Olivier Roustant
email: rous-tant@emse.fr

Franck Barthe
email: franck.barthe@math.univ-toulouse.fr

Bertrand Iooss
email: bertrand.iooss@edf.fr

Poincaré inequalities on intervalsapplication to sensitivity analysis

Keywords: Poincaré inequality, Spectral gap, Truncated distribution, Kummer's functions, Sobol-Hoeffding decomposition, Sobol indices, Derivative-based global sensitivity measures, Finite elements

The development of global sensitivity analysis of numerical model outputs has recently raised new issues on 1-dimensional Poincaré inequalities. Typically two kind of sensitivity indices are linked by a Poincaré type inequality, which provide upper bounds of the most interpretable index by using the other one, cheaper to compute. This allows performing a lowcost screening of unessential variables. The efficiency of this screening then highly depends on the accuracy of the upper bounds in Poincaré inequalities. The novelty in the questions concern the wide range of probability distributions involved, which are often truncated on intervals. After providing an overview of the existing knowledge and techniques, we add some theory about Poincaré constants on intervals, with improvements for symmetric intervals. Then we exploit the spectral interpretation for computing exact value of Poincaré constants of any admissible distribution on a given interval. We give semi-analytical results for some frequent distributions (truncated exponential, triangular, truncated normal), and present a numerical method in the general case. Finally, an application is made to a hydrological problem, showing the benefits of the new results in Poincaré inequalities to sensitivity analysis.

Introduction

Motivation

This research is motivated by sensitivity analysis of a costly numerical function f (x 1 , . . . , x d ) of independent random variables x i . We denote by µ = µ 1 ⊗• • •⊗µ d the distribution of x = (x 1 , . . . , x d ) and assume that f (x) ∈ L 2 (µ). Such problem occurs for instance in computer experiments, where a physical phenomenon is studied with a complex numerical code ( [START_REF] De Rocquigny | Uncertainty in industrial practice[END_REF]). One important question is screening, i.e. to identify unessential variables, which can be done by computing several criteria, called sensitivity indices. Among them, the variance-based indices -or Sobol indices, are often prefered by practitioners due to their easy interpretation ( [START_REF] Iooss | A review on global sensitivity analysis methods[END_REF]). They are defined on the Sobol-Hoeffding decomposition ( [START_REF] Hoeffding | A class of statistics with asymptotically normal distributions[END_REF], [START_REF] Efron | The jackknife estimate of variance[END_REF], [START_REF] Sobol | Sensitivity estimates for non linear mathematical models[END_REF]),

f (x) = I⊆{1,...,d} f I (x I ) = f 0 + 1≤i≤d f i (x i ) + 1≤i<j≤d f i,j (x i , x j ) + . . .
where x I is the subvector extracted from x whose coordinates belong to I. The terms f I satisfy non-simplification conditions E(f I (x I )|x J ) = 0 for all strict subset J ⊂ I, equivalent to:

f I (x I )dµ i (x i ) = 0 (1.1)
for all i ∈ I and all x I . Such conditions imply orthogonality, and lead to the variance decomposition:

Var(f (x)) = I⊆{1,...,d}

Var(f I (x I ))

The "total Sobol index" S T i is then defined as the ratio of variance of f (x) explained by x i (potentially with other variables):

S T i = Var I i f I (x I ) /D
where D = Var(f (x)). Thus one can decide that x i is not influential if S T i is less than (say) 5%.

Despite their nice interpretation, Sobol indices require numerous computations for their estimation. Then another global sensitivity index, called DGSM (Derivative-based Global Sensitivity Measure), can advantageously be used as a proxy [START_REF] Sobol | Derivative-based global sensitivity measures and the link with global sensitivity indices[END_REF][START_REF] Kucherenko | Monte carlo evaluation of derivative-based global sensitivity measures[END_REF]. Defined by

ν i = ∂f (x) ∂x i 2 dµ(x)
for i = 1, . . . , d, they are cheaper to compute, especially when the gradient of f is available (e.g. as output of an evaluation of a complex numerical code). As shown in [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF], Sobol indices and DGSM are connected by a 1-dimensional Poincaré-type inequality:

S T i ≤ C(µ i )ν i /D (1.2)
where C(µ i ) is a Poincaré constant for µ i . Recall that µ i satisfies a Poincaré inequality if the energy of any centered function is controlled by the energy of its derivative: For all g satisfying gdµ i = 0 there exists C(µ i ) > 0 s.t.

g 2 dµ i ≤ C(µ i ) g 2 dµ i (1.3)
Inequality 1.2 is easily obtained by applying the Poincaré inequality 1.3 to the functions x i → I i f I (x I ), which are centered for any choice of x I with i ∈ I (condition 1.1), and integrating with respect to the other variables x j (j = i). This allows performing a "low-cost" screening based on the upper bound of Sobol indices in 1.2 instead of Sobol indices directly. Hence, one can decide that x i is not influential if C(µ i ) νi D is less than (say) 5%.

The efficiency of this low-cost screening strongly depends on the accuracy of the Poincaré-type inequality 1.2, and this motivates the investigation of 1dimensional Poincaré inequalities. Notice that there is a large variety of probability distributions used in sensitivity analysis. They are often linked to prior knowledge about the range of variation of physical parameters. As a consequence most of them are continuous with finite support, possibly obtained by truncation. The most frequent probability density function (pdf) are: Uniform, (truncated) Gaussian, triangular, (truncated) lognormal, (truncated) exponential, (truncated) Weibull, (truncated) Gumbel (see for example [START_REF] De Rocquigny | Uncertainty in industrial practice[END_REF]). Less frequently, it can be found: (inverse) Gamma, Beta, trapezoidal, generalized extreme value.

It is also important to notice that one may not easily reduce the problem to the case of the uniform distribution, by applying the standard reincreasing arrangement technique. Indeed by denoting F i the cdf of µ i , an idea would be to consider the function g(u 1 , . . . , u d ) = f F -1 1 (u 1 ), . . . , F -1 d (u d ) where the u i = F i (x i ) are independent and uniform on [0, 1]. Then the total Sobol indices of f (x 1 , x 2 ) and g(u 1 , u 2 ) are equal. However, the derivatives of the transformations F -1 i can be large, and the DGSM computed on g may be large and even infinite. For instance, if f (x 1 , x 2 ) = x 1 + x 2 with x 1 , x 2 i.i.d. N (0, 1), the DGSM of g(u 1 , u 2 ) = Φ -1 (u 1 ) + Φ -1 (u 2 ) is equal, for each variable u i , to:

ν g = 1 0 [(Φ -1 ) (s)] 2 ds = 1 0 1 φ(Φ -1 (s)) 2 ds = R 1 φ(t) dt = R e t 2 /2 dt = +∞
On the other hand, the DGSM ν f of f is equal to 1 for each x i , and the corresponding upper bound is C P (N (0, 1))

ν f D = 1 2
, which is here exactly equal to the total Sobol index of x i .

To conclude about motivations in sensitivity analysis, it is worth mentioning that the idea of low-cost screening can be extended to higher-order interactions, and also depends on the accuracy of 1-dimensional Poincaré inequalities [START_REF] Roustant | Crossed-derivative based sensitivity measures for interaction screening[END_REF]. It allows screening out useless interactions and discovering additive structures in f .

Aim and plan

Our aim is to bridge the gap between industrial needs coming from sensitivity analysis problems and the theory of Poincaré inequalities, by providing an accessible introduction to Poincaré inequalites for non-specialists, and by using and developping the theory in order to deal with the specific situations motivated by low-cost screening.

In Section 2, we present the general background on Poincaré inequalities, together with the main techniques avaible to establish them (Muckenhoupt's criterion, perturbation and transportation methods, spectral methods). Most of these techniques provide upper bounds on the Poincaré constant for large classes of measures.

In the literature of 1-dimensional Poincaré inequalities, the exact value of the Poincaré constant is known for very few measures, such as: Uniform, Gaussian (which are classical, see e.g. [START_REF] Ané | Sur les inégalités de Sobolev logarithmiques[END_REF]), exponential [START_REF] Bobkov | Poincare's inequalities and talagrand's concentration phenomenon for the exponential distribution[END_REF] or logistic [START_REF] Barthe | Isoperimetry and stability of hyperplanes for product probability measures[END_REF]. More measures are needed for sensitivity analysis, as well as their restrictions to intervals. These situations were not studied specifically so far (apart from the Gaussian pdf, for which optimal constants are known on the whole real line and on intervals formed by successive zeros of Hermite polynomials, see e.g. [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF]). Section 3 provides further results which allow to deal with the family of restrictions of a given probability measure to intervals. In the case of symmetric measures and intervals, we derive new improvements.

Section 4 deals with the exact constant for some distributions required by sensitivity analysis. We present new semi-analytical results for the optimal constant of the triangular, truncated normal and truncated exponential. The optimal constant is obtained as the first zero of an analytic function. We also provide a general numerical algorithm that converges to the optimal Poincaré constant.

Finally we come back to our initial motivation and apply this research to two test-cases (Section 5). It is observed that the new results significantly improve on the exisiting ones, and now allow performing a low-cost screening of Sobol indices based on upper bounds computed by DGSMs.

For the ease of readability, the most technical proofs are postponed to the Appendix.

Background on Poincaré inequalities

In this section, we provide a quick survey of the main simple techniques allowing to derive Poincaré inequalities for probability measures on the real line. We often make regularity assumptions on the measures. This allows to avoid technicalities, without reducing the scope for realistic applications. Indeed, all the measures we are interested in are supported on an open interval (a, b), and have a positive continuous, piece-wize continuouly differentiable density. Moreover, when a is finite, they are monotonic on a neighborhood of a (and similarly for b).

Definitions

Consider an open interval of the real line Ω = (a, b) with -∞ ≤ a < b ≤ +∞. A locally integrable function f : Ω → R is weakly differentiable if there exists a locally integrable function g : Ω → R such that for all functions φ of class C ∞ with compact support in Ω:

Ω f (t)φ (t)dt = - Ω g(t)φ(t)dt.
Then g is a.e. uniquely determined (more precisely two functions with this property coincide almost everywhere), it is called the weak derivative of f and denoted by f .

