On Nine Duality Principles and Related Convex Dual Formulations Through a D.C. Approach for Non-Convex Optimization, Some New Corrections

This article develops duality principles and respective convex dual formulations through a D.C. approach applicable to some originally non-convex primal variational formulations. More specifically, in a first step, we develop applications to a Ginzburg- Landau type equation. The results are obtained through basic tools of functional analysis, calculus of variations, duality and optimization theory in infinite dimensional spaces. It is worth emphasizing we have obtained a convex dual variational formulation suitable for a large class of similar models in the calculus of variations. 1 On Nine Duality Principles and Related Convex Dual Formulations Through a D.C. Approach for Non-Convex Optimization, Some New Corrections Fabio Silva Botelho Department of Mathematics Federal University of Santa Catarina, UFSC Florian´ opolis, SC - Brazil

This article develops duality principles and respective convex dual formulations through a D.C. approach applicable to some originally non-convex primal variational formulations. More specifically, in a first step, we develop applications to a Ginzburg-Landau type equation. The

are obtained through basic tools of functional analysis, calculus of variations, duality and optimization theory in infinite dimensional spaces. It is worth emphasizing we have obtained a convex dual variational formulation suitable for a large class of similar models in the calculus of variations.

Duality principle, Ginzburg-Landau system in superconductivity, D.C. approach, convex dual formulation, calculus of variations. MSC: 49N15 1 Introduction This article develops a duality principle applicable to a large class of models in the calculus of variations. We present applications to a Ginzburg-Landau type equation through a D.C. approach. We recall the so-called D.C. approach refers to a difference between two convex functionals. More specifically, we obtain a convex dual formulation suitable for an appropriate optimiza- tion of a concerning primal functional. Remark 1.1. In particular in this version we present some important corrections on section 10. 1 It is worth mentioning the results on duality theory here addressed and developed are inspired mainly in the approaches of J.J.Telega, W.R. Bielski and co-workers presented in the articles [1, 2, 3, 4]. Other main reference is the D.C. approach found in the article by Toland, [5]. Moreover, details on the Sobolev spaces involved may be found in [6] and basic theoretical

in superconductivity may be found in [7, 8]. Similar results and models are addressed in [9, 10, 11, 12, 13, 14]. Basic results on convex analysis are addressed in [15, 16]. Finally, other related results may be found in [17, 18]. Now we start to describe the primal variational formulation for the Ginzburg-Landau model in superconductivity in question. Let Ω ⊂ R3 be an open, bounded and connected set with a regular (Lipschitzian) boundary by ∂Ω. Define the Ginzburg-Landau type functional J : V → R, by J(u) = γ 2 Z Ω ∇u · ∇u dx + α 2 Z Ω (u2 − β)2 dx − ⟨u, f⟩L2. (1) Here, V = W 1,2 0 (Ω), γ > 0, α > 0, β > 0 and f ∈ L∞(Ω). We also denote Y = Y ∗ = L2(Ω). 2 The main duality principle and related dual varia- tional formulation In this section we develop in details the main duality principle and respective convex dual variational formulation for the model in question. We highlight some similar results have been obtained in the preprint [20]. Fix K3 = 5 and K > 0 such that K ≫ max{K3, γ, α, β, 1/γ, 1/α, β, ∥f ∥∞}. Moreover, define the functionals F1 : V → R and F2 : V × Y ∗ → R by F1(u) = γ 2 Z Ω ∇u · ∇u dx + K 2 Z Ω u2 dx − ⟨u, f⟩L2 (2) and F2(u, v∗ 0) = −⟨u2, v∗ 0⟩L2 + K 2 Z Ω u2 dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx. (3) 2 Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ √ K/8} and D∗ = {v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ (3/2)KK3} . Also, we define the polar functionals F ∗ 1 : [Y ∗] → R and F ∗ 2 : [Y ∗]2 → R by F ∗ 1 (v∗ 1) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + f)2 −γ∇2 + K dx (4) and F ∗ 2 (v∗ 1, v∗ 0) = sup u∈V {⟨u, v∗ 1⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (v∗ 1)2 −2v∗ 0 + K dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx, (5) if v∗ 0 ∈ B∗. Furthermore, define the functional J ∗ : D∗ × B∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1) + F ∗ 2 (v∗ 1, v∗ 0). and the exactly penalized functional J ∗ 1 : D∗ × B∗ → R by J ∗ 1 (v∗ 1, v∗ 0) = J ∗(v∗ 1, v∗ 0) −100K2 2 −  v∗ 1 + f −γ∇2 + K  + v∗ 1 −2v∗ 0 + K 2 0,2 . (6) Let (ˆv∗ 1, ˆv∗ 0) ∈ D∗ × B∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0) = 0. From this and from the Legendre transform proprieties, for u0 = ˆv∗ 1 −2ˆv∗ 0 + K ∈ V, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0) = 0. 