A Minimaxmax Problem for Improving the Torsional Stability of Rectangular Plates
aa r X i v : . [ m a t h . A P ] F e b A MINIMAXMAX PROBLEM FOR IMPROVING THETORSIONAL STABILITY OF RECTANGULAR PLATES
ELVISE BERCHIO, DAVIDE BUOSO, FILIPPO GAZZOLA, AND DAVIDE ZUCCO
Abstract.
We use a gap function in order to compare the torsional per-formances of different reinforced plates under the action of external forces.Then, we address a shape optimization problem, whose target is to minimizethe torsional displacements of the plate: this leads us to set up a minimaxmax problem, which includes a new kind of worst-case optimization. Two kinds ofreinforcements are considered: one aims at strengthening the plate, the otheraims at weakening the action of the external forces. For both of them, westudy the existence of optima within suitable classes of external forces andreinforcements. Our results are complemented with numerical experimentsand with a number of open problems and conjectures. Introduction
When pedestrians cross a footbridge or the wind hits a suspension bridge, thedeck undergoes oscillations, which can be of three different kinds. The longitudinaloscillations, in the direction of the bridge, are usually harmless because bridgesare planned to withstand them. The lateral oscillations, which move the deckhorizontally away from its axis, may become dangerous, if the pedestrians walksynchronously; see the recent events at the London Millennium Bridge [1–3] andalso earlier dramatic historical events [4, § § Mathematics Subject Classification.
Key words and phrases.
Shape optimization; worst-case optimization; torsional instability;plates; bridges.Elvise Berchio (corresponding author), Politecnico di Torino, Torino, [email protected] Buoso, Universidade de Lisboa, Lisboa, Portugal. [email protected] Gazzola, Politecnico di Milano Milano, Italy. fi[email protected] Zucco, Universit`a di Torino, Torino, Italy. [email protected]. is continuous because for planar domains the energy space embeds into continu-ous functions. This does not occur in higher space dimensions or for lower orderproblems. The continuity of the solution is a crucial feature since it enables us touse the so-called gap function introduced in [6]. The gap function measures thedifference of the vertical displacements on the two free edges of the plate and istherefore a measure of its torsional response. The number measuring the max-imal gap is given by the maximum over the free edges of the gap function; see(7). Clearly, the maximal gap depends on the force through the Euler-Lagrangeequation satisfied by the solution and one is led to seek the force which yields thelargest torsional displacement. This gives a measure of the risk that the bridgecollapses. In order to lower this risk, one may try different ways of reinforcing thedeck.Imagine that one has a certain amount of stiff material (e.g., steel) and hasto decide where to place it within the plate in order to lower the maximal gapand, in turn, the torsional displacements. This material should occupy a properopen subset of the plate. In literature this kind of problem has been tackledin several ways; we refer to [7–9] for related problems on the torsion of a bar.Since this shape optimization problem is completely new, we choose two differentstrategies: we first assume that the stiff material reinforces a part of the plate,then we assume that it acts directly on the force and weakens it by a factorinvolving the characteristic function of the region occupied by the material and aconstant measuring the strength of the stiff material. Reinforcing the plate meansthat we add the stiff material in critical parts of the plate in order to increasethe energy necessary to bend it. Weakening the force means that we place some“aerodynamic damper” in order to reduce the action of the external force. Thesekinds of minimization problems naturally lead to homogenization [10], see also [9]for a stiffening problem for the torsion of a bar. Homogenization would lead tooptimal designs with reinforcements scattered throughout the structure, namelydesigns impossible to implement for engineers. And since the design of the stiffstructure should be usable for engineers, homogenization must be avoided and theclass of admissible geometries for the reinforcements should be sufficiently small.In this respect, we mention the paper by Nazarov-Sweers-Slutskij [11], where only“macro” reinforcements are considered, although in a fairly different setting. Thestructural optimization problem that we tackle may be seen as the “dual problem”of the one considered in the seminal work by Michell [12], see also updated resultsin [13, Chapter 4]: our purpose is to determine the best performance of the stiffmaterial by maintaining the cost whereas Michell aimed to determine the cheapeststiff material by maintaining the performance.For both the two mentioned ways of introducing the reinforcement, our purposeis to optimize the maximal gap. We will introduce suitable classes, for both theforce and the reinforcement, in which to set up the optimization problem. First weseek the “worst” forces for a given reinforcement. This number yields the maximalgap that may occur. Then, we seek the “best” reinforcements, which minimize theeffect of the forces. We are then led to solve a minimaxmax problem . The existenceof a maximal force and of a minimal reinforcement depends on how wide the classesare. In this paper, for the forces we mainly deal with the classes of Lebesgue
MINIMAXMAX PROBLEM 3 functions or of the dual of the energy space, while, concerning the reinforcements,we restrict our attention to simple designs, that may be appropriate for engineeringapplications: cross-type reinforcements, tiles of rectangular shapes, networks ofbounded length, and general Lipschitz domains, see Definition 3.1.This minimaxmax problem can also be seen as a worst-case optimization prob-lem , since one is interested in minimizing the worst value of a functional amongall possible designs. An extended presentation of worst-case optimization prob-lems in structural mechanics can be found in [14]; see also [15] for a worst-caseoptimization problem of a compliance functional in the Lebesgue space.This paper is organized as follows. In Section 2, we introduce rigorously thegap function with the minimaxmax problem. In Section 3, we identify suitableclasses for which the minimaxmax problem admits a solution (i.e., worst forceswith best reinforcements). In Section 4, we discuss symmetry properties of worstforces in the case of symmetric reinforcements. In Section 5, we investigate theworst force acting on a plate with no reinforcement. In Section 6, we analyzethe effects of cross-type reinforcements, while, in Section 7, we consider moregeneral polygonal-type reinforcements. In both cases we solve numerically theminimaxmax problem. Sections 8 to 11 are dedicated to the proofs of our results.Finally, Section 12 contains the conclusions on the work done.2.
Variational Setting and Reinforcements for the Plate
Up to scaling, in the following we may assume that the plate Ω has length π and width 2 ℓ with 2 ℓ ≪ π so that Ω =]0 , π [ × ] − ℓ, ℓ [ ⊂ R . According to theKirchhoff-Love theory [16,17] (see also [18] for a modern presentation), the energy E of a vertical deformation u of the plate Ω subject to a load f may be computedthrough the functional(1) E ( u ) := Z Ω (cid:18) (∆ u ) − σ )( u xy − u xx u yy ) − f u (cid:19) dxdy , where σ is the Poisson ratio and satisfies 0 < σ <
1. This implies that thequadratic part of the energy E is positive. For the partially hinged plate underconsideration, the functional E should be minimized in the space H ∗ (Ω) := n v ∈ H (Ω) : v = 0 on { , π }× ] − ℓ, ℓ [ o ;since Ω is a planar domain, one has the embedding H (Ω) ⊂ C (Ω), and thecondition on { , π }× ] − ℓ, ℓ [ is satisfied pointwise. By [5, Lemma 4.1] we knowthat H ∗ (Ω) is a Hilbert space when endowed with the scalar product( u, v ) H ∗ := Z Ω [∆ u ∆ v + (1 − σ )(2 u xy v xy − u xx v yy − u yy v xx )] dxdy and associated norm k u k H ∗ = ( u, u ) H ∗ , which is equivalent to the usual norm in H (Ω), that is, k u k H = k u k L + k D u k L . We also define H − ∗ (Ω) as the dualspace of H ∗ (Ω) and we denote by h· , ·i the corresponding duality. If f ∈ L (Ω)then the functional E is well-defined in H ∗ (Ω), while if f ∈ H − ∗ (Ω) we need toreplace R Ω f u with h f, u i . ELVISE BERCHIO, DAVIDE BUOSO, FILIPPO GAZZOLA, AND DAVIDE ZUCCO
Assume that the plate Ω is reinforced with a stiff material which occupies anopen region D ⊂ Ω and that D belongs to a certain class D , while f belongsto some space F of admissible forcing terms. We consider two possible ways ofreinforcing the plate: either we stiffen the plate by increasing the cost of thebending energy, or we add an aerodynamic damper by weakening the force. Thismodifies the original energy (1) into the two following ways:(2) E ( u ) := Z Ω (cid:20) (1 + dχ D ) (cid:18) (∆ u ) − σ )( u xy − u xx u yy ) (cid:19) − f u (cid:21) dxdy and(3) E ( u ) := Z Ω (cid:20) (∆ u ) − σ )( u xy − u xx u yy ) − f u dχ D (cid:21) dxdy , where χ D is the characteristic function of D and d > H ∗ (Ω).When dealing with E , for any D ⊂ Ω open, we introduce the bilinear form(4) ( u, v ) D := Z D [∆ u ∆ v + (1 − σ )(2 u xy v xy − u xx v yy − u yy v xx )] dxdy so that ( u, v ) Ω = ( u, v ) H ∗ . Then, for all f ∈ H − ∗ (Ω) the minimizer of E satisfiesthe weak Euler-Lagrange equation(5) ( u f,D , v ) H ∗ + d ( u f,D , v ) D = h f, v i ∀ v ∈ H ∗ (Ω) , which has no strong counterpart due to the lack of regularity of the term (1 + dχ D )that prevents an integration by parts.On the other hand, due to the lack of regularity of the term (1 + dχ D ), E is not defined for all f ∈ H − ∗ (Ω), but it is well-defined for any f ∈ L (Ω); inthis case the minimizer satisfies the equation ( u f,D , v ) H ∗ = R Ω fv dχ D dxdy , for all v ∈ H ∗ (Ω), which may also be written in its strong form:(6) (1 + dχ D )∆ u = f , in Ω ,u = u xx = 0 , on { , π }× ] − ℓ, ℓ [ ,u yy + σu xx = u yyy + (2 − σ ) u xxy = 0 , on ]0 , π [ ×{− ℓ, ℓ } . Since 0 < σ <
1, both E and E admit a unique critical point in H ∗ (Ω), theirabsolute minimum. The minimizer may be different for E and E but we willdenote both of them by u f,D since it will always be clear which functional weare dealing with. As we have just seen, the solution u f,D satisfies a weak Euler-Lagrange equation for E (but not a strong one) while it satisfies a strong Euler-Lagrange equation for E (and not a merely weak one).Assume that some classes F and D of admissible f and D are given. Take f ∈ F , D ∈ D , and the minimizer u f,D ∈ H ∗ (Ω) ⊂ C (Ω) of E or E , thencompute its gap function with its maximal gap :(7) G f,D ( x ) := u f,D ( x, ℓ ) − u f,D ( x, − ℓ ) , G ∞ f,D := max x ∈ [0 ,π ] |G f,D ( x ) | . MINIMAXMAX PROBLEM 5
In this way we have defined the map G ∞ f,D : F × D → [0 , ∞ [ with ( f, D )
7→ G ∞ f,D . Given D ∈ D , we first seek the worst f ∈ F such that(8) G ∞ D := max f ∈F G ∞ f,D = max f ∈F max x ∈ [0 ,π ] |G f,D ( x ) | , and then the best D ∈ D such that(9) G ∞ := min D ∈D G ∞ D = min D ∈D max f ∈F max x ∈ [0 ,π ] |G f,D ( x ) | . This is our minimaxmax problem . In the next sections we analyze some classes F and D , where (8) and (9) admit a solution. Note that G ∞ = G ∞ ( F , D ) is monotonewith respect to both the classes F and D but with opposite monotonicity.3. Existence Results for the Minimaxmax Problem
We determine some classes F and D of admissible forces and reinforcements forwhich (8) and (9) admit a solution. The proofs are given in Section 8. We firstshow that G ∞ D , as in (8), is well-defined for some choices of the class F . Theorem 3.1.
