Regularity of Weak Solution to Parabolic Fractional p-laplacian

Regularity of Weak Solution to Parabolic Fractional p-laplacian Lan Tang at BCAM Seminar July 18th, 2012

Table of contents 1 1. Introduction 1.1. Background 1.2. Some Classical Results for Local Case 2 2.1. Linear Case 2.2. Fractional p-laplacian 3 4 Open Problems

1.1. Background 1.1. Background 1.2. Some Classical Results for Local Case Let 1 < p < and D is a bounded domain in R n. We consider the variational problem min{ u p dx : u W 1,p 0 (D)}. D Then the minimizer (socalled p-laplacian) satisfies the equation div( u p 2 u) = 0 in D. And we also consider the corresponding parabolic p-laplacian on [a, b] D: u t div( u p 2 u) = 0, for (t, x) [a, b] D.

1.1. Background 1.2. Some Classical Results for Local Case 1.2. Some Classical Results for Local Case For the p Laplacian (p > 1): div( u p 2 u) = 0. Ural ceva (68) proved that for p > 2, the weak solutions of p-laplace Equation have Hölder continuous derivatives and Evans (82) gave another proof for this result. Lewis (83) and DiBenedetto (83) exteded this result to the case 1 0. Remark: In general, the weak solutions to p-laplacian do not have better regularity than C 1,α.

1.1. Background 1.2. Some Classical Results for Local Case For the parabolic p-laplacian: u t ( u p 2 u) = 0 in [a, b] Ω. DiBenedetto (86) proved that any weak solution is locally Hölder continuous for p > 2. For the singular case 1 < p < 2, Chen- DiBenedetto (92) proved that any essentially bounded weak solution is locally Hölder continuous.

2.1. Linear Case 2.2. Fractional p-laplacian In this section we consider the variational problem in non local form (p > 1 and 0 < s < 1): u(x) u(y) p min{ D D x y n+ps dxdy : u W s,p (D) and u = g on R n \D} Then the minimizer u is a fractional p-laplacian and it satisfies the integral equation in weak sense: D u(x) u(y) p 2 (u(x) u(y)) x y n+ps dy = 0

2.1. Linear Case 2.1. Linear Case 2.2. Fractional p-laplacian For the simplest case p = 2, 0 < s < 1, the equation would be reduced to u(x) u(y) dy = 0, u Hs R n x y n+2s It is equivalent to ( ) s u(x) = 0. By Caffarelli and Silvestre (2007) (An Extension Problem Related to the Fractional Laplacian, CPDE, 32 (8): 1245-1260), u can be extended to u : R n+1 R such that u (x, 0) = u(x) and where a = 1 2s. div( y a u ) = 0 in R n+1,

2.1. Linear Case 2.2. Fractional p-laplacian In Fabes, Kenig and Serapioni (1982) ( The local regularity of solutions of degenerate elliptic equations. CPDE 7(1): 77-116. ), they study the degenerate elliptic equation div(a(x) v) = 0 where cw(x) ξ 2 ξ T A(x)ξ Cw(x) ξ 2 for any ξ R n and w(x) > 0 is an A 2 -weight function i. e. there exists a uniform constant C > 0 such that sup B 1 B B w(x)dx 1 w(x) 1 dx C. B B They prove that v is Hölder continuous. Obviously, y a A 2, then u C α and so is u.

2.1. Linear Case 2.2. Fractional p-laplacian For the case of measurable coefficients, i.e. the integral kernel K(x, y) satisfies : λ x y (n+2s) K(x, y) Λ x y (n+2s) for two uniform positive constants 0 < λ Λ < and for any x y, K(x, y) = K(y, x). The equation is R n K(x, y)(u(x) u(y))dy = 0 Silvestre (2006) and Kassmann (2009) proved that any bounded weak solution is Hölder continuous. (1. Luis Silvestre. Hölder estimates for solutions of integro differential equations like the fractional Laplacian. Indiana U. Math. J. 55 (2006), 1155-1174. 2. Moritz Kassman. A priori estimates for integro-differential operators with measurable kernels. Calc. Var. PDEs, 34(1):1-21.)

2.2. Fractional p-laplacian 2.1. Linear Case 2.2. Fractional p-laplacian The equation is R n K(x, y) u(x) u(y) p 2 (u(x) u(y))dy = 0 where the integral kernel K(x, y) satisfies : λ x y (n+ps) K(x, y) Λ x y (n+ps) for two uniform positive constants 0 < λ Λ < and for any x y, K(x, y) = K(y, x).

2.1. Linear Case 2.2. Fractional p-laplacian Theorem A. Any bounded weak solution is locally Hölder continuous for p > 2 and ps 2, 0 < ps µ > 0 (2) where µ are two positive constants. Then u 1 γ in B 1/2 for some γ > 0. Remark: Proof of Lemma 2.1. is similar to that of Lemma 3.3 and we will give the details there.