3 Observe that ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂(v∗ 1)2 = − 1 −γ∇2 + K + 1 −2ˆv∗ 0 + K −100K2  − 1 −γ∇2 + K + 1 −2v∗ 0 + K 2 < 0. (7) Moreover, ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂(v∗ 0)2 = − 1 α + 4u2 −2v∗ 0 + K −100K2  4u2 (−2v∗ 0 + K)2  + 100K2O  1 K3  < 0, (8) and ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂v ∗ 1 ∂v ∗ 0 = −100K2  2u −2v∗ 0 + K  − 1 −γ∇2 + K + 1 −2v∗ 0 + K  + 2u −2v∗ 0 + K +100K2O  1 K3  , (9) in D∗ × B∗. Thus, by direct computation, we may obtain det  ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0) ∂v ∗ 1 ∂v ∗ 0 2 > 0, in D∗ × B∗. From such results, we may infer that J ∗ 1 is concave in D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0). On the other hand   ∂  J(u) + 100K2 2 ∥u − u0∥2 0  ∂u   |u=u0 = ∂J (u) ∂u |u=u0 = 0, 4 so that by an evident convexity, we have obtained J(u0) = inf u∈V  J(u) + 100K2 2 ∥u − u0∥2 0  . Joining the pieces, we have got, J(u0) = inf u∈V1  J(u) + 100K2 2 ∥u − u0∥2 0,2  = J ∗ 1 (ˆv∗ 1, ˆv∗ 0) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0). (10) . The objective of this section is complete. 3 Another duality principle Let 0 < ε ≪ 1. Define also in this section K3 = 3 and K = 5 and assume γ, α, β, ∥f ∥∞ = O(1). Define the functionals F1 : V → R, F2 : V × Y ∗ → R and F3 : V → R by F1(u) = ε 2 Z Ω u2 dx − ⟨u, f⟩L2, (11) F2(u, v∗ 0) = ⟨u2, v∗ 0⟩L2 + 5γ 2 Z Ω ∇u · ∇u dx − ε 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (12) and F3(u) = 4γ 2 Z Ω ∇u · ∇u dx. Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ K}, D∗ = {v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ K} and E∗ = {z∗ ∈ Y ∗ : ∥z∗∥∞ ≤ K and [(−γ∇2)−1z∗]f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗] → R and F ∗ 2 : [Y ∗]2 → R by F ∗ 1 (v∗ 1) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + f)2 ε dx (13) 5 F ∗ 2 (v∗ 1, v∗ 0) = sup u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 −5γ∇2 + 2v∗ 0 − ε dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (14) if v∗ 0 ∈ B∗. and F ∗ 3 (z∗) = sup w∈L2 {⟨u, z∗⟩L2 − F3(w)} = 1 8 Z Ω (z∗)2 −γ∇2 dx. (15) Furthermore, define the functional J ∗ : D∗ × B∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1) − F ∗ 2 (v∗ 1, z∗, v∗ 0) + F ∗ 3 (z∗). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 0, z∗) +1 6 p −γ∇2  v∗ 1 −γ∇2 + 2v∗ 0 − ε + z∗ −4γ∇2  2 0,2 . (16) Let (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ −4γ∇2 ∈ V, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. Observe that ∂2J ∗ 1 (v∗ 1, v∗ 0, z∗) ∂(z∗)2 = − 1 −5γ∇2 + 2v∗ 0 − ε + 1 −4γ∇2 + 1 48(−γ∇2) = −γ∇2 + 2v∗ 0 − ε (−5γ∇2 + 2v∗ 0 − ε)(−4γ∇2) + 1 48(−γ∇2) = 68(−γ∇2) + 52(2v∗ 0) + 52ε γ ∇2 (−5γ∇2 + 2v∗ 0 − ε)(−4γ∇2)48 . (17) 6 Here we assume that 67(−γ∇2) + 52(2ˆv∗ 0) > 0, so that, since J ∗ 1 is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗). Moreover, we assume γ, α, β are such that ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2 < − 1 10 Id, (18) and ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 ≈ O (1), (19) in D∗ × B∗. From such assumptions and results, since ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2 = −O(1/ε), we have that det  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 2 > 0, in D∗ × B∗. From such results, we may infer that J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) is concave in (v∗ 1, v∗ 0) on D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (20) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) +1 6 p −γ∇2  −u0 + z∗ −4γ∇2  2 0,2 , (21) 7 ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = −4γ∇2u, we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − F3(u) +1 6 p −γ∇2 (−u0 + u) 2 0,2 ≤ sup v∗ 0 ∈Y ∗ {F1(u) + F2(u, v∗ 0) − F3(u)} +1 6 p −γ∇2 (−u0 + u) 2 0,2 = J(u) + 1 6 p −γ∇2 (−u0 + u) 2 0,2 , (22) ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + 1 6 p −γ∇2 (−u0 + u) 2 0,2  = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (23) The objective of this section is complete. 