For a given open set D ⊂ Ω and p ∈ ]1 , + ∞ ] , the maximizationproblems (10) max (cid:8) G ∞ f,D : f ∈ H − ∗ (Ω) with k f k H − ∗ = 1 (cid:9) , ( for E ) , (11) max (cid:8) G ∞ f,D : f ∈ L p (Ω) with k f k L p = 1 (cid:9) , ( for both E and E ) , admit a solution (in the considered space). Then, we turn to problem (9). We introduce some classes D for which it isguaranteed the existence of a solution. Definition 3.1 (Classes of admissible reinforcements) . (a) Cross-type reinforce-ments : for
N, M ∈ N , µ ∈ ]0 , π/ N [ , ε ∈ ]0 , ℓ/M [ , x i ∈ [ µ, π − µ ] for i = 1 , . . . , N with x i +1 − x i > µ for i N − , and y j ∈ [ − ℓ + ε, ℓ − ε ] for j = 1 , . . . , M with y j +1 − y j > ε for j M − , define C := n D ⊂ Ω : D = (cid:16) N [ i =1 ] x i − µ, x i + µ [ × ] − ℓ, ℓ [ (cid:17) ∪ (cid:16) M [ j =1 ]0 , π [ × ] y j − ε, y j + ε [ (cid:17)o . (b) Tiles of rectangular shapes : for N ∈ N and ε ∈ ]0 , ℓ [ , define T := n D ⊂ Ω : D = N [ i =1 R i , R i ⊂ Ω is an open rectangle with inradius > ε o . (c) Networks of bounded length : for ε ∈ ]0 , ℓ [ and L > , define N := (cid:8) D ⊂ Ω : D = Σ ε where Σ ⊂ Ω is closed, connected, H (Σ) L (cid:9) . Here H denotes the one-dimensional Hausdorff measure of a set, and Σ ε repre-sents the ε -tubular neighborhood of Σ , namely the set of points in Ω at distance to Σ less than ε .(d) Lipschitz trusses : for ε ∈ ]0 , ℓ [ , define L := (cid:8) D ⊂ Ω : D open with the inner ε -cone property (cid:9) . ELVISE BERCHIO, DAVIDE BUOSO, FILIPPO GAZZOLA, AND DAVIDE ZUCCO
We recall that by the inner ε -cone property we mean that at every point x of theboundary ∂D there is some truncated cone from x with an opening angle ε andradius ε inside D . Notice that some of these classes are monotone with respect to set inclusion,namely
C ⊂ T ⊂ L , for suitable choices of the parameters
N, ε, µ, L . Theorem 3.2.
For a given κ ∈ ]0 , πℓ [ the minimization problem (12) min {G ∞ D : D ∈ D with | D | = κ } , admits a solution whenever the class D is one of those introduced in Definition 3.1(with the parameters chosen so as to satisfy the area constraint). Symmetric Framework for the Minimaxmax Problem
Whenever the class D of the minimaxmax problem (9) reduces to symmetricreinforcements, the class F can be reduced without changing the problem. Wesay that a set D ⊂ Ω is symmetric with respect to the midline (or, for short,symmetric) if ( x, y ) ∈ D if and only if ( x, − y ) ∈ D , for all ( x, y ) ∈ Ω. Then, weintroduce the subspaces of even and odd functions with respect to y : H E (Ω) := { u ∈ H ∗ (Ω) : u ( x, − y ) = u ( x, y ) ∀ ( x, y ) ∈ Ω } ,H O (Ω) := { u ∈ H ∗ (Ω) : u ( x, − y ) = − u ( x, y ) ∀ ( x, y ) ∈ Ω } . We first notice that(13) H E (Ω) ⊥ H O (Ω) , H ∗ (Ω) = H E (Ω) ⊕ H O (Ω) . For all u ∈ H ∗ (Ω) we denote by u e ∈ H E (Ω) and u o ∈ H O (Ω) its componentsaccording to this decomposition, namely u e ( x, y ) = u ( x,y )+ u ( x, − y )2 and u o ( x, y ) = u ( x,y ) − u ( x, − y )2 . The orthogonal projections P E : H ∗ (Ω) → H E (Ω) and P O : H ∗ (Ω) → H O (Ω) are defined onto these subspaces as P E u := u e and P O u := u o , for every u ∈ H ∗ (Ω). Then, we define: H − E (Ω) := { f ∈ H − ∗ (Ω) : h f, v i = 0 ∀ v ∈ H O (Ω) } ,H − O (Ω) := { f ∈ H − ∗ (Ω) : h f, v i = 0 ∀ v ∈ H E (Ω) } . In particular, H O (Ω) ⊆ ker f for every f ∈ H − E (Ω) and H E (Ω) ⊆ ker f forevery f ∈ H − O (Ω). Moreover, H − ∗ (Ω) = H − E (Ω) ⊕ H − O (Ω), that is for every f ∈ H − ∗ (Ω) there exists a unique couple ( f e , f o ) ∈ H − E (Ω) × H − O (Ω) such that f = f e + f o ; with f e := f ◦ P E and f o := f ◦ P O . As usual, we endow H − ∗ (Ω)with the norm k f k H − ∗ := sup k v k H ∗ =1 h f, v i , and we observe that(14) k f k H − ∗ = max (cid:8) k f e k H − ∗ , k f o k H − ∗ (cid:9) ∀ f ∈ H − ∗ (Ω) . The next result shows that if the reinforcement D is symmetric with respect tothe midline then the worst forces f , whose existence is ensured by Theorem 3.1,can be sought in the class of odd distributions. MINIMAXMAX PROBLEM 7
Theorem 4.1.
Assume that D ⊂ Ω is open and symmetric with respect to the x -axis. Then, (10) is equivalent to max {G ∞ f,D : f ∈ H − O (Ω) , k f k H − ∗ = 1 } . More-over, if f ∈ H − O (Ω) is such that k f k H − ∗ = 1 and G ∞ f,D = max n G ∞ g,D : g ∈ H − ∗ (Ω) with k g k H − ∗ = 1 o , then there exist infinitely many g ∈ H − ∗ (Ω) such that P E g = 0 , k g k H − ∗ = 1 , and G ∞ g,D = G ∞ f,D . Theorem 4.1 states that, for a symmetric reinforcement D (possibly D = ∅ as forthe free plate), the maximization of the gap function can be restricted to the classof odd distributions. But Theorem 4.1 does not state that only odd f attain themaximum. And indeed, G ∞ f,D is not sensitive to the addition of some φ ∈ H − E (Ω)to f , provided that the total norm is not exceeded. An interesting open problemis to determine whether there exists a unique f ∈ H − O (Ω) maximizing G ∞ f,D (up toa sign change). We expect the answer to depend on D , in particular on possibleadditional symmetry properties of D . We prove Theorem 4.1 in Section 9.Next, we have the following L p -version of Theorem 4.1. Theorem 4.2.
Assume that D ⊂ Ω is open and symmetric with respect to the x -axis and let p ∈ ]1 , ∞ ] . Then, problem (11) is equivalent to the maximizationproblem max n G ∞ f,D : f ∈ L p (Ω) , f odd in y , k f k L p = 1 o .If < p < ∞ , then any maximizer is necessarily odd with respect to y .If p = ∞ and an odd maximizer f satisfies | f ( x, y ) | < on a subset of Ω ofpositive measure, then there exist infinitely many maximizers g ∈ L ∞ (Ω) such that g e = 0 and k g k L ∞ = 1 . Theorem 4.2 states that, for a symmetric reinforcement D , the maximization ofthe gap function can be restricted to the class of odd functions. Moreover, differ-ently from Theorem 4.1, if 1 < p < ∞ it says that only odd functions f attain themaximum. On the other hand, in the case p = ∞ oddness may fail, provided thatthere exists an odd maximizer satisfying the somewhat strange property statedin Theorem 4.2: the reason of this assumption will become clear in the proof ofTheorem 4.2 given in Section 9.5. Worst Cases on the Free Plate
In this section, we consider the free plate with no reinforcement ( D = ∅ ) so that E and E coincide, and we study problem (10). For simplicity, for all f ∈ H − ∗ (Ω),we set G f ( x ) = G f, ∅ ( x ) and G ∞ f = G ∞ f, ∅ .Following the suggestion of Theorem 4.1, for any z ∈ ]0 , π [ we focus on the odd distribution(15) T z := δ ( z,ℓ ) − δ ( z, − ℓ ) ∈ H − O (Ω) , where δ P is the Dirac delta with mass concentrated at P ∈ Ω. Let u z ∈ H ∗ (Ω)be the unique solution of the equation ( u z , v ) H ∗ = h T z , v i , for all v ∈ H ∗ (Ω). Bythe Riesz Theorem, this means that u z is the representative of T z and therefore(by taking v = u z ) k T z k H − ∗ = k u z k H ∗ = h T z , u z i = G T z ( z ) /
2. This enables us to
ELVISE BERCHIO, DAVIDE BUOSO, FILIPPO GAZZOLA, AND DAVIDE ZUCCO normalize T z and introduce the distribution T z := √ T z √ G Tz ( z ) such that k T z k H − ∗ = 1.For any integer m , set(16) Υ m := sinh ( mℓ ) m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] . In Section 11 we prove the following result.
Proposition 5.1.
For all x, z ∈ ]0 , π [ we have G T z ( x ) = 4 π (1 − σ ) ∞ X m =1 Υ m sin( mz ) sin( mx ) , G T z ( x ) = √ G T z ( x ) p G T z ( z ) . Let us explain how Proposition 5.1 suggests a conjecture for the solution of(10) when D = ∅ . Let Υ m be as in (16) and consider the function Φ definedas Φ( x ) := P ∞ m =1 Υ m sin ( mx ), for every x ∈ [0 , π ]. Note that Φ( x ) > x ∈ ]0 , π [ and(17) Φ(0) = Φ( π ) = 0 , Φ( π ) = ∞ X k =0 Υ k +1 , Φ ′ ( π ) = 0 , Φ ′′ ( π ) < . Some numerical computations and (17) suggest that Φ achieves its maximum at x = π/ π ) > Φ( x ) ∀ x = π . Moreover, by H¨older’s inequality, Proposition 5.1, and condition (18), for every x, z ∈ ]0 , π [ (cid:12)(cid:12) G T z ( x ) (cid:12)(cid:12) = 4 √ π (1 − σ ) p G T z ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X m =1 Υ m sin( mz ) sin( mx ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) √ π (1 − σ ) p G T z ( z ) ∞ X m =1 p Υ m | sin( mz ) | p Υ m | sin( mx ) | √ π (1 − σ ) p G T z ( z ) (cid:18) ∞ X m =1 Υ m sin ( mz ) (cid:19) · (cid:18) ∞ X m =1 Υ m sin ( mx ) (cid:19) = 2 √ p π (1 − σ ) Φ( x ) √ p π (1 − σ ) Φ( π ) . Note that the above application of the H¨older inequality yields a strict inequalitywhenever z = x . Therefore, after taking the maximum over [0 , π ] we deduce that G ∞ T z < √ √ π (1 − σ ) Φ( π ) / for every z = π and that for z = π the equality holds G ∞ T π/ = G T π/ ( π ) = √ √ π (1 − σ ) Φ( π ) / . Hence, if (18) holds, then we would inferthat for all z ∈ ]0 , π [ we have G ∞ T z G ∞ T π/ with equality if and only if z = π/ π/ , ± ℓ ). A numerical support of this MINIMAXMAX PROBLEM 9 fact is provided by Table 1 below. The values collected there have been obtainedusing the software Mathematica, approximating the Fourier series for G ∞ T z up tothe 10,000-th term. Table 1.
