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A Poincaré type inequality with three constraints
Gisella Croce, Antoine Henrot
To cite this version:
Gisella Croce, Antoine Henrot. A Poincaré type inequality with three constraints. 2021. �hal-
03236684�
GISELLA CROCE AND ANTOINE HENROT
Abstract. In this paper, we consider a problem in calculus of variations motivated by a quantitative isoperimetric inequality in the plane. More precisely, the aim of this article is the computation of the minimum of the variational problem
u∈Winf ˆπ
−π
[(u0)2−u2]dθ
ˆπ
−π
|u|dθ 2
whereu∈ Wis aH1(−π, π) periodic function, with zero average on (−π, π) and orthogonal to sine and cosine.
1. Introduction
In this article we are interested in the following variational problem :
u∈W
inf ˆ
π−π
[(u
0)
2− u
2]dθ ˆ
π−π
|u|dθ
2where W denotes the subspace of functions in the Sobolev space H
1(−π, π) that are 2π-periodic, satisfying the following constraints:
(L1) ˆ
π−π
u(θ) dθ = 0 (L2)
ˆ
π−π
u(θ) cos(θ) dθ = 0 (L3)
ˆ
π−π
u(θ) sin(θ) dθ = 0.
Our aim is to compute the value of the minimum and to identify the minimizer. The difficulty comes here from the nonlinear term in the denominator together with the three constraints.
This problem is motivated from a shape optimization problem and more precisely from a quantitative isoperimetric inequality. Indeed, for any open bounded set of R
n, let us introduce the isoperimetric deficit:
δ(Ω) = P (Ω) − P (B)
P (B) , (1)
where |B| = |Ω|. Let the barycentric asymmetry be defined by:
λ
0(Ω) = |Ω∆B
xG|
|Ω|
where B
xGis the ball centered at the barycentre x
G= 1
|Ω|
ˆ
Ω
x dx of Ω and such that |Ω| = |B
xG|.
Fuglede proved in [9] that there exists a positive constant (depending only on the dimension n) such that δ(Ω) ≥ C(n) λ
20(Ω), for any convex subsets Ω of R
n. (2) Now, the constant C(n) is unknown (as it is the case in most quantitative inequalities like (2)) and it would be interesting to find the best constant. This leads to consider the minimization of the ratio
G
0(Ω) = δ(Ω) λ
20(Ω)
among convex compact sets in the plane, in particular. In the study of this minimization problem, one is led to exclude sequences converging to the ball in the Hausdorff metric. The strategy is to prove that
1991Mathematics Subject Classification. (2010) 28A75, 49J45, 49J53, 49Q10, 49Q20.
Key words and phrases. calculus of variations, Euler equation, Poincar´e type inequality . 1
on these sequences G
0is greater than 0.406 which is the value of G
0(S) where S is a precise set with the shape of a stadium, as computed in [1].
If a convex planar set E has barycenter in 0, it can be parametrized in polar coordinates with respect to 0, as
E = {y ∈ R
2: y = tx(1 + u(x)), x ∈ S
1, t ∈ [0, 1]} , (3) where u is a Lipschitz periodic function. Then the shape functional G
0(E) can be written as a functional H of the function u describing E, as follows :
G
0(E) = H(u) = π 2
ˆ
π−π
hp (1 + u)
2+ u
0(θ)
2− 1 i dθ
1 2
ˆ
π−π
|(1 + u)
2− 1|dθ
2. (4) The constraints of area (fixed equal to π without loss of generality) and barycentre in 0 read in terms of a periodic u ∈ H
1(−π, π) as:
(NL1) 1 2π
ˆ
π−π
(1 + u)
2dθ = 1;
(NL2) ˆ
π−π
cos(θ)[1 + u(θ)]
3dθ = 0;
(NL3) ˆ
π−π
sin(θ)[1 + u(θ)]
3dθ = 0.
The computation of the minimum of H , under the constraints (NL1), (NL2) and (NL3), seems very difficult. However, for sequences of sets converging to the ball in the Hausdorff metric, the limit of
m
ε:= inf{H (u), kuk
L∞= ε, u ∈ H
1(−π, π) periodic, satisfying (NL1), (NL2), (NL3)}
as ε → 0, equals the limit of the shape functional G
0for these sequences. Thus, a possible strategy consists in estimating from below the minimum of H by a simpler functional, namely its linearization.
Define
m = inf
u∈W
ˆ
π−π
[(u
0)
2− u
2]dθ ˆ
π−π
|u|dθ
2(5) where W is the space of periodic H
1(0, 2π) functions satisfying the constraints:
(L1) ˆ
π−π
u dθ = 0 (L2)
ˆ
π−π
u cos(θ) dθ = 0 (L3)
ˆ
π−π
u sin(θ) dθ = 0.
In [3] we proved that
lim inf
ε→0
m
ε≥ π
4 m. (6)
The aim of our article is to compute m and to prove that lim inf
ε→0
m
ε≥ π
4 m > 0.406.
We can easily get an estimate from below of m by using the Cauchy-Schwarz inequality ˆ
π−π
|u|dθ
2≤ 2π ˆ
π−π
u
2dθ.
