A Generalized Radial BrezisNirenberg Problem Rafael D. Benguria - - PowerPoint PPT Presentation

Rafael D. Benguria Instituto de F´ ısica, PUC Santiago, Chile The Analysis of Complex Quantum Systems: Large Coulomb Systems and Related Matters CIRM, Luminy, Marseille October 23, 2019.

A Generalized Radial Brezis–Nirenberg Problem

This is joint work with Soledad Benguria Department of Mathematics University of Wisconsin Madison, WI, USA Talk partially based on the manuscript: RB and S. Benguria, A Generalized Radial Brezis–Nirenberg Problem, preprint, 2019.

Motivation (The Lane–Emden equation): The equation −∆u = up (1) for u > 0 in a ball of radius R in R3, with Dirichlet boundary conditions, is called, in physics, the Lane–Emden equation of index p. It was introduced in 1869 by Homer Lane, who was interested in computing both the temperature and the density of mass on the surface of the Sun. Unfortunately Stefan’s law was unknown at the time (Stefan published his law in 1879). Instead, Lane used some experimental results of Dulong and Petit and Hopkins on the rate of emission of radiant energy by a heated surface, and he got the value of 30,000 degrees Kelvin for the temperature of the Sun, which is too big by a factor of

• 5. Then he used his value of the temperature together with the solution of (1)

with p = 3/2, to estimate the density u near the surface.

5,778 K

Motivation (The Lane–Emden equation): After the Lane–Emden equation was introduced, it was soon realized that it only had bounded solutions vanishing at R if the exponent is below 5. In fact, for 1  p < 5 there are bounded solutions, which are decreasing with the distance from the center. In 1883, Sir Arthur Schuster constructed a bounded solution of the Lane–Emden equation in the whole R3 vanishing at infinity. This equation

• n the whole R3, with exponent p = 5 plays a major role in mathematics. It

is the Euler–Lagrange equation equation that one obtains when minimizing the quotient R (ru)2 dx R u6 dx 1/3 . (1) This quotient is minimized if u(x) = 1/(|x|2 + m2)1/2. The minimizer is unique modulo multiplications by a constant, and translations. This function u(x), is precisely the function determined by A. Schuster, up to a multiplicative con-

• stant. Inserting this function u back in (1), gives the classical Sobolev inequality

(S. Sobolev 1938), R (ru)2 dx R u6 dx 1/3 3(π 2 )4/3, (2) for all functions in D1(R3).

The Brezis–Nirenberg problem on RN

In 1983 Brezis and Nirenberg considered the nonlinear eigenvalue problem, −∆u = λu + |u|4/(n−2)u, with u ∈ H1

0(Ω), where Ω is bounded smooth domain in Rn, with n ≥ 3.

Among other results, they proved that if n ≥ 4, there is a positive solution

• f this problem for all λ ∈ (0, λ1) where λ1(Ω) is the first Dirichlet eigenvalue
• f Ω. They also proved that if n = 3, there is a µ1(Ω) > 0 such that for any

λ ∈ (µ1, λ1), the nonlinear eigenvalue problem has a positive solution. Moreover, if Ω is a ball, µ1 = λ1/4.

The Brezis–Nirenberg problem on RN

For positive radial solutions of this problem in a (unit) ball, one is led to an ODE that still makes sense when n is a real number rather than a natural number. Precisely this problem with 2 ≤ n ≤ 4, was considered by E. Jannelli, The role played by space dimension in elliptic critical problems, J. Differential Equations, 156 (1999), pp. 407–426. Among other things Jannelli proved that this problem has a positive solution if and only if λ is such that j−(n−2)/2,1 < √ λ < j+(n−2)/2,1, where jν,k denotes the k–th positive zero of the Bessel function Jν.