Numerical values of 10 × G ∞ T z and 10 × G ∞ T z (with ℓ = π/
150 and σ = 0 . z π π π π π π π π π π × G ∞ T z .
809 659 .
067 695 .
691 739 .
38 792 .
677 859 .
592 946 .
815 1066 .
21 1238 .
29 1429 . × G ∞ T z .
326 21 .
354 23 .
854 27 .
012 31 .
123 36 .
686 44 .
609 56 .
687 76 .
596 102 . It is evident that the worst case is attained for z = π and that the map z
7→ G ∞ T z is increasing on [0 , π/
2] (note that it is symmetric with respect to π/ G ∞ T z .6. Weakening the Force with Cross-Type Reinforcements
In this section, we minimize the energy E given in (3) finding the explicitsolution and, in turn, the explicit gap function for particular choices of forces f and reinforcements D . We take symmetric cross-type reinforcements D ∈ C (seeDefinition 3.1) with one horizontal arm and 2 N + 1 vertical arms for some non-negative integer N . More precisely, fix 0 < µ < (2 N +1) π N +1) , 0 < ε < ℓ , (where thefirst condition prevents overlapping of vertical arms) and consider the set (19) D Nε,µ := (cid:0) ]0 , π [ × ] − ε, ε [ (cid:1) N +1 [ i =1 (cid:16)(cid:16) πi N +2 − µ N +1 , πi N +2 + µ N +1 (cid:17) × ] − ℓ, ℓ [ (cid:17) . We will drop the subscripts in D Nε,µ in order to lighten the notation, writing themwhen needed to avoid confusion. To compare the effect of the reinforcements onthe torsional instability, we are keeping the area of the set D N fixed, indeed wehave | D N | = 2 πε + 4 µ ( ℓ − ε ) for any N . Furthermore, for g ∈ L (]0 , π [) and α > α N (since this simplifies some computations), following the suggestion ofTheorem 4.2, we consider the odd function(20) f α ( x, y ) := R α sinh( αy ) g ( x )with R α := α C g (cosh( αℓ ) − and C g := R π | g ( x ) | dx , so that k f α k L = 1. We define(21) β m := γ m Υ m C g (1 − σ ) and ω m := γ m C g (1+ σ ) sinh( mℓ ) cosh( mℓ )+(1 − σ ) mℓ (1 − σ ) m [(3+ σ ) sinh( mℓ ) cosh( mℓ )+(1 − σ ) mℓ ] , where the coefficients Υ m are as defined in (16), and (22) γ m := 2 π Z π g ( x ) sin( mx ) dx − dπ (1+ d ) 2 N +1 X i =1 Z πi N +2 + µ N +1 πi N +2 − µ N +1 g ( x ) sin( mx ) dx . Then, we obtain an explicit form for the gap function corresponding to problem(6) with f = f α and D = D N , and we analyze its asymptotic behavior as α → + ∞ . Theorem 6.1.
Let α > with α N , let u α be the unique solution of (6) with f = f α and D = D N , let G α be as in (7) with u f,D = u α . As α → + ∞ , G α ( x ) = P ∞ m =1 β m ( α ) sin( mx ) converges uniformly on [0 , π ] to the func-tion G ( x ) := P ∞ m =1 β m sin( mx ) , where the Fourier coefficients β m ( α ) are so that β m ( α ) = β m − ω m α + o (cid:0) α (cid:1) , with β m and ω m > given in (21) . In Section 10 we prove Theorem 6.1. We derive the explicit value of β m ( α )in (57). Furthermore, we show that G ( x ) is the gap function corresponding to asolution of the limit problem (58).We exploit Theorem 6.1 to numerically solve the minimaxmax problem (9).More precisely, we fix ℓ = π/
150 and σ = 0 . f α in (20) with g ( x ) = sin( nx ) for n = 1 , . . . ,
10 and we call f n its H − ∗ (Ω) limit as α → + ∞ (seeLemma 10.2) and G n the corresponding gap function. Then, we consider(23) F = { f , ..., f } and D = { D , ..., D } . The results are summarized in Table 2, in terms of the maximal gap G ∞ n . Thenumerical values in Table 2 have been obtained using the software Mathematica,approximating the Fourier series for G n up to the 250-th term. Table 2.
Numerical values of 10 ×G ∞ n , with ℓ = π/ σ = 0 . d = 2, g ( x ) = sin( nx ), D = D N , µ = 0 . µ = 0 . × G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ ∅ .
444 16 .
357 7 . . . . . . . . D .
113 15 .
980 14 .
249 4 . .
422 2 . . . . . D .
964 13 .
158 6 . . . . .
284 1 . . . D .
292 13 .
979 5 . . . . . . . . D .
839 13 .
892 6 . . . . . . . . D .
135 14 .
080 6 . . . . . . . . D .
320 14 .
050 6 . . . . . . . . × G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ G ∞ ∅ .
444 16 .
357 7 . . . . . . . . D .
707 14 .
748 16 .
541 5 . . . . . . . D .
544 11 .
277 5 . . . . .
807 1 . .
949 3 . D .
602 12 .
383 5 . . . . . . . . D .
473 12 .
287 5 . . . . . . . . D .
950 12 .
559 5 . . . . . . . . D .
251 12 .
526 5 . . . . . . . . Several comments are in order. First we notice that, as expected from the state-ment of Theorem 6.1, the results do not depend on ε . Moreover, µ = 0 . µ = 0 . . MINIMAXMAX PROBLEM 11
It is worth noting that there is no monotonicity of G ∞ n with respect to either thenumber of branches, or to the frequency of sin( nx ), nor to the reinforcement thick-ness µ . Also, we observe that each forcing term has its own “best truss” yieldinga minimal maximal gap: the pattern is quite clear and it follows a descendingdiagonal in the two Tables 2. Basically, we see that D n − (i.e., the cross with2 n − g ( x ) = sin( nx ) and the reason isthat the plate is reinforced in the points where g attains either a maximum or aminimum; we did not display all the related lines but the same pattern holds trueuntil N = 10. In particular, if n = 2 we know that D is the best reinforcementsince there are parts of the truss under the two extremal points of g ( x ) = sin(2 x ),see the left-hand picture in Figure 1 where we depict the longitudinal behavior of g ( x ) = sin(2 x ) and the truss D (black spots on the horizontal axis). xy xy Figure 1.
The forces g ( x ) = sin(2 x ) (left) and g ( x ) = sin(7 x )(right) with the truss D .We remark that some trusses aggravate the torsional instability, i.e., they in-crease the maximal gap G ∞ n : this is due to a bad combination between the shapeof the forcing term g and the location of D . For example, we observe that thereinforcement D improves the performance when g ( x ) = sin( nx ) with n = 1 , n the torsional performance is worse than that of theunstiffened plate (with D = ∅ ). We also observe that there are some “anomalousvalues” of G ∞ n , see e.g., the values corresponding to D and n = 7 or n = 9:they are considerably larger than the other values in the same column and thereason is again that the place where D acts interacts badly with g . In particular,we notice that both sin(7 x ) and sin(9 x ) have the same sign in correspondence of x = π , π , π that are the centers of the three vertical arms of D ; in particular, g ( x ) = sin(7 x ) = ⇒ g ( π ) = g ( π ) = − √ , g ( π ) = −
1, see the right picture inFigure 1.Next, we exploit Theorem 6.1 to solve analytically the maxmax problem (8)when D reduces to one horizontal bar (including the free plate). In general, max-imizing a Fourier series is a tricky problem that can be solved only for particularchoices of the coefficients, see e.g., [20]. This is why we focus on the set Γ offunctions g = g ( x ) satisfying one of the following: • g ( x ) = sin( mx ) with m ∈ N . • g ( x ) = X m > N e γ m sin( mx ) with { e γ m } m ∈ N ⊂ ℓ and N ∈ N large enough. • g ( x ) = sin( mx ) + sin(3 mx ) with m ∈ N . • g ( x ) = N X m =1 sin((2 m − x ) with N ∈ N sufficiently large.Then, we define the class F Γ := n f : f = lim α →∞ f α in H − ∗ (Ω) , with f α as in (20) and g ∈ Γ o and, in Section 10, we prove the following. Theorem 6.2.
Let F = F Γ be as above and assume that D =]0 , π [ × ] − ε, ε [ forsome < ε < ℓ (i.e., µ = 0 in (19) ). Then, the solution of the maxmax problem (8) is given by [lim α →∞ R α sinh( αy )] sin( x ) , where the limit is in H − ∗ (Ω) . Theorem 6.2 states that the worst case as α → ∞ corresponds to the function g ( x ) = sin( x ). However, g ( x ) = sin( x ) seems not to be the worst case in general:to see this, compare the values of G ∞ given in Table 2 with G ∞ T π/ given in Table 1.7. Weakening Resonant Forces with Polygonal Reinforcements
In this section, we intend to study numerically the gap function (7) and therelated minimaxmax problem (9) in the case the class F contains some “resonant-type” force f and the class D contains “not-so-nice” domains D ∈ L (see Definition3.1). Hence, we minimize the energy (3).Throughout this section, we fix ℓ = π/
150 and σ = 0 . √ mℓ ) > σ / (2 − σ ) √ mℓ so that m m e m of ∆ with the boundary conditions in(30) having m − x -direction and the corresponding eigenvalue ν m are known; see [5]. Notice that m < ν m < (cid:0) m + π ℓ (cid:1) . A detailed analysis of thevariation of all the eigenvalues under domain deformations was performed in [21].We aim to study the effect of a reinforcement D when the force f is at resonance,namely proportional to a torsional (odd) eigenfunction: we take f = e m ( x, y ). Forthese functions f we then deal with problem (6) and we seek the best shape of thereinforcement D in order to lower the maximal gap G ∞ D . We numerically studyproblem (9) within classes of forces (with m from 1 to 5) and of reinforcements D of sets composed by two parallel strips, by triangles, by squares, and by hexagonsas in Figure 2:(24) F = { e , ..., e } and D = { Strips, Triangles, Squares, Hexagons } . The black lines are the thick stiffening trusses D put below the plate and theirtotal area is constant. More precisely, the first plate is reinforced by two paralleltrusses of width X = π ≈ . πX .The three remaining shapes all have two parallel trusses of width W = π alongthe free edges of the plate for a total area of 2 πW , while the remaining area of2 π ( X − W ) is distributed in connecting transverse trusses which generate somepolygons all along the plate, see again Figure 2. The triangular transverse truss iscomposed of 74 vertical segments having length 2 ℓ − W and 75 oblique segmentshaving length ( π/ − W ) √
2, both having width 0 . MINIMAXMAX PROBLEM 13
Figure 2.