Let µ be a probability measure on Ω, and f : Ω → R be a Borel measurable function. Recall that the variance of f for µ is defined as

Var µ (f ) = inf a∈R Ω (f -a) 2 dµ. Obviously Var µ (f ) = +∞ if f 2 dµ = +∞. When f 2 dµ < +∞, it holds Var µ (f ) = f 2 dµ - f dµ 2 = f -f dµ 2 dµ.
Definition 1. Let µ(dt) = ρ(t)dt be an absolutely continuous probability measure Ω. We say that µ verifies a Poincaré inequality on Ω if there exists a constant C < +∞ such that for all f weakly differentiable functions f on Ω:

Var µ (f ) ≤ C Ω (f ) 2 dµ. (2.1)
In this case, the smallest possible constant C above is denote C P (µ), it is refered to as the Poincaré constant of the measure µ.

Observe that the above integrals are always defined, with values in [0, +∞]. Roughly speaking, a Poincaré inequality expresses in a quantitative way that a function with a small weak derivative, measured in the sense of µ, has to be close to a constant function again in the sense of µ.

Remark 1. Weakly differentiable functions are exactly the functions which admit (in their equivalence class for a.e. equality) a continuous version which satisfies (see e.g. [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF], [START_REF] Brézis | Analyse fonctionnelle[END_REF])

∀x, y ∈ Ω, f (y) = f (x) + y x f (t) dt.
Such functions are also called (locally) absolutely continuous. Their variations can be recovered by integrating their weak derivatives. It is therefore plain that they provide a good setting for Poincaré inequalities. On the contrary, it is not possible to work just with a.e. differentiable functions: for instance the famous Cantor function (a.k.a. the Devil's stairs) increases from 0 to 1 but is a.e. differentiable with zero derivative. However, everywhere differentiable functions with locally integrable derivative are weakly differentiable, see e.g. [START_REF] Rudin | Real and complex analysis[END_REF].

To be very precise, we should have denoted the Poincaré constant as C P (µ, Ω). Indeed, we could also consider the probability measure µ on Ω as acting on any larger open interval Ω . The restriction to Ω of weakly differentiable functions on Ω are specific weakly differentiable function on Ω (they are continuous at boundary points). Therefore, satisfying a Poincaré inequality on Ω is formally a more demanding property. From now on, we will assume that Ω is the interior of the convex hull of the support of µ, which is consistent with the notation C P (µ).

Obviously, some measures cannot satisfy a Poincaré inequalities, for instance the uniform measure on (0, 1) ∪ (2, 3) (one can choose a differentiable function f on (0, 3) which is equal to 0 on (0, 1) and to 1 on [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF][START_REF] Ané | Sur les inégalités de Sobolev logarithmiques[END_REF]. Then (f ) 2 dµ = 0).

Next let us present an equivalent definition of Poincaré inequalities, in a more convenient analytic setting. Let L 2 (µ) be the set of (equivalence classes for a.e. equality of) measurable functions on Ω such that Ω f 2 dµ < +∞. We denote . L 2 (µ) , or simply . , its associated norm: f = Ω f 2 dµ 1/2 . The first weighted Sobolev space H 1 µ (Ω) is defined by

H 1 µ (Ω) = {f ∈ L 2 (µ) such that f ∈ L 2 (µ)},
where f is the weak derivative of f . It is well known that H 1 µ (Ω) is an Hilbert space with the norm f 2

H 1 µ (Ω) = f 2 + f 2 .
More generally, for an integer ≥ 1, the space H µ (Ω) is defined by:

H µ (Ω) = {f ∈ L 2 (µ) such that for all k ≤ , f (k) ∈ L 2 (µ)}
where f (k) is the k-th weak derivative of f . An important particular case is when Ω is bounded and there exists two positive constants m, M such that

∀t ∈ Ω, 0 < m ≤ ρ(t) ≤ M
Then if Leb denotes the Lebesgue measure on Ω, it holds that L 2 (µ) = L 2 (Leb) and H µ (Ω) = H Leb (Ω) for all positive integers , and the norms on these spaces are equivalent when µ is replaced by Leb. In other words, the weighted Sobolev spaces are equivalent to the usual Sobolev spaces. This remark will allow us to use several results which are available for Leb and a bounded Ω. Firstly, H 1 µ (µ) then contains functions which are continuous on Ω and piecewise C 1 on Ω. Secondly, the spectral theory of elliptic problems (see e.g. [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF], Chap. 7) will then be valid. We now come back to the general case. Definition 2. Let µ(dt) = ρ(t)dt be an absolutely continuous probability measure on an open interval Ω, such that ρ > 0 almost everywhere on Ω. We say that µ admits a Poincaré inequality on Ω if there exists a constant C < +∞ such that for all f in H 1 µ (Ω) verifying Ω f dµ = 0, we have:

Ω f 2 dµ ≤ C Ω (f ) 2 dµ (2.2)
The best possible constant C is denoted C P (µ). If there exists f opt , a centered function of H 1 µ (Ω), such that (2.2) is an equality for C = C P (µ), we say that the inequality is saturated by f opt .

This definition means Var µ (f ) ≤ C(µ) Ω (f ) 2 dµ for weakly differentiable functions f such that f and f are square integrable for µ. Hence Definition 1 is formally stronger as it involves general weakly differentiable functions. Nevertheless, the Poincaré inequality of Definition 2 implies the one of Definition 1: first it is enough to consider functions f with square integrable weak derivative f , then one can apply the Poincaré inequality to certain truncations of f , which belong to H 1 µ (Ω), in order to show that f is also necessarily square integrable for µ.

Poincaré inequalities are often stated in terms of the ratio of energies of a function and its derivative:

Definition 3. Let f in H 1 µ (Ω) with f 2 dµ > 0.
The Rayleigh ratio of f is:

J(f ) = f 2 f 2 = Ω f 2 dµ Ω f 2 dµ (2.3)
Thus µ admits a Poincaré inequality if and only if the Rayleigh ratio admits a positive lower bound over the subspace of centered functions

H 1,c µ (Ω) = {f ∈ H 1 µ (Ω), Ω f dµ = 0}. In that case, C P (µ) = inf f ∈H 1,c µ (Ω)-{0} J(f ) -1 = sup f ∈H 1,c µ (Ω)-{0} J(f ) -1
The expression of C P (µ) as a supremum makes it easy to compute lower bounds, by choosing an appropriate test function f . An example is given below:

Proposition 1. C P (µ) ≥ Var µ , with equality if µ = N (0, 1).

Proof. The lower bound is obtained for f (x) = x -E µ . The equality case is well-known (see e.g. [START_REF] Ané | Sur les inégalités de Sobolev logarithmiques[END_REF]).

A general criterion

The class of probability measures that admit a Poincaré inequality has been completely characterized in the work of Muckenhoupt [START_REF] Muckenhoupt | Hardys inequality with weights[END_REF]. See also [START_REF] Bobkov | Hardy Type Inequalities via Riccati and Sturm-Liouville Equations[END_REF] for refinements.

Theorem 1 ( [START_REF] Muckenhoupt | Hardys inequality with weights[END_REF]). Let µ(dt) = ρ(t)dt be a probability measure on Ω = (a, b).

Let m be a median of µ and define:

A -= sup a<x<m µ (a, x) m x 1 ρ(t) dt, A + = sup m<x<b µ (x, b) x m 1 ρ(t) dt,
with the convention 0 • ∞ = 0. Then µ admits a Poincaré inequality iff A -and A + are finite, and in this case

1 2 max(A -, A + ) ≤ C P (µ) ≤ 4 max(A -, A + ). (2.4)
This result explains how to estimate, up to a multiplicative factor, the Poincaré constant by a simpler quantity. As explained in [START_REF] Barthe | Interpolated inequalities between exponential and Gaussian, Orlicz hypercontractivity and isoperimetry[END_REF], it can be interpreted as a reduction from functions to sets (by decomposition functions according to their level sets): a Poincaré inequality is equivalent to a comparison between the measures of sets, and a notion of µ-capacity. In one dimension, a further reduction allows to restrict to increasing function vanishing at m, for which level sets for positive values are of the form (x, b).

Perturbation and transport

Given a measure verifying a Poincaré inequality, one may try to deform it into another measure still having a finite Poincaré constant. Many such results exist in the literature. Here we mention two fundamental ones: the bounded perturbation principle, and the Lipschitz transportation principle. They hold in very general settings, but for the purpose of this article, it is enough to state them on R. Lemma 1. Let µ be a probability measure on Ω ⊂ R. Let ψ : Ω → R be a bounded measurable function and let μ be the probability measure given by dμ = e ψ dµ/Z where Z is the normalizing constant. Then

C P (μ) ≤ e sup ψ-inf ψ C P (µ).
This fact is easily proved at the level of Poincaré inequalities by using Var μ = inf a∈R (f -a) 2 dμ and applying obvious bounds on ψ. Lemma 2. Let µ be an absolutely continuous probability measure on an open interval Ω ⊂ R. Let T : Ω → R be a Lipschitz map, (i.e. verifying that there exists L ∈ R such that for all x, y ∈ Ω, |T (x) -T (y)| ≤ L|x -y|) . Assume that T µ, the image measure of µ by T , is also absolutely continuous. Then

C P (T µ) ≤ T 2 Lip C P (µ),
where T Lip is the smallest possible value of the constant L above.

Proof. It is convenient here to work with locally Lipschitz functions (i.e. functions which are Lipschitz on any compact interval). They are weakly differentiable with locally bounded derivative (see e.g. [START_REF] Brézis | Analyse fonctionnelle[END_REF]). However one can also define pointwise and as a whole their "absolute value of the derivative" as

(f ) = Var ν (g). Observe that f = g • T is also locally Lipschitz and verifies |f |(x) ≤ L|g |(T (x)) for all x. Consequently |f | 2 dµ ≤ L 2 (|g |(T (x))) 2 dµ(x) = L 2 |g | 2 dν.
Hence we have proved that Var ν (g) ≤ C P (µ)L 2 |g | 2 dν. The result follows.

In the case of probability measures µ, ν on the real line, and when µ has no atoms, a natural map which pushes µ forward to ν is the monotonic map T := F -1 ν • F µ (here F -1 ν stands for the generalized left inverse). It remains to estimate the Lipschitz norm of T . Lemma 3. Let µ and ν be probability measures on R. Assume that µ(dt) = ρ µ (t)dt where ρ µ is positive and continous on (a µ , b µ ) (-∞ ≤ a µ , b µ ≤ +∞) and vanishes outside. Let us make the same structural assumption for ν. Then T :=

F -1 ν •F µ is well defined and differentiable on (a µ , b µ ), with T = ρ µ /ρ ν •F -1 ν •F µ . Consequently its Lipschitz norm is T Lip = sup (aµ,bµ) ρ µ ρ ν • F -1 ν • F µ = sup (0,1) ρ µ • F -1 µ ρ ν • F -1 ν = sup (aν ,bν ) ρ µ • F -1 µ • F ν ρ ν .
The previous two lemmas can be applied when µ(dx) = exp(-|x|)dx/2, the double exponential measure. In this case ρ ν •F -1 ν (t) = min(t, 1-t) and C P (µ) = 4. This is how Bobkov and Houdré [START_REF] Bobkov | Isoperimetric constants for product probability measures[END_REF] established the following estimate:

C P (ν) ≤ 4 sup x∈(aν ,bν ) min(F ν (x), 1 -F ν (x)) ρ ν (x) 2 .
(2.5)

These authors also deduced from this approach that for log-concave probability measures ν on R, with median m, C P (ν) ≤ 1/ρ ν (m) 2 .

