Abstract
Our aim in this paper is to investigate the existence and uniqueness result for a class of mixed variational problems. They are governed by two variational inequalities. By applying the saddle-point theory, we obtain the existence of solutions to mixed variational problems. Finally, some frictional contact problems are given to illustrate our main results.
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1 Introduction
The abstract problem in this paper is a class of mixed variational problems governed by two variational inequalities, with a bilinear function and functional which is convex and lower semicontinuous. Considering such kinds of variational problems sets the functional background in the study of elastic contact problems with unilateral constraints and nonmonotone interface laws. For very recent work, see [1–3]. Our results improve the results in [4–7], which consider a bilinear functional.
Let X be a Hilbert space, Y a reflexive Banach space and Λ be a nonempty, closed and convex subset Y. Let \(A:X\rightarrow X\), \(B:X\times\Lambda\rightarrow\mathbb{R}\), \(\phi :X\to\mathbb{R}\) be given maps to be specified later, and \(w,f\in X\) be fixed. We consider the mixed variational problems of following forms.
Problem 1
Find \((u,\lambda)\in X\times\Lambda\) such that
Recently, a lot of work was devoted to the modeling in contact mechanics. Weak formulations of contact problems involve the theory of variational inequalities and the theory of saddle-point problems; see e.g. [7–13]. The purpose of this paper is to investigate the weak solvability of a unilateral frictionless contact problem using a technique with dual Lagrange multipliers. The weak formulations with dual Lagrange multipliers allow one to write efficient algorithms in order to approximate the weak solutions. In the present work, the behavior of the materials is described by using the subdifferential of a proper, convex, lower semicontinuous functional and the contact is modeled with Signorini’s condition with zero gap. The results extend and improve the results obtained in [5, 6], where a unilateral frictionless contact model for nonlinearly elastic materials is analyzed.
In this paper, by applying the saddle-point theory, we obtain the existence of solutions to the mixed variational problems. The main results are that Problem 1 has a unique solution \((u,\lambda)\in X\times \Lambda\) and the solution \((u,\lambda)\) is Lipschitz continuous with respect to f and ω. Then we apply them to find that our model has a unique weak solution and the weak solution is Lipschitz continuous with respect to the data.
The rest of this paper is organized as follows. In Section 2, we will introduce some useful preliminaries and necessary materials. Section 3 is devoted to proving the existence and uniqueness results. In the last section, we give some examples of friction contact problems to illustrate our main results.
2 Preliminaries
In this section, we will introduce some basic preliminaries which are used throughout this paper. Denote by \(\|\cdot\|_{X}\) and \((\cdot ,\cdot)_{X}\) the norm and inner product of the Hilbert space X the inner product of X, respectively.
Definition 2
Let K be a nonempty subset of a Banach space X. A function \(f:K\rightarrow\mathbb{R}\) is said to be
-
(i)
convex on K if for every finite subset \(\{ x_{1},\ldots,x_{n}\}\subset K\) and \(\{\lambda_{1},\ldots,\lambda_{n}\} \subset\mathbb{R}_{+}\) such that \(\sum_{i=1}^{n}\lambda_{i}=1\) and \(\sum_{i=1}^{n}\lambda_{i}x_{i}\in K\), then
$$ f\Biggl(\sum_{i=1}^{n}\lambda_{i}x_{i} \Biggr)\leq\sum_{i=1}^{n} \lambda_{i}f(x_{i}); $$ -
(ii)
concave on K if −f is convex on K;
-
(iii)
upper semicontinuous (u.s.c.) at \(x_{0}\), if, for any sequence \(\{x_{n}\}_{n\geq1}\subset K\) with \(x_{n}\rightarrow x_{0}\), we have
$$ \limsup_{n\rightarrow\infty}f(x_{n})\leq f(x_{0}); $$ -
(iv)
lower semicontinuous (l.s.c.) at \(x_{0}\), if, for any sequence \(\{x_{n}\}_{n\geq1}\subset K\) with \(x_{n}\rightarrow x_{0}\), we have
$$ \liminf_{n\rightarrow\infty}f(x_{n})\geq f(x_{0}). $$f is said to be u.s.c. (l.s.c.) on K if f is u.s.c. (l.s.c.) at x for all \(x\in K\).
