Navier-Stokes Equation · 2018. 3. 4. · T. Muthukumar Navier-Stokes. Kinetic Equations Understanding the dynamics of a physical matter (solid, liquid and gas) has been part of human

Navier-Stokes Equation

T. [email protected]

Indian Institute of Technology-Kanpur

24 & 27 Feb 2018

T. Muthukumar Navier-Stokes

Clay Problem

In the year 2000 Clay Mathematics Institute stated sevenimportant unsolved problems for the millennium.The Poincare conjecture was solved in 2003.One of the remaining six problems includes a problem onNavier-Stokes equation.The purpose of this talk to understand the problem, itsdifficulty, what is known and unknown.


Kinetic Equations

Understanding the dynamics of a physical matter (solid, liquidand gas) has been part of human interest from antiquity.The dynamics of fluids (liquid and gas) is governed byinteraction of atoms/molecules (microscopic description)described by Boltzmann equations.

The collision dominated regime can be approximatelymodelled by Navier-Stokes and Euler Equations.

How accurate is the macroscopic description of a microscopicdynamics? Does it break down? If yes, when?


Kinetic Equations

Understanding the dynamics of a physical matter (solid, liquidand gas) has been part of human interest from antiquity.The dynamics of fluids (liquid and gas) is governed byinteraction of atoms/molecules (microscopic description)described by Boltzmann equations.The collision dominated regime can be approximatelymodelled by Navier-Stokes and Euler Equations.



Kinetic Equations

Understanding the dynamics of a physical matter (solid, liquidand gas) has been part of human interest from antiquity.The dynamics of fluids (liquid and gas) is governed byinteraction of atoms/molecules (microscopic description)described by Boltzmann equations.The collision dominated regime can be approximatelymodelled by Navier-Stokes and Euler Equations.



Navier-Stokes Equations

Let u(x , t) = (ui (x , t))n1 be the velocity of the fluid,

and p(x , t) be the pressure of the fluid at position x and timet.Then by Newton’s second law of motion, for 1 ≤ i ≤ n,

fi + ν∆ui −∂p∂xi

= ∂ui∂t +

n∑j=1

uj∂ui∂xj

in Ω× [0,T ).

(fi ) is the external force exerted on the fluid and ν > 0 is theviscosity of the fluid.A compact notation for the system of n equations is

f + ν∆u −∇p = ∂u∂t + (u · ∇)u in Ω× [0,T ).




and p(x , t) be the pressure of the fluid at position x and timet.

Then by Newton’s second law of motion, for 1 ≤ i ≤ n,


= ∂ui∂t +

n∑j=1

uj∂ui∂xj

in Ω× [0,T ).


f + ν∆u −∇p = ∂u∂t + (u · ∇)u in Ω× [0,T ).






= ∂ui∂t +

n∑j=1

uj∂ui∂xj

in Ω× [0,T ).

(fi ) is the external force exerted on the fluid and ν > 0 is theviscosity of the fluid.

A compact notation for the system of n equations is

f + ν∆u −∇p = ∂u∂t + (u · ∇)u in Ω× [0,T ).






= ∂ui∂t +

n∑j=1

uj∂ui∂xj

in Ω× [0,T ).


f + ν∆u −∇p = ∂u∂t + (u · ∇)u in Ω× [0,T ).


Viscosity and Reynolds number

Viscosity (ν) is the elasticity counterpart for fluids.

Reynolds number is the ratio

Re :=∣∣∣∣∣ u · ∇uν(∆u + 1

3∇(∇ · u))

∣∣∣∣∣ .The viscosity ν is inversely propotional to the Reynoldsnumber.The situation ν = 0 or Re =∞ is called the Euler equation.


Viscosity and Reynolds number

Viscosity (ν) is the elasticity counterpart for fluids.Reynolds number is the ratio

Re :=∣∣∣∣∣ u · ∇uν(∆u + 1

3∇(∇ · u))

∣∣∣∣∣ .The viscosity ν is inversely propotional to the Reynoldsnumber.The situation ν = 0 or Re =∞ is called the Euler equation.


Incompressibility Condition

If fluid is incompressible then the divergence free condition issatisfied:

div(u) :=n∑

i=1

∂ui∂xi

= 0 in Ω× [0,T ).