Actually, we can improve on (2.5) by choosing the logistic measure µ(dx) = e x (1+e x ) 2 dx instead of the double exponential measure in the proof. Indeed, in this case ρ µ • F -1 µ (t) = t(1 -t) is smaller than min(t, 1 -t), while the Poincaré constant is the same C P (µ) = 4, as shown in [START_REF] Barthe | Isoperimetry and stability of hyperplanes for product probability measures[END_REF]. Consequently

F ν (x)(1 -F ν (x)) ρ ν (x) 2 .
Remark 2. We have mentioned in the introduction that rewriting the sensitivity analysis questions in terms of uniform variables does not provide good results for our purposes. This approach amounts to consider the monotone transport from uniform variables to the variables of interest, and this transport is not Lipschitz for unbounded variables. The results of the present section show that rewriting the problem in terms of logistic variables is more effective.

Spectral interpretation

When µ has a continuous density that does not vanish on a compact interval [a, b], then µ admits a Poincaré constant on [a, b], as follows from the Muckenhoupt condition (Theorem 1) or from the bounded perturbation principle (Lemma 1). Furthermore, modulo a small additional regularity condition, the Poincaré inequality is saturated and the Poincaré constant is related to a spectral problem, as detailed now.

(P1) Find f ∈ H 1 µ (Ω) s.t. J(f ) = f 2 f 2 is minimum under f dµ = 0 (P2) Find f ∈ H 1 µ (Ω) s.t. f , g = λ f, g ∀g ∈ H 1 µ (Ω) (P3) Find f ∈ H 2 µ (Ω) s.t. f -V f = -λf and f (a) = f (b) = 0
Then the eigenvalue problems (P2) and (P3) are equivalent, and their eigenvalues form an increasing sequence (λ k ) k≥0 of non-negative real numbers that tends to infinity. Moreover 0 is an eigenvalue, and all eigenvalues are simple. The eigenvectors (u k ) k≥0 form a Hilbert basis of L 2 (µ) and verify:

u k (x) = λ k ρ(x) b x u k (t)ρ(t)dt (2.6)
In particular, up to a modification with Lebesgue measure zero, the

u k 's are C 1 on [a, b] and if ρ is of class C k , u k is of class C k+1 .
Furthermore when λ = λ 1 , the first positive eigenvalue, (P 2) and (P 3) are equivalent to (P 1) and the minimum of (P 1) is attained for f = u 1 . In other words, the optimal Poincaré constant is C P (µ) = 1/λ 1 and the inequality is saturated by a non-zero solution of (P3). Finally, u 1 is strictly monotonic.

Although the equivalence between (P1) and (P3) is well-known, it was not easy to find a self-contained proof. Several arguments can be found in the literature of functional inequalities, and especially in [START_REF] Bobkov | Hardy Type Inequalities via Riccati and Sturm-Liouville Equations[END_REF] for mixed Dirichlet and Neumann boundary conditions f (a) = f (b) = 0. We also found complementary arguments in the literature of numerical analysis, for instance in [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF], for the uniform distribution. Since neither of them are fully corresponding to our setting, we have inserted below a proof of Theorem 2, making a synthesis of the ideas of the two literature fields. In particular, we adopt the Hilbertian point of view chosen in [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF], which is central for estimating the optimal Poincaré constant, as seen in Section 4.2, and adapt the short and elementary proofs of [START_REF] Bobkov | Hardy Type Inequalities via Riccati and Sturm-Liouville Equations[END_REF] coming under more general principles (maximum principle, regularity theorem).

Proof of Theorem 2. Since Ω is bounded an ρ is continuous on Ω, there exists two positive real numbers m, M such that:

∀t ∈ Ω, 0 < m ≤ ρ(t) ≤ M
Then, as mentioned in § 2.1, L 2 (µ) and all weighted Sobolev spaces H µ (Ω) are topologically unchanged when we replace µ by the Lebesgue measure on Ω. This allows using the spectral theory of elliptic problems (see e.g. [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF], Chap. 7) on usual Sobolev spaces with bounded Ω.

More precisely, let us consider problem (P2), and denote

H = L 2 (µ) and V = H 1 µ (Ω).
Then the injection V ⊂ H is compact, and V is dense in H. (P2) can be written as an eigenvalue problem of the form:

a(f, g) = λ f, g ∀g ∈ V (2.7)
where a(f, g) = f , g and ., . denotes the scalar product on H. In order to apply spectral theory, a should be coercive with respect to V, i.e. ∃C > 0 s.t.

a(f, f ) ≥ C f, f H 1 µ (Ω)
for all f , which is not true here. A convenient way to overcome this issue is to consider the equivalent shifted problem (see [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF], §8.2.):

α(f, g) = (λ + 1) f, g ∀g ∈ V (2.8)
where α(f, g) = f , g + f, g is the usual scalar product on V. It is coercive on V. Then we can apply Theorem 7.3.2 in [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF]: The possible eigenvalues form an increasing sequence (λ k + 1) k≥0 of non-negative values that tends to infinity and the eigenvectors (u k ) k≥0 form a Hilbert basis of L 2 (µ). Further, from (2.7) with f = g, we have λ k ≥ 0. Now remark that 0 is an eigenvalue and the corresponding eigenspace is spanned by the constant function u 0 = 1. Indeed,

taking g = f in (2.2) leads to b a f (x) 2 e -V (x) dx = 0 and f is a constant function.
Now let us prove the equivalence between (P2) and (P3). We restrict the presentation to λ > 0 since the case λ = 0 is direct, using a Sturm-Liouville form of (P3), i.e. (f V ) = -λf . Formally the link comes from an integration by part of the left hand side of (P2):

b a f g ρ = f (b)ρ(b)g(b) -f (a)ρ(a)g(a) - b a (f ρ) g (2.9)
Since this must be equal to λ b a f gρ for all g ∈ H 1 µ (Ω), we should have f (a) = f (b) = 0 and (f ρ) = -λf ρ which is problem (P3). Actually this method, read in the reverse sense, shows that (P3) implies (P2). Indeed if f ∈ H 2 µ (Ω) then f ρ ∈ H 1 µ (Ω) and the integration by part is valid. However, to see that (P2) implies (P3), an argument of regularity must be used since f is only assumed to belong to H 1 µ (Ω). This is achieved by proving that, µ-almost surely,

f (x) = λ ρ(x) b x f (t)ρ(t)dt.
(2.10)

Indeed, this implies that f ρ is C 1 on [a, b] and (2.9) is valid. Furthermore, (2.10) also implies f (a) = f (b) = 0 and this gives (P3). The proof of (2.10) is postponed to the appendix. Now let us prove that all eigenvalues are simple. Notice that, from (2.10), all solutions of (P3) are locally Lipschitz, up to a modification of Lebesgue measure 0. Thus when λ is an eigenvalue, the vector space of solutions of the second order linear differential equation f -V f = -λf has dimension 2. Adding Neumann conditions, the dimension is equal to 1. Indeed, let f and g be two solutions of (P 3). Then f (b) = 0 otherwise f would be identically null by uniqueness of the Cauchy problem (since f (b) = 0). Hence there exists a real number c such that

g(b) = cf (b). Then consider h = g -cf . We have h(b) = h (b) = 0 implying, by unicity of the Cauchy problem, that h = 0, i.e. g = cf .
In this Hilbertian framework, the equivalence between (P2) and (P1) is easily seen

. Indeed, if f ∈ H 1 µ (Ω) verifies f dµ = 0 then f is orthogonal to 1 and hence f = +∞ k=1 f k u k . Then, it holds: J(f ) = f 2 f 2 = α(f, f ) f 2 -1 = +∞ k=1 λ k f 2 k +∞ k=1 f 2 k ≥ λ 1
with equality iff f is proportional to u 1 . This also proves that the minimum in (P1) is attained precisely on the 1-dimensional space spanned by u 1 .

Finally we report to the Appendix the proof of monotonicity properties as well as connection to variation calculus. Remark 3. Another solution to make coercive the bilinear form a(f, g) = f , g is to project onto the space of centered functions H 1,c µ (Ω), as done in [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF]. Indeed, on that space, coercivity is equivalent to the existence of a Poincaré constant, which can be proved independently by the bounded perturbation principle as mentioned above. However, this seems a less convenient setting for the numerical method developed in Section 4.2 since the 'hat' functions used do not belong to H 1,c µ (Ω). Definition 4 (Spectral gap). The smallest strictly positive eigenvalue

λ 1 of Lf = f -V f with Neumann boundary conditions f (a) = f (b) = 0 is called
Neumann spectral gap, or simply spectral gap, and denoted by λ(µ).

We conclude this section by a proposition showing that, under regularity conditions on V , the Neumann problem (P3) can be written as a Dirichlet problem.

Proposition 2. If V is of class C 2 , then (P3) is equivalent to: (P4) Find h ∈ H 2 µ (Ω) s.t. h -V h -V h = -λh and h(a) = h(b) = 0
Proof. Define L and K the operators

Lf = f -V f Kf = f -V f -V f
Let f be a solution of (P3). Since V is of class C 2 then by Theorem 2, f ∈ H 3 (µ). Now define h = f . We have: K(f ) = (Lf ) = -λf and h is solution of (P4).

Conversely, let h be a solution of (P4) and let f be the primitive function of h such that f dµ = 0. Then (Lf ) = -λf and there exists c ∈ R such that

Lf = -λf + c. Remarking that b a L(f )e -V = b a
(f e -V ) = 0, the centering condition implies c = 0 which proves that f is a solution of (P3).

More techniques

We conclude this quick overview by briefly quoting two other methods.