We now recall some elements of convex analysis. For more details, we refer to [10, 14, 15].
Definition 3
Let A and B be nonempty sets. A pair \((u,\lambda)\in C\times D\) is said to be a saddle point of a functional \(\mathcal{L}:C\times D\rightarrow\mathbb{R}\) if
In this paper, we apply the following result to prove our existence result.
Theorem 4
Let X, Y be reflexive Banach spaces, and let \(A\subset X\), \(B\subset Y\) be nonempty, closed, convex subsets. Assume that a functional \(\mathcal{L}:C\times D\rightarrow\mathbb{R}\) satisfies the following conditions:
-
(i)
\(v\mapsto\mathcal{L}(v,\mu)\) is convex and l.s.c. for all \(\mu\in D\),
-
(ii)
\(\mu\mapsto\mathcal{L}(v,\mu)\) is concave and u.s.c. for all \(v\in C\).
Moreover, assume that
-
(iii)
C is bounded or
$$ \lim_{\|v\|_{X}\rightarrow\infty,v\in C}\mathcal{L}(v,\mu _{0})=\infty\quad \textrm{for~some } \mu_{0}\in D, $$ -
(iv)
D is bounded or
$$ \lim_{\|\mu\|_{Y}\rightarrow\infty,\mu\in D}\inf_{v\in C}\mathcal {L}(v, \mu_{0})=-\infty. $$
Then the functional \(\mathcal{L}\) has at least one saddle point.
We also recall the following definition.
Definition 5
Let X be a Hilbert space and \(\varphi:X\to\mathbb{R}\). φ is said to be Gâteaux differentiable at \(u\in X\), if there exists an element \(\nabla\varphi(u)\in X\) such that
The element \(\nabla\varphi(u)\) which satisfies the relation above is unique and is called the gradient of φ at u. \(\varphi:X\to\mathbb{R}\) is said to be Gâteaux differentiable if it is Gâteaux differentiable at every point of X. In this case, the operator \(\nabla\varphi(u): X \to X\) that maps every element \(u\in X\) into the element \(\nabla\varphi(u)\) is called the gradient operator of φ. The convexity of Gâteaux differentiable functions can be characterized as follows.
Lemma 6
Let X be a Hilbert space and \(\varphi:X\to\mathbb{R}\) be Gâteaux differentiable. Then φ is convex if and only if
3 Existence and uniqueness result
We will make the following hypotheses to obtain the existence and uniqueness result for Problem 1:
We will prove the following existence and uniqueness result.
Theorem 7
Assume that (3), (4), (5), (6) are satisfied. Then, for any \(\omega,f\in X\), Problem 1 has a unique solution \((u,\lambda)\in X\times\Lambda\).
Firstly, we give the following equivalence which can be deduced in the proof of Theorem 4 in [6].
Lemma 8
Problem 1 is equivalent to the following problem:
Find \((u,\lambda)\in X\times\Lambda\) such that
The proof of Theorem 7 is carried out in several steps. Let \(\eta\in X\) be given and consider the following auxiliary problem.
Problem 9
Given \(w,f\in X\), find \((u_{\eta},\lambda_{\eta})\in X\times\Lambda \) such that
Define now the operator \(T:X\rightarrow X\) by
where \(u_{\eta}\in X\) denotes the unique solution of Problem 9.
Lemma 10
The operator T has a unique fixed point.
Proof
From Lemma 23 in the Appendix we know that Problem 9 has a unique solution \((u_{\eta},\lambda _{\eta})\in X\times\Lambda\). Let \(\eta_{1},\eta_{2}\in X\) and \((u_{\eta_{1}},\lambda_{\eta _{1}}), (u_{\eta_{2}},\lambda_{\eta_{2}})\in X\times\Lambda\) be the unique solutions of Problem 9 corresponding to \(\eta _{1}\), \(\eta_{2}\), respectively. From (7) we have
From (8) it follows that
Then
and hence
We use the hypothesis (3)(b) to obtain
It is clear that \(m_{A}\leq L_{A}\) and hence \(0<1-\frac{3m_{A}^{2}}{4L_{A}^{4}}<1\). Consequently,
which implies that the operator T is a contraction. By applying the Banach contraction principle, we deduce that there exists a unique \(\eta^{*}\in X\) such that \(\eta^{*}=T\eta^{*}\). This completes the proof of the lemma. □
We now have all the ingredients to provide the proof of the main result in this section.