Qualitative patterns of the trusses in D .2 ℓ − W and width W . Finally, the hexagonal transverse truss is composed of17 Y -shaped components, alternating upwards and downwards, complemented bytwo segments at the opposite ends of the plate (playing the role of the obliquebranches of Y ), whose measures are ℓ − W for the length of the vertical legs and Z = 0 . L of Definition 3.1, have their own motivation. The first one is the mostnatural, putting reinforcements only on the two free edges. The triangular truss isthe most frequently used by engineers. The third one is also natural, putting thesimplest transverse connections between the free edges. Finally, a truss composedof regular hexagons was shown to have better bending performances in [22] wherethe “boundary effects” were neglected. In fact, what really counts is to haveangles of size 2 π/
3, as in irrigation or traffic problems , see [23–25]. Let us alsomention that it has been known since the 19th century that soap bubbles reach anequilibrium on flat surfaces when the angles between three adjacent bubbles arealways 2 π/
3, see [26]. This angle has the peculiarity to “optimize the distances”and it is therefore interesting to measure its performance also in stiffening trusses.The numerical values for the maximal gap are reported in Table 3.
Table 3.
Numerical values of 10 ×G ∞ e m ,D for the different polyg-onal reinforcements D and resonant forces e m (with ℓ = π/ σ = 0 .
2, and d = 2). e e e e e ∅ .
629 21.811 14.537 10.899 8.7147Strips 25 .
448 6 . . . . .
363 7 . . . . .
946 6 . . . . .
875 7 . . . . D that we have introduced here could be enlarged by consideringalso other geometries for D . Regarding the hexagonal design, we actually stud-ied different positions of the intersections in the Y-shaped elements. The results contained in Table 3 are given for elements where the intersections occur on themidline of the plate, hence with the vertical branch having length ℓ , while we per-formed computations also for cases where the vertical branch is longer or shorterthan ℓ . Even though one might expect the gap functions to be monotone or tohave a unique minimum point (with respect to the length of the vertical branch),this does not occur, the behavior being very specific depending on the particularresonant force e m considered. In some cases, the maximal gap exits the range wesaw in Table 3: for e and e the gap function is always bounded by that of thesquares and that of the triangles, while for e the branch b of length b = 2 ℓ/ e the cases b = 4 ℓ/ , ℓ/ e the case b = 23 ℓ/
20 performs worse thanthe triangles while the cases b = 4 ℓ/ , ℓ/
20 are better than the squares.8.
Proofs of the Existence Results
We first prove the continuity of the map defined in (7). We recall that in allthe cases considered for the class F , the weak* topology coincides with the weaktopology, except when F = L ∞ (Ω). Proposition 8.1.
Let F be either H − ∗ (Ω) (for E ) or L p (Ω) with p ∈ ]1 , + ∞ ] (forboth E and E ). Let also D be a class of open subdomains of Ω closed with respectto the L topology. Then the map G ∞ f,D : F × D → [0 , ∞ [ with ( f, D )
7→ G ∞ f,D is sequentially continuous when F is endowed with the weak* topology and D isendowed with the L topology.Proof. Let { ( f n , D n ) } n ⊂ F × D be such that ( f n , D n ) → ( f, D ) as n → + ∞ ,hence f n ⇀ ∗ f in F and χ D n → χ D in L as n → + ∞ . We denote by u = u f,D and u n = u f n ,D n the corresponding solutions of (5). Recalling (4), (5) with f = f n and D = D n reads(25) ( u n , v ) H ∗ + d ( u n , v ) D n = h f n , v i ∀ v ∈ H ∗ (Ω) . Since f n ⇀ ∗ f in F , the above equality with v = u n yields k u n k H ∗ C forsome C >
0. In particular, u n ⇀ ¯ u up to a subsequence in H ∗ (Ω) for some¯ u ∈ H ∗ (Ω). Next, by adding and subtracting d ( u n , v ) D in (25), we obtain that,for every v ∈ H ∗ (Ω)(26) ( u n , v ) H ∗ + d ( u n , v ) D + d ( u n , v ) D n \ D − d ( u n , v ) D \ D n = h f n , v i . Since χ D n → χ D in L (Ω) yields | D n △ D | → n → + ∞ , we deduce that | ( u n , v ) D n \ D | C k v k H ( D n \ D ) = o (1) as n → + ∞ and similarly ( u n , v ) D \ D n = o (1). By this, passing to the limit in (26), we conclude that (¯ u, v ) H ∗ + d (¯ u, v ) D = h f, v i for all v ∈ H ∗ (Ω); hence ¯ u ≡ u . Furthermore, from the compactness ofthe embedding H ∗ (Ω) ⊂ C (Ω), we obtain u n → u in C (Ω). In terms of thegap functions, this means that G f n ,D n ( x ) converges uniformly to G f,D ( x ) as n → + ∞ over [0 , π ]. In particular, G ∞ f n ,D n → G ∞ f,D as n → + ∞ . This concludes theproof. (cid:3) MINIMAXMAX PROBLEM 15
Theorem 3.1.
Fix D ⊂ Ω. If { f n } ⊂ H − ∗ (Ω) is a maximizing sequence for (10),since k f n k H − ∗ = 1, up to a subsequence, we have f n ⇀ f in H − ∗ (Ω). By Propo-sition 8.1, max (cid:8) G ∞ f,D : f ∈ H − ∗ (Ω), k f k H − ∗ = 1 (cid:9) = G ∞ f,D . Moreover, it must be k f k H − ∗ = 1. Otherwise, if k f k H − ∗ <
1, set e f = f / k f k H − ∗ and by linearity weget G ∞ e f,D = G ∞ f,D / k f k H − ∗ > G ∞ f,D , a contradiction that proves the first part ofTheorem 3.1.Now, let { f n } ⊂ L p (Ω) be a maximizing sequence for (11) such that k f n k L p = 1.Up to a subsequence and for some f , we have f n ⇀ f in L p (Ω) if 1 < p < ∞ and f n ⇀ ∗ f in L ∞ (Ω). In particular, by lower semicontinuity of the norms withrespect to these convergences, k f k L p k f n k L p = 1. Moreover, by Proposition 8.1,we have max (cid:8) G ∞ f,D : f ∈ L p (Ω) with k f k L p = 1 (cid:9) = G ∞ f,D . Finally, the proof that k f k L p = 1 follows by arguing as above. (cid:3) Theorem 3.2.
Using the Direct Method of the Calculus of Variations, it is sufficientto find a topology for which the functional D
7→ G ∞ D defined in (10) is lowersemicontinuous while the class of admissible sets D is compact. For this purposewe use the L -convergence of sets, namely the L -convergence of the characteristicfunctions associated to the sets. Indeed, by its definition (10) and the continuityproved in Proposition 8.1, it follows that the functional G ∞ is lower-semicontinuouswith respect to the L -convergence of sets. Therefore, it remains to prove that theclasses introduced in Definition 3.1 are compact with respect to this convergence:we do it for each class.(a) Consider a sequence of crosses { D n } in C : by the Bolzano-Weierstrass The-orem the sequences of points { x in } and { y in } converge, up to subsequences, tosome x i ∈ [ µ, π − µ ], i = 1 , . . . , N , and some y j ∈ [ − ℓ + ε, ℓ − ε ], j = 1 , . . . , M ,respectively. By the Lebesgue Dominated Convergence Theorem, it turns out that | D n △ D | → n → ∞ where D is the cross (cid:16) N [ i =1 ( x i − µ, x i + µ × ] − ℓ, ℓ [ (cid:17) ∪ (cid:16) M [ j =1 ]0 , π [ × ] y j − ε, y j + ε [ (cid:17) ;this means that χ D n → χ D in L (Ω) as n → ∞ . Moreover, | D | = κ , thanks tothe area constraint. Therefore, the class C with area constraint is compact withrespect to the L -convergence of sets.(b) To each rectangle R ⊂ Ω we associate its four vertices V ( R ) , . . . , V ( R )in such a way that V ( R ) is the upper-right vertex (i.e., the one with largest y -coordinate in the case such a vertex is unique, otherwise the one with largest x -coordinate) and the remaining V i ( R ) are ordered clockwise. Consider a sequenceof rectangles { R n } all having inradius at least ε : by the Bolzano-Weierstrass The-orem, up to extracting a subsequence (that we do not relabel), the sequenceof vertices { V ( R n ) } converges to some point V ∈ Ω. Up to extracting a fur-ther subsequence, the sequence of vertices { V ( R n ) } also converges to some point V ∈ Ω. Repeating this argument for the remaining vertices, we infer that eachof the four sequences of vertices { V i ( R n ) } converges, up to subsequences, to somepoint V i ∈ Ω (for i = 1 , , , R be the open convex hull of the four points V , . . . , V ; since, by construction, the scalar product of two consecutivesides is ( V i ( R n ) V i +1 ( R n ) , V i +1 ( R n ) V i +2 ( R n )) = 0 for i = 1 , , ,
4, where we set V ( R ) := V ( R ) and V ( R ) := V ( R ), passing to the limit as n → ∞ , and usingthe continuity of the scalar product it follows that ( V i V i +1 , V i +1 V i +2 ) = 0 (for i = 1 , , ,
4, where V := V and V := V ). Moreover, since the distance betweentwo consecutive vertices of R n is larger than 2 ε for all n , also the inradius of Ris at least ε . This implies that the set R is an open rectangle having the distinctvertices V i ( R ) = V i (for i = 1 , , , | R n △ R | → n → ∞ ; this means that χ R n → χ R in L (Ω) as n → ∞ .Then take a sequence of sets D n ∈ T with D n = ∪ Ni =1 R in . Using the argumentabove, up to subsequences, we have that | R in △ R i | → n → ∞ for somerectangles R i all having inradius at least ε . Hence, χ R in → χ R i in L (Ω) for all i = 1 , ..., N and, in turn, χ D n → χ D in L (Ω). The area constraint yields that | D | = κ . Therefore, the class T with area constraint is compact with respect tothe L -convergence of sets.(c) Let Σ n be a sequence of closed connected sets with H (Σ n ) L . Fromthe Blaschke Selection Theorem and the Go lab Theorem (see e.g. [27, Theorem4.4.17]), up to a subsequence we know that Σ n → Σ with respect to the Hausdorffdistance, where Σ is a closed and connected set with H (Σ) L . Then the distancefunction to Σ n converges to the distance function to Σ uniformly on Ω. This, withthe fact that the Lebesgue measure of the set ∂K ε = { x ∈ Ω : dist K = ε } is zero,implies that K εn converges in L to K ε (see [28]).(d) Using again [28, Theorem 2.4.10] we obtain the compactness with respectto the L convergence of the space L with area constraint. (cid:3) Proofs of the Symmetry Results
Theorem 4.1.
