ONE-DIMENSIONAL PARABOLIC p LAPLACIAN EQUATION. Youngsang Ko. 1. Introduction. We consider the Cauchy problem of the form (1.1) u t = ( u x p 2 u x

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1 Kangweon-Kyungki Math. Jour. 7 (999), No. 2, pp ONE-DIMENSIONAL PARABOLIC p LAPLACIAN EQUATION Youngsang Ko Abstract. In this paper we establish some bounds for solutions of parabolic one dimensional p-laplacian equation.. Introduction We consider the Cauchy problem of the form (.) u t = ( u p 2 u ) in S = R [0, ) where p > 2. Equations like (.) were studied by many authors and arise in different physical situations, for the detail see [7]. An important quantity of the study of equation (.) is the local velocity of propagation V (, t), whose epression in terms of u can be obtained by writing the equation as a conservation law in the form In this way we get u t + (uv ) = 0. V = v v p 2, where the nonlinear potential v(, t) is (.2) v = p p 2 u p 2 p. and by direct computation v satisfies (.3) v t = (p 2)v v p 2 v + v p. Received December 3, Mathematics Subject Classification: 35J. Key words and phrases: p-laplacian, free boundary.

2 40 Youngsang Ko In [7], it was shown that V satisfies V 2(p )t, which can also be written as (.4) (v v p 2 ) 2(p )t. Without loss of generality we may consider the case where u 0 vanishes on R and is a continuous positive function, at least, on an interval (0, a) with a > 0. Let P [u] = {(, t) S : u(, t) > 0} be the positivity set of a solution u. Then P [u] is bounded to the left in (, t)-plane by the left interface curve = ζ(t)[7], where ζ(t) = inf{ R : u(, t) > 0}. Moreover there is a time t [0, ), called the waiting time, such that ζ(t) = 0 for 0 t t and ζ(t) < 0 for t > t. It is shown [7] that t is finite(possibly zero) and ζ(t) is a nonincreasing C function on (t, ). For the interface of the porous medium equation { u t = (u m ) in R n [0, ), u(, 0) = u 0 on R n much more is known. D. G. Aronson and J. L. Vazquez [2] and independently K. Höllig and H. O. Kreiss [8] showed the interfaces are smooth after the waiting time. S. Angenent [] showed that the interfaces are real analytic after the waiting time. In dimensions n > 2, L. A. Caffarelli and N. J. Wolanski [4] showed under some nondegeneracy conditions on the initial data, the interface can be described by a C, function when t > T, for some T > 0. Very recently, P. Daskalopoulos and R. Hamilton [6] showed the interface is smooth when 0 < t < T, for some T > 0. On the other hand much less is known for the parabolic p-laplacian equation. For dimensions n > 2, H. Choe and J. Kim [5] showed, under some nondegeneracy conditions on the initial data, the interface is Lipschitz continuous and one of the authors [9] improved this result, showing that, under the same hypotheses, the interface is a C, surface after some time. In [2], Aronson and Vazquez established C regularity of the interfaces by establishing the bounds for v (k) for k 2, where v = m m um

3 One-dimensional parabolic p Laplacian equation 4 represents the pressure of the gas flow through a porous medium, while u represents the density. In this paper we establish bounds for v k, k = 2, 3, near the interface after the waiting time, where v is the solution of (.3). 2. The Upper and Lower Bounds for v Let q = ( 0, t 0 ) be a point on the left interface, so that 0 = ζ(t 0 ), v(, t 0 ) = 0 for all ζ(t 0 ), and v(, t 0 ) > 0 for all sufficiently small > ζ(t 0 ). We assume the left interface is moving at q. Thus t 0 > t. We shall use the notation R δ,η = R δ,η (t 0 ) = {(, t) R 2 : ζ(t) < ζ(t) + δ, t 0 η t t 0 + η}. Proposition 2.. Let q be the point as above. Then there eist positive constants C, δ and η depending only on p, q and u such that v C in R δ,η/2. Proof. From (.4) we have, v. But from 2(p ) 2 v p 2 t Lemma 4.4 in [7] v is bounded away and above from zero near q where u(, t) > 0. Proposition 2.2. Let q = ( 0, t 0 ) be as before. Then there eist positive constants C 2, δ and η depending only on p, q and u such that v C 2 in R δ,η/2. Proof. From Theorem 2 and Lemma 4.4 in [7] we have (2.) ζ (t 0 ) = v v p 2 = v p and (2.2) v t = v p = a on the moving part of the interface { = ζ(t), t > t }. Choose ɛ > 0 such that (2.3) (p )a 5pɛ 4[(p 2) 2 + (p ) 2 ](a + ɛ)ɛ. Then by Theorem 2 in [7], there eists a δ = δ(ɛ) > 0 and η = η(ɛ) (0, t 0 t ) such that R δ,η P [u], (2.4) (a ɛ) p < v < (a + ɛ) p

