Functional inequalities and applications B. Al TAKI Sorbonne - - PowerPoint PPT Presentation

Functional inequalities and applications

• B. Al TAKI

Sorbonne Universit´ e and INRIA-Paris July 31, 2019 CEMRACS’19 Geophysical Fluids, Gravity Flows

• B. Al Taki

Functional inequalities and applications CEMRACS’19 1 / 22

Outline

◮ Part I: Weighted Sobolev inequalities

◮ Applications to geophysics models: Lake equations ◮ Difficulties...

◮ Part II: Logarithmic Sobolev inequalities

◮ Application to gas dynamic system. ◮ Diffusive capillary models of Korteweg type.

• B. Al Taki

Functional inequalities and applications CEMRACS’19 2 / 22

Importance of weighted spaces

Poincar´ e-Wirtinger’s inequality: let f : Ω ⊂ Rn − → R smooth, compactly supported

• Ω

|u − ¯ u|p dx ≤ C

• Ω

|∇u|p dx ¯ u = 1 |Ω|

• Ω

|u| dx. For p = 2, this inequality is the key tool of solving (P1)

• −∆u

= f in Ω, u = g

• n ∂Ω.

for Ω smooth domain and f , g such that f ∈ H−1(Ω), g ∈ H1/2(∂Ω). But: What happens when f / ∈ H−1(Ω)??

• B. Al Taki

Functional inequalities and applications CEMRACS’19 3 / 22

Importance of weighted spaces

Poincar´ e-Wirtinger’s inequality: let f : Ω ⊂ Rn − → R smooth, compactly supported

• Ω

|u − ¯ u|p dx ≤ C

• Ω

|∇u|p dx ¯ u = 1 |Ω|

• Ω

|u| dx. For p = 2, this inequality is the key tool of solving (P1)

• −∆u

= f in Ω, u = g

• n ∂Ω.

for Ω smooth domain and f , g such that f ∈ H−1(Ω), g ∈ H1/2(∂Ω). But: What happens when f / ∈ H−1(Ω)??

• B. Al Taki

Functional inequalities and applications CEMRACS’19 3 / 22

Example 1: (P2) −∆u = − div f in Ω, u = 0,

• n ∂Ω.

For x ∈ Ω, we define f and dM as follows f (x) = |x|−N/p′ x ∈ Ω dM := dist(x, M) = |x| =

• x2

1 + x2 2 + . . . + x2 N.

Notice that f p′

Lp′ (Ω) =

• Ω

|f (x)|p′ dx =

• Ω

|x|−N dx = const 1 r−1 dr = ∞, However f p′

Lp′ (Ω,dε

M) =

• Ω

|f (x)|p′|x|ε dx =

• Ω

|x|−N+ε dx = const 1 r−1+ε dr < ∞. f / ∈ Lp′(Ω) but f ∈ Lp′(Ω, dε

M) =

⇒ possibility of solution in W 1,p(Ω, dǫ

m)?

Remark Details about Laplacian equation in weighted spaces can be found in the works of Farwig, Sohr,.

• B. Al Taki

Functional inequalities and applications CEMRACS’19 4 / 22

For Ω smooth domain, consider − div(b∇u) = b f in Ω, u = 0

• n ∂Ω.

Variational problem: find u such that a(u, v) :=

• Ω

∇u∇vb dx =

• Ω

f · vb dx := L(v) v ∈ H1

0(Ω)

◮ If 0 < c1 ≤ b(x) ≤ c2 < ∞ a(u, u) =

• Ω

|∇u|2b dx ≥ c1uH1

0 (Ω)

⇒ Coercivity on Sob. space L(v) =

• Ω

f · vb dx ≤ c2f L2(Ω)vL2(Ω) ⇒ Continuity on Sob. spaces Lax Milgram’s Theo. ⇒ existence. of sol. in H1

0(Ω).

◮ When b(x) → 0 or b(x) → ∞ when x → ∂Ω, We lose the coercivity and continuity in Sobolev spaces What about existence of sol. in weighted Sob. spaces.??