Let f ∈ H − ∗ (Ω) be such that k f k H − ∗ = 1 and consider the solution u f ∈ H ∗ (Ω) of (5). Since D is symmetric, following the decomposition (13) wemay rewrite (5) as(27) ( u ef , v e ) H ∗ + ( u of , v o ) H ∗ + d ( u ef , v e ) D + d ( u of , v o ) D = h f e , v e i + h f o , v o i , for all v ∈ H ∗ (Ω). Moreover, by (7), we have G f,D ( x ) = u of ( x, ℓ ) − u of ( x, − ℓ ) andalso that G ∞ f,D = max x ∈ [0 ,π ] (cid:12)(cid:12) u of ( x, ℓ ) − u of ( x, − ℓ ) (cid:12)(cid:12) . In particular, if f o = 0 then u o = 0 and G ∞ f,D = 0 so that f cannot be a maximizer for G ∞ f,D . Hence, by (14),there exists 0 < α α = k f o k H − ∗ k f k H − ∗ = 1. Consider now theproblem ( w, v ) H ∗ + d ( w, v ) D = α h f o , v i for all v ∈ H ∗ (Ω). By linearity and by (27),its solution is w = u o /α , then G foα ,D ( x ) = α G f,D ( x ) and G ∞ foα ,D = α G ∞ f,D > G ∞ f,D . Hence, we have shown that for all f ∈ H − ∗ (Ω) such that k f k H − ∗ = 1, thereexists g ∈ H − O (Ω) such that k g k H − ∗ = 1 ( g = f o /α ) and G ∞ g,D > G ∞ f,D . Thisproves the first part of Theorem 4.1.The remaining part of Theorem 4.1 follows the inverse path. Let f be as in thestatement and take any φ ∈ H − E (Ω) such that k φ k H − ∗
1. Then, put g = f + φ MINIMAXMAX PROBLEM 17 so that g o = f and g e = φ . By (14) we have k g k H − ∗ = 1. By slightly modifyingthe arguments above we see that G ∞ g,D = G ∞ f,D . (cid:3) For the proof of Theorem 4.2 we need the following result.
Lemma 9.1.
Let p ∞ and a > . If φ ∈ L p (] − a, a [) then (28) k φ o k L p k φ k L p . Moreover:– if p = 1 then the inequality in (28) is strict if and only if | φ o ( x ) | < | φ e ( x ) | ina subset of ] − a, a [ of positive measure;– if < p < ∞ then the inequality in (28) is strict if and only if φ is not odd( φ φ o );– if p = ∞ then the inequality in (28) is strict if and only if for any { x n } ⊂ ] − a, a [ such that | φ ( x n ) | → k φ k L ∞ , one has lim inf n | φ e ( x n ) | > ; in particular, if φ ∈ C [ − a, a ] , then the inequality is strict if and only if φ e ( x ) = 0 in every point x where | φ | attains its maximum.Proof. Since φ o ( x ) = φ ( x ) − φ ( − x )2 , the inequality (28) follows from the Minkowskiinequality and the symmetry of ] − a, a [.If p = 1, then the Minkowski inequality, just used to obtain (28), reads Z a − a | φ o ( x ) | dx = 12 Z a − a | φ ( x ) − φ ( − x ) | dx Z a − a | φ ( x ) | + | φ ( − x ) | dx = Z a − a | φ ( x ) | dx so that it reduces to an equality if and only if0 > φ ( x ) φ ( − x ) = h φ e ( x ) + φ o ( x ) ih φ e ( − x ) + φ o ( − x ) i = φ e ( x ) − φ o ( x ) for a.e. x ∈ ] − a, a [. This means that | φ e ( x ) | | φ o ( x ) | for a.e. x ∈ ] − a, a [. Sincethis is a necessary and sufficient condition, the statement for p = 1 is proved.If p ∈ ]1 , + ∞ [, the Minkowski inequality is itself obtained via an application ofH¨older’s inequality and equality holds if and only if the two involved functions aremultiples of each other. In the present situation, this means that φ ( x ) = αφ ( − x )for some α = α ( p ) < x ∈ ] − a, a [. The only possibility is that α = −
1, which means that φ = φ o and φ e ≡
0. Since this is a necessary andsufficient condition, also the statement for p > p = ∞ , we claim that equality holds in (28) if and only if there exists { x n } ⊂ ] − a, a [ such that | φ ( x n ) | → k φ k L ∞ and φ e ( x n ) →
0. Indeed, if such a sequenceexists, then | φ o ( x n ) | = | φ ( x n ) − φ e ( x n ) | → k φ k L ∞ which proves that k φ o k L ∞ = k φ k L ∞ . Conversely, if equality holds then there exists { x n } ⊂ ] − a, a [ such that φ o ( x n ) → k φ k L ∞ . This yields φ ( x n ) − φ e ( x n ) = φ o ( x n ) → k φ k L ∞ and φ e ( x n ) − φ ( − x n ) = − φ o ( − x n ) = φ o ( x n ) → k φ k L ∞ , which proves that φ e ( x n ) → | φ ( ± x n ) | would tend to exceed k φ k L ∞ . The claim is so provedand therefore the strict inequality occurs in the opposite situation: this proves thefirst statement.In the case, where φ ∈ C ([ − a, a ]), the sequences just used to prove the state-ment may be replaced by their limits. (cid:3) Theorem 4.2.
For every p ∈ ]1 , ∞ ], (28) combined with the argument in the proofof Theorem 4.1 yields that a maximizer f can be sought as an odd function.If 1 < p < ∞ , by contradiction, let f ∈ L p (Ω) such that k f k L p = 1 be anon-odd maximizer. Since f f o , by Lemma 9.1 k f o k L p < k f k L p = 1. Take now f = f o / k f o k L p , recalling that f e plays no role in the value of the gap function,we obtain k f k L p = 1 and G ∞ f,D = G ∞ f,D / k f o k L p > G ∞ f,D , a contradiction.If p = ∞ , take an odd function f ∈ L ∞ (Ω) such that k f k L ∞ = 1 and G ∞ f,D =max n G ∞ φ,D : φ ∈ L ∞ (Ω) with k φ k L ∞ = 1 o . If | f ( x, y ) | < ω ⊂ Ωof positive measure, take any even function h such that h ( x, y ) ≡ \ ω and | h ( x, y ) | < − | f ( x, y ) | in ω . Then, g = f + h is not odd and satisfies k g k L ∞ = 1, g e = h = 0, and G ∞ g,D = G ∞ f,D (by linearity since G ∞ h,D = 0). (cid:3) Proofs of Theorems 6.1 and 6.2
We prove Theorem 6.1 and Theorem 6.2 in several steps. Let g ∈ L (]0 , π [), α > α N and(29) k α ( x, y ) = K α e αy g ( x ) , K α := α C g sinh( αℓ ) and C g := Z π | g ( x ) | dx , so that k k α k L = 1. Let h α := k α dχ DN ; we first focus on the auxiliary problem (30) ∆ w = h α , in Ω ,w = w xx = 0 , on { , π }× ] − ℓ, ℓ [ ,w yy + σw xx = w yyy + (2 − σ ) w xxy = 0 , on ]0 , π [ ×{− ℓ, ℓ } . Indeed, if w α solves (30), then a multiple of its odd part u α ( x, y ) := R α K α w oα ( x, y ) = sinh( αℓ )2(cosh( αℓ ) −
1) ( w α ( x, y ) − w α ( x, − y ))solves problem (6) with f = f α and D = D N . Moreover, if G α ( x ) is the gapfunction corresponding to w α , then R α K α G α ( x ) is the gap function corresponding to u α . Therefore, since R α K α = 1 + 2 e − αℓ + o ( e − αℓ ) as α → + ∞ , the limit of the gapfunction corresponding to u α and the asymptotic behavior of the correspondingcoefficients are exactly the same as those for w α .Now, we focus on the explicit solution of the auxiliary problem. We expand g ∈ L (]0 , π [) in a Fourier series(31) g ( x ) = ∞ X m =1 e γ m sin( mx ) , e γ m = 2 π Z π g ( x ) sin( mx ) dx . Then, if we set(32) I N := N +1 [ i =1 (cid:18) πi N + 2 − µ N + 1 , πi N + 2 + µ N + 1 (cid:19) , MINIMAXMAX PROBLEM 19 for every x ∈ ]0 , π [ and every y ∈ ] − ℓ, − ε [ ∪ ] ε, ℓ [, we have(33) g ( x )1 + dχ D N ( x, y ) = g ( x )1 + dχ I N ( x ) = ∞ X m =1 γ m sin( mx ) , where the coefficients γ m are as defined in (22), while if we set b γ m = e γ m d , we have g ( x )1+ dχ DN ( x,y ) = g ( x )1+ d = P ∞ m =1 b γ m sin( mx ) for all x ∈ ]0 , π [ and y ∈ ] − ε, ε [.In the sequel, we will need the following constants (only depending on m , α and ε ): F ( ε ) := α ( α − m ) sinh( mε )+2 m cosh( mε )2 m + ( α − m ) m sinh( mε ) − α cosh( mε )2 m ε ,F ( ε ) := − α ( α − m ) cosh( mε )+2 m sinh( mε )2 m − ( α − m ) m cosh( mε ) − α sinh( mε )2 m ε ,F ( ε ) := ( α − m ) α cosh( mε ) − m sinh( mε )2 m ,F ( ε ) := − ( α − m ) α sinh( mε ) − m cosh( mε )2 m , (34) F ± ( ε ) := F ( ε ) ± e − αε F ( − ε ) , F ± ( ε ) := F ( ε ) ± e − αε F ( − ε ) ,F ± ( ε ) := F ( ε ) ± e − αε F ( − ε ) , F ± ( ε ) := F ( ε ) ± e − αε F ( − ε ) , (35)(36) a = a ( m, α, ε ) := K α e αε γ m − b γ m ( m − α ) ,G := − a (cid:8) (1 − σ ) m cosh( mℓ ) F +1 ( ε ) + m [2 cosh( mℓ ) + (1 − σ ) mℓ sinh( mℓ )] F +4 ( ε )+(1 − σ ) m sinh( mℓ ) F − ( ε ) + m [2 sinh( mℓ ) + (1 − σ ) mℓ cosh( mℓ )] F − ( ε ) (cid:9) G := a (cid:8) (1 − σ ) m cosh( mℓ ) F − ( ε ) − m [(1+ σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] F − ( ε )+(1 − σ ) m sinh( mℓ ) F +1 ( ε ) − m [(1+ σ ) sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] F +4 ( ε ) (cid:9) ,G := − a (cid:8) (1 − σ ) m cosh( mℓ ) F − ( ε ) + m [2 cosh( mℓ ) + (1 − σ ) mℓ sinh( mℓ )] F − ( ε )+(1 − σ ) m sinh( mℓ ) F +2 ( ε ) + m [2 sinh( mℓ ) + (1 − σ ) mℓ cosh( mℓ )] F +3 ( ε ) (cid:9) ,G := a (cid:8) (1 − σ ) m cosh( mℓ ) F +2 ( ε ) − m [(1+ σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] F +3 ( ε )+(1 − σ ) m sinh( mℓ ) F − ( ε ) − m [(1+ σ ) sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] F − ( ε ) (cid:9) . (37) Then, we set C := m cosh( mℓ ) (cid:18) K α γ m σm − α m − α sinh( αℓ )+ G (cid:19) +sinh( mℓ ) (cid:18) αK α γ m (2 − σ ) m − α m − α cosh( αℓ )+ G (cid:19) m [(3+ σ ) sinh( mℓ ) cosh( mℓ )+(1 − σ ) mℓ ] ,D := m sinh( mℓ ) (cid:18) K α γ m σm − α m − α cosh( αℓ )+ G (cid:19) +cosh( mℓ ) (cid:18) αK α γ m (2 − σ ) m − α m − α sinh( αℓ )+ G (cid:19) m [(3+ σ ) sinh( mℓ ) cosh( mℓ ) − (1 − σ ) mℓ ] ,A := D m [(1+ σ ) sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] − αK α γ m (2 − σ ) m − α m − α sinh( αℓ ) − G (1 − σ ) m sinh( mℓ ) ,B := C m [(1+ σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] − αK α γ m (2 − σ ) m − α m − α cosh( αℓ ) − G (1 − σ ) m cosh( mℓ ) , (38) and A := A + aF ( ε ) , A := A + ae − αε F ( − ε ) ,B := B + aF ( ε ) , B := B + ae − αε F ( − ε ) ,C := C + aF ( ε ) , C := C + ae − αε F ( − ε ) ,D := D + aF ( ε ) , D := D + ae − αε F ( − ε ) . (39) The following statement allows us to determine the explicit solution of (30).
