ON D. Y. GAO AND R. W. OGDEN’S PAPER “MULTIPLE SOLUTIONS TO NON-CONVEX VARIATIONAL PROBLEMS WITH IMPLICATIONS FOR PHASE 1 TRANSITIONS AND NUMERICAL COMPUTATION” 1 0 2 by M. D. VOISEI n (Towson University, Department of Mathematics, Towson, MD-21252, U.S.A.) a J C. ZA˘LINESCU 8 ‡ 1 (University “Al.I.Cuza” Ia¸si, Faculty of Mathematics, 700506-Ia¸si, Romania, and Institute of Mathematics Octav Mayer, Ia¸si, Romania) ] C [Received 19 February 2010. Revise 19 February 2010] O . h t Summary a m In this note we prove that a recent result stated by D. Y. Gao and R. W. Ogden on global minimizers and local extrema in a phase transition problem is false. Ourgoal [ isachievedbyprovidingathoroughanalysisofthecontextandresultinquestionand 1 counter-examples. v 4 3 5 1. Introduction 3 The optimization problem we have in focus is introduced on (5, p. 505) where one says . 1 “The primal variational problem (1.1) for the soft device can be written in the form 0 1 1 ( ) : min P (u)= 1µu2 + 1ν 1u2 αu 2 dx F(u) , (3.2)” :1 whePrse (seue∈(U5s,(cid:26)p.s501)) Z0 h2 x 2 (cid:0)2 x− x(cid:1) i − (cid:27) v 1 i X “F(u)= fudx+σ1u(1) (2.8)”, Z0 r and (see (5, p. 505)) a “ = u (0,1) u 4(0,1), u(0)=0 . (3.1)” s x U ∈L | ∈L In Section 2 we explain the natural interpretation for the definition of . (cid:8) (cid:9) s U As mentionedon(5,p.498),“µ, ν andαarepositive materialconstants”,and“wefocus mainly on the case for which να2 > 2µ” (see (5, p. 499)). Moreover (see (5, p. 498)), “To make the mixing of phases more dramatic, we introduce a distributed axial loading (body force) f [0,1] per unit length of I”. These assumptions will be in force throughout this ∈C article. Therefore, from 1 “σ(x)= f(s)ds+σ (2.12)” 1 Zx one obtains that σ 1[0,1] and ∈C ‡ ([email protected]) Q. Jl Mech. Appl. Math. (1999) 52 (1), 1–16 (cid:13)c Oxford University Press 1999 2 m. d. voisei ET AL. 1 “F(u)= σ(x)u dx. (2.13)” x Z0 Furthermore (see (5, p. 501)), one says “... we obtain the Gao–Strang total complementary energy Ξ(u,ζ) (16) for this non-convex problem in the form 1 1 Ξ(u,ζ)= = 1u2(ζ +µ) αu ζ 1ν 1ζ2 dx fudx σ u(1), (2.7)”. ··· 2 x − x − 2 − − − 1 Z0 Z0 In the text above (1(cid:2)6) is our reference (6). (cid:3) In (5, pp. 501,502)one obtains “the so-calledpure complementary energy functional (7, 17) 1 1 (σ+αζ)2 Psd(ζ)=−2 µ+ζ +ν−1ζ2 dx, (2.14) Z0 (cid:18) (cid:19) which is well defined on the dual feasible space = ζ 2 ζ(x)+µ=0, ζ(x)> 1να2, x [0,1] .” Sa ∈L | 6 −2 ∀ ∈ References (7, 17) above are our references (1) and (2). (cid:8) (cid:9) Probably, by “well defined on ... ” the authors of (5) mean that Pd(ζ) R for every Sa s ∈ ζ . Note that a ∈S (σ+αζ)2 β2 +ν 1ζ2 = +2αβ+α2(ζ+µ)+ν 1ζ2, (1.1) − − µ+ζ ζ+µ where (see (5, p. 502)) “β(x)=σ(x) µα, η =(να2 2µ)3/27ν. (2.21)” − − and β 1[0,1]. Let us set B := s [0,1] β(s)=0 , Bc :=[0,1] B . ∈C 0 { ∈ | } 0 \ 0 Let ζ 2 := 2[0,1] and set E := x [0,1] ζ(x)+µ = 0 . In the sequel we use ζ ∈ L L { ∈ | } the convention 0/0 := 0, which agrees with the convention 0 ( ) := 0 used in measure theory. With this convention in mind, from (1.1), we obtain t·ha±t∞Pd(ζ) R if and only if β2 1 := 1[0,1] which implicitly provides that β2 is well-definsed a∈lmost everywhere ζ+µ ∈ L L ζ+µ (a.e. for short), i.e., E B is negligible. ζ 0 \ Consider β2 A := ζ 2 1 A := ζ 2 ζ(x)+µ=0 for a.e. x Bc . 1 ∈L | ζ+µ ∈L ⊂ 2 ∈L | 6 ∈ 0 (cid:26) (cid:27) (cid:8) (cid:9) The setA isthe greatestsubsetof 2 forwhichPd(ζ) R. Noticethatζ 2 makes β2 1 L s ∈ ∈L ζ+µ be well-defined iff ζ A . Also, note that A . 2 a 2 Denote by λ the L∈ebesgue measure on RS. F⊂or ζ A we have 1 ∈ 1 (σ+αζ)2 1 Pd(ζ)= +ν 1ζ2 dx ν 1µ2λ(E ). (1.2) s −2 µ+ζ − − 2 − ζ Z[0,1]\Eζ(cid:18) (cid:19) Notice that in the trivial case β =0 we have A =A = 2 and so Pd is well-defined on 1 2 L s because in this case Pd is well-defined on 2. Sa s L Proposition 1.1. If β =0 then A and Pd is not well-defined on . 6 Sa 6⊂ 1 s Sa Proof. Because β = 0 and β 1[0,1], there exist γ > 0 and 0 6 a < b 6 1 such that 6 ∈ C β2(x)>γ foreveryx [a,b]. Considerζ(x):=x a µforx (a,b)andζ(x):=1 µfor ∈ − − ∈ − about global minimizers and local extrema in phase transition 3 x [0,1] (a,b). Then ζ(x) > µ> 1να2 for every x [0,1] and ζ 2; hence ζ . ∈ \ − −2 ∈ ∈L ∈Sa Note that ζ / A since β2 γ >0 and 1 1 dx b dx =+ . ∈ 1 ζ+µ ≥ ζ+µ 0 ζ(x)+µ ≥ a x a ∞ − In the sequel Pd is understood as being dRefined on A . R s 1 Assume for the rest of this section that β = 0. This yields that λ(Bc) > 0 since β is 6 0 continuous. The next result is surely known. We give the proof for easy reference. Lemma 1.2. Assume that α < , where (α ) [0, ). Then there exists a n>1 n ∞ n n>1 ⊂ ∞ non-decreasing sequence (β ) (0, ) with β and α β < . Pn n>1 ⊂ ∞ n →∞ n>1 n n ∞ Proof. Because the series n>1αn is convergent, the sequencPe (Rn) converges to 0, where Rn := ∞k=n+1αn. HencPe there exists an increasing sequence (nk)k>1 ⊂ N∗ such that R < 2 k for all k > 1 and n > n . Consider β := 1 for n 6 n and β := k for n P− k n 1 n n <n6n . Clearly, (β ) is non-decreasing and limβ = . Moreover, k k+1 n n ∞ nm+1 n1 m nk+1 n1 m nk+1 α β = α + α β 6 α + k α p p p p p p p Xp=1 Xp=1 Xk=1p=Xnk+1 Xp=1 kX=1 p=Xnk+1 m ∞ ∞ ∞ 6 α + kR 6 α + k2 k < . p nk p − ∞ p=1 k=1 p=1 k=1 X X X X Therefore, the series α β is convergent. n>1 n n Let us denote the aPlgebraic interior (or core) of a set by “core”. Proposition 