4 42 Youngsang Ko and (2.5) vv (a ɛ) 2 p ɛ in R δ,η. Then from (2.4) we have (2.6) (a ɛ) p ( ζ) < v(, t) < (a + ɛ) p ( ζ) in R δ,η and (2.7) (a + ɛ) < ζ (t) < (a ɛ) in [t, t 2 ] where t = t 0 η and t 2 = t 0 + η. We set (2.8) ζ (t) = ζ(t ) b(t t ) where b = a + 2ɛ. Then clearly ζ(t) > ζ (t) in (t, t 2 ]. On P [u], w v satisfies L(w) = w t (p 2)v v p 2 w (3p 4) v p 2 v w [(p 2) 2 + 2(p ) 2 ] v p 2 w 2 = 0. 3(p 2) 2 v v p 4 v ww (p 2) 2 (p 3)v v p 4 w 3 We shall construct a barrier for w in R δ,η of the form φ(, t) where and will be decided later. By a direct computation we have L(φ) = where ζ(t) + ζ (t), ( ζ) 2 {ζ (p 2)v v p 2 2 ζ + (3p 4) v p 2 v } + ( ζ ) 2 {ζ (p 2)v v p 2 2 ζ + (3p 4) v p 2 v } [(p 2) 2 + 2(p ) 2 ] v p 2 φ 2 + Ḡ Ḡ = 3(p 2) 2 vv v p 4 φφ (p 2) 2 (p 3)v v p 4 φ 3 = (p 2) 2 v v p 4 ( φ 3v [ ( ζ) + 2 ( ζ ) ] (p 3)[ 2 ζ + ) ζ ]2.

5 One-dimensional parabolic p Laplacian equation 43 If we choose and satisfying v p 3 ma(, ), then Ḡ 0 in R δ,η. Now set Ā = and B =. Then we have ( ζ) 2 ( ζ ) 2 { L(φ) Ā ζ + v p 2 2 { (p 2)v ζ + (3p 4)v } 2[(p 2) 2 + 2(p ) 2 ]} + B { ζ + v p 2 2 { (p 2)v ζ + (3p 4)v } 2[(p 2) 2 + 2(p ) 2 ]} } {(p Ā )a (5p 7)ɛ 2[(p 2) 2 + 2(p ) 2 ](a + ɛ) p 2 p + B } {(p )a (5p 6)ɛ 2[(p 2) 2 + 2(p ) 2 ](a + ɛ) p 2 p. Set and 0 < (2.9) = (p )a (5p 7)ɛ 2[(p 2) 2 + 2(p ) 2 ](a + ɛ) p 2 p (p )a (5p 6ɛ) 2[(p 2) 2 + 2(p ) 2 ](a + ɛ) p 2 p. = 0 Then from (2.3) > 0 and L(φ) 0 in R δ,η for all (0, 0 ] and. Let us now compare w and φ on the parabolic boundary of R δ,η. In view of (2.5) and (2.6) we have and in particular v ɛ(a ɛ) p ζ in R δ,η ɛ(a ɛ) p v (ζ(t) + δ, t) in [t, t 2 ]. δ By the mean value theorem and (2.7) we have for some τ (t, t 2 ) Now set ζ(t) + δ ζ (t) = δ + (a + 2ɛ)(t t ) + ζ (τ)(t t ) δ + 3ɛ(t t ) δ + 6ɛη. η min{η(ɛ), δ(ɛ)/6ɛ}.

