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Lagrange multipliers in elastic–plastic torsion problem for nonlinear monotone operators
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ScienceDirect J. Differential Equations 259 (2015) 817–837 www.elsevier.com/locate/jde
Lagrange multipliers in elastic–plastic torsion problem for nonlinear monotone operators S. Giuffrè a,∗ , A. Maugeri b , D. Puglisi b a D.I.I.E.S., Mediterranean University of Reggio Calabria, Loc. Feo di Vito, 89122 Reggio Calabria, Italy b Department of Mathematics and Computer Science, University of Catania, Viale A. Doria, 6, Catania, 95125, Italy
Received 30 June 2014 Available online 3 March 2015
Abstract The existence of Lagrange multipliers as a Radon measure is ensured for an elastic–plastic torsion problem associated to a nonlinear strictly monotone operator. A regularization of this result, namely the existence of Lp Lagrange multipliers, is obtained under strong monotonicity assumption on the operator. Moreover, the relationships between elastic–plastic torsion problem and the obstacle problem are investigated. Finally, an example of the so-called “Von Mises functions” is provided, namely of solutions of the elastic–plastic torsion problem, associated to nonlinear monotone operators, which are not obtained by means of the obstacle problem in the case f = constant. © 2015 Elsevier Inc. All rights reserved. MSC: 35J87; 65K10; 49N15 Keywords: Elastic–plastic torsion; Lagrange multipliers; Variational inequalities; Nonlinear monotone operators; Radon measure; Von Mises functions
1. Introduction
Rn
Let ⊂ Rn be an open bounded convex set with Lipschitz boundary ∂ and let a(p) : Rn → be an operator of class C 2 , strictly monotone, namely
* Corresponding author.
E-mail addresses: [email protected] (S. Giuffrè), [email protected] (A. Maugeri), [email protected] (D. Puglisi). http://dx.doi.org/10.1016/j.jde.2015.02.019 0022-0396/© 2015 Elsevier Inc. All rights reserved.
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(a(P ) − a(Q), P − Q) > 0
∀P , Q ∈ Rn , P = Q.
(1)
We are aimed at the investigation of the existence of Lagrange multipliers associated to the following nonlinear problem
Find u ∈ K :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K,
(2)
where K= v
∈ W01,∞ () :
n ∂v 2 i=1
∂xi
≤ 1 a.e. on .
In particular, we are able to prove that the Lagrange multiplier is always a Radon measure when the operator is strictly monotone, whereas the Lagrange multiplier is a Lp function when the operator is strongly monotone, namely there exists ν > 0, such that (a(P ) − a(Q), P − Q) > ν P − Q 2
∀P , Q ∈ Rn , P = Q.
(3)
Our motivation for studying this type of problems is that these settings appear as the most natural and ultimate ones (existence and regularity of solutions are ensured in [2]). The results of the paper have been achieved, using delicate tools of nonlinear partial differential equations and a new theory of infinite dimensional duality developed by the authors in [5–8,18–20], which has revealed itself effective also in nonlinear case. The classical theory of duality does not work in an infinite dimensional setting, when the interior of the ordering cone of the sign constraints is empty and this new theory overcomes this difficulty. As in the linear case, also in the nonlinear case we are able to find the Von Mises functions, namely functions which clarify how the elastic–plastic solutions work. Problem (2) is strictly related to the elastic–plastic torsion problem. According to Von Mises [28] (see also [21,25]), the elastic–plastic torsion problem of a cylindrical bar with cross section is to find a function ψ(x) which vanishes on the boundary ∂ and, together with its first derivatives, is continuous on ; nowhere on the gradient of ψ must have an absolute value (modulus) less than or equal to a given positive constant τ ; whenever in the strict inequality holds, the function ψ must satisfy the differential equation ψ = −2μθ , where the positive constants μ and θ denote the shearing modulus and the angle of twist per unit length respectively. The elastic–plastic torsion problem and its relationships with obstacle problem have been deeply investigated in years 1965–1980 only for the Laplacian (see W. Ting [25–27] for n = 2 and H. Brezis [3] for the multidimensional case). Later on these studies have been resumed, with particular regards to existence and properties of Lagrange multipliers (see for linear operators [8,13,15,16,18,19], for generalized Lagrange multipliers [4]). Instead the results contained in the paper are concerned with the elastic–plastic torsion problem for nonlinear operators. In detail, when the operator is strictly monotone, we show that there exists μ ∈ (L∞ ())∗ such that
S. Giuffrè et al. / J. Differential Equations 259 (2015) 817–837
⎧ y ≥ 0 ∀y ∈ L∞ (), y ≥ 0 a.e. in ; μ, ⎪
⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ ⎨ μ, − 1 = 0; ∂xi i=1 ⎪ n n ⎪ ⎪ ∂u ∂ϕ ∂ϕ ⎪ ⎪ a (Du) − f ϕ dx = μ, −2
⎪ i ⎩ ∂x ∂x i i ∂xi i=1
819
∀ϕ ∈ W01,∞ ().
i=1
Moreover, when the operator is strongly monotone, we show that there exists μ ∈ Lp () such that ⎧ μ ≥ 0 a.e. in ⎪ ⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ = 0 a.e. in ⎨μ 1 − ∂xi i=1 ⎪ n ⎪ ⎪ ∂ai (Du) ⎪ ⎪ + f = μ a.e. in . ⎪ ⎩ ∂xi i=1
The paper is organized as follows: in Section 2 we present the main results of the paper, in Section 3 we remind some definitions and results that we need in the sequel, in Section 4 we prove our results under strict monotonicity assumption, namely, in a first step, if f is a constant, we prove an equivalence between problem (2) and the obstacle problem: Find u ∈ Kδ :
n i=1
ai (Du)
∂v ∂u − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ Kδ ,
(4)
where Kδ = v ∈ W01,∞ () : |v(x)| ≤ δ(x) = dist(x, ∂) a.e. on (see Theorem 1), and then we obtain the existence of Lagrange multipliers associated to Problem (2), with f ∈ Lp (), p > 1, as a Radon measure (see Theorem 2). In Section 5 we prove a regularization of previous results under strong monotonicity assumption (3) and assuming f to be a positive constant, namely we obtain the existence of a Lagrange multiplier for variational inequality (2) as a Lp function (see Theorem 3) and a relationship between the elastic-region E = {x ∈ : |∇u| < 1} and the region where u does not touch the obstacle δ(x) (see Theorem 4). Finally in Section 6 we provide an example of solution of the elastic–plastic torsion problem associated to a nonlinear monotone operator and in Section 7 we summarize our results and future work. 2. Main results Let ⊂ Rn be an open bounded convex set with Lipschitz boundary ∂ and a be an operator of class C 2 . First we suppose the operator to satisfy strict monotonicity assumption (1). In a first theorem we prove the equivalence between elastic–plastic torsion problem and obstacle problem. It is well known that δ(x) ∈ W01,∞ ().
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Theorem 1. Under the above assumptions on and a, if a(0) = 0 and f ≡ const., the solution u of (2) coincides with the solution of (4). Second we may prove the existence of Lagrange multipliers for problem (2) as a Radon measure. Theorem 2. Under the above assumptions on and a, let f ∈ Lp (), p > 1, and u ∈ K be the solution to (2). Then there exists μ ∈ (L∞ ())∗ such that ⎧ y ≥ 0 ∀y ∈ L∞ (), y ≥ 0 a.e. in ; μ, ⎪
⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ ⎨ μ, − 1 = 0; ∂xi i=1 ⎪ n n ⎪ ⎪ ∂u ∂ϕ ∂ϕ ⎪ ⎪ a (Du) − f ϕ dx = μ, −2
∀ϕ ∈ W01,∞ (). ⎪ i ⎩ ∂xi ∂xi ∂xi i=1
(5)
i=1
Moreover we are able to prove the following regularization theorem concerning Lagrange multipliers, under strong monotonicity assumption. Theorem 3. Under the same assumptions on as above, let a satisfy strong monotonicity assumption (3), with a(0) = 0, let f be a positive constant and u ∈ K ∩ W 2,p () be the solution to problem (2). Then there exists μ ∈ Lp () such that ⎧ μ ≥ 0 a.e. in ⎪ ⎪ n ⎪ ⎪ ∂u 2 ⎪ ⎪ = 0 a.e. in ⎨μ 1 − ∂xi i=1 ⎪ n ⎪ ⎪ ∂ai (Du) ⎪ ⎪ + f = μ a.e. in . ⎪ ⎩ ∂xi
(6)
i=1
Of course, as for the linear problem, it is easy to prove that, if u ∈ K and there exists μ satisfying (6), then u is also the solution to problem (2). Let us notice that, if u is the solution to problem (2), in virtue of Theorem 1, it is the solution of problem (4). In particular, since f ≡ const. > 0, a is monotone and a(0) = 0, following the same arguments as in [22], it is possible to prove that u is the solution of the problem Find u ∈ K1 :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K1 ,
(7)
where K1 = v ∈ W01,∞ () : 0 ≤ v(x) ≤ δ(x) a.e. on . Finally, we are able to prove that the elastic region coincides with the set where u does not touch the obstacle, namely
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Theorem 4. Under the same assumptions on , a and f as in Theorem 3, setting I = {x ∈ : u(x) = δ(x)}, = {x ∈ : u(x) < δ(x)} it results P = {x ∈ : |∇u| = 1} = I, E = {x ∈ : |∇u| < 1} = . Let us remark that problem (4) in [9] can be studied by means of conditions (6) and the techniques used in order to prove Theorem 3. Moreover the techniques used in the paper may be applied to unilateral contact boundary value problem and to weighted traffic equilibrium problem in non pivot Hilbert spaces (see [11,12]). Theorems 2 and 3 are proved using strong duality theory, which is an important tool in order to study Lagrange multipliers. To this end, in this section we recall some classical and recent results of strong duality theory, that will be used in what follows. 3. Duality theory First we recall the following strong duality property in the classical sense and its relationship with the saddle points of the Lagrange functional (see [17,14]). Let S be a nonempty subset of a real linear space X; (Y, · ) be a partially ordered real normed space with ordering cone C, with C ∗ = {λ ∈ Y ∗ : λ, y ≥ 0 ∀y ∈ C} the dual cone of C, Y ∗ topological dual of Y . Theorem 5 (Strong duality property). Let ψ : S → R be a given objective functional; g : S → Y be a given constraint mapping; let the composite mapping (ψ, g) : S → R × Y be convex-like with respect to product cone R+ × C in R × Y . Let the constraint set be given as K := {v ∈ S : g(v) ∈ −C} which is assumed to be nonempty. Let the ordering cone C have a nonempty interior int(C). If the primal problem min
g(v)∈−C v∈S
ψ(v)
(8)
is solvable and the generalized Slater condition is satisfied, namely there is a vector vˆ ∈ S with g(v) ˆ ∈ −int(C), then the dual problem max inf [ψ(v) + μ(g(v))]
μ∈C ∗ v∈S
(9)
is also solvable and the extremal values of the two problems are equal. Moreover, if u is the optimal solution to problem (8) and μ ∈ C ∗ is a solution of the problem (9), it results μ(g(u)) = 0.
