The influence of roughness in the equilibrium problem in lubrication with imposed load

Abstract

In this article we study a lubricated system consisting on a slider moving over a smooth surface and a known external force (the load) applied upon the slider. The slider moves at constant velocity and close proximity to the surface and the gap is filled by an incompressible fluid (the lubricant). At the equilibrium, the position of the slider presents one degree of freedom to be determined by the balance of forces acting on the system: the load and the total force exerted by the pressure of the lubricant. The pressure distribution is described by a variational inequality of elliptic type known as Swift–Stieber model and based on Reynolds equation. The distance h between the surfaces in a two dimensional domain Ω is given by $\begin{matrix} h_{η} (x_{1}, x_{2}, y) = h_{0} (x_{1}, x_{2}) + h_{1} (y) + η, (x_{1}, x_{2}) \in Ω, y \in [0, 1] \end{matrix}$ where $h_{0} (x_{1}, x_{2}) \sim | x_{1} |^{α}$ for $α > 0$ and $h_{1} (y) \sim | y - y_{0} |^{β}$ for y being the homogenization variable.

The main result of the article quantify the influence of the roughness in the load capacity of the mechanism in the following way: $\begin{matrix} If \{\begin{matrix} α < \frac{3}{γ} & for 0 < γ ⩽ 2 \\ α < min {\frac{1}{γ - 2}, \frac{3}{γ}} & for γ > 2 \end{matrix} \end{matrix}$ then, the mechanism presents finite load capacity, i.e. ${lim}_{η \to 0} \int_{Ω} p_{η} < \infty$ . Infinite load capacity is obtained for $γ > 1$ and $α > 2 / (γ - 1)$ . A one dimensional particular case is given for $γ > 3 / 2$ with infinite load capacity.

Keywords

Lubrication Reynolds variational inequality homogenization inverse problem

1. Introduction

Lubricants have been used from the beginning of the history to avoid contact between surfaces in relative motion, as journal bearings in carriage or in boats rudder among ancients applications. Industrial revolution brought a large number of applications to improve machine efficiency and tribology became also an area of scientific study. In 1886, O. Reynolds proposed a Partial Differential Equation, todays known as Reynolds Equation, to model the phenomenon.

The equation describes the behavior of a fluid, the lubricant, filling the gap between two surfaces in close proximity and relative motion. The model can be obtained by applying asymptotic expansions technics to Navier–Stokes equations under the usual assumption in the fluid domain: one dimension of the fluid domain is far smaller than the other two. Such assumption allows us to simplify Navier–Stokes equation to one elliptic equation in a two dimensional domain eliminating the smaller dimension of the domain. See for instance Nazarov [15] and Bayada and Chambat [2] for a rigorous proofs of the asymptotic convergence. The model that Reynolds heuristically deduce in 1886 for an incompressible fluid, is a linear elliptic equation where the solution describes the distribution of the pressure “p” of the fluid. If we denote by h the distance between the surfaces and by $(U_{1}, V_{1})$ its velocity, the equation reads $\begin{matrix} - div (\frac{h^{3}}{6 ν} \nabla p) = - div (h U_{1}, h V_{1}) - \frac{\partial h}{\partial t}, (x_{1}, x_{2}) \in Ω, t > 0, \end{matrix}$ where ν describes the viscosity of the fluid and Ω is the two dimensional domain. For a prescribed bounded function h and a given boundary condition $p_{0}$ , the problem has a unique solution $p \in H^{1} (Ω)$ satisfying $p = p_{0}$ in $\partial Ω$ . Once the pressure of the fluid is known, the behavior of the fluid is given by the two main components of the velocity $(u, v)$ $\begin{matrix} u = \frac{1}{2 ν} \frac{\partial p}{\partial x_{1}} z (z - h) + U_{1} (1 - \frac{z}{h}), v = \frac{1}{2 ν} \frac{\partial p}{\partial x_{2}} z (z - h) + V_{1} (1 - \frac{z}{h}), \end{matrix}$ while the third component w verifies $w = 0$ when $\frac{\partial h}{\partial t} = 0$ .

Nevertheless, the model proposed by O. Reynolds may presents some unrealistic solutions when the pressure p becomes negative, and it happens for instance for some functions h with a concave geometries. Starting with Sommerfield at the beginning of the XX century, several authors have modified the original model of Reynolds to avoid unrealistic situations and to consider cavitation, one of the most relevant phenomena in lubrication. Cavitation, the formation of vapor cavities in the lubricant caused by null pressure, produces wear and important lost of efficiency and even spoil the system. Cavitation is not considered in the original model of O. Reynolds. Modeling cavitation has been an important issue in lubrication throughout the 20th century where many different models have been considered (see for instance Bayada and Chambat [1] and Section 3 in the survey Bayada and Vázquez [4] for more details). Concerning cavitation, we highlight the model proposed by Elrod–Adams in [12], where an extra unknown is introduced to measure the concentration of the lubricant.

In this paper we consider the variational inequality to describe the distribution of the pressure of the lubricant in a two dimensional domain Ω. The model, also known as Swift–Stieber model (see Stieber [17] and Swift [18]), is introduced in the 30’s as an obstacle problem where the obstacle $ψ \equiv 0$ . Because its easy numerical implementation became one of the most used models in the last decades (see Bayada and Vázquez [4] and S. Martin [14] for more details). The model considers two different regions, the first one, with a complete fluid film where $p > 0$ where the Reynolds equation is satisfied, and a second region, where cavitation occurs and $p = 0$ . We assume that h is independent of t and $p_{0} = 0$ , then, the solution to the Reynolds cavitation model is a nonnegative function $p \in H_{0}^{1} (Ω)$ , satisfying $\begin{array}{l} \int_{Ω} \frac{h^{3}}{6 ν} \frac{\partial p}{\partial x_{1}} (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + \int_{Ω} \frac{h^{3}}{6 ν} \frac{\partial p}{\partial x_{2}} (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \\ ⩾ U_{1} \int_{Ω} h (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + U_{2} \int_{Ω} h (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \end{array}$ for any nonnegative function $ϕ \in H_{0}^{1} (Ω)$ . Clearly, in the region where $p > 0$ , the solution satisfies the Reynolds equation and the imposed condition of non-negativity guarantees that the pressure is in the set of physically admissible solutions.