Proof of Theorem 7
(i) For the existence, let \(\eta^{*}\in X\) be the fixed point of the operator T. We write (7) and (8) for \(\eta=\eta^{*}\) and observe that \(u_{\eta^{*}}=T\eta^{*}=\eta^{*}\). Hence, we conclude that the function \((u_{\eta^{*}},\lambda_{\eta ^{*}})\in X\times\Lambda\) is a solution to Problem 1.
The uniqueness of a solution to Problem 1 is proved directly. Let \((u_{1},\lambda_{1}),(u_{2},\lambda_{2})\in X\times \Lambda\) be two solutions. By Lemma 8 we deduce that
Taking \(v=u_{1}-u_{2}\) in the above inequality we obtain
From (2) it follows that
Combining (4)(b), (5)(c), (11) and (12), we conclude that \(u_{1}=u_{2}\). Moreover, from (10) it follows that
which implies that \(\lambda_{1}=\lambda_{2}\). This completes the proof of the theorem. □
Next, we consider some stability results.
Theorem 11
Assume that (3), (4), (5), (6) are satisfied. If \((u_{1},\lambda_{1}),(u_{2},\lambda_{2})\in X\times \Lambda\) are two solutions of Problem 1 corresponding to \(f_{1},f_{2}\in X\) with \(w=0_{X}\), then there exists \(C=C(L_{A},m_{A},\alpha_{B},L_{\phi})>0\) such that
Proof
From (3)(b) it follows that
and hence
Since
from (5)(b) we have
Combining (14) and (15) we get (13). □
Theorem 12
Assume that (3), (4), (5), (6) are satisfied. If \((u_{1},\lambda_{1}),(u_{2},\lambda_{2})\in X\times \Lambda\) are two solutions of Problem 1 corresponding to \(w_{1}, f_{1}\in X\) and \(w_{2},f_{2}\in X\) (\(w_{i}\neq0_{X}\), \(i=1,2\)), then there exists \(C_{1}=C_{1}(L_{A},m_{A},L_{B},\alpha_{B},L_{\phi})>0\) such that
Proof
From (3)(c) it follows that
and hence
where \(k_{1}\), \(k_{2}\) are strictly positive real constants. Choosing \(k_{1}\), \(k_{2}\) such that
we have
4 Contact problems
In this section, we consider some elastic frictional problems to illustrate our main results.
The elastic body occupies an open bounded connected set \(\Omega\subset \mathbb{R}^{d}\) (\(d=1,2,3\)) with boundary \(\Gamma=\partial\Omega\), assume to be Lipschitz continuous. Assume that Γ consists of three sets \(\overline{\Gamma}_{1}\), \(\overline{\Gamma}_{2}\) and \(\overline{\Gamma}_{3}\), with mutually disjoint relatively open sets \(\Gamma_{1}\), \(\Gamma_{2}\) and \(\Gamma_{3}\), such that \(\operatorname{meas}(\Gamma_{1})>0\).
We use the notation \(\boldsymbol {x}= (x_{i})\) for a generic point in \(\Omega\cup\Gamma\) and \(\boldsymbol {\nu }= (\nu_{i})\) for the outward unit normal at Γ. The indices i, j run between 1 and d and, unless stated otherwise, the summation convention over repeated indices is used. The notation \(\mathbb{S}^{d}\) stands for the space of second order symmetric tensors on \(\mathbb{R}^{d}\). On the spaces \(\mathbb{R}^{d}\) and \(\mathbb{S}^{d}\) we use the inner products and the Euclidean norms defined by
respectively. For a vector field, notation \(u_{\nu}\) and \(\boldsymbol {u}_{\tau }\) represent the normal and tangential components of u on Γ given by \(u_{\nu}=\boldsymbol {u}\cdot \boldsymbol {\nu }\) and \(\boldsymbol {u}_{\tau}=\boldsymbol {u}-u_{\nu} \boldsymbol {\nu }\). Also, \(\sigma_{\nu}\) and \(\boldsymbol {\sigma }_{\tau}\) represent the normal and tangential components of the stress field σ on the boundary, i.e. \(\sigma_{\nu}=(\boldsymbol {\sigma }\boldsymbol {\nu })\cdot \boldsymbol {\nu }\) and \(\boldsymbol {\sigma }_{\tau }=\boldsymbol {\sigma }\boldsymbol {\nu }-\sigma_{\nu} \boldsymbol {\nu }\).