Proposition 10.1.
Assume that g ∈ L (]0 , π [) satisfies (31) . For α > with α N , let K α be as in (29) . Then, the unique solution of (30) is given by w α ( x, y ) = w ( x, y ) , in ]0 , π [ × ] ε, ℓ [ ,w ( x, y ) , in ]0 , π [ × ] − ε, ε [ ,w ( x, y ) , in ]0 , π [ × ] − ℓ, − ε [ . with, for i = 1 , . . . , , w i ( x, y ) := ∞ X m =1 (cid:0) ( A i + C i y ) cosh( my )+( B i + D i y ) sinh( my )+ e αy K α γ im ( m − α ) (cid:1) sin( mx ) and the constants A i = A i ( m, α, ε ) , B i = B i ( m, α, ε ) , C i = C i ( m, α, ε ) , and D i = D i ( m, α, ε ) as defined in (38) and (39) , while γ m = γ m = γ m and γ m = b γ m .Proof. In order to solve the problem, we split the domain Ω into three rectangles: R :=]0 , π [ × ] ε, ℓ [, R :=]0 , π [ × ] − ε, ε [, R :=]0 , π [ × ] − ℓ, − ε [, so that we obtain (40) ∆ w = K α e αy g ( x )(1 + dχ I N ( x )) − , in R ,w = ( w ) xx = 0 , on { , π }× ] ε, ℓ [ , ( w ) yy + σ ( w ) xx = ( w ) yyy + (2 − σ )( w ) xxy = 0 , on ]0 , π [ ×{ ℓ } , (41) ( ∆ w = K α e αy g ( x )(1 + d ) − , in R ,w = ( w ) xx = 0 , on { , π }× ] − ε, ε [ , (42) ∆ w = K α e αy g ( x )(1 + dχ I N ( x )) − , in R ,w = ( w ) xx = 0 , on { , π }× ] − ℓ, ε [ , ( w ) yy + σ ( w ) xx = ( w ) yyy + (2 − σ )( w ) xxy = 0 , on ]0 , π [ ×{− ℓ } , where I N is as defined in (32). We also have to add the junction conditions: (43) w = w , ( w ) y = ( w ) y , ( w ) yy = ( w ) yy , ( w ) yyy = ( w ) yyy , in ]0 , π [ ×{ ε } (44) w = w , ( w ) y = ( w ) y , ( w ) yy = ( w ) yy , ( w ) yyy = ( w ) yyy , in ]0 , π [ ×{− ε } Let φ ∈ H (]0 , π [) be the unique solution of(45) (cid:26) φ ′′′′ ( x ) + 2 α φ ′′ ( x ) + α φ ( x ) = g ( x )(1 + dχ I N ( x )) − , x ∈ ]0 , π [ ,φ (0) = φ ( π ) = φ ′′ (0) = φ ′′ ( π ) = 0 . By (33), and recalling that α N , φ may be written as φ ( x ) = ∞ X m =1 γ m ( m − α ) sin( mx ) , x ∈ ]0 , π [ , with the γ m as defined in (22). Moreover, φ ′′ ∈ H (]0 , π [) is given by φ ′′ ( x ) = − ∞ X m =1 γ m m ( m − α ) sin( mx ) , x ∈ ]0 , π [ , MINIMAXMAX PROBLEM 21 and this series converges in H (]0 , π [) and, hence, uniformly. We will also needthe constants ζ m := γ m σm − α ( m − α ) , ζ m := γ m (2 − σ ) m − α ( m − α ) . Let us now restrict our attention to R . By the system (45), we have that∆ [ K α e αy φ ( x )] = K α e αy g ( x )(1+ dχ I N ( x )) − . Hence, if we introduce the auxiliaryfunction v ( x, y ) := w ( x, y ) − K α e αy φ ( x ) with w solving (40), we see that v solves (46) ∆ v = 0 , in R ,v = ( v ) xx = 0 , on { , π }× ] ε, ℓ [ , ( v ) yy + σ ( v ) xx = − K α e αℓ [ α φ + σφ ′′ ] , on ]0 , π [ ×{ ℓ } , ( v ) yyy + (2 − σ )( v ) xxy = − K α αe αℓ [ α φ + (2 − σ ) φ ′′ ] , on ]0 , π [ ×{ ℓ } . We seek solutions of (46) by separating variables, namely we seek functions Y m = Y m ( y ) such that v ( x, y ) = P ∞ m =1 Y m ( y ) sin( mx ) solves (46). Then∆ v ( x, y ) = ∞ X m =1 [( Y m ) ′′′′ ( y ) − m ( Y m ) ′′ ( y ) + m ( Y m )( y )] sin( mx ) , and the equation in (46) yields(47) ( Y m ) ′′′′ ( y ) − m ( Y m ) ′′ ( y ) + m Y m ( y ) = 0 , for y ∈ ] ε, ℓ [ . The solutions of (47) are linear combinations of cosh( my ), sinh( my ), y cosh( my ), y sinh( my ), that is, Y m ( y ) = ( A + C y ) cosh( my ) + ( B + D y ) sinh( my ), where A , B , C , D will be determined by imposing the boundary conditions in (43)and (46). By differentiating we obtain ( Y m ) ′ ( y ) = ( A m + D + C my ) sinh( my ) + ( B m + C + D my ) cosh( my ) + sinh( my ) , ( Y m ) ′′ ( y ) = ( A m + 2 D m + C m y ) cosh( my ) + ( B m + 2 C m + D m y ) sinh( my ) , ( Y m ) ′′′ ( y ) = ( A m + 3 D m + C m y ) sinh( my ) + ( B m + 3 C m + D m y ) cosh( my ) . The two boundary conditions on ]0 , π [ ×{ ε, ℓ } , see (46), become respectively ∞ X m =1 [( Y m ) ′′ ( ℓ ) − σm Y m ( ℓ )] sin( mx ) = − K α e αℓ [ α φ ( x ) + σφ ′′ ( x )] , ∞ X m =1 [( Y m ) ′′′ ( ℓ ) − (2 − σ ) m ( Y m ) ′ ( ℓ )] sin( mx ) = − K α e αℓ [ α φ ( x ) + (2 − σ ) φ ′′ ( x )] , for all x ∈ ]0 , π [. Hence, from the Fourier expansion of φ , we deduce that ( Y m ) ′′ ( ℓ ) − σm Y m ( ℓ ) = K α e αℓ ζ m , ( Y m ) ′′′ ( ℓ ) − (2 − σ ) m ( Y m ) ′ ( ℓ ) = K α e αℓ α ζ m . By pluggingthis information into the explicit form of the derivatives of Y m we find the system (1 − σ ) m cosh( mℓ ) A + m [2 cosh( mℓ ) + (1 − σ ) mℓ sinh( mℓ )] D + (1 − σ ) m sinh( mℓ ) B + m [2 sinh( mℓ ) + (1 − σ ) mℓ cosh( mℓ )] C = K α e αℓ ζ m , − (1 − σ ) m cosh( mℓ ) B + m [(1+ σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] C − (1 − σ ) m sinh( mℓ ) A + m [(1+ σ ) sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] D = K α e αℓ α ζ m . Similarly, for R we introduce the function v ( x, y ) := w ( x, y ) − K α e αy φ ( x ),with w solving (42), and we see that v satisfies (48) ∆ v = 0 , in R ,v = ( v ) xx = 0 , on { , π }× ] − ℓ, − ε [ , ( v ) yy + σ ( v ) xx = − K α e − αℓ [ α φ + σφ ′′ ] , on ]0 , π [ ×{− ℓ } , ( v ) yyy + (2 − σ )( v ) xxy = − K α αe − αℓ [ α φ + (2 − σ ) φ ′′ ] , on ]0 , π [ ×{− ℓ } . By separating variables, we seek functions Y m = Y m ( y ) so that the function v ( x, y ) = P ∞ m =1 Y m ( y ) sin( mx ) solves (48). Then∆ v ( x, y ) = ∞ X m =1 [( Y m ) ′′′′ ( y ) − m ( Y m ) ′′ ( yY m ) + m Y m ( y )] sin( mx )and then Y m ( y ) = A cosh( my ) + B sinh( my ) + C y cosh( my ) + D y sinh( my ).As with R , we are then led to the system (1 − σ ) m cosh( mℓ ) A + m [2 cosh( mℓ ) + (1 − σ ) mℓ sinh( mℓ )] D − (1 − σ ) m sinh( mℓ ) B − m [2 sinh( mℓ ) + (1 − σ ) mℓ cosh( mℓ )] C = K α e − αℓ ζ m , − (1 − σ ) m cosh( mℓ ) B + m [(1+ σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] C +(1 − σ ) m sinh( mℓ ) A − m [(1+ σ ) sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] D = K α e − αℓ α ζ m . Finally, let e γ m be the Fourier coefficients of g , see (31), and e φ ∈ H (]0 , π [)be defined as e φ ( x ) = P ∞ m =1 e γ m ( m − α ) sin( mx ), for every x ∈ ]0 , π [. For R weintroduce the auxiliary function v ( x, y ) := w ( x, y ) − K α e αy e φ ( x )1+ d with e φ as aboveand w solving (41), and we see that v satisfies(49) (cid:26) ∆ v = 0 in R v = ( v ) xx = 0 on { , π }× ] − ε, ε [ . We seek again solutions of (49) by separating variables, namely we seek functions Y m = Y m ( y ) such that v ( x, y ) = P ∞ m =1 Y m ( y ) sin( mx ) solves (49). Then∆ v ( x, y ) = ∞ X m =1 [( Y m ) ′′′′ ( y ) − m ( Y m ) ′′ ( y ) + m Y m ( y )] sin( mx )and then Y m ( y ) = A cosh( my ) + B sinh( my ) + C y cosh( my ) + D y sinh( my ) . In this case we have no boundary conditions to use as a constraint. Instead, weimpose the junction conditions (43) and (44) by which we get the relations (39).Combining (39) with (48) and (49), we obtain a 4 × A , B , C , D which decouples into the following 2 × (cid:26) (1 − σ ) m cosh( mℓ ) A + m [2 cosh( mℓ ) + (1 − σ ) mℓ sinh( mℓ )] D = K α ζ m cosh( αℓ ) + G − (1 − σ ) m sinh( mℓ ) A + m [(1+ σ )sinh( mℓ ) − (1 − σ ) mℓ cosh( mℓ )] D = αK α ζ m sinh( αℓ )+ G and (cid:26) (1 − σ ) m sinh( mℓ ) B + m [2sinh( mℓ ) + (1 − σ ) mℓ cosh( mℓ )] C = K α ζ m sinh( αℓ )+ G − (1 − σ ) m cosh( mℓ ) B + m [(1+ σ )cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ )] C = αK α ζ m cosh( αℓ )+ G where the G i = G i ( m, α, ε ) are defined in (37). The solutions of the above systemsare given in (38) and, combined with (39), allow us to write the explicit form of w , w and w . MINIMAXMAX PROBLEM 23