• B. Al Taki

Functional inequalities and applications CEMRACS’19 5 / 22

If we define ”formally” Vb as follows Vb =

• v measurable s.t
• Ω

|∇v|2 b dx < ∞

• and suppose that

∇ · L2

b(Ω) :=

• Ω

|∇ · |2 b dx define a norm. Then we can verify that

• a(·, ·) is bilinear and coercive,i.e.,

a(u, u) =

• Ω

|∇u|2 b dx ≥ u2

Vb

• L(·) is linear and continuous, i.e.,

|L(u)| =

• Ω

f · v b dx

• ≤ Cb1/2f L2(Ω)b1/2uL2(Ω)
• B. Al Taki

Functional inequalities and applications CEMRACS’19 6 / 22

◮ To finish the proof, it remains to verify that

• ∇ · L2

b(Ω) is a norm

• Vb endowed with the ”norm” ∇ · L2

b(Ω) is a Hilbert space

In other words, the question is: What conditions on the weight function b guarantee the validity of these two statements Among the most known weights, we found the Muckenhoupt weight. That means b ∈ Aq ⇐ ⇒ sup

Q

1 |Q|

• Q

b dx 1 |Q|

• Q

b−1/q−1 dx q−1 < ∞, b ∈ L1

loc(Ω)

This family of weights ensure that W n,p

b

(Ω) admits similar properties as W n,p(Ω). ◮ Some examples in Aq: b(x) = |x − x0|α

• r

b(x) = ρ(x)α := dist(x, ∂Ω)α, −(n − 1) < α < (n − 1)(q − 1) Ref.

• B. O. Turessson, Nonlinear Potential Theory and Weighted Sobolev Spaces (2000).
• B. Al Taki

Functional inequalities and applications CEMRACS’19 7 / 22

Importance of Muckenhoupt weight

Ω =]0, 1[ and u smooth s.t. u|∂Ω = 0. Then if b ∈ Aq, then we have 1 |u|q b dx ≤ 1 |u′|q b dx.

• Proof. Since u|∂Ω = 0, then we can write

u(x) = x u′(y) dy hence |u|q b = b

• x

u′(y) dy

• q

. Hence 1 |u|q b dx = 1

• b
• x

u′(y) dy

• q

dx ≤ C 1

• b
• 1

|u′(y)| dy

• q

dx ≤ C 1 b dx

• 1

|u′(y)| b1/q b−1/q dy

• q

≤ C 1 b dx 1 b−1/q−1 dx q−1 1 |u′(y)|q b dx

• .
• B. Al Taki

Functional inequalities and applications CEMRACS’19 8 / 22

Difficulties

♣ Trace: Characterization when b = ρα(x) = distα(x, ∂Ω), −1 < α < q − 1 (⇒ b ∈ Aq) T 1,q

b

: W 1,q

b

(Ω) − → W 1− 1+α

q

,q(∂Ω) bounded linear operator

♣ Regular solution: Ω =]0, 1[, b = ρα, −1 < α < 1 and ρ(x) = dist(x, ∂Ω) −∂x(b(x)∂xu) = f , in Ω, u = 0,

• n ∂Ω,

Straightforward computation yields to ∂2

x u = − b′(x)

(b(x))2 x f (z) dz + 1 b(x) f (x) − b′(x) (b(x))2 c. However ρ∂2

x u ∈ L2(Ω) ⇐

⇒ 1

• 1

ρα(x) 2 dx < ∞ = ⇒ α < 1 2 . ♣ Higher dimension? Big issue! Tangential regularity = ⇒ Normal regularity?

• B. Al Taki

Functional inequalities and applications CEMRACS’19 9 / 22

Lake equations

Theo. ∂t(buµ) + div(buµ ⊗ uµ)−2µ div(bD(uµ) + b div uµ I) + b∇pµ = 0 div(buµ) = 0 Navier boundary condition buµ · n = 0 (bD(uµ) · n + b div uµ I · n) · τ = ηu · τ. ◮ If b ∈ W 1,∞

loc (Ω)

with b = ρα 0 < α < 1 (⇒ b ∈ A2) = ⇒ ∃! uµ sol. with uµ ∈ L∞(0, T; L2

b(Ω)) ∩ L2(0, T; Vb) ∩ C((0, T), Hb − weak)

Vb =

• v ∈ H1

b(Ω), bv · n = 0, div(bv) = 0

• .

◮ Moreover if b = ρ(x)α, 0 < α < 1/2, (⇒ b ∈ A3/2) for x ∈ V (∂Ω), = ⇒ we can find pµ with ∇pµ ∈ W −1,∞(0, T; H−1

b

(Ω)) Work in progress with C. Lacave (Institute of Fourier) ◮ Regular solutions? ◮ What happen when µ tends to zero?