The first one plays a major role in the study of multidimensional functional inequalities. It has its roots in early works by Hörmander and Brascamp-Lieb, and was developped by Bakry and his collaborators who introduced and exploited a notion of curvature and dimension for diffusion generators, see e.g. [START_REF] Ané | Sur les inégalités de Sobolev logarithmiques[END_REF]. The simplest instance of these results is as follows:

Theorem 3. Let µ be a probability measure on an open interval Ω with dµ(x) = e -V (x) 1 Ω (x) dx where V is twice continuously differentiable. If there exists r > 0 such that for all x ∈ Ω, V (x) ≥ r then C P (µ) ≤ 1/r.
The estimate is exact for the standard Gaussian measure.

The second result is known as Chen's variational formula [START_REF] Chen | Analytic proof of dual variational formula for the first eigenvalue in dimension one[END_REF]. Its main advantage is to express the Poincaré constant as an infimum (while, as already mentioned, its definition is given as a supremum). This allows to obtain upper bounds rather easily. We state it in a different form, which appeared in [START_REF] Djellout | Lipschitzian norm estimate of one-dimensional Poisson equations and applications[END_REF].

Theorem 4. Let dµ(x) = e -V (x) 1 [a,b] (x)dx be a probability measure on a seg- ment, with V twice continuously differentiable on [a, b]. Then λ(µ) = 1 C P (µ) = sup g >0 inf (a,b) (-Lg) g ,
where Lf = f -V f and the supremum is taken over three times differentiable functions on (a, b), having positive first derivative.

As an example of application, Chen's formula recovers the following classical statement, which often allows to calculate the spectral gap. It is the reciprocal of the fact that a solution of the spectral problem (P3) corresponding to the smallest positive eigenvalue is necessarily strictly monotonic (one part of Theorem 2).

Corollary 2. Under the framework and notations of Section 2.4, if f is a strictly monotonic solution of problem (P3)

Lf = -λf, f (a) = f (b) = 0
for some λ > 0, then λ is the spectral gap. The same is true for problem (P4) replacing 'monotonic' by 'non vanishing'.

Proof. By definition of the spectral gap, λ(µ) ≤ λ. Conversely, up to a change of sign we can assume that f > 0 on (a, b). Then Chen's formula immediately gives λ(µ) ≥ inf (a,b) (-Lf ) f = λ. The last part comes from the equivalence between (P3) and (P4): If h is a non vanishing solution of (P4), then the primitive function of h such that f dµ = 0 is a strictly monotonic solution of (P3) (see the proof of Prop. 2).

Poincaré inequalities on intervals

This section gathers tools for the study of the Poincaré constants of the restrictions of a given measures to subintervals.

General results

The Poincaré constant cannot increase by restriction to a subinterval: Consequently, the map defined on intervals by

I → C P (µ |I ) is non-decreasing. Proof. Let f ∈ H 1 (µ |I ).
Set Ω = (α, β) and I = (α , β ). If α < α then by hypothesis, ρ is bounded from above and below in a neighborhood of α , which ensures that f has a limit at α , and can be continuously extended by a constant on (α, α ]. The same applies if β < β. This allows to build a weakly differentiable function f on Ω which concides with f on I and has zero derivative outside I. The result easily follows:

Var µ |I (f ) = inf a I (f -a) 2 dµ µ(I) ≤ inf a Ω ( f -a) 2 dµ µ(I) ≤ C P (µ) Ω ( f ) 2 dµ µ(I) = C P (µ) I (f ) 2 dµ µ(I) = C P (µ) I (f ) 2 dµ |I .
The previous result helps proving a continuity type property with respect to the support: Lemma 5. Consider again the framework of Lemma 4, and let I be a family and increasing subintervals such that I ↑ Ω when → 0. Typically, I = (a + , b -) for Ω = (a, b), I = (a, a + 1/ ) for Ω = (a, +∞) and so on. Then the Poincaré constant on Ω is the limit of the Poincaré constant on subintervals:

C P (µ) = lim →0 C P (µ |I )
Proof. Firstly, by Lemma 4, the sequence C P (µ |I ) is increasing and bounded by C P (µ). Hence C P (µ) ≥ sup

>0 C P (µ |I ) = lim →0 C P (µ |I ). Conversely, let f ∈ H 1 µ (Ω).
Observe that, by Lebesgue theorem,

µ(I ) = I ρ(t)dt → →0 µ(Ω) = 1
and for all µ-integrable function g:

E µ (g) = Ω g(t)ρ(t)dt = lim →0 I g(t)ρ(t)dt = lim →0 I g(t) ρ(t) µ(I ) dt = lim →0 E µ |I (g).
Applying this result to f , f 2 and f 2 , it holds:

Var µ (f ) Ω f 2 dµ = lim →0 Var µ |I (f ) I f 2 dµ |I ≤ lim →0 C P (µ |I ) Since it is true for all f ∈ H 1 µ (Ω), we get C P (µ) ≤ lim →0 C P (µ |I )

Improvements for symmetric settings

In general, the Poincaré constant of a restriction to an interval is at most the one of the initial measure. A better result is available in symmetric settings: Lemma 6. Let µ be a probability measure on R. Assume µ(dx) = ρ µ (x)dx where ρ µ is a unimodal function with a maximum at x = 0. Let I ⊂ R be a nonempty interval and ν := µ(•∩I)/µ(I). Assume in addition that µ(R -) = ν(R -).

Then the monotonic transport from µ to ν is µ(I)-Lipschitz. Consequently

C P (ν) ≤ µ(I) 2 C P (µ).
Proof. By hypothesis, F ν (0) = F µ (0). Hence T (0) = 0 and since T is increasing, T (x) and x have the same sign. We claim that for all x (in the interior of the support of µ), |T (x)| ≤ |x|. Indeed for x ≥ 0, this inequality amount to F µ (x) ≤ F ν (x). The latter is obvious when x > sup(I) since in this case F ν (x) = 1. For 0 ≤ x ≤ sup(I), the claim boils down to

µ(R -) + x 0 ρ µ ≤ ν(R -) + x 0 ρ µ µ(I) , which is obvious using µ(R -) = ν(R -). For x ≤ 0, the claim amounts to x ≤ T (x), that is F ν (x) ≤ F µ (x). This is obvious if x < inf(I). If inf(I) ≤ x ≤ 0, the inequality amounts to ν(R -) - 0 x ρ µ µ(I) ≤ ν(R -) - 0 x ρ µ ,
which is easily veryfied. Eventually, for x in the interior of the support of µ,

T (x) = ρ µ (x) ρ ν (T (x)) = µ(I) ρ µ (x) ρ µ (T (x)) ≤ µ(I),
since on R + , 0 ≤ T (x) ≤ x and ρ µ is non-increasing, and on R -, x ≤ T (x) ≤ 0 and ρ µ is non-decreasing.

Our goal here is to get finer results when the underlying measure is even. A first natural observation is that extremal (or almost extremal) functions for the Poincaré inequality can be found among odd functions. Lemma 7. Let µ be an even probability measure on R. Assume that it is supported on a symmetric interval, and is absolutely continuous, with a positive continuous density in the interior of this interval. Then for any non constant function f ∈ H 1 (µ), there exists an odd function g ∈ H 1 (µ) with

(f ) 2 dµ Var µ (f ) ≥ (g ) 2 dµ Var µ (g) • Proof. Let (-b, b
) denote the interior of the support of µ. By hypothesis, f (0) is well defined and by the variational definition of the variance

(f ) 2 dµ Var µ (f ) ≥ (f ) 2 dµ (f -f (0)) 2 dµ = 0 -b (f ) 2 dµ + b 0 (f ) 2 dµ 0 -b (f -f (0)) 2 dµ + b 0 (f -f (0)) 2 dµ .
Using the inequality u+v x+y ≥ min( u x , v y ), which is valid for non-negative numbers u, v, x, y such that x + y > 0 (and with the convention that v/0 = +∞), we get that (f )

2 dµ Var µ (f ) ≥ min 0 -b (f ) 2 dµ 0 -b (f -f (0)) 2 dµ , b 0 (f ) 2 dµ b 0 (f -f (0)) 2 dµ
.

In order to conclude, we consider the odd functions g, h defined on (-b, b), such that for all x ∈ [0, b), g(x) = f (x) -f (0), and for all x ∈ (-b, 0], h(x) = f (x) -f (0). By symmetry of µ,

0 -b (f ) 2 dµ 0 -b (f -f (0)) 2 dµ = (h ) 2 dµ Var µ (h) , and b 0 (f ) 2 dµ b 0 (f -f (0)) 2 dµ = (g ) 2 dµ
Var µ (g) •

Next, we derive from the above proposition a perturbation principle for the Neumann spectral gap. It should be compared with the "bounded perturbation principle" that we recalled in the previous section. x) dx be an even probability measure, with continuous potential V . Let ϕ : (-b, b) → R + be an even function. Assume that ϕ is non-increasing on [0, b) and that μ defined by dμ = ϕ dµ is a probability measure. Then C P (μ) ≤ C P (µ).

Proposition 3. Let dµ(x) = 1 (-b,b) e -V (
Proof. By the previous lemma, it is enough to establish a Poincaré inequality for μ for odd functions only. Let f be an odd weakly differentiable function. By Lemma 4, for any a ∈ (0, b), C P (µ |(-a,a) ) ≤ C P (µ). Since f is odd and µ is even, (-a,a) f dµ = 0, and the inequality Var

µ |(-a,a) (f ) ≤ C P (µ) (f ) 2 dµ |(-a,a)
can be rewritten as

f 2 1 (-a,a) dµ ≤ C P (µ) (f ) 2 1 (-a,a) dµ.
Next, for all x ∈ (-b, b),

ϕ(x) = +∞ 0 1 t≤ϕ(x) dt = +∞ 0 1 It (x) dt,
where I t = ϕ -1 [t, +∞) . Our hypotheses on ϕ ensure that I t is empty, or is a symmetric interval included in (-b, b). Hence, if it is not empty, it is equal, up to a negligible set to (-a, a) for some a. Therefore we know that

f 2 1 It dµ ≤ C P (µ) (f ) 2 1 It dµ,
is true for all t ≥ 0. But, by definition, pointwise, ϕ = R + 1 It . Thus integrating the latter inequality over t yields

f 2 ϕ dµ ≤ C P (µ) (f ) 2 ϕ dµ.
This can be rewritten as Var μf ≤ C P (µ) (f ) 2 dμ.

As an example of application, using V ≡ 0 and φ(x) = e -x 2 /2 , one can bound the Poincaré constant of the Gaussian measure restricted to [-b, b] by the one of the uniform measure on that interval:

C P (N (0, 1)|[-b, b]) ≤ 4b 2 π 2 .

