6. Degenerate nonlocal Equation of Porous Medium Type (Joint work
We consider the extension problem.
∇(ya∇v∗) =0 iny>0 lim
y→0yavy∗(x,y,t) =(vm1)t(x,0,t) x∈Ω v∗(x,0,t) =0 onRn\Ω
(13)
fora=1−2σ. Since the diffusion coefficientsD(v) =|v|1−m1 goes to infinity asv→0, we need to control the oscillation ofvfrom below. Hence, we consider the new function w∗derived fromv∗such thatw∗(x,y,t) =M−v∗(x,y,t+t0)with
M=M(t0) =supt≥t0>0v∗. After showing the bound ofL∞-norm ofv, we know that the solution satisfies
v∗(·,t)≤M(t0)<∞ (t≥t0).
From this, we get to a familiar situation:
∇(ya∇w∗) =0 iny>0
− lim
y→0+ya∇yw∗(x,y) =h
(M−w∗)m1i
t(x,0) x∈Ω
w∗(x,0,t) =M onRn\Ω.
(14)
Theorem
(From Ln+2σn−2σ to L∞) Let v(x,t)be a function in L∞(0,T;Ln2n−2σ(Ω))∩L2(0,T; ˙Hσ0(Rn)), then
sup
x∈Ω
|v(x,T)| ≤C∗ kv0k
Ln2n−2σ(Ω)
T2mn−(n−mn2σ)(1+m) for some constant C∗>0.
For the second theorem, we need better control ofv.
Theorem (H ¨older regularity of fractional FDE)
For x0= (x01,· · ·,x0n), we define Qr(x0,t0) = [x0i −r,x0i +r]n×[t0−r2σ,t0], for t0>r2σ>0. Assume now that[x0i −r,x0i+r]n⊂Ωand v(x,t)is bounded in
Rn×[t0−r2σ,t0], then there exist constantsγandβin(0,1)that can be determined a priori only in terms of the data, such that v is Cβin Qγr(x0,t0).
Properties
1 (Scaling Invariance)
u0(x0,t0) =Lm2σ−1T−m1−1u x0
L,t0 T
2 (L1-contraction) Z
Ω
hu(x,t)−˜u(x,t)i
+dx≤ Z
Ω
hu(x, τ)−˜u(x, τ)i
+dx (15)
As a consequence,
ku(t)−u(t)˜ k1≤ ku0−˜u0k1. (16)
3 Comparison Principle
4 Mass conservation in Cauchy problem.R
Rnu0(x)dx=R
Rnu(x,t)dxfort>0.
5 (Extinction in Finite Time in the bounded domain with zero data) There exists T∗>0 such thatu(·,t) =0 for allt≥T∗, i.e.,
lim
t→T∗
ku(·,t)k∞=0
6 (No waiting time) Even foru0(x)having a compact support,u(x,t)>0 fort>0.
Estimates on Finite Extinction Time
Lemma
Whenn+2n−2σσ <m<1, there exists a positive constant C such that the solution v=um of (M.P)satisfies
T∗−t≤C Z
Rn
vm+1m (x,t)dx
!1+m1−m .
Lemma
Whenn+2n−2σσ <m<1, the solution v satisfies Z
Rn
vm+1m (x,t)dx≤
1− t T∗
1+m1−mZ
Rn
vm+1m (x,0)dx. (17)
Proof of Theorem for L
∞-norm
Multiplying the equation by the functionvk = (v−Ck)+and integrating in space,Rn, we have
1 m
Z
Rn
d dt
"Z vk
0
(ξ+Ck)m1−1ξdξ
# dx+
Z
Rn
vk[(−4)σvk]dx≤0 (18) since
(Ck +vk)m1vk = (Ck+ξ)m1ξ
ξ=vk ξ=0 =
Z vk
0
d dξ
h(Ck+ξ)m1ξi dξ.
After some computation, we haveUk ≤2C(m+1) R
Ωvk1+mm (s)dx 1+m2m
for Uk =sup
t≥Tk
Z
Ω
v
1 m+1
k dx+ (m+1)
Z ∞
Tk
kvkk2
H˙σdt. By applying time average and H ¨older regularity,
Uk ≤C0 21+m4m
1+(1+m)σmn
− 2 1+m
!k
t1−
1−m 1+m 0 N1+m4m
1+(1+m)σmn
−2
U
2m 1+m
1+(1+m)σmn
k−1
Proof of Theorem for C
α-norm
1 For the technical reason, we consider the equation for thew=maximum−v.
2 Wr consider local Energy estimate for theextension problemsince the values on the local region has large influence from the whole domain. Technically the multiple of the cut-off functionηandwis hard to applied the nonlocal operator.
And the extension problem is easy to handle since it is a local equation. But the cut-off functionis only depends onx, notz. Still it can be played as an cut-off function in(x,z)variable for a little bit larger level set. And we need to useintrinsic scale, for example the scaled parabolic cylinder is
Qk(θ0) =BRk×
−θ−0αRk2σ,0
3 If localL2-norm is uniformly small forusuch that|u|<1 in a parabolic cylinder,
|u|<1/2 in a half cylinder as the control forL∞-norm.
4 For general case, there is a timet∗, when the solution is close to the supremum with nontrivial measure. Through DiBenedetto’s trick, we can show the solution is close to the supremum with nontrivial measure fort∗<t<0.
5 Use De Giorgi’s isoperimetric inequality to show that the upper level set is
uniformly small after uniform step. Then the localL∞-estimate says the supremum decreases with a uniform amount.
Application to the asymptotic behavior for the parabolic flows
Theorem
Under the above assumptions on u0and m, we have the following property near the extinction time of a solution u(x,t): for any sequence{u(x,tn)}, we have a
subsequence tnk →T∗and aϕ(x)such that lim
k→∞(T∗−tnk)−1/(1−m)
u(x,tnk)−U(x,tnk;T∗) →0
uniformly in compact subset ofΩfor U(x,t;T∗) = (T∗−t)1/(1−m)ϕ1/m(x)whereϕis a eigen-function of fully nonlinear equation
(−4)σϕ= 1
1−mϕm1 inΩ
ϕ=0 onRn\Ω
ϕ >0 inΩ.
7.Geometry of the Ground State (Joint work with Sunghoon Kim)
(−4)12ϕ=λϕp inΩ
ϕ >0 inΩ
ϕ=0 onRn\Ω.
(19) The main question we address is motivated by the following conjecture:
Conjecture
Letϕσbe the ground state eigenfunction for the symmetric stable processes of index 0< σ <1killed upon leaving the interval I= (−1,1). Thenϕσis concave on I.
Rodrigo Ba ˜nuelos, Tadeusz Kulczycki, Pedro J. M ´endez-Hern ´andez.On the shape of the ground state eigenfunction for stable processes.Potential Anal. 24(2006), no. 3
The solutionuof
(−4)σu+ut =0 (x,t)∈Rn×[0,∞) u(x,t)>0 x∈Rn
u(x,t) =0 x∈Rn\Ω u(x,0) =g(x) x∈Rn
is the trace,u∗(x,0,t), ofu∗satisfying the followingextension problem:
4xu∗+u∗,zz=0 inRn×R+×[0,∞) u∗z(x,0,t) =u∗t(x,0,t) x∈Ω u∗(x,0,t) =0 x∈Rn\Ω u∗(x,0,0) =g(x) x∈Rn.
This relation betweenuandu∗is very useful to discuss the geometric properties.
Theorem
LetΩbe a convex bounded domain, and let u0≥0be a continuous and bounded initial function. If(u0)−n+12 is strictly convex, then the solution u∗is power-convex in the space variable x for all t,z>0, i.e., Dx2(u∗)−n+12 ≥0.
Corollary
Under the same condition, the stationary profileϕ(x)of u(x,t)is power-convex, i.e., D2x(ϕ(x))−n+12 ≥0.
Lemma
IfΩis smooth and strictly convex, thenϕ(x)is strictly power-convex: there exists a constant c1>0such that
Dx2(ϕ(x))−n+12 ≥c1I. The constant c1depends onϕandΩ.
Parabolic Approximation and Nontriviality of the Limit
Lemma (Approximation Lemma) For every u0∈H˙
1 2
0(Ω), we have
|eλ1tu(x,t)−a1ϕ(x)| ≤Ce−(λ2−λ1)t (20) and
keλ1tu(x,t)−a1ϕ(x)k
Ckx(Ω)≤CKe−(λ2−λ1)t (21) for k=1,2,· · ·.
Put
eλ1tu(x,t)−a1ϕ(x) =e−(λ2−λ1)tη(x,t).
Then,η(x,t)satisfies the equationηt+ (−4)12η=λ2η. The regularity onηgives the conclusion.
Global Regularity and Interior Regularity
Lemma
If u∗is a solution of the problem, then the function u∗is bounded and ku∗kL∞(Rn×R+)≤sup
x∈Rn
u0(x). ku∗tkL∞
(Rn×R+)≤Cku0kH1/2(Rn) ku∗zkL∞(Ω)≤C at z=0
Lemma
If u∗is a solution of the problem , and if|u∗| ≤1on the cylinder B1+×[0,1], Then
|u∗xi| ≤C, |ux∗ixi| ≤C
|u∗zz| ≤C (i=1,· · ·,n)on the cylinder B1+/2×h1
2,1i
for some constant C>0.
(i) (Bernstein Computation) Find a sub-solution to apply maximum principle:
X=A u∗2+η2(u∗xi)2 where A is a large number andηis a function satisfying
η(x,z,t) = ez
2t
2 + t
2 z+t2 4
!
−1
!
(1− |x|2−z2). Observe(ux∗
i)2≤XonB1+/2×h
1 2,1i
. (ii) Apply the same method for
X =A u∗x
i
2+η2(u∗x
i,xi)2
Boundary Regularity
Lemma (Boundary Regularity foru)
LetΩbe a convex subset ofRnwith n>1and u is the solution. Then,
u2(x,t)−c0dist(x, ∂Ω)
≤C0(dist(x, ∂Ω))1+γ, t>0 near the boundary∂Ωfor some uniform constant C0and0< γ <1.