Then, a Wirtinger-type inequality (or Parseval formula) shows that
u∈W
inf ˆ
π−π
[(u
0)
2− u
2]dθ 2π
ˆ
π−π
u
2dθ
≥ 3 2π .
Unfortunately this estimate on m is not sufficient to prove the above desired inequality. Moreover, at first
glance, one could think that cos 2θ is a minimizer of the functional in (5), as for the above Wirtinger-type
inequality written for functions satisfying (L1), (L2) and (L3), but this is not true. We will prove the
following result:
Theorem 1.1. Let m be defined by (5). Then m = 1
2(4 − π) and the minimizer u is the odd and π periodic function defined on [0, π/2] by u(θ) = cos θ + sin θ − 1.
A direct consequence of our theorem is that, by (6), lim inf
ε→0
m
ε≥ 0.457 > 0.406.
The strategy to prove our result is the following. It is immediate to see that the minimization problem (5) has a solution. Thus we write the Euler equation that any minimizer u satisfies. For that purpose, we introduce the three Lagrange multipliers, related to the three constraints (L1), (L2) and (L3) : they can be written as a function of sgn(u). Moreover we will get an explicit expression of u on any nodal domain, as a function of the endpoints of this interval. This will be done in Section 2. Technically, we will need to prove that the length of any nodal domain is less than π: this is used in the final argument.
It turns out that it is more difficult than expected and this is done completely in Section 3. After that, the main idea consists in proving that the Lagrange multipliers are all zero and that the nodal domains have all the same length (see Section 4). Finally, in Section 5, we will prove that the number of nodal domains is precisely four and find the explicit expression of u given in the above Theorem.
We remark that the minimization problem (5) is a variant of the Wirtinger inequality :
inf
u∈H1per(−π,π):´π
−πu=0
ˆ
π−π
|u
0|
2dθ ˆ
π−π
|u|
2dθ
= 1
because of the constraints (L2) and (L3) and the L
1(−π, π) norm at the denominator. In the literature one can find various generalizations of the Wirtinger inequality, without our constraints (L2) and (L3).
In the series of papers [2], [4], [5], [6], [7], [10], [11], [12], [14], [15], the authors consider different norms of u
0and u and on the mean value of u, namely
inf
u∈Wper1,p(−π,π):´π
−π|u|r−2u=0
ˆ
π−π
|u
0|
pdθ ˆ
π−π
|u|
qdθ
for all values of p, q, r greater than 1. We also mention [8] in which the authors study, in any dimension N ≥ 1, the inequality
inf
u∈W1,p(Ω):´
Ω|u|p−2uω=0
ˆ
Ω
|∇u|
pω ˆ
Ω
|u|
pω
with a positive log-concave weight ω, on a convex bounded domain Ω ⊂ R
N.
In our article, the Rayleigh quotient that we minimise is not ”too nonlinear” as in these cited papers.
The difficulty comes from the orthogonality to sine and cosine.
2. Preliminaries
2.1. Existence of a minimizer and Euler equation. The existence of a minimizer of the functional in (5) follows easily from the direct methods in the calculus of variations:
Proposition 2.1. The minimization problem (5) has a solution u. The Euler equation satisfied by u is
− u
00− u = m · sgn(u) + λ
0+ λ
1cos θ + λ
2sin θ . (7) where the Lagrange multipliers are given by
λ
0= − m 2π
ˆ
π−π
sgn(u(θ))dθ λ
1= − m
π ˆ
π−π
sgn(u(θ)) cos θdθ λ
2= − m
π ˆ
π−π
sgn(u(θ)) sin θdθ.
(8)
In particular, the function u is C
1.
The proof of this proposition is straightforward. In particular, the expression of the Lagrange multi- pliers is obtained by integrating the Euler equation after multiplication by 1, cos θ, sin θ. Moreover, the C
1regularity of the function u comes from the Euler equation that shows that its second derivative is L
∞, implying that u ∈ W
2,∞⊂ C
1.
Notice that one can assume that ˆ
π−π
|u| = 1 (9)
by homogeneity. It will be done in all the paper.
Remark 2.2. Using (8) we see that
m + λ
0= m 2π
ˆ
π−π
(1 − sgn(u(θ)))dθ > 0 (10)
and
− m + λ
0= − m 2π
ˆ
π−π
(1 + sgn(u(θ)))dθ < 0 . (11)
Up to a translation on θ, we can assume that one Lagrange multiplier is zero. Indeed, by periodicity, replacing u(θ) by u(θ + a) amounts to replace λ
2by cos aλ
2− sin aλ
1. Thus we can choose a such that λ
2= 0. Therefore in the sequel, we will assume:
the Lagrange multiplier λ
2= 0. (12)
We introduce the measure of the sets where u is positive and negative:
`
+= |{x ∈ (−π, π) : u(x) ≥ 0}|, `
−= |{x ∈ (−π, π) : u(x) < 0}| (13) With these notations, we can rewrite m + λ
0and m − λ
0:
m + λ
0= m`
−π , m − λ
0= m`
+π = m
π (2π − `
−). (14)
Note also that if u(x) is a minimizer of our problem, then −u(x) or u(−x) or −u(−x) are also minimizers.
Therefore, without loss of generality, we can assume, from now on, that `
+≥ `
−.
2.2. Expression of the solution. As usual, we call nodal domain, each interval on which u has a constant sign. By periodicity, there is an even number of nodal domains. Moreover, we will prove below (see Proposition 2.3) that there is at least four nodal domains. The main difficulty will be to prove that there are exactly four nodal domains with same length. It will be a consequence that the Lagrange multipliers are all zero and will be done in the last section.
On each nodal domain, we can integrate the Euler equation and get an explicit expression of the solution. We are going to write explicitly u on two consecutive intervals [a, b], [b, c], where
u(a) = u(b) = u(c) = 0 and
u ≥ 0 in [a, b], u ≤ 0 in [b, c] .
Here we assume the lengths of the intervals not equal to π. The case of an interval of length π will be considered in section 3.
By integrating the Euler equation on [a, b] and using that u(a) = u(b) = 0, we find u(x) = A
0cos x + B
0sin x − (m + λ
0) − λ
12 x sin x, x ∈ [a, b] (15)
where
A
0= (m + λ
0) cos(
a+b2) cos(
b−a2) − λ
12
(b − a) sin a sin b sin(b − a) , B
0= (m + λ
0) sin(
a+b2)
cos(
b−a2) + λ
12
b sin b cos a − a sin a cos b sin(b − a) .
(16)
By integrating the Euler equation on [b, c], and using that u(b) = u(c) = 0, we find u(x) = A
1cos x + B
1sin x − (−m + λ
0) − λ
12 x sin x, x ∈ [b, c]
where
A
1= (−m + λ
0) cos(
c+b2) cos(
c−b2) − λ
12
(c − b) sin c sin b sin(c − b) , B
1= (−m + λ
0) sin(
b+c2)
cos(
c−b2) + λ
12
c sin c cos b − b sin b cos c sin(c − b) .
(17)
Now we can obtain another expression of the solution on [b, c] using the C
1regularity of u in b. This gives
A
1cos b + B
1sin b = λ
0− m + λ
12 b sin b and
(A
0− A
1) sin b = (B
0− B
1) cos b . We then get a different expression for A
1and B
1:
A
1= (λ
0− m) cos b + λ
12 b sin b cos b − B
0sin b cos b + A
0sin
2b B
1= (λ
0− m) sin b + λ
12 b sin
2b + B
0cos
2b − A
0sin b cos b . Replacing A
0, B
0of formulas (16) in the above expressions of A
1and B
1, one gets:
A
1= λ
0cos(
a+b2)
cos(
b−a2) − m cos(
3b−a2) cos(
b−a2) − λ
12 (b − a) sin a sin b sin(b − a) B
1= λ
0sin(
a+b2)
cos(
b−a2) − m sin(
3b−a2) cos(
b−a2) + λ
12
b sin b cos a − a sin a cos b sin(b − a) .
(18)
Expressions (17) and (18) of A
1give λ
0sin b sin(
c−a2)
cos(
b−a2) cos(
c−b2) − m sin b sin(
a+c−2b2) cos(
b−a2) cos(
c−b2) = λ
12 sin b
(b − a) sin a
sin(b − a) − (c − b) sin c sin(c − b)
. Let us assume now that b 6= 0. If `
1= b − a and `
2= c − b, this equality can be written as
λ
0sin `
1+ `
22 − m sin `
2− `
12 = λ
12 cos `
12 cos `
22 `
1sin `
1sin a − `
2sin `
2sin c
. (19)
Let us assume now that b = 0. Expressions (17) and (18) of B
1give (−m + λ
0) tan c
2 + λ
12 c = (λ
0+ m) tan a 2
+ λ
12 a that is,
λ
0sin c − a
2
− m sin a + c
2
+ λ
12 (c − a) cos c 2
cos a 2
= 0 This is exactly equation (19) written in the case b = 0.
As a first consequence of the previous formulae, let us prove the following Proposition 2.3. A minimizer u has at least four nodal domains.
Proof. Assume, by contradiction, that the minimizer has only two nodal domains, say (a, b) ⊂ [−π, π]
where u is positive and its complement where u is negative. First of all, from λ
2= 0 and Proposition 2.1, we infer
0 = − ˆ
a−π
sin xdx + ˆ
ba
sin xdx − ˆ
πb
sin xdx = 2(cos a − cos b).
Therefore a = −b. Using this equality, we deduce from (10) and (11) in the same way that m + λ
0= 2m
π (π − b) and − m + λ
0= − 2mb π and from (8),
λ
12 = − 2m π sin b.
We observe that b 6=
π2, otherwise the function u(x) cos x would be positive on (−π, π) contradicting the constraint (L2). Formulae (16) and (18) give
A
0= 2m
π cos b π − b − b sin
2b
, B
0= 0 and
A
1= 2m π cos b
(π − b) sin
2b − b
, B
1= −2m sin b.
Since B
0= 0, we deduce from the expression of u on this interval (see (15)) that u is even on (−b, b). In
particular, since u is C
1, one has u
0(−b) = −u
0(b). Therefore, on the intervals [b −2π, −b] and [b, −b + 2π],
where u is negative, u solves the same ordinary differential equation, with u(b) = u(−b) = 0 and opposite
values for the derivative in b, −b. This implies that u is even in the whole interval (−π, π).