Exact values of Poincaré constants

In this section, we show how to derive a Poincaré constant on an interval Ω either as a first zero of a function expressed analytically (semi-analytical approach), or by a full numerical method based on finite elements. We consider the setting of the spectral theorem (Thm. 2): Ω bounded and ρ > 0 on Ω (with ρ continuous on Ω and piecewise C 1 on Ω). The general case can be derived from it by the continuity type property of Lemma 5. Hence, if Ω is unbounded or ρ vanishes at the boundary of Ω, then the Poincaré constant is the limit of Poincaré constants on bounded subintervals on which ρ > 0.

Semi-analytical results

This section presents some examples where the spectral gap is obtained in a closed-form expression or by numerically searching a zero of a function expressed analytically. They are summarized in Table 1. The most famous one is about the uniform distribution on [a, b], corresponding to a spectral problem for the Laplacian operator, f = -λf . Another well-known result is about the truncated normal, for the intervals formed by consecutive zeros of Hermite polynomials. After recalling this result, we show how to extend it for a general interval and develop the case of truncated (double) exponential and triangular distributions. Without loss of generality, we will consider only one value for the parameters, remarking that the general result is obtained by rescaling (Lemma 2 applied to T and T -1 when T is an affine function).

In all these examples, solutions of the second-order differential equation can be expressed analytically as a linear combination of odd and even known functions. The whole spectrum then corresponds to the zeros of one determinant obtained with boundary conditions, as detailed in the following proposition.

Prob. Dist. µ Support C P (µ) Form of f opt (x) Uniform -1 2 , 1 2 1/π 2 sin(πx) N (0, 1) R 1 x [r n,i , r n,i+1 ] 1/(n + 1) H n+1 (x) [a, b] see Prop. 8 related to Kummer's functions Double exp. R 4 × [a, b] ⊆ R + 1 4 + π b-a 2 -1 e x/2 cos π b-a x + φ General case See Prop. 5 Triangular [-1, 1] r -2 1 ≈ 0.1729 sign(x)J 0 (r 1 (1 -|x|))
Table 1. Examples of analytical results of the spectral problem with Neumann conditions. We have used the following notations: Hn is the Hermite polynomial of degree n and r n,i its i th zero, J 0 is the Bessel function of the first kind with ν = 0, and r 1 is its first root.

Proposition 4. Let f λ , g λ a basis of solutions of the second-order differential equation u -V u = -λu. Then the spectral gap on [a, b] is the first zero of

d(λ) = det f λ (a) g λ (a) f λ (b) g λ (b) = f λ (a)g λ (b) -f λ (b)g λ (a)
Furthermore if V is even and a = -b, the spectral gap is the first zero of λ → g λ (b).

Remark 4. This result is immediately adapted to the equivalent spectral problem with Dirichlet conditions (P4), u -V u -V u = -λu. In that case, with similar notations, the spectral gap is the first zero of

d(λ) = det f λ (a) g λ (a) f λ (b) g λ (b) = f λ (a)g λ (b) -f λ (b)g λ (a)
and if a = -b with even V , it is the first zero of λ → f λ (b).

Proof of Proposition 4. If the interval is symmetric and V even, then by Lemma 7 a solution can be found among odd functions, and we can assume that u = g λ . The Neumann conditions are then equivalent to g λ (b) = 0, which proves the result.

In the general case, let u be a non-zero solution of the spectral problem. Then u can be written as u = Af + Bg, for some A, B ∈ R. The boundary conditions lead to the linear system Xc = 0, with c = A B and

X = f λ (a) g λ (a) f λ (b) g λ (b) .
Now there exists a non zero solution if and only if the system is degenerated, i.e. d(λ) = 0. This gives the whole spectrum and, in particular, the spectral gap is the first zero. 1 , approximately equal to 0.1729, saturated by f (x) = sign(x)J 0 (r 1 (1 -|x|)).

Remark 5. The result, involving Bessel functions, shows a connection with the spectrum of the Laplacian on the unit disk D. Actually, using the symmetry of the problem, we can see that the spectral problem for the triangular distribution is the same than for the Laplacian on D restricted to radial solutions, as investigated e.g. in [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF], § 8.1.1d. Details are given in the appendix.

Truncated normal distribution

Let us consider the standard Gaussian distribution dγ(

x) = 1 √ 2π e -V (x) with V (x) = x 2
2 . The spectral problem (P3) associated to Poincaré inequality is relative to the operator Lf = f -V f :

Lf = f -xf
The spectral gap is known on every interval formed by successive roots of Hermite polynomials (see e.g. [START_REF] Dautray | Mathematical Analysis and Numerical Methods for Science and Technology[END_REF]), as detailed in Prop. 7. Recall that Hermite polynomials H n are the orthonormal basis of polynomials for L 2 (dγ) with unitary highest coefficient:

∞ -∞ H n (x)H m (x)dγ(x) = δ n,m
. The first ones are:

H 0 (x) = 1 H 1 (x) = x H 2 (x) = x 2 -1 H 3 (x) = x 3 -3x
and the following can be obtained by the recursive relation:

H n+1 (x) = xH n (x)- H n (x)
. They are also shown to verify the identity H n+1 (x) = (n + 1)H n (x) which proves, together with the recursion formula, that H n is solution of the differential equation Lf = -nf .

Proposition 7 (Spectral gap of the normal distribution, specific case). Let n ≥ 1, x 1 < • • • < x n the roots of the n-th Hermite polynomial H n . Let I be one of the intervals:

(-∞, x 1 ], [x 1 , x 2 ],. . . ,[x n-1 , x n ], [x n , +∞). Then the (Neumann) spectral gap on I is n + 1.
Proof. On each I, the function H n is of constant sign and vanishes on the boundary. Since H n+1 = (n+1)H n then inside I, H n+1 is strictly monotonic and H n+1 = 0 on the boundary. Furthermore, H n+1 is solution of Lf = -(n + 1)f . Hence, by Corollary 2, n + 1 is the spectral gap.

In the general case, the spectral gap of the truncated normal distribution is related to hypergeometric series, and more specifically to confluent hypergeometric series or Kummer's function ( [START_REF] Abramowitz | Handbook of Mathematical Functions[END_REF], Section 13). The Kummer's function M a1,b1 (z) = 1 F 1 (a 1 ; b 1 ; z) is a series p≥0 x p with x 0 = 1 satisfying

x p+1 x p = (p + a 1 )z (p + b 1 )(p + 1) (4.1)
Proposition 8 (Spectral gap of the truncated normal distribution). Let a < b, and define:

h 0 (λ, t) = M 1-λ 2 ; 1 2 t 2 2 h 1 (λ, t) = M 2-λ 2 ; 3 2 t 2 2
Notice that h 0 and h 1 generalize Hermite polynomials: When λ is an odd (resp. even) positive integer, then t → h 0 (λ, t) (resp. t → th 1 (λ, t)) is equal, up to a multiplicative constant, to the Hermite polynomial of degree λ -1; In particular they have the same zeros ([1], formulas 13.6.17

Numerical method

In this section we present a numerical method to estimate the spectral gap. The technique is well-known in the literature of elliptic problems (see e.g. [START_REF] Raviart | Introduction à l'analyse numérique des équations aux dérivés partielles[END_REF] or [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF]), and we adapt it to our case. We consider the framework and notations of Section 2.4. The spectral problem (P3),

f -V f = -λf, f (a) = f (b) = 0
can be solved numerically by using finite element methods.

The first step is to consider the variational problem (P2), rewritten under the form (see the Proof of Theorem 2):

α(f, g) = (λ + 1) f, g ∀g ∈ H 1 µ (Ω) (4.2)
where α(f, g) = f , g + f, g is coercive on H 1 µ (Ω), and to show its equivalence to (P3). This was done in Theorem 2.

The second step is to observe that (4.2) can be solved algebraically in a finite-dimensional space. Denote G h a finite-dimensional space of H 1 µ (Ω), where h > 0 is a discretization parameter, and let (g i ) 0≤i≤n be a basis of G h . Typically, G h = P 1 , the space of Lagrange finite elements, composed of piecewise linear functions on [a, b]. A basis is formed by the 'hat' functions g i represented in Figure 1. A solution in G h of (4.2) is given by

0 h g 0 1 (n-1)h g n (i-1)h ih (i+1)h g i (i+2)h g i+1 ... ...
f h = n i=0 f h,i g i
and the weak formulation in G h can be written in the matricial form:

K h f h = (λ + 1)M h f h (4.3)
where f h is the vector of (f h,i ) 0≤i≤n , and K h and M h are called respectively 'rigidity matrix' and 'mass matrix', defined by:

K h = (α(g i , g j )) 0≤i,j≤n M h = ( g i , g j ) 0≤i,j≤n
Notice that M h and K h are symmetric and positive definite. Then the problem can be reexpressed in a standard form by using the Choleski decomposition of

M h = L h L T h
, where L h is a lower triangular matrix. Indeed, denoting

K h = L -1 h K h (L T h ) -1 and f h = L T f h , (4.
3) is written:

K h f h = (λ + 1) f h
Thus λ and f h are obtained by performing the eigen decomposition of the symmetric matrix K h . Finally f h is deduced from the relation

f h = (L T h ) -1 f h .
The last step is to observe that the solutions of the finite-dimensional weak formulation (4.3) converge to the solutions of (P2), and equivalently (P3), when h → 0, as stated in [START_REF] Raviart | Introduction à l'analyse numérique des équations aux dérivés partielles[END_REF] (Theorem 6.5.1.), [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF] ( § 6.2.2.), or [START_REF] Babuska | Eigenvalue Problems[END_REF] ( § 8, Eq. 8.43 and § 10.1.2. Eq. 10.22), with a speed of convergence linked to the regularity of the solutions. These results are expressed for the Lebesgue measure on [a, b], but are still valid under the assumptions of Theorem 2 since for all integer , H µ (Ω) is then equal to H Leb (Ω) with equivalent norms (see the proof of Theorem 2). In our situation, the space spanned by the first two eigenfunctions lies in H +1 µ (Ω), with ≥ 1 (see Theorem 2). Hence the smallest strictly positive eigenvalue of (4.3), i.e. the second one, converges to the spectral gap at the speed O(h 2 ), and -since it is a simple eigenvalue -the corresponding eigenvector converges to a function saturating the Poincaré inequality at the speed O(h ).

Remark 6 (Numerical improvements). We briefly mention two well-known improvements for finite elements algorithms, which apply to our case (see e.g. [START_REF] Allaire | Numerical Analysis and Optimization: An Introduction to Mathematical Modelling and Numerical Simulation[END_REF]). Firstly, it is possible to replace the computations of integrals in M h by quadrature formulas ('mass lumping' speed-up technique). This allows obtaining a diagonal matrix and L h is simply the square root of its diagonal terms. The convergence result is still valid. Secondly, if the solution is regular enough (typically f ∈ H 3 µ (Ω)), more regular basis functions can be used (such as P 2 finite elements). The speed of convergence is higher, at the price of a higher computational cost.