The incompressibility condition signifies that the volume ofthe fluid does not change during the flow! Roughly speaking,fluids moving close/above the speed of sound are consideredcompressible.NSE is given the parabolic boundary conditions: Theunknowns (u, p) satisfies

ut + (u · ∇)u +∇p = f + ν∆u in Ω× (0,T )div(u) = 0 in Ω× (0,T )u(x , t) = 0 in ∂Ω× (0,T )u(x , 0) = u0(x) on Ω,

For consistency we assume div(u0) = 0.




div(u) :=n∑

i=1

∂ui∂xi

= 0 in Ω× [0,T ).

The incompressibility condition signifies that the volume ofthe fluid does not change during the flow! Roughly speaking,fluids moving close/above the speed of sound are consideredcompressible.

NSE is given the parabolic boundary conditions: Theunknowns (u, p) satisfies


For consistency we assume div(u0) = 0.




div(u) :=n∑

i=1

∂ui∂xi

= 0 in Ω× [0,T ).

The incompressibility condition signifies that the volume ofthe fluid does not change during the flow! Roughly speaking,fluids moving close/above the speed of sound are consideredcompressible.NSE is given the parabolic boundary conditions: Theunknowns (u, p) satisfies


For consistency we assume div(u0) = 0.T. Muthukumar Navier-Stokes

Conflict between Transport and Dissipation term

The non-linear term (u · ∇)u is the convective materialderivative term describing ‘inertial acceleration’.

The nonlinear term (inertial acceleration) and viscosity term(dissipation) play opposite roles during the flow giving rise toa (possible) ‘turbulence’.If ν∆u >> (u · ∇)u then it is kind of heat equation; oneexpects to see linear, non-turbulent behaviour (Globalregularity exist). Velocity << inverse of spatial scale. In twodimensions this can be established and validates NSE.If ν∆u << (u · ∇)u then one expects to see nonlinear,turbulent behaviour and blow-up. Velocity >> inverse ofspatial scale.



The non-linear term (u · ∇)u is the convective materialderivative term describing ‘inertial acceleration’.The nonlinear term (inertial acceleration) and viscosity term(dissipation) play opposite roles during the flow giving rise toa (possible) ‘turbulence’.

If ν∆u >> (u · ∇)u then it is kind of heat equation; oneexpects to see linear, non-turbulent behaviour (Globalregularity exist). Velocity << inverse of spatial scale. In twodimensions this can be established and validates NSE.If ν∆u << (u · ∇)u then one expects to see nonlinear,turbulent behaviour and blow-up. Velocity >> inverse ofspatial scale.



The non-linear term (u · ∇)u is the convective materialderivative term describing ‘inertial acceleration’.The nonlinear term (inertial acceleration) and viscosity term(dissipation) play opposite roles during the flow giving rise toa (possible) ‘turbulence’.If ν∆u >> (u · ∇)u then it is kind of heat equation; oneexpects to see linear, non-turbulent behaviour (Globalregularity exist). Velocity << inverse of spatial scale. In twodimensions this can be established and validates NSE.

If ν∆u << (u · ∇)u then one expects to see nonlinear,turbulent behaviour and blow-up. Velocity >> inverse ofspatial scale.



The non-linear term (u · ∇)u is the convective materialderivative term describing ‘inertial acceleration’.The nonlinear term (inertial acceleration) and viscosity term(dissipation) play opposite roles during the flow giving rise toa (possible) ‘turbulence’.If ν∆u >> (u · ∇)u then it is kind of heat equation; oneexpects to see linear, non-turbulent behaviour (Globalregularity exist). Velocity << inverse of spatial scale. In twodimensions this can be established and validates NSE.If ν∆u << (u · ∇)u then one expects to see nonlinear,turbulent behaviour and blow-up. Velocity >> inverse ofspatial scale.