By (L2), 0 = ˆ
π0
u(x) cos xdx, that is, 0 = A
0b + sin b cos b
2 + A
1π 2 − A
1b + sin b cos b
2 − B
12 sin
2b + λ
18 π − 2m sin b.
Replacing the above expressions of A
0, A
1, B
1yields 0 = m sin b
− 3
2 + (π − 2b) tan(b)
.
Setting x =
π2− b we are led to the equation x/ tan x = 3/4. Since x 7→ x/ tan x is even and decreasing from 1 to 0, this equation has two solutions in (−
π2,
π2) yielding the only two possible values b ' 0.726 and b ' 2.4155.
Since we have chosen the normalization ˆ
π−π
|u(x)|dx = 1, we get 1
2 = ˆ
π0
|u(x)|dx = ˆ
b0
u(x)dx − ˆ
πb
u(x)dx = 4m π
(π − 2b) tan b + sin
2b − b(π − b) .
Plugging the value of b obtained above, we would get m = 2.1003. Now, the minimum of the functional defined in (5) must be less or equal to the functional evaluated in the function u described in Theorem 1.1, that is, m ≤ 1
2(4 − π) . This gives a contradiction with the value of m found above.
3. The length of the nodal intervals cannot be greater than π
In this section we prove that the length of any nodal interval of the solution u is strictly less than π.
We argue by contradiction: we will use different analytic tools to get the desired contradiction, mainly by considering the integral of u on a nodal domain.
We assume that there exists a nodal interval (a, b) of length ` greater than π. Without loss of generality, we can assume that
• u > 0 on (a, b);
• a ∈ [−π, 0] and b ∈ (0, π] (the function x 7→ u(x + π) is also a minimizer satisfying λ
2= 0);
• a + b
2 ≤ 0 (since u(−x) is also a minimizer); this implies a ≤ − π 2 .
In the sequel we will call negative interval (resp. positive interval) any interval where u is negative (resp.
positive).
On a negative interval (a
j, b
j) of length `
j6= π, we have ˆ
bjaj
u(x)dx = (−m + λ
0)
2 tan `
j2 − `
j+ λ
12
2 sin `
j2 cos a
j+ b
j2 1 + `
jsin `
j, (20)
while, on a positive interval (a
k, b
k) of length `
k6= π, we have ˆ
bkak
u(x)dx = (m + λ
0)
2 tan `
k2 − `
k+ λ
12
2 sin `
k2 cos a
k+ b
k2 1 + `
ksin `
k. (21)
In the case of a nodal domain of length π, let (a, a + π) be such an interval where we suppose u > 0. Now the Euler equation
−u
00− u = m + λ
0+ λ
1cos x on (a, a + π) u(a) = 0, u(a + π) = 0
has not a unique solution since 1 is an eigenvalue on the interval. Moreover, by the Fredholm alternative, the right-hand side of the equation must be orthogonal to the eigenfunction sin(x − a) providing the relation
m + λ
0= λ
14 π sin a. (22)
Lemma 3.1. The Lagrange multiplier λ
1is negative.
Proof. We first study the case where (a, b) has length ` > π. Let us analyse equation (21). We recall that m + λ
0> 0 by (10); for ` > π, both terms 2 tan(`/2) − ` and 1 + `/ sin ` are negative. If λ
1≥ 0, the integral is negative: this is a contradiction with the sign of u on (a, b).
In the case of a nodal domain of length π, one has λ
1< 0 by equation (22), since m + λ
0> 0 and
sin a < 0.
Lemma 3.2. The Lagrange multiplier λ
1satisfies
λ
12
≤ 2 sin `
−2 m
π = 2
`
−sin `
−2 (m + λ
0) < m + λ
0, (23) where `
−is the measure of {x : u(x) < 0}.
Proof. Let us introduce the two numbers:
`
b−= |{t > b, u(t) < 0}|, `
a−= |{t < a, u(t) < 0}| . Obviously `
a−+ `
b−= `
−. By (10), m + λ
0= m`
−π . Now, since λ
1< 0 by the previous lemma,
λ
12
= m
2π ˆ
π−π
sign(u) cos tdt
and ˆ
π−π
sign(u) cos tdt = ˆ
a−π
sign(u) cos tdt + ˆ
ba
cos tdt + ˆ
πb
sign(u) cos tdt.
By the bathtub principle (see [13]), the value of ˆ
a−π
sign(u) cos tdt is maximum when we choose sign(u) =
−1 on the left, namely on (−π, −π + `
b−] (because the cos is increasing on [−π, a]) and similarly for the last integral. Therefore, we get that
πλ
1m
≤ −
ˆ
−π+`b−−π
cos tdt + ˆ
π+`a−−π+`b−
cos tdt − ˆ
ππ−`a−
cos tdt = 2(sin `
a−+ sin `
b−). (24) Since
sin `
a−+ sin `
b−= 2 sin `
−2
cos
`
a−− `
b−2
≤ 2 sin `
−2
≤ `
−we finally get estimate (23), using (10) and (24).
Let us introduce the following quantity :
A = |λ
1/2|
m + λ
0. (25)
Our strategy to prove that there cannot be an interval of length greater than π consists in proving that A <
π2or A cos(
a+b2) <
2π. Indeed, in the case where b−a = ` > π, these estimates imply that
ˆ
b audt < 0, that is, a contradiction. This is due to the following Lemma 3.3. In the case where b − a = ` = π, one has A = 2
π| sin a| by (22). Therefore proving that A <
π2provides a contradiction in this case too.
Lemma 3.3. The function g : ` 7→ ` − 2 tan(`/2) +
π4sin(`/2)[1 + `/ sin `] is positive on (π, 2π).
Proof. Indeed g is positive if and only if k(t) = t cos t − sin t +
π1[2t + sin(2t)] is negative on
π2, π . Observe that k(
π2) = 0 and k(π) < 0. Now, the derivative of k, k
0(t) = −t sin(t) +
π2[1 + cos(2t)], is the difference between two functions which intersect in only one point t
0. Since k
0is negative near
π2and positive near π, k is minimal at t
0and therefore k < 0 on
π2, π
.
We are now going to find some estimates on A, taking into account the midpoint of a nodal interval (a
j, b
j), that we will denote by m
j.
Lemma 3.4. Let (a
j, b
j) be a negative interval. If m
j∈ / h
− π 2 , π
2 i , then
A ≤ l
2j(2π − `
−)
12`
−| cos(m
j)| . (26)
For a positive interval (a
k, b
k), whose midpoint m
k∈
− π 2 , π
2
one has
A ≤ l
k212 cos(m
k) . (27)
Proof. For a negative interval (a
j, b
j), as soon as a
j+ b
j2 ∈ / h
− π 2 , π
2 i
the fact that ˆ
bjaj
u < 0 implies
|λ
1/2|
m − λ
0≤ h(`
j)
| cos(
aj+b2 j)|
where h(x) = 2 tan(
x2) − x
2 sin(
x2) 1 +
sinxx. We claim that for x ∈ [0, π], one has h(x) ≤
x122. By using (10), (11), (14) and this bound on h we have the following estimate on A:
A ≤ l
2j(2π − `
−) 12`
−| cos(
aj+b2 j)| .
We now prove our claim, that is, the bound on h. The statement is equivalent to the positivity of f (x) =
x122(x + sin x) −2 sin(x/2) + x cos(x/2) in [0, π]. Observe that f (0) = 0. The result will follow if we prove that the derivative of f is positive, that is, k(x) = 3x + 2 sin(x) + x cos(x) − 6 sin(x/2) is positive.
We remark that k(0) = 0, k(
π2) > 0, k(π) > 0. We will split the analysis into two cases.
(1) If x ∈ [0,
π2] it is easy to see that k
00< 0 and therefore k(x) ≥ 0 in [0,
π2].
(2) In the case where x ∈ [
π2, π], k is decreasing. Indeed it is easy to see that k
0is negative, since k
0is convex, k
0(π) = 0, k
0(
π2) < 0.
In the same way, using (21), we can prove that, on a positive interval (a
k, b
k) (m
kbeing its midpoint):
A ≤ l
2k12 cos(m
k) ,
when m
k∈ (−
π2,
π2).
Lemma 3.5. There are no negative intervals on the ”left” of I
0= (a, b) whose midpoint is between −π/2 and π/2. There is at most one negative interval on the ”right” of I
0whose midpoint lies between −π/2 and π/2.
Proof. Since a ≤ − π
2 , there are no negative intervals on the ”left” of I
0whose midpoint is between − π 2 and π
2 .
We prove the second statement by contradiction. If there were two negative intervals, say I
1and I
3with a positive interval I
2= (a
2, b
2) between them, its midpoint would satisfy m
2≤ π − `
22 . Moreover, since b
2≤ π
2 , and 0 ≤ b < a
2= b
2− `
2≤ π
2 − `
2, we infer `
2≤ π
2 . Now using (27), we see that A ≤ `
2212 cos(m
2) ≤ `
2212 sin(`
2/2) .
Now, it is immediate to check that x 7→ x
2/ sin(x/2) is increasing. Therefore the previous inequality would imply for `
2≤
π2: A ≤ π
2/(24 √
2) <
2πimplying ˆ
b2a2
udt < 0, that is, a contradiction.
We are now going to find a lower bound for `
−, the measure of {u < 0}. We know that ˆ
{u>0}
u(x)dx + ˆ
{u<0}
u(x)dx = 0, while ˆ
π−π
|u(x)|dx = ˆ
{u>0}
u(x)dx − ˆ
{u<0}
u(x)dx = 1. Therefore ˆ
{u>0}
u(x)dx = − ˆ
{u<0}
u(x)dx = 1
2 . (28)
On all (but possibly one) negative intervals I
jwhose midpoint m
j∈ / h
− π 2 , π
2 i
, we have ˆ
Ij
−u(x)dx ≤ (m − λ
0)[2 tan(`
j/2) − `
j]. (29) The following proposition gives a lower bound for `
−.
Proposition 3.6. `
−≥ 1.55.
Proof. There is only one possibility between
• there is a negative interval of length `
j≥ 1.6;
• the length of all negative intervals is less than 1.6.
The first possibility implies that `
−≥ 1.6 > 1.55. We need to study the second possibility. We will split the analysis into two cases: in the first case we will assume that no negative interval has its midpoint in [−
π2,
π2]. In the second case we will assume that the midpoint of one interval, say I
1, is before
π2. This is possible due to Lemma 3.5.
(1) Assume that no negative interval has its midpoint in h
− π 2 , π
2 i
. We observe that 2 tan(x/2) − x ≤ 0.45x
34 , 0 ≤ x ≤ 0.8 .
Using this inequality in (29) and summing up on all negative intervals I
j, we get 1
2 = X
j
ˆ
Ij
(−u(x))dx ≤ 0.45(m − λ
0) 4
X
j
`
3j.
Now, the maximization of the convex function t ∈ R
+→ t
3and the fact that the sum of the lengths `
jof all negative intervals is `
−give
X
j
`
3j≤ `
3−.
Moreover, consider the functional defined by (5). Using as a test function the explicit function given in Theorem (1.1) we get an estimate of m:
m ≤ 1
2(4 − π) . (30)
Recalling that m − λ
0= m
π (2π − `
−) by (14), we end up with 1
2 ≤ 0.45(2π − `
−) 8π(4 − π) `
3−.
This polynomial inequality provides finally the inequality `
−≥ 1.74.
(2) Assume that the midpoint of one interval, say I
1, is before
π2. In the above computation of X
j
ˆ
Ij
(−u(x))dx, we need to add the following positive term
P = − λ
12 2 sin `
12
cos m
11 + `
1sin `
1(31) coming from the integral on I
1(see (20), second term of the right hand side). In other words
1
2 ≤ 0.45(2π − `
−)
8π(4 − π) `
3−+ P . (32)
We are going to distinguish two cases, according to the values of `
1:
(a) Assume `
1≥ 0.228. In this case, we look at the first negative interval I
−1on the left of I
0. Since b ≤ π − `
12 ≤ π
2 − 0.114 and ` ≥ π, we have m
−1= a − `
−12 = b − ` − `
−12 ≤ −0.114 − π + `
−12 .
If `
−1≥ π
2 , then `
−≥ π
2 + 0.228 > 1.75. If, on the contrary, `
−1< π
2 , then m
−1≥ − 5π 4 necessarily (otherwise a = m
−1+ `
−12 < −π which is impossible). Therefore,
| cos(m
−1)| ≥ min ( √
2 2 ,
cos
0.114 + π + `
−12
)
:= C.
Using this estimate and `
−≥ 0.228 + `
−1in (26) we get A ≤ `
2−1(2π − 0.228 − `
−1)
12(0.228 + `
−1)C .
As a function of `
−1, the right-hand side is increasing and for `
−1≤ 1.322 we get the inequality A ≤ 0.636 < 2
π , that is, a contradiction. Therefore, in this case, we deduce
`
−1≥ 1.322 and `
−≥ `
1+ `
−1≥ 1.55.
(b) Assume that `
1≤ 0.228. Lemma 3.2 and inequalities (14) and (30) give
− λ
12 ≤ m + λ
0= m `
−π ≤ `
−2π(4 − π) .
We can use these inequalities to estimate the positive term P defined by (31) : P ≤ `
−2π(4 − π) 2 sin(0.114)
1 + 0.228 sin(0.228)
. Therefore we get from (32)
1
2 ≤ 0.45(2π − `
−)
8π(4 − π) `
3−+ `
−2π(4 − π) 2 sin(0.114)
1 + 0.228 sin(0.228)
. This polynomial inequality implies `
−≥ 1.55.
We are now in position to prove the main result of this section.
Theorem 3.7. There is no nodal interval of length greater than π.
Proof. We will divide the proof in two cases: the solution u has at least 6 nodal intervals and the solution u has only 4 nodal intervals.
(1) We are going to prove that there cannot be 6 nodal intervals or more. We recall that a ≤ − π 2 . We divide the proof into two cases according to the values of b.
(a) Assume that b ≥ π
2 . We will consider the two negative intervals next to I
0= (a, b), namely I
−1on the left to I
0and I
1on the right, and prove that both have lengths greater than 1.17.
(i) Assume by contradiction that `
−1≤ 1.17. On one hand m
−1= a − `
−12 ≤ − π 2 − `
−12 . On the other hand, m
−1≥ − 5π
4 (otherwise a = m
−1+ `
−12 < −π which is impossible).
Therefore we have
| cos(m
−1)| ≥ min 1
√ 2 ,
cos
π + `
−12
= min 1
√ 2 , sin `
−12
. Plugging this estimate in (26) and using Proposition 3.6 yields
A ≤ `
2−1(2π − 1.55) 12 · 1.55 · min n
√1
2
, sin(`
−1/2) o . (33)
Now the right-hand side is increasing in `
−1and its value for `
−1= 1.17 is less than 0.631 < 2
π that is absurd, proving the claim.
(ii) Assume by contradiction that `
1≤ 1.17. In this case we are going to use that b ≥ π to estimate m
1: 2
π 2 + `
12 ≤ b + `
12 = m
1= b + `
12 < π + 1.17 2 < 5π
4 . One can now repeat the above argument to get a contradiction.
We observe that `
−≥ `
−1+ `
1≥ 2.34.
With the same arguments of the two steps above, we can prove that `
−1≥ π
2 and `
1≥ π 2 . Indeed it is sufficient to replace 1.17 by π
2 and 1.55 by 2.34 in the estimate of A. This gives A ≤ 0.49 < 2
π , that is a contradiction. We deduce that `
−1≥ π
2 and `
1≥ π
2 that implies
`
−≥ π: this is a contradiction, since `
+> π.
We have thus proved that in the case b ≥ π
2 one cannot have 6 or more nodal domains.
(b) Assume that b < π
2 . One can prove that `
−1≥ 1.17, following exactly the same argument as
in step (i) above. Now since we do not know the position of m
1, we work with the intervals
I
1= [a
1, b
1], I
2= [a
2, b
2], I
3= [a
3, b
3], with b
1= a
2, b
2= a
3.
(i) Assume that b
2≤ π
2 . Observe that 0 < m
2≤ π 2 − `
22 where we have used that b > 0.
Using (27) we get
A ≤ `
2212 sin(`
2/2) . Since b
2≤ π
2 and 0 ≤ b < a
2= b
2− `
2≤ π
2 − `
2, one has `
2≤ π
2 . Therefore the right hand side of the above estimate of A is less than 2
π , that is a contradiction.
(ii) Assume that b
2= a
3≥ π
2 . We start by proving that `
−≥ 1.93. Assume by contra- diction that `
−≤ 1.93. On one hand we have m
3= a
3+ `
32 ≥ π 2 + `
32 . On the other hand we claim that m
3< 5π
4 . We have m
3= b + `
1+ `
2+ `
32 . Observe that
• b < π
2 by hypothesis,
• `
2≤ π − 1.55 since `
2+ ` ≤ `
+= 2π − `
−≤ 2π − 1.55,
• `
1+ `
32 ≤ 0.76 since `
−1+ `
1+ `
32 ≤ `
−≤ 1.93 and `
−1≥ 1.17.
Thus m
3< 3π
2 − 1.55 + 0.76 < 5π
4 . We deduce from the previous bounds on m
3that:
| cos m
3| ≥ min 1
√ 2 ,
cos
π + `
32
which provides, as in case (a)(i), `
3≥ 1.17.
We deduce that `
−≥ `
−1+ `
3> 1.93.
We are going to prove that `
−1≥ 1.532. Assume that this is not the case. Using the same argument as in step (a)(i), one has a contradiction from the following estimate :
A ≤ 1.532
2(2π − 1.93) 12 · 1.93 · min n
√1
2
, sin(1.532/2) o < 2
π . (34)
It is easy to prove that `
−> 2.67. Indeed, if this is not the case, replacing 1.93 by 2.67 in (34), we get a contradiction since A is still less than
π2.
We are going to prove that `
−1>
π2. Assume that this is not the case. By the same argument as in step (a)(i), one has a contradiction from estimate
A ≤ (π/2)
2(2π − 2.67) 12 · 2.67 · √
2/2 < 2 π . We are going to prove that `
3≥ π
2 . Assume that this is not true. Recall that `
−< π.
As above, on one hand we have m
3= a
3+ `
32 ≥ π 2 + `
32 . On the other hand m
3< 5π 4 . Indeed m
3= b + `
1+ `
2+ `
32 . Observe that
• b < π
2 by hypothesis,
• `
2≤ π − 2.67
• `
1+ `
32 ≤ π − π 2
Thus m
3< 2π − 2.67 < 5π
4 . We deduce from the previous bounds on m
3that:
| cos m
3| ≥ 1
√ 2 . This gives a contradiction since A ≤ (π/2)
2(2π − 2.67)
12 · 2.67 · √
2/2 < 2 π . Therefore `
3> π
2 .
This last estimate implies π ≥ `
−> `
−1+ `
3> π. Therefore there cannot be 6 or more nodal intevals.
(2) It remains to consider the case of 4 nodal intervals: I
−1= I
3, I
1(two negative ones) and I
0, I
2(two positive ones). The strategy for the proof is similar, but now we are going to work with a
more explicit expression of λ
1, λ
2. Without loss of generality, we can assume that the lengths
satisfy 0 < `
1≤ `
3< π (up to replacing u(x) by u(−x)). Moreover, as before, we can get the lower bound `
3≥ 1.17 (see (b), (ii)). Let us rewrite the four intervals as I
0= (a
0, a
1), I
1= (a
1, a
2), I
2= (a
2, a
3), I
−1= I
3= (a
3− 2π, a
0) and m
0, m
1, m
2, m
3are the midpoints of each interval. We can assume −π ≤ m
3≤ − π
2 by Lemma 3.5. From λ
2= 0 we have 0 =
ˆ
π−π
sign(u(x)) sin xdx = 2 (cos a
0− cos a
1+ cos a
2− cos a
3) . (35) Gathering − cos a
1+ cos a
2on the one hand and cos a
0− cos a
3on the other hand, the right hand side of (35) can be rewritten
sin `
32 sin m
3+ sin `
12 sin m
1= 0. (36)
In the same way, coming back to the definition of λ
1, see (8), we have λ
12 = 2m π
sin `
32 cos m
3+ sin `
12 cos m
1. (37)
Observing that cos m
3< 0, by (36) cos m
3can be rewritten as cos m
3= −
s
1 − sin
2`21sin
2m
1sin
2`23. (38)
By using (37), (38) and (10), the quantity A defined in (25) can be rewritten as a function of
`
1, `
3, m
1as
A = A(`
1, `
3, m
1) = 2 q
sin
2`23− sin
2`21sin
2m
1− sin
`21cos m
1`
1+ `
3. (39)
We need to distinguish the cases m
1≤ π
2 and m
1> π
2 . In the first case, we will prove that we can assume `
1≤ π
6 while, in the second case, we can assume `
1≤ 0.62.