6 44 Youngsang Ko Since ɛ satisfies (2.3) and is given by (2.9) it follows that φ(ζ + δ, t) 2δ (p )a (5p 6ɛ) 4[(p 2) 2 + 2(p ) 2 ](a + ɛ) p 2 p δ Moreover from (3.5) and (2.9) φ(, t ) ζ(t ) (a + ɛ) p δ p ɛ(a ɛ) > ζ(t ) ɛ v on [t, t 2 ] > v (, t ) on (ζ(t ), ζ(t ) + δ]. Let Γ = {(, t) R 2 : = ζ(t), t t t 2 }. Clearly Γ is a compact subset of R 2. Fi (0, 0 ). For each point s Γ there is an open ball B s centered at s such that (vv )(, t) (a ɛ) p in B s P [u]. In view of (2.6) we have φ(, t) ζ v (, t) in B s P [u]. Since Γ can be covered by a finite number of these balls it follows that there is a γ = γ() (0, δ) such that φ(, t) w(, t) in R δ,η. Thus for every (0, 0 ), φ is a barrier for w in R δ,η. By the comparison principle for parabolic equations [0] we conclude that v (, t) ζ(t) + ζ (t) in R δ,η, where is given by (2.9) and (0, 0 ) is arbitrary. Now let 0 to obtain v (, t) ζ 2 in R. ɛη 3. Bounds for ( ) 3 v In this section we find the estimates of the derivatives of the form ( ) 3 v (3) v.

7 One-dimensional parabolic p Laplacian equation 45 By a direct computation we have, (3.) L 3 (v (3) ) = v (3) where A = (p 2)v p t (p 2)vv p 2 v (3) (A + B)v (3) Cv (3) D(v (3) ) 2 v 3 (p 2) 2 (p 3)(p 4)vv p 5 v 4 = 0 Ev p 3 B = (3p 4)v p + (p 2) 2 vv p 3 v, + 3(p 2) 2 vv p 3 v, C = v v p 2 {(3p 4)(p ) + 2[(p 2) 2 + 2(p ) 2 ] + 6(p 2) 2 (p 3)vv 2 v + 3(p 2) 2 }, D = 3(p 2) 2 vv p 3, E = [(p 2) 2 + 2(p ) 2 ](p 2) + (p 2) 2 (p 3). Suppose that q = ( 0, t 0 ) is a point on the left interface for which (2.) holds. Fi ɛ (0, a) and take δ 0 = δ 0 (ɛ) > 0 and η 0 = η(ɛ) (0, t 0 t ) such that R 0 R δ0,η 0 (t 0 ) P [u] and (2.5) holds. Thus we also have (2.6) and (2.7) in R 0. Then by rescaling and interior estimate we have Proposition 3.. There are constants K R +, δ (0, δ 0 ), and η (0, η 0 ) depending only on p,q and C 2 such that v (3) K (, t) in R δ,η. ζ(t) Proof. Set δ = min{ 2δ 0 3, 2sη 0}, η = η 0 δ 4s, and define R(, t) {(, t) R 2 : < λ2, t λ4s } < t t for (, t) R δ,η, where s = a + ɛ and λ = ζ(t). Then (, t) R δ,η implies that R(, t) R 0. Since δ 0 3δ, λ < δ and ζ is nonincreasing, 2 we have and t 0 η 0 = t 0 η λ 4s < t < t 0 + η < t 0 + η 0 λ 2 = + ζ(t) 2 = + ζ(t) 2 > ζ(t 0 + η 0 )

8 46 Youngsang Ko ζ(t 0 η) + δ + λ 2 < ζ(t 0 η 0 ). Also observe that for each (, t) R δ,η, R(, t) lies to the right of the line = ζ(t)+s(t t). Net set = λξ + and t = λτ +t. The function W (ξ, τ) v (λξ +, λτ + t) = v (, t) satisfies the equation { W τ = (p 2) v } λ vp 2 W ξ + (3p 4)v p W (3.2) in the region +[2(p 2) 2 vv p 3 +λ[(p 2) 2 (p 3)vv p 4 B v (p 2)v p ]W ξ ξ (v ) 3 (p 2)v p 2 (v ) 2 ] {(ξ, τ) R 2 : ξ 2, 4s < τ 0 }, and W C 2 in B. In view of (2.6) and (2.7) and (a ɛ) p ζ(t) λ v(, t) λ (a + ɛ) p ζ(t) λ ζ(t) ζ(t) ζ(t) + s(t t) ζ(t) + λ 4. Therefore λ 4 = λ 2 ζ(t) λ 4 ζ(t) + λ 2 ζ(t) = 3λ 2 which implies (a ɛ) p v 3(a + ɛ) p. 4 λ 2 Hence by (2.4) equation (3.2) is uniformly parabolic in B. Moreover, it follows from Proposition 2.2 that W satisfies all of the hypotheses of Theorem 5.3. of [0]. Thus we conclude that there eists a constant K = K(a, p, C 2 ) > 0 such that W (0, 0) ξ K; that is, v (3) (, t) K λ. Since (, t) R δ,η is arbitrary, this proves the proposition.

9 One-dimensional parabolic p Laplacian equation 47 We now turn to the barrier construction. If γ (0, δ) we will use the notation R γ δ,η = Rγ δ,η (t 0) {(, t) R 2 : ζ(t)+γ ζ(t)+δ, t 0 η t t 0 +η}. Proposition 3.2. Let R δ,η be the region constructed in the proof of Proposition 2.2 with (3.3) 0 < δ < For (, t) R γ δ,η, let (3.4) φ γ (, t) (p )a p 2(p 2) 2 K. ζ(t) γ/3 + ζ (t) where ζ is given by (2.8), and and are positive constant less than K/2. Then there eist δ (0, δ ) and η (0, η ) depending only on a, p and C 2 such that L 3 (φ γ ) 0 in R γ δ,η for all γ (0, δ). Proof. Choose ɛ such that (3.5) 0 < ɛ < (p )a 3p 23. There eist δ 2 (0, δ ) and η (0, η ) such that (2.4), (2.6) and (2.7) hold in R δ2,η. Fi γ (0, δ 2 ). For (, t) R γ δ 2,η, we have } 2(p L 3 (φ 3 ) = {ζ 2)vvp 2 ( ζ γ/3) 2 ζ γ/3 + A + B } 2(p + {ζ 2)vvp 2 + A + B ( ζ ) 2 ζ Cφ 3 D(φ 3 ) 2 Ev p 3 v 3 (p 2) 2 (p 3)(p 4)vv p 5 where A, B, C, D and E are as before. From (2.6), together with the fact that ζ ζ γ/3 we have v ζ v ζ γ/3 (a + ɛ) p (a + ɛ) p ζ ζ γ/3 γ γ γ/3 = 3 (a + ɛ) p. 2 v 4

10 48 Youngsang Ko From (3.3), we have (3.6) D, D < DK (p )a (p )ɛ < DK Then since C is bounded and from (2.4) and (2.6), we have L 3 (φ 3 ) { (p )a (7p )ɛ C Y 2D E Y 2 } Y 2 { + (p )a (7p 0)ɛ C ( ζ ) ( ζ ) 2 2D E ( ζ ) 2 } { (p )a 3p 23 ɛ δ Y 2 2 ( C E Y } ) { (p )a ( ζ ) 2 2 3p 2 2 ɛ δ 2 ( C E } ζ ) where Y = ζ γ/3 and E = E v p 3 v. 3 Since ɛ satisfies (3.5) we can choose δ = δ 2 (ɛ, p, a, C 2 ) > 0 so small that L 3 (φ 3 ) 0 in R γ δ,η. Remark 3.. From (3.6) the Proposition 3.2 will be true for any, (0, K). Proposition 3.3. (Barrier Transformation). Let δ and η be as in Proposition 3.2 with the additional restriction that (3.7) η < δ 6ɛ, where ɛ is as in Proposition 3.2. constant (3.8) v (3) (, t) Then v (3) also satisfies Suppose that for some nonnegative ζ(t) + ζ (t) (3.9) v (3) (, t) 2/3 ζ(t) + + 2/3 ζ (t) in R δ,η. in R δ,η. Proof. By Remark 3., for any γ (0, δ) since + 2/3 K the function 2/3 φ 3 (, t) = ζ γ/ /3 ζ