Illustrations

We illustrate the results of the previous paragraphs on the truncated double exponential and truncated normal distributions. For the former, the semianalytical result has been used. The numerical method was also tested, and gives a so good approximation that the difference between the two methods could not be seen in Figure 2. For the latter, however, the numerical method based on finite elements has been prefered, rather than searching for the first zero of Kummer's functions. Indeed, in practice, it is not easy to find numerically the first zero of a given function. In particular, it was not possible to justify theoretically the empirical observation that the first zero is the only zero lying in the interval ]0, σ -2 I [, where σ 2 I is the variance of the truncated distribution on I. On the other hand, the numerical method of Section 4.2 is theoretically sound. Implementations have been done with the R software [START_REF] Core | R: A Language and Environment for Statistical Computing[END_REF] and the R package 'orthopolynom' [START_REF] Novomestky | orthopolynom: Collection of functions for orthogonal and orthonormal polynomials[END_REF].

Truncated exponential distribution

Denote by µ the double exponential distribution. Figure 2 shows the Poincaré constant of µ truncated on I = [a, b]. In the general case, we plotted the contour lines of C P (µ) as a function of F µ (a) = P (X ≤ a) and 1 -F µ (b) = P (X ≥ b), where X ∼ µ. In this representation, symmetric intervals correspond to the diagonal y = x, intervals (-∞, 0] to y = 1 2 , and intervals [0, +∞) to x = 1 2 . We also added the upper bound of Prop. 5, which is an exact bound in the regions b ≤ 0 and a ≥ 0, i.e. 1 -F µ (b) ≥ 1 2 and F µ (a) ≥ 1 2 . For symmetric intervals, C P (µ) is represented as a function of the mass µ(I), as well as two upper bounds and the lower bound corresponding to the variance of µ on I, σ 2

I = 2 1 -e -b 1 + b + b 2 2 /F µ (b)
. We can see that the upper bound obtained by symmetric transport 4µ(I) 2 (Lemma 6) is quite large here, except for strong truncations. The upper bound of Prop. 5 is much sharper. The lower bound is also very sharp, except for small truncations, since it tends to 2 when µ(I) tends to 1 whereas the Poincaré constant tends to 4.

Truncated normal distribution

Similar plots have been produced for the normal distribution N (0, 1), truncated on I = [a, b], gathered in Figure 3. In the symmetric case, the upper bound obtained by transport µ(I) 2 (Lemma 6) looks globally accurate, especially for strong truncations. The lower bound given by the variance of the truncated distribution σ 2 I = 1 -2 bφ(b) µ(I) is even sharper. In order to visualize the link to Hermite polynomials, we have added the points corresponding to intervals [-b, b] where -b, b are two successive zeros of Hermite polynomials of degree 2n, up to degree 100. Recall that in that case C P (µ) = 1/(2n + 1).

In the general case, we have added the intervals [r n,i-1 , r n,i ] corresponding to consecutive zeros of Hermite polynomials of degree n (up to n = 100), associated to Poincaré constants equal to 1/(n + 1). We can observe that the whole set of Hermite consecutive zeros poorly fill the space of possible intervals, except for very small symmetric intervals (located around the diagonal x + y = 1). The Poincaré constants obtained with their extensions, namely Kummer's functions, thus provide a useful improvement in many cases. q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q 0.0 0.2 0.4 0.6 0. C opt σ I 2 q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q

Applications

In this section, we perform some global sensitivity analysis by computing the DGSM-based upper bounds of the total Sobol indices (see Equation 1.2), on two hydraulic applications based on different models. For these applications, the common problem under investigation is the assessment of the water level in the terminal section of a watercourse in case of flood. The water level is evaluated as a function of the discharge and other physical parameters which will be detailed hereinafter. Uncertainties on these physical parameters are modelled by probability distributions, chosen from expert knowledge and in accordance with the empirical distributions obtained during measurement campaigns. Flood phenomenon is governed by the Saint Venant shallow water equations, connecting the water level localized in space and time, the discharge, the water section, the lateral inflows, the slope and the head losses due to the friction between the water body and the riverbed. The first application relies on a coarse simplification of these equations while the second application will be based on a 1D solving model.

The computations of Poincaré constants have been done using the R package 'sensitivity' [START_REF] Pujol | sensitivity: Global Sensitivity Analysis of Model Outputs[END_REF] of the R software [START_REF] Core | R: A Language and Environment for Statistical Computing[END_REF]. In this package, the function 'Poincar-eConstant()' allows to compute the upper bounds given by the double exponential transport and logistic transport, while the function 'PoincareOptimal()' provides the Poincaré constant by solving numerically the spectral problem.

First study on a simplified flood model

This application aims at simulating the height of a river and compares it to the height of a dyke that protects industrial facilities. This academic model has been introduced for a pedagogical purpose in several methodological papers (see for example [START_REF] Iooss | A review on global sensitivity analysis methods[END_REF]), and used for illustrating DGSM-based sensitivity analysis in [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF] and [START_REF] Roustant | Crossed-derivative based sensitivity measures for interaction screening[END_REF].

In the case of steady flow, with no inflows, and large rectangular section, assuming that the classical Manning-Strickler formulation is used for the head losses, we obtain the closed-form solution:

S = H + Z v -H d -C b with H =   Q BK s Zm-Zv L   0.6 (5.1)
where the output variable S is the maximal annual overflow (in meters), H is the maximal annual height of the river (in meters), Q is the river flowrate, K s the Strickler's friction coefficient, Z m and Z v the upstream and downstream riverbed levels (w.r.t. a fixed reference), L and B are the length and width of the water section, H d is the dyke height and C b is the bank level. Table 2 gives the probability distributions of the model input variables which are supposed to be independent. Notice that, as in the rest of this paper, we have used here the notation N (µ, σ 2 ) where σ is the standard deviation, contrarily to [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF] which uses N (µ, σ). Hence the standard deviation of K s is equal to 8. Table 3 gives the Poincaré constants and two upper bounds for the different probability distributions used in this test case (the uniform distribution is not represented as the result is well known). For the sake of interpretation, we have considered scaled distributions. Indeed, a simple change of variable (which can also be viewed as a linear transport) shows that:

C P N (µ, σ 2 )|[a, b] = σ 2 C P N (0, 1) a -µ σ , b -µ σ , C P (G(µ, β)|[a, b]) = β 2 C P G(0, 1) a -µ β , b -µ β , C P (T (a, b)) = b -a 2 2 C P (T (-1, 1)),
where the notation |I means that the distribution is truncated on the interval I. We can see the strong decrease factor between the Poincaré constant and the upper bounds based on double exponential transport (gain factor around 6) or logistic transport (gain factor around 2). Figure 4 gives the final global sensitivity analysis results for this test case. It is based on the numerical values obtained in [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF]. In particular, the DGSM ν i (i = 1, . . . , 8) have been computed via a sample of the output derivatives coming from a low discrepancy sequence of the inputs of size 10 000. The previous results of [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF] (given here in Fig. 4, left) have shown that the DGSM-based upper bounds can be used for a screening purpose (i.e. identifying non influential inputs which have a total Sobol index close to zero), but are useless for quantifying the effects of the influential inputs (because of their non-informative high values). In contrary, our new results (Fig. 4, right) show that the DGSM-based upper bounds give a reliable information on the real influence of the inputs (in terms of contribution to the model output variance) thanks to their closeness to total Sobol indices. The hierarchy of influence given by the DGSM-based upper bounds is the same than for total Sobol indices: the flowrate Q is the most influential input, followed by H d , Z v and K s . The remaining four inputs are not active in the model. We can also note that screening is now fully efficient as C b can be judged as non influential (by choosing for example a threshold of 5% of the total variance for the sensitivity index). Remark 7. For the DGSM-based upper bound of K s , our value (Table 3, left) is different from the one of [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF] (Table 5). Indeed, an error is present in [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF] where the multiplicative factor 2π has been omitted (see also the erratum in [START_REF] Roustant | Crossed-derivative based sensitivity measures for interaction screening[END_REF], Remark 3). Hence, in [START_REF] Lamboni | Derivative-based global sensitivity measures: General links with Sobol' indices and numerical tests[END_REF], 0.198 has to be replaced by 1.244.

Application on a 1D hydraulic model

The Mascaret computer code ( [START_REF] Goutal | MAS-CARET: a 1-D open-source software for flow hydrodynamic and water quality in open channel networks[END_REF]) is a computer code based on a 1D solver of the Saint Venant equations, allowing engineers to calculate water height for river flood events. The studied case, taken from [START_REF] Petit | Couplage entre indices à base de dérivées et mode adjoint pour l'analyse de sensibilité globale[END_REF] and illustrated in Figure 5, is the French Vienne river in permanent regime whose input data are an upstream flowrate, a downstream water level, physical parameters (Strickler coefficients) and geometrical data. The geometrical data consist of 12 transverse river profiles. In summary, the model includes the following random input variables, assumed independent:

• 12 Strickler's friction coefficients of the main channel K s,c (noted K 1 s,c , . . . , K 12 s,c ) whose distributions are uniform on [START_REF] Iooss | Some new insights in derivative-based global sensitivity measures[END_REF]40],

• 12 Strickler's friction coefficients of the flood plain K s,p (noted K 1 s,p , . . . , K 12 s,p ) whose distributions are uniform on [START_REF] Brézis | Analyse fonctionnelle[END_REF][START_REF] Rudin | Real and complex analysis[END_REF],

• 12 slope perturbations dZ (noted dZ 1 , . . . , dZ 12 ) whose distributions are standard Gaussian, truncated on [-3, 3],

• and one discharge value Q whose distribution is Gaussian with zero mean, standard deviation 50 and truncated on [-150, 150].