1.3. Assume that β = 0. Then coreA is empty. In particular, coreA = 2 1 6 core = . a S ∅ Proof. Let ζ A be fixed. Then there exists a sequence (B ) of pairwise disjoint 2 n n>1 ∈ Lebesguemeasurablesets(evenintervals)suchthatBc = B andλ(B )>0forn>1 0 ∪n>1 n n 2 2 (see e.g. (7, p. 42)). We have that ζ(x)+µ dx= ζ(x)+µ dx< , and n>1 Bn B0c ∞ so,fromthe previouslemma,there existsa non-decreasingsequence(β ) (0, )with P R (cid:12) (cid:12) R (cid:12) n n>1(cid:12)⊂ ∞ β and (cid:12) (cid:12) (cid:12) (cid:12) n →∞ 2 β ζ(x)+µ dx< . (1.3) n ∞ n>1 ZBn X (cid:12) (cid:12) (cid:12) (cid:12) Define u : [0,1] R by u(x) := √β (ζ(x)+µ) for x B and u(x) := 0 for x B . n n 0 → − ∈ ∈ From (1.3) we have that u 2. Moreover, for every δ >0 there exists a sufficiently large ∈L N >1 such that t=βN−1/2 ∈(0,δ) and ζ+tu∈/ A2; this happens because BN ⊂{x∈B0c | 1/2 ζ(x)+βN− u(x)+µ=0}andλ(BN)>0. Weprovedthatζ 6∈coreA2. HencecoreA2 =∅. Onpage502of(5)itissaidthat“Thecriticalityconditionwith respectto ζ leadsto the ... ‘dual algebraic equation’ (DAE) for ... (2.14) ..., namely 2ν 1ζ+α2 (µ+ζ)2 =(σ µα)2. (2.16)” − − To our knowledge, one can speak about Gˆateaux differentiability of a function f : E (cid:0) (cid:1) ⊂ 4 m. d. voisei ET AL. X Y, with X,Y topological vector spaces, at x E only if x is in the core of E. As we hav→e seen above, Pd(ζ) R only for ζ A and co∈reA = . s ∈ ∈ 1 1 ∅ So what is the precise critical point notion for Pd so that, when using that notion, one s gets (5, (2.16)), other than just formal computation? Taking into account the comment (see (5, p. 502)) “It should be pointed out that the integrand in each of Pd(ζ) and Pd(ζ) has a singularity at ζ = µ, which explains the s h − exclusion ζ = µ in the definition of ”, we must point out that there is an important a 6 − S difference between the condition ζ = µ (as measurable functions) and ζ(x) = µ a.e. on 6 − 6 − [0,1] since it is known that ζ = µ means that ζ(x)= µ on a set of positive measure. 6 − 6 − Alternatively, from the above considerations, 2 µ is a (nonempty) open set, while L \{− } the set A := ζ 2 λ(E )=0 has, as previously seen, empty core (in particular has 3 ζ ∈L | empty interior). (cid:8) (cid:9) The quoted text from (5, p. 502) continues with: “In fact, it turns out that, in general, ζ = µ does not correspondto a critical point of either Pd(ζ) or Pd(ζ). Exceptionally, we − s h mayhaveζ(x)= µforsomex (0,1),but thisis alwaysassociatedwithσ(x)=µα. Itis − ∈ therefore important to note that when (2.16) holds, the integrand in (2.14) and (2.15) can be written as 2α(σ+αζ)+ν 1ζ(3ζ +2µ), (2.17) − andwhenζ = µ (andσ =µα) this reduces toν 1µ2, andthe singularityin the integrand − − is thus