After cavitation, one of the most important topics concerning the applications of the field is the “Load problem” where a known external force is applied to the mechanism. In that case, the distance between the lubricated surfaces is unknown and h may presents some degrees of freedom. If we only allow vertical displacement of the upper surface, $\begin{matrix} h = h_{0} + α \end{matrix}$ where $h_{0}$ is a given function describing the geometry between the surfaces which satisfies $\begin{matrix} min_{x \in Ω} {h_{0}} = 0 \end{matrix}$ and α represents the vertical displacement.

In this article, we only consider the equilibrium position, i.e. where the force exerted by the pressure balances the external application of the force. The external force F is assumed to be constant and known while the force exerted by the pressure is given in terms of the integral of the pressure, therefore the solution has to satisfies $\int_{Ω} p d x = F$ .

Since only vertical displacements of the upper surface are allowed, the problem consists on determine the parameter α. It is known that $\int_{Ω} p d x$ is a continuous function of α, so, the existence of equilibrium positions depends on $\begin{matrix} lim_{α \to \infty} \int_{Ω} p d x, and lim_{α \to 0} \int_{Ω} p d x . \end{matrix}$ Since the first limit is 0, the load capacity, defined by $\begin{matrix} sup_{α > 0} \int_{Ω} p d x, \end{matrix}$ depends on the second limit. Different cases have been considered in the literature, where the contact points i.e. $\begin{matrix} x \in Ω, such that h_{0} (x) = 0, \end{matrix}$ and the geometry of $h_{0}$ close to such points, determine the cited second limit. If the second limit is infinity, we have that for any given force $F > 0$ there exists an equilibrium of the problem. In the case, i.e. the second limit is finite, we only have a range of admisible forces F where the problem has a solution.

We list bellow a non exhaustive selection of the existing results in the literature.

In Ciuperca, Hafidi and Jai, [7], the problem is studied in a two dimensional domain for the line contact case where $h_{0} (x_{1}, x_{2}) \sim | x_{1} - d_{1} |^{α}$ and also for the point contact case, where $h_{0} (x_{1}, x_{2}) \sim {(| x_{1} - d_{1} |^{2} + | x_{2} - d_{2} |^{2})}^{α / 2}$ .

The authors obtain infinite load capacity for $α ⩾ 1$ for the line contact case and for $α ⩾ \frac{3}{2}$ for the point contact case. If $α < \frac{3}{4}$ in the line contact case and $α < \frac{4}{3}$ in point contact case, the result in [7] shows finite load capacity. Notice that for $α \in [\frac{3}{4}, 1)$ for the line contact and $α \in [\frac{4}{3}, \frac{3}{2})$ for the contact point are not considered in [7].

In Buscaglia, Ciuperca, Hafidi and Jai [5], the authors consider a function h with two degrees of freedom defined as follows $\begin{matrix} h = h_{0} (x_{1}, x_{2}) + η + θ x_{1} \end{matrix}$ where η defines the vertical displacement and θ an angular movement. The existence of equilibrium positions for $n = 1$ and $n = 2$ is obtained for any positive force “F” under assumptions $\begin{matrix} h_{0} (x_{1}, x_{2}) \sim {(1 - x_{1})}^{α}, α > 1 \end{matrix}$ and $\begin{matrix} \frac{\partial}{\partial x_{1}} (h_{0} (x_{1}, x_{2}) + η + θ x_{1}) ⩽ 0 . \end{matrix}$ The previous assumptions garantee that the contact between the surfaces at the limit case is in the form studied in Ciuperca, Hafidi and Jai [7] for the line contact and $α > 1$ .

The evolution problem have been also studied in Díaz and Tello [11] for the flat case, i.e. $h_{0} = constant$ . In [11], the behavior of the surface depends on the integral of the external force, i.e. $\begin{matrix} \int_{0}^{t} F (s) d s . \end{matrix}$ The case $h = h_{0} (x_{1}, x_{2}) + η (t)$ is considered in Ciuperca and Tello [10], for the line contact case, i.e. $\begin{matrix} \{\begin{matrix} h_{0} (0, x_{2}) = 0, h (x_{1}, x_{2}) > 0 & for (x_{1}, x_{2}) \in Ω, x_{1} \neq 0, \\ h_{0} (x_{1}, x_{2}) \sim | x_{1} |^{α} & when x_{1} \to 0 for some α ⩾ 1 \end{matrix} \end{matrix}$ and for the point contact case $\begin{array}{l} h_{0} (0, 0) = 0 and h_{0} (x_{1}, x_{2}) > 0 in Ω - {0} . \\ h_{0} (x) \sim {(x_{1}^{2} + x_{2}^{2})}^{α / 2}, when x \to 0 for some α ⩾ \frac{3}{2} . \end{array}$

In [10], the authors obtained the existence of a unique global solution for $α ⩾ \frac{3}{2}$ for the line contact case and $α ⩾ 2$ for the point contact case. For the other range of parameters i.e. ( $α < \frac{3}{2}$ in Line contact case and $α < 2$ in point contact case) there exists examples where solution collapses at finite time, i.e. $h \to 0$ as $t \to T$ for some $T < \infty$ .

In Ciuperca, Jai and Tello [9] the load problem is studied for the Elrod–Adams model where a two dimensional surface, defined by $h_{0} (x_{1}, x_{2}) \sim | x_{1} - γ |^{α}$ for some $α > 1$ , moves at constant velocity over a plane. In that case, the existence of equilibrium for any given force F applied to the upper surface is established.

In Ciuperca, Jai and Tello [8], a given force is applied to a journal bearing system modelized by a variational inequality, we proved the existence of equilibrium position for any given force F applied ortogonally to the journal axis which moves parallel to the exterior cylinder. Recently, in Lombera [13] the misaligned case has been studied obtaining finite load capacity for the misaligned case.

The cited works consider the load problem for smooth surfaces, nevertheless the effects of the roughness in the load capacity of the mechanism have not been analysed from a theoretical point of view. In particular, as the thickness of the gap decreases, the roughness becomes more influent in the pressure distribution and as we show in this article, it determines the threshold values of the forces and geometries which presents finite load capacity.

The influence of roughness in lubrication have been considered in the last decades with an extensive literature in the problem. We highlight the works of Christensen and Tonder [6] and Patir and Cheng [16] in the decade of 70’s. In [6], a periodic roughness is given and a deterministic approach is used to obtain a modified Reynolds like equation, while in [16], the model is obtained starting from an stochastic distribution of roughness.

By using a rigorous mathematical homogenization approach, Bayada and Faure [3] presented a modified Reynolds equations as a limit when the roughness period tends to zero. See for instance the survey Bayada and Vázquez [4] and Martin [14] for more details.

As we have seen above, three of the main topics in Lubrication are cavitation, load capacity and roughness. In this article we consider these three main topics, describing cavitation through a variational inequality (the Swift–Stieber model). The motivation of this study is to quantify the influence of the roughness in the load problem. In that direction, we obtain the threshold values in the geometry of the surfaces and roughness which distinguish finite or infinity load capacity.

Setting of the problem. We consider a two-dimensional homogenized model of lubrication where the roughness is taken into account. The problem is posed in a bounded domain of the form $\begin{matrix} Ω = (a_{1}, b_{1}) \times (a_{2}, b_{2}) \end{matrix}$ with $\begin{matrix} a_{1} < 0 ⩽ b_{1}, a_{2} < 0 ⩽ b_{2} . \end{matrix}$ In order to define the distance between the lubricated surfaces we introduce the following function $\begin{matrix} h_{0} : \overline{Ω} \to [0, \infty) . \end{matrix}$ We assume that $h_{0} \in C^{1} (\overline{Ω})$ and if roughness is not taken into account, we are in the case “line contact”, that is $\begin{matrix} h_{0} (x) > 0, if x \neq (0, x_{2}) \end{matrix}$ and $\begin{matrix} h_{0} (x) = 0, if x = (0, x_{2}) . \end{matrix}$ The roughness on one of the lubricated surfaces is given by $\begin{matrix} h_{1} : [0, 1] \to [0, + \infty), \end{matrix}$ for $\begin{matrix} (1) & h_{1} \in C^{1} ([0, 1]) \end{matrix}$ satisfying $h_{1} ⩾ 0$ .

Now, for any fixed $η > 0$ we consider the function $\begin{matrix} h_{η} : \overline{Ω} \times [0, 1] \to (0, \infty) \end{matrix}$ given by $\begin{matrix} h_{η} (x_{1}, x_{2}, y) = h_{0} (x_{1}, x_{2}) + h_{1} (y) + η \end{matrix}$ for $x = (x_{1}, x_{2}) \in Ω$ , $y \in [0, 1]$ and $η > 0$ .

For any $k \in Z$ we introduce the functions $\begin{matrix} \overline{h_{η}^{k}} : Ω \to (0, \infty) \end{matrix}$ defined by $\begin{matrix} \overline{h_{η}^{k}} = \int_{0}^{1} h_{η}^{k} (x, y) d y, for any x \in \overline{Ω} . \end{matrix}$

Remark 1.
Observe that for $k < 0$ it is possible that $\overline{h_{η}^{k}} (0, x_{2})$ goes to $+ \infty$ when η goes to 0.

In order to describe the variational inequality (Swift–Stieber model), we introduce the following Hilbert spaces: $\begin{matrix} V : = {u \in H^{1} (Ω), such that u (a_{1}, x_{2}) = u (b_{1}, x_{2}) = 0, and periodic in x_{2}} \end{matrix}$ and $\begin{matrix} V_{+} : = {u \in V, u ⩾ 0} . \end{matrix}$ Notice that the boundary conditions of the solution are
Dirichlet homogeneous in $x_{1}$ ;

Periodic boundary conditions in $x_{2}$ .
We also assume that $\begin{matrix} (2) & h_{0} is a periodic function in x_{2} . \end{matrix}$
Remark 2.
We may also consider Dirichlet homogeneous boundary conditions in $\partial Ω$ . In that case we have to replace V by $H_{0}^{1} (Ω)$ and remove the periodicity of $h_{0}$ in $x_{2}$ .

Then, for any fixed $η > 0$ we have the following variational inequality:

Find $p \in V_{+}$ such that $\begin{array}{l} \int_{Ω} {(\overline{h_{η}^{- 3}})}^{- 1} \frac{\partial p}{\partial x_{1}} (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + \int_{Ω} {(\overline{h_{η}^{- 3}})}^{- 1} \frac{\partial p}{\partial x_{2}} (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \\ (3) & ⩾ U_{1} \int_{Ω} \frac{\overline{h_{η}^{- 2}}}{\overline{h_{η}^{- 3}}} (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + U_{2} \int_{Ω} \overline{h_{η}} (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \end{array}$ for all $ϕ \in V_{+}$ , where $U_{1} > 0$ and $U_{2} \in R$ are given normalized velocities.

Notice that the total force exerted by the pressure of the fluid is $\int_{Ω} p d x$ .
Remark 3.
It is well known that for any $η > 0$ , problem (3) admits a unique solution $p \in V_{+}$ .

The main question that we study in this work is the behavior of $\int_{Ω} p d x$ when $η \to 0$ . Notice that the pressure p depends on η, but for simplicity of notation we omite the index η in p.

We want to find out whether the integral $\int_{Ω} p d x$ is bounded when $η \to 0$ . We need some extra assumptions of the behavior of $h_{0} (x_{1}, x_{2})$ (respectively $h_{1}$ ) when $x_{1} \to 0$ (repectively $y \to y_{0}$ ).

We assume that there exists $α > 0$ and $β > 1$ such that $\begin{array}{l} (4) & h_{0} (x_{1}, x_{2}) = | x_{1} |^{α} {\tilde{h}}_{0} (x_{1}, x_{2}) and \frac{\partial h_{0}}{\partial x_{1}} (x_{1}, x_{2}) = x_{1} | x_{1} |^{α - 2} h_{01} (x_{1}, x_{2}) \\ (5) & h_{1} (y) = | y - y_{0} |^{β} {\tilde{h}}_{1} (y) \end{array}$ where ${\tilde{h}}_{0}, h_{01}$ and ${\tilde{h}}_{1}$ are bounded functions and far from 0, i.e. there exists two positive constants $c_{1}$ and $c_{2}$ independent of η such that $\begin{array}{l} 0 < c_{1} < c_{2}, \\ c_{1} ⩽ {\tilde{h}}_{0} (x), {\tilde{h}}_{01} (x) ⩽ c_{2}, \forall x \in \overline{Ω}, \\ c_{1} ⩽ {\tilde{h}}_{1} ⩽ c_{2}, y \in [0, 1] . \end{array}$ The case without roughness, i.e. $h_{1} = 0$ , which implies $\begin{matrix} \overline{h_{η}^{k}} = {(h_{0} (x_{1}) + η)}^{k} \end{matrix}$ have been studied in Ciuperca, Hafidi and Jai [7] where infinite load capacity is obtained for $α ⩾ 1$ and finite load capacity for $α \in (0, 1)$ .

The article is organized as follows. In Section 2 we consider some preliminaries concerning the coefficients of the Reynolds equation with roughness to be applied in the subsequent sections. In Section 3 we consider the finite load capacity case, the result is obtained for α and γ satisfying $\begin{matrix} \{\begin{matrix} α < \frac{3}{γ} & if 0 < γ ⩽ 2 \\ α < min {\frac{1}{γ - 2}, \frac{3}{γ}} & if γ > 2 \end{matrix} \end{matrix}$ the results are inclosed in Theorem 1. In Section 4, the infinite load capacity case is presented for $γ > 1$ and $α > 2 / (γ - 1)$ (see Theorem 2). Finally, we present a particular case for a one dimensional domain and $γ > 3 / 2$ with infinite load capacity.
2. Preliminaries

In this section we consider some preliminar estimates of the coefficients of the homogenized equation. We begin by the following lemma.

Lemma 1.
There exist positive constants $c_{3}$ , $c_{4}$ and δ independent of η satisfying $0 < c_{3} < c_{4}$ , $δ > 0$ such that for $h_{0} (x) + η < δ$ we have
$c_{3} ⩽ \overline{h_{η}^{k}} (x) ⩽ c_{4}$ for $k \in N$ .

$c_{3} {(h_{0} (x) + η)}^{1 / β - k} ⩽ \overline{h_{η}^{- k}} ⩽ c_{4} {(h_{0} (x) + η)}^{1 / β - k}$ for $k \in N$ with $k > \frac{1}{β}$ .

$\begin{matrix} \frac{\partial}{\partial x_{1}} [\frac{\overline{h_{η}^{- 2}}}{\overline{h_{η}^{- 3}}}] = \frac{\partial h_{0}}{\partial x_{1}} E (x, η) \end{matrix}$ with $1 ⩽ E (x, η) ⩽ c_{4}$ .

Proof.

This is a consequence of positivity of $h_{1}$ and assumptions in $h_{0}$ .

We introduce $\begin{matrix} I_{1} = {y \in (0, 1) : h_{1} (y) ⩽ h_{0} (x) + η}, \end{matrix}$ since $h_{1} (y) \sim {(y - y_{0})}^{β}$ , there exists a constant $c_{5} > 0$ such that $\begin{matrix} meas (I_{1}) ⩾ c_{5} {[h_{0} (x) + η]}^{\frac{1}{β}} . \end{matrix}$ Then $\begin{matrix} \int_{0}^{1} \frac{d y}{{[h_{0} (x) + h_{1} (y) + η]}^{k}} ⩾ \int_{I_{1}} \frac{d y}{2^{k} {(h_{0} (x) + η)}^{k}} ⩾ c_{5} 2^{- k} {(h_{0} (x) + η)}^{\frac{1}{β} - k} \end{matrix}$ which ends the inequality on the left hand side.

