilon (u_\varepsilon (t,x))\right) dx=\frac{(2-\varepsilon )(1-\varepsilon )^2}{2}L,\\ E_\varepsilon [v_\varepsilon ](t)&=\int _0^L\left( \frac{(\partial _tv_\varepsilon (t,x))^2+(\partial _{xx}^2v_\varepsilon (t,x))^2}{2}+\Phi _\varepsilon (v_\varepsilon (t,x))\right) dx=\frac{(2-\varepsilon )(1+\varepsilon )}{2}L, \end{aligned}$$

respectively.

Example 8.2

For every \(\varepsilon >0\), the solutions \(u_\varepsilon \) and \(v_\varepsilon \) of the two following problems

$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _{tt}^2u_\varepsilon =-\partial _{xxxx}^4u_\varepsilon -\Phi '(u_\varepsilon ),&{}\quad t>0,\,\,0<x<L,\\ \partial _{xx}^2u_\varepsilon (t,0)=\partial _{xx}^2u_\varepsilon (t,L)=0,&{}\quad t>0,\\ \partial _{xxx}^3u_\varepsilon (t,0)=\partial _{xxx}^3u_\varepsilon (t,L)=0,&{}\quad t>0,\\ u_\varepsilon (0,x)=1+\varepsilon ,&{}\quad 0<x<L,\\ \partial _tu_\varepsilon (0,x)=\varepsilon ,&{}\quad 0<x<L, \end{array}\right. } \end{aligned}$$
(8.3)
$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _{tt}^2v_\varepsilon =-\partial _{xxxx}^4v_\varepsilon -\Phi '(v_\varepsilon ),&{}\quad t>0,\,\,0<x<L,\\ \partial _{xx}^2v_\varepsilon (t,0)=\partial _{xx}^2v_\varepsilon (t,L)=0,&{}\quad t>0,\\ \partial _{xxx}^3v_\varepsilon (t,0)=\partial _{xxx}^3v_\varepsilon (t,L)=0,&{}\quad t>0,\\ v_\varepsilon (0,x)=1-\varepsilon ,&{}\quad 0<x<L,\\ \partial _tv_\varepsilon (0,x)=0,&{}\quad 0<x<L, \end{array}\right. } \end{aligned}$$
(8.4)

are

$$\begin{aligned} u_\varepsilon (t,x)=\varepsilon t+1+\varepsilon ,\quad v_\varepsilon (t,x)=(1-\varepsilon )\cos (\sqrt{2}t). \end{aligned}$$

We have

$$\begin{aligned}&\left\| u_\varepsilon (0,\cdot )-v_\varepsilon (0,\cdot )\right\| _{L^2(0,L)}+\left\| \partial _tu_\varepsilon (0,\cdot )-\partial _tv_\varepsilon (0,\cdot )\right\| _{L^2(0,L)}=3\varepsilon \sqrt{L},\\&\lim _{t\rightarrow \infty }u_\varepsilon (t,x)=\infty ,\quad \limsup _{t\rightarrow \infty }v_\varepsilon (t,x)=1-\varepsilon . \end{aligned}$$

Moreover, as \(\varepsilon \rightarrow 0\),

$$\begin{aligned} u_\varepsilon (t,x)\rightarrow 1,\quad v_\varepsilon (t,x)\rightarrow \cos (\sqrt{2}t). \end{aligned}$$

The energies associated to (8.3) and (8.4) are

$$\begin{aligned} E[u_\varepsilon ](t)&=\int _0^L\left( \frac{(\partial _tu_\varepsilon (t,x))^2+(\partial _{xx}^2u_\varepsilon (t,x))^2}{2}+ \Phi (u_\varepsilon (t,x))\right) dx=\frac{\varepsilon ^2+2}{2}L,\\ E[v_\varepsilon ](t)&=\int _0^L\left( \frac{(\partial _tv_\varepsilon (t,x))^2+(\partial _{xx}^2v_\varepsilon (t,x))^2}{2}+\Phi (v_\varepsilon (t,x))\right) dx=(1-\varepsilon )^2L, \end{aligned}$$

respectively.

Example 8.3

Consider the function

$$\begin{aligned} u(t,x)= {\left\{ \begin{array}{ll} \sqrt{2}\sin (\sqrt{2}t),&{}\quad \text {if}\,\, 0\le t\le \frac{\pi }{4\sqrt{2}},\\ \sqrt{2} t+1-\frac{\pi }{4},&{}\quad \text {if}\,\, t\ge \frac{\pi }{4\sqrt{2}}. \end{array}\right. } \end{aligned}$$
(8.5)

Clearly, u solves the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _{tt}^2u=-\partial _{xxxx}^4u-\Phi '(u),&{}\quad t>0,\,\,x\in (0,L),\\ \partial _{xx}^2u(t,0)=\partial _{xx}^2u(t,L)=0,&{}\quad t>0,\\ \partial _{xxx}^3u(t,0)=\partial _{xxx}^3u(t,L)=0,&{}\quad t>0,\\ u(0,x)=0,&{}\quad x\in (0,L),\\ \partial _tu(0,x)=2,&{}\quad x\in (0,L), \end{array}\right. } \end{aligned}$$

but

$$\begin{aligned} u\in C^1([0,\infty )\times [0,L]){\setminus } C^2([0,\infty )\times [0,L]). \end{aligned}$$

Indeed

$$\begin{aligned} \lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^- }u\left( t,x\right) =1,\quad&\lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^+}u\left( t,x\right) =1,\\ \lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^-}\partial _tu\left( t,x\right) =\sqrt{2},\quad&\lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^+} \partial _tu\left( t,x\right) =\sqrt{2},\\ \lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^-}\partial _{tt}^2u\left( t,x\right) =-2,\quad&\lim _{t \rightarrow \frac{\pi }{4\sqrt{2}}^+ } \partial _{tt}^2u\left( t,x\right) =0. \end{aligned}$$

The energy associated to (8.5) is

$$\begin{aligned} E[u](t)=\int _0^L\left( \frac{(\partial _tu(t,x))^2+(\partial _x u(t,x))^2}{2}+ \Phi (u(t,x))\right) dx=2L, \end{aligned}$$

for every \(t\ge 0\).