(10)
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Theorem 6. Under the same assumptions as above, suppose the ordering cone C to be closed. Then a point (u, μ) ∈ S × C ∗ is a saddle point of the Lagrange functional L(v, μ) = ψ(v) + μ(g(v)), namely L(u, μ) ≤ L(u, μ) ≤ L(v, μ), ∀v ∈ S, ∀μ ∈ C ∗ , if and only if u is a solution of the primal problem (8), μ is a solution of the dual problem (9) and the extremal values of the two problems are equal. In many infinite dimensional settings the above classical duality theory cannot be applied, since the ordering cone has an empty interior. In these cases it is possible to use a very recent duality theory, obtained using new separation theorems based on the notion of quasi-relative interior. To this aim we first present the problem formulation and then we recall the preliminary concepts and the results that we shall use in the sequel. Let X be a real normed space and S be a nonempty convex subset of X; let (Y, · ) be a real normed space partially ordered by a convex cone C, with C ∗ = {λ ∈ Y ∗ : λ, y ≥ 0 ∀y ∈ C} the dual cone of C, Y ∗ topological dual of Y , and let (Z, · Z ) be a real normed space with topological dual Z ∗ . Given a point x ∈ X and a subset M of X, the set TM (x) := h ∈ X : h = lim νn (xn − x), νn > 0, xn ∈ M ∀n ∈ N, lim xn = x n
n
is called the tangent cone to M at x. Let ψ : S → R and g : S → Y be two convex functions and let h : S → Z be an affine-linear function. Let us consider the problem min ψ(x),
(11)
x∈K
where K = {x ∈ S : g(x) ∈ −C, h(x) = θZ } and the dual problem max inf {ψ(x) + λ, g(x) + μ, h(x) }.
λ∈C ∗ x∈S μ∈Z ∗
(12)
λ and μ are the so-called Lagrange multipliers, associated to the sign constraints and to equality constraints, respectively. Moreover, as it is well-known, it always results: max inf {ψ(x) + λ, g(x) + μ, h(x) } ≤ min ψ(x),
λ∈C ∗ x∈S μ∈Z ∗
x∈K
(13)
and, if problem (11) is solvable and in (13) the equality holds, we say that the strong duality between primal problem (11) and dual problem (12) holds. In order to obtain the strong duality, delicate conditions, called “constraints qualification conditions”, have to be satisfied. In the infinite dimensional settings the next assumption, the so-called Assumption S, results to be a necessary and sufficient condition for the strong duality (see [5–7,20]).
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Definition 1 (Assumption S). We shall say that Assumption S is fulfilled at a point x0 ∈ K if it results TM (14) (0, θY , θZ ) ∩ ]−∞, 0[ × {θY } × {θZ } = ∅, where = {(ψ(x) − ψ(x0 ) + α, g(x) + y, h(x)) : x ∈ S \ K, α ≥ 0, y ∈ C}. M The following Theorem holds (see Theorem 1.1 in [7] for the proof): Theorem 7. Under the above assumptions on ψ , g, h and C, if problem (11) is solvable and Assumption S is fulfilled at the extremal solution x0 ∈ K, then also problem (12) is solvable, the extreme values of both problems are equal, namely, if (x0 , λ, μ) ∈ K × C ∗ × Z ∗ is the optimal point of problem (12), ψ(x0 ) = min ψ(x) = max∗ inf {ψ(x) + λ, g(x) + μ, h(x) } x∈K
λ∈C x∈S μ∈Z ∗
= ψ(x0 ) + λ, g(x0 ) + μ, h(x0 )
(15)
and, it results: λ, g(x0 ) = 0. Let us recall that the following one is the so-called Lagrange functional ∀x ∈ S, ∀λ ∈ C ∗ , ∀μ ∈ Z ∗ .
L(x, λ, μ) = ψ(x) + λ, g(x) + μ, h(x) ,
(16)
Using the Lagrange functional, (15) may be rewritten ψ(x0 ) = min ψ(x) = L(x0 , λ, μ) = max∗ inf L(x, λ, μ). λ∈C x∈S μ∈Z ∗
x∈K
By means of Theorem 7, it is possible to show the usual relationship between a saddle point of the Lagrange functional and the solution of the constraint optimization problem (11) (see Theorem 5 in [6] for the proof). Theorem 8. Let us assume that the assumptions of Theorem 7 are satisfied. Then x0 ∈ K is a minimal solution to problem (11) if and only if there exist λ ∈ C ∗ and μ ∈ Z ∗ such that (x0 , λ, μ) is a saddle point of the Lagrange functional (16), namely L(x0 , λ, μ) ≤ L(x0 , λ, μ) ≤ L(x, λ, μ),
∀x ∈ S, λ ∈ C ∗ , μ ∈ Z ∗
and, moreover, it results that λ, g(x0 ) = 0.
(17)
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In [18] the authors introduce the following assumption and prove that it is still a necessary and sufficient condition for the strong duality, also in the case that equality constraints are not affine-linear. Definition 2 (Assumption S ). We shall say that Assumption S is fulfilled at a point x0 ∈ K if it results TM (0, θY , θZ ) ∩ (]−∞, 0[ × {θY } × {θZ }) = ∅,
(18)
where M = {(ψ (x0 )(x − x0 ) + α, g(x0 ) + g (x0 )(x − x0 ) + y, h (x0 )(x − x0 ) : x ∈ S \ K, α ≥ 0, y ∈ C}. Theorem 9. Let ψ : S −→ R be a given convex functional, let g : S −→ Y be a convex mapping with ψ, g having a directional derivative at x0 ∈ K solution to the problem (8) in every direction x − x0 with arbitrary x ∈ S. Suppose h : S −→ Z be linear. Then Assumption S is fulfilled at the minimal point x0 ∈ K if and only if the strong duality holds. Very recently, in [19] a very general tool has been introduced, in which no conditions are required on the map ψ and on the constrains g, h in order to get the strong duality. Let ψ : S −→ R be a functional, g : S −→ Y and h : S −→ Z be any maps. Let us define ϕ : Y × Z −→ R by ϕ(y, z) =
inf
x∈S g(x)∈y−C h(x)=z
ψ(x).
Definition 3 (Assumption NES). We shall say that the Assumption NES is fulfilled for the triple (ψ, g, h) with respect to K if and only if the subdifferential of the convex map ϕ at θY ×Z is non empty, i.e. ∂ϕ(θY ×Z ) = ∅. In [19] the following equivalence was proved. Theorem 10. Strong duality is equivalent to Assumption NES.
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4. Strictly monotone nonlinear operators In order to obtain the equivalence between Problems (2) and (4), let us prove the following proposition. Proposition 11. Under the same assumptions of Theorem 1, the solution u of problem (4) satisfies
|u(x) − u(y)| |u|1 = sup y ≤ |δ(x)|1 = 1. : x, y, ∈ , x = |x − y| Proof. We argue in a similar way as done in [22] in the case a(Du) = u. Let v˜ be the extension by zero of v ∈ W01,∞ () to the whole space Rn . If u is the solution to (4), consider uh (x) = u(x ˜ + h) − u(x) ˜ − |h|
∀x, h ∈ R n
(19)
and u+ ˜ + h) − u(x) ˜ − |h|, 0}. h (x) = max{u(x
(20)
If we define v(x) = max{u(x), ˜ u(x) ˜ + uh (x)} = u(x) ˜ + u+ h (x) and w(x) = min{u(x), ˜ u(x) ˜ − uh (x − h)} = u(x) ˜ − u+ h (x − h), since −δ(x) ≤ w(x) ≤ u(x) ˜ ≤ v(x) ≤ δ(x) a.e. on , it results that v/ and w/ belong to Kδ . Then we may use v/ and w/ as admissible functions in (4), and taking into account that n u+ (x) = u+ h h (x − h) = 0 for x ∈ R \ , we get
Rn i=1
−
n
n
Rn i=1
∂u+ h (x) ai (D u(x)) ˜ dx ≥ ∂xi
f u+ h (x)dx,
Rn
∂u+ h (x − h) ai (D u(x)) ˜ dx ≥ − ∂xi
Rn
f u+ h (x − h)dx.
(21)
(22)
Making the change of variables x → x + h in (22), and since f is constant, we have
n
Rn i=1
ai (D u(x ˜ + h))
∂u+ h (x) dx ≤ ∂xi
Rn
f u+ h (x)dx.
(23)
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Adding (21) and (23), we get
n
Rn i=1
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂u+ h (x) dx ≤ 0. ∂xi
(24)
If we set Eh+ = {x ∈ R n : u+ ˜ + h) − u(x) ˜ − |h| ≥ 0}, it follows h (x) = u(x
n
Eh+ i=1
+
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x n
Rn \Eh+ i=1
∂u+ h (x) dx ∂xi
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂u+ h (x) dx ≤ 0. ∂xi
Bearing in mind (19), (20), it means
n
Eh+ i=1
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂ u(x ˜ + h) ∂ u(x) ˜ − dx ≤ 0. ∂xi ∂xi
(25)
+ In virtue of strict monotonicity assumption (1), inequality (25) implies u+ h = 0 in Eh and then
u(x ˜ + h) − u(x) ˜ ≤ |h|
∀x, h ∈ Rn ,
what we wished to prove. 2 Now, we are able to prove Theorem 1, namely the equivalence between Problems (2) and (4). Proof of Theorem 1. Let v ∈ K. For each x ∈ , let x0 ∈ ∂ such that δ(x) = |x − x0 |. It results |v(x) − v(x0 )| = |v(x)| = |∇v(x)| ˜ |x − x0 | ≤ |x − x0 | = δ(x), namely v ∈ Kδ . Then the inclusion K ⊂ Kδ
(26)
always holds. Now, let us suppose v ∈ Kδ be a solution of (4). Since, by Proposition 11, we have that v ∈ K, hence (26) trivially implies that v is solution of (2). By uniqueness of the solutions of (2) and (4) respectively, one concludes that they coincide. 2 If we content to obtain a Radon measure that plays the role of a Lagrange multiplier, we are able to prove the existence of such a Lagrange multiplier associated to an elastic–plastic torsion problem related to a strictly monotone nonlinear operator of second order. In fact, Theorem 2 means that, if we consider a solution u to variational inequality (2), then conditions (5) are satisfied.
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From conditions (5) it follows that, if u belongs to the elastic region E, μ ≡ 0 and then u is a (Du) solution of the elliptic equation Au = f a.e. in , where A = − ni=1 ∂ai∂x and, in particular, i a solution of (2) solves the elastic–plastic torsion problem. Conversely it is easily proved that, if u ∈ K satisfies conditions (5), then u solves variational inequality (2). Proof of Theorem 2. The proof is based on the following steps. First we rewrite variational inequality (2) as the minimum problem min ψ(v) = ψ(u) = 0
(27)
v∈K
where ψ(v) =
n
i=1
∂v ∂u ai (Du) − ∂xi ∂xi
− f (v − u)dx .