Now we discuss parabolic fractional p-laplacian. Let us assume that n 2, 0 < s < 1 and p > 1. The measurable function K : R R n R n R satisfies that K(t, x, y) = K(t, y, x) for any x y and the following : there are two positive constants λ and Λ such that λ x y n+ps K(t, x, y) Λ x y n+ps, x, y Rn. Usually, the parabolic fractional p-laplacian is like the following form: u t (x, t)+ K(t, x, y) u(x) u(y) p 2 (u(x) u(y))dy = 0 ( ) R n

In the following we will introduce the weak solution of ( ): Definition We say w C([a, b]; L 2 (R n )) L p ([a, b]; W s,p (R n )) to be a weak solution to the parabolic fractional p-laplacian if and only if η C0 (R n ) and t [a, b], the following holds: K(t, x, y) w(x) w(y) p 2 (w(x) w(y))η(x)dxdy R n R n = w t (x)η(x)dx. R n

For the special case: 0 < s < 1 and p = 2, then the equation becomes u t (x, t) + K(t, x, y)(u(x) u(y))dy = 0, ( ) R n λ where x y n+2s K(t, x, y) Λ and x y n+2s K(t, x, y) = K(t, y, x) for x y. L. Caffarelli, C. Chan and A. Vasseur recently proved that any weak solution to ( ) with initial value in L 2 (R n ) is uniformly bounded and Hölder continuous. ( see Regularity theory for parabolic nonlinear integral operators, J. Amer. Math. Soc. 24(3), 849-869, 2011)

We assume that n 2, 0 < ps 2. The measurable function K : R n R n R is symmetric in x, y for any x y and satisfies the following estimate: there are two positive constants λ and Λ such that λ x y n+ps K(t, x, y) Λ, x, y Rn x y n+ps Usually, the parabolic fractional p-laplacian is like the following form: u t (x, t) + K(x, y) u(x) u(y) p 2 (u(x) u(y))dy = 0. R n

Theorem B. Any weak solution to the parabolic fractional p-laplacian above with the initial data in L p (R n ) is locally Hölder continuous. For the proof, there are two main steps to go: (1) if the initial data for w is bounded in L p (R n ), then it is essentially bounded. (2) we need to give some De Giorgi oscillation lemma for w. Firstly, we have Lemma 3.1 Any weak solution w C([a, b]; L 2 (R n )) L p ([a, b]; W s,p (R n )) to the parabolic fractional p-laplacian is uniformly bounded on [t 0, b] R n for any a < t 0 < b. if the initial data for w is bounded in L p (R n ).

For simplicity, we may assume that a = 2 and b = 0. Then w C([ 2, 0]; L 2 (R n )) L p ([ 2, 0]; W s,p (R n )) and Lemma 3.1. can be reduced to the following Lemma: Lemma 3.2 There is a constant ɛ 0 (0, 1) depending only on n, p, s, λ and Λ such that any weak solution w : [ 2, 0] R n R of parabolic fractional p-laplacian, the following is true: If w( 2, x) p dx ɛ 0, R n then w(t, x) 1/2 on [ 1, 0] R n.

Outline of Proof: Let T k = ( 1 2 k ), λ k = 1/2 2 k 1 ) and w k = (w λ k ) +. We define U k = sup (x, t)dx + 0 t T k 0 T k R n w 2 k R n R n K(x, y) w k (t, x) w k (t, y) p dtdxdy. Then by Sobolev imbedding theorem and interpolation inequality, we can get U k C k U 1+α k 1 where C > 1 and 0 < α < 1 are two uniform constants.

Therefore there exists some positive constant δ 0, such that if U 1 δ 0, then lim k 0 U k = 0. And U 1 2 p 2 R n w(x, 2) p dx Thus if we let ɛ 0 = δ 0 /2 p 2 and w(x, 2) p dx ɛ 0, then R n w(t, x) 1/2 on [ 1, 0] R n.

Now we will give the De Giorgi oscillation lemma: Lemma 3.3. Let w be a weak solution and w 1 on [ 2, 0] R n. If the following condition holds: {w < 0} ([ 2, 1] B 1 ) µ > 0, where µ is a constant. Then there is a small number θ > 0 such that w 1 θ in [ 1, 0] B 1.

Proof. We define the function m : [ 2, 0] R as m(t) = t 2 c 0 {x : w(t, x) < 0} B 1 e C 1(t s) ds where c 0 and C 1 are two positive constants which will be determined later. In fact, c 0 is chosen to be very small, and C 1 is very large, then m(t) can be very small: m(t) 2ω nc 0 C 1, t [ 2, 0] where ω n is the volume of the unit ball in R n.