• B. Al Taki

Functional inequalities and applications CEMRACS’19 10 / 22

Logarithmic Sobolev inequalities

Sobolev inequality: let f : Rn − → R smooth, compactly supported

Rn |f |p dx

2/p ≤ C

• Rn |∇f |2 dx,

p = 2n n − 2 (> 2) n ≥ 3 = ⇒ 2 p log

Rn |f |p dx

• ≤ C log

Rn |∇f |2 dx

• Assume
• Rn f 2 dx = 1, Jensen’s inequality for f 2 dx

log

Rn |f |p dx

• = log

Rn |f |p−2f 2 dx

• ≥

Rn log

• |f |p−2

f 2 dx

• p − 2

p

• Rb f 2 log f 2 dx ≤ log

Rn |∇f |2 dx

• Form of logarithmic sobolev inequality
• B. Al Taki

Functional inequalities and applications CEMRACS’19 11 / 22

Different forms and applications

(Euclidean) Logarithmic Sobolev inequality

• Rn f 2 log f 2 dx ≤ n

2 log 2 nπe

• Rn |∇f |2 dx
• Rn f 2 dx = 1

dx − → dµ(x) = e−|x|2/2 dx (2π)n/2 µ standard Gaussian probability measure on Rn change f 2 into f 2e−|x|2/2 f : Rn − → R, smooth s.t.

• Rn f 2 dµ = 1
• Rn f 2 log f 2 dµ ≤ 2
• Rn |∇f |2 dµ

(Gaussian) Logarithmic Sobolev inequality

• B. Al Taki

Functional inequalities and applications CEMRACS’19 12 / 22

Different forms and applications

(Euclidean) Logarithmic Sobolev inequality

• Rn f 2 log f 2 dx ≤ n

2 log 2 nπe

• Rn |∇f |2 dx
• Rn f 2 dx = 1

dx − → dµ(x) = e−|x|2/2 dx (2π)n/2 µ standard Gaussian probability measure on Rn change f 2 into f 2e−|x|2/2 f : Rn − → R, smooth s.t.

• Rn f 2 dµ = 1

(entropy)

• Rn f log f dµ ≤ 2
• Rn f |∇ log f |2 dµ

(Fisher information) (Gaussian) Logarithmic Sobolev inequality

• B. Al Taki

Functional inequalities and applications CEMRACS’19 13 / 22

Interest of Logarithmic Sobolev Inequalities

Many applications: ◮ Quantum Navier-Stokes Superfluids, quantum semiconductors, weakly interacting Bose gases,... ◮ Navier-Stokes-Korteweg liquid-vapour flows including phase transitions ◮ Compressible Navier-Stokes equations Gas dynamics ◮ Derrida-Lebowitz-Speer-Spohn equations: quantum semiconductor ◮ Dispersive N-S eq.([Sone, Birkh¨ auser Boston, Inc., ’02]) In some situation, Asymp. analysis of the Boltz. eq. for NS and Euler eqs. ◮ Probability (work by M. Ledoux, D. Bakry) ◮ ...

• B. Al Taki

Functional inequalities and applications CEMRACS’19 14 / 22

Quantum Navier-Stokes ∂tρ + div(ρu) = 0 ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(ρD(u)) = −ερ∇ ∆√ρ

√ρ

• ,

Bohm’s identity ρ∇ ∆√ρ √ρ

• = div(ρ∇2 log ρ)

Need to estimates −

• Ω

ρ∇ ∆√ρ √ρ

• · (u + ∇ log ρ) dx = d

dt

• Ω

|∇√ρ|2 +

• Ω

ρ|∇2 log ρ|2 dx. Question: Information from

• Ω

ρ|∇2 log ρ|2 dx ≥

• Ω

|∇ρ1/4|4 dx Global weak sol. of Quantum Navier-Stokes, Jungel (SIAM, 2010) ⇓ Compressible Navier-Stokes (ε = 0) ∂tρ + div(ρu) = 0 ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(ρD(u)) = 0, Global weak sol. of compressible Navier-Stokes, A. Vasseur, C. Yu (Invent. Math. 2016)