(28)
Then, setting S = X = W01,∞ (), Y = L∞ (), ψ(v) : W01,∞ () → R n ∂v 2 g(v) = − 1 : W01,∞ () → L∞ (), ∂xi i=1
the ordering cone in Y is C = {w ∈ L∞ () : w(t) ≥ 0 a.e. in } and in this case it is possible to prove that intC = ∅ and the generalized Slater condition is verified (see [15,16]). So we are in position to apply Theorems 5 and 6. Consequently, if u is a solution of (2), then of the problem (27), there exists μ ∈ C ∗ solution of the dual problem max inf [ψ(v) + μ, g(v) ]
(29)
μ∈C ∗ v∈S
and (u, μ) is a saddle point of the so called Lagrange functional L(v, μ) = ψ(v) + μ, g(v) , ∀v ∈ W01,∞ (), ∀μ ∈ C ∗ , namely L(u, μ) ≤ L(u, μ) ≤ L(v, μ), ∀v ∈ W01,∞ (), ∀μ ∈ C ∗ . From the right hand side of (30) it follows, in virtue of (10), ∀v ∈ W01,∞ () L(v, μ) =
n
i=1
∂v ∂u ai (Du) − ∂xi ∂xi
≥ L(u, μ) = μ, g(u) = 0.
− f (v − u) dx + μ, g(v)
(30)
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It follows that, for every v ∈ W01,∞ (), 0≤
n
=
i=1
n
=
i=1
n
∂v ∂u ai (Du) − ∂xi ∂xi
i=1
∂v ∂xi
n ∂v ∂v + μ, ∂xi ∂xi i=1
− f (v − u) dx + μ, g(v) − μ, g(u)
n ∂u 2 ∂v 2 − − f (v − u) dx + μ, ∂xi ∂xi i=1 n ∂u ∂u ∂u ∂v − − − f (v − u) dx + μ, ∂xi ∂xi ∂xi ∂xi i=1 ∂u − . ∂xi
∂v ∂u ai (Du) − ∂xi ∂xi ai (Du)
Choosing v = u + ϕ, ∀ϕ ∈ W01,∞ (), we obtain n n ∂u ∂ϕ ∂ϕ ∂ϕ ∂u ∂ϕ ai (Du) − f ϕ dx + μ, + + μ, ∂xi ∂xi ∂xi ∂xi ∂xi ∂xi i=1 i=1 i=1 n n n ∂u ∂ϕ ∂ϕ ∂ϕ 2 = ai (Du) − f ϕ dx + 2 μ, + μ, ≥ 0. ∂xi ∂xi ∂xi ∂xi
n
i=1
i=1
i=1
Choosing v = u − ϕ, ∀ϕ ∈ W01,∞ (), we obtain n n ∂u ∂ϕ ∂ϕ ∂ϕ ∂u ∂ϕ ai (Du) − f ϕ dx + μ, − + μ, ∂xi ∂xi ∂xi ∂xi ∂xi ∂xi i=1 i=1 i=1 n n n ∂u ∂ϕ ∂ϕ ∂ϕ 2 = ai (Du) − f ϕ dx + 2 μ, − μ, ≤ 0. ∂xi ∂xi ∂xi ∂xi
n
i=1
i=1
i=1
Then, considering the test function ε ϕ, ε > 0 in both the inequalities and letting ε tend to zero, we get: n
i=1
n ∂u ∂ϕ ∂ϕ ai (Du) − f ϕ dx = μ, −2
∂xi ∂xi ∂xi
∀ϕ ∈ W01,∞ ().
(31)
i=1
Then from (10), (31), taking into account that C ∗ = {μ ∈ (L∞ ())∗ : μ(y) ≥ 0 ∀y ∈ L∞ (), y(x) ≥ 0 a.a. x ∈ } we obtain conditions (5).
2
Moreover, (see [10,29]), if μ ∈ (L∞ ())∗ , μ can be expressed by a Radon’s integral with respect to the finitely additive measure :
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μ(v) =
v(x)(dx).
is finitely additive, has a bounded total variation and is absolutely continuous with respect to the Lebesgue measure, that is m(B) = 0 implies (B) = 0. From this properties of μ and conditions (5) it is possible to prove that the solution of variational inequality (2) is also a solution of the elastic–plastic torsion problem and vice versa. 5. Regularization of Lagrange multipliers for strongly monotone nonlinear operators In the previous section we proved Theorem 2, namely the existence of a Lagrange multiplier in the dual of L∞ , applying classical duality theory in the case X = S = W01,∞ (). In order to regularize the previous result, namely to prove the existence of Lp Lagrange multipliers, we should apply strong duality theory in the case X = S = H01 (), but in this case the ordering cone C = {w ∈ L1 () : w(t) ≥ 0 a.e. in } has an empty interior, then the classical strong duality theory cannot be applied. It is necessary to use a very recent duality theory, obtained using new separation theorems based on the notion of quasi-relative interior. In this case it is possible to use this theory, since the quasi-relative interior of the ordering cone is nonempty. Then, in this section, under strong monotonicity assumption (3) we prove the existence of Lp Lagrange multipliers. Let us consider variational inequality (2) under assumption (3) and let u ∈ K be the solution to (2). From the regularity results in [2] it follows that, if f ∈ Lp (), 1 < p < ∞, u belongs to W 2,p () ∩ K. In particular, if p > n, Du belongs to C 0,α (). From Theorem 1, it follows that u is a solution to problem (4). Since strong monotonicity holds and u is regular, it also solves the problem Find u ∈ Kδ1 :
n
ai (Du)
i=1
∂v ∂u − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ Kδ1 ,
(32)
where Kδ1 = v ∈ H01 () : |v(x)| ≤ δ(x) = dist(x, ∂) a.e. on . Moreover, since f is positive and a monotone, with a(0) = 0, u is also the solution to Find u ∈ K2 :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K2 ,
(33)
where K2 = v ∈ H01 () : 0 ≤ v(x) ≤ δ(x) a.e. in . Proof of Theorem 3. Let us rewrite problem (33) as an optimization problem. Let us set ψ(v) = (Au − f ) (v − u) dx, v ∈ K2 ,
(34)
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where A=−
n ∂ai (Du)
∂xi
i=1
.
As we already observed, u ∈ K2 ∩ W 2,p () is the solution of (33) and let us remark that min ψ(v) = ψ(u) = 0.
(35)
v∈K2
Now let us show that the optimization problem (35) fulfills Assumption S. To this end, noting that X = Y = L2 (), C = C ∗ = v ∈ L2 () : v(x) ≥ 0 a.e. in , g(v) = v − δ, we must show that, if we have l, θL2 () = lim νn (ψ(vn ) + αn ), lim νn (vn − δ(x) + yn )
n
n
with {νn } a positive real sequence, αn ∈ R+ , vn ∈ L2 () \ K2 , limn ψ(vn ) + αn = 0, limn (vn − δ(x) + yn ) = 0 in L2 (), yn ∈ C, then l must be nonnegative. We have lim νn ψ(vn ) = lim νn (Au − f )(vn − u) dx n
n
n
(Au − f )(vn − δ(x) + yn ) dx +
= lim νn
= lim
(Au − f )(δ(x) − yn − u) dx
(Au − f )νn (vn − δ(x) + yn ) dx + νn
n
(Au − f )(δ(x) − yn − u) dx .
Now, since δ(x) − yn ≤ δ(x), taking into account that lim νn (vn − δ(x) + yn ) = 0 in L2 () n
and (Au − f )(δ(x) − yn − u) dx ≥ 0
we get l ≥ 0. Then the strong duality holds. Now, letting L(v, μ) =
(Au − f )(v − u) dx +
μ(v − δ) dx,
v ∈ L2 (), μ ∈ C,
(36)
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by Theorem 8, there exists μ ∈ C such that μ(u − δ) dx = 0
(37)
and L(u, μ) ≤ L(u, μ) ≤ L(v, μ)
∀v ∈ L2 (), ∀μ ∈ C.
(38)
Considering the right-hand side of (38), we get
(Au − f )(v − u) dx +
μ(v − δ) dx ≥ 0
∀v ∈ L2 ().
(39)
Making use of (37), we can rewrite (39) as (Au − f + μ)(v − u) dx ≥ 0
∀v ∈ L2 (),
(40)
from which we get Au − f + μ = 0 a.e. in ,
(41)
μ(x)(u(x) − δ(x)) = 0 a.e. in .
(42)
and from (37) we also have
In order to obtain conditions (6), it is enough to show Theorem 4. Proof of Theorem 4. Consider the set I = {x ∈ : u(x) = δ(x)}. Since a.e. in we have u(x) ≤ δ(x), each point x ∈ I is a local maximum point, then ∇(u − δ)(x) = 0, and recalling that |∇δ| = 1, also |∇u| = 1, namely x ∈ P . Hence E = \ P ⊂ \ I = , i.e. E ⊂ . Now we show that = {x ∈ : u(x) < δ(x)} ⊂ E = {x ∈ : |∇u(x)| < 1}. From conditions (41) and (42) it follows that, a.e. in , μ = 0 and Au = f with Au = −
n ∂ai (Du) i=1
∂xi
=−
a.e. in ,
(43)
n ∂ai (Du) ∂ 2 u . ∂pj ∂xj ∂xi
i,j =1
We follow the method used in [2, Lemma III.10], taking into account that Au ∈ L∞ ( ) and f is a constant function. As a consequence, u ∈ W 3,p ( ), ∀p, 1 < p < ∞. Differentiating (43) with respect to xk , we obtain 2 n n ∂ ∂ai (Du) ∂ 3u ∂ai (Du) ∂ u + =0 ∂xk ∂pj ∂xj ∂xi ∂pj ∂xj ∂xi ∂xk
i,j =1
i,j =1
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Multiplying by n i,j,k=1
∂u ∂xk
and summing with respect to k, we get
∂ ∂xk
∂ai (Du) ∂pj
n ∂ai (Du) ∂ 3u ∂u ∂ 2 u ∂u + = 0. ∂xj ∂xi ∂xk ∂pj ∂xj ∂xi ∂xk ∂xk
(44)
i,j,k=1
It results ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
n
=
i,j,k=1
+
∂ ∂xi
n i,j,k=1
∂ai (Du) ∂pj
n ∂ai (Du) ∂ 2 u ∂ 2u ∂u ∂ 2 u + ∂xk ∂xk ∂xj ∂pj ∂xk ∂xj ∂xk ∂xi i,j,k=1
∂ai (Du) ∂u ∂ 3u . ∂pj ∂xk ∂xj ∂xi ∂xk
(45)
From (44) it follows 2 n n ∂ai (Du) ∂ 3u ∂u ∂ ∂ai (Du) ∂ u ∂u =− , ∂pj ∂xj ∂xi ∂xk ∂xk ∂xk ∂pj ∂xj ∂xi ∂xk
i,j,k=1
(46)
i,j,k=1
then, substituting in (45), it results ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
=
n ∂ai (Du) ∂ 2 u ∂ 2u ∂pj ∂xk ∂xj ∂xk ∂xi
i,j,k=1
+
∂ ∂xi
∂ai (Du) ∂pj
∂ ∂ 2u − ∂xk ∂xj ∂xk
∂ai (Du) ∂pj
From monotonicity assumption it follows ∂ai (Du) i,j
∂pj
ξi ξj > 0
then ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
∀ξ ∈ Rn \ {0},
∂ 2u ∂xj ∂xi
∂u . ∂xk
(47)
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2 n ∂ ∂ ∂ai (Du) ∂ 2u ∂ai (Du) ∂ u ∂u ≥ − ∂xi ∂pj ∂xk ∂xj ∂xk ∂pj ∂xj ∂xi ∂xk i,j,k=1
=
n i,j,k=1
n n ∂ 2 ai (Du) ∂ 2 u ∂ 2u ∂ 2 ai (Du) ∂ 2 u ∂ 2u ∂u − = 0. (48) ∂pj ∂pl ∂xl ∂xi ∂xk ∂xj ∂pj ∂pl ∂xk ∂xl ∂xj ∂xi ∂xk l=1
l=1
Therefore the operator
−B(ϕ) =
n n 1 ∂ai (Du) ∂ 2 ϕ 1 ∂ ∂ai (Du) ∂ϕ + ≤0 2 ∂pj ∂xi ∂xj 2 ∂xi ∂pj ∂xj i,j =1
a.e. in .