π π/2

j0,1 = 2.4048 . . . The Brezis–Nirenberg problem on RN

The Bakry-´ Emery Laplacian. We think of this problem as a further variant of the Brezis-Nirenberg prob- lem, namely we study (1) with the Laplacian replaced by a particular form of a weighted Laplacian, or if one prefers, the drift Laplacian. The interest on weighted Laplacians originated in the early 1980’s for different reasons coming from physics, geometry, and probability. Depending on the context, a weighted Laplacian is often called the Witten Lapla- cian (Witten, 83) or the Bakry-´ Emery Laplacian (Bakry-Emery, 85).

The Bakry-´ Emery Laplacian. During the past decade there has been a growing interest in studying the spectral properties of weighted Laplacians or drift Laplacians. A Bakry–´ Emery manifold, denoted by the triple (M, g, φ) is a complete Rie- mannian manifold (M, g) together with some function φ 2 C2(M) where the measure on M is the weighted measure exp(φ) dVg. The Bakry–´ Emery Lapla- cian ∆φ associated with such a manifold is given by ∆φ = ∆ rφ · r which is self-adjoint with respect to the inner product associated with the weighted measure. Here ∆ is the standard Laplace-Beltrami operator and r is the gradient operator on the Bakry–´ Emery manifold. Weighted Laplacians were also introduced, in a different context, by Chavel and Feldman 1991.

The Bakry-´ Emery Laplacian. Typically, in the Bakry-´ Emery Laplacian, the potential φ is smooth, and so is the drift term. However, singular drifts have also been considered in the literature. In fluid mechanics a weighted Laplacian with a singular drift is rather common, but typically the drift is divergence free, in other words, away from the singularities the potential φ is harmonic. Recently (J. Reine und Angew. Math., 2018) A. Grigoryan, S.–X. Ouyang, and

• M. R¨
• ckner considered a drift with a (singular) potential of the form φ(x) =

|x|−α with α > 0. In that case the singular drift has, generically, a nonzero divergence.

The Brezis–Nirenberg problem for the weighted Laplacian with a Singular Drift In our case we consider the Brezis-Nirenberg problem for the weighted Laplacian in Rn (n ≥ 3), the singular drift derives from a potential of the form φ(x) = δ log(|x|). Notice that in this case the weighted measure, exp(−φ) dVg becomes |x|−δ dx where dx is the standard Lebesgue measure in Rn. Thus the weighted measure can be thought of as the Lebesgue measure on a space of an effective fractional dimension d ≡ n − δ, a fact that we use intensively in the proofs of

• ur theorems.

The Bakry-´ Emery Laplacian. In our case, when we consider the Brezis-Nirenberg problem for the weighted Laplacian in Rn (n 3), the singular drift derives from a potential of the form (x) = log(|x|). Notice that in this case the weighted measure, exp() dVg becomes |x|−δ dx where dx is the standard Lebesgue measure in Rn. Thus the weighted measure can be thought of as the Lebesgue measure on a space of an effective fractional dimension d ⌘ n . First consider the problem, ∆u + ~ x |x|2 · ru = u + |u|4/(n−2−δ)u, (1) with u 2 H1

0(Ω), and Ω is the ball in Rn, n 3, centered at the origin.

Because of Hardy’s inequality, the operator with the singular drift one considers

• n the left side of (1) is a positive operator provided < (n 2)/2. Notice that

the critical Sobolev exponent on the right side of (1) depends on the parameter that characterizes the singular drift. In terms of the “effective dimension” the critical Sobolev exponent is given by the standard form, (d + 2)/(d 2), a remark which is important in the proofs of our theorems. We are interested in the range of values of and for which (1) admits positive radial smooth solutions.

Euclidean Case. Theorem [RB, S. Benguria, 2017]. When Ω ⊂ Rn, with n ≥ 3, is a unit ball centered at the origin we have, i) If n = 3 and δ ∈ (−1, 1/2), (1) has a unique positive radial solution u ∈ H1

0(Ω)

provided j2

−(1−δ)/2,1 < λ < j2 (1−δ)/2,1,

and no positive radial solutions for λ outside that range. If n = 3 and δ ≤ −1, then (1) has a unique positive radial solution u ∈ H1

0(Ω) provided

0 < λ < j2

(1−δ)/2,1,

and no positive radial solutions outside that range. ii) If n = 4 and δ ∈ (0, 1), (1) has a unique positive radial solution u ∈ H1

0(Ω)

provided j2

−(2−δ)/2,1 < λ < j2 (2−δ)/2,1,

and no positive radial solutions for λ outside that range. If n = 4 and δ ≤ 0, then (1) has a unique positive radial solution u ∈ H1

0(Ω) provided

0 < λ < j2

(2−δ)/2,1,

and no positive radial solutions outside that range.

Euclidean Case. Theorem [RB, S. Benguria, 2017]. (Cont’d): iii) If n = 5 and δ ∈ (1, 3/2), (1) has a unique positive radial solution u ∈ H1

0(Ω)

provided j2

−(3−δ)/2,1 < λ < j2 (3−δ)/2,1,

and no positive radial solutions for λ outside that range. If n = 5 and δ ≤ 1, then (1) has a unique positive radial solution u ∈ H1

0(Ω) provided

0 < λ < j2

(3−δ)/2,1,

and no positive radial solutions outside that range. Next we consider the Brezis-Nirenberg problem with a singular drift on geodesic balls of Sn (n ≥ 3).

The Brezis–Nirenberg problem on SN

We consider the nonlinear eigenvalue problem, −∆Snu = λu + |u|4/(n−2)u, with u ∈ H1

0(Ω), where Ω is a geodesic ball in Sn. In dimension 3, Bandle

and Benguria (JDE, 2002) proved that for λ > −3/4 this problem has a unique positive solution if and only if π2 − 4θ2

1

4θ2

1

< λ < π2 − θ2

1

θ2

1

where θ1 is the geodesic radius of the ball.

The Brezis–Nirenberg problem on SN

NP SP θ θ/2 x’ x r

The Brezis–Nirenberg problem on S3

Having our notation in place, the Brezis–Nirenberg problem for the singular drift Laplacian on Sn is given by, ∆Snu + δ cos θ sin θ ˆ θ · ru = λu + |u|4/(n−2−δ)u, (1) where u 2 H1

0(D), and D is a geodesic ball in Sn, n 3. Equation (1) involves

a weighted Laplace–Beltrami operator with a singular drift deriving from the potential φ(θ) = δ log sin θ, δ 2 R. As in the Euclidean case, because of the appropriate Hardy Inequality, the operator on the left side of (1) is positive– definite provided δ < (n 2)/2. In analogy with the Euclidean case, here the weighted measure, exp(φ) dVg becomes sin θ−δ dµ where dµ is the invariant measure in Sn. Hence, again in this case, the weighted measure can be thought

• f as the measure on a space of an effective fractional dimension d ⌘ n δ.

The Brezis–Nirenberg problem on SN

The Brezis–Nirenberg problem on SN

For positive radial solutions of this problem one is led to an ODE that still makes sense when d is a real number rather than a natural number. Our main result is the following: Theorem: [RB, S. Benguria] (Nonlinear Analysis, 2017) For any 2 < d < 4, i) If ≥ −d(d − 2)/4 and 0 ≤ ✓1 ≤ ⇡, the boundary value problem, in the interval (0, ✓1), with u0(0) = u(✓1) = 0 has a positive solution if and only if is such that 1 4[(2`2 + 1)2 − (d − 1)2] < < 1 4[(2`1 + 1)2 − (d − 1)2] where `1 (respectively `2) is the first positive value of ` for which the associated Legendre function P(2d)/2

`

(cos ✓1) (respectively P(d2)/2

`

(cos ✓1)) vanishes. ii) If ≤ −d(d − 2)/4 and 0 ≤ ✓1 ≤ ⇡/2, the boundary value problem, in the interval (0, ✓1), with u0(0) = u(✓1) = 0 does not have a positive solution.

The Brezis–Nirenberg problem on SN

Strategy of the Proof: For the nonexistence of solutions: i) Use a Rellich–Pohozaev’s type argument for values of λ below the lower bound. ii) Multiply the ODE by the first eigenfunction of the Dirichlet problem to rule

• ut the values of λ larger than the upper bound.

For the Existence part, use a variational characterization of λ and a Brezis–Lieb lemma (or, alternatively, a concentration compactness argument).

The analogous problem on HN: For N = 3, this was treated by Silke Stapelkamp on her Ph. D. Thesis (U. Basel, 2001). For 2 ≤ N ≤ 4, this was considered by Soledad Benguria, “The solution gap

• f the Brezis–Nirenberg problem on the hyperbolic space”, Monatsheste f¨

ur Mathematik, 2016.

General Brezis–Nirenberg Problem Here we consider a generalization of the radial Brezis-Nirenberg problem. Namely, let R ∈ (0, ∞) and let a be such that

• 1. a ∈ C3[0, R]
• 2. a(0) = a00(0) = 0;
• 3. a(x) > 0 for all x ∈ (0, R); and
• 4. lim

x!0

a(x) x = 1.

General Brezis-Nirenberg Problem Given n ∈ (2, 4), we study the existence of positive solutions u ∈ H1

0(Ω) of

−u00(x) − (n − 1)a0(x) a(x) u0(x) = λu(x) + u(x)p (1) with boundary condition u0(0) = u(R) = 0. Here, as in the original problem, p = (n + 2)/(n − 2) is the critical Sobolev exponent. Notice that the radial Brezis-Nirenberg problem on the Euclidean space corre- sponds to taking a(x) = x; on the hyperbolic space, to taking a(x) = sinh(x); and on the spherical space, to taking a(x) = sin(x), with R ≤ π/2.

General Brezis-Nirenberg Problem

• Theorem. [Existence of positive solutions].

For any 2 < n < 4 the boundary value problem −u00(x) − (n − 1)a0(x) a(x) u0(x) = λu(x) + u(x)

n+2 n−2

(1) with u0(0) = u(R) = 0, and x ∈ (0, R) has a positive solution if λ ∈ (µ1, λ1). Here, λ1 is the first positive eigenvalue of y00 + a0 a y0 + λ − α2 ✓a0 a ◆2 + αa00 a ! y = 0 (2) with boundary conditions limx!0 y(x)xα = 1, where α = (2−n)/2, and y(R) =

• 0. And µ1 is the first positive eigenvalue of (2) with boundary conditions limx!0 y(x)xα =

1 and y(R) = 0.

General Brezis-Nirenberg Problem

• Theorem. [Non-existence],

There is no positive solution to problem (BN) if λ ≥ λ1, or if λ ∈ [N ⇤, µ1], where N ⇤ = sup

0<x<R

⇢α2 a2 (a02 − 1) − αa00 a

• .

Moreover, (BN) has no solution if λ ≤ M ⇤, where M ⇤ = inf

0<x<R

⇢ α2 a00 a − α 2 ✓a000 a0 + a00 a ◆ . Notice that in the cases that have already been studied, N ⇤ and M ⇤ coincide. In fact, in the Euclidean case, N ⇤ = M ⇤ = 0; in the hyperbolic case, N ⇤ = M ⇤ = n(n − 2)/4; and in the spherical case, N ⇤ = M ⇤ = −n(n − 2)/4.

General Brezis-Nirenberg Problem Finally, we obtain a uniqueness result under an additional assumption on a. Namely, we show the following: Theorem [Uniqueness]. Suppose that a satisfies the condition a0a00−aa000 ≥ 0 for all x ∈ (0, R). Then if λ > N ⇤, problem (BN) has at most one positive solution.

• Remark. In the previously studied cases (i.e, Euclidean, hyperbolic, spherical),

we have that a0a00−aa000 = 0 for all x. so the above condition is trivially satisfied.

General Brezis-Nirenberg Problem: Existence The solutions to (BN) correspond to minimizers of the quotient Q(u) = !n Z R u02an1 dx − !n Z R u2an1 dx !n Z R up+1an1 dx ! n−2

n

(1) Here, !n represents the surface area of the unit sphere in n-dimensions, and is explicitly given by !n = 2⇡

n 2 /Γ

n

2

• .

We will begin by showing that there exists a choice of cutoff function ', with '(0) = 1 and '0(0) = '(R) = 0, such that if we define u✏(x) = '(x) (✏ + x2)

n−2 2 ,

(2) then Q(u✏) < Sn if > µ1. Here Sn is the Sobolev constant. To do so, we start by obtaining estimates for each of the three integrals in the quotient Q(u✏). We use a classical argument due to Lieb: if Q(u✏) < Sn, and < 1, then there exists a solution to equation (BN).

General Brezis-Nirenberg Problem: Existence

• Lemma. There exists a cutoff function ' with '(0) = 1 and '0(0) = '(R) = 0,

such that if we define u✏(x) = '(x) (✏ + x2)

n−2 2 ,

(1) then ≥ µ1 implies that Q(u✏) < Sn, where Sn is the Sobolev constant. One gets, Q(u✏) = Sn + ✏

n−2 2 (Kn!n) 2−n 2 !nT(') + O(✏),

where T(') ≡ Z R '02an1 x2n4 dx+(n−1)(n−2) Z R '2an1 x2n2 ✓a0 a x − 1 ◆ dx− Z R '2an1 x2n4 dx. (2) Therefore, it suffices to show that there exists a choice of ' for which T(') is negative when > µ1.

General Brezis-Nirenberg Problem: Existence Making the change of variables ϕ = y xn2a

2−n 2

we can write the Euler equation associated to F(ϕ) (the first two terms of T(ϕ)) as, y00 + a0 a y0 + αa00 a − α2 ✓a0 a ◆2 + µ ! y = 0. (1) Since by hypothesis ϕ(0) = 1, and since ϕ(x) = y(x)xn2a(x)

2−n 2 , it must

follow that limx!0 y(x)xα = 1. This is precisely the boundary condition defin- ing µ1.

General Brezis-Nirenberg Problem: Non Existence To prove that there are no solutions for λ ≥ λ1 is standard. To prove that there are no solutions when N ⇤ ≤ λ ≤ µ1, one uses a Rellich– Pohozaev argument. Let g be a smooth non-negative function such that g(0) = g0(0) = 0 and g(R) > 0, and u a positive solution of (BN). Then, one has the following virial theorem: 1 2a2n2(R)u0(R)2g(R) = Z R u2A dx + Z R up+1B dx, (1)

General Brezis-Nirenberg Problem: Non Existence 1 2a2n2(R)u0(R)2g(R) = Z R u2A dx + Z R up+1B dx, (1) with, A[g] ≡ a2n2 " g000 4 + g00 ✓a0 a ◆ 3(n − 1) 4 + g0 λ + (n − 1)(2n − 3) 4 ✓a0 a ◆2 + (n − 1) 4 ✓a00 a ◆! +λ(n − 1) ✓a0 a ◆ g

• ;

(2) and B[g] ≡ 1 2g0a2n2 + (ga2n2)0 p + 1 . (3) Our goal is to show there is choice of function g such that A[g] ≡ 0, and if λ ≤ µ1, then B[g] < 0. Since the left-hand-side of the first equation is positive, we will obtain a contradiction, thus concluding there are no solutions for values

• f λ in this range.

General Brezis-Nirenberg Problem: Non-Existence. The goal is achieved by taking: g(x) = a2n(x)y1(x)y2(x), where y1 and y2 are linearly independent solutions to y00 + ✓a0 a ◆ y0 + ky = 0, (1) where, as before, k = α ✓a00 a ◆ − α2 ✓a0 a ◆2 + λ. (2) Then A[g] ≡ 0. and B[g] < 0, and we are done.

General Brezis-Nirenberg Problem: Uniqueness For proving uniqueness when µ1 < λ < λ1 one uses the classical techniques by Kwong an Li (1992).

Thank you!