For t [ 1, 0], m(t) c 0 e 2C 1 {w < 0} ([ 2, 1] B 1 ) c 0 e 2C 1 µ. And we let η : R n R be defined as follows: η C0 (Rn ) and 1 for x 1 η(x) = 0 for x 2 non increasing in x for all x R n We will prove the following claim: w(t, x) 1 m(t)η(x) on [ 2, 0] R n.

This claim implies Lemma 3.3: In fact, if the claim holds true, then And for t [ 1, 0], w(t, x) 1 m(t) on [ 1, 0] B 1. m(t) c 0 e 2C 1 {w < 0} ([ 2, 1] B 1 ) c 0 e 2C 1 µ, thus w(t, x) 1 c 0 e 2C 1 µ on [ 1, 0] B 1.

Next, we will prove this claim by contradiction. If the cliam was not true, then there is some point (t, x) [ 2, 0] R n, such that w(t, x) + m(t)η(x) > 1 Now let (t 0, x 0 ) be the point which realizes the maximum of w(t, x) + m(t)η(x) on [ 2, 0] R n. Then w t (t 0, x 0 ) m (t 0 )η(x 0 ). In the following, we let w(x) = w(t 0, x). If w(y) 0, then obviously w(x 0 ) + m(t 0 )η(x 0 ) w(y) m(t 0 )η(y) 1 m(t 0 ) 1 2ω nc 0 C 1.

Thus R n K(x 0, y) w(x 0 ) w(y) p 2 (w(x 0 ) w(y))dy m(t 0 ) p 1 R n K(x 0, y) η(x 0 ) η(y) p 2 (η(x 0 ) η(y))dy + C (1 2ω nc 0 C 1 ) p 1 {y B 1 : w(t 0, y) 0} Since ps < p 1, then K(x 0, y) η(x 0 ) η(y) p 2 (η(x 0 ) η(y))dy C 2 R n where C 2 is the constant which depends only on n, p, s, λ, Λ.

Hence, we have m (t 0 )+C 2 m(t 0 ) p 1 C (1 2ω nc 0 C 1 ) p 1 {y B 1 : w(t 0, y) 0}. which implies that c 0 {y B 1 : w(t 0, y) 0} C 1 m(t 0 ) + C 2 m(t 0 ) p 1 C (1 2ω nc 0 C 1 ) p 1 {y B 1 : w(t 0, y) 0}

If we choose the constants c 0 and C 1 in the following way: 4ω n c 0 < C 1 c 0 e 2C 1 < 1 4µ 2c 0 < C(1 2ω nc 0 C 1 2C 2 C 1 ) p 1 Then m(t 0 ) = 0 and w(t 0, x 0 ) > 1 which contradicts the assumption of Lemma 3.1: w(t, x) 1 on [ 2, 0] R n.

From Lemma 3.3, we will have the following: Corallary 3.4. Let w be a weak solution to the parabolic fractional p-laplacian and w 1 on [ 1, 0] R n. Then max w 1 θ Q 1/4 where θ is the same constant as that in Lemmma 3.1. and Q r = [ r ps, 0] B r, r > 0.

Now we prove that any weak solution to (1) is Hölder continuous. Firstly, we have the following lemma: Lemma 3.5. Let w : [ 2, 0] R n R be a weak solution to the parabolic fractional p-laplacian and without loss of generality, we assume that w 1 on [ 1, 0] B 1 (0). Let ω = sup Q 1 Then for any natural number n, sup w inf w (1 K n K θ)n ω n w inf Q 1 w. where K n = ( ( 1 4 )ps, 0] B rn (0), r n = ( 1 4 )n (1 θ) n(p 2) ps from Lemma 3.2. and θ is

Proof. We prove it by induction. Obviously, K 1 = ( (1/4) ps, 0] B r1 (0) ( (1/4) ps, 0] B 1/4 (0) Hence by Lemma 3.3, we have sup w inf w (1 θ)ω. K 1 K 1 So the statement is true for n = 1. Now we assume that the statement is true for n = m where m is some positive integer. We define w as follows: w(t, x) = w((1/4)mps t, r m x) (1 θ) m. By our assumption, we have that sup w inf w ω. Q 1 Q 1

And by the argument for n = 1, we get sup K 1 Let us go back to w, then we have w inf w (1 θ)ω. K 1 sup w inf w (1 θ)m+1 ω. K m+1 K m+1 Hence the statement is also true for n = m + 1. Therefore for any natural number n, sup w inf w (1 K n K θ)n ω. n

From Lemma 3.5, we also can get the following proposition: Proposition 3.6. Let w : [ 2, 0] R n R be a weak solution and without loss of generality, we assume that w 1 on [ 1, 0] B 1 (0). Then there exist constants γ > 1 and α > 0 such that for any 0 < ρ < 1, we have sup w inf w γωρ α. Q ρ Q ρ

Open Problems Open Problems 1. Any higher regularity for the weak solution? 2. Can we remove the conditions ps 2?