First we have
c1(dist(x, ∂Ω))12 ≤u(x,t)≤c2(dist(x, ∂Ω))12, near the boundary∂Ω. (22) ustays away from the upper bound or lower bound at an interior point with a fixed amountδ. Harnack estimate in the interior gives a ball whereustays away from the upper or lower bound. Such improvement in a ball with uniform measure will make improvement all the way to the boundary by comparing super- or sub-solution having the improvement. Then we have improvement.
c3
dist x, ∂Ω1
4
12
≤u1
4(x,t)≤c4
dist x, ∂Ω1
4
12
, 3t1
4 ≤t≤t1
for some constantsc3<c4such that
Hence, we get, forv=u2, u−n+12
νν= v−n+11
νν=− 1
n+1
vvνν−n+2
n+1
vν2 v2n+3n+1
≈ (n+2)vν2 (n+1)2v2n+3n+1
→ ∞ asx∈Ω→x0∈∂Ω. (23)
Lemma (Estimate foru∗near the infinity)
Assume that u(x,t)is a bounded and integrable function with a compact support. Let ˆu(x,z,t)be the convolution of u and Poisson kernel p of the Harmonic extension:
u∗(x,z,t) = Z
Rn
p(x, ξ,z)u(ξ,t)dξ where
p(x, ξ,z) = cnz (|x−ξ|2+z2)n+12
, z>0, x, ξ∈Rn, cn= Γn+1
2
πn+12 . Then, there exist some constant c1and R1such thatu is power-convex in the spaceˆ variable x for all t>0on{(x,z) :|x|>c1R1,0<z<R1}, i.e.,
Dx2h
(u∗)−n+12 i
≥0 ∀|x|>c1R1,0<z<R1. (24)
Approximation
For any 0< δ < δ0, let us consider the solution˜uδof
4x˜uδ+ ˜uδ,zz=0 (x,t)∈B1 δ
×R+, 0<z< 1 δ u˜δ,z(x,0,t) =˜uδ,t(x,0,t) (x,t)∈Ω×R+
˜uδ(x,0,t) =δξ(x,t) x∈B1 δ\Ω
˜
uδ(x,z,t) =δ (x,z)∈∂B1 δ
×[0,1/δ] u˜δ(x,1/δ,t) =δζ(x) (x,t)∈B1
δ ×R+ u˜δ(x,z,0) =u(x,z) x∈B1
δ,0≤z≤1 δ
(25)
Let’s define the positive functionsζ(x)∈C0∞(B1
δ)satisfying (ζ)−n+12 : bounded and strictly convex andξ(x,t)∈C∞({B1
δ
\Ω} ×R+)such that
ξ(x,t) =1 onx∈∂Ω,
ξ(x,t) =δ onx∈∂B1 δ,
, δ χ δξχ
!−n+12
.
Lemma
LetΩbe a convex bounded domain and letu˜0≥0be a continuous and bounded initial function. If(u0)−n+12 is strictly convex, then the solutionu˜δis power-convex in the space variable x for all t>0, i.e., Dx2h
(˜uδ)−n+12 i
≥0.
Letw= (˜uδ)−n+12 . Thenwsatisfies 0=w24(x,z)w−(n+3)
2 w|∇(x,z)w|2 inB1
δ ×[0,1/δ]×R+ (26)
and
wz(x,0,t)−wt(x,0,t) =0 x∈B1
δ (27)
To estimate the minimum of the second derivatives, we look at the quantity wαα(x0,z0,t) +εψ(t) = inf
(x,z)∈B1 δ
×[0,1δ],s∈[0,t]
inf
eβ∈Rnx,|eβ|=1
wββ(x,z,s) +εψ(s)
(1≤β≤n), which is taken along a directioneα∈Rnx,(|eα|=1), in which the minimum of the second directional derivative is achieved.
Case 1.x0∈∂B1
δ andz0>0. Nondegneracy of the gradient on the boundary implies a contradiction.
Case 2.We consider the case :z0=0 andx0∈∂Ω. Boundary estimate for nonlocal equations implieswααi∞and a contradiction.
Case 3.We next deal with the casex0∈Ωandz0=0.
Zw(x,z,t) = inf
eβ∈Rnwββ(x,z,t) +εψ(t)
0≥(Zw)t =wαα,t(x1,0,t1) +εψt(t1) = (Zw)z+εψt(t1), and 0≤(Zw)z =wαα,z(x1,0,t1).
Hence, by choosing the functionψ(t)withψt(t)>0,we get a contradiction.
Case 4.Finally, let us consider the case thatx0∈B1
δ and 0<z0<1/δ. Z =w,˜αβ˜η˜αηβ˜+εψ(t), e˜α, eβ˜∈Rn+1
(x,z)
with
ηβ˜(x,z) =δα˜β+cαyβ˜+1
2cαcγ˜yγ˜yβ˜.
Selectcαandψ(t)so thatZhas a contradiction at the maximum point.
Finally selectψ(t)which satisfies the condition in the cases to have a contradiction.