Properties of the Solutions

If (u, p) is a solution of NSE with f = 0 and Ω = Rn then

(Translation) (uc , pc) is also a solution, for any c ∈ Rn, whereuc(x , t) := u(x − ct, t) + c and pc(x , t) := p(x − ct, t).(Rotation) (uθ, pθ) is also a solution, for any orthogonalmatrix θ ∈ O(n), where uθ(x , t) := θT u(θx , t) andpθ(x , t) := p(θx , t).(Dilation) (uλ, pλ) is also a solution, for any λ > 0, whereuλ(x , t) := λ−1u(λ−1x , λ−2t) andpλ(x , t) := λ−2p(λ−1x , λ−2t).The dilation property signifies that, for large λ > 1, thesolution uλ ‘magnifies’ the behaviour of u at fine space scale1/λ and time scale 1/λ2.



If (u, p) is a solution of NSE with f = 0 and Ω = Rn then(Translation) (uc , pc) is also a solution, for any c ∈ Rn, whereuc(x , t) := u(x − ct, t) + c and pc(x , t) := p(x − ct, t).

(Rotation) (uθ, pθ) is also a solution, for any orthogonalmatrix θ ∈ O(n), where uθ(x , t) := θT u(θx , t) andpθ(x , t) := p(θx , t).(Dilation) (uλ, pλ) is also a solution, for any λ > 0, whereuλ(x , t) := λ−1u(λ−1x , λ−2t) andpλ(x , t) := λ−2p(λ−1x , λ−2t).The dilation property signifies that, for large λ > 1, thesolution uλ ‘magnifies’ the behaviour of u at fine space scale1/λ and time scale 1/λ2.



If (u, p) is a solution of NSE with f = 0 and Ω = Rn then(Translation) (uc , pc) is also a solution, for any c ∈ Rn, whereuc(x , t) := u(x − ct, t) + c and pc(x , t) := p(x − ct, t).(Rotation) (uθ, pθ) is also a solution, for any orthogonalmatrix θ ∈ O(n), where uθ(x , t) := θT u(θx , t) andpθ(x , t) := p(θx , t).

(Dilation) (uλ, pλ) is also a solution, for any λ > 0, whereuλ(x , t) := λ−1u(λ−1x , λ−2t) andpλ(x , t) := λ−2p(λ−1x , λ−2t).The dilation property signifies that, for large λ > 1, thesolution uλ ‘magnifies’ the behaviour of u at fine space scale1/λ and time scale 1/λ2.



If (u, p) is a solution of NSE with f = 0 and Ω = Rn then(Translation) (uc , pc) is also a solution, for any c ∈ Rn, whereuc(x , t) := u(x − ct, t) + c and pc(x , t) := p(x − ct, t).(Rotation) (uθ, pθ) is also a solution, for any orthogonalmatrix θ ∈ O(n), where uθ(x , t) := θT u(θx , t) andpθ(x , t) := p(θx , t).(Dilation) (uλ, pλ) is also a solution, for any λ > 0, whereuλ(x , t) := λ−1u(λ−1x , λ−2t) andpλ(x , t) := λ−2p(λ−1x , λ−2t).

The dilation property signifies that, for large λ > 1, thesolution uλ ‘magnifies’ the behaviour of u at fine space scale1/λ and time scale 1/λ2.



If (u, p) is a solution of NSE with f = 0 and Ω = Rn then(Translation) (uc , pc) is also a solution, for any c ∈ Rn, whereuc(x , t) := u(x − ct, t) + c and pc(x , t) := p(x − ct, t).(Rotation) (uθ, pθ) is also a solution, for any orthogonalmatrix θ ∈ O(n), where uθ(x , t) := θT u(θx , t) andpθ(x , t) := p(θx , t).(Dilation) (uλ, pλ) is also a solution, for any λ > 0, whereuλ(x , t) := λ−1u(λ−1x , λ−2t) andpλ(x , t) := λ−2p(λ−1x , λ−2t).The dilation property signifies that, for large λ > 1, thesolution uλ ‘magnifies’ the behaviour of u at fine space scale1/λ and time scale 1/λ2.


Function Spaces for Solutions

For open connected Ω ⊂ Rn

C∞c,σ(Ω) := v ∈ [C∞c (Ω)]n | div(v) = 0.

J(Ω) := C∞c,σ(Ω)‖∇‖2 .J2

0,σ(Ω) is the L2 closure of v ∈ [C∞c (Ω)]n | div(v) = 0;

In general, J(Ω) ⊆ v ∈ [C∞c (Ω)]n‖∇‖2 | div(v) = 0.