removed.” This shows that the convention we used (namely 0/0=0), our interpretation for Pd(ζ), s and formula (1.2) are in agreement with the authors of (5) point of view. 2. Problem reformulation Everyuin isrepresentedbyanabsolutelycontinuousfunctionon[0,1]withu(0)=0and s U u 4(0,1). More accurately, = u W1,4(0,1) u(0)=0 . In a different notation, x s ∈ L U ∈ | denoting by p the space p[0,1], we have L L (cid:8) (cid:9) x u v 4, x [0,1]:u(x)= v(t)dt. s ∈U ⇐⇒∃ ∈L ∀ ∈ Z0 So, the problem ( ) above becomes s P 1 ( ): minP (v)= 1µv2+ 1ν 1v2 αv 2 σv dx Ps v∈L4 s Z0 h2 2 (cid:0)2 − (cid:1) − i b b and Ξ becomes 1 Ξ(v,ζ)= 1v2(ζ+µ) αvζ 1ν 1ζ2 σv dx (v 4, ζ 2). (2.1) 2 − − 2 − − ∈L ∈L Z0 (cid:2) (cid:3) b x Note that P (u)=P (v), Ξ(u,ζ)=Ξ(v,ζ), for u(x)= v(t)dt, x [0,1]. s s 0 ∈ R b b about global minimizers and local extrema in phase transition 5 It is easy to see that P and Ξ are Fr´echet differentiable and s 1 dP (vb)(h)=b µv+ν 1v2 αv (v α) σ hdx, s 2 − − − Z0 (cid:2) 1(cid:0) (cid:1) (cid:3) b dΞ(,ζ)(v)(h)= [v(ζ +µ) αζ σ]hdx, · − − Z0 1 dbΞ(v, )(ζ)(k)= 1v2 αv ν 1ζ kdx, · 2 − − − Z0 (cid:2) (cid:3) for v,h 4 and ζ,k b2. Therefore, ∈L ∈L P (v)=µv+ν 1v2 αv (v α) σ 4/3, ∇ s 2 − − − ∈L Ξ(,ζ)(v)=(cid:0)v(ζ+µ) (cid:1) αζ σ 4/3, (2.2) b∇ · − − ∈L ∇bΞ(v,·)(ζ)= 12v2−αv−ν−1ζ ∈L2. Moreover, b 1 d2P (v)(h,k)= µ+ν 3v2 3αv+α2 hkdx (v,h,k 4). (2.3) s 2 − ∈L Z0 (cid:2) (cid:0) (cid:1)(cid:3) b Hence v 4 is a critical point of P if and only if s ∈L µv+ν 1v2 αv (v α) σ =0, (2.4) b2 − − − (cid:0) (cid:1) and (v,ζ) 4 2 is a critical point of Ξ if and only if ∈L ×L v(ζ +µ) αζ σ =0, 1v2 αv ν 1ζ =0. (2.5) − − b 2 − − − From the expression of Ξ we observe that Ξ(v, ) is concave on 2 for every v 4; · L ∈ L furthermore, Ξ(,ζ) is convex (concave) for those ζ 2 with ζ > µ (ζ 6 µ). · ∈L − − b b Lemma 2.1. Let v 4 and set b ∈L ζ :=ν 1v2 αv . (2.6) v 2 − (cid:0) (cid:1) Then ζ 2, dΞ(v,.)(ζ)=0 iff ζ =ζ , and v v ∈L b sup Ξ(v,ζ)=Ξ(v,ζv)=Ps(v). (2.7) ζ 2 ∈L b b b Proof. The facts thatfor (v,ζ) 4 2 we haveζ =ζ iffdΞ(v,.)(ζ)=0andζ 2 are v v ∈L ×L ∈L straightforward. Equality (2.7) is due to the fact that every critical point (namely ζ =ζ ) v of a concave function (namely Ξ(v, )) is a global maximum pboint of that function. · Consider the set b β σ αµ A := ζ 2 4 = ζ 2 − 4 , 0 ∈L | ζ+µ ∈L ∈L | ζ+µ ∈L (cid:26) (cid:27) (cid:26) (cid:27) 6 m. d. voisei ET AL. More precisely, ζ A iff ζ 2, E B , and β 4([0,1] E ). ∈ 0 ∈L ζ ⊂ 0 ζ+µ ∈L \ ζ For ζ 2 with E B set ζ 0 ∈L ⊂ σ+αζ β v := =α+ . (2.8) ζ ζ+µ ζ+µ More precisely v (x)=α+ β(x) for x [0,1] E and v (x)=α for x E . Notice that ζ ζ(x)+µ ∈ \ ζ ζ ∈ ζ ζ A iff v 4. 