To complete the proof of Proposition 10.1 we show that the series defining w α converges in the next lemma. (cid:3) Lemma 10.1. If g ∈ L (]0 , π [) , then the series defining w α in Proposition 10.1converges uniformly in Ω up to the second derivative.Proof. We start by studying the uniform convergence of the series which defines w . We have | w | ∞ X m =1 (cid:16) e mℓ | A + B | + ℓ | C + D | ) + e − mε | A − B | + ℓ | C − D | ) + e αℓ K α γ m ( m − α ) (cid:17) . From the following relations m tanh( mℓ ) G a + 2 G a = 1cosh( mℓ ) [(1 − σ ) m F − + 2 m F − + m (1 − σ ) ℓF +4 )] − ( σ + 3) m [ F − cosh( mℓ ) + F +4 sinh( mℓ )] , mℓ ) G a + 2 mG a = 1cosh( mℓ ) [ − (1 − σ ) m F − − (1 − σ ) m ℓF +3 + m (1 + σ ) F − )] − ( σ + 3) m [ F − cosh( mℓ ) + F +3 sinh( mℓ )] , and after some lengthy calculations, as m → + ∞ , we get A + B = − a ( F ( ε ) + F ( ε )) + K γ m e − mℓ m + o (cid:18) γ m e − mℓ m (cid:19) ,C + D = − a ( F ( ε ) + F ( ε )) + K γ m e − mℓ m + o (cid:18) γ m e − mℓ m (cid:19) , (50)for some K , K ∈ R \ { } . Hence, as m → ∞ , we have (51) A + B = K γ m e − mℓ m + o (cid:16) γ m e − mℓ m (cid:17) , C + D = K γ m e − mℓ m + o (cid:16) γ m e − mℓ m (cid:17) . On the other hand, as m → + ∞ , we have A − B = K ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) ,C − D = K ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) , (52)for some K , K ∈ R \ { } . Hence, since as m → + ∞ we have a ( F ( ε ) − F ( ε )) = − K α εe αε ( γ m − b γ m ) e mε m + o (cid:18) ( γ m − b γ m ) e mε m (cid:19) ,a ( F ( ε ) − F ( ε )) = K α e αε ( γ m − b γ m ) e mε m + o (cid:18) ( γ m − b γ m ) e mε m (cid:19) , we conclude that A − B = − K α εe αε ( γ m − b γ m ) e mε m + o (cid:18) ( γ m − b γ m ) e mε m (cid:19) ,C − D = K α e αε ( γ m − b γ m ) e mε m + o (cid:18) ( γ m − b γ m ) e mε m (cid:19) . The above equalities, together with (51) and the summability of the coefficients e γ m and b γ m , prove the uniform convergence of the series defining w up to thesecond derivative.Next,we consider w . We estimate | w | ∞ X m =1 (cid:16) e mε | A + B | + ε | C + D | )+ e mε | A − B | + ε | C − D | )+ e αε K α b γ m ( m − α ) (cid:17) . Since as m → + ∞ we have a ( F ( ε ) + F ( ε )) = K α εe αε ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) ,a ( F ( ε ) + F ( ε )) = K α e αε ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) , from (50) we deduce that A + B = − K α εe αε ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) ,C + D = − K α e αε ( γ m − b γ m ) e − mε m + o (cid:18) ( γ m − b γ m ) e − mε m (cid:19) . This, jointly with (52) and the summability of the coefficients e γ m , b γ m , proves theuniform convergence of the series defining w up to the second derivative.The computations for w are similar to those for w and we omit them. (cid:3) Now, we focus on the limiting behavior of w α as α → + ∞ . We first determinethe limit of h α . Lemma 10.2.
Let C g be as in (29) and I N be as in (32) . As α → + ∞ we havethat h α → h in H − ∗ (Ω) , where h ∈ H − ∗ (Ω) is defined as follows h h, v i = Z π g ( x ) C g (1 + dχ I N ( x )) v ( x, ℓ ) dx ∀ v ∈ H ∗ (Ω) . The proof is a consequence of an integration by parts, similar to that of Lemma11.1 below and therefore we omit it. Next, we set(53) w ( x, y ) = ∞ X m =1 (cid:0) ( A + C y ) cosh( my ) + ( B + D y ) sinh( my ) (cid:1) sin( mx ) , where A = A ( m ), B = B ( m ), C = C ( m ), D = D ( m ) are given by A := γ m ((1 − σ ) mℓ sinh( mℓ ) + 2 cosh( mℓ ))2 C g (1 − σ ) m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) − (1 − σ ) mℓ ] ,B := γ m ((1 − σ ) mℓ cosh( mℓ ) + 2 sinh( mℓ ))2 C g (1 − σ ) m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] ,C := − γ m sinh( mℓ )2 C g m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] ,D := − γ m cosh( mℓ )2 C g m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) − (1 − σ ) mℓ ] , with γ m as in (22). The following lemma holds. MINIMAXMAX PROBLEM 25
Lemma 10.3.
Let w α be the unique solution of (30) , w be as in (53) and h be asin Lemma 10.2. Then, as α → + ∞ , we have that w α ( x, y ) → w ( x, y ) in H ∗ (Ω) and w is the unique solution of the problem (54) ( w, v ) H ∗ = h h, v i ∀ v ∈ H ∗ (Ω) . Proof.
Let A , B , C , D be as defined in (38), let A , B , C , D be as definedin (53), we get A → A , B → B , C → C , D → D as α → + ∞ . Moreover,from (29), (34), and (36), as α → + ∞ , we have K α = αe − αℓ C g + o ( αe − αℓ ) , a = e − α ( ℓ − ε ) ( γ m − b γ m ) C g α + o (cid:18) e − α ( ℓ − ε ) α (cid:19) , and F ( ε ) = sinh( mε ) − mε cosh( mε )2 m α + o ( α ) , F ( ε ) = cosh( mε )2 m α + o ( α ) ,F ( ε ) = mε sinh( mε ) − cosh( mε )2 m α + o ( α ) , F ( ε ) = − sinh( mε )2 m α + o ( α ) . Hence, recalling the definition of the constants A i , B i , C i , D i given in (38) and(39), for i = 1 , ,
3, we deduce that A i → A , B i → B , C i → C , D i → D , as α → + ∞ . Therefore, by exploiting the summability of the coefficients in the seriesdefining w α , we conclude that w α → w a.e. in Ω, as α → + ∞ .On the other hand, by Lemma 10.2, there holds h α → h in H − ∗ (Ω) as α → + ∞ .Let w h ∈ H ∗ (Ω) be the unique solution of (54), so that clearly k w α − w h k H ∗ k h α − h k H − ∗ , w α → w h in H ∗ (Ω) and uniformly in Ω. Hence, w ≡ w h and thisconcludes the proof of the lemma. (cid:3) Proof of Theorem 6.1.
By Proposition 10.1, the gap function corresponding to w α is G α ( x ) = w ( x, ℓ ) − w ( x, − ℓ ) = P ∞ m =1 ξ m ( α ) sin( mx ) with the coefficients ξ m ( α ) := ( A − A ) cosh( mℓ ) + ( B + B ) sinh( mℓ ) + ( C + C ) ℓ cosh( mℓ )+( D − D ) ℓ sinh( mℓ ) + αγ m C g ( m − α ) , (55)while the gap function corresponding to w can be written as the function G ( x ) = w ( x, ℓ ) − w ( x, − ℓ ) = P ∞ m =1 β m sin( mx ) with the coefficients β m = (cid:0) B sinh( mℓ ) + 2 C ℓ cosh( mℓ ) (cid:1) sin( mx ) = 2 γ m Υ m C g (1 − σ ) , (56)where the coefficients Υ m are as defined in (16). Since, by Lemma 10.3, thefunction w α converges to w uniformly in Ω, we havemax x ∈ [0 ,π ] |G α ( x ) − G ( x ) | = max x ∈ [0 ,π ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X m =1 ξ m ( α ) sin( mx ) − ∞ X m =1 β m sin( mx ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) max x ∈ [0 ,π ] | w α ( x, ℓ ) − w ( x, ℓ ) | + max x ∈ [0 ,π ] | w α ( x, − ℓ ) − w ( x, − ℓ ) | , and so the right-hand side converges to zero. Next, we specify the asymptotic behavior of ξ m ( α ). To this end, by Proposition10.1, we note that, as α → + ∞ , the following estimates hold A − A = a ( F ( ε ) − e − αε F ( − ε )) = O ( e − α ( ℓ − ε ) ) + O ( e − α ( ℓ + ε ) ) ,B + B = 2 B + a ( F ( ε ) + e − αε F ( − ε )) = 2 B + O ( e − α ( ℓ − ε ) ) + O ( e − α ( ℓ + ε ) ) ,C + C = 2 C + a ( F ( ε ) + e − αε F ( − ε )) = 2 C + O ( e − α ( ℓ − ε ) ) + O ( e − α ( ℓ + ε ) ) ,D − D = a ( F ( ε ) − e − αε F ( − ε )) = O ( e − α ( ℓ − ε ) ) + O ( e − α ( ℓ + ε ) ) ,B = B − γ m ((1 + σ ) cosh( mℓ ) − (1 − σ ) mℓ sinh( mℓ ))2 C g (1 − σ ) m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] 1 α + o (cid:18) α (cid:19) ,C = C − γ m cosh( mℓ )2 C g m [(3 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] 1 α + o (cid:18) α (cid:19) . Recalling the definition of ξ m ( α ), we conclude that, as α → + ∞ , ξ m ( α ) = 2 B sinh( mℓ ) + 2 C ℓ cosh( mℓ ) − (1 + σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓC g (1 − σ ) m [(3+ σ ) sinh( mℓ ) cosh( mℓ ) + (1 − σ ) mℓ ] γ m α + o (cid:18) α (cid:19) = β m − ω m α + o (cid:18) α (cid:19) with ω m as in (21).In view of the discussion at the beginning of this section, the statement ofTheorem 6.1 now follows simply by setting(57) β m ( α ) := R α K α ξ m ( α )with ξ m ( α ) as given in (55). Notice that we still denote by G α ( x ) the gap functionscorresponding to u α . Furthermore, by Lemma 10.3, u α ( x, y ) → u ( x, y ) in H ∗ (Ω)where u is the odd part of the unique solution of problem (54). Namely, u is thethe unique solution of the problem(58) ( u, v ) H ∗ = h h o , v i ∀ v ∈ H ∗ (Ω)where h o is the odd part of h as defined in Lemma 10.2. (cid:3) Proof of Theorem 6.2.