The classical model for the process is as follows.
Problem 13
Find a displacement field \(\boldsymbol {u}\colon\Omega\to\mathbb{R}^{d}\) and a stress field \(\boldsymbol {\sigma }\colon\Omega\to\mathbb{S}^{d}\) such that
Now, we describe (18)–(23) in the following. Equation (18) represents the elastic constitutive law. Equation (19) is the equation of equilibrium, where \(\boldsymbol {f}_{0}\) denotes the density of the body forces, (20) is the displacement homogeneous boundary condition which means that the body is fixed on \(\Gamma_{1}\), and (21) is the traction boundary condition with surface tractions of density \(\boldsymbol {f}_{2}\) acting on \(\Gamma_{2}\). Conditions (22) and (23), given on the contact surface \(\Gamma_{3}\), represent the contact and the friction law, respectively. Here ζ is the friction bound.
In the rest of the paper we use standard notation for Lebesgue and Sobolev spaces and, in addition, we use the spaces V and \(\mathcal{H}\) defined by
Here and below we still denote by v the trace of an element \(\boldsymbol {v}\in H^{1}(\Omega;\mathbb{R}^{d})\). The space \(\mathcal{H}\) will be endowed with the Hilbertian structure given by the inner product
and the associated norm \(\|\cdot\|_{\mathcal{H}}\). On the space V we consider the inner product
and the associated norm \(\|\cdot\|_{V}\). Recall that, since \(\operatorname{meas}(\Gamma_{1}) > 0\), it follows that V is a real Hilbert space. Moreover, by the Sobolev trace theorem, we have
where \(\|\gamma\|\) is the norm of the trace operator \(\gamma :V\rightarrow L^{2}(\Gamma_{3};\mathbb{R}^{d})\).
In the study of Problem 13 we assume that the following conditions hold:
Finally, we assume that the densities of body forces and surface tractions have the regularity
We define an operator \(A:V\rightarrow V\) by
for all \(\boldsymbol {u},\boldsymbol {v}\in V\). Moreover, we define an element \(\boldsymbol {f}\in V^{*}\) by
for all \(\boldsymbol {v}\in V\).
Next, we introduce the space of admissible displacement fields U defined by
Then \((U,(\cdot,\cdot)_{U})\) is a Hilbert space, where
From Proposition 2.1 and Remark 2.2 in [5] we know that \(\gamma (U)\) can be organized as a Hilbert space. Let \(D^{T}\) be the dual space of \(\gamma(U)\). We define \(\boldsymbol {\lambda }\in D^{T}\) by
where \(\langle\cdot,\cdot\rangle_{T}\) denotes the duality pairing between \(D^{T}\) and \(\gamma(U)\). Furthermore, we define \(B:U\times D^{T}\rightarrow\mathbb{R}\) by
Let us introduce the following subset of \(D^{T}\):
Obviously, \(\boldsymbol {\lambda }\in\Lambda\). Moreover, from (23) it follows that
and
Then, combining (18)–(23), from Problem 4.1 in [5], Problem 13 can be written as the following variational formulation:
Problem 14
Find \((\boldsymbol {u},\boldsymbol {\lambda })\in U\times\Lambda\) such that
Now, we give the following existence, uniqueness and dependence results.