4 One more duality principle Let 0 < ε ≪ 1. Define also in this section K3 = 3, K ≫ max{K3, γ, α, β} where we assume, after multiplication by a suitable constant, γ, α, ∥f ∥∞ = O(100). We highlight that multiplying γ, α, ∥f ∥∞ by a constant does not change any critical point since, up to such a multiplying constant, the Euler Lagrange equations keep the same. Define the functionals F1 : V → R, F2 : V × Y ∗ → R and F3 : V → R by F1(u) = ε 2 Z Ω u2 dx − ⟨u, f⟩L2, (24) 8 F2(u, v∗ 0) = ⟨u2, v∗ 0⟩L2 + γ 2 Z Ω ∇u · ∇u dx + K 2 Z Ω u2 dx − ε 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (25) and F3(u) = K 2 Z Ω u2 dx. Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 4√ K/10}, D∗ = n v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ 4√ K o and E∗ = {z∗ ∈ Y ∗ : ∥z∗∥∞ ≤ KK3 and z∗f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗] → R and F ∗ 2 : [Y ∗]2 → R by F ∗ 1 (v∗ 1) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + f)2 ε dx (26) F ∗ 2 (v∗ 1, v∗ 0) = sup u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 −γ∇2 + 2v∗ 0 + K − ε dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (27) if v∗ 0 ∈ B∗. and F ∗ 3 (z∗) = sup w∈L2 {⟨u, z∗⟩L2 − F3(w)} = 1 2K Z Ω (z∗)2 dx. (28) Furthermore, define the functional J ∗ : D∗ × B∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1) − F ∗ 2 (v∗ 1, z∗, v∗ 0) + F ∗ 3 (z∗). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 0, z∗) +10 2 v∗ 1 −γ∇2 + 2v∗ 0 − ε + z∗ K 2 0,2 . (29) 9 Let (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ K ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. Observe that ∂2J ∗ 1 (v∗ 1, v∗ 0, z∗) ∂(z∗)2 = − 1 −γ∇2 + 2v∗ 0 + K − ε + 1 K + 10 K2 = −γ∇2 + 2v∗ 0 − ε (−γ∇2 + 2v∗ 0 + K − ε)(K) + 10 K2 = (−γ∇2 + 2v∗ 0 − ε + 10)K + 10(−γ∇2 + 2v∗ 0 − ε) (−γ∇2 + 2v∗ 0 + K − ε)10K2 . = O (−γ∇2 + 2v∗ 0 − ε + 10) 10K2  + O(1/K3) (30) Here we assume that (−γ∇2) + (2ˆv∗ 0) + 9 > 0, so that, since J ∗ 1 is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗). We also assume γ, α, β are such that ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2 = − 1 α + O  10 1002  < − 1 2α (31) 10 and ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 ≈ O  10 1002  , (32) in D∗ × B∗. From such assumptions and results, since ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2 = −O(1/ε), we have that det  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 2 > 0, in D∗ × B∗. From such results, we may infer that J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) is concave in (v∗ 1, v∗ 0) on D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (33) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) +10 2  −u0 + z∗ K  2 0,2 , (34) ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = Ku, we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − F3(u) +10 2 ∥(−u0 + u)∥2 0,2 ≤ sup v∗ 0 ∈Y ∗ {F1(u) + F2(u, v∗ 0) − F3(u)} +10 2 ∥(−u0 + u)∥2 0,2 = J(u) + 10 2 ∥(−u0 + u)∥2 0,2 , (35) 11 ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + 10 2 ∥(−u0 + u)∥2 0,2  = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (36) The objective of this section is complete. 5 A fourth duality principle Define K3 = 3, K ≫ max{K3, γ, α, β} and assume γ, α, β, ∥f ∥∞ = O(1). Define the functionals F1 : V × Y ∗ → R, F2 : V × Y ∗ → R and F3 : V → R by F1(u, v∗ 0) = γ 2 Z Ω ∇u · ∇u dx + K 2 Z Ω u2 dx +⟨u2, v∗ 0⟩L2 − ⟨u, f⟩L2, (37) F2(u, v∗ 0) = K 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (38) and F3(u) = 2K 2 Z Ω u2 dx. Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 4√ K/8}, D∗ = {v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ (3/2)KK3} and E∗ = {z∗ ∈ Y ∗ : ∥z∗∥∞ ≤ 2KK3 and z∗f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗]2 → R, F ∗ 2 : [Y ∗]2 → R and F ∗ 3 : Y ∗ → R, by F ∗ 1 (v∗ 1, v∗ 0) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u, v∗ 0)} = 1 2 Z Ω (v∗ 1 + f)2 −γ∇2 + 2v∗ 0 + K dx (39) 12 F ∗ 2 (v∗ 1, v∗ 0, z∗) = sup u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 K dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (40) if v∗ 0 ∈ B∗. and F ∗ 3 (z∗) = sup w∈L2 {⟨w, z∗⟩L2 − F3(w)} = 1 4K Z Ω (z∗)2 dx. (41) Furthermore, define the functional J ∗ : B∗ × D∗ × E∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1, v∗ 0) − F ∗ 2 (v∗ 1, v∗ 0, z∗) + F ∗ 3 (z∗). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 0, z∗) −100K2 2 − v∗ 1 + f −γ∇2 + 2v∗ 0 + K + v∗ 1 K 2 0,2 +2K 2 − v∗ 1 K + z∗ 2K 2 0,2 . (42) Let (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ × E∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ 2K ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. Observe that 13 ∂2J ∗ 1 (v∗ 1, v∗ 0, z∗) ∂(z∗)2 = − 1 K + 1 2K +2  1 2K 2 = − 1 2K + 1 2K = 0. (43) so that, since J ∗ 1 (up to an approximate convex regularization) is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗). Moreover, for (v∗ 1, v∗ 0) ∈ D∗ × B∗, denoting u = (v∗ 1 + f) −γ∇2 + 2v∗ 0 + K , we have ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2 = − 1 −γ∇2 + 2v∗ 0 + K − 1 K + 2 K −100K2  − 1 −γ∇2 + 2v∗ 0 + K + 1 K 2 = −2K + γ∇2 − 2v∗ 0 (−γ∇2 + 2v∗ 0 + K)K + 2 K −100K2  − 1 −γ∇2 + 2v∗ 0 + K + 1 K 2 = (−γ∇2 + 2v∗ 0) (−γ∇2 + 2v∗ 0 + K)K −100K2  − 1 −γ∇2 + 2v∗ 0 + K + 1 K 2 < 0 (44) ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2 = − 1 α − 4u2 −γ∇2 + 2v∗ 0 + K −100K2 4u2 (−γ∇2 + 2v∗ 0 + K)2 − 100K2 O(1/K3) < 0, (45) 14 and ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 = 2u −γ∇2 + 2v∗ 0 + K − 100K2  − 1 −γ∇2 + 2v∗ 0 + K + 1 K  2u (−γ∇2 + 2v∗ 0 + K) −100K2 O(1/K3). (46) From such assumptions and results, by direct computation, we may obtain det  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 2 > 0, in D ∗ ×B∗. From such results, we may infer that J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) is concave in (v∗ 1, v∗ 0) on D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (47) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u, ˆv∗ 0) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) +2K 2  −u0 + z∗ 2K  2 0,2 , (48) ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = 2Ku, we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u, ˆv∗ 0) + F2(u, ˆv∗ 0) − F3(u) +2K 2 ∥(−u0 + u)∥2 0,2 ≤ sup v∗ 0 ∈Y ∗ {F1(u, v∗ 0) + F2(u, v∗ 0) − F3(u)} +2K 2 ∥(−u0 + u)∥2 0,2 = J(u) + K ∥(−u0 + u)∥2 0,2 , (49) 15 ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1 n J(u) + K ∥(−u0 + u)∥2 0,2 o = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (50) The objective of this section is complete. 6 A fifth duality principle Define K3 = 3 and K ≫ max{K3, α, γ, β, 1/α, 1/γ, 1/β}. Define the functionals F1 : V → R, F2 : V × Y ∗ → R and F3 : V → R, by F1(u) = γ 2 Z Ω ∇u · ∇u dx −⟨u, f⟩L2, (51) F2(u, v∗ 0) = ⟨u2, v∗ 0⟩L2 + K 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (52) F3(u) = K 2 Z Ω u2 dx Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 8√ K}, D∗ = n v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ 8√ K o and E∗ = {(z∗ ∈ [Y ∗] : ∥z∗∥∞ ≤ KK3 and z∗f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗] → R, F ∗ 2 : [Y ∗]3 → R and F ∗ 3 : Y ∗ → R, by F ∗ 1 (v∗ 1) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + f)2 −γ∇2 dx (53) 16 F ∗ 2 (v∗ 1, v∗ 0, z∗) = sup u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 2v∗ 0 + K dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (54) if v∗ 0 ∈ B∗, and F ∗ 3 (z∗) = sup w∈L2 {⟨w, z∗⟩L2 − F3(w)} = 1 2K Z Ω (z∗)2 dx. (55) Furthermore, define the functional J ∗ : B∗ × D∗ × E∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1) − F ∗ 2 (v∗ 1, v∗ 0, z∗) + F ∗ 3 (z∗). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 0, z∗) + 1 12αK2 3 −γ∇2  z∗ K  + 2v∗ 0  z∗ K  − f 2 0,2 (56) Let (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ × E∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ K ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. Observe that ∂2J ∗ 1 (v∗ 1, v∗ 0, z∗) ∂(z∗)2 = − 1 2v∗ 0 + K + 1 K + 1 6αK2 3 (−γ∇2 + 2v∗ 0)2 K2 . (57) 17 in D∗ × B∗ × E∗. Assume ˆv∗ 0 ∈ B∗ is such that ∂2J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∂(z∗)2 > 0. Therefore, since J ∗ 1 is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗). Moreover, for (v∗ 1, v∗ 0) ∈ D∗ × B∗, we have ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2 = − 1 −γ∇2 − 1 2v∗ 0 + K < 0 (58) and denoting u = −v∗ 1 + ˆz∗ 2v∗ 0 + K , we have ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2 = − 1 α − 4u2 2v∗ 0 + K + 2 3αK2 3  z∗ K 2 ≤ − 1 α − 4u2 2v∗ 0 + K + 2 3α = − 1 3α − 4u2 2v∗ 0 + K < 0, (59) and ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 = 2u 2v∗ 0 + K . (60) From such assumptions and results, by direct computation, we may obtain det  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 2 > 0, (61) in D∗ × B∗. 18 From such results, we may infer that J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) is concave in (v∗ 1, v∗ 0) on D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (62) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) + 1 12αK2 3 −γ∇2  z∗ K  + 2ˆv∗ 0  z∗ K  − f 2 0,2 (63) ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = Ku, we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − F3(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 ≤ sup v∗ 0 ∈Y ∗ {F1(u) + F2(u, v∗ 0) − F3(u)} + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 = J(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 , (64) ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f