C∞c,σ(Ω) := v ∈ [C∞c (Ω)]n | div(v) = 0.

J(Ω) := C∞c,σ(Ω)‖∇‖2 .

J20,σ(Ω) is the L2 closure of v ∈ [C∞c (Ω)]n | div(v) = 0;





C∞c,σ(Ω) := v ∈ [C∞c (Ω)]n | div(v) = 0.

J(Ω) := C∞c,σ(Ω)‖∇‖2 .J2






C∞c,σ(Ω) := v ∈ [C∞c (Ω)]n | div(v) = 0.

J(Ω) := C∞c,σ(Ω)‖∇‖2 .J2




Helmholtz Decomposition and Leray Projection Operator

For open connected Ω ⊂ Rn, we have[L2(Ω)]n = J2

0,σ(Ω)⊕ (J20,σ(Ω))⊥ where

(J20,σ(Ω))⊥ := v ∈ [L2(Ω)]n | ∃p ∈ L2

loc(Ω) s.t. v = ∇p.For bounded Lipschitz domains Ω

L2σ(Ω) = v ∈ [L2(Ω)]n | div(v) = 0 and v · n |∂Ω= 0;

L2σ(Ω)⊥ := v ∈ [L2(Ω)]n | ∃p ∈ L2(Ω) s.t. v = ∇p.

The projection operator P : [L2(Ω)]n → L2σ(Ω) is a bounded linear

operator called the Helmholtz or Leray Projection.


Integrability of the non-linear term

For u, v ∈ H10 (Ω) and w ∈ C∞c (Ω), by Holder’s inequality, we have∫

Ωu · ∇v · w dx ≤ ‖uw‖2‖∇v‖2

≤ ‖u‖2r‖w‖2s‖∇v‖2

for r , s ≥ 1 such that 1/r + 1/s = 1. Using the continuousimbedding

H10 (Ω) ⊂

Lr (Ω), r ∈ [1,∞) n = 2Lr (Ω), r ∈ [2, 2n

n−2 ] n ≥ 3

the best choice is r = nn−2 and, hence, its conjugate s = n

2 . Thus,∫Ω

u · ∇v · w dx ≤ ‖u‖ 2nn−2‖w‖n‖∇v‖2

≤ C‖∇u‖2‖∇w‖2‖∇v‖2.

The last inequality is valid only for n ≤ 4 because, for n ≥ 5,n /∈ [2, 2n

n−2 ].


Integrability of the non-linear termFor u, v ∈ H1

0 (Ω) and w ∈ C∞c (Ω), by Holder’s inequality, we have∫Ω

u · ∇v · w dx ≤ ‖uw‖2‖∇v‖2

≤ ‖u‖2r‖w‖2s‖∇v‖2

for r , s ≥ 1 such that 1/r + 1/s = 1.

Using the continuousimbedding

H10 (Ω) ⊂

Lr (Ω), r ∈ [1,∞) n = 2Lr (Ω), r ∈ [2, 2n

n−2 ] n ≥ 3


2 . Thus,∫Ω

u · ∇v · w dx ≤ ‖u‖ 2nn−2‖w‖n‖∇v‖2

≤ C‖∇u‖2‖∇w‖2‖∇v‖2.


n−2 ].




u · ∇v · w dx ≤ ‖uw‖2‖∇v‖2

≤ ‖u‖2r‖w‖2s‖∇v‖2


H10 (Ω) ⊂

Lr (Ω), r ∈ [1,∞) n = 2Lr (Ω), r ∈ [2, 2n

n−2 ] n ≥ 3


2 . Thus,∫Ω

u · ∇v · w dx ≤ ‖u‖ 2nn−2‖w‖n‖∇v‖2

≤ C‖∇u‖2‖∇w‖2‖∇v‖2.


n−2 ].




u · ∇v · w dx ≤ ‖uw‖2‖∇v‖2

≤ ‖u‖2r‖w‖2s‖∇v‖2


H10 (Ω) ⊂

Lr (Ω), r ∈ [1,∞) n = 2Lr (Ω), r ∈ [2, 2n

n−2 ] n ≥ 3


2 . Thus,∫Ω

u · ∇v · w dx ≤ ‖u‖ 2nn−2‖w‖n‖∇v‖2

≤ C‖∇u‖2‖∇w‖2‖∇v‖2.


n−2 ].T. Muthukumar Navier-Stokes

Properties of the nonlinear term

If u ∈ J(Ω) then(a) ‖u ⊗ u‖3 ≤ C‖∇u‖2

2,(b) ‖u · ∇u‖3/2 ≤ C‖∇u‖2

2,(c) u · ∇u = div(u ⊗ u),(d) (u · ∇v) · v = 1

2 u · ∇|v |2,(e)

∫Ω u · ∇v · v = 0,

(f)∫

Ω u · ∇v · w = −∫

Ω u · ∇w · v .

Using (c) the NSE can be rewritten asut − ν∆u + div(u ⊗ u) +∇p = f in QT

div(u) = 0 in QTu = 0 on ∂Ω× (0,T ),

u(x , 0) = u0 in Ω.


Properties of the nonlinear term

If u ∈ J(Ω) then(a) ‖u ⊗ u‖3 ≤ C‖∇u‖2

2,(b) ‖u · ∇u‖3/2 ≤ C‖∇u‖2

2,(c) u · ∇u = div(u ⊗ u),(d) (u · ∇v) · v = 1

2 u · ∇|v |2,(e)

∫Ω u · ∇v · v = 0,

(f)∫

Ω u · ∇v · w = −∫

Ω u · ∇w · v .Using (c) the NSE can be rewritten as

ut − ν∆u + div(u ⊗ u) +∇p = f in QTdiv(u) = 0 in QT

u = 0 on ∂Ω× (0,T ),u(x , 0) = u0 in Ω.


Formal Energy Estimate

Theorem (Energy Estimate)If a divergence free solution exists for NSE then

ddt ‖u‖

22 + ν‖∇u‖2

2 ≤C(Ω)ν‖f ‖2

2.

For all t ∈ [0,T ]

‖u(t)‖22 ≤ e−Cνt‖u0‖2

2 + C(Ω)ν2 (1− eCνt)‖f ‖2

2

andν

∫ T

0‖∇u(t)‖2

2 ≤ C(Ω, ‖u0‖2, ‖f ‖2, ν,T ).


Formal Proof of Energy Estimate

Formally multiply u both sides of NSE and integrate in spacevariable

12

ddt ‖u‖

22 + ν‖∇u‖2

2 =∫

Ωf · u

By incompressibility condition and property (d) the pressureand nonlinear term vanish.

Therefore, by Young’s inequality,

12

ddt ‖u‖

22+ν‖∇u‖2

2 ≤ C(Ω)‖f ‖2‖∇u‖2 ≤C(Ω)ν‖f ‖2

2+ν

2‖∇u‖22

Apply Gronwall inequality to obtain the estimate in timevariable. The estimate on gradient follows as a consequence.The estimate are necessary condition for existence of solutions.It turns out they are sufficient for existence of weak solutions.




12

ddt ‖u‖

22 + ν‖∇u‖2

2 =∫

Ωf · u

By incompressibility condition and property (d) the pressureand nonlinear term vanish.Therefore, by Young’s inequality,

12

ddt ‖u‖

22+ν‖∇u‖2

2 ≤ C(Ω)‖f ‖2‖∇u‖2 ≤C(Ω)ν‖f ‖2

2+ν

2‖∇u‖22





12

ddt ‖u‖

22 + ν‖∇u‖2

2 =∫

Ωf · u

By incompressibility condition and property (d) the pressureand nonlinear term vanish.Therefore, by Young’s inequality,

12

ddt ‖u‖

22+ν‖∇u‖2

2 ≤ C(Ω)‖f ‖2‖∇u‖2 ≤C(Ω)ν‖f ‖2

2+ν

2‖∇u‖22



Leray-Hopf Solution

A function u : QT → R is said to be weak Leray-Hopf solution toNSE if

u ∈ L∞(0,T ; J20,σ(Ω)) ∩ L2(0,T ; J(Ω)),

the map t 7→∫

Ω u(x , t) · v(x) dx is continuous on [0,T ] foreach v ∈ [L2(Ω)]n,For all φ ∈ C∞c,σ(QT ), u satisfies∫

QT(−u · φt −∇φ : (u ⊗ u) + ν(∇u : ∇φ)− f · φ) = 0,

As t → 0+, ‖u(t)− u0‖22 → 0.

For all t ∈ [0,T ],

12‖u(t)‖2

2 +ν∫ t

0

∫Ω|∇u|2 dx ds ≤ 1

2‖u0‖22 +∫ t

0

∫Ω

(f ·u) dx ds.


Existence

Theorem (Existence)(Leray 1932-34) Let n = 2 and n = 3. If f ∈ L2(0,T ; J∗(Ω)) andu0 ∈ J2

0,σ(Ω) then there exists at least one weak Leray-Hopfsolution.

In general, uniqueness is not known!Observe that Leray proved the existence of ‘such’ solutionbefore the concept of distribution and weak solutions wereintroduced.


Strong Solution

DefinitionA weak Leray-Hopf solution u is called a strong solution if∇u ∈ L∞(0,T ; L2(Ω)).

Suppose u is a strong solution then formally12

ddt ‖∇u‖2

2 + ν‖∆u‖22 +

∫Ω

(u · ∇)u · (−∆u) =∫

Ωf · (−∆u)

The pressure term vanishes by the incompressibility condition.ByHolder’s inequality,∫

Ωu · ∇u · (−∆u) dx ≤ ‖u‖4‖∇u‖4‖∆u‖2.

But by interpolation inequality (Ladyzhenskaya) for w ∈ H10 (Ω),

‖w‖4 ≤

21/4‖w‖1/22 ‖∇w‖1/2

2 n = 2c‖w‖1/4

2 ‖∇w‖3/42 n = 3.


Strong Solution



ddt ‖∇u‖2

2 + ν‖∆u‖22 +

∫Ω

(u · ∇)u · (−∆u) =∫

Ωf · (−∆u)

The pressure term vanishes by the incompressibility condition.

ByHolder’s inequality,∫

Ωu · ∇u · (−∆u) dx ≤ ‖u‖4‖∇u‖4‖∆u‖2.


‖w‖4 ≤

21/4‖w‖1/22 ‖∇w‖1/2

2 n = 2c‖w‖1/4

2 ‖∇w‖3/42 n = 3.


Strong Solution



ddt ‖∇u‖2

2 + ν‖∆u‖22 +

∫Ω

(u · ∇)u · (−∆u) =∫

Ωf · (−∆u)


Ωu · ∇u · (−∆u) dx ≤ ‖u‖4‖∇u‖4‖∆u‖2.


‖w‖4 ≤

21/4‖w‖1/22 ‖∇w‖1/2

2 n = 2c‖w‖1/4

2 ‖∇w‖3/42 n = 3.


Strong Solution



ddt ‖∇u‖2

2 + ν‖∆u‖22 +

∫Ω

(u · ∇)u · (−∆u) =∫

Ωf · (−∆u)


Ωu · ∇u · (−∆u) dx ≤ ‖u‖4‖∇u‖4‖∆u‖2.


‖w‖4 ≤

21/4‖w‖1/22 ‖∇w‖1/2

2 n = 2c‖w‖1/4

2 ‖∇w‖3/42 n = 3.


Why is it easy in Two dimensions?

If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ Cy2. Rewrite y ′

y ≤ Cy and

y ≤ Ce∫ T

0 y . Since∫

y is finite, e∫

y is finite.

Theorem (Uniqueness)(O. Ladyzhenskaya) Let n = 2. If f ∈ L2(0,T ; J∗(Ω)) andu0 ∈ J2

0,σ(Ω) then the weak Leray-Hopf solution is unique.

Theorem (Regularity)Let n = 2, Ω is a bounded domain with smooth boundary,f ∈ L2(QT ) and u0 ∈ J(Ω). If u is the unique weak Leray-Hopfsolution of NSE then u ∈ H2,1(QT ) whereH2,1(QT ) := v ∈ L2(0,T ; H2(Ω) | vt ∈ L2(QT ) and∇u ∈ C([0,T ]; L2(Ω)). Further, there exists p ∈ L2(0,T ; H1(Ω)such that ut + u · ∇u − ν∆u = f −∇p and div(u) = 0 a.e. in QT .




y ≤ Cy and

y ≤ Ce∫ T

0 y . Since∫

y is finite, e∫

y is finite.







y ≤ Cy and

y ≤ Ce∫ T

0 y . Since∫

y is finite, e∫

y is finite.







y ≤ Cy and

y ≤ Ce∫ T

0 y . Since∫

y is finite, e∫

y is finite.