Set µ = 0 and g ( x ) = sin( nx ) for some positive integer n ,then the coefficients (22) become γ n = 1 while γ m = 0 if m = n . Furthermore,since C g = R π | sin( nx ) | dx = 2, by Theorem 6.1 the corresponding gap function is(59) G n ( x ) = Υ n sin( nx ) , where Υ n := Υ n / (1 − σ ) with the Υ n as defined in (16). Hence, G ∞ n = Υ n . Thethesis follows by showing that G ∞ < Υ for every g ∈ Γ. Let us consider separatelythe four cases in the set Γ. • Let g ( x ) = sin( mx ) for some positive integer m . By (59) we know that G ∞ m = Υ m . Since 0 < σ <
1, some lengthy computations reveal that d Υ m dm < m Υ m is strictly decreasing and max m G ∞ m = Υ . • For given N ∈ N and { e γ m } m ∈ N ⊂ ℓ ( N ), let g ( x ) = P m > N e γ m sin( mx ).Since sup m | e γ m | C g /π and P ∞ m =1 Υ m converges, by Theorem 6.1 we infer theexistence of N ∈ N sufficiently large such that G ∞ π P m > N Υ m < Υ . MINIMAXMAX PROBLEM 27 • Let m be a positive integer, and take g ( x ) = sin( mx ) + sin(3 mx ). By theprosthaphaeresis formulas, g ( x ) = 4 sin( mx ) cos ( mx ) and we compute the valueof C g in (29). By putting x k = ( k/m ) π with k = 0 , ..., m , for m = 2 n we have C g = n − X j =0 Z x j +1 x j (cid:0) sin( mx ) + sin(3 mx ) (cid:1) dx − Z x j +2 x j +1 (sin( mx ) + sin(3 mx )) dx ! , while for m = 2 n + 1 we have C g = n − X j =0 "Z x j +1 x j (cid:0) sin( mx ) + sin(3 mx ) (cid:1) dx − Z x j +2 x j +1 (cid:0) sin( mx ) + sin(3 mx ) (cid:1) dx + Z x n +1 x n (cid:0) sin( mx ) + sin(3 mx ) (cid:1) dx . In any case, we get that C g = 8 /
3. Furthermore, by Theorem 6.1 we have |G ( x ) | = 2 C g | (cid:0) Υ m sin( mx ) + Υ m sin(3 mx ) (cid:1) | (cid:0) Υ m + Υ m (cid:1) (cid:0) Υ + Υ (cid:1) , where in the last step we exploit the monotonicity of the map n Υ n . Finally,some lengthy computations reveal that 3 F < F , hence G ∞ Υ holds also for g ( x ) = sin( mx ) + sin(3 mx ). • Take g ( x ) = g N ( x ) = P Nm =1 sin((2 m − x ) for N ∈ N sufficiently large, to befixed later. It is known that g N ( x ) = sin ( N x ) / sin( x ) for x ∈ ]0 , π [, see [20, p.73].Hence, C g N = Z π | g N ( x ) | dx = Z π g N ( x ) dx = 2 N X m =1 m − > log(2 N − . By this and Theorem 6.1 we deduce that G ∞ C g N X m =1 Υ m N − N X m =1 F m ( ℓ ) Υ for N sufficiently large, since the last summation converges when N → + ∞ . Thisconcludes the proof of Theorem 6.2. (cid:3) (cid:3) Proof of Proposition 5.1
For z ∈ ]0 , π [, 0 < η < min { z, π − z } , and α >
0, we take g ( x ) = χ [ z − η,z + η ] ( x )in (20) and we set f α,η ( x, y ) := R α,η sinh( αy ) χ [ z − η,z + η ] ( x ) with the constant R α,η := α η (cosh( αℓ ) − , so that k f α,η k L = 1. Let us establish the first ingredientfor the proof of Proposition 5.1. Lemma 11.1.
Let D = ∅ and let G α,η and G T z ( T z as in (15) ) be the gap functionscorresponding to the solutions of (5) with f = f α,η and f = T z , respectively. As ( α − , η ) → (0 , we have that G α,η ( x ) → G T z ( x ) uniformly on [0 , π ] .Proof. We first claim that f α,η → T z in H − ∗ (Ω) as ( α − , η ) → (0 , ( α − ,η ) → (0 , Z Ω f α,η ( x, y ) v ( x, y ) dxdy = v ( z,ℓ ) − v ( z, − ℓ )2 , ∀ v ∈ H ∗ (Ω) . Take v ∈ H ∗ (Ω) and compute Z Ω f α,η ( x, y ) v ( x, y ) dxdy = R α,η Z π χ [ z − η,z + η ] ( x ) (cid:18)Z ℓ − ℓ sinh( αy ) v ( x, y ) dy (cid:19) dx = η (cosh( αℓ ) − Z z + ηz − η (cid:16)(cid:2) cosh( αy ) v ( x, y ) (cid:3) ℓ − ℓ − Z ℓ − ℓ cosh( αy ) v y ( x, y ) dy (cid:17) dx = cosh( αℓ )cosh( αℓ ) − v ( ξ z , ℓ ) − v ( ξ z , − ℓ )2 − αℓ ) − Z ℓ − ℓ cosh( αy ) v y ( θ z , y ) dy, for some z − η < ξ z , θ z < z + η . Now we observe that, uniformly with respect to η , lim α − → R ℓ − ℓ cosh( αy )cosh( αℓ ) − v y ( θ z , y ) dy = 0, by the Lebesgue Dominated Conver-gence Theorem and thus (60) follows. From this we infer that the correspondingsolutions converge in H ∗ (Ω) and then the proof can be completed by arguing asin Proposition 8.1. (cid:3) In view of Lemma 11.1, the proof of Proposition 5.1 follows once we have provedthe following statement.
Lemma 11.2.
Assume z ∈ ]0 , π [ , < η < min { z, π − z } and α > with α N .As ( α − , η ) → (0 , , the gap function G α,η ( x ) corresponding to the solution of (6) with f = f α,η and D = ∅ converges to π (1 − σ ) P ∞ m =1 Υ m sin( mz ) sin( mx ) uniformlyon [0 , π ] , with the Υ m as defined in (16) .Proof. The explicit form of the gap function G α,η ( x ) follows from Theorem 6.1by replacing f α with f α,η and D N with ∅ . Namely, we assume µ = ε = 0 in(19) and g ( x ) = χ [ z − η,z + η ] ( x ) in (20), hence C g = 2 η . With these specifications,by (39), we have A = A , B = B = B , C = C = C , and D = D ,while γ m = e γ m in (31). Hence, by (55) and (57), we it follows that the function G α,η ( x ) = P ∞ m =1 β m ( α, η ) sin( mx ) and β m ( α, η ) = sinh( αℓ )cosh( αℓ ) − (cid:18) B sinh( mℓ ) + 2 C ℓ cosh( mℓ ) + α e γ m η ( m − α ) (cid:19) , with B and C as in (38) and e γ m = πm sin( mz ) sin( mη ). By noting that e γ m ( η )2 η = mz ) π + o ( η ) as η → α − , η ) → (0 ,
0) we obtain β m ( α, η ) = (cid:0) e − αℓ + o ( e − αℓ ) (cid:1) (cid:18) mz ) π + o ( η ) (cid:19) (cid:18) m − σ + o (cid:18) α (cid:19)(cid:19) = 4Υ m sin( mz ) π (1 − σ ) + o (1) , with Υ m as defined in (16). This completes the proof of the lemma. (cid:3) Conclusions, Perspectives and Open Problems
With possible applications to the deck of a footbridge or a suspension bridge,this paper deals with the problem of minimizing the torsional displacements ofpartially hinged reinforced plates. We showed that the gap function (7) is ex-tremely useful to measure the torsional instability and that it gives hints on howto compare the torsional performances of different plates through the minimaxmax
MINIMAXMAX PROBLEM 29 problem (9), namely a robust shape optimization in the worst case setting. Thedemonstrated existence results prove that the problem is well-defined and, in somecases, it also allows to find properties of the worst force. On the other hand, somemeaningful problems prove themselves to be very difficult to handle and optimalelements are hardly characterizable. This led us to provide some conjectures thatwe now motivate in detail.There are other classes D where the minimum problem (9) admits a solutionbut the ones in Definition 3.1 appear particularly appropriate for engineering ap-plications. A quite general class of admissible domains, where it is possible todefine problem (9), is that of measurable sets with uniformly bounded De Giorgiperimeter and fixed area. In this setting, thanks to a compactness result for BVfunctions, the existence of a solution is still guaranteed. Therefore, we point outthat it could be interesting to know if such a solution has enough regularity andgeometrical properties to belong to one of those classes of Definition 3.1.When D = ∅ , namely in the free plate case, in Proposition 5.1 we provide theexplicit representation of the gap function when the load is concentrated on theboundary. The same statement seems out of reach for more general load, that is forodd distributions such as δ ( z,w ) − δ ( z, − w ) , with z ∈ ]0 , π [ and w ∈ [0 , ℓ [. Nevertheless,it is reasonable to expect that the gap function amplifies whenever w → ℓ and z remains fixed. For this reason, we expect T π/ to be the worst case among allpossible normalized couples of odd concentrated loads. This leads to the following. Conjecture 12.1.
When D = ∅ , T π/ and − T π/ are the unique maximizers ofthe worst case problem (10) . The worst case problem (10) may also be set up in different (smaller) classes ofloads such as L p -spaces and one has the maximization problem (11), see Theorem3.1. We have no guess about the possible solutions of (11) when p ∈ ]1 , ∞ [. We alsosuspect that there exists no maximizer for (11) in L (Ω); see Section 6. Moreover,we believe that the strange property stated in Theorem 4.2 for p = ∞ may not befulfilled since we expect the following. Conjecture 12.2.
Let p = ∞ . For every D ⊂ Ω , the unique maximizers of theproblem (11) are the odd function f ( x, y ) = y/ | y | , y = 0 , and its opposite − f . Finally, as concerns the most ambitious goal to solve the minimaxmax problem(9), we conclude by stating two conjectures which are supported by numericalcomputations. More precisely, Table 2 in Section 6 suggests the following.
Conjecture 12.3.
Let F and D be as in (23). The optimum of the minimaxmaxproblem (9) is the couple ( f , D ) . Table 3 in Section 7 shows that the least G ∞ D is obtained for strips, then squares,hexagons, while the largest G ∞ D is obtained for triangles. This is somehow surpris-ing since squares are expected to be in between triangles and hexagons. Moreover,Table 3 suggests the following. Conjecture 12.4.
Let F and D be as in (24). The optimum of the minimaxmaxproblem (9) is the couple ( e , Strips).
Acknowledgements.
The authors are grateful to J.B. Kennedy for his kind revi-sion of the use of the English Language within the present paper. The first, second,and fourth authors are partially supported by the Research Project FIR (Futuroin Ricerca) 2013
Geometrical and qualitative aspects of PDE’s . The third authoris partially supported by the PRIN project
Equazioni alle derivate parziali di tipoellittico e parabolico: aspetti geometrici, disuguaglianze collegate, e applicazioni .The four authors are members of the Gruppo Nazionale per l’Analisi Matematica,la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di AltaMatematica (INdAM).