2 0,2 = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (65) The objective of this section is complete. 19 7 A sixth duality principle Define K3 = 3, K ≫ max{K3, α, γ, β, 1/α, 1/γ, 1/β} and 0 < ε ≪ 1.. Define the functionals F1 : V → R, F2 : V × Y ∗ → R and F3 : V → R, by F1(u) = ε 2 Z Ω u2 dx −⟨u, f⟩L2, (66) F2(u, v∗ 0) = γ 2 Z Ω ∇u · ∇u dx + ⟨u2, v∗ 0⟩L2 + K 2 Z Ω u2 dx − ε 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (67) F3(u) = K 2 Z Ω u2 dx Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 8√ K}, D∗ = n v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ 8√ K o and E∗ = {(z∗ ∈ [Y ∗] : ∥z∗∥∞ ≤ KK3 and z∗f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗] → R, F ∗ 2 : [Y ∗]3 → R and F ∗ 3 : Y ∗ → R, by F ∗ 1 (v∗ 1) = sup u∈V {⟨u, v∗ 1⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + f)2 ε dx (68) F ∗ 2 (v∗ 1, v∗ 0, z∗ 2) = sup u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 −γ∇2 + 2v∗ 0 + K − ε dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (69) if v∗ 0 ∈ B∗, and F ∗ 3 (z∗ 1) = sup w∈L2 {⟨w, z∗ 1⟩L2 − F3(w)} = 1 2K Z Ω (z∗ 1)2 dx. (70) 20 Furthermore, define the functional J ∗ : B∗ × D∗ × E∗ → R by J ∗(v∗ 1, v∗ 0, z∗) = −F ∗ 1 (v∗ 1) − F ∗ 2 (v∗ 1, v∗ 0, z∗) + F ∗ 3 (z∗). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 0, z∗) + 1 12αK3 3 −γ∇2  z∗ K  + 2v∗ 0  z∗ K  − f 2 0,2 (71) Let (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ × E∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ K ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = 0. Observe that ∂2J ∗ 1 (v∗ 1, v∗ 0, z∗) ∂(z∗)2 = − 1 2v∗ 0 + K − ε + 1 K + 1 6αK2 3 (−γ∇2 + 2v∗ 0)2 K2 . (72) in D∗ × B∗ × E∗. Assume ˆv∗ 0 ∈ B∗ is such that ∂2J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ∂(z∗)2 > 0. Therefore, since J ∗ 1 is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗). Moreover, for (v∗ 1, v∗ 0) ∈ D∗ × B∗, we have ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2 = −1 ε − 1 −γ∇2 + 2v∗ 0 + K − ε < 0 (73) 21 and denoting u = −v∗ 1 + ˆz∗ −γ∇2 + 2v∗ 0 + K − ε , we have ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2 = − 1 α − 4u2 −γ∇2 + 2v∗ 0 + K − ε + 2 3αK2 3  z∗ K 2 ≤ − 1 α − 4u2 −γ∇2 + 2v∗ 0 + K − ε + 2 3α = − 1 3α − 4u2 −γ∇2 + 2v∗ 0 + K − ε < 0, (74) and ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 = 2u −γ∇2 + 2v∗ 0 + K − ε . (75) From such assumptions and results, by direct computation, we may obtain det  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 0∂v ∗ 1  =  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 1)2  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂(v∗ 0)2  −  ∂2J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) ∂v ∗ 1 ∂v ∗ 0 2 > 0, (76) in D∗ × B∗. From such results, we may infer that J ∗ 1 (v∗ 1, v∗ 0, ˆz∗) is concave in (v∗ 1, v∗ 0) on D∗ × B∗, so that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (77) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) + 1 12αK2 3 −γ∇2  z∗ K  + 2ˆv∗ 0  z∗ K  − f 2 0,2 (78) 22 ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = Ku, we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) ≤ F1(u) + F2(u, ˆv∗ 0) − F3(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 ≤ sup v∗ 0 ∈Y ∗ {F1(u) + F2(u, v∗ 0) − F3(u)} + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 = J(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f 2 0,2 , (79) ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + 1 12αK2 3 −γ∇2u + 2ˆv∗ 0u − f
2 0,2 = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 0, z∗) ) (80) The objective of this section is complete. 8 A duality principle for the complex Ginzburg- Landau system Let Ω ⊂ R3 be an open, bounded and connected set with a regular (Lipschitzian) boundary by ∂Ω. In this section, generically we denote, ⟨h1, h2⟩L2 = Re Z Ω h1, h∗ 2 dx  , ∀h1, h2 ∈ L2(Ω; C) with similar notations for vectorial cases. Here, h∗ 2 denotes the complex conjugate of h2 and Re[z] denotes the real part of z ∈ C. 23 Now, define a complex Ginzburg-Landau type functional J : V → R, by J(u) = γ 2 Z Ω |∇u − iρAu|2 dx + α 2 Z Ω (u2 − β)2 dx − ⟨u, f⟩L2 + 1 8π Z Ω | Curl A − B0|2 dx, (81) Here, V1 = W 1,2 0 (Ω; C), V2 = W 1,2(Ω; R3), and V = V1 × V2. Moreover, γ > 0, α > 0, β > 0 and f ∈ L∞(Ω; C). We also denote Y = Y ∗ = L2(Ω), Y2 = Y ∗ 2 = L2(Ω; C) and Y1 = Y ∗ 1 = L2(Ω; C3). Here u ∈ W 1,2 0 (Ω; C) denotes the local density proportion of super-conducting electrons in the superconductive sample Ω . Also, A : W 1,2(Ω; R3) is a magnetic potential and B0 ∈ L2(Ω; R3) denotes an external magnetic field. Define K3 = 3, K ≫ max{K3, α, γ, β, 1/α, 1/γ, 1/β}. Define the functionals F1 : V → R, F2 : V1 × Y ∗ → R, F3 : V1 → R and F4 : V × Y ∗ 1 → R, by F1(u) = γ 2 Z Ω |∇u − iρAu|2 dx (82) F2(u, v∗ 0) = ⟨|u|2, v∗ 0⟩L2 + K 2 Z Ω |u|2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx. (83) F3(u) = K Z Ω |u|2 dx and F4(u, A, v∗ 1) = ⟨∇u − iρAu, v∗ 1⟩L2 + K 2 Z Ω |u|2 dx − ⟨u, f⟩L2. (84) 24 Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 8√ K}, D∗ 1 = n v∗ 1 ∈ Y ∗ 1 : ∥v∗ 1∥∞ ≤ 8√ K o D∗ 2 = {v∗ 2 ∈ Y ∗ 2 : ∥v∗ 2∥∞ ≤ (3/2)KK3} E∗ = {z∗ ∈ Y ∗ 2 : ∥z∗∥∞ ≤ 2KK3, in Ω} and assuming the Gauge of London, E1 = {A ∈ V2 : div A = 0, in Ω and A · n = 0, on ∂Ω}, Moreover, define the polar functionals F ∗ 1 : Y ∗ 1 → R, F ∗ 2 : Y ∗ 2 × Y ∗ × Y ∗ 2 → R, F ∗ 3 : Y ∗ 2 → R and F ∗ 4 : Y ∗ 2 × Y ∗ 1 × Y ∗ 2 × V2 → R by F ∗ 1 (v∗ 1) = sup u∈V  ⟨w, v∗ 1⟩L2 − γ 2 Z Ω |w|2 dx  = 1 2γ Z Ω |v∗ 1|2 dx (85) F ∗ 2 (v∗ 2, v∗ 0, z∗) = sup u∈V {⟨u, −v∗ 2 + z∗/2⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω | − v∗ 2 + z∗/2|2 2v∗ 0 + K dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (86) if v∗ 0 ∈ B∗, F ∗ 3 (z∗) = sup w∈L2 {⟨w, z∗⟩L2 − F3(w)} = 1 4K Z Ω |z∗|2 dx. (87) and F ∗ 4 (v∗ 2, v∗ 1, z∗, A) = sup u∈V {⟨u, v∗ 2 + z∗/2⟩L2 − F2(u, A, v∗ 1)} = 1 2 Z Ω |v∗ 2 + z∗/2 + div v∗ 1 + iρA · v∗ 1 + f |2 K dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (88) Furthermore, defining F5(A) = 1 8π Z Ω | Curl A − B0|2 dx, 25 define also the functional J ∗ : D∗ 1 × D∗ 2 × B∗ × E∗ × E1 → R by J ∗(v∗ 2, v∗ 1, v∗ 0, z∗, A) = −F ∗ 1 (v∗ 1) − F ∗ 2 (v∗ 2, v∗ 0, z∗) + F ∗ 3 (z∗) − F ∗ 4 (v∗ 2, v∗ 1, z∗, A) + F5(A). and the exactly penalized functional J ∗ 1 : D∗ 1 × D∗ 2 × B∗ × E∗ × E1 → R by J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, z∗, A) = J ∗(v∗ 2, v∗ 1, v∗ 0, z∗, A) + K1 2 − div v∗ 1 − iρAv∗ 1 + 2v∗ 0  z∗ 2K  − f 2 0,2 (89) Let (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) ∈ D∗ 1 × D∗ 2 × B∗ × E∗ × E1 be such that δJ ∗(ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ 2K ∈ V1, we obtain δJ(u0, ˆA) = 0, J(u0, A) = J ∗(ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA), and δJ ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = 0. Here K1 > 0 is the largest positive (in fact close) real constant such that J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, ˆz∗, ˆA) is concave in ( v∗ 2, v∗ 1, v∗ 0) in D∗ 1 × D∗ 2 × B∗. Observe that J ∗ 1 is quadratic in ( z∗, A). Here we assume K1 > 0 is also such that det ( ∂2J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) ∂(z∗) ∂A ) > 0. Therefore, since J ∗ 1 is quadratic in ( z∗, A), we obtain J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = inf (z∗,A)∈E∗×E1 J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, z∗, A). Also, from the previous mentioned concavity, J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = sup (v∗ 2 ,v∗ 1 ,v∗ 0)∈D∗ 2 ×D∗ 1 ×B∗ J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, ˆz∗, ˆA). From these previous results and a standard Saddle Point Theorem, we have got J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = sup (v∗ 2 ,v∗ 1 ,v∗ 0)∈D∗ 2 ×D∗ 1 ×B∗  inf (z∗,A)∈E∗×E1 J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, z∗, A)  = inf (z∗,A)∈E∗×E1 ( sup (v∗ 2 ,v∗ 1 ,v∗ 0)∈D∗ 2 ×D∗ 1 ×B∗ J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, z∗, A) ) (90) 26 Finally, observe that J(u0, ˆA) = J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) ≤ F1(u, A) + F2(u, ˆv∗ 0) − ⟨u, z∗⟩L2 + F ∗ 3 (z∗) + K 2 Z Ω |u|2 dx +F5(A) + K1 2 − div ˆv∗ 1 − iρAˆv∗ 1 + 2ˆv∗ 0  z∗ 2K  − f 2 0,2 (91) ∀u ∈ V1, z∗ ∈ E∗, A ∈ E1. In particular, for z∗ = 2Ku, we obtain J(u0, ˆA) = J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) ≤ F1(u, A) + F2(u, ˆv∗ 0) − F3(u) + K 2 Z Ω |u|2 dx +F5(A) + K1 2 ∥− div ˆv∗ 1 − iρAˆv∗ 1 + 2ˆv∗ 0u − f ∥2 0,2 ≤ sup v∗ 0 ∈Y ∗  F1(u) + F2(u, v∗ 0) − F3(u) + K 2 Z Ω |u|2 dx  +F5(A) + K1 2 ∥− div ˆv∗ 1 − iρAˆv∗ 1 + 2ˆv∗ 0u − f ∥2 0,2 = J(u, A) + K1 2 ∥− div ˆv∗ 1 − iρAˆv∗ 1 + 2ˆv∗ 0u − f ∥2 0,2 , (92) ∀u ∈ V1. Joining the pieces, we have got J(u0, ˆA) = inf (u,A)∈V1×E1  J(u) + K1 2 ∥− div ˆv∗ 1 − iρAˆv∗ 1 + 2ˆv∗ 0u − f ∥2 0,2  = J ∗ 1 (ˆv∗ 2, ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆA) = sup (v∗ 2 ,v∗ 1 ,v∗ 0)∈D∗ 2 ×D∗ 1 ×B∗  inf (z∗,A)∈E∗×E1 J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, z∗, A)  = inf (z∗,A)∈E∗×E1 ( sup (v∗ 2 ,v∗ 1 ,v∗ 0)∈D∗ 2 ×D∗ 1 ×B∗ J ∗ 1 (v∗ 2, v∗ 1, v∗ 0, z∗, A) ) (93) The objective of this section is complete. 