On the jerky crack growth in elastoplastic materials

Abstract

The purpose of this paper is to show that in elastoplastic materials cracks can grow only in an intermittent way. This result is rigorously proved in the framework of a simplified model.

1 Introduction

In this paper we give a contribution to the mathematical derivation of the properties of the quasistatic crack growth in elastoplastic materials. The study of this subject has a long history (see, e.g., [11, 14, 15]). Our aim is to obtain a precise mathematical result in a simplified model where perfect plasticity interacts with crack growth. In particular, under suitable assumptions we prove that cracks are piecewise constant in time.

In our simplified model the reference configuration $\varOmega$ is a bounded connected open subset of ${\mathbb {R}}^2$ with Lipschitz boundary. We consider only the antiplane case, so that the displacement u is a function from $\varOmega$ into ${\mathbb {R}}$. We assume that the cracks and the plastic slips may occur only on a prescribed segment $\varGamma$, whose interior is contained in $\varOmega$ and whose end-points belong to $\partial \varOmega$. It is not restrictive to assume that $\varGamma :=\{(x,0): a\le x\le b\}$ for some $a<b$.

Since there is no plastic part in $\varOmega \setminus \varGamma$, the displacement u belongs to $H^1(\varOmega \setminus \varGamma )$ and the elastic energy is given by

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy. \end{aligned}$$

We assume that at each time the crack has the form $\varGamma _a^s:=\{(x,0): a\le x\le s\}$ for some $a\le s\le b$ and that the energy spent to produce it is equal to $s-a$. On $\varGamma _s^b:=\{(x,0): s\le x\le b\}$ the plastic slip is determined by the jump of the displacement:

$$\begin{aligned}{}[u]=u^+-u^-, \end{aligned}$$

where $u^+$ and $u^-$ are the traces of u on $\varGamma$ from above and from below. The plastic dissipation distance between the current displacement u and a previous displacement $u_0$ is given by

$$\begin{aligned} \int _{\varGamma _s^b}|[u]-[u_0]|\,dx. \end{aligned}$$

The evolution is driven by a time-dependent Dirichlet boundary condition $u=w(t)$ imposed on a prescribed Borel subset $\partial _D\varOmega$ of $\partial \varOmega$. We first consider the incremental formulation. Given a subdivision $0=t_0< t_1<\cdots<t_{n-1}<t_n=T$ of the interval [0, T], for $i=1,\ldots ,n$ let $(u_i,s_i)$ be a solution of the incremental minimum problem for the pair (u, s):

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega \setminus \varGamma )\\ u=w(t_i)\ \text {on }\partial _D\varOmega \\ s_{i-1}\le s\le b \end{array}}\Big \{ \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy+s+\int _{\varGamma _{s}^b}|[u]-[u_{i-1}]|dx\Big \}.\nonumber \end{aligned}$$

As in [6] we can prove (Theorem 2.5) that, passing to a subsequence, the piecewise constant interpolation of $(u_i,s_i)$ converges, as the fineness of the subdivision tends to zero, to a quasistatic evolution, i.e., a pair (u, s) which satisfies the following conditions:

• (a) (irreversibility) s is nondecreasing on [0, T];

• (b) (equilibrium) for every $t\in [0,T]$ we have $u(t)\in H^1(\varOmega \setminus \varGamma )$, $u(t)=w(t)$ on $\partial _D\varOmega$, and

  $$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+s(t)\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+\int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u(t)]|dx, \end{aligned}$$

  for every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$, with ${\hat{u}}=w(t)$ on $\partial _D\varOmega$, and every ${\hat{s}}\in [s(t),b]$;

• (c) (energy-dissipation inequality) for every $t_1,t_2\in [0,T]$, with $t_1<t_2$, we have

  $$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_2)|^2dxdy+s(t_2)-s(t_1)+\int _{\varGamma _{s(t_2)}^b}|[u(t_2)]-[u(t_1)]|dx\\ \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau , \end{aligned}$$

  where $\dot{w}$ is the time-derivative of w.

As in [6] we can obtain (Theorem 2.9) an energy-dissipation balance, using a suitable notion of dissipation (Definition 2.6). Therefore our notion of quasistatic evolution is formulated within the framework of rate-independent processes developed in [12, 13]. When no plastic slip is present, i.e., $[u(t)]=0$ on $\varGamma _{s(t)}^b$, this evolution agrees, in the antiplane case, with the variational solution of the crack growth problem introduced in [9] and studied in [2].

The main result of our paper (Theorem 4.1) is that, if (u, s) satisfies hypotheses (a)-(c), and w satisfies suitable continuity conditions, then s is piecewise constant. In other words, the crack growth is jerky. This behaviour is in agreement with the numerical simulations in [1] and with many experimental results (see, e.g., [7, 10]). As a consequence of well-known results on perfect plasticity (Theorem 4.14), from this property of s we deduce (Theorem 4.15) that u is piecewice absolutely continuous with values in $H^1(\varOmega \setminus \varGamma )$. A concluding example (Theorem 5.1) shows that, in general, s is not constant.

A numerical study of the simplified model of the present paper will appear in [5].

2 Formulation of the problem

The reference configuration is a bounded connected open set $\varOmega \subset {\mathbb {R}}^2$ with Lipschitz boundary $\partial \varOmega$. On a prescribed Borel subset $\partial _D\varOmega$ of $\partial \varOmega$ we shall impose a time-dependent Dirichlet boundary condition. On its complement $\partial \varOmega \setminus \partial _D\varOmega$ we shall consider the homogeneous Neumann boundary condition.

In our simplified model we assume that the cracks and the plastic slips may occur only on a prescribed segment $\varGamma :=\{(x,0): a\le x\le b\}$ contained in ${\overline{\varOmega }}$, with $(a,0),\,(b,0)\in \partial \varOmega$ and $(x,0)\in \varOmega$ for every $a<x<b$. For every $a\le s_1\le s_2\le b$ we set $\varGamma _{s_1}^{s_2}:=\{(x,0): s_1\le x\le s_2\}$.

We assume that there exists an open neighbourhood U of $\varGamma$ in ${\mathbb {R}}^2$ such that $U\cap (\varOmega \setminus \varGamma )$ is the union of two disjoint connected open sets $U^+$ and $U^-$ with Lipschitz boundary. We also assume that for every $a<x<b$ we have $(x,y)\in U^\pm$ whenever |y| is small and $\pm y>0$. Let $\varOmega ^\pm$ be the connected component of $\varOmega \setminus \varGamma$ containing $U^\pm$. Note that under our hypotheses we have $\varOmega \setminus \varGamma =\varOmega ^+\cup \varOmega ^-$ and that it may happen that $\varOmega ^+=\varOmega ^-$, if $\varOmega$ is not simply connected (see Fig. 1) We set $\partial ^\pm \varOmega :=\partial \varOmega ^\pm \setminus \varGamma$ and $\partial ^\pm _D\varOmega :=\partial _D\varOmega \cap \partial \varOmega ^\pm$. We assume that

$$\begin{aligned} \partial ^+_D\varOmega \quad \hbox {and} \quad \partial ^-_D\varOmega \quad \hbox {have positive one-dimensional measure.} \end{aligned}$$

(2.1)

Fig. 1

Examples of sets $\varOmega$, $\partial _D\varOmega$, and $\varGamma$

Since we are dealing with the antiplane case, the displacement $u=u(x,y)$ is a scalar function belonging to $H^1(\varOmega \setminus \varGamma )$. An admissible crack will be a segment of the form $\varGamma _a^s$ for some $a\le s\le b$. Given a displacement $u\in H^1(\varOmega \setminus \varGamma )$, the jump of u across $\varGamma$ is given by

$$\begin{aligned}{}[u]=u^+-u^-, \end{aligned}$$

where $u^+$ is the trace on the side of $\varGamma$ corresponding to $y>0$, and $u^-$ is the trace on the opposite side.

The Dirichlet boundary condition will be prescribed through a function

$$\begin{aligned} w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0})) \end{aligned}$$

(2.2)

for a suitable $s_0\in [a,b)$.

Definition 2.1

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. A quasistatic evolution with boundary value w on $\partial _D\varOmega$ is a pair (u, s), with $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ measurable and $s:[0,T]\rightarrow [s_0,b]$, that satisfies the following conditions:

• (a) (irreversibility) s is nondecreasing;

• (b) (equilibrium) for every $t\in [0,T]$ we have $u(t)=w(t)$ on $\partial _D\varOmega$ and

  $$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+s(t)\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+\int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u(t)]|dx, \end{aligned}$$

  for every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$, with ${\hat{u}}=w(t)$ on $\partial _D\varOmega$, and every ${\hat{s}}\in [s(t),b]$;

• (c) (energy-dissipation inequality) for every $t_1,t_2\in [0,T]$, with $t_1<t_2$, we have

  $$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_2)|^2dxdy+s(t_2)-s(t_1)+\int _{\varGamma _{s(t_2)}^b}|[u(t_2)]-[u(t_1)]|dx\\ \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned}$$

Remark 2.2

Taking ${\hat{u}}=w(t)$ and ${\hat{s}}=b$ in condition (b) above, by (2.2) we obtain that there exists a constant $M_1>0$ such that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy\le M_1\quad \text {for every }t\in [0,T]. \end{aligned}$$

(2.3)

Together with the measurability of $t\mapsto u(t)$ this implies that the last integral in condition (c) above is well defined. Moreover, since $u(t)=w(t)$ on $\partial _D\varOmega$, by (2.1)–(2.3) there exists a constant $M_0>0$ such that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }| u(t)|^2dxdy\le M_0\quad \text {for every }t\in [0,T]. \end{aligned}$$

(2.4)

Remark 2.3

Let us now comment on the term

$$\begin{aligned} \int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau \end{aligned}$$

(2.5)

which appears in the energy-dissipation balance. The Euler equation for the equilibrium condition gives that u(t) is harmonic in $\varOmega \setminus \varGamma$ for every $t\in [0,T]$. Moreover, if u is sufficiently regular, the equilibrium condition implies that $\frac{\partial u(t)}{\partial \nu }=0$ on $\partial \varOmega \setminus \partial _D\varOmega$, where $\nu$ is the outward unit normal to $\partial \varOmega$, $\big (\frac{\partial u(t)}{\partial y})^+=\big (\frac{\partial u(t)}{\partial y}\big )^-=0$ on $\varGamma _a^{s(t)}$, and $\big (\frac{\partial u(t)}{\partial y})^+=\big (\frac{\partial u(t)}{\partial y}\big )^-$ on $\varGamma _{s(t)}^b$ (the last property follows easily from (3.8) and (3.9), proved below in a more general setting). Therefore, since $(\dot{w})^+(\tau )=(\dot{w})^-(\tau )$ on $\varGamma _{s(t)}^b$ (by our assumption on w and s), integrating by parts we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy=\int _{\partial _D\varOmega }\frac{\partial u(\tau )}{\partial \nu }\dot{w}(\tau )dS, \end{aligned}$$

where S is the line-measure on $\partial _D\varOmega$. Thus (2.5) equals

$$\begin{aligned} \int _{t_1}^{t_2}\Big (\int _{\partial _D\varOmega }\frac{\partial u(\tau )}{\partial \nu }\dot{w}(\tau )dS\Big )d\tau . \end{aligned}$$

(2.6)

Since $\frac{\partial u(\tau )}{\partial \nu }$ represents the force acting on the boundary, (2.6) represents the work done by this force in the interval $[t_1,t_2]$.

Remark 2.4

The previous remark suggests that Definition 2.1 does not change if w is replaced by another function $w_*\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$ such that

$$\begin{aligned} w(t)=w_*(t)\quad \text {on }\partial _D\varOmega . \end{aligned}$$

This is actually true without any additional regularity assumption. Indeed, if (u, s) is a quasistatic evolution for w, then

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy=\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}_*(\tau )dxdy\quad \text {for a.e.}\ \tau \in [0,T]. \end{aligned}$$

This follows from Lemma 3.1, since $\dot{w}(\tau )-\dot{w}_*(\tau )\in H^1(\varOmega \setminus \varGamma )$, $\dot{w}(\tau )-\dot{w}_*(\tau )=0$ on $\partial _D\varOmega$, and $[\dot{w}(\tau )-\dot{w}_*(\tau )]=0$ on $\varGamma _{s(\tau )}^b$.

The following result shows the existence of a quasistatic evolution with prescribed initial data.

Theorem 2.5

Let $T>0$, $s_0\in [a,b)$, $u_0\in H^1(\varOmega \setminus \varGamma )$, and let $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. Assume that $u_0=w(0)$ on $\partial _D\varOmega$ and

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_0|^2dxdy+s_0\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+\int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u_0]|dx, \end{aligned}$$

for every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$, with ${\hat{u}}=w(0)$ on $\partial \varOmega$, and every $s_0\le {\hat{s}}\le b$. Then there exists a quasistatic evolution with boundary value w on $\partial _D\varOmega$, satisfying the initial conditions $u(0)=u_0$ and $s(0)=s_0$.

To prove the theorem it is convenient to introduce the notion of dissipation, which is a particular case of the one considered in [6, Section 2.3].

Definition 2.6

Let $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ and $s:[0,T]\rightarrow [a,b]$. The dissipation of (u, s) on the interval $[t_1,t_2]\subset [0,T]$ is defined as:

$$\begin{aligned} {\mathrm {Diss}}(u(\cdot ),s(\cdot );t_1,t_2):=\sup \sum _{i=1}^k\Big (s(\tau _i)-s(\tau _{i-1})+\int _{\varGamma _{s(\tau _i)}^b}|[u(\tau _i)]-[u(\tau _{i-1})]|dx\Big ) \end{aligned}$$

where the supremum is taken over all finite partitions $t_1=\tau _0\le \tau _1\le \cdots \le \tau _k=t_2$.

Proof of Theorem 2.5. The proof is a simplified version of the proof of [6, Theorem 2.5]. We fix a sequence of subdivisions $(t^i_n)_{0\le i\le n}$ with

$$\begin{aligned}&0=t^0_n<t^1_n<\cdots<t^{n-1}_n<t^n_n=T, \end{aligned}$$

(2.7)

$$\begin{aligned}&\displaystyle \lim _{n\rightarrow \infty }\max _{1\le i\le n}(t^i_n-t^{i-1}_n)=0. \end{aligned}$$

(2.8)

For every n we set $u^0_n=u_0$, $s^0_n=s_0$, and for every $i=1,\ldots ,n$ we define inductively $(u^i_n,s^i_n)$ as a solution of the incremental minimum problem

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega \setminus \varGamma )\\ u=w^i_n\ \text {on }\partial _D\varOmega \\ s^{i-1}_n\le s\le b \end{array}}\Big \{ \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy+s+\int _{\varGamma _{s}^b}|[u]-[u^{i-1}_n]|dx\Big \}, \end{aligned}$$

(2.9)

where $w_n^i:=w(t_n^i)$.

Note that, by the triangle inequality, from (2.9) we obtain that $u_n^i$ satisfies

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy+s_n^i\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+\int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u_n^i]|dx, \end{aligned}$$

(2.10)

for every $s_n^i\le {\hat{s}}\le b$ and every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$ with ${\hat{u}}=w_n^i$ on $\partial _D\varOmega$.

To estimate $u_n^i$ we compare $(u_n^i,s_n^i)$ with $(w^i_n,s_n^{i})$ in the minimum problem (2.10) and we obtain

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla w_n^i|^2dxdy+\int _{\varGamma _{s_n^i}^b}|[u_n^i]|dx\le C_1+\int _{\varGamma _{s_n^i}^b}|[u_n^i]|dx,\nonumber \\ \end{aligned}$$

(2.11)

for a suitable constant $C_1>0$ independent of i and n . By the Trace Inequality there exists a constant $C_2>0$ independent of i and n such that

$$\begin{aligned} \int _{\varGamma }|[u_n^i]|dx\le C_2\Big (\int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\Big )^{1/2}+C_2\Big (\int _{\varOmega \setminus \varGamma }|u_n^i|^2dxdy\Big )^{1/2}. \end{aligned}$$

Since $u_n^i=w_n^i$ on $\partial _D\varOmega$, by (2.1), (2.2), and the Poincaré Inequality there exists a constant $C_3>0$ independent of i and n, such that

$$\begin{aligned} \int _{\varGamma }|[u_n^i]|dx\le C_3\Big (\int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\Big )^{1/2}+C_3. \end{aligned}$$

Therefore (2.11) gives

$$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\le C_3\Big (\int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\Big )^{1/2}+C_3+C_1 \end{aligned}$$

which implies that there exists a constant $C_4>0$ independent of i and n such that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy\le C_4. \end{aligned}$$

(2.12)

We now compare $(u_n^i,s_n^i)$ with $(u_n^{i-1}+w^i_n-w^{i-1}_n,s_n^{i-1})$ in the minimum problem (2.9) and we obtain

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^i|^2dxdy+s_n^i-s_n^{i-1}+\int _{\varGamma _{s_n^i}^b}|[u_n^i]-[u^{i-1}_n]|dx\\&\quad \le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^{i-1}|^2dxdy+\int _{t_n^{i-1}}^{t_n^i}\Big ( \int _{\varOmega \setminus \varGamma }\nabla u_n^{i-1}\nabla \dot{w}(t)dxdy\Big )dt+R_n^i, \end{aligned}$$

where

$$\begin{aligned} R_n^i:=\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla (w_n^i-w_n^{i-1})|^2dxdy. \end{aligned}$$

Iterating this inequality for every $0\le i<j\le n$ we obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^j|^2dxdy+\sum _{h=i+1}^j\Big (s_n^h-s_n^{h-1}+\int _{\varGamma _{s_n^h}^b}|[u_n^h]-[u^{h-1}_n]|dx\Big )\\&\quad \le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n^{i}|^2dxdy+\sum _{h=i+1}^j\int _{t_n^{h-1}}^{t_n^h}\Big ( \int _{\varOmega \setminus \varGamma }\nabla u_n^{h-1}\nabla \dot{w}(\tau )dxdy\Big )d\tau +R_n, \end{aligned} \end{aligned}$$

(2.13)

where $R_n:=\sum _{i=1}^nR_n^i$. Since $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$ we have that $R_n\rightarrow 0$.

Let $u_n(t)$, $s_n(t)$, and $w_n(t)$ be the piecewise constant interpolations of $u_n^i$, $s_n^i$, and $w_n^i$ defined by

$$\begin{aligned} u_n(t):=u_n^{i-1},\quad s_n(t):=s_n^{i-1},\quad w_n(t):=w_n^{i-1}\qquad \text {for } t_n^{i-1}\le t<t_n^{i}. \end{aligned}$$

(2.14)

Note that by (2.12) we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }|\nabla u_n(t)|^2dxdy\le C_4\quad \text { for every }t\in [0,T]\text { and every }n. \end{aligned}$$

(2.15)

Inequality (2.13) can be rewritten as

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_n^j)|^2dxdy+{\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );t_n^i,t_n^j)\\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_n^i)|^2dxdy+\int _{t_n^i}^{t_n^j}\Big (\int _{\varOmega \setminus \varGamma }\nabla u_n(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau +R_n. \end{aligned}$$

Since the function $t\mapsto \big (\int _{\varOmega \setminus \varGamma }|\nabla \dot{w}(t)|^2dxdy\big )^{1/2}$ is integrable, using (2.8) and (2.15) we deduce from the previous inequality that there exists ${\tilde{R}}_n\rightarrow 0$ such that for every $0\le t_1<t_2\le T$ we have

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_2)|^2dxdy+{\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );t_1,t_2) \\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u_n(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau +{\tilde{R}}_n. \end{aligned} \end{aligned}$$

(2.16)

In particular, by (2.15) this inequality implies that ${\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );0,T)$ is bounded uniformly with respect to t and n. To continue the proof we need the following lemmas.

Given a set A, let $\chi _A$ be its characteristic function, defined by $\chi _A(x):=1$ if $x\in A$ and $\chi _A(x):=0$ if $x\notin A$.

Lemma 2.7

Assume that $\Vert u_n(t)\Vert _{H^1(\varOmega \setminus \varGamma )}$ and ${\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );0,T)$ are bounded uniformly with respect to t and n. Then there exist a subsequence of $(u_n, s_n)$, not relabelled, a nondecreasing function $s:[0,T]\rightarrow [a,b]$, and a function $g:[0,T]\rightarrow L^1(\varGamma )$ such that

$$\begin{aligned}&s_n(t)\rightarrow s(t), \end{aligned}$$

(2.17)

$$\begin{aligned}&[u_n(t)]\chi _{\varGamma _{s_n(t)}^b}\rightarrow g(t)\chi _{\varGamma _{s(t)}^b}\quad \text {strongly in }L^1(\varGamma ), \end{aligned}$$

(2.18)

for every $t\in [0,T]$.

Proof

The statement on the convergence of $s_n$ is a consequence of Helly’s Theorem. Let D be a countable dense subset of [0, T]. By a diagonal argument we can find a subsequence of $u_n$, not relabelled, and a bounded function $v:D\rightarrow H^1(\varOmega \setminus \varGamma )$ such that $u_n(t)\mathop {\rightharpoonup }v(t)$ weakly in $H^1(\varOmega \setminus \varGamma )$ for every $t\in D$. This implies that

$$\begin{aligned}{}[u_n(t)]\rightarrow [v(t)] \quad \text {strongly in }L^2(\varGamma ) \end{aligned}$$

(2.19)

for every $t\in D$.

To prove (2.18) for every $t\in [0,T]$ we introduce the nondecreasing functions $V_n:[0,T]\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} V_n(t):={\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );0,t). \end{aligned}$$

(2.20)

By Helly’s Theorem there exist a subsequence, not relabelled, and a nondecreasing function V such that $V_n(t)\rightarrow V(t)$ for every $t\in [0,T]$.

Let $t_0\in (0,T)$ be a continuity point for both V and s. For every $\varepsilon >0$ there exists $\delta >0$ such that $|V(t)-V(t_0)|<\varepsilon$ and $|s(t)-s(t_0)|<\varepsilon$ for every $t\in [0,T]$ with $|t-t_0|<\delta$. Let $t\in D$ with $t_0<t<t_0+\delta$. Then $V_n(t)\rightarrow V(t)<V(t_0)+\varepsilon$ and $V_n(t_0)\rightarrow V(t_0)$. By Definition 2.6 it follows that

$$\begin{aligned} \int _{\varGamma _{s_n(t)}^b}|[u_n(t)]-[u_n(t_0)]|dx\le {\mathrm {Diss}}(u_n(\cdot ),s_n(\cdot );t_0,t)\le V_n(t)-V_n(t_0)<\varepsilon \end{aligned}$$

(2.21)

for n large enough. Moreover, since $\Vert u_n(t)\Vert _{H^1(\varOmega \setminus \varGamma )}$ is uniformly bounded, there exists a constant $C>0$ such that $\Vert [u_n(t_0)]\Vert _{L^2(\varGamma )}\le C$ for every n. This implies that

$$\begin{aligned} \begin{aligned}&\int _\varGamma |[u_n(t_0)]\chi _{\varGamma _{s_n(t)}^b}-[u_n(t_0)]\chi _{\varGamma _{s_n(t_0)}^b}|dx\\&\quad \le \int _{\varGamma _{s_n(t_0)}^{s_n(t)}}|[u_n(t_0)]|dx\le \Vert [u_n(t_0)]\Vert _{L^2(\varGamma )}(s_n(t)-s_n(t_0))^{1/2}<C\varepsilon ^{1/2} \end{aligned} \end{aligned}$$

(2.22)

for sufficiently large n. By the triangle inequality (2.21) and (2.22) give

$$\begin{aligned} \int _\varGamma |[u_n(t)]|\chi _{\varGamma _{s_n(t)}^b}-[u_n(t_0)]\chi _{\varGamma _{s_n(t_0)}^b}|dx<\varepsilon +C\varepsilon ^{1/2}, \end{aligned}$$

and this inequality, together with (2.19), implies that $[u_n(t_0)]\chi _{\varGamma _{s_n(t_0)}^b}$ is a Cauchy sequence in $L^1(\varGamma )$, hence it converges to a function $g(t_0)\in L^1(\varGamma )$. Since $s_n(t_0)\rightarrow s(t_0)$ we have $g(t_0)=g(t_0)\chi _{\varGamma _{s(t_0)}^b}$.

Therefore (2.18) holds for all continuity points of both V and s. Since the set of all other points is at most countable, we can apply again the diagonal argument to extract a further subsequence along which (2.18) holds for all t. $\square$

Lemma 2.8

For every $w\in H^1(\varOmega \setminus \varGamma )$, $s\in [a,b]$, and $g\in L^1(\varGamma )$ let $u^w_{s,g}$ be the unique solution of the minimum problem

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega \setminus \varGamma )\\ u=w\ \text {on }\partial _D\varOmega \end{array}}\Big \{ \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy+\int _{\varGamma _{s}^b}|[u]-g|dx\Big \}. \end{aligned}$$

(2.23)

Let $w_n,w\in H^1(\varOmega \setminus \varGamma )$, $s_n,s\in [a,b]$, and $g_n,g\in L^1(\varGamma )$ be such that

$$\begin{aligned}&w_n\rightarrow w\quad \text {strongly in }H^1(\varOmega \setminus \varGamma ), \end{aligned}$$

(2.24)

$$\begin{aligned}&s_n\rightarrow s,\end{aligned}$$

(2.25)

$$\begin{aligned}&g_n\chi _{\varGamma _{s_n}^b}\rightarrow g\chi _{\varGamma _s^b}\quad \text {strongly in }L^1(\varGamma ). \end{aligned}$$

(2.26)

Then $u^{w_n}_{s_n,g_n}\rightarrow u^w_{s,g}$ strongly in $H^1(\varOmega \setminus \varGamma )$.

Proof

Note that the uniqueness of the solution to (2.23) follows easily from the strict convexity of the functional with respect to $\nabla u$, using (2.1).

We set $u_n:=u^{w_n}_{s_n,g_n}$ and $u:=u^w_{s,g}$. From the minimality of $u_n$ we have

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n|^2dxdy+\int _{\varGamma _{s_n}^b}|[u_n]-g_n|dx\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla w_n|^2dxdy+\int _{\varGamma _{s_n}^b}|[w_n]-g_n|dx, \end{aligned}$$

which gives the boundedness of $u_n$ in $H^1(\varOmega \setminus \varGamma )$, thanks to (2.1). Hence there exist a subsequence, not relabelled, and a function $v\in H^1(\varOmega \setminus \varGamma )$ with $v=w$ on $\partial _D\varOmega$, such that $u_n\mathop {\rightharpoonup }v$ weakly in $H^1(\varOmega \setminus \varGamma )$. Using lower semicontinuity it is easy to prove that v solves (2.23), hence $v=u$. By the arbitrariness of the subsequence we conclude that the whole sequence $u_n$ converges to u weakly in $H^1(\varOmega \setminus \varGamma )$. To prove the strong convergence we first observe that $[u_n]\rightarrow [u]$ strongly in $L^2(\varGamma )$ and

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n|^2dxdy+\int _{\varGamma _{s_n}^b}|[u_n]-g_n|dx\\&\quad \le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla (u+w_n-w)|^2dxdy+\int _{\varGamma _{s_n}^b}|[u+w_n-w]-g_n|dx, \end{aligned}$$

by minimality. By (2.242.25)–(2.26) this implies

$$\begin{aligned} \limsup _n \int _{\varOmega \setminus \varGamma }|\nabla u_n|^2dxdy\le \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy, \end{aligned}$$

which, together with the weak convergence, gives $\nabla u_n\rightarrow \nabla u$ strongly in $L^2(\varOmega \setminus \varGamma ;{\mathbb {R}}^2)$. Taking (2.1) into account, this implies the strong convergence of $u_n$ to u. $\square$

For every $\alpha ,\beta \in {\mathbb {R}}$ we set $\alpha \vee \beta :=\max \{\alpha ,\beta \}$ and $\alpha \wedge \beta :=\min \{\alpha ,\beta \}$.

Proof of Theorem 2.5(continuation) Let s and g be the functions given by Lemma 2.7 and for every $t\in [0,T]$ let u(t) be the solution of the minimum problem

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega \setminus \varGamma )\\ u=w(t)\ \text {on }\partial _D\varOmega \end{array}}\Big \{ \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy+\int _{\varGamma _{s(t)}^b}|[u]-g(t)|dx\Big \}. \end{aligned}$$

Let us prove that $t\mapsto u(t)$ from [0, T] into $H^1(\varOmega \setminus \varGamma )$ is measurable. It is enough to show that $t\mapsto u(t)$ is continuous at every continuity point $t_0\in (0,T)$ of both V and s, where V is defined as in the proof of Lemma 2.7. Let us fix such a point $t_0$ and a sequence $t_k\rightarrow t_0$. Taking into account Lemma 2.8, since $u(t_k)=u^{w(t_k)}_{s(t_k),g(t_k)}$, it is sufficient to prove that

$$\begin{aligned} g(t_k)\chi _{\varGamma _{s(t_k)}^b}\rightarrow g(t_0)\chi _{\varGamma _{s(t_0)}^b}\quad \text {strongly in }L^1(\varGamma ). \end{aligned}$$

(2.27)

By Definition 2.6 and (2.20) we have

$$\begin{aligned}&\int _\varGamma |[u_n(t_k)]\chi _{\varGamma _{s_n(t_k)}^b}-[u_n(t_0)]\chi _{\varGamma _{s_n(t_0)}^b}|dx\nonumber \\&\quad \le \int _{\varGamma _{s_n(t_0\vee t_k)}^b}|[u_n(t_k)]-[u_n(t_0)]|dx +\int _{\varGamma _{s_n(t_0\wedge t_k)}^{s_n(t_0\vee t_k)}}|[u_n(t_0\wedge t_k)]|dx\\&\quad \le V_n(t_0\vee t_k)-V_n(t_0\wedge t_k)+\Vert [u_n(t_0\wedge t_k)]\Vert _{L^2(\varGamma )}|s_n(t_k)-s_n(t_0)|^{1/2}. \end{aligned}$$

Since $\Vert u_n(t)\Vert _{H^1(\varOmega \setminus \varGamma )}$ is bounded uniformly with respect to n and t, there exists a constant $C>0$ such that

$$\begin{aligned} \int _\varGamma |[u_n(t_k)]\chi _{\varGamma _{s_n(t_k)}^b}-[u_n(t_0)]\chi _{\varGamma _{s_n(t_0)}^b}|dx{\le } V_n(t_0\vee t_k)-V_n(t_0\wedge t_k)+C|s_n(t_k)-s_n(t_0)|^{1/2}. \end{aligned}$$

Passing to the limit as $n\rightarrow \infty$ along a suitable subsequence and using Lemma 2.7, we obtain

$$\begin{aligned} \int _\varGamma |g(t_k)\chi _{\varGamma _{s(t_k)}^b}-[g(t_0)]\chi _{\varGamma _{s(t_0)}^b}|dx\le V(t_0\vee t_k)-V(t_0\wedge t_k)+C|s(t_k)-s(t_0)|^{1/2}. \end{aligned}$$

Since V and s are continuous in $t_0$, this gives (2.27) and concludes the proof of the measurability of $t\mapsto u(t)$.

We now prove the equilibrium condition (b) in Definition 2.1. By (2.10) and (2.14), for every t and n we have that

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n(t)|^2dxdy+s_n(t)\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+ \int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u_n(t)]|dx\,\nonumber \\ \end{aligned}$$

(2.28)

for every $s_n(t)\le {\hat{s}}\le b$ and every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$ with ${\hat{u}}=w_n(t)$ on $\partial _D\varOmega$. In particular, taking ${\hat{s}}=s_n(t)$, we see that $u_n(t)$ satisfies the minimum problem (2.23) with $w=w_n(t)$, $s=s_n(t)$, and $g_n=[u_n(t)]$. Since $w_n(t)\rightarrow w(t)$ strongly in $H^1(\varOmega \setminus \varGamma _a^{s_0})$, $s_n(t)\rightarrow s(t)$, and $[u_n(t)]\chi _{\varGamma _{s_n(t)}^b}\rightarrow g(t)\chi _{\varGamma _{s(t)}^b}$ strongly in $L^1(\varGamma )$, by Lemma 2.8 we have

$$\begin{aligned} u_n(t)\rightarrow u(t) \quad \text {strongly in }H^1(\varOmega \setminus \varGamma ) \end{aligned}$$

(2.29)

for every $t\in [0,T]$.

We now fix $t\in [0,T]$, $s(t)\le {\hat{s}}\le b$, and ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$ with ${\hat{u}}=w(t)$ on $\partial \varOmega$. We have to prove that

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+s(t)\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+{\hat{s}}+ \int _{\varGamma _{{\hat{s}}}^b}|[{\hat{u}}]-[u(t)]|dx.\qquad \end{aligned}$$

(2.30)

Let ${\hat{s}}_n:={\hat{s}}\vee s_n(t)$ and ${\hat{u}}_n:={\hat{u}}+w_n(t)-w(t)$. Since ${\hat{u}}_n=w_n(t)$ on $\partial _D\varOmega$ and ${\hat{s}}_n\ge s_n(t)$, by (2.28) we have

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u_n(t)|^2dxdy+s_n(t)\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}_n|^2dxdy+{\hat{s}}_n+ \int _{\varGamma _{{\hat{s}}_n}^b}|[{\hat{u}}_n]-[u_n(t)]|dx.\nonumber \\ \end{aligned}$$

(2.31)

Since $u_n(t)\rightarrow u(t)$ and ${\hat{u}}_n\rightarrow {\hat{u}}$ strongly in $H^1(\varOmega \setminus \varGamma )$ by (2.29), while $s_n(t)\rightarrow s(t)$ and ${\hat{s}}_n\rightarrow {\hat{s}}$, we can pass to the limit in (2.31) and we obtain (2.30), which gives the equilibrium condition (b) in Definition 2.1.

We conclude by proving now the energy-dissipation inequality (c) in Definition 2.1. By (2.16) and by Definition 2.6 we have

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_2)|^2dxdy+s_n(t_2)-s_n(t_1)+\int _{\varGamma _{s_n(t_2)}^b}|[u_n(t_2)]-[u_n(t_1)]|dx\\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u_n(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u_n(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau +{\tilde{R}}_n. \end{aligned}$$

for every $0\le t_1\le t_2\le T$ and for every n. By (2.29) we can pass to the limit and obtain condition (c). $\square$

The following theorem shows that the notion of evolution according to Definition 2.1 can be expressed equivalently by using the notion of dissipation introduced in Definition 2.6. This shows the analogy with the definition used in [6].

Theorem 2.9

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. A pair (u, s) is a quasistatic evolution with boundary value w on $\partial _D\varOmega$ if and only if $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ is measurable, $s:[0,T]\rightarrow [s_0,b]$, conditions (a) and (b) of Definition 2.1 are satisfied, and one of the following two conditions holds:

(c$'$):

(energy-dissipation inequality starting from 0) for every $t\in [0,T]$ we have

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+{\mathrm {Diss}}(u(\cdot ),s(\cdot );0,t)\\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(0)|^2dxdy+\int _{0}^{t}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned}$$

(c$''$):

(energy-dissipation balance) for every $t_1,t_2\in [0,T]$ with $t_1<t_2$, we have

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_2)|^2dxdy+{\mathrm {Diss}}(u(\cdot ),s(\cdot );t_1,t_2)\\&\quad =\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(t)\nabla \dot{w}(t)dxdy\Big )dt. \end{aligned}$$

Proof

Let (u, s) be a quasistatic evolution with boundary value w on $\partial _D\varOmega$. By (c) and Definition 2.6 we obtain

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_2)|^2dxdy+{\mathrm {Diss}}(u(\cdot ),s(\cdot );t_1,t_2)\\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma } \nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau , \end{aligned}$$

which clearly implies (c${}'$).

To prove that (a)&(b)&(c${}'$)$\implies$(c${}''$) we argue as in the proof of the energy balance in [6, Section 6] and we obtain

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+{\mathrm {Diss}}(u(\cdot ),s(\cdot );0,t)\\&\quad \ge \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(0)|^2dxdy+\int _{0}^{t}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau ) \nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned}$$

This inequality, together with (c${}'$) gives (c${}''$) for $t_1=0$. The general case for (c${}''$) follows by additivity.

The implication (c${}''$)$\implies$(c) is an immediate consequence of Definition 2.6. $\square$

3 Some auxiliary results

In this section we prove a characterization of the solutions of the minimum problems considered in Lemma 2.8, which are connected with the equilibrium condition (b) in Definition 2.1. This is obtained by means of a suitable weak formulation of their boundary conditions on $\varGamma$. In the last part of the section we present a technical result that will be crucial in the proof of our main result in Sect. 4.

It is convenient to introduce the notation

$$\begin{aligned} H^1_{0,D}(\varOmega \setminus \varGamma ):=\{u\in H^1(\varOmega \setminus \varGamma ): u=0\text { on }\partial _D\varOmega \}. \end{aligned}$$

(3.1)

We also set

$$\begin{aligned} \partial ^\pm U:=\partial U^\pm \setminus \varGamma , \end{aligned}$$

(3.2)

where U and $U^\pm$ are the open sets introduced at the beginning of Sect. 2.

Lemma 3.1

Let $w\in H^1(\varOmega \setminus \varGamma )$, $s\in [a,b]$, $g\in L^1(\varGamma )$, and let u be the minimiser of

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega \setminus \varGamma )\\ u=w\ \text {on }\partial _D\varOmega \end{array}}\Big (\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy+\int _{\varGamma _s^b}|[u]-g|dx\Big ). \end{aligned}$$

(3.3)

Then there exists $\psi \in L^\infty (\varGamma )$, with $\psi =0$ a.e. on $\varGamma _a^s$ and $|\psi |\le 1$ a.e. on $\varGamma _s^b$, such that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi \, dxdy=\int _{\varGamma }\psi [\varphi ] \,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

(3.4)

Proof

Let $\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )$. Since $u+\varepsilon \varphi =w$ on $\partial _D\varOmega$ for every $\varepsilon \in {\mathbb {R}}$, by minimality

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla (u+\varepsilon \varphi )|^2dxdy+\int _{\varGamma _s^b}|[u]-g+\varepsilon [\varphi ]|dx\\&\quad -\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy -\int _{\varGamma _s^b}|[u]-g|dx\ge 0. \end{aligned}$$

Developing the square and using the triangle inequality we get

$$\begin{aligned} \frac{\varepsilon }{2} \int _{\varOmega \setminus \varGamma }|\nabla \varphi |^2dxdy+\int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi dxdy+\int _{\varGamma _s^b}|[\varphi ]|dx\ge 0 \end{aligned}$$

for every $\varepsilon >0$. Taking the limit as $\varepsilon \rightarrow 0+$ we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi dxdy\ge -\int _{\varGamma _s^b}|[\varphi ]|dx. \end{aligned}$$

Using the same inequality also for $-\varphi$, we deduce that

$$\begin{aligned} \Big |\int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi \, dxdy\Big |\le \int _{\varGamma _s^b}|[\varphi ]| dx \end{aligned}$$

(3.5)

for every $\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )$. Given $\varphi \in H^1(U^+)$ with $\varphi =0$ on $\partial ^+U$, we can extend it by 0 and we obtain a function in $H^1_{0,D}(\varOmega \setminus \varGamma )$. Therefore (3.5) gives

$$\begin{aligned} \Big |\int _{U^+}\nabla u\nabla \varphi dxdy\Big |\le \int _{\varGamma _s^b}|\varphi ^+|dx \end{aligned}$$

(3.6)

for every $\varphi \in H^1(U^+)$ with $\varphi =0$ on $\partial ^+U$, where $\varphi ^+$ denotes the trace of $\varphi$ on $\varGamma$ from above. Moreover, (3.5) gives also

$$\begin{aligned} \int _{U^+}\nabla u\nabla \varphi \,dxdy+\int _{U^-}\nabla u\nabla \varphi dxdy=0 \end{aligned}$$

(3.7)

for every $\varphi \in H^1_0(U\cap \varOmega )$.

Let $\mu$ be the distribution on $U\cap \varOmega$ defined by

$$\begin{aligned}\langle \mu ,\varphi \rangle :=\int _{U^+}\nabla u\nabla \varphi \,dxdy=-\int _{U^-}\nabla u\nabla \varphi \, dxdy\end{aligned}$$

for every $\varphi \in C_c^\infty (U\cap \varOmega )$. By (3.6) it is easy to prove that there exist $\psi \in L^\infty (\varGamma )$, with $\psi =0$ a.e. on $\varGamma _a^s$ and $|\psi |\le 1$ a.e. on $\varGamma _s^b$, such that

$$\begin{aligned} \langle \mu ,\varphi \rangle =\int _{\varGamma }\psi \varphi \,dx\quad \text {for every }\varphi \in C_c^\infty (U\cap \varOmega ). \end{aligned}$$

By density

$$\begin{aligned} \int _{U^+}\nabla u\nabla \varphi \, dxdy=\int _{\varGamma }\psi \varphi \,dx\quad \text {and}\quad \int _{U^-}\nabla u\nabla \varphi \, dxdy=-\int _{\varGamma }\psi \varphi \,dx \end{aligned}$$

for every $\varphi \in H^1_0(U\cap \varOmega )$.

Given $\varphi \in H^1(U^+)$ with $\varphi =0$ on $\partial ^+U$, we can extend it to a function belonging to $H^1_0(U\cap \varOmega )$. Therefore

$$\begin{aligned} \int _{U^+}\nabla u\nabla \varphi \, dxdy=\int _{\varGamma }\psi \varphi ^+ \,dx \end{aligned}$$

(3.8)

for every $\varphi \in H^1(U^+)$ with $\varphi =0$ on $\partial ^+U$. Similarly we prove that

$$\begin{aligned} \int _{U^-}\nabla u\nabla \varphi \, dxdy=-\int _{\varGamma }\psi \varphi ^- \,dx \end{aligned}$$

(3.9)

for every $\varphi \in H^1(U^-)$ with $\varphi =0$ on $\partial ^-U$, where $\varphi ^-$ denotes the trace of $\varphi$ on $\varGamma$ from below. By taking the sum we get

$$\begin{aligned} \int _{U^+\cup U^-}\nabla u\nabla \varphi \, dxdy=\int _{\varGamma }\psi [\varphi ] \,dx \end{aligned}$$

(3.10)

for every $\varphi \in H^1(U^+\cup U^-)$ with $\varphi =0$ on $\partial ^+U\cup \partial ^-U$.

Let $\omega _k$ be a sequence in $C^\infty _c(U\cap \varOmega )$, with $0\le \omega _k\le 1$, such that $\omega _k\rightarrow 1$ a.e. on $\varGamma$. Given $\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )$ we set $\varphi _k:=\omega _k\varphi$ and ${\hat{\varphi }}_k:=(1-\omega _k)\varphi$. Then $\varphi _k\in H^1(U^+\cup U^-)$, $\varphi _k=0$ on $\partial ^+U\cup \partial ^-U$, and ${\hat{\varphi }}_k\in H^1_{0,D}(\varOmega \setminus \varGamma )$. Moreover $[\varphi _k]\rightarrow [\varphi ]$ strongly in $L^1(\varGamma )$. Since $\varphi =\varphi _k+{\hat{\varphi }}_k$ we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi \, dxdy =\int _{U^+\cup U^-}\nabla u\nabla \varphi _k dxdy + \int _{\varOmega \setminus \varGamma }\nabla u\nabla {\hat{\varphi }}_k dx{dy}. \end{aligned}$$

(3.11)

By (3.10) we have

$$\begin{aligned} \int _{U^+\cup U^-}\nabla u\nabla \varphi _k dxdy \rightarrow \int _{\varGamma }\psi [\varphi ] \,dx, \end{aligned}$$

(3.12)

while (3.5) gives

$$\begin{aligned} \Big |\int _{\varOmega \setminus \varGamma }\nabla u\nabla {\hat{\varphi }}_k\, dxdy\Big |\le \int _{\varGamma _s^b}|[{\hat{\varphi }}_k]| \,dx\rightarrow 0. \end{aligned}$$

(3.13)

Equality (3.4) follows from (3.11)–(3.13). $\square$

Lemma 3.2

Let $v,\, w\in H^1(\varOmega \setminus \varGamma )$, let $s\in [a,b]$, let $g:=[v]$, and let u be the minimiser of (3.3). Then the function $\psi$ introduced in Lemma 3.1 satisfies $\psi =-1$ a.e. on $\{[u]> g\}\cap \varGamma _s^b$ and $\psi =1$ a.e. on $\{[u]< g\}\cap \varGamma _s^b$.

Proof

Since $U^-$ has Lipschitz boundary, there exist ${\tilde{u}}, \tilde{v}\in H^1(U\cap \varOmega )$ such that ${\tilde{u}}=u$ and ${\tilde{v}}=v$ in $U^-$. Let ${\hat{u}}, {\hat{v}}\in H^1(U\cap (\varOmega \setminus \varGamma ))$ be defined by ${\hat{u}}:=u-{\tilde{u}}$ and ${\hat{v}}:=v-{\tilde{v}}$, so that ${\hat{u}}^+=[{\hat{u}}]=[u]$ and ${\hat{v}}^+=[{\hat{v}}]=[v]=g$ on $\varGamma$, while ${\hat{u}}={\hat{v}}=0$ in $U^-$.

Let $A:=\{[u]>[v]\}\cap \varGamma _s^b=\{{\hat{u}}^+>{\hat{v}}^+\}\cap \varGamma _s^b$. To prove that $\psi =-1$ a.e. on A it is enough to show that

$$\begin{aligned} \int _A\psi \varphi ^+dx+\int _A\varphi ^+dx=0 \end{aligned}$$

(3.14)

for every $\varphi \in H^1(U^+)$ with $\varphi =0$ on $\partial ^+U$. Let us fix such a $\varphi$ and for every k let $\varphi _k:=(\varphi \wedge (k\omega ))\vee (-k\omega )\in H^1(U^+)$, where $\omega :=({\hat{u}}-{\hat{v}})\vee 0$. We extend $\varphi _k$ to $\varOmega \setminus \varGamma$ by setting $\varphi _k=0$ on $\varOmega \setminus (\varGamma \cup U^+)$. Since $\varphi _k=0$ on $\partial ^+U$, the extended function satisfies $\varphi _k\in H^1_{0,D}(\varOmega \setminus \varGamma )$. For every $\varepsilon$ with $|\varepsilon |<\frac{1}{k}$ we have $|\varepsilon [\varphi _k]|\le |\varepsilon \varphi _k^+|\le \omega ^+=({\hat{u}}^+-{\hat{v}}^+)\vee 0=([u]-[v])\vee 0$ a.e. on $\varGamma _s^b$. It follows that

$$\begin{aligned} \int _{\varGamma _s^b}|[u]-[v]+\varepsilon [\varphi _k]|dx-\int _{\varGamma _s^b}|[u]-[v]|dx=\int _{\varGamma _s^b}\varepsilon [\varphi _k]dx=\int _{\varGamma _s^b}\varepsilon \varphi _k^+dx. \end{aligned}$$

By the minimality of u, repeating the argument at the beginning of the proof of Lemma 3.1 we obtain

$$\begin{aligned} \frac{\varepsilon ^2}{ 2} \int _{\varOmega \setminus \varGamma }|\nabla \varphi _k|^2dxdy+\varepsilon \int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi _kdxdy+\int _{\varGamma _s^b}\varepsilon \varphi _k^+dx\ge 0 \end{aligned}$$

for every $\varepsilon \in (-\frac{1}{k},\frac{1}{k})$. Taking the derivative at $\varepsilon =0$ and using (3.4) we obtain

$$\begin{aligned} \int _{\varGamma _s^b}\psi \varphi _k^+\,dx+\int _{\varGamma _s^b}\varphi _k^+\,dx=0. \end{aligned}$$

Since $\{\varphi _k^+\ne 0\}\cap \varGamma _s^b\subset A$, we obtain

$$\begin{aligned} \int _A\psi \varphi _k^+\,dx+\int _A\varphi _k^+\,dx=0. \end{aligned}$$

Passing to the limit as $k\rightarrow \infty$ we obtain (3.14).

The proof on the set $\{[u]< g\}\cap \varGamma _s^b$ is similar. $\square$

Lemma 3.3

Let $w\in H^1(\varOmega \setminus \varGamma )$, $s\in [a,b]$, $g\in L^1(\varGamma )$, and let $u\in H^1(\varOmega \setminus \varGamma )$ with $u=w$ on $\partial _D\varOmega$. Suppose that there exists $\psi \in L^\infty (\varGamma )$ satisfying (3.4) such that

$$\begin{aligned}&\psi =0\text { a.e. on }\varGamma _a^s, \end{aligned}$$

(3.15)

$$\begin{aligned}&\psi =-1\text { a.e. on }\{[u]> g\}\cap \varGamma _s^b, \end{aligned}$$

(3.16)

$$\begin{aligned}&\psi =1 \text { a.e. on }\{[u]< g\}\cap \varGamma _s^b, \end{aligned}$$

(3.17)

$$\begin{aligned}&|\psi |\le 1 \text { a.e. on }\{[u]=g\}\cap \varGamma _s^b. \end{aligned}$$

(3.18)

Then u is the minimiser of (3.3).

Proof

Let us fix $v\in H^1(\varOmega \setminus \varGamma )$ with $v=w$ on $\partial _D\varOmega$ and let $\varphi :=v-u$. Then $\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )$. For every $\varepsilon \in [0,1]$ we define

$$\begin{aligned} f(\varepsilon ):=\frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla (u+\varepsilon \varphi )|^2dxdy+\int _{\varGamma _s^b}|[u]-g+\varepsilon [\varphi ]|dx, \end{aligned}$$

and we set

$$\begin{aligned} f'_r(0):=\lim _{\varepsilon \rightarrow 0+}\frac{f(\varepsilon )-f(0)}{\varepsilon }. \end{aligned}$$

By convexity the limit exists and

$$\begin{aligned} f(1)-f(0)\ge f'_r(0). \end{aligned}$$

(3.19)

Since

$$\begin{aligned} f(1)-f(0)= & {} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla v|^2dxdy+\int _{\varGamma _s^b}|[v]-g|dx\\&- \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u|^2dxdy-\int _{\varGamma _s^b}|[u]-g|dx, \end{aligned}$$

by (3.19) the minimality is proved if we show that

$$\begin{aligned} f'_r(0)\ge 0. \end{aligned}$$

(3.20)

By taking the derivative with respect to $\varepsilon$ in the first term of the definition of f we obtain

$$\begin{aligned} f'_r(0)= \int _{\varOmega \setminus \varGamma }\nabla u\nabla \varphi \,dxdy +\lim _{\varepsilon \rightarrow 0+}\frac{1}{\varepsilon }\int _{\varGamma _s^b}\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )dx. \end{aligned}$$

(3.21)

By (3.16) on $\{[u]> g\}\cap \varGamma _s^b$ we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+}\frac{1}{\varepsilon }\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )=[\phi ]=-\psi [\varphi ]. \end{aligned}$$

(3.22)

By (3.17) on $\{[u]< g\}\cap \varGamma _s^b$ we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+}\frac{1}{\varepsilon }\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )=-[\phi ]=-\psi [\varphi ]. \end{aligned}$$

(3.23)

Finally, by (3.18) on $\{[u]=g\}\cap \varGamma _s^b$ we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+}\frac{1}{\varepsilon }\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )=|[\varphi ]|\ge -\psi [\varphi ]. \end{aligned}$$

(3.24)

By the triangle inequality we have

$$\begin{aligned} \frac{1}{\varepsilon }\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )\ge -|[\varphi ]| \end{aligned}$$

for every $\varepsilon \in (0,1]$. We can now apply the Fatou Lemma and from (3.22)–(3.24) we obtain

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+}\frac{1}{\varepsilon }\int _{\varGamma _s^b}\big (|[u]-g+\varepsilon [\varphi ]|-|[u]-g|\big )dx\ge -\int _{\varGamma _s^b}\psi [\varphi ]\,dx.\nonumber \end{aligned}$$

(3.25)

Using this inequality, together with (3.4), (3.15), and (3.21), we obtain (3.20). $\square$

The following technical result will be used in the proof of Lemma 4.5, which is crucial to obtain our main result on the jerky crack growth. Let us fix a sequence $\varOmega _k$ of open subsets of $\varOmega$ with boundary of class $C^\infty$ such that $\varOmega _k\subset \subset \varOmega _{k+1}$ for every k and $\varOmega \setminus \varGamma =\cup _k\varOmega _k$. For every k we set (see Fig. 2)

$$\begin{aligned} S_k:=\varOmega \setminus ({\overline{\varOmega }}_k\cup \varGamma ). \end{aligned}$$

(3.26)

Fig. 2

The sets $\varOmega _k$ and $S_k$

We now prove the convergence to zero in $L^\infty$ of the sequence of harmonic functions $z_k$ on $S_k$ which satisy the homogeneous Dirichlet condition on $\partial \varOmega _k$, the homogeneous Neumann condition on $\partial \varOmega$, and the nonhomogeneous boundary condition $\frac{\partial z_k}{\partial \nu }=1$ on both sides of $\varGamma$. For every $R>0$ let $B_R=\{(x,y)\in {\mathbb {R}}^2:x^2+y^2<R^2\}$ and $B_R^{\pm }=\{(x,y)\in B_R: \pm y>0\}$.

Lemma 3.4

For every k let $S_k$ be as in (3.25) and let $z_k$ be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} z_k\in H^1(S_k),\quad z_k=0\text { a.e.\ on }\partial \varOmega _k, \\ \displaystyle \int _{S_k} \nabla z_k\nabla \varphi \,dxdy=\int _\varGamma (\varphi ^++\varphi ^-)\,dx\\ \text {for every }\varphi \in H^1(S_k)\text { with }\varphi =0\text { a.e. on }\partial \varOmega _k. \end{array}\right. } \end{aligned}$$

(3.27)

We extend $z_k$ by setting $z_k:=0$ in $\varOmega _k$. Then $z_k\rightarrow 0$ strongly in $L^\infty (\varOmega \setminus \varGamma )$.

Proof

We first prove that

$$\begin{aligned} z_k\rightarrow 0 \quad \hbox {strongly in }H^1(\varOmega \setminus \varGamma ). \end{aligned}$$

(3.28)

By taking $\varphi :=z_k$ in (3.26) we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } |\nabla z_k|^2dxdy=\int _{S_k} |\nabla z_k|^2dxdy=\int _\varGamma (z_k^++z_k^-)\,dx. \end{aligned}$$

(3.29)

Since $z_k=0$ in $\varOmega _k$, the Trace Inequality, together with the Poincaré Inequality, gives a constant $c>0$ such that

$$\begin{aligned} \int _\varGamma (z_k^++z_k^-)\,dx\le c \Big ( \int _{\varOmega \setminus \varGamma } |\nabla z_k|^2dxdy \Big )^{1/2} \end{aligned}$$

for k large enough. Together with (3.28) this implies that $\nabla z_k$ is bounded in $L^2(\varOmega \setminus \varGamma )$, hence $z_k$ is bounded in $H^1(\varOmega \setminus \varGamma )$. Since $z_k=0$ in $\varOmega _k$, we deduce that $z_k\rightharpoonup 0$ weakly in $H^1(\varOmega \setminus \varGamma )$. This implies that $z_k^++z_k^-\rightarrow 0$ strongly in $L^2(\varGamma )$, and (3.28) gives $\nabla z_k\rightarrow 0$ strongly in $L^2(\varOmega \setminus \varGamma )$. Since $z_k=0$ in $\varOmega _k$, this proves (3.27).

By the maximum principle we have

$$\begin{aligned} z_k\ge 0\quad \hbox {in }S_k. \end{aligned}$$

(3.30)

Indeed, if we take $\varphi :=z_k\wedge 0$ in (3.26) we obtain

$$\begin{aligned} \int _{S_k} |\nabla (z_k\wedge 0)|^2dxdy= \int _{S_k} \nabla z_k\nabla (z_k\wedge 0)\,dxdy=\int _\varGamma ((z_k^+\wedge 0)+ (z_k^-\wedge 0))\,dx\le 0. \end{aligned}$$

This inequality, together with the boundary condition on $\partial \varOmega _k$, implies that $z_k\wedge 0=0$ in $S_k$, which proves (3.29). Since $z_k\in C^\infty (S_k\cup \partial \varOmega _k)$ by the regularity theory of elliptic equations, (3.29) implies that $\frac{\partial z_k}{\partial \nu }\le 0$ on $\partial \varOmega _k$, where $\nu$ is the outer unit normal to $S_k$. Hence

$$\begin{aligned} \int _{S_k} \nabla z_k\nabla \varphi \,dxdy= \int _{\partial \varOmega _k}\frac{\partial z_k}{\partial \nu }\varphi \,ds\le 0 \end{aligned}$$

(3.31)

for every $\varphi \in H^1_0(\varOmega \setminus \varGamma )$ with $\varphi \ge 0$ in $\varOmega \setminus \varGamma$.

Let us prove that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } \nabla z_k\nabla \varphi \,dxdy\le \int _\varGamma (\varphi ^++\varphi ^-)\,dx \end{aligned}$$

(3.32)

for every $\varphi \in H^1(\varOmega \setminus \varGamma )$ with $\varphi \ge 0$. Let us fix such a $\varphi$ and let $\omega \in C^\infty _0(\varOmega \setminus \varGamma )$ with $\omega \ge 0$ in $\varOmega \setminus \varGamma$ and $\omega =1$ in ${\overline{\varOmega }}_k$. Then we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } \nabla z_k\nabla \varphi \,dxdy=\int _{\varOmega \setminus \varGamma } \nabla z_k\nabla (\omega \varphi )\,dxdy+ \int _{\varOmega \setminus \varGamma } \nabla z_k\nabla ((1-\omega )\varphi )\,dxdy\qquad \end{aligned}$$

(3.33)

By (3.30) we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } \nabla z_k\nabla (\omega \varphi )\,dxdy\le 0. \end{aligned}$$

(3.34)

Since $(1-\omega )\varphi =0$ on $\partial \varOmega _k$ and $(1-\omega )\varphi ^\pm =\varphi ^\pm$ on $\varGamma$, by (3.26) we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } \nabla z_k\nabla ((1-\omega )\varphi )\,dxdy=\int _\varGamma (\varphi ^++\varphi ^-)\,dx. \end{aligned}$$

(3.35)

Inequality (3.31) follows from (3.32)–(3.34).

By the maximum principle we have

$$\begin{aligned} \Vert z_k\Vert _{L^\infty (\varOmega \setminus \varGamma )}\le \Vert z_k^++z_k^-\Vert _{L^\infty (\varGamma )}. \end{aligned}$$

(3.36)

Indeed, if $M:=\Vert z_k^++z_k^-\Vert _{L^\infty (\varGamma )}$ and we take $\varphi :=(z_k-M)\vee 0$ in (3.31) we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma } |\nabla ((z_k-M)\vee 0)|^2dxdy=\int _{\varOmega \setminus \varGamma } \nabla z_k\nabla ((z_k-M)\vee 0)\,dxdy\le 0, \end{aligned}$$

which, together with the boundary condition on $\partial \varOmega _k$, implies that $(z_k-M)\vee 0=0$ in $\varOmega \setminus \varGamma$. This proves (3.35).

Therefore, to prove the lemma it is enough to show that $z_k^++z_k^-\rightarrow 0$ in $L^\infty (\varGamma )$. We shall prove only that $z_k^+\rightarrow 0$ in $L^\infty (\varGamma )$, since the result for $z_k^-$ can be proved in the same way. Let us prove first that $z_k$ is uniformly small in the intersection between $U^+$ and a suitable neighbourhood of (a, 0). Since $U^+$ has Lipschitz boundary, there exist an open neighbourhood V of (a, 0), a constant $R>0$, and a bi-Lipschitz map $\varPhi :B_R\rightarrow V$ such that $\varPhi (B_R^+)=U^+\cap V$. To simplify the exposition we assume $a=0$. Since part of the boundary of $U^+$ near $(a,0)=(0,0)$ is rectilinear, we may assume that there exists $\alpha >0$ such that $\varPhi$ is the identity map in the sector $\{(x,y)\in B_R:0\le y<\alpha x\}$ and that $(b,0)\notin B_R$.

Let $v_k(x,y):=z_k(\varPhi (x,y))$. By (3.31) and by well known properties of elliptic equations, there exists a symmetric $2{\times }2$ matrix $(a_{ij})$ of functions in $L^\infty (B_R^+)$, satisfying the uniform ellipticity condition, such that

$$\begin{aligned} \sum _{i,j=1}^2 \int _{B_R^+} a_{ij}\partial _jv_k\partial _i\varphi \,dxdy\le \int _{\varGamma _0^{R}} \varphi \,dx \end{aligned}$$

(3.37)

for every $\varphi \in H^1(B_R^+)$ with $\varphi \ge 0$ in $B_R^+$ and $\varphi =0$ on $\partial ^+ B_R:=\partial B_R\cap \partial B_R^+$, where $\partial _1=\frac{\partial }{\partial x}$ and $\partial _2=\frac{\partial }{\partial y}$.

Let $H:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be the Heaviside function defined by $H(x)=1$ for $x>0$ and $H(x)=0$ for $x<0$. Since

$$\begin{aligned} \int _{B_R^+}H\partial _2\varphi \,dxdy=- \int _{\varGamma _0^R} \varphi \,dx \end{aligned}$$

for every $\varphi \in H^1(B_R^+)$ with $\varphi =0$ on $\partial ^+ B_R$, from (3.36) we obtain that

$$\begin{aligned} \sum _{i,j=1}^2 \int _{B_R^+} a_{ij}\partial _jv_k\partial _i\varphi \,dxdy\le - \int _{B_R^+}H\partial _2\varphi \,dxdy \end{aligned}$$

(3.38)

for every $\varphi \in H^1(B_R^+)$ with $\varphi \ge 0$ in $B_R^+$ and $\varphi =0$ on $\partial ^+B_R$.

For every $(x,y)\in B_R^-$ we define $v_k(x,y):=v_k(x,-y)$, $a_{ij}(x,y):=a_{ij}(x,-y)$ for $i=j$, $a_{ij}(x,y):=-a_{ij}(x,-y)$ for $i\ne j$. Note that $v_k\in H^1(B_R)$, $a_{ij}\in L^\infty (B_R)$, and that the matrix $(a_{ij})$ is uniformly elliptic in $B_R$. Moreover, we define $F\in L^\infty (B_R)$ as

$$\begin{aligned} F(x,y):= {\left\{ \begin{array}{ll} -H(x) &{} \hbox {if }(x,y)\in B_R^+, \\ H(x) &{} \hbox {if }(x,y)\in B_R^-. \end{array}\right. } \end{aligned}$$

For every $\varphi \in H^1(B_R^-)$, with $\varphi \ge 0$ in $B_R^-$ and $\varphi =0$ on $\partial ^-B_R:=\partial B_R\cap \partial B_R^-$, we have

$$\begin{aligned}&\displaystyle \sum _{i,j=1}^2 \int _{B^-} a_{ij}\partial _j v_k\partial _i\varphi \,dxdy = \sum _{i,j=1}^2 \int _{B^+} a_{ij}\partial _j v_k\partial _i{\hat{\varphi }}\,dxdy\, \\&\displaystyle \int _{B_R^-} F\partial _2\varphi \,dxdy=- \int _{B_R^+} H\partial _2{\hat{\varphi }}\,dxdy. \end{aligned}$$

where ${\hat{\varphi }}(x,y):=\varphi (x,-y)$. Therefore (3.37) yields

$$\begin{aligned} \sum _{i,j=1}^2 \int _{B_R} a_{ij}\partial _j v_k\partial _i\varphi \,dxdy\le \int _{B_R} F\partial _2\varphi \,dxdy, \end{aligned}$$

(3.39)

for every $\varphi \in H^1_0(B_R)$ with $\varphi \ge 0$.

Given $0<r<R$, let $v^{(r)}$ be the solution of the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} v^{(r)}\in H^1_0(B_r), \\ \displaystyle \sum _{i,j=1}^2 \int _{B_r} a_{ij}\partial _j v^{(r)}\partial _i\varphi \,dxdy= \int _{B_r} F\partial _2\varphi \,dxdy\quad \text {for every }\varphi \in H^1(B_r). \end{array}\right. } \end{aligned}$$

(3.40)

Then $v_k= v^{(r)}+ v_k^{(r)}$, where $v_k^{(r)}\in H^1(B_r)$ and

$$\begin{aligned} \sum _{i,j=1}^2 \int _{B_r} a_{ij}\partial _j v_k^{(r)}\partial _i\varphi \,dxdy\le 0, \end{aligned}$$

(3.41)

for every $\varphi \in H^1_0(B_r)$ with $\varphi \ge 0$.

By (3.39) and by the global estimates for solutions of Dirichlet problems for elliptic equations with bounded measurable coefficients (see [16, Théorème 4.2]) for every $p>2$ there exists a constant $K_p>0$, independent of r, such that

$$\begin{aligned} \sup _{B_r} | v^{(r)}|\le K_p \Vert F\Vert _{L^p(B_r)} r^{1-\frac{2}{p}}. \end{aligned}$$

(3.42)

By (3.40) and by the local estimates for sub-solutions of elliptic equations with bounded measurable coefficients (see [16, Théorème 5.1]) there exists a constant $K>0$, independent of k, such that

$$\begin{aligned} \sup _{ B_{r/2}} v_k^{(r)}\le K\Big ( \frac{1}{\pi r^2}\int _{B_r} | v_k^{(r)}|^2 dxdy\Big )^{1/2} . \end{aligned}$$

Since $v_k= v^{(r)}+ v_k^{(r)}$, from these inequalities we get

$$\begin{aligned} \sup _{ B_{r/2}} v_k \le K\Big ( \frac{1}{\pi r^2}\int _{B_r} | v_k|^2 dxdy\Big )^{1/2} + K_p( K+1) \Vert F\Vert _{L^p(B_r)} r^{1-\frac{2}{p}} . \end{aligned}$$

(3.43)

Since $v_k(x,y):=z_k(\varPhi (x,y))\ge 0$ on $B_R^+$, by (3.27) we have that $v_k\rightarrow 0$ strongly in $L^2(B_R^+)$ and by (3.42) we have

$$\begin{aligned} \limsup _{k\rightarrow \infty } \sup _{V_{r/2}} |z_k| \le K_p( K+1) \Vert F\Vert _{L^p(B_r)} r^{1-\frac{2}{p}} , \end{aligned}$$

where $V_{r/2}:= \varPhi (B^+_{r/2})$. Therefore, for every $\varepsilon >0$ there exist $k_0$ and a neighbourhood W of (a, 0) such that

$$\begin{aligned} \sup _{W\cap U^+} |z_k| \le \varepsilon \end{aligned}$$

(3.44)

for every $k\ge k_0$. In a similar way we can prove the same result in a neighbourhood of (b, 0).

For every $a<x<b$ the local estimates at the boundary for solutions to Neumann problems, together with (3.27), imply that there exist $k_0$ and a neighbourhood W of (x, 0) such that (3.43) holds. By a covering argument we conclude that $z_k^+\rightarrow 0$ in $L^\infty (\varGamma )$. $\square$

We now use the previous lemma to show that the displacement u corresponding to a quasistatic evolution is bounded in $L^\infty$ provided the same property holds for the boundary value w.

Corollary 3.5

Let $T>0$, $s_0\in [a,b)$, $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$, and let (u, s) be a quasistatic evolution with boundary value w on $\partial _D\varOmega$ according to Definition 2.1. Assume that w(t) is bounded in $L^\infty (\varOmega )$ uniformly with respect to t. Then there exists a constant $M>0$ such that

$$\begin{aligned} \Vert u(t)\Vert _{L^\infty (\varOmega \setminus \varGamma )}\le M \end{aligned}$$

(3.45)

for every $t\in [0,T]$.

Proof

Let us fix k and let $\varOmega _k$, $S_k$, and $z_k$ be as in Lemma 3.4. Since u(t) is harmonic in $\varOmega \setminus \varGamma$, by (2.4) and by the Mean Value Theorem there exists a constant $M_k$ such that

$$\begin{aligned} \max _{{\overline{\varOmega }}_k}|u(t)|\le M_k \end{aligned}$$

(3.46)

for every $t\in [0,T]$. It is not restrictive to assume that

$$\begin{aligned} M_k\ge \Vert w(t)\Vert _{L^\infty (\varOmega )}\quad \text {for every }t\in [0,T]. \end{aligned}$$

(3.47)

Using the standard argument that leads to the maximum principle we now prove that

$$\begin{aligned} |u(t)|\le M_k+z_k \quad \hbox {in }S_k. \end{aligned}$$

(3.48)

By the equilibrium condition (b) in Definition 2.1 and by Lemma 3.1 for every $t\in [0,T]$ there exists $\psi (t)\in L^\infty (\varGamma )$, with $\Vert \psi (t)\Vert _{L^\infty (\varGamma )}\le 1$, such that

$$\begin{aligned} \int _{S_k}\nabla u(t)\nabla \varphi \, dxdy=\int _{\varGamma }\psi (t) [\varphi ] \,dx \end{aligned}$$

for every $\varphi \in H^1(S_k)$ with $\varphi =0$ on $\partial \varOmega _k\cup \partial _D\varOmega$. By (3.26) we have

$$\begin{aligned} \int _{S_k} \nabla (M_k+ z_k)\nabla \varphi \,dxdy=\int _\varGamma (\varphi ^++\varphi ^-)\,dx \end{aligned}$$

for every $\varphi \in H^1(S_k)$ with $\varphi =0$ on $\partial \varOmega _k$. Subtracting the first equality from the second one we get

$$\begin{aligned} \int _{S_k}\nabla (M_k+ z_k-u(t))\nabla \varphi \, dxdy=\int _{\varGamma }\big ((1-\psi (t))\varphi ^++(1+ \psi (t))\varphi ^-\big ) \,dx \end{aligned}$$

(4.1)

for every $\varphi \in H^1(S_k)$ with $\varphi =0$ on $\partial \varOmega _k\cup \partial _D\varOmega$. Let us take $\varphi :=(M_k+z_k -u(t))\wedge 0$. Since $z_k=0$ on $\partial \varOmega _k$ and $M_k-u(t)\ge 0$ on $\partial \varOmega _k$ by (3.45), we have that $\varphi =0$ on $\partial \varOmega _k$. Since $z_k\ge 0$ on $\partial _D\varOmega$ by (3.29) and $M_k -u(t)= M_k -w(t)\ge 0$ on $\partial _D\varOmega$ by (3.46), we have also $\varphi =0$ on $\partial _D\varOmega$. Therefore (3.48) gives

$$\begin{aligned} \int _{S_k}\nabla (M_k+z_k-u(t))\nabla \big ( (M_k+z_k-u(t))\wedge 0\big )\, dxdy\le 0. \end{aligned}$$

This gives $(M_k+z_k-u(t))\wedge 0=0$ in $S_k$, which implies $u(t)\le M_k+z_k$ in $S_k$. In the same way we prove that $-u(t)\le M_k+z_k$, obtaining (3.47). This inequality together with (3.45) yields (3.44), since $z_k\in L^\infty (S_k)$ by Lemma 3.4. $\square$

4 The jerky growth of the cracks

In this section we prove the main result of the paper: under suitable continuity assumptions on the boundary datum w, for every quasistatic evolution (u, s) the nondecreasing function s is piecewise constant. In other words, the crack grows only through sudden jumps. More precisely, we obtain the following result.

Theorem 4.1

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))\cap C^0([0,T];L^\infty (\varOmega ))$. Let (u, s) be a quasistatic evolution with boundary value w on $\partial _D\varOmega$, according to Definition 2.1. Then there exist a finite number of times $t_0, t_1,\ldots , t_m$, with $0= t_0<t_1<\cdots<t_{m-1}<t_m= T$, and a finite number $s_1, s_2, \ldots , s_m$ of elements of $[s_0,b]$, with $s_0\le s_1<s_2<\cdots<s_{m-1}<s_m\le b$, such that for every $j=1,\ldots ,m$ we have $s(t)=s_j$ for every $t\in (t_{j-1},t_j)$.

Remark 4.2

The previous theorem does not exclude that $s_1\ne s(0)$, i.e., the constant value of s(t) in the interval $[0,t_1]$ might be different from s(0). This means that a jump of the crack might occur at $t=0$. However, if we take $u(0)=0$, the energy-dissipation condition (c) in Definition 2.1 gives

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy+s(t)-s(0)+\int _{\varGamma _{s(t)}^b}\!\!\!\!\!|[u(t)]|dx\nonumber \\&\quad \le \int _{0}^{t}\!\!\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau , \end{aligned}$$

(4.2)

which implies, by (2.3) and Theorem 4.1,

$$\begin{aligned} s_1-s(0)=s(t)-s(0)\le M_1^{1/2}\int _{0}^{t}\Big (\int _{\varOmega \setminus \varGamma }|\nabla \dot{w}(\tau )|^2dxdy\Big )^{1/2}d\tau , \end{aligned}$$

for every $t\in (0,t_1)$. Taking the limit as $t\rightarrow 0+$ we obtain that $s_1=s(0)$. Therefore, Theorem 4.1 implies that, if $u(0)=0$, then $s(t)=s(0)$ for every $t\in [0,t_1)$.

We now fix the notation we are going to use in the lemmas that will lead to the proof of Theorem 4.1. Let (u, s) be a quasistatic evolution with boundary value w on $\partial _D\varOmega$, according to Definition 2.1. For every $t_1$, $t_2\in [0,T]$, with $t_1<t_2$, we define

$$\begin{aligned} \begin{aligned}&\omega _{1,2}=\omega (t_1,t_2):=\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(t)\nabla \dot{w}(t)dxdy\Big )dt\\&-\frac{1}{2}\int _{\varOmega \setminus \varGamma }(\nabla u(t_2)+\nabla u(t_1))(\nabla w(t_2)-\nabla w(t_1))dxdy. \end{aligned} \end{aligned}$$

(4.3)

Note that $\omega _{1,2}$ can be interpreted as the difference between the integral on $[t_1,t_2]$ of the function $t\mapsto \int _{\varOmega \setminus \varGamma }\nabla u(t)\nabla \dot{w}(t)dxdy$ and its approximation obtained by replacing $\nabla u(t)$ with $(\nabla u(t_2)+\nabla u(t_1))/2$.

To simplify the notation we set

$$\begin{aligned} u_i=u(t_i),\ w_i=w(t_i),\ s_i=s(t_i). \end{aligned}$$

(4.4)

By the equilibrium condition (b) we can apply Lemma 3.1 and we obtain that for $i=1,2$ there exists $\psi _i\in L^\infty (\varGamma )$ such that

$$\begin{aligned}&\psi _i=0\hbox { a.e. on }\varGamma _a^{s_i}\hbox { and }|\psi _i|\le 1\hbox { a.e. on }\varGamma _{s_i}^b, \end{aligned}$$

(4.5)

$$\begin{aligned}&\displaystyle \int _{\varOmega \setminus \varGamma }\nabla u_i\nabla \varphi \, dxdy=\int _{\varGamma }\psi _i [\varphi ] \,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

(4.6)

The first step in the proof of Theorem 4.1 is given by the following result.

Lemma 4.3

Under the assumptions of Theorem 4.1, let $0\le t_1<t_2\le T$, and let $u_i$, $w_i$, $s_i$, $\psi _i$, and $\omega _{1,2}$ be as in (4.2)–(4.5). Then

$$\begin{aligned}&\frac{1}{2}\int _{\varGamma _{s_1}^{s_2}}\!\! \!\! \psi _1[u_2-u_1]\,dx\nonumber \\&\quad +\frac{1}{2}\int _{\varGamma _{s_2}^b} \!\! \!\! (\psi _1+\psi _2)[u_2-u_1]\,dx+s_2-s_1+ \int _{\varGamma _{s_2}^b} \!\! \!\! |[u_2-u_1]| dx\le \omega _{1,2}. \end{aligned}$$

(4.7)

Moreover, there exists a constant M, independent of $t_1$, $t_2$, $s_1$, and $s_2$, such that

$$\begin{aligned}&\displaystyle -\frac{1}{2}\int _{\varGamma _{s_1}^{s_2}}|[u_2-u_1]|dx+s_2-s_1\le \omega _{1,2} \end{aligned}$$

(4.8)

$$\begin{aligned}&\displaystyle \frac{1}{2}\int _{\varGamma _{s_2}^b}(\psi _1+\psi _2)[u_2-u_1]\,dx+\int _{\varGamma _{s_2}^b}|[u_2-u_1]|\,dx\le M(s_2-s_1)+\omega _{1,2}. \end{aligned}$$

(4.9)

Proof

By the energy-dissipation inequality (condition (c) in Definition 2.1) we have

$$\begin{aligned}&\displaystyle \frac{1}{2}\int _{\varOmega \setminus \varGamma }(\nabla u_2+\nabla u_1)(\nabla u_2-\nabla u_1)dxdy+s_2-s_1+\int _{\varGamma _{s_2}^b}|[u_2]-[u_1]|dx\nonumber \\&\displaystyle \le \int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(t)\nabla \dot{w}(t)dxdy\Big )dt. \end{aligned}$$

(4.10)

We set $\varphi :=(u_2-u_1)-(w_2-w_1)$ and observe that $\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )$ and that $[\varphi ]=[u_2-u_1]$ on $\varGamma _{s_1}^b$. By (4.4) and (4.5) we obtain

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_2(\nabla u_2-\nabla u_1)dxdy\nonumber \\&\quad =\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_2\nabla \varphi \,dxdy+\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_2(\nabla w_2-\nabla w_1)dxdy\\&\quad =\frac{1}{2}\int _{\varGamma _{s_2}^b}\psi _2[u_2-u_1]\,dx+\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_2(\nabla w_2-\nabla w_1)dxdy. \nonumber \end{aligned}$$

(4.11)

In the same way we obtain

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_1(\nabla u_2-\nabla u_1)dxdy\nonumber \\&\quad =\frac{1}{2}\int _{\varGamma _{s_1}^b}\psi _1[u_2-u_1]\,dx+\frac{1}{2}\int _{\varOmega \setminus \varGamma }\nabla u_1(\nabla w_2-\nabla w_1)dxdy. \end{aligned}$$

This equality, together with (4.2), (4.9), and (4.10), gives (4.6).

Since $|\psi _i|\le 1$ a.e. on $\varGamma$ for $i=1,2$, we have $\frac{1}{2}(\psi _1+\psi _2)[u_2-u_1]+|[u_2-u_1]|\ge 0$ and $\psi _1[u_2-u_1]\ge -|[u_2-u_1]|$ a.e. on $\varGamma$. Therefore (4.6) implies (4.7).

By Corollary 3.5 there exists a constant M, independent of $t_1$, $t_2$, $s_1$, and $s_2$, such that

$$\begin{aligned} |[u_2-u_1]|\le 2M \quad \text { on }\varGamma . \end{aligned}$$

Since $|\psi _1|\le 1$ a.e. on $\varGamma _{s_1}^b$, this implies that

$$\begin{aligned} \frac{1}{2}\int _{\varGamma _{s_1}^{s_2}}\psi _1[u_2-u_1]\,dx+s_2-s_1\ge ( -M+1)(s_2-s_1)\ge -M(s_2-s_1). \end{aligned}$$

This inequality, together with (4.6), gives (4.8). $\square$

To continue the proof of Theorem 4.1 we want to show that under suitable assumptions on $t_1,t_2,s_1,s_2$ we have

$$\begin{aligned} |[u_2-u_1]|\le 1\quad \hbox { a.e. on }\varGamma . \end{aligned}$$

(4.12)

This inequality, together with (4.7), gives

$$\begin{aligned} s_2-s_1\le 2\omega _{1,2}, \end{aligned}$$

(4.13)

which is an important intermediate result in the proof. The next lemma is the first step in the proof of (4.11).

Lemma 4.4

Under the assumptions of Theorem 4.1, for every $\varepsilon >0$ there exists $\delta >0$ such that

$$\begin{aligned}&\Vert w_2-w_1\Vert _{L^\infty (\varOmega )}<\varepsilon ,\\&\Vert u_2-u_1\Vert _{H^1(\varOmega \setminus \varGamma )}<\varepsilon , \end{aligned}$$

(4.14)

whenever

$$\begin{aligned}&0\le t_1<t_2\le T \quad \hbox {and}\quad t_2-t_1<\delta , \end{aligned}$$

(4.15)

$$\begin{aligned}&s_0\le s_1\le s_2\le b \quad \hbox {and}\quad s_2-s_1<\delta , \end{aligned}$$

(4.16)

where $u_i$, $w_i$, and $s_i$, $i=1,2$, are defined as in (4.3).

Proof

Given a pair of sequences $t^n_1$, $t^n_2\in [0,T]$, with $t^n_1<t^n_2$, we set $u^n_i:=u(t^n_i)$, $w^n_i:=w(t^n_i)$, $s^n_i:=s(t^n_i)$, and $\omega ^n_{1,2}:=\omega (t^n_1,t^n_2)$, where $\omega$ is defined in (4.2). To prove the lemma it is enough to show that

$$\begin{aligned}&w^n_2-w^n_1\rightarrow 0 \quad \text {strongly in }L^\infty (\varOmega ), \end{aligned}$$

(4.17)

$$\begin{aligned}&u^n_2-u^n_1\rightarrow 0 \quad \text {strongly in }H^1(\varOmega \setminus \varGamma ), \end{aligned}$$

(4.18)

assuming that $t^n_2-t^n_1\rightarrow 0$ and $s^n_2-s^n_1\rightarrow 0$. The convergence of $w^n_2-w^n_1$ follows from the fact that $w\in C^0([0,T];L^\infty (\varOmega ))$. Note that by Remark 2.2, the convergence $t^n_2-t^n_1\rightarrow 0$ implies that $\omega ^n_{1,2}\rightarrow 0$, since $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. By compactness we may also assume that there exists $t_*\in [0,T]$ and $s_*\in [s_0,b]$ such that $t^n_1\rightarrow t_*$, $t^n_2\rightarrow t_*$, $s^n_1\rightarrow s_*$, and $s^n_2\rightarrow s_*$.

By Remark 2.2 a suitable subsequence satisfies $u^n_i\mathop {\rightharpoonup }u^*_i$ weakly in $H^1(\varOmega \setminus \varGamma )$ for some $u^*_i\in H^1(\varOmega \setminus \varGamma )$, for $i=1,2$. This implies in particular that $[u^n_i]\rightarrow [u^*_i]$ strongly in $L^2(\varGamma )$. Since $w^n_i\rightarrow w_*:=w(t_*)$ strongly in $H^1(\varOmega \setminus \varGamma _{a}^{s_0})$, from the minimality of $u^n_i$ and Lemma 2.8 we deduce that $u^n_i\rightarrow u^*_i$ strongly in $H^1(\varOmega \setminus \varGamma )$ and that

$$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u^*_i|^2\,dxdy\le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla v|^2\,dxdy+\int _{\varGamma _{s_*}^b}|[v]-[u^*_i]|dx \end{aligned}$$

for every $v\in H^1(\varOmega \setminus \varGamma )$ with $v=w_*$ on $\partial _D\varOmega$.

By the Euler condition (see Lemma 3.1) we obtain that there exist $\psi ^n_i\in L^\infty (\varGamma )$, with $|\psi ^n_i|\le 1$ a.e. on $\varGamma$ and $\psi ^*_i\in L^\infty (\varGamma )$, with $|\psi ^*_i|\le 1$ a.e. on $\varGamma$, such that for $i=1,2$ we have

$$\begin{aligned}&\int _{\varOmega \setminus \varGamma }\nabla u^n_i\nabla \varphi \, dxdy=\int _{\varGamma }\psi ^n_i [\varphi ] \,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma )\hbox { and every }n,\nonumber \\&\int _{\varOmega \setminus \varGamma }\nabla u^*_i\nabla \varphi \, dxdy=\int _{\varGamma }\psi ^*_i [\varphi ] \,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

(4.19)

Therefore the convergence of $u^n_i$ to $u^*_i$ in $H^1(\varOmega \setminus \varGamma )$ implies that $\psi ^n_i\mathop {\rightharpoonup }\psi ^*_i$ weakly${}^*$ in $L^\infty (\varGamma )$.

Since $\frac{1}{2}(\psi ^n_1+\psi ^n_2)[u^n_2-u^n_1]+|[u^n_2-u^n_1]|\ge 0$ (which follows from the fact that $|\psi ^n_i|\le 1$) and $\omega _{1,2}^n\rightarrow 0$, by (4.8) we have

$$\begin{aligned} \frac{1}{2}\int _{\varGamma }(\psi ^n_1+\psi ^n_2)[u^n_2-u^n_1]\,dx+\int _{\varGamma }|[u^n_2-u^n_1]|\,dx\rightarrow 0. \end{aligned}$$

This implies that

$$\begin{aligned} \frac{1}{2}\int _{\varGamma }(\psi ^*_1+\psi ^*_2)[u^*_2-u^*_1]\,dx+\int _{\varGamma }|[u^*_2-u^*_1]|\,dx=0. \end{aligned}$$

Since $\frac{1}{2}(\psi ^*_1+\psi ^*_2)[u^*_2-u^*_1]+|[u^*_2-u^*_1]|\ge 0$, we deduce that $\frac{1}{2}(\psi ^*_1+\psi ^*_2)[u^*_2-u^*_1]+|[u^*_2-u^*_1]|=0$ a.e. on $\varGamma$. Using the inequality $|\psi ^*_i|\le 1$ a.e. on $\varGamma$, we obtain $\psi ^*_1=\psi ^*_2$ on $\{[u^*_2-u^*_1]\ne 0\}$.

As $u^*_1=u^*_2=w^*$ on $\partial _D\varOmega$, we have $u^*_2-u^*_1\in H^1_{0,D}(\varOmega \setminus \varGamma )$. By (4.19) we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla (u^*_2-u^*_1)\nabla \varphi \, dxdy=\int _{\varGamma }(\psi ^*_2-\psi ^*_1) [\varphi ] \,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

Taking $\varphi =u^*_2-u^*_1$ we deduce that $\nabla u^*_1=\nabla u^*_2$, which implies $u^*_1= u^*_2$, since $u^*_1=u^*_2=w^*$ on $\partial _D\varOmega$ and (2.1) holds. Therefore the strong convergence of $u^n_i$ to $u^*_i$ implies (4.18). Since the result does not depend on the subsequence, (4.18) holds for the whole sequence. $\square$

We now complete the proof of (4.11).

Lemma 4.5

Under the assumptions of  Theorem 4.1, there exists $\delta _0>0$ such that (4.11) holds whenever

$$\begin{aligned}&0\le t_1<t_2\le T \quad \hbox {and}\quad t_2-t_1<\delta _0, \end{aligned}$$

(4.20)

$$\begin{aligned}&s_0\le s_1\le s_2\le b \quad \hbox {and}\quad s_2-s_1<\delta _0, \end{aligned}$$

(4.21)

where $u_i$ and $s_i$, $i=1,2$, are defined as in (4.3).

Proof

Let $\varOmega _k$ be a sequence of open subsets of $\varOmega$ with boundary of class $C^\infty$ such that $\varOmega _k\subset \subset \varOmega _{k+1}$ for every k and $\varOmega \setminus \varGamma =\cup _k\varOmega _k$. For every k let $S_k$ and $z_k$ be defined by (3.25) and (3.26). By Lemma 3.4 there exists $k_0$ such that

$$\begin{aligned} \Vert z_{k_0}\Vert _{L^\infty (S_{k_0})}\le 1/8. \end{aligned}$$

(4.22)

We fix $\rho >0$ such that $B_\rho (x_0,y_0)\subset \varOmega \setminus \varGamma$ for every $(x_0,y_0)\in {\overline{\varOmega }}_{k_0}$. By the Mean Value Theorem we have

$$\begin{aligned} |(u_2-u_1)(x_0,y_0)|\le \frac{1}{\pi \rho ^2}\int _{B_\rho (x_0,y_0)}|u_2-u_1|\,dxdy\le \frac{1}{\pi ^{1/2} \rho }\Vert u_2-u_1\Vert _{L^2(\varOmega \setminus \varGamma )} \qquad \end{aligned}$$

(4.23)

for every $(x_0,y_0)\in {\overline{\varOmega }}_{k_0}$. We now fix $0<\varepsilon _0<1/4$ such that $\varepsilon _0/(\pi ^{1/2}\rho )<1/4$. The constant $\delta$ given by Lemma 4.4 for $\varepsilon =\varepsilon _0$ will be denoted by $\delta _0$. If (4.20) and (4.21) hold, by (4.23) we have

$$\begin{aligned} |(u_2-u_1)(x_0,y_0)|< \tfrac{1}{4} \quad \hbox {for every }(x_0,y_0)\in {\overline{\varOmega }}_{k_0}. \end{aligned}$$

(4.24)

Using the standard argument that leads to the maximum principle we now prove that

$$\begin{aligned} |u_2-u_1|\le \tfrac{1}{4}+2z_{k_0} \quad \hbox {in }S_{k_0}. \end{aligned}$$

(4.25)

By (4.5) we have

$$\begin{aligned} \int _{S_{k_0}}\nabla (u_2-u_1)\nabla \varphi \, dxdy=\int _{\varGamma }(\psi _2-\psi _1) [\varphi ] \,dx \end{aligned}$$

for every $\varphi \in H^1(S_{k_0})$ with $\varphi =0$ on $\partial \varOmega _{k_0}\cup \partial _D\varOmega$. By (3.26) we have

$$\begin{aligned} \int _{S_{k_0}}\nabla (\tfrac{1}{4}+2 z_{k_0})\nabla \varphi \,dxdy=\int _\varGamma 2(\varphi ^++\varphi ^-)\,dx \end{aligned}$$

for every $\varphi \in H^1(S_{k_0})$ with $\varphi =0$ on $\partial \varOmega _{k_0}$. Subtracting the terms of the first equality from those of the second one we get

$$\begin{aligned} \int _{S_{k_0}}\nabla (\tfrac{1}{4}+2 z_{k_0}-u_2+u_1)\nabla \varphi \, dxdy=\int _{\varGamma }\big ((2-\psi _2+\psi _1)\varphi ^++(2+ \psi _2-\psi _1)\varphi ^-\big ) \,dx \end{aligned}$$

(4.26)

for every $\varphi \in H^1(S_{k_0})$ with $\varphi =0$ on $\partial \varOmega _{k_0}\cup \partial _D\varOmega$. Let us take $\varphi :=(\frac{1}{4}+2z_{k_0} -u_2+u_1)\wedge 0$. Since $z_{k_0}=0$ on $\partial \varOmega _{k_0}$ and $\frac{1}{4} -u_2+u_1\ge 0$ on $\partial \varOmega _{k_0}$ by (4.24), we have that $\varphi =0$ on $\partial \varOmega _{k_0}$. Since $z_{k_0}\ge 0$ on $\partial _D\varOmega$ by (3.29) and $\frac{1}{4} -u_2+u_1=\frac{1}{4} -w_2+w_1\ge 0$ on $\partial _D\varOmega$ by (4.13), we have also $\varphi =0$ on $\partial _D\varOmega$. Therefore (4.26) gives

$$\begin{aligned} \int _{S_{k_0}}\nabla (\tfrac{1}{4}+2z_{k_0}-u_2+u_1)\nabla \big ( (\tfrac{1}{4}+2z_{k_0}-u_2+u_1)\wedge 0\big )\, dxdy\le 0. \end{aligned}$$

This gives $(\frac{1}{4}+2z_{k_0}-u_2+u_1)\wedge 0=0$ in $S_{k_0}$, which implies $u_2-u_1\le \frac{1}{4}+2z_{k_0}$ in $S_{k_0}$. In the same way we prove that $u_1-u_2\le \frac{1}{4}+2z_{k_0}$, obtaining (4.25).

By (4.22), (4.24), and (4.25) we obtain $|u_2-u_1|\le \frac{1}{2}$ in $\varOmega \setminus \varGamma$. This implies that $|u_2^+-u_1^+|\le \frac{1}{2}$ and $|u_2^--u_1^-|\le \frac{1}{2}$ on $\varGamma$, which give (4.11). $\square$

Inequality (4.11), together with (4.7), gives the following result.

Corollary 4.6

Under the assumptions of Theorem 4.1, let $\delta _0>0$ be the constant introduced in Lemma 4.5. Let $t_1,\,t_2\in [0,T]$, with $t_1<t_2$, and let $s_1$, $s_2$, and $\omega _{1,2}$ be defined by (4.2) and (4.3). If (4.20) and (4.21) hold, then (4.12) is satisfied.

We now consider an interval $[\tau _1,\tau _2]$, with no restriction on its length. Iterating estimate (4.12) on the intervals of a suitable subdivision we obtain an estimate on the difference $s(\tau _2)-s(\tau _1)$.

Lemma 4.7

Under the assumptions of Theorem 4.1, let $\delta _0>0$ be the constant introduced in Lemma 4.5 and let $[\tau _1,\tau _2]\subset [0,T]$. Suppose that there exists a finite subdivision $\tau _1=t_0<t_1<\cdots <t_m=\tau _2$ of the interval $[\tau _1,\tau _2]$ such that

$$\begin{aligned} t_j-t_{j-1}<\delta _0\quad \text { and }\quad s(t_j)-s(t_{j-1})<\delta _0 \end{aligned}$$

(4.27)

for every $j=1,\ldots ,m$. Then

$$\begin{aligned} s(\tau _2)-s(\tau _1)\le 2\sum _{j=1}^m\omega (t_{j-1},t_j), \end{aligned}$$

(4.28)

where $\omega$ is defined by (4.2).

Proof

It is enough to apply Corollary 4.6 to each interval $[t_{j-1}, t_j]$. $\square$

To conclude the proof of Theorem 4.1 we have to show that, under suitable assumptions, it is possible to find a subdivision such that (4.27) holds and the right-hand side of (4.28) is arbitrarily small. We shall see (Corollary 4.11) that the latter property is related to the approximation of a Lebesgue integral by its Riemann sums.

As for (4.27), it is clear that the second inequality follows from an estimate on $t_j-t_{j-1}$ when s is continuous. The following lemma shows that this happens even if s is discontinuous provided it is nondecreasing and its jumps have an amplitude less than $\delta _0$. For every nondecreasing real-valued function s defined on an interval $[\tau _1,\tau _2]$ and for every $t\in [\tau _1,\tau _2]$ we define

$$\begin{aligned} {[}s](t):=s(t+)-s(t-), \end{aligned}$$

where $s(t+)$ and $s(t-)$ are the right and left limits of s at t, with the convention $s(\tau _1-)=s(\tau _1)$ and $s(\tau _2+)=s(\tau _2)$.

Lemma 4.8

Let $\tau _1<\tau _2$ and let $s:[\tau _1,\tau _2]\rightarrow {\mathbb {R}}$ be a nondecreasing function. Let $\delta _0>0$ be such that $[s](t)<\delta _0$ for every $t\in [\tau _1,\tau _2]$. Then there exists $\eta _0\in (0,\delta _0]$ such that

$$\begin{aligned} s(t_2)-s(t_1)<\delta _0\quad \text { for every }t_1,t_2\in [\tau _1,\tau _2]\ \text {with}\ 0<t_2-t_1<\eta _0. \end{aligned}$$

Proof

Let J be the set of jump points of s, which is at most countable. Let us prove that

$$\begin{aligned} \sup _{t\in J}[s](t)<\delta _0. \end{aligned}$$

(4.29)

This is trivial if the supremum is zero. Otherwise we fix $0<\delta _1<\sup _{t\in J}[s](t)$ and we observe that

$$\begin{aligned} \sup _{t\in J}[s](t)= \max _{t\in F_1}[s](t)<\delta _0, \end{aligned}$$

where $F_1$ is the finite set defined by $F_1:=\{t\in [\tau _1,\tau _2]: [s](t)>\delta _1\}$. This concludes the proof of (4.29).

Let $\delta _2$ be such that

$$\begin{aligned} \sup _{t\in J}[s](t)+2\delta _2<\delta _0. \end{aligned}$$

(4.30)

Let us decompose s as $s=s_j+s_c$, where $s_j$ is the pure jump component of s defined by

$$\begin{aligned} s_j(t)=s(t)-s(t-)+\sum _{\tau \in J,\tau < t}[s](\tau ), \end{aligned}$$

(4.31)

while $s_c$ is its continuous component.

Let $\eta _1>0$ be such that

$$\begin{aligned} s_c(t_2)-s_c(t_1)<\delta _2 \end{aligned}$$

(4.32)

whenever $0<t_2-t_1<\eta _1$. On the other hand there exists a finite set $F_2\subset J$ such that

$$\begin{aligned} \sum _{t\in J\setminus F_2}[s](t)<\delta _2. \end{aligned}$$

(4.33)

Since $F_2$ is finite, the distance between any two distinct points in $F_2$ is larger than some constant $\eta _2>0$.

Set $\eta _0:=\eta _1\wedge \eta _2\wedge \delta _0$ and let $t_1, t_2\in [\tau _1,\tau _2]$ with $0<t_2-t_1<\eta _0$. First of all we note that $[t_1,t_2]$ contains at most one point $\tau \in F_2$. Then, by (4.31)–(4.33), we have

$$\begin{aligned} s(t_2)-s(t_1)=s_j(t_2)-s_j(t_1)+s_c(t_2)-s_c(t_1)\le \delta _2+[s](\tau )+\delta _2. \end{aligned}$$

By (4.30) the conclusion follows. $\square$

Combining Lemmas 4.7 and 4.8 we obtain the following result.

Lemma 4.9

Under the assumptions of Theorem 4.1, let $\delta _0>0$ be the constant given by Lemma 4.5, and let F be the finite set defined by

$$\begin{aligned} F:=\{t\in [0,T]: [s](t)\ge \delta _0\}. \end{aligned}$$

Let $\tau _1,\tau _2\in [0,T]$, with $\tau _1<\tau _2$, be such that

$$\begin{aligned}{}[\tau _1,\tau _2]\cap F=\emptyset , \end{aligned}$$

and let $\tau _1=t_0<t_1<\cdots <t_m=\tau _2$ be a subdivision of the interval $[\tau _1,\tau _2]$ such that

$$\begin{aligned} t_j-t_{j-1}<\eta _0 \end{aligned}$$

(4.34)

for every $j=1,\ldots ,m$, where $\eta _0$ is the constant introduced in Lemma 4.8 corresponding to $\delta _0$. Then (4.28) holds.

Proof

By Lemma 4.8 the inequality (4.34) implies the second condition in (4.27), so that the conclusion follows from Lemma 4.7. $\square$

The following proposition, related to the approximation of a Lebesgue integral by suitable Riemann sums, will be used to show that the right-hand side of (4.28) can be made arbitrarily small by a suitable choice of the subdivision.

Proposition 4.10

Let H be a Hilbert space, let $\tau _1<\tau _2$, and let $f,g:[\tau _1,\tau _2]\rightarrow H$ be Bochner integrable functions. Assume that there exists a constant $M>0$ such that $\Vert f(t)\Vert \le M$ for every $t\in [\tau _1,\tau _2]$, where $\Vert \cdot \Vert$ denotes the norm in H. Then for every integer $k\ge 1$ there exists a subdivision $\tau _1=t_0^k<t_1^k<\cdots <t_{m_k}^k=\tau _2$ such that $t_j^k-t_{j-1}^k\le \frac{1}{k}$ for every $1\le j\le m_k$ and

$$\begin{aligned} \begin{aligned} \int _{\tau _1}^{\tau _2}(f(t),g(t))dt&=\lim _{k\rightarrow \infty }\sum _{j=1}^{m_k}\int _{t_{j-1}^k}^{t_j^k}(f(t_j^k),g(t))dt\\&=\lim _{k\rightarrow \infty }\sum _{j=1}^{m_k}\int _{t_{j-1}^k}^{t_j^k}(f(t_{j-1}^k),g(t))dt, \end{aligned} \end{aligned}$$

(4.35)

where $(\cdot ,\cdot )$ denotes the scalar product in H.

Proof

A direct proof of (4.35) can be obtained by adapting the proof in [8, page 63]. We provide here a short proof based on [4, Lemma 4.12], which guarantees for every $k\ge 1$ the existence of a subdivision $\tau _1=t_0^k<t_1^k<\cdots <t_{m_k}^k=\tau _2$ such that $t_j^k-t_{j-1}^k\le \frac{1}{k}$ for every $1\le j\le m_k$ and

$$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j=1}^{m_k}\int _{t_{j-1}^k}^{t_j^k}\Vert f(t)-f(t_j^k)\Vert dt\rightarrow 0. \end{aligned}$$

(4.36)

Let us define $F_k:[\tau _1,\tau _2)\rightarrow H$ by

$$\begin{aligned} F_k(t):=f(t)-\sum _{j=1}^{m_k}f(t_j^k)\chi _{[t_{j-1}^k,t_j^k)}(t)=\sum _{j=1}^{m_k}(f(t)-f(t_j^k))\chi _{[t_{j-1}^k,t_j^k)}(t). \end{aligned}$$

By (4.36) we have $F_k\rightarrow 0$ in $L^1([\tau _1,\tau _2); H)$. Since $\Vert F_k(t)\Vert \le 2M$ for every $t\in [\tau _1,\tau _2)$ and g is Bochner integrable, we obtain that

$$\begin{aligned} \int _{\tau _1}^{\tau _2}(F_k(t),g(t))dt\rightarrow 0, \end{aligned}$$

(4.37)

which gives the first equality in (4.35). The second one can be proved in the same way. $\square$

Corollary 4.11

Under the assumptions of Theorem 4.1, let $\tau _1,\tau _2\in [0,T]$, with $\tau _1<\tau _2$, and let $\omega$ be defined by (4.2). Then there exists a sequence of subdivisions $\tau _1=t_0^k<t_1^k<\cdots <t_{m_k}^k=\tau _2$ such that $t_j^k-t_{j-1}^k\le \frac{1}{k}$ for every $1\le j\le m_k$ and

$$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j=1}^{m_k}\omega (t_{j-1}^k,{t_j^k})=0. \end{aligned}$$

Proof

It is enough to apply the previous proposition with $X:=L^2(\varOmega \setminus \varGamma ;{\mathbb {R}}^2)$, $f(t):=\nabla u(t)$, and $g(t):=\nabla \dot{w}(t)$. $\square$

Proof

Let $\delta _0>0$ be the constant introduced in Lemma 4.5, let $\eta _0>0$ be the constant introduced in Lemma 4.8 related to $\delta _0$, and let F be the finite set defined by

$$\begin{aligned} F:=\{t\in [0,T]: [s](t)\ge \delta _0\}\cup \{0,T\}. \end{aligned}$$

Let $\tau _1,\tau _2\in [0,T]$ be such that $\tau _1<\tau _2$ and $[\tau _1,\tau _2]\cap F=\emptyset .$ By Corollary 4.11, for every $\varepsilon >0$ we can find a finite subdivision $\tau _1=t_0<t_1<\cdots <t_m=\tau _2$ of the interval $[\tau _1,\tau _2]$ such that $t_j-t_{j-1}<\eta _0$ for every $j=1,\ldots ,m$ and

$$\begin{aligned} 2\sum _{j=1}^m\omega (t_{j-1},t_j)<\varepsilon . \end{aligned}$$

By Lemma 4.9 we obtain $s(\tau _2)-s(\tau _1)<\varepsilon$. By the arbitrariness of $\varepsilon$ we deduce that $s(\tau _2)\le s(\tau _1)$ and by monotonicity we deduce that s is constant on the interval $[\tau _1,\tau _2]$. It follows that s is constant in each connected component of $[0,T]\setminus F$. This concludes the proof. $\square$

To prove the regularity of u on $[0,T]\setminus \{t_0,t_1,\ldots , t_m\}$, it is convenient to introduce a different notion of quasistatic evolution in which the crack does not grow.

Definition 4.12

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. A quasistatic evolution with fixed crack and boundary value w on $\partial _D\varOmega$ is a function $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ such that

($\mathrm a_0$):

(measurability) $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ is measurable;

($\mathrm b_0$):

(equilibrium) for every $t\in [0,T]$ we have $u(t)=w(t)$ on $\partial _D\varOmega$ and

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy\le \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla {\hat{u}}|^2dxdy+\int _{\varGamma _{s_0}^b}|[{\hat{u}}]-[u(t)]|dx, \end{aligned}$$

for every ${\hat{u}}\in H^1(\varOmega \setminus \varGamma )$ with ${\hat{u}}=w(t)$ on $\partial _D\varOmega$;

($\mathrm c_0$):

(energy-dissipation inequality) for every $t_1,t_2\in [0,T]$, with $t_1<t_2$, we have

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_2)|^2dxdy+\int _{\varGamma _{s_0}^b}|[u(t_2)]-[u(t_1)]|dx\\&\quad \le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned}$$

Remark 4.13

Taking ${\hat{u}}=w(t)$ in condition ($\mathrm b_0$) above we obtain that there exists a constant $M>0$ such that

$$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy\le M+\int _{\varGamma _{s_0}^b}|[u(t)]|dx\quad \text {for every }t\in [0,T]. \end{aligned}$$

(4.38)

By (2.1) and (2.2), the Trace Inequality, combined with the Poincaré Inequality, implies that there exists a constant $c>0$ such that

$$\begin{aligned} \int _{\varGamma _{s_0}^b}|[u(t)]|dx\le c\Big (\int _{\varOmega \setminus \varGamma }|\nabla u(t)|^2dxdy\Big )^{1/2}+c. \end{aligned}$$

This inequality and (4.38) imply that $\nabla u(t)$ is bounded in $L^2$ uniformly with respect to t. Together with the measurability of $t\mapsto u(t)$ this ensures that the last integral in condition ($\mathrm c_0$) above is well defined.

Theorem 4.14

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$, and let $u:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ be a quasistatic evolution with fixed crack and boundary value w. Then $u\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$ and

$$\begin{aligned} \Big (\int _{\varOmega \setminus \varGamma }|\nabla u(\tau _2)-\nabla u(\tau _1)|^2dxdy\Big )^{1/2} \le \int _{\tau _1}^{\tau _2}\Big (\int _{\varOmega \setminus \varGamma }|\nabla \dot{w}(\tau )|^2dxdy\Big )^{1/2}d\tau \end{aligned}$$

(4.39)

for every $\tau _1,\tau _2\in [0,T]$ with $\tau _1<\tau _2$.

Proof

The proof is taken from [3, Theorem 5.2], with obvious simplifications. Let us fix $\tau _1,\tau _2\in [0,T]$ with $\tau _1<\tau _2$. From the energy-dissipation condition ($\mathrm c_0$) between $\tau _1$ and $\tau _2$ we obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(\tau _2)|^2dxdy+\int _{\varGamma _{s_0}^b}|[u(\tau _2)]-[u(\tau _1)]|dx \\&\le \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(\tau _1)|^2dxdy+\int _{\tau _1}^{\tau _2}\Big (\int _{\varOmega \setminus \varGamma }\nabla u(\tau )\nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned} \end{aligned}$$

(4.40)

The Euler equation corresponding to the equilibrium condition ($\mathrm b_0$) of Definition 4.12 (see Lemma 3.1) implies that

$$\begin{aligned} -\int _{\varOmega \setminus \varGamma }\nabla u(\tau _1)\nabla \varphi \, dxdy\le \int _{\varGamma _{s_0}^b}| [\varphi ] |\,dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

Taking $\varphi :=u(\tau _2)-u(\tau _1)-(w(\tau _2)-w(\tau _1))$ we obtain

$$\begin{aligned} \begin{aligned}&-\int _{\varOmega \setminus \varGamma }\nabla u(\tau _1)\nabla u(\tau _2)\,dxdy+ \int _{\varOmega \setminus \varGamma }|\nabla u(\tau _1)|^2dxdy\\&\quad \le -\int _{\varOmega \setminus \varGamma }\nabla u(\tau _1)(\nabla w(\tau _2)-\nabla w(\tau _1))\,dxdy + \int _{\varGamma _{s_0}^b}|[u(\tau _2)]-[u(\tau _1)]|dx, \end{aligned} \end{aligned}$$

(4.41)

where we have used the fact that $[w(\tau _1)]=[w(\tau _2)]=0$ on $\varGamma _{s_0}^b$. Adding (4.40) and (4.41) we get

$$\begin{aligned} \frac{1}{2}\int _{\varOmega \setminus \varGamma }|\nabla u(\tau _2)-\nabla u(\tau _1)|^2dxdy \le \int _{\tau _1}^{\tau _2}\Big (\int _{\varOmega \setminus \varGamma }(\nabla u(\tau )-\nabla u(\tau _1))\nabla \dot{w}(\tau )dxdy\Big )d\tau . \end{aligned}$$

Since this holds for every $\tau _2\in (\tau _1,T]$, by the Gronwall Inequality we obtain (4.39) for every $\tau _2\in (\tau _1,T]$. This inequality, together with the integrability of the function $\tau \mapsto \big (\int _{\varOmega \setminus \varGamma }|\nabla \dot{w}(\tau )|^2dxdy\big )^{1/2}$, implies that $\nabla u\in AC([0,T];L^2(\varOmega \setminus \varGamma ;{\mathbb {R}}^2))$. Since $u(t)=w(t)$ on $\partial _D\varOmega$, by (2.1) and the Poincaré Inequality we conclude that $u\in AC([0,T];H^1(\varOmega \setminus \varGamma ))$. $\square$

Theorem 4.15

Under the assumptions of Theorem 4.1, for every $j=1,\ldots ,m$ there exists $u^j\in AC([t_{j-1},t_j];H^1(\varOmega \setminus \varGamma ))$ such that $u(t)=u^j(t)$ for every $t\in (t_{j-1},t_j)$.

Proof

Let us fix $1\le j\le m$. By Theorem 4.1, we have $s(t)=s_j$ for every $t\in (t_{j-1},t_j)$. Therefore for every $\tau _1,\tau _2\in (t_{j-1},t_j)$ with $\tau _1<\tau _2$, the function u is a quasistatic evolution with fixed crack in the sense of Definition 4.12 on the interval $[\tau _1,\tau _2]$.

By Theorem 4.14 we obtain (4.39) for every $[\tau _1,\tau _2]\subset (t_{j-1},t_j)$. This shows that the restriction of u to the open interval $(t_{j-1},t_j)$ can be extended to an absolutely continuous function $u^j:[t_{j-1},t_j]\rightarrow H^1({\varOmega \setminus \varGamma })$. $\square$

Remark 4.16

Besides the assumptions of Theorem 4.1, suppose also that $w(0)=u(0)=0$ and that $s(0)=s_0$. Then there exists $u^1\in AC([0,t_1];H^1(\varOmega \setminus \varGamma ))$ such that $u(t)=u^1(t)$ for every $t\in [0,t_1)$. Indeed, (4.1) implies that $u(t)\rightarrow 0$ strongly in $H^1(\varOmega \setminus \varGamma )$ as $t\rightarrow 0+$.

Theorem 4.17

Let $T>0$, $s_0\in [a,b)$, and $w\in AC([0,T];H^1(\varOmega \setminus \varGamma _a^{s_0}))$. Let $u_1,u_2$ be two quasistatic evolutions with fixed crack and boundary condition w on $\partial _D\varOmega$. If $u_1(0)=u_2(0)$ then $u_1(t)=u_2(t)$ for every $t\in [0,T]$.

Proof

The proof is taken from [3, Theorem 5.9], with obvious simplifications. Since $u_2\in AC([0,T];H^1(\varOmega \setminus \varGamma ))$ by Theorem 4.15, from the energy-dissipation condition ($\mathrm c_0$) (dividing by $t_2-t_1$, and passing to the limit as $t_1,t_2\rightarrow t$), we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u_2(t)\big (\nabla \dot{u}_2(t)-\nabla \dot{w}(t)\big )dxdy\le -\int _{\varGamma _{s_0}^b}|[\dot{u}_2(t)]|dx \end{aligned}$$

(4.42)

for a.e. $t\in (0,T)$.

On the other hand, for every $t\in [0,T]$, the Euler equation (see Lemma 3.1) for the equilibrium condition ($\mathrm b_0$) for $u_1$ gives that there exists $\psi _1(t)\in L^\infty (\varGamma _{s_0}^b)$ with $|\psi _1(t)|\le 1$, such that

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\nabla u_1(t)\nabla \varphi \,dxdy= \int _{\varGamma _{s_0}^b}\psi _1(t)[\varphi ]dx\quad \text {for every }\varphi \in H^1_{0,D}(\varOmega \setminus \varGamma ). \end{aligned}$$

(4.43)

Since $u_2(t)=w(t)$ on $\partial _D\varOmega$ for every $t\in [0,T]$ and $u_2\in AC([0,T];H^1(\varOmega \setminus \varGamma ))$, we have that $\dot{u}_2(t)-\dot{w}(t)\in H^1_{0,D}(\varOmega \setminus \varGamma )$ for a.e. $t\in (0,T)$. Using $\varphi =-(\dot{u}_2(t)-\dot{w}(t))$ in (4.43) we obtain

$$\begin{aligned} -\int _{\varOmega \setminus \varGamma }\nabla u_1(t)\big (\nabla \dot{u}_2(t)-\nabla \dot{w}(t)\big )dxdy =-\int _{\varGamma _{s_0}^b}\psi _1(t)[\dot{u}_2(t)]dx \end{aligned}$$

(4.44)

for a.e. $t\in (0,T)$. Since $|\psi _1(t)|\le 1$, adding (4.42) and (4.44) we get

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\big (\nabla u_2(t)-\nabla u_1(t)\big )\big (\nabla \dot{u}_2(t)-\nabla \dot{w}(t)\big )dxdy\le 0 \end{aligned}$$

(4.45)

for a.e. $t\in (0,T)$.

In a similar way we obtain

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\big (\nabla u_1(t)-\nabla u_2(t)\big )\big (\nabla \dot{u}_1(t)-\nabla \dot{w}(t)\big )dxdy\le 0 \end{aligned}$$

(4.46)

for a.e. $t\in (0,T)$. Adding (4.45) and (4.46) we have

$$\begin{aligned} \int _{\varOmega \setminus \varGamma }\big (\nabla u_2(t)-\nabla u_1(t)\big )\big (\nabla \dot{u}_2(t)-\nabla \dot{u}_1(t)\big )dxdy\le 0 \end{aligned}$$

(4.47)

for a.e. $t\in (0,T)$. This implies that the absolutely continuous function $t\mapsto \int _{\varOmega \setminus \varGamma }|\nabla u_2(t)-\nabla u_1(t)|^2$ has a nonpositive derivative a.e. in [0, T], thus it is nonincreasing. Since it is 0 at $t=0$ we conclude that $\nabla u_1(t)=\nabla u_2(t)$ for every $t\in [0,T]$. Since $u_1(t)=u_2(t)=w(t)$ on $\partial _D\varOmega$ we deduce that $u_1(t)=u_2(t)$ for every $t\in [0,T]$ by (2.1). $\square$

The following corollary is an immediate consequence of Remark 4.16 and Theorem 4.17.

Corollary 4.18

Besides the assumptions of Theorem 4.1, suppose also that $w(0)=u(0)=0$ and that $s(0)=s_0$. Then u is uniquely determined in the interval $[0,t_1)$.

5 An example

In this section we describe an example of quasistatic evolution (u, s) where s is not constant. We consider here

$$\begin{aligned} \varOmega :=(a,b)\times (-h,h),\quad \varGamma :=[a,b]\times \{0\},\quad \partial _D\varOmega :=[a,b]\times \{-h,h\}, \end{aligned}$$

(5.1)

for some $h>0$. Therefore we have $\varOmega ^+=(a,b)\times (0,h)$ and $\partial ^+_D\varOmega =[a,b]\times \{h\}$. The boundary condition at time $t\in [0,T]$ will be $u(t)=t$ on $[a,b]\times \{h\}$ and $u(t)=-t$ on $[a,b]\times \{-h\}$. This leads to the following choice for $w(t)\in H^1(\varOmega )$:

$$\begin{aligned} w(t)(x,y):=t\frac{y}{h}. \end{aligned}$$

(5.2)

Let $z_0\in H^1(\varOmega ^+)$ be the solution of the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta z_0=0 &{}\text {in }\varOmega ^+,\\ z_0=0 &{}\text {if }y=h,\\ \frac{ \partial z_0}{\partial x}=0 &{} \text {if } x=a\text { and }x=b,\\ \frac{ \partial z_0}{\partial y}=0 &{} \text {if } a<x<s_0\text { and }y=0,\\ \frac{ \partial z_0}{\partial y}=1 &{} \text {if } s_0<x<b\text { and }y=0.\\ \end{array}\right. } \end{aligned}$$

(5.3)

We shall prove that $z_0\in C^0({\overline{\varOmega }}^+)$ (see Remark 5.6 and Lemma 5.7 below). We define $z_0$ in $\varOmega ^-$ by

$$\begin{aligned} z_0(x,y):=-z_0(x,-y)\quad \text {for every }(x,y)\in \varOmega ^-. \end{aligned}$$

Theorem 5.1

Let $\varOmega$, $\varGamma$, $\partial _D\varOmega$, and w be as in (5.1) and (5.2). Let $T>0$, $s_0\in (a,b)$, and let (u, s) be a quasistatic evolution with boundary condition w on $\partial _D\varOmega$ and initial conditions $u(0)=0$ and $s(0)=s_0$. Assume that

$$\begin{aligned} T>T_*:=-\inf _{\varOmega ^+}z_0\quad \text {and}\quad \int _{\varOmega ^+}|\nabla z_0|^2dxdy+s_0>b, \end{aligned}$$

(5.4)

where $z_0$ is defined by (5.3). Then s(t) takes at least two distinct values in two nondegenerate intervals.

Remark 5.2

Since $z_0(x,y)\rightarrow y$ as $s_0\rightarrow a+$, the second inequality in (5.4) is surely satisfied if $h>1$ and $s_0$ is sufficiently close to a.

To prove Theorem 5.1 we shall construct a quasistatic evolution $u_*$ with fixed crack and boundary condition w such that $u_*(0)=0$ and

$$\begin{aligned} u_*(t)={\left\{ \begin{array}{ll} t+z_0 \qquad \text { in }\varOmega ^+\\ -t+z_0 \qquad \text {in }\varOmega ^- \end{array}\right. } \end{aligned}$$

(5.5)

for every $t> T_*$. If we had $s(t)=s_0$ for every $t\in [0,T]$, by the uniqueness result proved in Theorem 4.17 we would have $u(t)=u_*(t)$ for every $t\in [0,T]$. On the other hand, we shall see that, if $\int _{\varOmega ^+}|\nabla z_0|^2dxdy+s_0>b$ and condition (5.5) holds, then $(u_*(t),s_0)$ does not satisfy the equilibrium condition (b) in Definition 2.1. This contradiction shows that s cannot be constantly equal to $s_0$.

The construction of $u_*$ requires a careful analysis of the properties of the solutions of some auxiliary minimum problems. Due to the symmetry of the data we shall work in $\varOmega ^+$. This is justified by the following remark.

Remark 5.3

Since w(t) is odd with respect to y, a function $u_*:[0,T]\rightarrow H^1(\varOmega \setminus \varGamma )$ is a quasistatic evolution with fixed crack and boundary condition w if and only if it is odd with respect to y and satisfies the following conditions

($\mathrm a_0$):

(measurability) $u_*:[0,T]\rightarrow H^1(\varOmega ^+)$ is measurable;

($\mathrm b_0$):

(equilibrium) for every $t\in [0,T]$ we have $u_*(t)=t$ on $\partial ^+_D\varOmega$, and

$$\begin{aligned} \frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(t)|^2dxdy\le \frac{1}{2} \int _{\varOmega ^+}|\nabla {\hat{u}}|^2dxdy+\int _{\varGamma _{s_0}^b}|{\hat{u}}^+- u_*^+(t)|dx \end{aligned}$$

(5.6)

for every ${\hat{u}}\in H^1(\varOmega ^+)$ with ${\hat{u}}=t$ on $\partial _D^+\varOmega$.

($\mathrm c_0$):

(energy-dissipation inequality) for every $t_1,t_2\in [0,T]$, with $t_1<t_2$, we have

$$\begin{aligned}&\frac{1}{2}\int _{\varOmega ^+}|\nabla u_*(t_2)|^2dxdy+\int _{\varGamma _{s_0}^b}|u_*^+(t_2)-u_*^+(t_1)|dx\\&\quad \le \frac{1}{2}\int _{\varOmega ^+}|\nabla u_*(t_1)|^2dxdy+\int _{t_1}^{t_2}\Big (\int _{\varOmega ^+}\frac{\partial u_*(\tau )}{\partial y} dxdy\Big )d\tau . \end{aligned}$$

Indeed, the oddness of $u_*(t)$ with respect to y follows from the uniqueness of the solutions of problems of the form (3.3) and from the oddness of the data.

To prove (5.5) we need a detailed study of the properties of the solutions of (5.6), which uses the Euler conditions introduced in Lemmas 3.1–3.3. This analysis requires the results of the following two lemmas, which give a precise description of the singularities of some solutions of the Laplace equation with suitable boundary conditions.

For every $R>0$ let $\varGamma ^-_R=(-R,0)\times \{0\}$, and $\varGamma ^+_R=(0,R)\times \{0\}$. In the next lemmas we identify the point (x, y) with the complex number $z=x+iy$.

Lemma 5.4

Let $R>0$ and let $u\in H^1(B_R^+)$ be such that $\varDelta u=0$ in $B_R^+$, $\frac{\partial u}{\partial y}=0$ on $\varGamma ^-_R$, and $u=0$ on $\varGamma ^+_R$. Let $S_0$ be defined by

$$\begin{aligned} S_0(z):={\mathfrak {Im}}(\sqrt{z}), \end{aligned}$$

where for $y\ge 0$ we use the determination of $\sqrt{z}$ such that $\sqrt{-1}=i$. Then

$$\begin{aligned} u=cS_0+u_{reg} \end{aligned}$$

(5.7)

for some $c\in {\mathbb {R}}$ and $u_{reg}\in C^1({\overline{B}}{}^+_r)$ for every $0<r<R$.

Proof

Using Schwarz symmetrization principle we may assume that u is harmonic in $B_R\setminus {\overline{\varGamma }}{}_R^-$ and satisfies the homogeneous Neumann boundary condition on both sides of $\varGamma _R^-$. By using the conformal map $z\mapsto \sqrt{z}$ we can write

$$\begin{aligned} u(z)=v(\sqrt{z}), \end{aligned}$$

(5.8)

where v is harmonic on $B_{\sqrt{R}}\cap \{(x,y):x>0\}$, belongs to $H^1(B_{\sqrt{R}}\cap \{(x,y):x>0\})$, and satisfies $\frac{\partial v}{\partial \nu }=0$ on $\{(0,y):-\sqrt{R}< y<\sqrt{R}\}$ and $v=0$ on $\{(x,0):0< x<\sqrt{R}\}$. We now extend v to the whole ball $B_{\sqrt{R}}$ by reflection and we obtain a function, still denoted by v, which is harmonic on $B_{\sqrt{R}}$ and satisfies $v=0$ on $\{(x,0):-\sqrt{R}< x<\sqrt{R}\}$.

Therefore, there exists a holomorphic function f defined on $B_{\sqrt{R}}$ such that

$$\begin{aligned} v(z)={\mathfrak {Im}}f(z)\quad \text {for every }z\in B_{\sqrt{R}}. \end{aligned}$$

(5.9)

We may assume $f(0)=0$. Since f is real on the real axis we can write

$$\begin{aligned} f(z)=\sum _{k=1}^\infty a_kz^k, \end{aligned}$$

where $a_k\in {\mathbb {R}}$ and the series converges uniformly on compact subsets of $B_{\sqrt{R}}$. Let g be the holomorphic function on $B_{\sqrt{R}}$ defined by

$$\begin{aligned} g(z)=\sum _{k=2}^\infty a_kz^k, \end{aligned}$$

Therefore (5.8) and (5.9) imply (5.7) with $c=a_1$ and

$$\begin{aligned} u_{reg}(z)={\mathfrak {Im}}\big (g(\sqrt{z})\big ). \end{aligned}$$

(5.10)

Let us fix $0<r<R$. It remains to prove that $u_{reg}\in C^1({\overline{B}}{}^+_r)$. Since

$$\begin{aligned} \nabla u_{reg}(x,y)=\Big ({\mathfrak {Im}}\big (\frac{g'(\sqrt{z})}{2\sqrt{z}}\big ),{\mathfrak {Re}}\big (\frac{g'(\sqrt{z})}{2\sqrt{z}}\big )\Big ) \end{aligned}$$

(5.11)

it is enough to prove that

$$\begin{aligned} z\mapsto \frac{g'(\sqrt{z})}{\sqrt{z}} \end{aligned}$$

(5.12)

is continuous on ${\overline{B}}{}^+_r$. Since

$$\begin{aligned} g'(z)=\sum _{k=2}^\infty ka_k z^{k-1}, \end{aligned}$$

(5.13)

the function $h(z):=g'(z)/z$ is holomorphic on $B_{\sqrt{R}}$. Therefore we have $g'(\sqrt{z})/\sqrt{z}=h(\sqrt{z})$, which gives the continuity of (5.12) and concludes the proof. $\square$

Lemma 5.5

Let $R>0$ and let $u\in H^1(B_R^+)$ be such that $\varDelta u=0$ in $B_R^+$, $\frac{\partial u}{\partial y}=0$ on $\varGamma ^-_R$, and $\frac{\partial u}{\partial y}=1$ on $\varGamma ^+_R$. Let $S_1$ be defined by

$$\begin{aligned} S_1(z):=\frac{1}{\pi }{\mathfrak {Re}}(z\log (-z)). \end{aligned}$$

Then $u=S_1+u_{reg}$ with $u_{reg}\in C^\infty ({\overline{B}}{}^+_r)$ for every $r<R$.

Proof

By direct computation we see that $S_1\in H^1(B^+_R)$, it is harmonic on $B^+_R$ and satisfies the boundary conditions $\frac{\partial S_1}{\partial y}=0$ on $\varGamma ^-_R$, and $\frac{\partial S_1}{\partial y}=1$ on $\varGamma ^+_R$. Therefore $u_{reg}:=u-S_1\in H^1(B^+_R)$, it is harmonic and satisfies the homogeneous Neumann boundary condition on $\varGamma ^-_R\cup \varGamma ^+_R$, and hence on $(-R,R)\times \{0\}$. The conclusion follows from the regularity theory for elliptic equations with Neumann boundary condition. $\square$

The quasistatic evolution $u_*(t)$ will be constructed by using the solutions of some auxiliary boundary value problems depending on a parameter $\sigma$, and then by choosing a particular value $\sigma _t$ of this parameter. For every $t\ge 0$ and for every $\sigma \in [s_0,b]$ we consider the solution $u^t_\sigma \in H^1(\varOmega ^+)$ of the problem (see Fig. 3)

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta u=0 &{} \text {in }\varOmega ^+,\\ u=t &{}\text {if }y=h,\\ \frac{\partial u}{\partial x}=0 &{} \text {if }x=a\text { or }x=b,\\ \frac{\partial u}{\partial y}=0 &{}\text {for }a<x<s_0 \text { and }y=0,\\ \frac{\partial u}{\partial y}=1 &{}\text {for }s_0<x<\sigma \text { and }y=0,\\ u=0 &{}\text {for }\sigma<x<b \text { and }y=0. \end{array}\right. } \end{aligned}$$

(5.14)

By the continuous dependence on the data, the function $u^t_\sigma$ is continuous in $H^1(\varOmega ^+)$ with respect to t and $\sigma$.

Fig. 3

The boundary value problem for $u^t_\sigma$

Remark 5.6

In the particular case $\sigma =b$ we have $u^t_b=t+z_0$ where $z_0\in H^1(\varOmega ^+)$ is the solution of (5.3).

The following two lemmas give some important properties of $u^t_\sigma$, which will be used in our construction of $u_*(t)$.

Lemma 5.7

For every $t\ge 0$ and $\sigma \in [s_0,b]$ we have $u^t_\sigma \in C^\infty ({\overline{\varOmega }}{}^+\setminus \{(s_0,0),(\sigma ,0)\})\cap C^0({\overline{\varOmega }}{}^+)$.

Proof

The result follows from the regularity theory for elliptic equations; the regularity near the vertices of the rectangle can be easily obtained by extending $u^t_\sigma$ through a suitable reflection, while the continuity at the points $(s_0,0)$ and $(\sigma ,0)$ follows from Lemmas 5.4 and 5.5. $\square$

Lemma 5.8

Let $t\ge 0$ and let $\sigma \in [s_0,b]$ be such that $u^t_\sigma \ge 0$ in $\varOmega ^+$. Then

$$\begin{aligned} \frac{\partial u^t_\sigma }{\partial x}\le 0\quad \text {in }\varOmega ^+. \end{aligned}$$

(5.15)

Proof

Let $\displaystyle v:=\frac{\partial u^t_\sigma }{\partial x}$. By Lemma 5.7 we have that $v\in C^\infty ({\overline{\varOmega }}{}^+\setminus \{(s_0,0),(\sigma ,0)\})$ and satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta v=0 &{}\text {in }\varOmega ^+,\\ v=0 &{}\text { if }y=h,\\ v=0 &{}\text { if }x=a\text { or }x=b,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } a<x<s_0\text { and }y=0,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } s_0<x<\sigma \text { and }y=0,\\ v=0 &{}\text {if }\sigma<x<b\text { and }y=0. \end{array}\right. } \end{aligned}$$

(5.16)

Case ${s_0=\sigma }$. Let us consider the behaviour of the function $u^t_\sigma$ near $(s_0,0)$. By Lemma 5.4 we can write

$$\begin{aligned} u^t_\sigma (x,y)=c\sqrt{\rho }\sin (\theta /2) +u_{reg}(x,y) \end{aligned}$$

(5.17)

for some constant c and some function $u_{reg}\in C^1(\overline{\varOmega }{}^+)$, where $\rho ,\theta$ are polar coordinates around $(s_0,0)$, with $\theta \in [0,\pi ]$. We observe that $0=u^t_\sigma (x,0)=u_{reg}(x,0)$ for every $s_0<x<b$. This implies that $u_{reg}(s_0,0)=0$ and $\frac{\partial u_{reg}}{\partial x}(s_0,0)=0$.

By (5.17) we have $u^t_\sigma (x,0)=c\sqrt{s_0-x} +u_{reg}(x,0)$ for every $a<x<s_0$, while the properties of $u_{reg}$ imply that $|u_{reg}(x,0)|\le M|x-s_0|$ for a suitable constant M. Hence the inequality $c<0$ would lead to $u^t_\sigma (x,0)<0$ for $x<s_0$, x close to $s_0$, in contradiction with the assumption $u^t_\sigma (x,0)\ge 0$. This shows that $c\ge 0$.

Since

$$\begin{aligned} v(x,y)=\frac{\partial u^t_\sigma }{\partial x}(x,y)=-\frac{c}{2\sqrt{\rho }}\sin (\theta / 2)+\frac{\partial u_{reg}}{\partial x}(x,y), \end{aligned}$$

(5.18)

we have

$$\begin{aligned} \limsup _{(x,y)\rightarrow (s_0,0)}v(x,y)\le 0. \end{aligned}$$

Therefore, if v is positive at some point of $\varOmega ^+$, by the maximum principle v attains its maximum on ${\overline{\varOmega }}{}^+$ at a point of $\partial \varOmega ^+\setminus \{(s_0,0)\}$ where v has a positive value. By (5.16) this point must be of the form $(x_0,0)$ with $a<x_0<s_0$. By the Hopf Maximum Principle we should have $\frac{ \partial v}{\partial y}(x_0,0)<0$, which contradicts (5.16). This shows that we must have $v\le 0$ in $\varOmega ^+$.

Case ${{ s_0<\sigma <b}}$. We have to study the behaviour of the function $u^t_\sigma$ near the points $(s_0,0)$ and $(\sigma ,0)$. By Lemma 5.5 and (5.14), near $(s_0,0)$ we have

$$\begin{aligned} u^t_\sigma (x,y)=\frac{1}{\pi }\big ((x-s_0)\log \rho - y (\theta -\pi )\big )+u_{s_0}^{reg}(x,y), \end{aligned}$$

(5.19)

where $u_{s_0}^{reg}$ is $C^\infty$ in a neighbourhood of $(s_0,0)$ in ${\overline{\varOmega }}{}^+$ and $\rho ,\theta$ are polar coordinates around $(s_0,0)$, with $\theta \in [0,\pi ]$. Hence

$$\begin{aligned} v(x,y)=\frac{1}{\pi }(\log \rho + 1)+\frac{\partial u_{s_0}^{reg}}{\partial x}(x,y), \end{aligned}$$

(5.20)

and this implies that

$$\begin{aligned} \lim _{(x,y)\rightarrow (s_0,0)}v(x,y)=-\infty . \end{aligned}$$

(5.21)

By Lemma 5.4 and (5.14), using polar coordinates $r,\phi$ around $(\sigma ,0)$, with $\phi \in [0,\pi ]$, we can write

$$\begin{aligned} u^t_\sigma (x,y)=c\sqrt{r}\sin (\phi /2) +u^{reg}_\sigma (x,y), \end{aligned}$$

(5.22)

where $c\in {\mathbb {R}}$ and $u_\sigma ^{reg}$ is $C^1$ in a neighbourhood of $(\sigma ,0)$ in ${\overline{\varOmega }}{}^+$. This gives

$$\begin{aligned} v(x,y)=-\frac{c}{2\sqrt{r}} \sin (\phi /2)+\frac{\partial u_\sigma ^{reg}}{\partial x}(x,y). \end{aligned}$$

(5.23)

Arguing as in the case $s_0=\sigma$ we can prove that $c\ge 0$ and that $u_\sigma ^{reg}(x,0)=0$ for every $\sigma<x<b$. Since $\frac{\partial u_\sigma ^{reg}}{\partial x}(\sigma ,0)=0$ we have

$$\begin{aligned} \limsup _{(x,y)\rightarrow (\sigma ,0)}v(x,y)\le 0. \end{aligned}$$

(5.24)

By (5.21) and (5.24), the subharmonic function $v\vee 0$ can be extended to a continuous function on $\overline{\varOmega }{}^+$ which satisfies

$$\begin{aligned} (v\vee 0)(s_0,0)=(v\vee 0)(\sigma ,0)=0. \end{aligned}$$

(5.25)

Therefore, if v is positive at some point of $\varOmega ^+$, by the maximum principle for subharmonic functions $v\vee 0$ attains its maximum on ${\overline{\varOmega }}{}^+$ at a point of $\partial \varOmega ^+$ where v has a positive value. By (5.16) and (5.25) this point must be of the form $(x_0,0)$ with $a<x_0<s_0$ or $s_0<x_0<\sigma$. By the Hopf Maximum Principle we should have $\frac{ \partial v}{\partial y}(x_0,0)<0$, which contradicts (5.16). This concludes the proof of (5.15) for $s_0<\sigma <b$.

Case ${{ \sigma =b}}$. In this case the only singular point of v is $(s_0,0)$ and we can repeat the argument of the previous case with obvious simplifications. $\square$

For $t\ge 0$ we define

$$\begin{aligned} \sigma _t:=\max \{\sigma \in [s_0,b]:u^t_\sigma \ge 0 \text { in }\varOmega ^+\}\quad \text {and}\quad u_*(t):=u_{\sigma _t}^{t}. \end{aligned}$$

(5.26)

The existence of the maximum follows easily from the continuous dependence of $u^t_\sigma$ on $\sigma$. It is easy to see that for $t=0$ we have $\sigma _0=s_0$.

The results of following three lemmas will be used to prove Lemma 5.12, which shows that $u_*(t)$ is a quasistatic evolution.

Lemma 5.9

Let $t\ge 0$. Then

$$\begin{aligned} 0\le \frac{\partial u_*(t)}{\partial y}(x,0)\le 1 \end{aligned}$$

(5.27)

for every $\sigma _t<x<b$.

Proof

It is not restrictive to assume $\sigma _t<b$. By (5.26) and Lemma 5.8 for every $y\in (0,h)$ the function $x\mapsto u_*(t)(x,y)$ is nonnegative and nonincreasing in (a, b). Since $u_*(t)(x,0)=0$ for $x\in (\sigma _t,b)$, the function $x\mapsto ( u_*(t)(x,y)- u_*(t)(x,0))/y$ is nonnegative and nonincreasing in $(\sigma _t,b)$ for every $y\in (0,h)$. Taking the limit as $y\rightarrow 0+$ we deduce that $x\mapsto \frac{\partial u_*(t)}{\partial y}(x,0)$ is nonnegative and nonincreasing in $(\sigma _t,b)$.

It remains to prove the second inequality in (5.27). If it is not satisfied, by the monotonicity of $x\mapsto \frac{\partial u_*(t)}{\partial y}(x,0)$ there exists $\varepsilon \in (0, b-\sigma _t)$ such that

$$\begin{aligned} \frac{\partial u_*(t)}{\partial y}(x,0)> 1\quad \text {for every }x\in (\sigma _t,\sigma _t+\varepsilon ). \end{aligned}$$

Let $\sigma \in (\sigma _t,\sigma _t+\varepsilon )$. We want to prove that $u^t_\sigma \ge u^t_{\sigma _t}$ in $\varOmega ^+$. Setting $v:=u^t_\sigma - u^t_{\sigma _t}$ we have that $v\in H^1(\varOmega ^+)$ and satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta v=0 &{}\text {in }\varOmega ^+,\\ v=0 &{}\text { if }y=h,\\ \frac{ \partial v}{\partial x}=0 &{}\text { if }x=a\text { or }x=b,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } a<x<s_0\text { and }y=0,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } s_0<x<\sigma _t\text { and }y=0,\\ \frac{ \partial v}{\partial y}<0 &{} \text {if } \sigma _t<x<\sigma \text { and }y=0,\\ v=0 &{}\text {if }\sigma<x<b\text { and }y=0. \end{array}\right. } \end{aligned}$$

Integrating by parts we obtain the weak formulation

$$\begin{aligned} \int _{\varOmega ^+}\nabla v\nabla \varphi \,dxdy=-\int _{\varGamma _{\sigma _t}^\sigma }\frac{ \partial v}{\partial y}\varphi ^+dx \end{aligned}$$

for every $\varphi \in H^1(\varOmega ^+)$ with $\varphi =0$ on $\varGamma _\sigma ^b\cup \partial _D^+\varOmega$. Taking $\varphi :=v\wedge 0$ we obtain

$$\begin{aligned} \int _{\varOmega ^+}|\nabla (v\wedge 0)|^2=\int _{\varOmega ^+}\nabla v\nabla (v\wedge 0)\,dxdy=-\int _{\varGamma _{\sigma _t}^\sigma }\frac{ \partial v}{\partial y}(v^+\wedge 0)dx\le 0, \end{aligned}$$

which gives $\nabla (v\wedge 0)=0$. Taking into account the boundary condition $v=0$ on $\partial _D^+\varOmega$ we get $v\wedge 0=0$ in $\varOmega ^+$. This implies $v\ge 0$, so that $u^t_\sigma \ge u^t_{\sigma _t}$ in $\varOmega ^+$. Therefore $u^t_\sigma \ge 0$ in $\varOmega ^+$, which contradicts the maximality of $\sigma _t$ (see (5.26)), thus concluding the proof of the second inequality in (5.27). $\square$

Lemma 5.10

For every $0\le t_1\le t_2$ we have $u_*(t_1)\le u_*(t_2)\text { in }\varOmega ^+.$

Proof

Let us fix $0\le t_1\le t_2$. By the maximum principle we have

$$\begin{aligned} u_*(t_1)=u_{\sigma _{t_1}}^{t_1}\le u_{\sigma _{t_1}}^{t_2}\quad \text { in }\varOmega ^+. \end{aligned}$$

(5.28)

By (5.26) this implies $\sigma _{t_1}\le \sigma _{t_2}$. Let $v:=u_*(t_2)-u_{\sigma _{t_1}}^{t_2}=u_{\sigma _{t_2}}^{t_2}-u_{\sigma _{t_1}}^{t_2}\in H^1(\varOmega ^+)$. By (5.26) we have $u_{\sigma _{t_2}}^{t_2}(x,0)\ge 0$ for $x\in (\sigma _{t_1},\sigma _{t_2})$, while by the last line in (5.14) we have $u_{\sigma _{t_1}}^{t_2}(x,0)=0$ for $x\in (\sigma _{t_1},\sigma _{t_2})$. Hence $v(x,0)\ge 0$ for $x\in (\sigma _{t_1},\sigma _{t_2})$. Thus v satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta v=0 &{}\text {in }\varOmega ^+\,\\ v=0 &{}\text {if }y=h,\\ \frac{ \partial v}{\partial x}=0 &{}\text {if }x=a\text { or }x=b,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } a<x<s_0\text { and }y=0,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } s_0<x<\sigma _{t_1}\text { and }y=0,\\ v\ge 0 &{} \text {if } \sigma _{t_1}<x<\sigma _{t_2}\text { and }y=0,\\ v=0 &{}\text {if }\sigma _{t_2}<x<b\text { and }y=0. \end{array}\right. } \end{aligned}$$

By using the Maximum Principle (see also the proof of Lemma 5.9) we can prove that $v\ge 0$ in $\varOmega ^+$. Together with (5.28), this concludes the proof. $\square$

Lemma 5.11

For every $0\le t_1\le t_2$ the function $u_*(t_2)$ is the solution of the minimum problem

$$\begin{aligned} \min _{\begin{array}{c} u\in H^1(\varOmega ^+)\\ u=t_2 \text {on }\partial _D^+\varOmega \end{array}}\Big (\frac{1}{2} \int _{\varOmega ^+}|\nabla u|^2dxdy+\int _{\varGamma _{s_0}^b}|u^+-u_*(t_1)^+|dx\Big ). \end{aligned}$$

(5.29)

Proof

Let us fix $0\le t_1\le t_2$. By (5.14), (5.26), and Lemma 5.9 the function $u_*(t_2)$ satisfies

$$\begin{aligned} \left| \frac{\partial u_*(t_2)}{\partial y}(x,0)\right| \le 1 \quad \text {for every }x\in (s_0,b)\setminus \{\sigma _{t_2}\}. \end{aligned}$$

Moreover, by (5.14) we have

$$\begin{aligned} \frac{\partial u_*(t_2)}{\partial y}(x,0)=1 \quad \text {for every }x\in (s_0,b) \text { such that }u_*(t_2)(x,0)>u_*(t_1)(x,0), \end{aligned}$$

since $\{x\in (s_0,b):u_*(t_2)(x,0)>u_*(t_1)(x,0)\}\subset \{x\in (s_0,b):u_*(t_2)(x,0)>0\}\subset (s_0,\sigma _{t_2})$. By Lemma 3.3 (applied to the odd extension of $u_*(t_2)$ to $\varOmega \setminus \varGamma$) these properties of $\frac{\partial u_*(t_2)}{\partial y}$ on $\varGamma$, together with the boundary conditions of (5.14), imply that $u_*(t_2)$ is the solution of (5.29). $\square$

Lemma 5.12

The odd extension to $\varOmega \setminus \varGamma$ of the function $u_*$ defined by (5.26) is a quasistatic evolution with fixed crack and boundary condition w on each interval $[0,{\hat{T}}]$ with ${\hat{T}}>0$.

Proof

Let us fix ${\hat{T}}>0$. By Lemma 5.11 for every t the odd extension of $u_*(t)$ to $\varOmega \setminus \varGamma$ is the solution of the minimum problem (2.23) with $w=w(t)$, with w(t) defined by (5.2), $s=s_0$, and $g=0$. From Lemma 2.8 we deduce that $u_*:[0,{\hat{T}}]\rightarrow H^1(\varOmega ^+)$ is continuous.

To conclude the proof we have to show that $u_*(t)$ satisfies also conditions (b${}_0$) and (c${}_0$) in Remark 5.3. Condition (b${}_0$) follows from Lemma 5.11. To prove (c${}_0$) we fix $\tau _1,\tau _2\in [0,{\hat{T}}]$, with $\tau _1<\tau _2$ and a sequence of subdivisions $\tau _1=t_0^k<t_1^k<\cdots <t_{m_k}^k=\tau _2$ such that $t_j^k-t_{j-1}^k\le \frac{1}{k}$ for every $1\le j\le m_k$. By Lemma 5.11 for every j we have

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(t_j^k)|^2dxdy+\int _{\varGamma _{s_0}^b}|u_*(t_j^k)^+-u_*(t_{j-1}^k)^+|dx\\&\quad \le \frac{1}{2} \int _{\varOmega ^+}|\nabla {\hat{u}}|^2dxdy+\int _{\varGamma _{s_0}^b}|{\hat{u}}^+-u_*(t_{j-1}^k)^+|dx \end{aligned}$$

for every ${\hat{u}}\in H^1(\varOmega ^+)$ with ${\hat{u}}=t_j^k$ on $\partial _D^+\varOmega$. Taking ${\hat{u}}=u_*(t_{j-1}^k)+w(t_j^k)-w(t_{j-1}^k)$, where w is defined by (5.2), and using Lemma 5.10 we obtain

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(t_j^k)|^2dxdy+\int _{\varGamma _{s_0}^b}\big (u_*(t_j^k)^+-u_*(t_{j-1}^k)^+\big )dx\\&\quad \le \frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(t_{j-1}^k)|^2dxdy+(t_j^k-t_{j-1}^k)\int _{\varOmega ^+}\frac{\partial u_*(t^k_{j-1})}{\partial y} dxdy+\frac{1}{2}(t_j^k-t_{j-1}^k)^2\frac{b-a}{h}. \end{aligned}$$

Summing for $j=1,\ldots ,m_k$ we obtain

$$\begin{aligned}&\frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(\tau _2)|^2dxdy+\int _{\varGamma _{s_0}^b}\big (u_*(\tau _2)^+-u_*(\tau _1)^+\big )dx\\&\quad \le \frac{1}{2} \int _{\varOmega ^+}|\nabla u_*(\tau _1)|^2dxdy+\sum _{j=1}^{m_k}(t_j^k-t_{j-1}^k)\int _{\varOmega ^+}\frac{\partial u_*(t^k_{j-1})}{\partial y} dxdy+\frac{1}{2}\frac{b-a}{kh}. \end{aligned}$$

Since $u_*:[0,{\hat{T}}]\rightarrow H^1(\varOmega ^+)$ is continuous, taking the limit as $k\rightarrow \infty$ we obtain (c${}_0$). $\square$

The following lemma will be used to prove Lemma 5.14, which shows that (5.5) holds.

Lemma 5.13

Let $z_0\in H^1(\varOmega ^+)$ be the solution of problem (5.3). Then for every $t\ge 0$ we have $u_*(t)\ge t+z_0$ in $\varOmega ^+$.

Proof

Let us fix $t\ge 0$ and let $v:=u_*(t)-(t+z_0)$. Then $v\in H^1(\varOmega ^+)$ and, by Lemma 5.9, it satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta v=0 &{}\text {in }\varOmega ^+,\\ v=0 &{}\text { if }y=h,\\ \frac{ \partial v}{\partial x}=0 &{}\text { if }x=a\text { or }x=b,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } a<x<s_0\text { and }y=0,\\ \frac{ \partial v}{\partial y}=0 &{} \text {if } s_0<x<\sigma _t\text { and }y=0,\\ \frac{ \partial v}{\partial y}\le 0 &{} \text {if } \sigma _t<x<b\text { and }y=0.\\ \end{array}\right. } \end{aligned}$$

Integrating by parts we obtain the weak formulation

$$\begin{aligned} \int _{\varOmega ^+}\nabla v\nabla \varphi \,dxdy=-\int _{\varGamma _{\sigma _t}^b}\frac{ \partial v}{\partial y}\varphi ^+dx \end{aligned}$$

for every $\varphi \in H^1(\varOmega ^+)$ with $\varphi =0$ on $\partial _D^+\varOmega$. Taking $\varphi :=v\wedge 0$ and arguing as in Lemma 5.9 we can prove that $v\ge 0$ in $\varOmega ^+$. Hence $u_*(t)\ge t+z_0$ in $\varOmega ^+$. $\square$

Lemma 5.14

Let $T_*:=-\inf _{\varOmega ^+}z_0$, where $z_0$ is the solution of (5.3). If $t>T_*$, then $u_*(t)=t+z_0$ in $\varOmega ^+$.

Proof

Let us fix $t>T_*$. By Lemma 5.13 there exists $\eta >0$ such that $u^t_{\sigma _t}=u_*(t)\ge t+z_0\ge \eta$ in $\varOmega ^+$, hence $u^t_{\sigma _t}(x,0)\ge \eta$ for every $x\in (\sigma _t,b)$. Since $u^t_{\sigma _t}(x,0)=0$ for every $x\in (\sigma _t,b)$ by the last condition in (5.14), we deduce that $\sigma _t=b$. The conclusion follows from Remark 5.6. $\square$

Proof of Theorem 5.1

By Theorem 4.1 and Remark 4.2 we have $s(t)=s_1=s_0$ for every $t\in [0,t_1)$. Moreover, if $t_1<T$ we have also $s(t)=s_2$ for every $t\in (t_1,t_2)$. To prove the theorem it is enough to show that $t_1<T$.

Assume, by contradiction, that $t_1=T$. Since $s(t)=s_0$ for every $t\in [0,T)$, the function u is a quasistatic evolution with fixed crack and boundary condition w on each interval $[0,{\hat{T}}]$ with $0<{\hat{T}}<T$. Let $u_*$ be the function defined by (5.26). By Lemma 5.12 the odd extension (with respect to y) of $u_*$ is a quasistatic evolution with fixed crack and boundary condition w on each interval $[0,{\hat{T}}]$ with ${\hat{T}}>0$. Since $u(0)=0=u_*(0)$, by the uniqueness result proved in Theorem 4.17 we have $u(t)=u_*(t)$ for every $t\in [0,T)$. Let us fix $t\in (T_*,T)$. By Lemma 5.14 we have $u_*(t)=t+ z_0$ in $\varOmega ^+$, which implies (5.5). Taking ${\hat{s}}=b$, ${\hat{u}}=t$ in $\varOmega ^+$, and ${\hat{u}}=-t$ in $\varOmega ^-$ in the equilibrium condition (b) of Definition 2.1, we obtain

$$\begin{aligned} \frac{1}{2} \int _{\varOmega \setminus \varGamma }|\nabla z_0|^2dxdy+s_0\le b, \end{aligned}$$

which contradicts the second inequality in (5.4). This proves that $t_1<T$ and concludes the proof of the theorem. $\square$