Theorem 15
Assume that (24) and (25) are satisfied. Then Problem 14 has a unique solution \((\boldsymbol {u},\boldsymbol {\lambda }) \in U\times\Lambda\). Moreover, if \((\boldsymbol {u}_{1},\boldsymbol {\lambda }_{1}),(\boldsymbol {u}_{2},\boldsymbol {\lambda }_{2})\in U\times\Lambda\) are two solutions of Problem 14 corresponding to \(\boldsymbol {f}^{1},\boldsymbol {f}^{2}\in U\), then there exists \(C_{2}=C_{2}(L_{\mathcal{A}},m_{\mathcal{A}},L_{\phi},\gamma)>0\) such that
Proof
The proof is analogous to Theorem 6.1 in [5]. From (24) and (25), it is easy to verify that (3) and (5) hold. Then, by applying Theorem 7 and Theorem 11, we can obtain the results. □
Problem 16
Find a displacement field \(\boldsymbol {u}\colon\Omega\to\mathbb{R}^{d}\) and a stress field \(\boldsymbol {\sigma }\colon\Omega\to\mathbb{S}^{d}\) such that
Let \(D^{S}\) be the dual space of \(\gamma(V)\). We define \(\boldsymbol {\lambda }\in D^{S}\) by
where \(\langle\cdot,\cdot\rangle_{S}\) denotes the duality pairing between \(D^{S}\) and \(\gamma(V)\). Furthermore, we define \(B:V\times D^{S}\rightarrow\mathbb{R}\) by
Let us introduce the following subset of \(D^{T}\):
Obviously, \(\boldsymbol {\lambda }\in\Lambda\). Moreover, from (31) it follows that
and
Then Problem 16 can be written as the following variational formulation.
Problem 17
Find \((\boldsymbol {u},\boldsymbol {\lambda })\in V\times\Lambda\) such that
We have the following result which is analogous to Theorem 6.2 in [5] and Theorem 15.
Theorem 18
Assume that (24) is satisfied. Then Problem 17 has a unique solution \((\boldsymbol {u},\boldsymbol {\lambda }) \in V\times\Lambda\). Moreover, if \((\boldsymbol {u}_{1},\boldsymbol {\lambda }_{1}),(\boldsymbol {u}_{2},\boldsymbol {\lambda }_{2})\in V\times\Lambda\) are two solutions of Problem 17 corresponding to \(\boldsymbol {f}^{1},\boldsymbol {f}^{2}\in V\), then there exists \(C_{3}=C_{3}(L_{\mathcal{A}},m_{\mathcal{A}},L_{\phi},\gamma)>0\) such that
Problem 19
Find a displacement field \(\boldsymbol {u}\colon\Omega\to\mathbb{R}^{d}\) and a stress field \(\boldsymbol {\sigma }\colon\Omega\to\mathbb{S}^{d}\) such that
Here \(g:\Gamma_{3}\to \mathbb{R}_{+}\) denotes the gap between the deformable body and the foundation, measured along the outward normal τ.
We keep (32) and (33). We assume that there exists \(g_{\mathrm{ext}}:\Omega\to \mathbb{R}\) such that
where \(\gamma':H^{1}(\Omega)\to L^{2}(\Gamma)\) is the well-known Sobolev trace operator. Moreover, we assume that the unit outward normal to \(\Gamma_{3}\), \(\boldsymbol {\nu }_{3}\), is a constant vector.
Then from (35) it follows that
and by (33) we have
Then Problem 19 can be written as the following variational formulation:
Problem 20
Find \((\boldsymbol {u},\boldsymbol {\lambda })\in V\times\Lambda\) such that
We have the following result which is analogous to Theorem 6.3 in [5] and Theorem 15.
Theorem 21
Assume that (24) is satisfied. Then Problem 20 has a unique solution \((\boldsymbol {u},\boldsymbol {\lambda }) \in V\times\Lambda\). Moreover, if \((\boldsymbol {u}_{1},\boldsymbol {\lambda }_{1}),(\boldsymbol {u}_{2},\boldsymbol {\lambda }_{2})\in V\times\Lambda\) are two solutions of Problem 20 corresponding to \(\boldsymbol {f}^{1},\boldsymbol {f}^{2},g_{\mathrm{ext}}\boldsymbol {\nu }_{3},g^{*}_{\mathrm{ext}}\boldsymbol {\nu }_{3}\in V\), then there exists \(C_{4}=C_{4}(L_{\mathcal {A}},m_{\mathcal{A}},L_{\phi},\gamma)>0\) such that
5 Conclusion
In this paper, we study a class of mixed variational problems governed by two variational inequalities with dual Lagrange multipliers. We prove the existence and uniqueness of solution to the mixed variational problem and apply it to obtain the solutions to some frictional contact problems whose weak formulation can be transferred to the mixed variational problem. We point out that considering such kinds of mixed variational problems sets the functional background in the study of many frictional contact problems. To conclude, our results improve many results with bilinear cases and can be further studied.