i,j =1
We may apply the maximum principle to operator −B, acting on |∇u|2 − 1 on , since the n ∂ ∂ai (Du) ∂ai (Du) coefficients aij = ∂pj and bj = i=1 ∂xi are bounded. Since is an open set, ∂pj ∂ = {x ∈ : |∇u| = 1} and, by the maximum principle, the value 1 cannot be attained at interior points and one concludes that |∇u(x)| < 1 for x ∈ , that is ⊂ E. 2 Finally, from Theorem 4 and (41), (42) we obtain conditions (6) and the proof of Theorem 3 is completed. 2 Remark 1. Another way to reach the strong duality is to verify Assumption NES. Indeed, in this particular setting, our map ϕ : L2 () −→ R is defined by ϕ(α) =
(Au − f )(v − u)dx.
(49)
ψ(v) = inf ψ(v) = ψ(u) = 0,
(50)
inf
v∈H01 () 0≤v≤δ+α
In particular, in virtue of (34), ϕ(θL2 () ) =
inf
v∈H01 () 0≤v≤δ
v∈K2
then it results ∂ϕ(θL2 () ) = ϕ ∗ ∈ L2 () : ϕ(α) ≥ ϕ ∗ , α ∀α ∈ L2 () .
(51)
For each v ∈ H01 (), 0 ≤ v ≤ δ + α, since (Au − f ) ≤ 0 on the set {x ∈ : u(x) = δ(x)} and (Au − f ) = 0 on the set {x ∈ : 0 < u(x) < δ(x)}, we have
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(Au − f )(v − u)dx =
{x∈: u(x)=δ(x)}
(Au − f )(v − u)dx
+ ≥ ≥
{x∈: 0
{x∈: u(x)=δ(x)}
{x∈: u(x)=δ(x)}
(Au − f )(v − u)dx
(Aδ − f )(v − δ)dx (Aδ − f ) · αdx
Thus, if we set μ = (Aδ − f ) · χ{x∈: u(x)=δ(x)} , μ, α =
(Aδ − f ) · χ{x∈: u(x)=δ(x)} · α dx,
∀α ∈ L2 (),
it follows that ϕ(α) ≥ μ, α
∀α ∈ L2 ().
(52)
From (50), (51) and (52) it follows μ ∈ ∂ϕ(θL2 () ). In any case, the strong duality holds. 6. Example In [8] an example of “Von Mises functions” has been shown, namely of solutions of the elastic–plastic torsion problem associated to a linear operator. The same arguments also fit very well with nonlinear strictly monotone operators, so we may obtain another example, as follows. We consider an operator a(p) : Rn → Rn , of class C 2 , strictly monotone, namely (a(p) − a(q), p − q) > 0
∀p, q ∈ Rn , p = q.
Let us consider ⊆ Rn with boundary ∂ ∈ C 2,1 = W 3,∞ and P = μ , where μ = {x ∈ : δ(x) = d(x, ∂) < μ}. As it is well known μ can be chosen in such a way that for every x ∈ μ there is a unique closest point from ∂ to x and δ(x) owns the same regularity of ∂ on μ . Then δ(x) ∈ W 3,p (P ), ∀p > 1, and its trace δ/∂P ∈ W 3−1/p,p (∂P ). Let f (x) =
n
Di ai (Dδ) − δ(x) a.e. in P
i=1
and w(x) ∈ W 3,p (E), ∀p > 1, the solution of
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⎧ n ⎪ ⎨
Di ai (Dw) = 0 a.e. in E ⎪ i=1 ⎩ w(x) = δ(x) on ∂E. As in Section 5 we can directly prove that, in E, G(Dw) = |Dw|2 − 1 verifies ∂ ∂ai (Dw) ∂ G(Dw) ≥ 0. ∂xi ∂pj ∂xj i,j
Then we may apply maximum principle to G(Dw), from which it follows |Dw| < 1 in E. The function u(x) ∈ W 2,p (), ∀p > 1,
δ(x) w(x)
x∈ P x∈ E
f (x) f˜(x) = 0
x∈ P x ∈ E,
u(x) = arises. Setting
it results n
Di ai (Du) − f˜(x) =
i=1
δ(x) x ∈ P 0 x ∈ E,
namely n i=1
Di ai (Du) − f˜(x) =
n ∂ ∂u χP (x) ∂xi ∂xi i=1
Moreover χP (x)(|∇u|2 − 1) = 0, and by means of maximum principle we can prove |∇u| ≤ 1, that is u is a solution of an elastic–plastic torsion problem.
a.e. in .
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7. Conclusions In this paper we study the existence of Lagrange multipliers associated to the elastic–plastic torsion problem for nonlinear monotone operators. We prove, under strict monotonicity assumption, the existence of a Lagrange multiplier as a Radon measure. Moreover in order to regularize this result, namely to obtain the existence of Lp Lagrange multipliers, we assume a constraint qualification condition (Assumption S), that is a sufficient and necessary condition in order that strong duality holds. Finally the relationships between elastic–plastic torsion problem and obstacle problem are investigated. Further work is to study the existence of Lagrange multipliers for the problem with non-constant gradient constraints (for the existence of solutions of variational and quasivariational inequalities with nonconstant gradient constraints or with curl constraint see [1,23,24]). References [1] A. Azevedo, F. Miranda, L. Santos, Variational and quasivariational inequalities with first order constraints, J. Math. Anal. Appl. 397 (2) (2013) 738–756. [2] H. Brezis, G. Stampacchia, Sur la régularité de la solution d’inéquations elliptiques, Bull. Soc. Math. France 96 (1968) 153–180. [3] H. Brezis, Moltiplicateur de Lagrange en Torsion Elasto-Plastique, Arch. Ration. Mech. Anal. 49 (1972) 32–40. [4] V. Chiadó-Piat, D. Percivale, Generalized Lagrange multipliers in elastoplastic torsion, J. Differential Equations 114 (1994) 570–579. [5] P. Daniele, S. Giuffrè, General infinite dimensional duality and applications to evolutionary network equilibrium problems, Optim. Lett. 1 (2007) 227–243. [6] P. Daniele, S. Giuffrè, G. Idone, A. Maugeri, Infinite dimensional duality and applications, Math. Ann. 339 (2007) 221–239. [7] P. Daniele, S. Giuffrè, A. Maugeri, Remarks on general infinite dimensional duality with cone and equality constraints, Commun. Appl. Anal. 13 (4) (2009) 567–578. [8] P. Daniele, S. Giuffrè, A. Maugeri, F. Raciti, Duality theory and applications to unilateral problems, J. Optim. Theory Appl. 162 (3) (2014) 718–734. [9] L. De Pascale, L. Evans, A. Pratelli, Integral estimates for transport densities, Bull. Lond. Math. Soc. 36 (2004) 383–395. [10] N. Dunford, J.T. Schwartz, Linear Operators, Part I: General Theory, Interscience Publishers, New York, 1958. [11] S. Giuffrè, Strong solvability of boundary value contact problems, Appl. Math. Optim. 51 (3) (2005) 361–372. [12] S. Giuffrè, S. Pia, Weighted traffic equilibrium problem in non pivot Hilbert spaces, Nonlinear Anal. 71 (12) (2009) e2054–e2061. [13] S. Giuffrè, A. Maugeri, New results on infinite dimensional duality in elastic–plastic torsion, Filomat 26 (5) (2012) 1029–1036. [14] S. Giuffrè, Elements of duality theory, in: Q.H. Ansari (Ed.), Topics in Nonlinear Analysis and Optimization, World Education, Delhi, 2012, pp. 251–267. [15] S. Giuffrè, A. Maugeri, Lagrange multipliers in elastic–plastic torsion, AIP Conf. Proc. Rodi 1558 (2013) 1801–1804. [16] S. Giuffrè, A. Maugeri, A measure-type Lagrange multiplier for the elastic–plastic torsion, Nonlinear Anal. 102 (2014) 23–29. [17] J. Jahn, Introduction to the Theory of Nonlinear Optimization, 3rd edition, Springer, Berlin, Heidelberg, New York, 2007. [18] A. Maugeri, D. Puglisi, A new necessary and sufficient condition for the strong duality and the infinite dimensional Lagrange multiplier rule, J. Math. Anal. Appl. 415 (2) (2014) 661–676. [19] A. Maugeri, D. Puglisi, A new necessary and sufficient condition for non-convex strong duality via subdifferential, Numer. Funct. Anal. Optim. 35 (2015) 1095–1112. [20] A. Maugeri, F. Raciti, Remarks on infinite dimensional duality, J. Global Optim. 46 (2010) 581–588. [21] W. Prager, P.G. Hodge, Theory of Perfectly Plastic Solids, Wiley, New York, 1951. [22] J.F. Rodriguez, Obstacle Problems in Mathematical Physics, North-Holland, Amsterdam, 1987.
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[23] J.F. Rodriguez, L. Santos, Quasivariational solutions for first order quasilinear equations with gradient constraint, Arch. Ration. Mech. Anal. 205 (2) (2012) 493–514. [24] L. Santos, Variational problems with non-constant gradient constraints, Port. Math. 59 (2) (2002) 205–248. [25] T.W. Ting, Elastic-plastic torsion of a square bar, Trans. Amer. Math. Soc. 113 (1966) 369–401. [26] T.W. Ting, Elastic-plastic torsion problem II, Arch. Ration. Mech. Anal. 25 (1967) 342–366. [27] T.W. Ting, Elastic-plastic torsion problem III, Arch. Ration. Mech. Anal. 34 (1969) 228–244. [28] R. Von Mises, Three remarks on the theory of the ideal plastic body, in: Reissner Anniversary Volume, Edwards, Ann Arbor, MI, 1949. [29] K. Yosida, Functional Analysis, Springer, Berlin, Heidelberg, New York, 1970.