To prove the righthand side inequality we consider $δ > 0$ such that there exists $η_{1}$ and $η_{2}$ small enough satisfying $\begin{matrix} h_{1} (y) ⩽ δ ⟺ y \in [y_{0} - η_{1}, y_{0} + η_{2}] = I_{2} . \end{matrix}$ Notice that by (5) such a δ exists.

Now, we have $\begin{matrix} \int_{[0, 1] - I_{2}} \frac{d y}{{[h_{0} (x) + h_{1} (y) + η]}^{k}} ⩽ \int_{[0, 1] - I_{2}} \frac{d y}{{[h_{0} (x) + δ + η]}^{k}} ⩽ \frac{1}{δ^{k}} . \end{matrix}$ On the other hand, we get $\begin{array}{l} \int_{I_{2}} \frac{d y}{{[h_{0} (x) + h_{1} (y) + η]}^{k}} & ⩽ \int_{I_{2}} \frac{d y}{{[c_{1} | y - y_{0} |^{β} + h_{0} (x) + η]}^{k}} \\ = \frac{1}{{(h_{0} (x) + η)}^{k}} \int_{I_{2}} \frac{d y}{{[1 + c_{1} \frac{| y - y_{0} |^{β}}{h_{0} (x) + η}]}^{k}} . \end{array}$ We now introduce the following change of variables $\begin{matrix} z = \frac{y - y_{0}}{{(h_{0} (x) + η)}^{1 / β}} . \end{matrix}$ Then, it results $\begin{matrix} \int_{I_{2}} \frac{d y}{| h_{0} (x) + h_{1} (y) + η |^{k}} ⩽ \frac{{(h_{0} (x) + η)}^{1 / β}}{{(h_{0} (x) + η)}^{k}} \int_{- {\tilde{η}}_{1}}^{\tilde{η_{2}}} \frac{d z}{{(1 + c_{1} | z |^{β})}^{k}} \end{matrix}$ with $\begin{matrix} {\tilde{η}}_{j} = \frac{η_{j}}{| h_{0} (x) + η |^{1 / β}}, for j = 1, 2 . \end{matrix}$ Since $β k > 1$ , we have $\begin{matrix} \int_{- {\tilde{η}}_{1}}^{\tilde{η_{2}}} \frac{d z}{| 1 + c_{1} | z |^{β} |^{k}} ⩽ \int_{- \infty}^{\infty} \frac{d z}{| 1 + c_{1} | z |^{β} |^{k}} < + \infty \end{matrix}$ which ends the proof of ii).

We have $\begin{matrix} \frac{\partial}{\partial x} (\frac{\overline{h^{- 2}}}{\overline{h^{- 3}}}) = \frac{\partial h_{0}}{\partial x} E (x, η) \end{matrix}$ where $\begin{matrix} E (x, η) = \frac{- 2 {(\overline{h^{- 3}})}^{2} + 3 \overline{h^{- 2}} \cdot \overline{h^{- 4}}}{{(\overline{h^{- 3}})}^{2}} . \end{matrix}$ Thanks to Cauchy–Swartz inequality, we obtain $\begin{matrix} {\overline{h^{- 3}}}^{2} ⩽ \overline{h^{- 2}} \cdot \overline{h^{- 4}}, \end{matrix}$ which implies $\begin{matrix} E (x, η) ⩾ 1 . \end{matrix}$ Now, thanks to ii), we obtain the desired upper bound on E and the proof ends.
□
Remark 4.
From Lemma 1 we deduce that in the variational inequality (3) the diffusion coefficients ${(\overline{h_{η}^{- 3}})}^{- 1}$ and $\overline{h_{η}^{3}}$ behave as ${[h_{0} (x) + η]}^{3 - 1 / β}$ and 1 respectively (for $h_{0} (x) + η$ small) while the coefficients $\frac{\overline{h_{η}^{- 2}}}{\overline{h_{η}^{- 3}}}$ and $\overline{h_{η}^{1}}$ behave as $h_{0} (x) + η$ and 1 respectively.

In the following, thanks to Remark 4, we consider a more general problem of the form $\begin{matrix} (6) & \{\begin{matrix} Find p \in V_{+} such that \\ \int_{Ω} A_{1} (x, η) \frac{\partial p}{\partial x_{1}} (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + \int_{Ω} A_{2} (x, η) \frac{\partial p}{\partial x_{2}} (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \\ ⩾ U_{1} \int_{Ω} B_{1} (x, η) (\frac{\partial ϕ}{\partial x_{1}} - \frac{\partial p}{\partial x_{1}}) + U_{2} \int_{Ω} B_{2} (x, η) (\frac{\partial ϕ}{\partial x_{2}} - \frac{\partial p}{\partial x_{2}}) \end{matrix} \end{matrix}$ where $A_{1}$ , $A_{2}$ , $B_{1}$ and $B_{2}$ satisfy the following assumptions $\begin{matrix} A_{1}, A_{2}, B_{1}, B_{2} \in C (\overline{Ω} \times [0, \infty)) \end{matrix}$ and periodic in $x_{2}$ . There exist real positive numbers γ, $c_{1}$ and $c_{2}$ satisfying $0 < c_{1} < c_{2}$ such that for any $x \in Ω$ and $η ⩾ 0$ we have

We also assume $U_{1} > 0$ and $U_{2} \in R$ . Remark 5.
If the data $h_{0}$ , $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ are independent of $x_{2}$ , then the solution of (3) is the corresponding solution of the one-dimensional problem.

3. The bounded case. Finite load capacity

In this section we give conditions on γ and α which guarantee the boundedness of $\int_{Ω} p d x$ for p being the solution of (6).