• B. Al Taki

Functional inequalities and applications CEMRACS’19 15 / 22

Quantum Navier-Stokes ∂tρ + div(ρu) = 0 ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(ρD(u)) = −ερ∇ ∆√ρ

√ρ

• ,

Bohm’s identity ρ∇ ∆√ρ √ρ

• = div(ρ∇2 log ρ)

Need to estimates −

• Ω

ρ∇ ∆√ρ √ρ

• · (u + ∇ log ρ) dx = d

dt

• Ω

|∇√ρ|2 +

• Ω

ρ|∇2 log ρ|2 dx. Question: Information from

• Ω

ρ|∇2 log ρ|2 dx ≥

• Ω

|∇ρ1/4|4 dx Global weak sol. of Quantum Navier-Stokes, Jungel (SIAM, 2010) ⇓ Compressible Navier-Stokes (ε = 0) ∂tρ + div(ρu) = 0 ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(ρD(u)) = 0, Global weak sol. of compressible Navier-Stokes, A. Vasseur, C. Yu (Invent. Math. 2016)

• B. Al Taki

Functional inequalities and applications CEMRACS’19 15 / 22

Navier-Stokes Korteweg ∂tρ + div(ρu) = 0, ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(µ(ρ)D(u)) + ∇(λ(ρ) div u) = ερ∇(

• K(ρ)∆(

ρ

• K(s)ds)),

Generalized Bohm’s identity (Bresch et al (2016)) ρ∇(

• K(ρ)∆(

ρ

• K(s)ds)) = div(F(ρ)∇∇ψ(ρ)) + ∇
• (F ′(ρ)ρ − F(ρ))∆ψ(ρ)
• with

√ρψ′(ρ) =

• K(ρ), F ′(ρ) =
• K(ρ)ρ.

Difficulty: find an estimate/sign of (s′(ρ) = µ′(ρ)/ρ)

• Ω

ρ∇(

• K(ρ)∆(

ρ

• K(s)ds)) ·
• u + ∇s(ρ)
• dx

= 1 2 d dt

• Ω

|∇

• K(ρ)|2 dx +
• Ω

F(ρ)∇∇ψ(ρ) : ∇∇s(ρ) dx +

• Ω

(F ′(ρ)ρ − F(ρ))∆ψ(ρ)) ∆s(ρ) dx Global weak sol. for s(ρ) =? ψ(ρ) =? ⇓ Compressible Navier-Stokes (ε = 0) ∂tρ + div(ρu) = 0 ∂t(ρu) + div(ρu ⊗ u) + ∇ργ − div(µ(ρ)D(u)) + ∇(λ(ρ) div u) = 0, Global weak sol. for µ(ρ) =? λ(ρ) =? (In progress)

• B. Al Taki

Functional inequalities and applications CEMRACS’19 16 / 22

Question: Sign of I :=

• Ω

F(ρ)∇∇ψ(ρ) : ∇∇s(ρ) dx +

• Ω

(F ′(ρ)ρ − F(ρ))∆ψ(ρ)) ∆s(ρ) dx We take particular case ψ(ρ) = ρn s(ρ) = ρm = ⇒ F(ρ) = ρn+1 (F ′(ρ)ρ − F(ρ))∆ψ(ρ)) = nρn+1 Thus I =

• Ω

ρn+1∇∇ρn : ∇∇ρm dx + n

• Ω

ρn+1∆ρn∆ρm dx Particular cases: ◮ Jungel, Matthes (n = m = 0)

• Ω

ρ|∇2 log ρ|2 dx ≥ c

• Ω

|∇ρ1/4|4 dx ◮ Bresch, Vasseur, Yu, (n = m)

• Ω

ρn+1|∇∇ρn|2 dx ≥ c

• Ω

|∇∇ρ

3n+1 2

|2 dx +

• Ω

|∇ρ

3n+1 4

|4 dx (−1 + 2 d < n < 1) ◮ A., (Key estimate of a publication about ghost effect system)

• Ω

ρn+1∇∇ρn : ∇∇ log ρ dx + n

• Ω

ρn+1∆ρn∆ log ρ dx ≥

• Ω
• ∆ρ

2n+1 2

2 dx (0 ≤ n ≤ 1/2)