9 An eighth duality principle Define K3 = 3 and K > 0, K1 > 0, K 2 > 0 such that K2 ≫ K1 ≫ K ≫ max{K3, α, γ, β, 1/α, 1/γ, 1/β}. Define the functionals F1 : V → R, F2 : V → R, F3 : V × Y ∗ → R and F4 : V × Y ∗ → R, by 27 F1(u) = γ 2 Z Ω ∇u · ∇u dx + K1 2 Z Ω u2 dx −⟨u, f⟩L2, (94) F2(u) = − K1 2 Z Ω u2 dx . (95) F3(u, v∗ 0) = K 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx and F4(u, v∗ 0) = −⟨u2, v∗ 0⟩L2 + K 2 Z Ω u2 dx Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 8√ K}, D∗ = {(v∗ 1, v∗ 2) ∈ Y ∗ × Y ∗ : ∥v∗ 1∥∞ ≤ (3/2)K1K3 and ∥v∗ 2∥∞ ≤ (3/2)K1K3} and E∗ = {z∗ ∈ Y ∗ : ∥z∗∥∞ ≤ (3/2)KK3 and z∗f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗]2 → R, F ∗ 2 : [Y ∗] → R, F ∗ 3 : [Y ∗]3 → R and F ∗ 4 : [Y ∗]2 → R, by F ∗ 1 (v∗ 1, z∗) = sup u∈V {⟨u, v∗ 1 + z∗⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + z∗ + f)2 −γ∇2 + K1 dx (96) F ∗ 2 (v∗ 2) = inf u∈V {⟨u, v∗ 2⟩L2 − F2(u)} = 1 2 Z Ω (v∗ 2)2 −K1 dx, (97) if v∗ 0 ∈ B∗, and F ∗ 3 (v∗ 1, v∗ 2, v∗ 0) = sup u∈V {⟨u, −v∗ 1 − v∗ 2⟩L2 − F3(u, v∗ 0)} = 1 2K Z Ω (v∗ 1 + v∗ 2)2 dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx. (98) 28 and F ∗ 4 (z∗) = sup u∈V {⟨u, z∗⟩L2 − F3(u, v∗ 0)} = 1 2 Z Ω (z∗)2 −2v∗ 0 + K dx. (99) Furthermore, define the functional J ∗ : D∗ × B∗ × E∗ → R by J ∗(v∗ 1, v∗ 2, v∗ 0, z∗) = −F ∗ 1 (v∗ 1, z∗) − F ∗ 2 (v∗ 2) − F ∗ 3 (v∗ 1, v∗ 2, v∗ 0) + F ∗ 4 (z∗, v∗ 0). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ → R by J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, z∗) = J ∗(v∗ 1, v∗ 2, v∗ 0, z∗) − K2 1 6K − v∗ 1 + z∗ + f −γ∇2 + K1 + v∗ 1 + v∗ 2 K 2 0,2 − K2 K2 1 2 − v∗ 1 + v∗ 2 K + v∗ 2 −K1 2 0,2 (100) Let (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) ∈ D∗ × B∗ × E∗ be such that δJ ∗(ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = 0. From this and from the Legendre transform proprieties, for u0 = ˆv∗ 2 −K1 ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = 0. Observe that for K2 ≫ K1 ≫ K > 0 as previously specified, we have that J ∗ 1 is concave in (v∗ 1, v∗ 2, v∗ 0) and convex in z∗, in D∗ × B∗ × E∗. Therefore, since J ∗ 1 is quadratic in z∗, we obtain J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = inf z∗∈E∗ J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, z∗). Moreover, from the mentioned concavity in ( v∗ 1, v∗ 2, v∗ 0), we have J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 2 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, ˆz∗). From these previous results and a standard Saddle Point Theorem, we have got 29 J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 2 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 2 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, z∗) ) (101) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) ≤ F1(u) − F ∗ 2 (ˆv∗ 2) − ⟨u, ˆv∗ 2⟩L2 − ⟨u, z∗⟩L2 + F3(u, ˆv∗ 0) + F ∗ 4 (z∗, ˆv∗ 0) (102) ∀u ∈ V1, z∗ ∈ E∗. In particular, for z∗ = −2ˆv∗ 0u + Ku and recalling that u0 = ˆv∗ 2 −K1 , we obtain J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) ≤ F1(u) − F ∗ 2 (ˆv∗ 2) + ⟨u, ˆv∗ 2⟩L2 +F3(u, ˆv∗ 0) − F4(u, ˆv∗ 0) ≤ F1(u) − F ∗ 2 (ˆv∗ 2) + ⟨u, ˆv∗ 2⟩L2 + sup v∗ 0 ∈Y ∗ {F3(u, v∗ 0) − F4(u, v∗ 0)} = J(u) + K1 2 ∥u − u0∥2 0,2 (103) ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + K1 2 ∥u − u0∥2 0,2  = J ∗ 1 (ˆv∗ 1, ˆv∗ 2, ˆv∗ 0, ˆz∗) = sup (v∗ 1 ,v∗ 2 ,v∗ 0)∈D∗×B∗  inf z∗∈E∗ J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, z∗)  = inf z∗∈E∗ ( sup (v∗ 1 ,v∗ 2 ,v∗ 0)∈D∗×B∗ J ∗ 1 (v∗ 1, v∗ 2, v∗ 0, z∗) ) (104) The objective of this section is complete. 10 A ninth duality principle Define K3 = 3 and let K > 0, K1 > 0 be such that K1 ≫ K ≫ max{K3, α, γ, β, 1/α, 1/γ, 1/β}. 30 Let 0 < ε ≪ max{1/K, α, γ}. Define the functionals F1 : V → R, F2 : V × Y ∗ → R, F3 : V → R and F4 : V → R, by F1(u) = γ 2 Z Ω ∇u · ∇u dx + K 2 Z Ω u2 dx −⟨u, f⟩L2, (105) F2(u, v∗ 0) = ⟨u2, v∗ 0⟩L2 + K 2 Z Ω u2 dx − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx, (106) F3(u) = (K1 + K) 2 Z Ω u2 dx (107) and F4(u) = (−K1 + K) 2 Z Ω u2 dx Define also, V1 = {u ∈ V : ∥u∥∞ ≤ K3 and uf ≥ 0, in Ω}, B∗ = {v∗ 0 ∈ Y ∗ : ∥2v∗ 0∥∞ ≤ 8√ K}, D∗ = {v∗ 1 ∈ Y ∗ : ∥v∗ 1∥∞ ≤ (3/2)K1K3} and E∗ 1 = {z∗ ∈ Y ∗ : ∥z∗∥∞ ≤ KK3 and z∗f ≥ 0, in Ω}. E∗ 1 = {z∗ 1 ∈ Y ∗ : ∥z∗ 1∥∞ ≤ K1K3 and z∗ 1f ≥ 0, in Ω}. Moreover, define the polar functionals F ∗ 1 : [Y ∗]2 → R, F ∗ 2 : [Y ∗]3 → R, F ∗ 3 : [Y ∗]2 → R and F ∗ 4 : [Y ∗]2 → R, by F ∗ 1 (v∗ 1, z∗) = sup u∈V {⟨u, v∗ 1 + z∗⟩L2 − F1(u)} = 1 2 Z Ω (v∗ 1 + z∗ + f)2 −γ∇2 + K dx (108) F ∗ 2 (v∗ 1, v∗ 0, z∗) = inf u∈V {⟨u, −v∗ 1 + z∗⟩L2 − F2(u, v∗ 0)} = 1 2 Z Ω (−v∗ 1 + z∗)2 2v∗ 0 + K dx + 1 2α Z Ω (v∗ 