Our goal in this test case is to update with the new DGSM bounds the work of [START_REF] Petit | Couplage entre indices à base de dérivées et mode adjoint pour l'analyse de sensibilité globale[END_REF] who compare DGSM bounds and Sobol indices. In this previous work, the derivatives of the model output with respect to the 37 inputs can be efficiently (with a cost independent of the number of inputs) computed by using the adjoint model of Mascaret (obtained by automatic differentiation, [START_REF] Griewank | Evaluating derivatives: Principles and techniques of automatic differentiation[END_REF]). From a Monte Carlo sample of the inputs of size n = 20 000, n output values (water height) and n output derivatives are obtained. This very large n has been used for a demonstrative purpose, while in other industrial studies n ranges from 100 to 1 000 ( [START_REF] Iooss | Some new insights in derivative-based global sensitivity measures[END_REF][START_REF] Touzany | Screening method using the derivative-based global sensitivity indices with application to reservoir simulator[END_REF]). From the sample of derivatives, the 37 DGSM ν i are then computed. [START_REF] Petit | Couplage entre indices à base de dérivées et mode adjoint pour l'analyse de sensibilité globale[END_REF] has shown that 32 inputs have DGSM-based upper bounds close to zero. Then, amongst the 37 inputs, only 5 are potentially active.

Table 4 gives our new results for these 5 remaining inputs. To compute the DGSM bound, we have also computed Var(f (x)) = 0.369 whose standard deviation (sd = 3.4e-3) is obtained by bootstrap, i.e. by resampling with replacement the Monte Carlo sample of output values. As inputs K 11 s,c and K 12 s,c follow a uniform distribution, we report only the optimal DGSM bounds, already used in [START_REF] Petit | Couplage entre indices à base de dérivées et mode adjoint pour l'analyse de sensibilité globale[END_REF]. For the 3 other inputs (dZ 11 , dZ 12 and Q), as in the previous application, a strong decrease factor is obtained for the bounds by using the logistic constant (gain factor around 4) and the optimal constant (gain factor around 6) instead of the double exponential constant. The new bounds now allow to identify the variable dZ 12 as inactive. Moreover, they provide the same ranking than total Sobol indices for the influential inputs. Notice that the optimal bounds are nearly equal to the total Sobol indices. The numerical errors, assessed by the standard deviations of the estimates, explain that the bound values are sometimes slightly smaller than the Sobol values.

In terms of interpretation, the sensitivity analysis results show that the hydraulic engineers have to concentrate the research efforts on the knowledge of the physical parameters K s and dZ at the 11th river profile, in order to be able to reduce the prediction uncertainty of the water level when a flood occurs. From a methodological point of view, the numerical model users are now able to perform a global sensitivity analysis, in the sense of variance decomposition, at a lower cost than before by using the DGSM-based technique. Since it is true for all g in L 2 (µ) this gives (2.10).

It remains to show that f , a minimizer of the Rayleigh ratio, is strictly monotonic. The main arguments can be found e.g. in [START_REF] Bobkov | Hardy Type Inequalities via Riccati and Sturm-Liouville Equations[END_REF]. Define g(x) = To see that f is strictly monotonic, consider again (2.10). Assume for instance that f is increasing (f ≥ 0), and let us prove that f is strictly increasing (f > 0). Then, as it is centered, the increasing function f is first ≤ 0 and then ≥ 0. From (2.10), it implies that f is first increasing from f (a) = 0 and then decreasing to f (b) = 0. Furthermore, as f is increasing and centered, we must have f (a) < 0 (otherwise f ≥ 0 on [a, b] and hence identically zero). By continuity, f < 0 in a neighborhood of a. Hence from (2.10) f is strictly increasing in a neighborhood of a. Similarly f is strictly decreasing in a neighborhood of b. Finally f > 0 on (a, b).

For completeness, we sketch the standard way to derive (P2) from (P1) by using calculus of variation. Without loss of generality we assume that f 2 dµ = 1. (P1) is then equivalent to: Define L(t, f, f ) = ( 1 2 f 2 -1 2 λ(f 2 -1) -βf )e -V (t) . The Lagrangian is:

J(f, λ, β) = b a L(t, f (t), f (t))dt
Considering a small deviation function g, we have The first order condition leads to cancel the integral, which gives here: This system must be singular otherwise f would be identically zero, hence det(M ) = 0. However, det(M ) = 1 4 (a -b + ab) < 0 since a < 0 < b. Proof of Proposition 6. For the triangular distribution T on Ω = (-1, 1), we have: µ(dt) = ρ(t)dt = e -V (t) dt with V (t) = -ln(1 -|t|). A simple application of Muckenhoupt criterion (2.4) shows that µ admits a Poincaré constant. However, we cannot directly apply Theorem 2 since ρ vanishes at the boundary of Ω. The idea is to consider an increasing sequence of subintervals I such that I ↑ Ω when → 0, for instance I = (-1 + , 1 -). Since T is a symmetric distribution on a symmetric support, a minimizer of the Rayleigh ratio can be found among odd functions (Lemma 7). Thus we can consider the spectral problem on I + = (0, 1 -): where J 0 and Y 0 are the Bessel functions of first and second type. J 0 is an even function of class C ∞ on R, whereas Y 0 is C ∞ on R * + with infinite limit at 0 + . By using the same argument as in Proposition 4, λ(µ |I ) is then the first zero of

J(f + g, λ, β) = J(f
f (t) - 1 1 -t f (
d(λ) = det   J 0 √ λ Y 0 √ λ J 0 √ λ Y 0 √ λ  
Now J (0) = 0, but Y 0 is not defined at 0 and xY 0 (x) → 2 π as x tends to zero ( [START_REF] Abramowitz | Handbook of Mathematical Functions[END_REF], Chapter 9, Eq. 9.1.9 and 9.1.28). So it is convenient to consider: This proves that λ(µ) is a zero of

d(λ, ) = d(λ) = det   J 0 √ λ Y 0 √ λ J 0 √ λ Y 0 √ λ  
d(λ, 0) = det J 0 √ λ Y 0 √ λ 0 2 π √ λ = 2 π √ λ J 0 √ λ .
Hence λ(µ) ≥ r 1 , where r 1 is the first zero of J 0 .

In order to prove the converse inequality, we compute the Rayleigh quotient of the following odd function defined on [-1, 1]: f (t) := sign(t)J 0 (r 1 (1 -|t|)) .

The expression of f has been deduced from y(x) = J 0 (r 1 x) by applying the transformation x = 1 -t, followed by a symmetrization. By construction, f and f are continuous at 0. Thus f is C 1 on Ω and piecewise C 2 on Ω. Furthermore, it verifies the following equation on [0, 1)

f -V f + r 2 1 f = 0, (5.3)
with boundary conditions f (0) = f (1) = 0. Since µ is even and f is odd, the Rayleigh quotient of f is simply

1 0 (f ) 2 ρ/ 1 0 f 2 ρ.
In order to calculate this quotient, we multiply Eq. 5.3 by f ρ and integrate by parts on [0, 1). However, since V = -ln(ρ) has a singularity at 1, we rather fix > 0 and work on I = [0, 1 -]. Observe that f and f ρ are C 1 on I . Starting from Eq. 5.3 and then integrating by parts, we obtain: Let us look for a series expansion of f , i.e. f (t) = n≥0 c n t n . Then, Let us split f into even an odd part, f (t) = f 0 (t)+tf 1 (t) with f 0 (t) = p≥0 u p t 2p and f 1 (t) = p≥0 v p t 2p , where u p = c 2p and v p = c 2p+1 . Then by definition of Kummer series (4.1), we recognize f 0 (t) = c 0 M 1-λ 2 ; 1 • The last item is direct, using that, by construction, h 0 (., 0) = h 1 (., 0) = 1.

f (t) = n≥1 nc n t n-1 ⇒ tf (t) = n≥0 nc n t n f (t) =

Corollary 1 .

 1 Let ν be a probability measure on (a, b) ⊂ R, with a positive continous density on (a, b). Then C P (ν) ≤ 4 sup x∈(a,b)

Theorem 2 .

 2 Let Ω = (a, b) be a bounded open interval of the real line, and assume that µ(dt) = ρ(t)dt = e -V (t) dt where V is continuous and piecewise C 1 on Ω = [a, b]. Consider the three following problems:

Lemma 4 .

 4 Let µ be a probability measure on R, of the form dµ(t) = 1 Ω (t)ρ(t) dt where Ω is an open interval on which ρ is continuous and positive. Let I ⊂ Ω be an open interval, and µ |I denote the probability measure obtained by restriction of µ to I (defined by µ |I (A) = µ(I ∩ A)/µ(I)). Then C P (µ |I ) ≤ C P (µ).

Truncated exponential distribution Proposition 5 .Proposition 6 .

 56 Let µ be the (double) exponential distribution on the real line, µ(dt) = 1 2 e -|t| dt, and let a < b. The Poincaré constant of the restriction of µ to [a, b] is C P (µ |[a,b] ) = 1 4 + ω 2 -1 , where: • If a, b have the same sign, ω = π |b-a| . The inequality is saturated by f (x) = e |x|/2 (-ω cos(ω(x -a)) + 1 2 sin(ω(x -a) • If a < 0 < b, then ω > π b-a . More precisely ω is the zero of x → cotan(|a|x) + cotan(|b|x) + 1/x in ]0, min(π/|a|, π/|b|)[. The inequality is saturated by f (x) = e |x|/2 A cos(ωx) + B -sgn(x) + 1 2 A ω sin(ωx) with A = ωcotan(ωa) -1 2 and B = 1 2 cotan(ωa) + ω. In particular if a = -b then ω is the zero of x → 2cotan(bx) + 1/x in ]0, π/b[, and the inequality is saturated by f (x) = e |x|/2 sin(ωx). Triangular distribution Let us consider the triangular distribution T supported by [-1, 1], µ(dt) = (1 -|t|)1 [-1,1] (t)dt. Denote by J 0 the Bessel function of the first kind with parameter ν = 0 ([1], Chapter 9), and r 1 its first root. Then the Poincaré constant for T is r -2

•••

  & 13.6.18). Denote N |[a,b] the standard normal distribution N (0, 1) truncated on [a, b]. Then its spectral gap λ(N |[a,b] ) is the first zero of the function d(.) = bh 0 (., a)h 1 (., b) -ah 0 (., b)h 1 (., a) Furthermore: If there exists λ such that a, b are two successive zeros of h 0 (λ, .) or two successive zeros of h 1 (λ, .) then λ(N |[a,b] ) = λ If a = -b then λ(N |[a,b] ) is the first zero of h 0 (., a). If a = 0 (resp. b = 0), then λ(N |[a,b] ) is the first zero of h 1 (., b) (resp. h 1 (., a)).

Figure 1 .

 1 Figure 1. Basis of finite elements P 1 on [0, 1]. The g i 's are hat functions for i = 1, . . . , n-1, truncated at the boundaries (i = 0 and i = n).