0 ζ ∈ ∈L In the sequel χ denotes the characteristic function of E [0,1], that is, χ (x) =1 for E E ⊂ x E and χ (x)=0 for x [0,1] E. E ∈ ∈ \ Lemma 2.2. For all ζ A and v 4 we have that dΞ(,ζ)(v +χ v) = 0 and Ξ(v + ∈ 0 ∈ L · ζ Eζ ζ χ v,ζ)=Pd(ζ). Eζ s b b Proof. According to (2.2), we have dΞ(,ζ)(v +χ v)=(v +χ v)(ζ +µ) αζ σ =χ v(ζ+µ)=0 ζ A , v 4. · ζ Eζ ζ Eζ − − Eζ ∀ ∈ 0 ∈L Since ζ A we havev =α andσ =αµ onE . Takinginto account(2.1), (1.2)andusing b 0 ζ ζ ∈ that outside E we have v2(ζ+µ)=(σ+αζ)v = (σ+αζ)2, we get ζ ζ ζ ζ+µ (σ+αζ)2 Ξ(v +χ v,ζ)= 1 +ν 1ζ2 dx ζ Eζ − 2 µ+ζ − Z[0,1]\Eζ(cid:18) (cid:19) b + αµ(α+v)− 12ν−1µ2−σ(α+v) dx=Psd(ζ). ZEζ (cid:0) (cid:1) In particular every ζ A is in the domain of Pd, that is, A A (which can be ∈ 0 s 0 ⊂ 1 observeddirectly,toosinceβ ). TheargumentaboveshowsthatΞ(,ζ)hasnocritical ∞ ∈L · points if ζ 2 A (due to the lack of regularity) and Ξ(,ζ) has an infinity of critical 0 ∈ L \ · points of the form v +χ v with v 4, if ζ A and λ(E )>0. b ζ Eζ ∈L ∈ 0 ζ Furthermore,forζ A ,ifζ+µ>0(ζ+µ60)thenv ibsaglobalminimum(maximum) 0 ζ ∈ pointofΞ(,ζ)becauseΞ(,ζ)isconvex(concave)andv isacriticalpointofΞ(,ζ). Hence ζ · · · b b inf Ξ(v,ζ) if ζ A and ζ > µ, b Psd(ζ)=( supvv∈∈LL44bΞ(v,ζ) if ζ ∈∈A00 and ζ 6−−µ. (2.9) Theorem 2.3. b (i)Let(v,ζ) 4 2beacriticalpointofΞ. Thenζ =ζ,v =(1 χ )v+αχ 4, ∈L ×L v ζ − Eζ Eζ ∈L visacriticalpointofP ,ζ A ,P (v)=Ξ(v,ζ)=Pd(ζ), 2ν 1ζ+α2 (µ+ζ)2 =(σ µα)2 s ∈ 0 s b s − − (i.e. ζ satisfies (5, (2.16))), and (cid:0) (cid:1) b b b 1 d2P (v)(h,k)=3 ζ ρ hkdx (2.10) s − Z0 (cid:0) (cid:1) for h,k 4, where b ∈L ρ:= 1 µ+να2 . (2.11) −3 (cid:0) (cid:1) about global minimizers and local extrema in phase transition 7 If, in addition, ζ > µ then − sup inf Ξ(v,ζ)= inf Ξ(v,ζ)=Ξ(v,ζ)=P (v)= inf P (v)=Pd(ζ)= sup Pd(ζ). s s s s ζ∈L2v∈L4 v∈L4 v∈L4 ζ∈A0,ζ>−µ (2.12) b b b b b In particular v is a global minimum of P on 4. s L (ii) If v 4 is a critical point of P then (v,ζ ) 4 2 is a critical point of Ξ. s v ∈L b ∈L ×L (iii) Assume that ζ is a measurable solution of 2ν 1ζ+α2 (µ+ζ)2 = (σ µα)2 and − b − b v 4. Then: ∈L (cid:0) (cid:1) (a) ζ A and (v ,ζ) 4 2. Moreover, 0 ζ ∞ ∞ ∈ ∈L ×L ⊂L ×L P (v +vχ )=Pd(ζ)+ 1ν (v2 α2+2ν 1µ)2dx (2.13) s ζ Eζ s 8 − − ZEζ b and (v +vχ ,ζ) is a critical point of Ξ iff P (v +vχ )=Pd(ζ) iff ζ Eζ s ζ Eζ s v2 α2+b2ν−1µb=0 a.e. in Eζ. (2.14) − In particular, (v ,ζ) is a critical point of Ξ iff λ(E )=0. ζ ζ (b) v +vχ is a critical point of P iff ζ Eζ s b v v2 α2b+2ν−1µ =0 a.e. in Eζ. − Proof. (i) Assume that (v,ζ)(cid:0) 4 2 is a c(cid:1)ritical point of Ξ. From (2.5) we see that ∈ L ×L ζ = ζ , v = (1 χ )v+αχ 4 which provides ζ A , v is a critical point of P , v ζ − Eζ Eζ ∈ L ∈ 0 s and 2ν 1ζ+α2 (µ+ζ)2 = (σ µα)2. Note that v +χ (bv α) = v. The equality − − ζ Eζ − b P (v)=Ξ(v,ζ)=Pd(ζ) is a consequence of Lemmas 2.1, 2.2. s (cid:0) (cid:1) s Taking into account (2.3) and the second equation in (2.5) we obtain that for h,k 4, ∈L b b 1 1 d2P (v)(h,k)= µ+ν 3ν 1ζ+α2 hkdx=3 ζ ρ hkdx. s − − Z0 Z0 (cid:2) (cid:0) (cid:1)(cid:3) (cid:0) (cid:1) Assume,inabddition,thatζ > µ. ThereforeΞ(,ζ)isconvexandPd(ζ)=inf Ξ(v,ζ) − · s v∈L4 (see (2.9)). Since v is a critical point it yields that v is a global minimum point of Ξ(,ζ). · Similarly, ζ is a global maximum point for the cboncave function Ξ(v, ). We get b · b Ξ(v,ζ)>Ξ(v,ζ)>Ξ(v,ζ) v 4, ζ b 2. ∀ ∈L ∀ ∈L This implies that b b b sup inf Ξ(v,ζ)> inf Ξ(v,ζ)=Ξ(v,ζ)= sup Ξ(v,ζ)> inf sup Ξ(v,ζ). ζ∈L2v∈L4 v∈L4 ζ∈L2 v∈L4ζ∈L2 b b b b b Since sup inf Ξ(v,ζ)6inf sup Ξ(v,ζ) (this happens for every function Ξ), ζ 2 v 4 v 4 ζ 2 we obtain∈toLgether∈Lwith (2.7) that ∈L ∈L b b b sup inf Ξ(v,ζ)= inf Ξ(v,ζ)=Ξ(v,ζ)=P (v)= inf P (v)=Pd(ζ). (2.15) s s s ζ∈L2v∈L4 v∈L4 v∈L4 b b b b b 8 m. d. voisei ET AL. In particular, v is a global minimum of P on 4. s L From (2.9) and (2.15) we have b Pd(ζ)= inf Ξ(v,ζ) sup inf Ξ(v,ζ) sup inf Ξ(v,ζ)=Pd(ζ). s v∈L4 ≤ζ∈A0,ζ>−µv∈L4 ≤ζ∈L2v∈L4 s The assertion (ii) fobllows directly from (2.4)band (2.5). b (iii) For given β 1[0,1] relation 2ν 1ζ+α2 (µ+ζ)2 = β2(x) (= (σ(x) µα)2) is a − ∈ C − polynomial equation in ζ. Let ζ : [0,1] R be such that ζ(cid:0)(x) is a solu(cid:1)tion of the previous equation for every → x [0,1], that is, ζ is a solution of (5, (2.16)). Because β2 is bounded (being continuous) ∈ we have that ζ is bounded. If, in addition, ζ is measurable then ζ 2. ∞ ∈L ⊂L (a) Note that, due to (5, (2.16)), E B and v = α+β/(µ+ζ) outside E whence ζ 0 ζ ζ ⊂ (v α)2 =2ν 1ζ+α2 ([0,1] E ). Thereforev 4.Thisshowsthatζ A . ζ − ∞ ζ ζ ∞ 0 − ∈L \ ∈L ⊂L ∈ Letv 4. Recallthatv +vχ =α+v,σ =αµ, ζ = µinsideE andv +vχ =v ∈L ζ Eζ − ζ ζ Eζ ζ outside E , and so ζ P (v +vχ )= 1µv2+ 1ν 1v2 αv 2 σv dx s ζ Eζ 2 ζ 2 2 ζ − ζ − ζ Z[0,1]\Eζh (cid:0) (cid:1) i b + 1µ(α+v)2+ 1ν 1(α+v)2 α(α+v) 2 αµ(α+v) dx. 2 2 2 − − ZEζh (cid:0) (cid:1) i (2.16) Taking into account that ζ(x) is a solution of the equation (5, (2.16)) and that for x [0,1] E one has ζ(x)+µ=0, one gets ζ ∈ \ 6 21µvζ2+ 21ν 21vζ2−αvζ 2−σvζ =−21(σζ++αµζ)2 − 21ν−1ζ2 on [0,1]\Eζ. (cid:0) (cid:1) A simple verification shows that 1µ(α+v)2+ 1ν 1(α+v)2 α(α+v) 2 αµ(α+v)= 1ν(v2 α2+2ν 1µ)2 1ν 1µ2. 2 2 2 − − 8 − − − 2 − Using the precedi(cid:0)ng equalities, from (2.