NAVIER-STOKES IN 2D - GLOBAL SOLUTIONS EXIST
I will just outline the argument, by often referring to the “easy results”. Start with
the problem
~ut − ∆~u + P[~u · ∇~u] = 0
(1)
~u(0, x) = f~ ∈ H s (R2 ), s > 0
First of all, there is a local solution at least up to time T0 = T0 (kf kH s ). Taking dot
product with ~u yields
1
∂t k~uk2L2 − h∆~u, ~ui + hP[~u · ∇~u], ~ui = 0
2
But since div(u) = 0, P~u = ~u and hence
hP[~u · ∇~u], ~ui = h~u · ∇~u, P~ui = h~u · ∇~u, ~ui =
Z
n Z
n
X
X
1
i
j j
=
u ∂i u u dx = −
∂i ui (uj )2 dx = 0
2
i,j=1
i,j=1
P
i
where in the last equality, we have used i ∂i u = div(~u) = 0. It follows that
1
∂t k~uk2L2 + k∇uk2L2 = 0
2
or (up to the existence time 0 < t < T0 )
Z t
2
k∇u(τ )k2L2 dτ = kf k2L2 .
(3)
k~u(t)kL2 + 2
(2)
0
In order to show globality of the solution, we need to show that some supercritical
norm (like H m , m > 0) CANNOT blow up in any finite time. Let us show that for
m = 1 and we would be done.
Differentiate the equation ∂k . We get
∂~k ut − ∆∂~k u + P[~u · ∇∂~k u] + P[∂~k u · ∇~u] = 0.
Now the vector ∂~k u is still divergence free, so we can take dot product with it and
use P∂~k u = ∂~k u. We get
1
∂t k∂~k uk2L2 + k∇∂~k uk2L2 + h~u · ∇∂~k u, ∂~k ui + h∂~k u · ∇~u, ∂~k ui = 0
(4)
2
As before
Z
n Z
n
X
X
1
i
j
j
~
~
h~u · ∇∂k u, ∂k ui =
u (∂i ∂k u )∂k u dx = −
div(~u)
(∂k uj )2 dx = 0.
2
i,j=1
j=1
whereas, we estimate the second term by Hölder’s and then Sobolev embedding
|h∂~k u · ∇~u, ∂~k ui| ≤ Ck∇~uk3L3 (R2 ) ≤ Ck~uk3Ḣ 4/3 (R2 )
1
2
NAVIER-STOKES IN 2D - GLOBAL SOLUTIONS EXIST
Use now Gagliardo-Nirenberg’s to further bound
2/3
1/3
k~ukḢ 4/3 ≤ Ck~ukḢ 1 k~ukḢ 2
It follows by Cauchy-Schwartz
1
|h∂~k u · ∇~u, ∂~k ui| ≤ Ck~uk2Ḣ 1 k~ukḢ 2 ≤ k~uk2Ḣ 2 + Ck~uk4Ḣ 1 .
2
Going back to (4), we get (after summation in k = 1, 2)
1
~ 2 2 + k~uk2 2 ≤ k~uk2 2 + Ck~uk4 1 .
∂t k∇uk
L
Ḣ
Ḣ
Ḣ
2
Thus
2
2
2
~
~
~
∂t k∇u(t)k
L2 ≤ Ck∇u(t)kL2 k∇u(t)kL2 .
2
~
Denoting I(t) = k∇u(t)k
2 , this is an inequality in the form
L
2
~
I 0 (t) ≤ CI(t)k∇u(t)k
L2 .
2
~
By Gronwall’s (i.e. dtd ln(I(t)) ≤ k∇u(t)k
L2 ) for every 0 < τ < t < T0 ,
Z t
Z t
2
2
2
~
~
k∇u(z)k
k∇u(z)k
ln(I(t)) − ln(I(τ )) ≤
2 dz ≤
2 dz ≤ kf k 2 .
L
τ
L
L
0
where in the last step, I have used the inequality (3) for the integral term. It follows
that
2
I(t) ≤ I(τ )ekf kL2 .
Now, since τ is a fixed number (and close to 0), I(τ ) < ∞. This implies that the
quantity I(t) CAN NEVER BLOW UP up to T0 . This means that the solution
persists globally (i.e. T0 = ∞.)