F 2 Figure 2 . 1 (

 221 Figure 2. Poincaré constant of the double exponential distribution µ truncated on [a, b]. Left: Contour plot as a function of Fµ(a) and 1 -Fµ(b), with the upper bound 1 4 + π b-a 2 -1 (red dashed lines). Right: Representation in the symmetric case for I = [-b, b], with two upper bounds and the lower bound given by the variance.

Figure 3 .

 3 Figure 3. Poincaré constant of the normal distribution N (0, 1) truncated on [a, b]. Left: Contour plot as a function of Φ(a) and 1 -Φ(b). Right: Representation in the symmetric case for I = [-b, b], with the upper bound µ(I) 2 and the lower bound given by the variance. Colored points correspond to Hermite polynomials: For a given degree, a same color is used.

Figure 4 .

 4 Figure 4. Comparison between total Sobol indices and DGSM-based upper bounds for the simplified flood model. Left: Results with double exponential transport; Right: New optimal bounds, obtained with Poincaré constants.

Figure 5 .

 5 Figure 5. Representation of a river cross section, showing the main channel (low flow channel) and the flood plain (high flow channel). Source: [25].

f

  (x)g(x)ρ(x)dx = g(a) )ρ(t)dt g (x)dx Recall that for λ > 0, b a f ρ = 0. Thus (P2) can be rewritten: b a f (x)g (x)ρ(x)dx = b a λ b x f (t)ρ(t)dt g (x)dx

2 [

 2 have g = |f | and thus (g ) 2 dµ = (f ) 2 dµ. Secondly, using the formula Var µ g = 1 g(y) -g(x)] 2 dµ(x)dµ(y), we have :)dµ(y) = Var µ fSince f is a minimizer of the Rayleigh ratio, the above inequality must be an equality, leading to y x |f (t)|dµ(t) = | y x f (t)dµ(t)| for all x, y in [a, b]. By continuity of f , there exists ∈ {-1, +1} ( = sgn[f (b)-f (a)]) such that |f | = f everywhere, hence f is monotonic.

  g + f g ) dµ = β b a gdµ Now, with g = 1 and using b a f dµ = 0, we get β = 0. This gives (P2). Now the Neumann conditions f (a) = f (b) = 0 provide the linear system M (A B) T = 0, with:

  t) + λf (t) = 0 with boundary conditions f (0) = f (1 -) = 0. By applying the change of variable f (t) = y(x) with x = 1 -t , we obtain the equivalent problem on ( , 1):y (x) + 1 x y (x) + λy(x) = 0with initial conditions y(1) = y ( ) = 0. This is a Bessel-type differential equation, which solution has the form (see e.g.[START_REF] Abramowitz | Handbook of Mathematical Functions[END_REF], Chapter 9):y(x) = AJ 0 x √ λ + BY 0 x √ λ

By lemma 5 ,

 5 λ(µ) = lim →0 λ(µ |I ) and d is continuous at (λ(µ), 0). Thus, it holds: 0 = d(λ(µ |I ), ) → →0 d(λ(µ), 0)

. 2 1 1 0 f 2 ρ = 1 0

 211 The last term tends to 0, when tends to 0. Thus by monotone convergence, r f 2 ρ. We have shown that the Rayleigh ratio of f is equal to r 2 1 . Therefore λ(µ) ≤ r2 1 and the proof is complete.Proof of Proposition 8. Here it is easier to consider the spectral problem with Dirichlet conditions (Problem (P4), see Prop. 2). It is given by:f (t) -tf (t) = -(λ -1)f (t) s.t. f (a) = f (b) = 0 (5.4)

n≥2 n(n - 1 )

 1 c n t n-2 = n≥0 (n + 2)(n + 1)c n+2 t nand solving (5.4) leads to finding a sequence (c n ) satisfying:

2 2•

 2 . Thus, with the notations of the Proposition,f (t) = c 0 h 0 (λ, t) + c 1 th 1 (λ, t)By applying Proposition 4 and Remark 4, we obtain that the spectral gap is the first zero of d(λ) and hence the first part of the Proposition. The symmetric case a = -b is also deduced. Furthermore: If there exists λ such that a, b are two successive zeros of h 0 (λ, .), then up to a change of sign, f = h 0 (λ, .) is a positive solution satisfying f (a) = f (b) = 0. Hence by Corollary 2, λ(N |[a,b] ) = λ. Analogous conclusion stands with h 1 .

  By a density argument it is enough to prove the Poincaré inequality for ν for all locally Lipschitz functions. Let g be a locally Lipschitz function on R. Then f := g • T is also locally Lipschitz on Ω, and Var µ (f ) ≤ C P (µ) |f | 2 dµ. Since ν is the image measure of µ by T , Var µ

	|f |(x) := lim sup y→x	|f (y) -f (x)| |y -x|	.
	By Rademacher's theorem, a locally Lipschitz function is differentiable almost
	everywhere, and the latter coincides a.e. with the absolute value of the derivative
	of f (x). By hypothesis, |f |(x) = |f (x)| will hold almost surely in the sense of
	µ and T µ as well.
	Set ν = T µ.

Table 2 .

 2 Input variables of the flood model and their probability distributions.

	Input	Description	Unit	Probability distribution
	X 1 = Q	Maximal annual flowrate m 3 /s	Gumbel G(1013, 558)
				truncated on [500, 3000]
	X 2 = Ks	Strickler coefficient	-	Normal N (30, 8 2 )
				truncated on [15, +∞[
	X 3 = Zv	River downstream level	m	Triangular T (49, 51)
	X 4 = Zm	River upstream level	m	Triangular T (54, 56)
	X 5 = H d	Dyke height	m	Uniform U [7, 9]
	X 6 = C b	Bank level	m	Triangular T (55, 56)
	X 7 = L	River stretch	m	Triangular T (4990, 5010)
	X 8 = B	River width	m	Triangular T (295, 305)

Table 3 .

 3 Poincaré constants and bounds for the scaled laws used in the simplified flood model.

	Prob. dist.	Upper bound	Upper bound	C P (µ) Lower bound
	µ	(db. exp. transp.) (logis. transp.)		Var(µ)
	T (-1, 1)	1	0.296	0.173	0.167
	N (0, 1 | [-1.87, +∞))	5.912	1.484	0.892	0.862
	G (0, 1 | [-0.92, +3.56])	6.956	2.418	1.257	1.012

Table 4 .

 4 Sensitivity indices for the Mascaret test case. S T gives the total Sobol index, ν the DGSM. Standard deviations of all these estimates are obtained by bootstrap and given in parentheses. Partial results are not shown for K 11 s,c , K 12 s,c , which are uniformly distributed.

	Inputs	K 11 s,c	K 12 s,c	dZ 11	dZ 12	Q
	S T	0.456	0.0159	0.293	0.015	0.239
		(2e-3)	(1e-4)	(1e-3)	(1e-4)	(1e-3)
	ν	5.695e-3 2.728e-4 1.089e-1 6.592e-3 3.553e-5
		(3e-5)	(4e-6)	(3e-4)	(9e-5)	(6e-8)
		By double exponential transport
	Upper bound for C P (µ)	-	-	6.249	6.249	15623.26
	Upper bound for S T	-	-	1.844	0.116	1.504
		-	-	(2e-3)	(2e-3)	(1.5e-2)
		By logistic transport
	Upper bound for C P (µ)	-	-	1.562	1.562	3905.815
	Upper bound for S T	-	-	0.461	0.028	0.376
		-	-	(4e-3)	(5e-4)	(4e-3)
		Optimal, with the Poincaré constant
	C P (µ)	40.528	40.528	0.976	0.976	2441.071
	Optimal bound for S T	0.625	0.029	0.288	0.017	0.235
		(2e-4)	(1e-5)	(3e-3)	(3e-4)	(2e-3)

  , λ, β) +

	a	b	g(t)	∂L ∂f	(t, f (t), f (t)) + g (t)	∂L ∂f	(t, f (t), f (t)) dt + O( 2 )

Acknowledgements

We are grateful to Fabrice Gamboa, who initiated this research. We thank EDF R&D/LNHE for providing the Mascaret test case and Sébastien Petit who has performed the computations on this model. We also thank Laurence Grammont and the members of the team 'Génie Mathématique & Industriel' for useful discussions. We acknowledge the participants of MascotNum 2016 & Mexico Mas-cotNum conferences for their feedbacks. Part of this research was conducted within the frame of the Chair in Applied Mathematics OQUAIDO, gathering partners in technological research (BRGM, CEA, IFPEN, IRSN, Safran, Storengy) and academia (CNRS, Ecole Centrale de Lyon, Mines Saint-Etienne, University of Grenoble, University of Nice, University of Toulouse) around advanced methods for Computer Experiments.

Appendix

End of proof of Theorem 2. Let us show (2.10), namely that µ-almost surely,

Optimal constants

Proof of Proposition 5. a) Let us start by the case 0 ≤ a < b. The spectral problem is given by:

The characteristic equation r 2 -r = -λ can be written as:

It is known that the spectral gap of the double exponential distribution on the whole real line is 1 4 [START_REF] Bobkov | Poincare's inequalities and talagrand's concentration phenomenon for the exponential distribution[END_REF]. By monotonicity of the spectral gap with respect to inclusion, we can deduce that λ ≥ 1 4 . It seems clear that this inequality must be strict, and this can be proved directly. Indeed, if λ = 1 4 , the general form of the solution is

and the Neumann conditions lead to the linear system M u = 0 with:

Since a = b, this implies A = B = 0 which is impossible. Hence λ > 1 4 , and the solution has the form f (t) = e t/2 (A cos(ωt) + B sin(ωt)). By applying Proposition 4, we deduce that the spectral gap is the first zero of b) The same proof is immediately adapted to the case a < b ≤ 0 by using the change of variables x → -x. c) Let us now assume that a < 0 < b. By the same argument as in a), we know that λ ≥ 1 4 . Let us temporarily admit that the case λ = 1 4 is impossible. Then for λ > 1 4 , the solution has the form:

where

and in particular f and f must be continuous at 0. Conversely, if this condition is satisfied, f will be of class ) is that f and f are continuous at 0. The first condition immediately implies A + = A -. For the second one, we have:

and continuity of f at 0 leads to B + -B -= -A/ω. Denoting B = B -, we have B + = B -A/ω, and f is rewritten:

The Neumann conditions f (a) = f (b) = 0 then provide a linear system in A, B which must be singular, leading to the condition det(M ) = 0 with:

Using the properties of the determinant, one can replace the first column C 1 by Finally, let us see that λ = 1 4 is impossible. Then the solution has the form:

As above, we express that f and f are continuous at 0. The first condition implies B + = B. For the second one, we have: