A Proofs of Section 2
A.5 Proof of Proposition 6
First of all, observe that Theorem 5 implies that the set ω2∩[0, π/2] has only one connected component. According to the symmetry property proved in Theorem 5 and since {χω2 = 1} ∩ [0, π/2] has only one component, we writeω2=
α, α+πL2
and rewrite the minimization problem (53) as a minimization problem with respect to the variableα. The functional to minimize is then
α∈h
To solve this minimization problem, it is more convenient to use the change of variable ξ = cos 2α+πL2
so that, after computation of the integrals, the minimization problem reduces to minJt(ξ) = 4t
The first and second derivative ofGtwith respect toξ are G0t(ξ) = 8 sin(πL)
Let us investigate the sign ofG00t(ξ) on [−ξ1,0]. Since (2πL−sin(πL)∈[0,2πL] whenevert∈[0,1], there holds
G00t(0) = 8 sin(πL) sin πL
2
((2πL−sin(πL))t−4πL)60.
Hence,
G00t(−ξ1) = 6 sin(πL) cos2 πL
2
−3(1−t) sin(πL) + (2πL−sin(πL))t−4πL
= 3
2sin(2πL) + (2πL−4 sin(πL))t−4πL60.
Indeed, the sign of G00t(−ξ1) can be obtained in the following way. Define the function f on [0, π] byf(u) = 32sin(2u)−4tsinu+ 2(t−2)u, where u plays the role ofπL. One has f0(u) = 6 cos2u−4tcosu+ 2t−7 andf00(u) = 4 sinu(t−3 cosu). Hencef0 is decreasing on [0,arccos(t/3)]
and increasing on [arccos(t/3), π]. Moreover, f0(0) =−2t−1 <0 and f0(arccos(t/3))<0 since t ∈ [0,1]. Since f0(π) = 6t−1, distinguishing between the two cases t 6 16 and t > 16, we obtain in the first case that f is decreasing on [0, π] and in the second case the existence of u1∈(0, π) such that f is decreasing on [0, u1] and then increasing on [u1, π]. Sincef(0) = 0 and f(π) = 2π(t−2)<0, in the two casesf is negative, which proves thatG00t(−ξ1)60.
Since G00t is a convex polynomial of degree 2 in the variableξ, the set {G00t 60} is connected and hence,G00t(ξ)60 for everyξ∈[−ξ1,0].
Let us next investigate the sign ofG0t(ξ) and of Gt(ξ) on [−ξ1,0]. First, we deduce from the sign ofG00t(ξ) on [−ξ1,0] thatG0tis decreasing on [−ξ1,0] and sinceG0t(0) = 8 sin(πL)(1−t)π2L2, it follows thatG0t(ξ)>0 for everyξ∈[−ξ1,0]. Hence,Gtis increasing on [−ξ1,0]. Using the facts thatGt(0) = 8 sin(πL)(1−t)π2L2>0, thatt7→Gt(−ξ1) is continuous, affine, has a positive limit whenttends to 1 and a negative one whent tends to 0, we deduce that
• ift < t0then there exists−ξt∈[−ξ1,0] such thatGt(ξ) is negative on [−ξ1,−ξt] and positive on [−ξt,0]; note moreover that−ξtis solution of the equationGt(ξ) = 0;
• ift>t0 thenGt(ξ)>0 for everyξ∈[−ξ1,0],
wheret0∈[0,1] is the unique solution of the equation Gt(−ξ1) = 0 where the unknown ist.
We deduce from all previous arguments that
• if t < t0 then Gt is decreasing on [−ξ1,−ξt] and increasing on [−ξt, ξ1], and thus −ξt = argmin{Jt(ξ), ξ∈[−ξ1, ξ1]};
• ift>t0 thenGtis increasing on [−ξ1, ξ1] and thus−ξ1= argmin{Jt(ξ), ξ∈[−ξ1, ξ1]}.
Let us study in particular the caset < t0. It follows from an implicit function argument that the mapt∈[0, t0]7→ −ξtis differentiable, since −ξtis solution of Gt(ξ) = 0. Setting g(t, ξ) =Gt(ξ), there holds moreover, fort∈(0, t0),
d(−ξt) dt (t) =−
∂g
∂t(t,−ξt)
∂g
∂ξ(t,−ξt)=−8 sin(πL) πL+ sin πL2 ξt
2 t∂G∂ξt(−ξt) ξt, and hence,
sign
d(−ξt) dt (t)
= sign(−ξt)<0,
for t > 0, since G0t(ξ) >0 on [−ξ1,0]. Observe also that ξ0 = cos 2α0+πL2
= 0, since α0 =
π
4(1−L). Defining αt as the inverse image of −ξt by the initial change of variable yields the statement.
Acknowledgment. The authors wish to thank Institut Henri Poincar´e (Paris, France) for pro-viding a very stimulating environment during the ”Control of Partial and Differential Equations and Applications” program in the Fall 2010.
The first author was partially supported by the ANR project GAOS “Geometric analysis of opti-mal shapes”.
The third author was partially supported by the Grant MTM2011-29306-C02-00 of the MICINN (Spain), project PI2010-04 of the Basque Government, the ERC Advanced Grant FP7-246775 NUMERIWAVES, the ESF Research Networking Program OPTPDE.
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