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Acknowledgements
The work was supported by the National Natural Science Foundation of China Grant No. 11361009 and High level innovation teams and distinguished scholars in Guangxi Universities.
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Appendix
Appendix
We will show the solvability of Problem 9 by using saddle-point theory with the functional \(\mathcal{L}_{\eta}:X\times\Lambda \rightarrow\mathbb{R}\),
where
Lemma 22
Problem 9 has a solution \((u_{\eta},\lambda_{\eta})\in X\times\Lambda\) if and only if \((u_{\eta},\lambda_{\eta})\) is a saddle point of the functional \(\mathcal{L}_{\eta}\).
Proof
For the necessity, let \((u_{\eta},\lambda_{\eta})\in X\times\Lambda \) be a solution of Problem 9. Inequality (8) implies that
Moreover, by combining (7) and (37), we have
Therefore,
For the sufficiency, let \((u_{\eta},\lambda_{\eta})\in X\times \Lambda\) be a saddle point of functional \(\mathcal{L}_{\eta}\). Note that
which implies (8). Next, we will prove (7). Since
we use (37) to see that, for all \(v\in X\),
For \(\forall t\in(0,1)\), \(v'\in X\) taking \(v=u_{\eta}+t(v'-u_{\eta})\) in the above inequality, we obtain
Then, since φ is convex, we obtain
Dividing by t and passing to the limit as \(t\rightarrow0\), we get (7). This completes the proof of the lemma. □
Lemma 23
Problem 9 has a unique solution \((u_{\eta},\lambda _{\eta})\in X\times\Lambda\).
Proof
By (5), it is easy to verify that the map \(v\mapsto\mathcal {L}_{\eta}(v,\mu)\) is convex and l.s.c. for all \(\mu\in\Lambda\), and \(\mu\mapsto\mathcal{L}_{\eta}(v,\mu)\) is concave and u.s.c. for all \(v\in X\). Since ϕ is convex and l.s.c., it admits an affine minorant, see e.g. [18, Proposition 5.2.25], that is, there are \(l\in X^{*}\) and \(b\in\mathbb{R}\) such that
Then we have
Therefore,
Next, we will prove that
From Lemma 4.2 and Corollary 4.6 in [19], for every \(\mu\in \Lambda\) the following inequality:
has a unique solution \(u_{\mu}\in X\). It is easy to verify that
Taking \(v=0_{X}\) in (39), we obtain
Hence,
By Lemma 8 we deduce that inequality (39) is equivalent to the following variational equation:
From (4)(b) we deduce that
and combining with (41) we obtain
It follows from (5) that
Therefore, there exists \(c>0\) such that
Then, combining (40) and (42), we deduce that there exists \(c'>0\) such that
Since \(\mu\in\Lambda\) is arbitrary, passing to the limit as \(\|\mu\| _{Y}\rightarrow\infty\) we get (38). By applying Theorem 4 we deduce that the functional \(\mathcal{L}\) has at least one saddle point, and then we conclude Problem 9 has at least one solution by applying Lemma 22.
Finally, we will show the uniqueness of the solution. In fact, let \((u_{\eta}^{1},\lambda_{\eta}^{1}),(u_{\eta }^{2},\lambda_{\eta}^{2})\in X\times\Lambda\) be two solutions of (41). Then we have
Choosing \(v=u_{\eta}^{2}-u_{\eta}^{1}\) in the above equation, we get
From (8) it follows that
Combining (5)(c), (43) and (44), we conclude that \(u_{\eta}^{1}=u_{\eta}^{2}\). Moreover, we have
By (4)(b), we conclude that \(\lambda_{\eta}^{1}=\lambda _{\eta}^{2}\). □
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Jiang, C. Existence and uniqueness result for a class of mixed variational problems. J Inequal Appl 2018, 95 (2018). https://doi.org/10.1186/s13660-018-1686-y
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DOI: https://doi.org/10.1186/s13660-018-1686-y