(cid:1)16) and (1.2) we obtain that (2.13) holds. A direct computation shows that (v +vχ ,ζ) is a critical point of Ξ if and only if ζ Eζ v2 α2+2ν 1µ=0 a.e.in E . Therefore the mentioned equivalencies are true. Moreover, − ζ − because να2 >2µ the last equivalence holds, too. b (b) Similarly, v +vχ is a criticalpointofP ifand onlyif v(v2 α2+2ν 1µ)=0 a.e. ζ Eζ s − − in E . ζ b Note the following direct consequences of the previous theorem: if v 4 is a critical point of P , then (v,ζ ) is a critical point of Ξ, ζ 2 is a s v v • ∈ L ∈ L solution of (5, (2.16)), and P (v)=Ξ(v,ζ )=Pd(ζ ); s v s v ifζ isameasurablesolutionof(5,b(2.16))andv 4satisfies(2.14)thenbζ =ζ • ∈L (vζ+vχEζ) and v +vχ is a global mbinimumbof P on 4; ζ Eζ s L itispossiblev +vχ tobeacriticalpointofP without(v +vχ ,ζ)beingacritical • ζ Eζ s ζ Eζ b point of Ξ; such a situation happens when v =0 and λ(E )>0. ζ b b about global minimizers and local extrema in phase transition 9 3. Discussion of (5, Th. 3) Based on the above considerations we discuss the result in (5, Th. 3); for completeness we also quote its proof. Recall that “β(x)=σ(x) αµ, η =(να2 2µ)3/27ν. (2.21)” − − “Theorem 3. (Global minimizer and local extrema) Suppose that the body force f(x) anddeadloadσ aregivenandthatσ(x)isdefinedby(2.12). Then,ifβ2(x)>η, x (0,1), 1 ∀ ∈ the DAE (2.16) has a unique solution ζ(x) > µ, which is a global maximizer of Pd over − s , and the corresponding solution u(x) is a global minimizer of P (u) over , a s s S U P (u)= minP (u)=maxPd(ζ)=Pd(ζ). (3.9) s s s s u∈Us ζ∈Sa If β2(x) 6 η, x (0,1), then (2.16) has three real roots ordered as in (3.5). Moreover, ∀ ∈ ζ (x) is a global maximizer of Pd(ζ) over the domain ζ > µ, the corresponding solution 1 s − u (x) is a global minimizer of P (u) over and 1 s s U P (u )= minP (u)= max Pd(ζ)=Pd(ζ ). (3.10) s 1 s s s 1 u∈Us ζ>−µ Forζ (x)andζ (x),thecorrespondingsolutionsu (x)andu (x)are,respectively,alocal 2 3 2 3 minimizer and a local maximizer of P (u), s P (u )= minP (u)= min Pd(ζ)=Pd(ζ ) (3.11) s 2 s s s 2 u∈U2 ζ3<ζ<−µ and P (u )= maxP (u)= max Pd(ζ)=Pd(ζ ), (3.12) s 3 s s s 3 u∈U3 −21να2<ζ<ζ2 where is a neighborhood of u , for j =2,3. j j U Proof. This theorem is a particular application of the general analytic solution obtained in (7, 14) following triality theory.” Note that (7, 14) are our references (1) and (3). Before discussing the previous result let us clarify the meaning of ζ and u (as well as i i ζ and u) appearing in the statement above. Actually these functions are introduced in the statement of (5, Th. 2): “Theorem2. (Closed-formsolutions)Foragivenbodyforcef(x)anddeadloadσ such 1 thatσ(x) isdefinedby(2.12),theDAE (2.16)hasatmostthreerealrootsζ (x), i=1,2,3, i given by (2.22)–(2.24) and ordered as ζ (x)> µ>ζ (x)>ζ (x)> 1να2. (3.5) 1 − 2 3 −2 For i=1, the function defined by x σ(s)+αζ (s) u (x)= i ds (3.6) i ζ (s)+µ Z0 i is a solution of (BVP1). For each of i=2,3, (3.6) is also a solution of (BVP1) provided ζ i is replaced by ζ for values of s [0,x) for which ζ (s) is complex. 