ScienceDirect J. Differential Equations 259 (2015) 817–837 www.elsevier.com/locate/jde
Lagrange multipliers in elastic–plastic torsion problem for nonlinear monotone operators S. Giuffrè a,∗ , A. Maugeri b , D. Puglisi b a D.I.I.E.S., Mediterranean University of Reggio Calabria, Loc. Feo di Vito, 89122 Reggio Calabria, Italy b Department of Mathematics and Computer Science, University of Catania, Viale A. Doria, 6, Catania, 95125, Italy
Received 30 June 2014 Available online 3 March 2015
Abstract The existence of Lagrange multipliers as a Radon measure is ensured for an elastic–plastic torsion problem associated to a nonlinear strictly monotone operator. A regularization of this result, namely the existence of Lp Lagrange multipliers, is obtained under strong monotonicity assumption on the operator. Moreover, the relationships between elastic–plastic torsion problem and the obstacle problem are investigated. Finally, an example of the so-called “Von Mises functions” is provided, namely of solutions of the elastic–plastic torsion problem, associated to nonlinear monotone operators, which are not obtained by means of the obstacle problem in the case f = constant. © 2015 Elsevier Inc. All rights reserved. MSC: 35J87; 65K10; 49N15 Keywords: Elastic–plastic torsion; Lagrange multipliers; Variational inequalities; Nonlinear monotone operators; Radon measure; Von Mises functions
1. Introduction
Rn
Let ⊂ Rn be an open bounded convex set with Lipschitz boundary ∂ and let a(p) : Rn → be an operator of class C 2 , strictly monotone, namely
* Corresponding author.
E-mail addresses: [email protected] (S. Giuffrè), [email protected] (A. Maugeri), [email protected] (D. Puglisi). http://dx.doi.org/10.1016/j.jde.2015.02.019 0022-0396/© 2015 Elsevier Inc. All rights reserved.
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(a(P ) − a(Q), P − Q) > 0
∀P , Q ∈ Rn , P = Q.
(1)
We are aimed at the investigation of the existence of Lagrange multipliers associated to the following nonlinear problem
Find u ∈ K :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K,
(2)
where K= v
∈ W01,∞ () :
n ∂v 2 i=1
∂xi
≤ 1 a.e. on .
In particular, we are able to prove that the Lagrange multiplier is always a Radon measure when the operator is strictly monotone, whereas the Lagrange multiplier is a Lp function when the operator is strongly monotone, namely there exists ν > 0, such that (a(P ) − a(Q), P − Q) > ν P − Q 2
∀P , Q ∈ Rn , P = Q.
(3)
Our motivation for studying this type of problems is that these settings appear as the most natural and ultimate ones (existence and regularity of solutions are ensured in [2]). The results of the paper have been achieved, using delicate tools of nonlinear partial differential equations and a new theory of infinite dimensional duality developed by the authors in [5–8,18–20], which has revealed itself effective also in nonlinear case. The classical theory of duality does not work in an infinite dimensional setting, when the interior of the ordering cone of the sign constraints is empty and this new theory overcomes this difficulty. As in the linear case, also in the nonlinear case we are able to find the Von Mises functions, namely functions which clarify how the elastic–plastic solutions work. Problem (2) is strictly related to the elastic–plastic torsion problem. According to Von Mises [28] (see also [21,25]), the elastic–plastic torsion problem of a cylindrical bar with cross section is to find a function ψ(x) which vanishes on the boundary ∂ and, together with its first derivatives, is continuous on ; nowhere on the gradient of ψ must have an absolute value (modulus) less than or equal to a given positive constant τ ; whenever in the strict inequality holds, the function ψ must satisfy the differential equation ψ = −2μθ , where the positive constants μ and θ denote the shearing modulus and the angle of twist per unit length respectively. The elastic–plastic torsion problem and its relationships with obstacle problem have been deeply investigated in years 1965–1980 only for the Laplacian (see W. Ting [25–27] for n = 2 and H. Brezis [3] for the multidimensional case). Later on these studies have been resumed, with particular regards to existence and properties of Lagrange multipliers (see for linear operators [8,13,15,16,18,19], for generalized Lagrange multipliers [4]). Instead the results contained in the paper are concerned with the elastic–plastic torsion problem for nonlinear operators. In detail, when the operator is strictly monotone, we show that there exists μ ∈ (L∞ ())∗ such that
S. Giuffrè et al. / J. Differential Equations 259 (2015) 817–837
⎧ y ≥ 0 ∀y ∈ L∞ (), y ≥ 0 a.e. in ; μ, ⎪
⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ ⎨ μ, − 1 = 0; ∂xi i=1 ⎪ n n ⎪ ⎪ ∂u ∂ϕ ∂ϕ ⎪ ⎪ a (Du) − f ϕ dx = μ, −2
⎪ i ⎩ ∂x ∂x i i ∂xi i=1
819
∀ϕ ∈ W01,∞ ().
i=1
Moreover, when the operator is strongly monotone, we show that there exists μ ∈ Lp () such that ⎧ μ ≥ 0 a.e. in ⎪ ⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ = 0 a.e. in ⎨μ 1 − ∂xi i=1 ⎪ n ⎪ ⎪ ∂ai (Du) ⎪ ⎪ + f = μ a.e. in . ⎪ ⎩ ∂xi i=1
The paper is organized as follows: in Section 2 we present the main results of the paper, in Section 3 we remind some definitions and results that we need in the sequel, in Section 4 we prove our results under strict monotonicity assumption, namely, in a first step, if f is a constant, we prove an equivalence between problem (2) and the obstacle problem: Find u ∈ Kδ :
n i=1
ai (Du)
∂v ∂u − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ Kδ ,
(4)
where Kδ = v ∈ W01,∞ () : |v(x)| ≤ δ(x) = dist(x, ∂) a.e. on (see Theorem 1), and then we obtain the existence of Lagrange multipliers associated to Problem (2), with f ∈ Lp (), p > 1, as a Radon measure (see Theorem 2). In Section 5 we prove a regularization of previous results under strong monotonicity assumption (3) and assuming f to be a positive constant, namely we obtain the existence of a Lagrange multiplier for variational inequality (2) as a Lp function (see Theorem 3) and a relationship between the elastic-region E = {x ∈ : |∇u| < 1} and the region where u does not touch the obstacle δ(x) (see Theorem 4). Finally in Section 6 we provide an example of solution of the elastic–plastic torsion problem associated to a nonlinear monotone operator and in Section 7 we summarize our results and future work. 2. Main results Let ⊂ Rn be an open bounded convex set with Lipschitz boundary ∂ and a be an operator of class C 2 . First we suppose the operator to satisfy strict monotonicity assumption (1). In a first theorem we prove the equivalence between elastic–plastic torsion problem and obstacle problem. It is well known that δ(x) ∈ W01,∞ ().
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Theorem 1. Under the above assumptions on and a, if a(0) = 0 and f ≡ const., the solution u of (2) coincides with the solution of (4). Second we may prove the existence of Lagrange multipliers for problem (2) as a Radon measure. Theorem 2. Under the above assumptions on and a, let f ∈ Lp (), p > 1, and u ∈ K be the solution to (2). Then there exists μ ∈ (L∞ ())∗ such that ⎧ y ≥ 0 ∀y ∈ L∞ (), y ≥ 0 a.e. in ; μ, ⎪
⎪ 2 n ⎪ ⎪ ∂u ⎪ ⎪ ⎨ μ, − 1 = 0; ∂xi i=1 ⎪ n n ⎪ ⎪ ∂u ∂ϕ ∂ϕ ⎪ ⎪ a (Du) − f ϕ dx = μ, −2
∀ϕ ∈ W01,∞ (). ⎪ i ⎩ ∂xi ∂xi ∂xi i=1
(5)
i=1
Moreover we are able to prove the following regularization theorem concerning Lagrange multipliers, under strong monotonicity assumption. Theorem 3. Under the same assumptions on as above, let a satisfy strong monotonicity assumption (3), with a(0) = 0, let f be a positive constant and u ∈ K ∩ W 2,p () be the solution to problem (2). Then there exists μ ∈ Lp () such that ⎧ μ ≥ 0 a.e. in ⎪ ⎪ n ⎪ ⎪ ∂u 2 ⎪ ⎪ = 0 a.e. in ⎨μ 1 − ∂xi i=1 ⎪ n ⎪ ⎪ ∂ai (Du) ⎪ ⎪ + f = μ a.e. in . ⎪ ⎩ ∂xi
(6)
i=1
Of course, as for the linear problem, it is easy to prove that, if u ∈ K and there exists μ satisfying (6), then u is also the solution to problem (2). Let us notice that, if u is the solution to problem (2), in virtue of Theorem 1, it is the solution of problem (4). In particular, since f ≡ const. > 0, a is monotone and a(0) = 0, following the same arguments as in [22], it is possible to prove that u is the solution of the problem Find u ∈ K1 :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K1 ,
(7)
where K1 = v ∈ W01,∞ () : 0 ≤ v(x) ≤ δ(x) a.e. on . Finally, we are able to prove that the elastic region coincides with the set where u does not touch the obstacle, namely
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Theorem 4. Under the same assumptions on , a and f as in Theorem 3, setting I = {x ∈ : u(x) = δ(x)}, = {x ∈ : u(x) < δ(x)} it results P = {x ∈ : |∇u| = 1} = I, E = {x ∈ : |∇u| < 1} = . Let us remark that problem (4) in [9] can be studied by means of conditions (6) and the techniques used in order to prove Theorem 3. Moreover the techniques used in the paper may be applied to unilateral contact boundary value problem and to weighted traffic equilibrium problem in non pivot Hilbert spaces (see [11,12]). Theorems 2 and 3 are proved using strong duality theory, which is an important tool in order to study Lagrange multipliers. To this end, in this section we recall some classical and recent results of strong duality theory, that will be used in what follows. 3. Duality theory First we recall the following strong duality property in the classical sense and its relationship with the saddle points of the Lagrange functional (see [17,14]). Let S be a nonempty subset of a real linear space X; (Y, · ) be a partially ordered real normed space with ordering cone C, with C ∗ = {λ ∈ Y ∗ : λ, y ≥ 0 ∀y ∈ C} the dual cone of C, Y ∗ topological dual of Y . Theorem 5 (Strong duality property). Let ψ : S → R be a given objective functional; g : S → Y be a given constraint mapping; let the composite mapping (ψ, g) : S → R × Y be convex-like with respect to product cone R+ × C in R × Y . Let the constraint set be given as K := {v ∈ S : g(v) ∈ −C} which is assumed to be nonempty. Let the ordering cone C have a nonempty interior int(C). If the primal problem min
g(v)∈−C v∈S
ψ(v)
(8)
is solvable and the generalized Slater condition is satisfied, namely there is a vector vˆ ∈ S with g(v) ˆ ∈ −int(C), then the dual problem max inf [ψ(v) + μ(g(v))]
μ∈C ∗ v∈S
(9)
is also solvable and the extremal values of the two problems are equal. Moreover, if u is the optimal solution to problem (8) and μ ∈ C ∗ is a solution of the problem (9), it results μ(g(u)) = 0.