0)2 dx + β Z Ω v∗ 0 dx, (109) 31 if v∗ 0 ∈ B∗, and F ∗ 3 (z∗, z∗ 1) = sup u∈V {⟨u, z∗ + z∗ 1⟩L2 − F3(u)} = 1 2(K1 + K) Z Ω (z∗ + z∗ 1)2 dx (110) and F ∗ 4 (z∗, z∗ 1) = inf u∈V {⟨u, z∗ − z∗ 1⟩L2 − F4(u)} = 1 2(−K∗ 1 + K) Z Ω (z∗ − z∗ 1)2 dx. (111) Furthermore, define the functional J ∗ : D∗ × B∗ × E∗ 1 × E∗ 2 → R by J ∗(v∗ 1, v∗ 0, z∗, z∗ 1) = −F ∗ 1 (v∗ 1, z∗) − F ∗ 2 (v∗ 1, v∗ 0, z∗) + F ∗ 3 (z∗, z∗ 1) + F ∗ 4 (z∗, z∗ 1). and the exactly penalized functional J ∗ 1 : D∗ × B∗ × E∗ 1 × E∗ 2 → R by J ∗ 1 (v∗ 1, v∗ 0, z∗, z∗ 1) = J ∗(v∗ 1, v∗ 0, z∗, z∗ 1) +(K2 1 − K2 − ε) 4K z∗ + z∗ 1 K1 + K − z∗ − z∗ 1 −K1 + K 2 0,2 . (112) Let (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) ∈ D∗ × B∗ × E∗ 1 × E∗ 2 be such that δJ ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = 0. From this and from the Legendre transform proprieties, for u0 = ˆz∗ + ˆz∗ 1 K1 + K = ˆz∗ 1 K1 = ˆz∗ K ∈ V1, we obtain δJ(u0) = 0, J(u0) = J ∗(ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1), and δJ ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = 0. Observe that for K1 ≫ K > 0 and ε > 0 as previously specified, we have ∂2J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) ∂(z∗)2 = − 1 −γ∇2 + K − 1 2ˆv∗ 0 + K + 1 K1 + K + 1 −K1 + K + (K2 1 − K2 − ε) 2K  1 K1 + K + 1 K1 − K 2 = − 1 −γ∇2 + K − 1 2ˆv∗ 0 + K + 1 K1 + K + 1 −K1 + K + (K2 1 − K2 − ε) 2K  2K1 K2 1 − K2 2 > 0. (113) 32 From such a result, since J ∗ 1 is quadratic in z∗, we have J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = inf z∗∈E1 J ∗ 1 (ˆv∗ 1, ˆv∗ 0, z∗, ˆz∗ 1). On the other hand, ∂2J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) ∂(z∗ 1)2 = + 1 K1 + K + 1 −K1 + K + (K2 1 − K2 − ε) 2K  1 K1 + K − 1 K1 − K 2 = − 2K K2 1 − K2 + (K2 1 − K2 − ε) 2K  2K K2 1 − K2 2 = − 2K K2 1 − K2 + 2K(K2 1 − K2 − ε) (K2 1 − K2)2 < 0, (114) so that J ∗ 1 is concave in ( v∗ 1, v∗ 0, z∗ 1) in D∗ × B∗ × E∗ 1 × E∗ 2. Thus, we have obtained J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = sup (v∗ 1 ,v∗ 0 ,z∗ 1)∈D∗×B∗×E∗ 2 J ∗ 1 (v∗ 1, v∗ 0, ˆz∗, z∗ 1). From such results and a standard Saddle Point Theorem we may infer that J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = inf z∗∈E1 ( sup (v∗ 1 ,v∗ 0 ,z∗ 1)∈D∗×B∗×E∗ 2 J ∗ 1 (v∗ 1, v∗ 0, z∗, z∗ 1) ) = sup (v∗ 1 ,v∗ 0 ,z∗ 1)∈D∗×B∗×E∗ 2  inf z∗∈E1 J ∗ 1 (v∗ 1, v∗ 0, z∗, z∗ 1)  . (115) Finally, observe that J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) ≤ F1(u) + F2(u, ˆv∗ 0) − 2⟨u, z∗⟩L2 + F ∗ 3 (z∗ + ˆz∗ 1) + ⟨u, z∗ − ˆz∗ 1⟩L2 − F4(u) + K2 1 2 z∗ + ˆz∗ 1 K1 + K − z∗ − ˆz∗ 1 −K1 + K 2 0,2 , (116) ∀u ∈ V1, z∗ ∈ E∗ 1 In particular, for z∗ ∈ E∗ 1 such that z∗ + ˆz∗ 1 = (K + K1)u and recalling that ˆz∗ 1 = K1u0, we obtain 33 J(u0) = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) ≤ F1(u) + F2(u, ˆv∗ 0) + F ∗ 3 (z∗ + ˆz∗ 1) − ⟨u, z∗ + ˆz∗ 1⟩L2 − F4(u) +(K2 1 − K2 − ε) 4K z∗ + ˆz∗ 1 K1 + K − z∗ − ˆz∗ 1 −K1 + K 2 0,2 = γ 2 Z Ω ∇u · ∇u dx + 4K2 1(K2 1 − K2 − ε) 4K(−K1 + K)2 ∥u − u0∥2 0,2 − ⟨u, f⟩L2 +⟨u2, ˆv∗ 0⟩L2 − 1 2α Z Ω (ˆv∗ 0)2 dx − β Z Ω ˆv∗ 0 dx ≤ γ 2 Z Ω ∇u · ∇u dx − ⟨u, f⟩L2 + 4K2 1(K2 1 − K2 − ε) 4K(−K1 + K)2 ∥u − u0∥2 0,2 + sup v∗ 0 ∈Y ∗  ⟨u2, v∗ 0⟩L2 − 1 2α Z Ω (v∗ 0)2 dx − β Z Ω v∗ 0 dx  = γ 2 Z Ω ∇u · ∇u dx + α 2 Z Ω (u2 − β)2 dx − ⟨u, f⟩L2 +4K2 1(K2 1 − K2 − ε) 4K(−K1 + K)2 ∥u − u0∥2 0,2 = J(u) + 4K2 1(K2 1 − K2 − ε) 4K(−K1 + K)2 ∥u − u0∥2 0,2, (117) ∀u ∈ V1. Joining the pieces, we have got J(u0) = inf u∈V1  J(u) + 4K2 1(K2 1 − K2 − ε) 4K(−K1 + K)2 ∥u − u0∥2 0,2  = J ∗ 1 (ˆv∗ 1, ˆv∗ 0, ˆz∗, ˆz∗ 1) = inf z∗∈E1 ( sup (v∗ 1 ,v∗ 0 ,z∗ 1)∈D∗×B∗×E∗ 2 J ∗ 1 (v∗ 1, v∗ 0, z∗, z∗ 1) ) = sup (v∗ 1 ,v∗ 0 ,z∗ 1)∈D∗×B∗×E∗ 2  inf z∗∈E1 J ∗ 1 (v∗ 1, v∗ 0, z∗, z∗ 1)  . (118) The objective of this section is complete. 11 Conclusion In this article, through a D.C. approach, we have developed duality principles and related convex dual variational formulations suitable for an originally non-convex primal ones. As a first application, we have set a duality principle and respective convex dual formulation for a Ginzburg-Landau type equation. We highlight the results here obtained are applicable to a large class of models in the calculus of variations, including some plate and shell non-linear theories, other models in superconduc- tivity, phase transition and micro-magnetism, among many others. 34 In a near future research we intend to apply such results to some of these mentioned related models. 1. Conflict of interest declaration: The author declares no conflict of interest concerning this article.

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