What Happens in Three Dimensions?

If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ C(‖u‖4

6 + α2)y .

Rewrite y ′y ≤ C(‖u‖4

6 + α2) and y ≤ y(0)eC∫ t

0 (‖u‖46+α2) dt .

The integrating factor is finite if∫ t

0‖u(t)‖4

6 dt <∞?

for every initial data.Now, that’s the Clay prize problem!



If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ C(‖u‖4

6 + α2)y .


6 + α2) and y ≤ y(0)eC∫ t

0 (‖u‖46+α2) dt .


0‖u(t)‖4

6 dt <∞?




If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ C(‖u‖4

6 + α2)y .


6 + α2) and y ≤ y(0)eC∫ t

0 (‖u‖46+α2) dt .


0‖u(t)‖4

6 dt <∞?




If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ C(‖u‖4

6 + α2)y .


6 + α2) and y ≤ y(0)eC∫ t

0 (‖u‖46+α2) dt .


0‖u(t)‖4

6 dt <∞?

for every initial data.

Now, that’s the Clay prize problem!



If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ C(‖u‖4

6 + α2)y .


6 + α2) and y ≤ y(0)eC∫ t

0 (‖u‖46+α2) dt .


0‖u(t)‖4

6 dt <∞?



What’s special about 6 and 4

TheoremA strong solution to 3D NSE exists iff Ladyzhenskaya-Prodi-Serrincondition is satisfied, i.e. u ∈ Lr (0,T ; Ls(Ω)) with r , s ≥ 1 and

2r + 3

s = 1.

Note that when s = 6 then r = 4 is a special case. The caser =∞ and s = 3 was settled by Escauriaza, Seregin and Sverak.

TheoremLet u and v be two weak Leray-Hopf solutions of NSE givenu0 ∈ J(Ω) and f ∈ L2(QT ). If one of the solution, say u, satisfiesthe L-P-S condition then u = v.




2r + 3

s = 1.

Note that when s = 6 then r = 4 is a special case.

The caser =∞ and s = 3 was settled by Escauriaza, Seregin and Sverak.





2r + 3

s = 1.






2r + 3

s = 1.




Unique LPS weak Solution

Open problem: We do not know if any weak Leray-Hopf weaksolution satisfies L-P-S condition.

The best we know (as on date) is that a Leray-Hopf weaksolution u ∈ Lr (0,T ; Ls(Ω)) with s, r ≥ 1 and

3s + 2

r = 32 > 1!


Unique LPS weak Solution

Open problem: We do not know if any weak Leray-Hopf weaksolution satisfies L-P-S condition.The best we know (as on date) is that a Leray-Hopf weaksolution u ∈ Lr (0,T ; Ls(Ω)) with s, r ≥ 1 and

3s + 2

r = 32 > 1!


Sobolev Inequality instead of Interpolation

Let n = 3. If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ Cy3 in addtion that∫

y <∞ gives the existence in short time as summarised below.

TheoremIf n = 3, f ∈ L2(QT ) and u0 ∈ J(Ω) then there exists T0 ∈ (0,T )such that NSE admits a strong solution in QT0 . Also, there existp ∈ L2(0,T ; H1(Ω)).

The case when f = 0, T0 has the lower bound

T0 ≥C(Ω)‖∇u0‖4

2

famously known as Leray’s estimate.


Sobolev Inequality instead of Interpolation

Let n = 3. If y(t) := α + ‖∇u(t)‖22 then y ′ ≤ Cy3 in addtion that∫

y <∞ gives the existence in short time as summarised below.

TheoremIf n = 3, f ∈ L2(QT ) and u0 ∈ J(Ω) then there exists T0 ∈ (0,T )such that NSE admits a strong solution in QT0 . Also, there existp ∈ L2(0,T ; H1(Ω)).

The case when f = 0, T0 has the lower bound

T0 ≥C(Ω)‖∇u0‖4

2

famously known as Leray’s estimate.