(10)
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Theorem 6. Under the same assumptions as above, suppose the ordering cone C to be closed. Then a point (u, μ) ∈ S × C ∗ is a saddle point of the Lagrange functional L(v, μ) = ψ(v) + μ(g(v)), namely L(u, μ) ≤ L(u, μ) ≤ L(v, μ), ∀v ∈ S, ∀μ ∈ C ∗ , if and only if u is a solution of the primal problem (8), μ is a solution of the dual problem (9) and the extremal values of the two problems are equal. In many infinite dimensional settings the above classical duality theory cannot be applied, since the ordering cone has an empty interior. In these cases it is possible to use a very recent duality theory, obtained using new separation theorems based on the notion of quasi-relative interior. To this aim we first present the problem formulation and then we recall the preliminary concepts and the results that we shall use in the sequel. Let X be a real normed space and S be a nonempty convex subset of X; let (Y, · ) be a real normed space partially ordered by a convex cone C, with C ∗ = {λ ∈ Y ∗ : λ, y ≥ 0 ∀y ∈ C} the dual cone of C, Y ∗ topological dual of Y , and let (Z, · Z ) be a real normed space with topological dual Z ∗ . Given a point x ∈ X and a subset M of X, the set TM (x) := h ∈ X : h = lim νn (xn − x), νn > 0, xn ∈ M ∀n ∈ N, lim xn = x n
n
is called the tangent cone to M at x. Let ψ : S → R and g : S → Y be two convex functions and let h : S → Z be an affine-linear function. Let us consider the problem min ψ(x),
(11)
x∈K
where K = {x ∈ S : g(x) ∈ −C, h(x) = θZ } and the dual problem max inf {ψ(x) + λ, g(x) + μ, h(x) }.
λ∈C ∗ x∈S μ∈Z ∗
(12)
λ and μ are the so-called Lagrange multipliers, associated to the sign constraints and to equality constraints, respectively. Moreover, as it is well-known, it always results: max inf {ψ(x) + λ, g(x) + μ, h(x) } ≤ min ψ(x),
λ∈C ∗ x∈S μ∈Z ∗
x∈K
(13)
and, if problem (11) is solvable and in (13) the equality holds, we say that the strong duality between primal problem (11) and dual problem (12) holds. In order to obtain the strong duality, delicate conditions, called “constraints qualification conditions”, have to be satisfied. In the infinite dimensional settings the next assumption, the so-called Assumption S, results to be a necessary and sufficient condition for the strong duality (see [5–7,20]).
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Definition 1 (Assumption S). We shall say that Assumption S is fulfilled at a point x0 ∈ K if it results TM (14) (0, θY , θZ ) ∩ ]−∞, 0[ × {θY } × {θZ } = ∅, where = {(ψ(x) − ψ(x0 ) + α, g(x) + y, h(x)) : x ∈ S \ K, α ≥ 0, y ∈ C}. M The following Theorem holds (see Theorem 1.1 in [7] for the proof): Theorem 7. Under the above assumptions on ψ , g, h and C, if problem (11) is solvable and Assumption S is fulfilled at the extremal solution x0 ∈ K, then also problem (12) is solvable, the extreme values of both problems are equal, namely, if (x0 , λ, μ) ∈ K × C ∗ × Z ∗ is the optimal point of problem (12), ψ(x0 ) = min ψ(x) = max∗ inf {ψ(x) + λ, g(x) + μ, h(x) } x∈K
λ∈C x∈S μ∈Z ∗
= ψ(x0 ) + λ, g(x0 ) + μ, h(x0 )
(15)
and, it results: λ, g(x0 ) = 0. Let us recall that the following one is the so-called Lagrange functional ∀x ∈ S, ∀λ ∈ C ∗ , ∀μ ∈ Z ∗ .
L(x, λ, μ) = ψ(x) + λ, g(x) + μ, h(x) ,
(16)
Using the Lagrange functional, (15) may be rewritten ψ(x0 ) = min ψ(x) = L(x0 , λ, μ) = max∗ inf L(x, λ, μ). λ∈C x∈S μ∈Z ∗
x∈K
By means of Theorem 7, it is possible to show the usual relationship between a saddle point of the Lagrange functional and the solution of the constraint optimization problem (11) (see Theorem 5 in [6] for the proof). Theorem 8. Let us assume that the assumptions of Theorem 7 are satisfied. Then x0 ∈ K is a minimal solution to problem (11) if and only if there exist λ ∈ C ∗ and μ ∈ Z ∗ such that (x0 , λ, μ) is a saddle point of the Lagrange functional (16), namely L(x0 , λ, μ) ≤ L(x0 , λ, μ) ≤ L(x, λ, μ),
∀x ∈ S, λ ∈ C ∗ , μ ∈ Z ∗
and, moreover, it results that λ, g(x0 ) = 0.
(17)
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In [18] the authors introduce the following assumption and prove that it is still a necessary and sufficient condition for the strong duality, also in the case that equality constraints are not affine-linear. Definition 2 (Assumption S ). We shall say that Assumption S is fulfilled at a point x0 ∈ K if it results TM (0, θY , θZ ) ∩ (]−∞, 0[ × {θY } × {θZ }) = ∅,
(18)
where M = {(ψ (x0 )(x − x0 ) + α, g(x0 ) + g (x0 )(x − x0 ) + y, h (x0 )(x − x0 ) : x ∈ S \ K, α ≥ 0, y ∈ C}. Theorem 9. Let ψ : S −→ R be a given convex functional, let g : S −→ Y be a convex mapping with ψ, g having a directional derivative at x0 ∈ K solution to the problem (8) in every direction x − x0 with arbitrary x ∈ S. Suppose h : S −→ Z be linear. Then Assumption S is fulfilled at the minimal point x0 ∈ K if and only if the strong duality holds. Very recently, in [19] a very general tool has been introduced, in which no conditions are required on the map ψ and on the constrains g, h in order to get the strong duality. Let ψ : S −→ R be a functional, g : S −→ Y and h : S −→ Z be any maps. Let us define ϕ : Y × Z −→ R by ϕ(y, z) =
inf
x∈S g(x)∈y−C h(x)=z
ψ(x).
Definition 3 (Assumption NES). We shall say that the Assumption NES is fulfilled for the triple (ψ, g, h) with respect to K if and only if the subdifferential of the convex map ϕ at θY ×Z is non empty, i.e. ∂ϕ(θY ×Z ) = ∅. In [19] the following equivalence was proved. Theorem 10. Strong duality is equivalent to Assumption NES.
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4. Strictly monotone nonlinear operators In order to obtain the equivalence between Problems (2) and (4), let us prove the following proposition. Proposition 11. Under the same assumptions of Theorem 1, the solution u of problem (4) satisfies
|u(x) − u(y)| |u|1 = sup y ≤ |δ(x)|1 = 1. : x, y, ∈ , x = |x − y| Proof. We argue in a similar way as done in [22] in the case a(Du) = u. Let v˜ be the extension by zero of v ∈ W01,∞ () to the whole space Rn . If u is the solution to (4), consider uh (x) = u(x ˜ + h) − u(x) ˜ − |h|
∀x, h ∈ R n
(19)
and u+ ˜ + h) − u(x) ˜ − |h|, 0}. h (x) = max{u(x
(20)
If we define v(x) = max{u(x), ˜ u(x) ˜ + uh (x)} = u(x) ˜ + u+ h (x) and w(x) = min{u(x), ˜ u(x) ˜ − uh (x − h)} = u(x) ˜ − u+ h (x − h), since −δ(x) ≤ w(x) ≤ u(x) ˜ ≤ v(x) ≤ δ(x) a.e. on , it results that v/ and w/ belong to Kδ . Then we may use v/ and w/ as admissible functions in (4), and taking into account that n u+ (x) = u+ h h (x − h) = 0 for x ∈ R \ , we get
Rn i=1
−
n
n
Rn i=1
∂u+ h (x) ai (D u(x)) ˜ dx ≥ ∂xi
f u+ h (x)dx,
Rn
∂u+ h (x − h) ai (D u(x)) ˜ dx ≥ − ∂xi
Rn
f u+ h (x − h)dx.
(21)
(22)
Making the change of variables x → x + h in (22), and since f is constant, we have
n
Rn i=1
ai (D u(x ˜ + h))
∂u+ h (x) dx ≤ ∂xi
Rn
f u+ h (x)dx.
(23)
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Adding (21) and (23), we get
n
Rn i=1
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂u+ h (x) dx ≤ 0. ∂xi
(24)
If we set Eh+ = {x ∈ R n : u+ ˜ + h) − u(x) ˜ − |h| ≥ 0}, it follows h (x) = u(x
n
Eh+ i=1
+
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x n
Rn \Eh+ i=1
∂u+ h (x) dx ∂xi
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂u+ h (x) dx ≤ 0. ∂xi
Bearing in mind (19), (20), it means
n
Eh+ i=1
˜ + h)) − ai (D u(x))) ˜ (ai (D u(x
∂ u(x ˜ + h) ∂ u(x) ˜ − dx ≤ 0. ∂xi ∂xi
(25)
+ In virtue of strict monotonicity assumption (1), inequality (25) implies u+ h = 0 in Eh and then
u(x ˜ + h) − u(x) ˜ ≤ |h|
∀x, h ∈ Rn ,
what we wished to prove. 2 Now, we are able to prove Theorem 1, namely the equivalence between Problems (2) and (4). Proof of Theorem 1. Let v ∈ K. For each x ∈ , let x0 ∈ ∂ such that δ(x) = |x − x0 |. It results |v(x) − v(x0 )| = |v(x)| = |∇v(x)| ˜ |x − x0 | ≤ |x − x0 | = δ(x), namely v ∈ Kδ . Then the inclusion K ⊂ Kδ
(26)
always holds. Now, let us suppose v ∈ Kδ be a solution of (4). Since, by Proposition 11, we have that v ∈ K, hence (26) trivially implies that v is solution of (2). By uniqueness of the solutions of (2) and (4) respectively, one concludes that they coincide. 2 If we content to obtain a Radon measure that plays the role of a Lagrange multiplier, we are able to prove the existence of such a Lagrange multiplier associated to an elastic–plastic torsion problem related to a strictly monotone nonlinear operator of second order. In fact, Theorem 2 means that, if we consider a solution u to variational inequality (2), then conditions (5) are satisfied.
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From conditions (5) it follows that, if u belongs to the elastic region E, μ ≡ 0 and then u is a (Du) solution of the elliptic equation Au = f a.e. in , where A = − ni=1 ∂ai∂x and, in particular, i a solution of (2) solves the elastic–plastic torsion problem. Conversely it is easily proved that, if u ∈ K satisfies conditions (5), then u solves variational inequality (2). Proof of Theorem 2. The proof is based on the following steps. First we rewrite variational inequality (2) as the minimum problem min ψ(v) = ψ(u) = 0
(27)
v∈K
where ψ(v) =
n
i=1
∂v ∂u ai (Du) − ∂xi ∂xi
− f (v − u)dx .