1 ∈ i Fora givent suchthatσ isdeterminedby(3.3) , oneofu (x), i=1,2,3,satisfies(3.4) 1 3 i 3 and hence solves (BVP2). Furthermore, P (u )=Pd(ζ ), i=1,2,3. (3.7)” s i s i Considering g : R R defined by g(ς) := 2ν 1ς+α2 (µ+ς)2, in fact, ζ (x) is the → − 1 (cid:0) (cid:1) 10 m. d. voisei ET AL. unique solution of the equation g(ς) = β2(x) on the interval [ µ, ), that is g(ζ (x)) = − ∞ 1 β2(x) and ζ (x) µ, while ζ (x) and ζ (x) are the unique solutions of the equation 1 ≥ − 2 3 g(ς)=β2(x)6η on [ρ, µ] and [ 1να2,ρ], respectively. We give this argument later on. − −2 Besidesthefactthatitisnotexplainedhow σ(s)+αζi(s) isdefinedinthecaseζ (s)+µ=0 ζ (s)+µ i i (which is possible if β(s) = 0) the only mention to u is in the following paragraph of the i proof of (5, Th. 2): “Foreachsolutionζ , i=1,2,3,thecorrespondingsolutionu isobtainedbyrearranging i i (2.10) in the form u = (σ+αζ)/(ζ +µ) and integrating. For a given t, the dead load σ x 1 is uniquely determined by (3.3) . Therefore, there is one u (x), i=1,2 or 3, satisfying the 3 i boundary condition u (1)=t, and this solves (BVP2).” i With our reformulation of the problem ( ), in the statements of (5, Th. 2, Th. 3) one s P mustreplace by 4, u byv := σ+αζi, u byv andP by P , being aneighborhoodof s i i s s j v , for j = 2,U3 (thisLis possible sinceζi+thµe operator v 4 u =U xv and its inverse j u v =u 4 arelinearcontinuousunderth∈eWL1,→4 tbopolog0yon∈ Us;whenceu s x s s U ∋ → ∈L R U ∈U is a local extrema for P iff the corresponding v 4 is a local extrema for P ). s s ∈L We agree that for τ2 > η the equation 2ν 1ς+α2 (µ+ς)2 = τ2 has a unique real − b solution ς > µ, while for 0 6 τ2 6 η the preceding equation has three real solutions 1 − (cid:0) (cid:1) ς ,ς ,ς with 1 2 3 1να2 6ς 6ρ6ς 6 µ6ς , −2 3 2 − 1 where ρ is given in Eq. (2.11). Indeed, let g :R R be defined by g(ς):= 2ν 1ς+α2 (µ+ς)2. Then g(ρ)=η and − → g′(ς)=2ν−1(ς +µ)2+2 2ν−1ς+α(cid:0)2 (µ+ς)=(cid:1)6ν−1(ς +µ)(ς ρ). − The behavior and graph of g are sho(cid:0)wed in Table(cid:1)s 1 and 2. ς 1να2 ρ µ + −∞ −2 − ∞ g (ς) + + + 0 0 + ′ − g(ς) 0 η 0 + −∞ ր ր ց ր ∞ Table 1 The behavior of g. Note that for τ =0 we have ς =ς = µ, ς = 1να2. For τ R consider also the fu1nctio2n − 3 −2 ∈ 1 τ2 h :R µ R, h (ς):= +2ατ +α2(ς+µ)+ν 1ς2 . τ τ − \{− }→ −2 ς +µ (cid:20) (cid:21) Note that h is the restriction to R µ of the continuous function hˆ : R R defined 0 0 \{− } → by ˆh (ς):= 1 α2(ς +µ)+ν 1ς2 ; clearly hˆ ( µ)= 1ν 1µ2. 0 −2 − 0 − −2 − Then (cid:2) (cid:3) 1 τ2 1g(ς) τ2 h′τ(ς)=−2 −(ς+µ)2 +α2+2ν−1ς =−2 (ς+−µ)2 ∀ς ∈R\{−µ}. (cid:18) (cid:19)