Existence of Strong Soln for Small data:3D

Theorem (Small data)If there exists a constant C(Ω) such that

arctan(‖∇u0‖2

2

)+ C(Ω)

(‖u0‖2

2 + ‖f ‖22

)<π

2

then there exists a strong solution to NSE. Also, there existp ∈ L2(0,T ; H1(Ω)).


Uniqueness of Strong Solution in Leray-Hopf Class

TheoremLet u and v be two strong solutions of NSE given u0 ∈ J(Ω) andf ∈ L2(QT ). Then u = v.

TheoremLet u and v be two weak Leray-Hopf solutions of NSE givenu0 ∈ J(Ω) and f ∈ L2(QT ). If one of them is a strong solutionthen u = v.

Theorem (Regularity)If u is a strong solution of NSE for given f ∈ L2(QT ) andu0 ∈ J(Ω) then u ∈ H2,1(QT ) and p ∈ L2(0,T ; H1(Ω).


Open Problems in Three DimensionsWhat is known in three dimensions?

Global existence of weak solutions.Uniqueness of weak solution in the LPS class.Short time existence of strong solutions.Uniqueness of strong solutions.

What are the open problems?The uniqueness of weak solution.Global existence of strong solution is open! (Clay millenniumproblem).Lesser known open problem: Above question is open even forthe case ν = 0, i.e. for the Euler equations.(Limit of NSE as ν → 0 for any dimension): For unboundeddomains, if Euler equation admits a solution in an intervalthen solution of NSE converges to a solution of Euler (Kato &P. Constantin). But for bounded domain the validity of similarstatement is still open?




What are the open problems?

The uniqueness of weak solution.Global existence of strong solution is open! (Clay millenniumproblem).Lesser known open problem: Above question is open even forthe case ν = 0, i.e. for the Euler equations.(Limit of NSE as ν → 0 for any dimension): For unboundeddomains, if Euler equation admits a solution in an intervalthen solution of NSE converges to a solution of Euler (Kato &P. Constantin). But for bounded domain the validity of similarstatement is still open?




What are the open problems?The uniqueness of weak solution.Global existence of strong solution is open! (Clay millenniumproblem).Lesser known open problem: Above question is open even forthe case ν = 0, i.e. for the Euler equations.(Limit of NSE as ν → 0 for any dimension): For unboundeddomains, if Euler equation admits a solution in an intervalthen solution of NSE converges to a solution of Euler (Kato &P. Constantin). But for bounded domain the validity of similarstatement is still open?


Suitable Weak Solution

The notion of suitable weak solution was introduced byCafarelli-Kohn-Nirenberg (1982).This was introduced to study the size of singularity sets ofweak solutions.They showed that the singularity sets are ‘small’ inspace-time, i.e. its Hausdorff measure is < 1! They cannotcontain a space-time curve.Still it is far from proving it is empty.Recent work approaches mainly point towards proving finitetime blow-up, in effect hinting the open problem to be innegation!


Vorticity Formulation of NSE

Let ω := ∇× u is the vorticity.

ωt − ν∆ω + (u · ∇)ω − (ω · ∇)u = ∇× f .

The term (ω · ∇)u is called the vorticity distortion.In 2D (ω · ∇)u = 0 and the vorticity satisfies

ωt − ν∆ω + (u · ∇)ω = ∇× f .

|ω|2 satisfies maximum principle.In 3D, (ω · ∇)u 6= 0. Thus, for large initial data ω0 thevorticity balance takes the form z ′ ∼ z2 giving rise to apossible “Blow-up”!


Euler Equations: ν = 0

Theorem (Lechtenstein (1925))(Existence in short time) Let u0 ∈ C 1,α then there exists T0 > 0and u ∈ C(0,T0; C 1,α) such that u uniquely solves the 3D Euler.

Theorem (Ebin-Marsden, Kato-Lai, Temam)Let u0 ∈ Hs , for s > 5/2, then there exists T0 > 0 andu ∈ C(0,T0; Hs) such that u uniquely solves the 3D Euler.

Theorem (Wiedemann (2011))A family (not unique) of global weak solutions exist for Euler 3D.

(Beale-Kato-Majda condition): If∫ T

0 ‖ω(t)‖∞ dt <∞ then wehave existence and uniqueness on the interval [0,T ]. Comparewith LPS condition!


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