(28)
Then, setting S = X = W01,∞ (), Y = L∞ (), ψ(v) : W01,∞ () → R n ∂v 2 g(v) = − 1 : W01,∞ () → L∞ (), ∂xi i=1
the ordering cone in Y is C = {w ∈ L∞ () : w(t) ≥ 0 a.e. in } and in this case it is possible to prove that intC = ∅ and the generalized Slater condition is verified (see [15,16]). So we are in position to apply Theorems 5 and 6. Consequently, if u is a solution of (2), then of the problem (27), there exists μ ∈ C ∗ solution of the dual problem max inf [ψ(v) + μ, g(v) ]
(29)
μ∈C ∗ v∈S
and (u, μ) is a saddle point of the so called Lagrange functional L(v, μ) = ψ(v) + μ, g(v) , ∀v ∈ W01,∞ (), ∀μ ∈ C ∗ , namely L(u, μ) ≤ L(u, μ) ≤ L(v, μ), ∀v ∈ W01,∞ (), ∀μ ∈ C ∗ . From the right hand side of (30) it follows, in virtue of (10), ∀v ∈ W01,∞ () L(v, μ) =
n
i=1
∂v ∂u ai (Du) − ∂xi ∂xi
≥ L(u, μ) = μ, g(u) = 0.
− f (v − u) dx + μ, g(v)
(30)
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It follows that, for every v ∈ W01,∞ (), 0≤
n
=
i=1
n
=
i=1
n
∂v ∂u ai (Du) − ∂xi ∂xi
i=1
∂v ∂xi
n ∂v ∂v + μ, ∂xi ∂xi i=1
− f (v − u) dx + μ, g(v) − μ, g(u)
n ∂u 2 ∂v 2 − − f (v − u) dx + μ, ∂xi ∂xi i=1 n ∂u ∂u ∂u ∂v − − − f (v − u) dx + μ, ∂xi ∂xi ∂xi ∂xi i=1 ∂u − . ∂xi
∂v ∂u ai (Du) − ∂xi ∂xi ai (Du)
Choosing v = u + ϕ, ∀ϕ ∈ W01,∞ (), we obtain n n ∂u ∂ϕ ∂ϕ ∂ϕ ∂u ∂ϕ ai (Du) − f ϕ dx + μ, + + μ, ∂xi ∂xi ∂xi ∂xi ∂xi ∂xi i=1 i=1 i=1 n n n ∂u ∂ϕ ∂ϕ ∂ϕ 2 = ai (Du) − f ϕ dx + 2 μ, + μ, ≥ 0. ∂xi ∂xi ∂xi ∂xi
n
i=1
i=1
i=1
Choosing v = u − ϕ, ∀ϕ ∈ W01,∞ (), we obtain n n ∂u ∂ϕ ∂ϕ ∂ϕ ∂u ∂ϕ ai (Du) − f ϕ dx + μ, − + μ, ∂xi ∂xi ∂xi ∂xi ∂xi ∂xi i=1 i=1 i=1 n n n ∂u ∂ϕ ∂ϕ ∂ϕ 2 = ai (Du) − f ϕ dx + 2 μ, − μ, ≤ 0. ∂xi ∂xi ∂xi ∂xi
n
i=1
i=1
i=1
Then, considering the test function ε ϕ, ε > 0 in both the inequalities and letting ε tend to zero, we get: n
i=1
n ∂u ∂ϕ ∂ϕ ai (Du) − f ϕ dx = μ, −2
∂xi ∂xi ∂xi
∀ϕ ∈ W01,∞ ().
(31)
i=1
Then from (10), (31), taking into account that C ∗ = {μ ∈ (L∞ ())∗ : μ(y) ≥ 0 ∀y ∈ L∞ (), y(x) ≥ 0 a.a. x ∈ } we obtain conditions (5).
2
Moreover, (see [10,29]), if μ ∈ (L∞ ())∗ , μ can be expressed by a Radon’s integral with respect to the finitely additive measure :
S. Giuffrè et al. / J. Differential Equations 259 (2015) 817–837
829
μ(v) =
v(x)(dx).
is finitely additive, has a bounded total variation and is absolutely continuous with respect to the Lebesgue measure, that is m(B) = 0 implies (B) = 0. From this properties of μ and conditions (5) it is possible to prove that the solution of variational inequality (2) is also a solution of the elastic–plastic torsion problem and vice versa. 5. Regularization of Lagrange multipliers for strongly monotone nonlinear operators In the previous section we proved Theorem 2, namely the existence of a Lagrange multiplier in the dual of L∞ , applying classical duality theory in the case X = S = W01,∞ (). In order to regularize the previous result, namely to prove the existence of Lp Lagrange multipliers, we should apply strong duality theory in the case X = S = H01 (), but in this case the ordering cone C = {w ∈ L1 () : w(t) ≥ 0 a.e. in } has an empty interior, then the classical strong duality theory cannot be applied. It is necessary to use a very recent duality theory, obtained using new separation theorems based on the notion of quasi-relative interior. In this case it is possible to use this theory, since the quasi-relative interior of the ordering cone is nonempty. Then, in this section, under strong monotonicity assumption (3) we prove the existence of Lp Lagrange multipliers. Let us consider variational inequality (2) under assumption (3) and let u ∈ K be the solution to (2). From the regularity results in [2] it follows that, if f ∈ Lp (), 1 < p < ∞, u belongs to W 2,p () ∩ K. In particular, if p > n, Du belongs to C 0,α (). From Theorem 1, it follows that u is a solution to problem (4). Since strong monotonicity holds and u is regular, it also solves the problem Find u ∈ Kδ1 :
n
ai (Du)
i=1
∂v ∂u − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ Kδ1 ,
(32)
where Kδ1 = v ∈ H01 () : |v(x)| ≤ δ(x) = dist(x, ∂) a.e. on . Moreover, since f is positive and a monotone, with a(0) = 0, u is also the solution to Find u ∈ K2 :
n i=1
∂v ∂u ai (Du) − ∂xi ∂xi
f (v − u)dx,
dx ≥
∀v ∈ K2 ,
(33)
where K2 = v ∈ H01 () : 0 ≤ v(x) ≤ δ(x) a.e. in . Proof of Theorem 3. Let us rewrite problem (33) as an optimization problem. Let us set ψ(v) = (Au − f ) (v − u) dx, v ∈ K2 ,
(34)
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where A=−
n ∂ai (Du)
∂xi
i=1
.
As we already observed, u ∈ K2 ∩ W 2,p () is the solution of (33) and let us remark that min ψ(v) = ψ(u) = 0.
(35)
v∈K2
Now let us show that the optimization problem (35) fulfills Assumption S. To this end, noting that X = Y = L2 (), C = C ∗ = v ∈ L2 () : v(x) ≥ 0 a.e. in , g(v) = v − δ, we must show that, if we have l, θL2 () = lim νn (ψ(vn ) + αn ), lim νn (vn − δ(x) + yn )
n
n
with {νn } a positive real sequence, αn ∈ R+ , vn ∈ L2 () \ K2 , limn ψ(vn ) + αn = 0, limn (vn − δ(x) + yn ) = 0 in L2 (), yn ∈ C, then l must be nonnegative. We have lim νn ψ(vn ) = lim νn (Au − f )(vn − u) dx n
n
n
(Au − f )(vn − δ(x) + yn ) dx +
= lim νn
= lim
(Au − f )(δ(x) − yn − u) dx
(Au − f )νn (vn − δ(x) + yn ) dx + νn
n
(Au − f )(δ(x) − yn − u) dx .
Now, since δ(x) − yn ≤ δ(x), taking into account that lim νn (vn − δ(x) + yn ) = 0 in L2 () n
and (Au − f )(δ(x) − yn − u) dx ≥ 0
we get l ≥ 0. Then the strong duality holds. Now, letting L(v, μ) =
(Au − f )(v − u) dx +
μ(v − δ) dx,
v ∈ L2 (), μ ∈ C,
(36)
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by Theorem 8, there exists μ ∈ C such that μ(u − δ) dx = 0
(37)
and L(u, μ) ≤ L(u, μ) ≤ L(v, μ)
∀v ∈ L2 (), ∀μ ∈ C.
(38)
Considering the right-hand side of (38), we get
(Au − f )(v − u) dx +
μ(v − δ) dx ≥ 0
∀v ∈ L2 ().
(39)
Making use of (37), we can rewrite (39) as (Au − f + μ)(v − u) dx ≥ 0
∀v ∈ L2 (),
(40)
from which we get Au − f + μ = 0 a.e. in ,
(41)
μ(x)(u(x) − δ(x)) = 0 a.e. in .
(42)
and from (37) we also have
In order to obtain conditions (6), it is enough to show Theorem 4. Proof of Theorem 4. Consider the set I = {x ∈ : u(x) = δ(x)}. Since a.e. in we have u(x) ≤ δ(x), each point x ∈ I is a local maximum point, then ∇(u − δ)(x) = 0, and recalling that |∇δ| = 1, also |∇u| = 1, namely x ∈ P . Hence E = \ P ⊂ \ I = , i.e. E ⊂ . Now we show that = {x ∈ : u(x) < δ(x)} ⊂ E = {x ∈ : |∇u(x)| < 1}. From conditions (41) and (42) it follows that, a.e. in , μ = 0 and Au = f with Au = −
n ∂ai (Du) i=1
∂xi
=−
a.e. in ,
(43)
n ∂ai (Du) ∂ 2 u . ∂pj ∂xj ∂xi
i,j =1
We follow the method used in [2, Lemma III.10], taking into account that Au ∈ L∞ ( ) and f is a constant function. As a consequence, u ∈ W 3,p ( ), ∀p, 1 < p < ∞. Differentiating (43) with respect to xk , we obtain 2 n n ∂ ∂ai (Du) ∂ 3u ∂ai (Du) ∂ u + =0 ∂xk ∂pj ∂xj ∂xi ∂pj ∂xj ∂xi ∂xk
i,j =1
i,j =1
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Multiplying by n i,j,k=1
∂u ∂xk
and summing with respect to k, we get
∂ ∂xk
∂ai (Du) ∂pj
n ∂ai (Du) ∂ 3u ∂u ∂ 2 u ∂u + = 0. ∂xj ∂xi ∂xk ∂pj ∂xj ∂xi ∂xk ∂xk
(44)
i,j,k=1
It results ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
n
=
i,j,k=1
+
∂ ∂xi
n i,j,k=1
∂ai (Du) ∂pj
n ∂ai (Du) ∂ 2 u ∂ 2u ∂u ∂ 2 u + ∂xk ∂xk ∂xj ∂pj ∂xk ∂xj ∂xk ∂xi i,j,k=1
∂ai (Du) ∂u ∂ 3u . ∂pj ∂xk ∂xj ∂xi ∂xk
(45)
From (44) it follows 2 n n ∂ai (Du) ∂ 3u ∂u ∂ ∂ai (Du) ∂ u ∂u =− , ∂pj ∂xj ∂xi ∂xk ∂xk ∂xk ∂pj ∂xj ∂xi ∂xk
i,j,k=1
(46)
i,j,k=1
then, substituting in (45), it results ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
=
n ∂ai (Du) ∂ 2 u ∂ 2u ∂pj ∂xk ∂xj ∂xk ∂xi
i,j,k=1
+
∂ ∂xi
∂ai (Du) ∂pj
∂ ∂ 2u − ∂xk ∂xj ∂xk
∂ai (Du) ∂pj
From monotonicity assumption it follows ∂ai (Du) i,j
∂pj
ξi ξj > 0
then ⎤ ⎡ n 1 ∂ ⎣ ∂ai (Du) ∂ |∇u|2 − 1 ⎦ 2 ∂xi ∂pj ∂xj j =1
∀ξ ∈ Rn \ {0},
∂ 2u ∂xj ∂xi
∂u . ∂xk
(47)
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2 n ∂ ∂ ∂ai (Du) ∂ 2u ∂ai (Du) ∂ u ∂u ≥ − ∂xi ∂pj ∂xk ∂xj ∂xk ∂pj ∂xj ∂xi ∂xk i,j,k=1
=
n i,j,k=1
n n ∂ 2 ai (Du) ∂ 2 u ∂ 2u ∂ 2 ai (Du) ∂ 2 u ∂ 2u ∂u − = 0. (48) ∂pj ∂pl ∂xl ∂xi ∂xk ∂xj ∂pj ∂pl ∂xk ∂xl ∂xj ∂xi ∂xk l=1
l=1
Therefore the operator
−B(ϕ) =
n n 1 ∂ai (Du) ∂ 2 ϕ 1 ∂ ∂ai (Du) ∂ϕ + ≤0 2 ∂pj ∂xi ∂xj 2 ∂xi ∂pj ∂xj i,j =1
a.e. in .