11 One-dimensional parabolic p Laplacian equation 49 satisfies L 3 (φ 3 ) 0 in R γ δ,η. On the other hand, on the parabolic boundary of R γ δ,η we have φ 3 v (3). In fact, for t = t and ζ + γ ζ + δ, with ζ = ζ(t ), we have φ 3 (, t ) = 2 ζ γ/ /3 > 4/3 + > v (3) (, t ) ζ ζ ζ while for = ζ + δ and t t t 2 we get, in view of (3.7), φ 3 (ζ + δ, t) 2/3 δ γ/3 + ζ + δ ζ + 2/3 δ + 6ɛη 2/3 δ + δ ζ + δ ζ + /3 v (3) (ζ + δ, t). δ Finally, for = ζ + γ, t t t 2 we have φ 3 (ζ + δ, t) = 2/3 γ γ/ /3 ζ + γ ζ γ + ζ + γ ζ v(3) (ζ + γ, t). By the comparison principle we get φ 3 v (3) in R γ δ.η for any γ (0, δ), and (3.9) follows by letting γ 0. Proposition 3.4. Let q = ( 0, t 0 ) be a point on the interface for which (2.) holds. Then there eist constants C 3, δ and η depending only on p, q and u such that ( ) 3 v C 3 in R δ,η/2. Proof. By Proposition 3. we have, by letting = 0, v (3) (, t) ζ 2 in R δ,η/2. ɛη Even though the equation (3.) is not linear for v (3), a lower bound can be obtained in a similar way. References [] S. Angenent, Analyticity of the interface of the porous media equation after the waiting time, Proc. Amer. Math. Soc. 02 (988), [2] D. G. Aronson and J. L. Vazquez, Eventual C -regularity and concavity for flows in one-dimensional porous media, Arch. Rational Mech. Anal. 99 (987),no.4,

12 50 Youngsang Ko [3] L. A. Caffarelli and A. Friedman, Regularity of the free boundary of a gas flow in an n-dimensional porous medium, Ind. Univ. Math. J. 29 (980), [4] L. A. caffarelli and N. J. Wolanski, C, regularity of the free boundary for the N-dimensional porous medium equation, Comm. Pure Appl. Math. 43 (990), [5] H. Choe and J. Kim, Regularity for the interfaces of evolutionary p-laplacian functions, Siam J. Math. Anal. 26 (4) (995), [6] P. Daskalopoulos and R. Hamilton, Regularity of the free boundary for the porous medium equation, J. A.M.S. (4) (998), [7] J. R. Esteban and J. L. Vazquez, Homogeneous diffusion in R with power-like nonlinear diffusivity, Arch. Rational Mech. Anal. 03(988) [8] K. Höllig and H. O. Kreiss, C -Regularity for the porous medium equation, Math. Z. 92 (986), [9] Y. Ko, C, regularity of interfaces for solutopns of the parabolic p-laplacian equation, Comm. in P. D. E., to appear. [0] O. A. Ladyzhenskaya, N.A. Solonnikov and N.N. Uraltzeva, Linear and quasilinear equations of parabolic type, Trans. Math. Monographs, 23, Amer. Math. Soc., Providence, R. I., 968. Department of Mathematics Kyonggi University Suwon, Kyonggi-do, Korea ysgo@@kuic.kyonggi.ac.kr