(a): Let us start with the first case m
1≤ π
2 and let us first look at the dependence of A with respect to m
1. The derivative of A(`
1, `
3, m
1) with respect to m
1has the same sign as
sin m
1
1 − cos m
1sin
`21q
sin
2`23− sin
2`21sin
2m
1
, that is, the same sign as
sin m
1sin
2`
32 − sin
2`
12
. The above quantity is positive, since 0 ≤ m
1≤ π
2 and `
3≥ `
1. Therefore, if m
1≤ π 2 A(`
1, `
3, m
1) ≤ A(`
1, `
3, π
2 ) = 2 q
sin
2`23− sin
2`21`
1+ `
3. (40)
Assume by contradiction that `
1> π
6 . We are going to prove that A < 2
π , thus reaching a contradiction. We set
x = `
3− `
12 and y = `
3+ `
12 and observe that 0 ≤ x < π
2 , 1.17
2 ≤ y < π
2 . The quantity A(`
1, `
3, π
2 ) can be rewritten as a function of x, y:
G(x, y) =
√ sin x sin y
y , (x, y) ∈ h 0, π
2 i ×
1.17 2 , π
2
.
The map x 7→ G(x, y) is increasing while y 7→ G(x, y) is decreasing on this set. We remark that the assumption `
1> π
6 is equivalent to y > x + π
6 . In the rectangle h 0, π
2 i ×
1.17 2 , π
2
this implies that x ∈ h 0, π
3
i . Now, it can be checked that
∀t ∈ h 0, π
3 i
, G(t, t + π 6 ) ≤ 2
π .
Indeed the function t 7→ G(t, t +
π6) is first increasing then decreasing and satisfies the above inequality for its maximum that is approximatively at 0.6627206. From the properties of G, we infer that G(x, y) < 2
π for all (x, y) such that y > x + π
6 , y ∈ [ 1.17 2 , π
2 ], x ∈ [0, π 3 ], that is, `
1> π
6 . We have thus proved that A < 2
π as soon as `
1> π
6 , that is a contradiction.
Therefore we can assume from now on that `
1≤ π
6 when m
1≤ π 2 .
(b): Let us now assume that m
1>
π2. Let us suppose first that m
1≥ 1.9, therefore | cos m
1| ≥
| cos 1.9| and, following (26) we infer
A(`
1, `
3, m
1) ≤ l
21(2π − `
1− `
3) 12(`
1+ `
3)| cos 1.9| .
Now this expression is decreasing in `
3and increasing in `
1, thus it is always less than its value for `
1= `
3< π
2 :
A(`
1, `
3, m
1) ≤ A(`
1, `
1, 1.9) = l
1(π − `
1)
12| cos 1.9| ≤ π
248| cos 1.9| < 2 π and the contradiction is obtained in this case. Therefore, we assume now π
2 ≤ m
1≤ 1.9.
Expressing m
3in terms of m
1, we have m
3= m
1− `
−2 − `
0≤ 1.9 − `
−2 − π, thus | sin m
3| ≤ sin(1.9 −
`2−) (recall that −π ≤ m
3≤ − π
2 ). By identity (36), we have
| sin 1.9| sin `
12 ≤ sin m
1sin `
12 = | sin m
3| sin `
32 ≤ sin `
32 sin
1.9 − `
−2
. This implies, by (38)
| cos m
3| = s
1 − sin
2`21sin
2m
1sin
2`23≥ s
1 − sin
2(1.9 −
`2−) sin
21.9 . Therefore, using (26) we can estimate A from above by
A ≤ `
23(2π − `
−) 12`
−s
1 − sin
2(1.9 −
`2−) sin
21.9
.
As a function of `
−the right-hand side is clearly decreasing. Now, if `
1≥ 0.62 (and then
`
−≥ 1.17 + 0.62), we can see that A < 2
π for any `
3∈ [1.17, π − `
1] giving the desired contradiction. Therefore, when m
1> π
2 , we can assume `
1≤ 0.62.
We use now the integral of u on the interval I
3. We recall that ˆ
I3
(−u(x))dx = m`
+π
2 tan `
32 − `
3− λ
12
2 sin `
32 cos m
31 + `
3sin `
3. Moreover 0 <
ˆ
I3
(−u(x))dx ≤ 1
2 by (28). We use the same idea as before, using the expression of λ
1/2 given in (37) and replacing sin
`21cos m
1by ± q
sin
2`21− sin
2`23sin
2m
3(with a + if m
1≤ π 2 and a − if π
2 ≤ m
1≤ 1.9), thanks to (36). This provides an expression of the integral as a function of the three variables `
1, `
3, m
3:
ˆ
I3