i,j =1
We may apply the maximum principle to operator −B, acting on |∇u|2 − 1 on , since the n ∂ ∂ai (Du) ∂ai (Du) coefficients aij = ∂pj and bj = i=1 ∂xi are bounded. Since is an open set, ∂pj ∂ = {x ∈ : |∇u| = 1} and, by the maximum principle, the value 1 cannot be attained at interior points and one concludes that |∇u(x)| < 1 for x ∈ , that is ⊂ E. 2 Finally, from Theorem 4 and (41), (42) we obtain conditions (6) and the proof of Theorem 3 is completed. 2 Remark 1. Another way to reach the strong duality is to verify Assumption NES. Indeed, in this particular setting, our map ϕ : L2 () −→ R is defined by ϕ(α) =
(Au − f )(v − u)dx.
(49)
ψ(v) = inf ψ(v) = ψ(u) = 0,
(50)
inf
v∈H01 () 0≤v≤δ+α
In particular, in virtue of (34), ϕ(θL2 () ) =
inf
v∈H01 () 0≤v≤δ
v∈K2
then it results ∂ϕ(θL2 () ) = ϕ ∗ ∈ L2 () : ϕ(α) ≥ ϕ ∗ , α ∀α ∈ L2 () .
(51)
For each v ∈ H01 (), 0 ≤ v ≤ δ + α, since (Au − f ) ≤ 0 on the set {x ∈ : u(x) = δ(x)} and (Au − f ) = 0 on the set {x ∈ : 0 < u(x) < δ(x)}, we have
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(Au − f )(v − u)dx =
{x∈: u(x)=δ(x)}
(Au − f )(v − u)dx
+ ≥ ≥
{x∈: 0
{x∈: u(x)=δ(x)}
{x∈: u(x)=δ(x)}
(Au − f )(v − u)dx
(Aδ − f )(v − δ)dx (Aδ − f ) · αdx
Thus, if we set μ = (Aδ − f ) · χ{x∈: u(x)=δ(x)} , μ, α =
(Aδ − f ) · χ{x∈: u(x)=δ(x)} · α dx,
∀α ∈ L2 (),
it follows that ϕ(α) ≥ μ, α
∀α ∈ L2 ().
(52)
From (50), (51) and (52) it follows μ ∈ ∂ϕ(θL2 () ). In any case, the strong duality holds. 6. Example In [8] an example of “Von Mises functions” has been shown, namely of solutions of the elastic–plastic torsion problem associated to a linear operator. The same arguments also fit very well with nonlinear strictly monotone operators, so we may obtain another example, as follows. We consider an operator a(p) : Rn → Rn , of class C 2 , strictly monotone, namely (a(p) − a(q), p − q) > 0
∀p, q ∈ Rn , p = q.
Let us consider ⊆ Rn with boundary ∂ ∈ C 2,1 = W 3,∞ and P = μ , where μ = {x ∈ : δ(x) = d(x, ∂) < μ}. As it is well known μ can be chosen in such a way that for every x ∈ μ there is a unique closest point from ∂ to x and δ(x) owns the same regularity of ∂ on μ . Then δ(x) ∈ W 3,p (P ), ∀p > 1, and its trace δ/∂P ∈ W 3−1/p,p (∂P ). Let f (x) =
n
Di ai (Dδ) − δ(x) a.e. in P
i=1
and w(x) ∈ W 3,p (E), ∀p > 1, the solution of
S. Giuffrè et al. / J. Differential Equations 259 (2015) 817–837
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⎧ n ⎪ ⎨
Di ai (Dw) = 0 a.e. in E ⎪ i=1 ⎩ w(x) = δ(x) on ∂E. As in Section 5 we can directly prove that, in E, G(Dw) = |Dw|2 − 1 verifies ∂ ∂ai (Dw) ∂ G(Dw) ≥ 0. ∂xi ∂pj ∂xj i,j
Then we may apply maximum principle to G(Dw), from which it follows |Dw| < 1 in E. The function u(x) ∈ W 2,p (), ∀p > 1,
δ(x) w(x)
x∈ P x∈ E
f (x) f˜(x) = 0
x∈ P x ∈ E,
u(x) = arises. Setting
it results n
Di ai (Du) − f˜(x) =
i=1
δ(x) x ∈ P 0 x ∈ E,
namely n i=1
Di ai (Du) − f˜(x) =
n ∂ ∂u χP (x) ∂xi ∂xi i=1
Moreover χP (x)(|∇u|2 − 1) = 0, and by means of maximum principle we can prove |∇u| ≤ 1, that is u is a solution of an elastic–plastic torsion problem.
a.e. in .
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7. Conclusions In this paper we study the existence of Lagrange multipliers associated to the elastic–plastic torsion problem for nonlinear monotone operators. We prove, under strict monotonicity assumption, the existence of a Lagrange multiplier as a Radon measure. Moreover in order to regularize this result, namely to obtain the existence of Lp Lagrange multipliers, we assume a constraint qualification condition (Assumption S), that is a sufficient and necessary condition in order that strong duality holds. Finally the relationships between elastic–plastic torsion problem and obstacle problem are investigated. Further work is to study the existence of Lagrange multipliers for the problem with non-constant gradient constraints (for the existence of solutions of variational and quasivariational inequalities with nonconstant gradient constraints or with curl constraint see [1,23,24]). References [1] A. Azevedo, F. Miranda, L. Santos, Variational and quasivariational inequalities with first order constraints, J. Math. Anal. Appl. 397 (2) (2013) 738–756. [2] H. Brezis, G. Stampacchia, Sur la régularité de la solution d’inéquations elliptiques, Bull. Soc. Math. France 96 (1968) 153–180. [3] H. Brezis, Moltiplicateur de Lagrange en Torsion Elasto-Plastique, Arch. Ration. Mech. Anal. 49 (1972) 32–40. [4] V. Chiadó-Piat, D. Percivale, Generalized Lagrange multipliers in elastoplastic torsion, J. Differential Equations 114 (1994) 570–579. [5] P. Daniele, S. Giuffrè, General infinite dimensional duality and applications to evolutionary network equilibrium problems, Optim. Lett. 1 (2007) 227–243. [6] P. Daniele, S. Giuffrè, G. Idone, A. Maugeri, Infinite dimensional duality and applications, Math. Ann. 339 (2007) 221–239. [7] P. Daniele, S. Giuffrè, A. Maugeri, Remarks on general infinite dimensional duality with cone and equality constraints, Commun. Appl. Anal. 13 (4) (2009) 567–578. [8] P. Daniele, S. Giuffrè, A. Maugeri, F. Raciti, Duality theory and applications to unilateral problems, J. Optim. Theory Appl. 162 (3) (2014) 718–734. [9] L. De Pascale, L. Evans, A. Pratelli, Integral estimates for transport densities, Bull. Lond. Math. Soc. 36 (2004) 383–395. [10] N. Dunford, J.T. Schwartz, Linear Operators, Part I: General Theory, Interscience Publishers, New York, 1958. [11] S. Giuffrè, Strong solvability of boundary value contact problems, Appl. Math. Optim. 51 (3) (2005) 361–372. [12] S. Giuffrè, S. Pia, Weighted traffic equilibrium problem in non pivot Hilbert spaces, Nonlinear Anal. 71 (12) (2009) e2054–e2061. [13] S. Giuffrè, A. Maugeri, New results on infinite dimensional duality in elastic–plastic torsion, Filomat 26 (5) (2012) 1029–1036. [14] S. Giuffrè, Elements of duality theory, in: Q.H. Ansari (Ed.), Topics in Nonlinear Analysis and Optimization, World Education, Delhi, 2012, pp. 251–267. [15] S. Giuffrè, A. Maugeri, Lagrange multipliers in elastic–plastic torsion, AIP Conf. Proc. Rodi 1558 (2013) 1801–1804. [16] S. Giuffrè, A. Maugeri, A measure-type Lagrange multiplier for the elastic–plastic torsion, Nonlinear Anal. 102 (2014) 23–29. [17] J. Jahn, Introduction to the Theory of Nonlinear Optimization, 3rd edition, Springer, Berlin, Heidelberg, New York, 2007. [18] A. Maugeri, D. Puglisi, A new necessary and sufficient condition for the strong duality and the infinite dimensional Lagrange multiplier rule, J. Math. Anal. Appl. 415 (2) (2014) 661–676. [19] A. Maugeri, D. Puglisi, A new necessary and sufficient condition for non-convex strong duality via subdifferential, Numer. Funct. Anal. Optim. 35 (2015) 1095–1112. [20] A. Maugeri, F. Raciti, Remarks on infinite dimensional duality, J. Global Optim. 46 (2010) 581–588. [21] W. Prager, P.G. Hodge, Theory of Perfectly Plastic Solids, Wiley, New York, 1951. [22] J.F. Rodriguez, Obstacle Problems in Mathematical Physics, North-Holland, Amsterdam, 1987.
S. Giuffrè et al. / J. Differential Equations 259 (2015) 817–837
837
[23] J.F. Rodriguez, L. Santos, Quasivariational solutions for first order quasilinear equations with gradient constraint, Arch. Ration. Mech. Anal. 205 (2) (2012) 493–514. [24] L. Santos, Variational problems with non-constant gradient constraints, Port. Math. 59 (2) (2002) 205–248. [25] T.W. Ting, Elastic-plastic torsion of a square bar, Trans. Amer. Math. Soc. 113 (1966) 369–401. [26] T.W. Ting, Elastic-plastic torsion problem II, Arch. Ration. Mech. Anal. 25 (1967) 342–366. [27] T.W. Ting, Elastic-plastic torsion problem III, Arch. Ration. Mech. Anal. 34 (1969) 228–244. [28] R. Von Mises, Three remarks on the theory of the ideal plastic body, in: Reissner Anniversary Volume, Edwards, Ann Arbor, MI, 1949. [29] K. Yosida, Functional Analysis, Springer, Berlin, Heidelberg, New York, 1970.