• B. Al Taki

Functional inequalities and applications CEMRACS’19 17 / 22

• Lemma. Ω ⊂ Rd, d = 2, 3, periodic domain and ρ positive smooth and n, m given in Figure 1,

then ∃ C1 > 0, C2 > 0 such that I =

• Ω

ρn+1∇∇ρn : ∇∇ρm dx + n

• Ω

ρn+1∆ρn∆ρm dx ≥ C1

• Ω
• ∇∇ρ

2n+m+1 2

2 + C2

• Ω

|∇ρ

2n+m+1 4

|4 dx. dimension 2 dimension 3

Figure:

• B. Al Taki

Functional inequalities and applications CEMRACS’19 18 / 22

• Proof. Simple computations, we obtain

J1 : =

• Ω

ρn+1∇∇ρn : ∇∇ρm dx =

• Ω

ρn+1∇ n θ ρn−θ∇ρθ : ∇ m θ ρm−θ∇ρθ dx = n m θ2

Ω

ρ2n+m−2θ+1|∇∇ρθ|2 dx−γ1

• Ω

ρ2n+m+1−θ∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx −γ2

• Ω

ρ2n+1+m−θ∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx + γ1 γ2

• Ω

ρ2n+m+1−θ(∇ρθ/2)4 dx

• with

γ1 = 4(θ − m) θ γ2 = 4(θ − n) θ . Now, let us choose θ such that θ = 2n + m + 1 2 . Thus the integral J1 becomes J1 := n m θ2

Ω

|∇∇ρθ|2 dx − (γ1 + γ2)

• Ω

∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx + γ1γ2

• Ω

(∇ρθ/2)4 dx

• B. Al Taki

Functional inequalities and applications CEMRACS’19 19 / 22

Similar computation give us J2 =

• Ω

ρn+1∆ρn∆ρm dx = n m θ2

Ω

|∆ρθ|2 dx − (γ1 + γ2)

• Ω

∆ρθ (∇ρθ/2)2 dx + γ1γ2

• Ω

(∇ρθ/2)4 dx

• .

Gathering J1 and J2 together and take in mind that

• Ω

|∇∇ρθ|2 dx =

• Ω

|∆ρθ|2 dx We infer with I = n m θ2

• (1 + n)
• Ω

|∇∇ρθ|2 dx + γ1γ2(1 + n)

• Ω

(∇ρθ/2)4 dx − (γ1 + γ2)

Ω

∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx + n

• Ω

∆ρθ (∇ρθ/2)2 dx

• .

We want to establish an estimate on −(γ1 + γ2)

Ω

∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx + n

• Ω

∆ρθ (∇ρθ/2)2 dx

• .
• B. Al Taki

Functional inequalities and applications CEMRACS’19 20 / 22

Remark that ρ∇∇ log ρ = ∇∇ρ − 4∇√ρ ⊗ ∇√ρ, ρ∆ log ρ = ∆ρ − 4(∇√ρ)2 This implies that

• Ω

ρ2|∇∇ log ρ|2 dx =

• Ω

|∇∇ρ|2 dx +

• Ω

|2∇√ρ|4 dx − 8

• Ω

∇∇ρ : ∇√ρ ⊗ ∇√ρ dx

• Ω

ρ2|∆ log ρ|2 dx =

• Ω

|∆ρ|2 dx +

• Ω

|2∇√ρ|4 dx − 8

• Ω

∆ρ (∇√ρ)2 dx Therefore −(γ1 + γ2)

Ω

∇∇ρθ : ∇ρθ/2 ⊗ ∇ρθ/2 dx + n

• Ω

∆ρθ (∇ρθ/2)2 dx

• = (γ1 + γ2)

8

• Ω

ρ2θ|∇∇ log ρθ|2 dx + n(γ1 + γ2) 8

• Ω

ρ2θ|∆ log ρθ|2 dx − (1 + n)(γ1 + γ2) 8

• Ω

|∇∇ρθ|2 dx − 2(1 + n)(γ1 + γ2)

• Ω

|∇ρθ/2|4 dx And remember that

• Ω

ρ2θ|∇2 log ρθ|2 dx + n

• Ω

ρ2θ|∆ log ρθ|2 dx ≥ c

• Ω

|2∇ρθ/2|4 dx.

• B. Al Taki

Functional inequalities and applications CEMRACS’19 21 / 22

Thank you for your attention

• B. Al Taki

Functional inequalities and applications CEMRACS’19 22 / 22