Theorem 1.
Under assumption ( 7a )–( 7e ) and α, γ satisfying $\begin{matrix} (8) & \{\begin{matrix} α < \frac{3}{γ} & if 0 < γ ⩽ 2 \\ α < min {\frac{1}{γ - 2}, \frac{3}{γ}} & if γ > 2, \end{matrix} \end{matrix}$ we have the existence of a constant c, independent of η, such that $\begin{matrix} \int_{Ω} p d x ⩽ c . \end{matrix}$
Proof.
We take $ϕ = 0$ and $ϕ = 2 p$ respectively in (6) to obtain $\begin{matrix} \int_{Ω} [A_{1} (x, η) {(\frac{\partial p}{\partial x_{1}})}^{2} + A_{2} (x, η) {(\frac{\partial p}{\partial x_{2}})}^{2}] d x = U_{1} \int_{Ω} B_{1} (x, η) \frac{\partial p}{\partial x_{1}} d x + U_{2} \int_{Ω} B_{2} (x, η) \frac{\partial p}{\partial x_{2}} d x . \end{matrix}$ We deduce $\begin{array}{l} \int_{Ω} [{(h_{0} + η)}^{γ} {(\frac{\partial p}{\partial x_{1}})}^{2} + {(\frac{\partial p}{\partial x_{2}})}^{2}] d x \\ ⩽ C \int_{Ω} {(h_{0} + η)}^{\frac{γ}{2}} | \frac{\partial p}{\partial x_{1}} | {(h_{0} + η)}^{- \frac{γ}{2}} B_{1} (x, η) d x + C \int_{Ω} | \frac{\partial p}{\partial x_{2}} | d x \\ ⩽ \frac{1}{2} \int_{Ω} {(h_{0} + η)}^{γ} {| \frac{\partial p}{\partial x_{1}} |}^{2} d x + \frac{1}{2} \int_{Ω} {| \frac{\partial p}{\partial x_{2}} |}^{2} d x + \tilde{c} (1 + \int_{Ω} {(h_{0} + η)}^{2 - γ} d x) . \end{array}$ From hypothesis (4) and (8) we deduce $\begin{matrix} (9) & \int_{Ω} {(h_{0} + η)}^{γ} {| \frac{\partial p}{\partial x_{1}} |}^{2} d x + \int_{Ω} {| \frac{\partial p}{\partial x_{2}} |}^{2} d x ⩽ C . \end{matrix}$ On the other hand we have $\begin{array}{l} \int_{Ω} p & = \int_{Ω} (- x_{1}) \frac{\partial p}{\partial x_{1}} d x = \int_{Ω} (- x_{1}) {(h_{0} + η)}^{\frac{γ}{2}} \frac{\partial p}{\partial x_{1}} {(h_{0} + η)}^{- \frac{γ}{2}} d x \\ ⩽ {[\int_{Ω} {(h_{0} + η)}^{γ} {| \frac{\partial p}{\partial x_{1}} |}^{2}]}^{\frac{1}{2}} {[\int_{Ω} x_{1}^{2} {(h_{0} + η)}^{- γ} d x]}^{\frac{1}{2}} . \end{array}$ with the help of (9) and thanks to assumptions (4) and (8), the proof ends. □
Remark 6.

For the homogenaized problem (3) from Remark 4, we have that $γ = 3 - \frac{1}{β}$ , then assumption (8) becomes $\begin{matrix} α < \frac{3 β}{3 β - 1} . \end{matrix}$

In the case without roughness (the case $h_{1} = 0$ ) we have $γ = 3$ then, the assumption (8) becomes $\begin{matrix} α < 1 \end{matrix}$ as in Ciuperca, Hafidi and Jai [7].

4. The unbounded case. Infinite load capacity

We introduce the following domain $\begin{matrix} Ω_{η} = (- 2 η^{1 / α}, - η^{1 / α}) \times (a_{2}, b_{2}) \end{matrix}$ and notice that $Ω_{η} \subset Ω$ for η small enough.

We denote by q the solution of the following elliptic equation on $Ω_{η}$ $\begin{matrix} (10) & \{\begin{matrix} Find q \in V_{η} such that \\ \frac{\partial}{\partial x_{1}} (A_{1} \frac{\partial q}{\partial x_{1}}) + \frac{\partial}{\partial x_{2}} (A_{2} \frac{\partial q}{\partial x_{2}}) = U_{1} \frac{\partial B_{1}}{\partial x_{1}} \end{matrix} \end{matrix}$ where $\begin{array}{l} V_{η} = & {ϕ \in H^{1} (Ω_{η}), such that ϕ is periodic in x_{2} and satisfy \\ ϕ (- 2 η^{1 / α}, x_{2}) = ϕ (- η^{1 / α}, x_{2}) = 0, for any a_{2} < x_{2} < b_{2}} . \end{array}$ It is clear that there exists a unique solution q to (10) which is a sub-solution of (6) (after extension by 0 on Ω) in the case $U_{2} = 0$ , that is $\begin{matrix} (11) & p ⩾ q, on Ω_{η} for U_{2} = 0 . \end{matrix}$ We have the following result.

Theorem 2.
Under assumption ( 7a )–( 7e ), for $\begin{matrix} U_{2} = 0, \end{matrix}$ $γ > 1$ and $\begin{matrix} (12) & α > \frac{2}{γ - 1}, \end{matrix}$ we obtain $\begin{matrix} \int_{Ω} p d x \to \infty as η \to \infty . \end{matrix}$ More precisely we have $\begin{matrix} \int_{Ω} p d x ⩾ C η^{1 + \frac{2}{α} - γ}, \end{matrix}$ with C a constant independent of η.
Proof.
Step 1. We first give a lower bound for $\begin{matrix} \int_{Ω_{η}} {(h_{0} + η)}^{γ} {(\frac{\partial ϕ}{\partial x_{1}})}^{2} d x . \end{matrix}$ Let us denote $\begin{matrix} V_{0 η} = {ϕ \in V_{η}, such that ϕ is independent of x_{2}} and V_{0 η}^{+} = {ϕ \in V_{0 η} : ϕ ⩾ 0} . \end{matrix}$ From (10) we have that for any $ϕ \in V_{0 η}^{+}$ $\begin{matrix} \int_{Ω_{η}} A_{1} \frac{\partial q}{\partial x_{1}} ϕ^{'} (x_{1}) = - \int_{Ω_{η}} V_{1} \frac{\partial B_{1}}{\partial x_{1}} ϕ d x . \end{matrix}$ Then, we deduce $\begin{matrix} - \int_{Ω_{e} t a} V_{1} \frac{\partial B_{1}}{\partial x_{1}} ϕ d x ⩽ C {[\int_{Ω_{η}} {(h_{0} + η)}^{γ} {(\frac{\partial q}{\partial x_{1}})}^{2}]}^{\frac{1}{2}} {[\int_{Ω_{η}} {(h_{0} + η)}^{γ} {| ϕ^{'} (x_{1}) |}^{2}]}^{\frac{1}{2}} \end{matrix}$ and therefore $\begin{matrix} (13) & \int_{Ω_{η}} {(h_{0} + η)}^{γ} {(\frac{\partial q}{\partial x_{1}})}^{2} ⩾ C sup_{ϕ \in V_{0 η}^{+} - {0}} \frac{{[\int_{Ω_{η}} (- \frac{\partial h_{0}}{\partial x_{1}}) ϕ]}^{2}}{\int_{Ω_{η}} {(h_{0} + η)}^{γ} | ϕ^{'} (x_{1}) |^{2}} . \end{matrix}$ Now, we take in (13) $\begin{matrix} ϕ = ϕ_{0} (\frac{x_{1}}{η^{1 / α}}) \end{matrix}$ with $ϕ_{0} \in D (- 2, - 1), ϕ_{0} ⩾ 0, ϕ_{0}$ not identically 0 and independent of η, to deduce that there exists a constant $C > 0$ such that $\begin{matrix} (14) & \int_{Ω_{η}} {(h_{0} + η)}^{γ} {(\frac{\partial q}{\partial x_{1}})}^{2} ⩾ C η^{2 + \frac{1}{α} - γ} . \end{matrix}$ Step 2.

From (10) we have $\begin{matrix} (15) & - \int_{Ω_{η}} V_{1} \frac{\partial B_{1}}{\partial x_{1}} q = \int_{Ω_{η}} A_{1} {| \frac{\partial q}{\partial x_{1}} |}^{2} + \int_{Ω_{η}} A_{2} {| \frac{\partial q}{\partial x_{2}} |}^{2} . \end{matrix}$ We notice that from maximum principle and assumption (4) and (7a)–(7e) we have $\begin{matrix} q ⩾ 0 on Ω_{η} . \end{matrix}$ We deduce that $\begin{matrix} \int_{Ω_{η}} (- \frac{\partial h_{0}}{\partial x_{1}}) q ⩾ C \int_{Ω} {(h_{0} + η)}^{γ} {| \frac{\partial q}{\partial x_{1}} |}^{2} \end{matrix}$ and thanks to assumption (4) and (14) we claim $\begin{matrix} \int_{Ω_{η}} q ⩾ C η^{1 + \frac{2}{α} - γ} . \end{matrix}$ Now, using (10) and the fact that $p ⩾ q$ on $Ω_{η}$ , since $p ⩾ 0$ on Ω, we obtain the wished result. □
Remark 7.

In the case of homogenaized problem (3) we have $γ = 3 - \frac{1}{β}$ , then assumption (12) becomes $\begin{matrix} α > \frac{2 β}{2 β - 1} . \end{matrix}$

In the case without roughness $(γ = 3)$ , assumption (12) becomes $α > 1$ .

We denote $\begin{matrix} α_{1} (γ) = \{\begin{matrix} \frac{3}{γ} & if 0 < γ ⩽ 2 \\ min {\frac{1}{γ - 2}, \frac{3}{γ}} & if γ > 2 \end{matrix} \end{matrix}$ (see Theorem 1) and $\begin{matrix} α_{2} (γ) = \frac{2}{γ - 1}, if γ > 1 (see Theorem 2). \end{matrix}$

Notice that for $γ > 1$ and $γ \neq 3$ we have $α_{1} (γ) < α_{2} (γ)$ and $α_{1} (γ) = α_{2} (γ) = 1$ for $γ = 3$ . Then for α such that $\begin{matrix} α_{1} (γ) ⩽ α ⩽ α_{2} (γ) \end{matrix}$ the behavior of $\int_{Ω} p d x$ remains undetermined.
5. A particular one dimensional case

In this point we consider a one dimensional case with $\begin{matrix} Ω = (a_{1}, 0) \end{matrix}$ and $\begin{matrix} \frac{\partial h_{0}}{\partial x_{1}} ⩽ 0, on Ω . \end{matrix}$ In this case, due to the maximum principle, the solution of (6) is in fact the solution of the equation $\begin{matrix} (16) & \{\begin{matrix} \frac{d}{d x} (A_{1} (x, η) \frac{d p}{d x}) = U_{1} \frac{d}{d x} B_{1} (x, η), \\ p (a_{1}) = p (0) = 0 . \end{matrix} \end{matrix}$ We remind that $A_{1}$ and $B_{1}$ satisfy assumptions (7a), (7c) and (7e).

As mentioned in Remark 7 iii), for $γ > 1$ , we cannot say anything about the behavior of $\int_{Ω} p d x$ as $η \to 0$ if $\begin{matrix} α \in [α_{1} (γ), α_{2} (γ)] . \end{matrix}$ In the following theorem we show that we can fill this gap for this particular case.

Theorem 3.

We assume that $γ > \frac{3}{2}$ . Then for any $\begin{matrix} α \in [α_{1} (γ), α_{2} (γ)] \end{matrix}$ we have that $\begin{matrix} \int_{Ω} p d x \to \infty, as η \to 0 . \end{matrix}$ Moreover there exists a constant $C ⩾ 0$ such that $\begin{matrix} \int_{Ω} p d x ⩾ C η^{γ - 1 - 2 / α}, if α_{1} (γ) ⩽ α < α_{2} (γ) \end{matrix}$ and $\begin{matrix} \int_{Ω} p d x ⩾ C log \frac{1}{η}, if α = α_{2} (γ) . \end{matrix}$

Proof.

We introduce the following change of unknown in (16) $\begin{matrix} x = η^{1 / α} z, \end{matrix}$ we have $\begin{matrix} p = η^{1 + 1 / α - γ} p_{0} \end{matrix}$ where $p_{0}$ is the solution to $\begin{matrix} (17) & \frac{d}{d z} ({((1 + {(- z)}^{α}) {\tilde{h}}_{1} (η^{1 / α} z))}^{γ} {\tilde{A}}_{1} (z, η) \frac{d p_{0}}{d z}) = \frac{d}{d z} ([(1 + {(- z)}^{α}) {\tilde{h}}_{1} (η^{1 / α} z)] {\tilde{B}}_{1} (z, η)), \end{matrix}$ on $\begin{matrix} Ω_{η} = (η^{- 1 / α} a_{1}, 0) \end{matrix}$ and satisfies the boundary condition $\begin{matrix} (18) & p_{0} = 0 on \partial Ω_{η} \end{matrix}$ for ${\tilde{A}}_{1}$ and ${\tilde{B}}_{1}$ satisfying $\begin{matrix} c_{1} ⩽ {\tilde{A}}_{1}, {\tilde{B}}_{1} ⩽ c_{2}, \end{matrix}$ with $c_{1}$ and $c_{2}$ given constants fullfilling $\begin{matrix} 0 < c_{1} ⩽ c_{2} . \end{matrix}$ Now, solving (17), (18), we deduce $\begin{matrix} \frac{d p_{0}}{d z} = E_{1} (η, z) - C_{η} E_{2} (η, z) \end{matrix}$ with $\begin{array}{l} E_{1} (η, z) = \frac{([(1 + {(- z)}^{α}) {\tilde{h}}_{1} (η^{1 / α} z)] {\tilde{B}}_{1} (z, η))}{{((1 + {(- z)}^{α}) {\tilde{h}}_{1} (η^{1 / α} z))}^{γ} {\tilde{A}}_{1} (z, η)}, \\ E_{2} (η, z) = \frac{1}{{((1 + {(- z)}^{α}) {\tilde{h}}_{1} (η^{1 / α} z))}^{γ} {\tilde{A}}_{1} (z, η)} \end{array}$ and $\begin{matrix} C_{η} = \frac{\int_{a_{1} η^{- 1 / α}}^{0} E_{1} (η, z) d z}{\int_{a_{1} η^{- 1 / α}}^{0} E_{2} (η, z) d z} . \end{matrix}$ Since $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} p_{0} (z) d z = - \int_{a_{1} η^{- 1 / α}}^{0} z \frac{d p_{0}}{d z} d z, \end{matrix}$ we can write $\begin{matrix} (19) & \int_{a_{1} η^{- 1 / α}}^{0} p_{0} (z) d z = \int_{a_{1} η^{- 1 / α}}^{0} (- z) E_{1} (η, z) d z - C_{η} \int_{a_{1} η^{- 1 / α}}^{0} (- z) c E_{2} (η, z) d z . \end{matrix}$ Now, we remark that $\begin{matrix} E_{1} (η, z) \sim {(- z)}^{α (1 - γ)} when z \to - \infty \end{matrix}$ and $\begin{matrix} E_{2} (η, z) \sim {(- z)}^{- α γ} when z \to - \infty . \end{matrix}$ We also have $\begin{array}{l} E_{1} (η, z) \sim 1 when z \to 0_{-} \\ E_{2} (η, z) \sim 1 when z \to 0_{-} . \end{array}$ We easily deduce, since $α ⩾ α_{1} (γ) > \frac{1}{γ - 1} > \frac{1}{γ}$ , that $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} E_{1} (η, z) d z \sim 1 as η \to 0 \end{matrix}$ and $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} E_{2} (η, z) d z \sim 1 as η \to 0, \end{matrix}$ which implies $C_{η} \sim 1$ as $η \to 0$ .

Now, we write $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} p_{0} (z) d z = \int_{a_{1} η^{- 1 / α}}^{0} (- z) (E_{1} (x, z) - C_{η} E_{2} (x, z)) d z \end{matrix}$ which implies $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} p_{0} (z) d z \sim log \frac{1}{η} for α = α_{2} (γ) \end{matrix}$ and $\begin{matrix} \int_{a_{1} η^{- 1 / α}}^{0} p_{0} (z) d z \sim C η^{γ - 1 - 2 / α} for α < α_{2} (γ) \end{matrix}$ and the proof ends. □

Footnotes

Acknowledgements

This work is partially supported by Project MTM2017-83391-P MICINN (Spain).

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