A Class of Functional Inequalities and their Applications to Fourth-Order Nonlinear Parabolic Equations
aa r X i v : . [ m a t h . A P ] O c t A CLASS OF FUNCTIONAL INEQUALITIES AND THEIR APPLICATIONSTO FOURTH-ORDER NONLINEAR PARABOLIC EQUATIONS
JIAN-GUO LIU AND XIANGSHENG XU
Abstract.
We study a class of fourth order nonlinear parabolic equations which include the thin-film equation and the quantum drift-diffusion model as special cases. We investigate these equationsby first developing functional inequalities of the type Z Ω u γ − α − β ∆ u α ∆ u β dx ≥ c Z Ω | ∆ u γ | dx, which seem to be of interest on their own right. Introduction
Let
T > R N with boundary ∂ Ω. We consider the existence of a solutionto the problem ∂ t u + div[ u n ∇ ( u α − ∆ u α )] = 0 in Ω T , (1.1) ∇ u · ν = u n ∇ ( u α − ∆ u α ) · ν = 0 on Σ T ,(1.2) u ( x,
0) = u ( x ) ≥ , (1.3)where Ω T = Ω × (0 , T ], Σ T = ∂ Ω × (0 , T ], ν is the unit outward normal to ∂ Ω, n, α ∈ (0 , ∞ ), and u = u ( x ) are given data whose precise assumptions will be made later.Fourth-order nonlinear parabolic equations arise in a variety of physical settings ([6], [8], [12],[20]). Two well-known examples are the thin film equation and the quantum drift-diffusion model,both of which are special cases of (1.1). In a typical thin film equation, we have that α = 1 , n > n = 1 , α = give us the quantum drift-diffusion equation . We haveseen extensive research work done on these two types of problems. We refer the reader to ([24],[22], [10], [26] [15]) and the references therein.The objective of our work is to present a unified mathematical approach to these two verydifferent physical problems. This is done via functional inequalities of the type(1.4) I ( u ) ≡ Z Ω u γ − α − β ∆ u α ∆ u β dx ≥ c Z Ω (∆ u γ ) dx for all u ∈ W γ , where(1.5) W γ = { u ≥ u γ ∈ W , (Ω) , ∇ u γ · ν = 0 on ∂ Ω } . Obviously, the validity of the above inequality depends on Ω , α, β and γ . We will focus on the casewhere Ω is bounded and convex. Then a result of [12] asserts that(1.6) Z Ω (∆ u γ ) dx ≥ Z Ω |∇ u γ | dx Mathematics Subject Classification.
Key words and phrases.
Existence, Nonlinear fourth order parabolic equations, Thin-film equation, Quantumdrift-diffusion model, Functional inequalities.Liu’s address: Department of Physics and Department of Mathematics, Duke University, Durham, NC 27708,
Email: [email protected]’s address: Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762.
Email : [email protected]. for all u ∈ W γ . Thus a slightly weaker version is the inequality(1.7) I ( u ) ≡ Z Ω u γ − α − β ∆ u α ∆ u β dx ≥ c Z Ω (cid:0) ∇ u γ (cid:1) dx for all u ∈ W γ . Several known inequalities are special cases of this. If β = 1 , α = γ = , then (1.7) is establishedfor box domains with sides parallel to the coordinate planes in [4] (also see [16]). It turns out[10, 22] that (1.7) is still valid if β = 1 , γ = α ∈ ( ( N − N +1 , ), and Ω is a bounded convex domain.The inequalities in [10, 22] are formulated in a measure-theoretic setting. See [27] for a more directapproach.The significance of this type of functional inequalities lies in the fact that the integrand on theleft-hand side of (1.4) can change signs. In essence, they are the nonlinear version of the G˚ardinginequality. To illustrate how they arise naturally in the study of fourth order nonlinear partialdifferential equations, we proceed to make some formal analysis of (1.1)-(1.3). That is, we assumethat u is a positive, smooth solution of (1.1). Use u β , where β >
0, as a test function in (1.1) toderive 1 β + 1 ddt Z Ω u β +1 dx + βn + β Z Ω u α − ∆ u α ∆ u n + β dx = 0 . (1.8)By (1.4), we have(1.9) Z Ω u α − ∆ u α ∆ u n + β dx ≥ c Z Ω (cid:16) ∆ u α + n + β − (cid:17) dx. For the moment, we ignore the restrictions under which the above inequality holds. We will addressthis issue in Section 2. Integrate (1.8) to obtain(1.10) max ≤ t ≤ T Z Ω u β +1 ( x, t ) dx + Z Ω T (cid:16) ∆ u α + n + β − (cid:17) dxds ≤ c. Our study of (1.4) is inspired by the integration by parts rule proved by Gianazza et al. [10] andby J¨ungel and Mattes [16]. We also refer the reader to [17] for the development of an algebraictechnique for dealing with such formulas. The framework we have developed here is also algebraicin nature, but it seems to be more direct and easier to use. This can best be illustrated by theapplication of our method to the standard thin film(1.11) ∂ t u + div ( u n ∇ ∆ u ) = 0 . In this case, the second integral in (1.8) becomes Z Ω ∆ u ∆ u n + β dx. This immediately puts us in a position to apply Lemma 2.5 in Section 2, from whence follows thatfor each β ∈ ( − n, − n ) there is a positive number c such that Z Ω ∆ u ∆ u n + β dx ≥ c Z Ω (cid:16) ∆ u n + β +12 (cid:17) dx. Of course, this result is well-known, see, e.g., [17] and the references therein. Also notice how easyit is for us to prove Lemma 2.5 in our framework. Most importantly, our method has led to thediscovery of Corollary 2.2 in Section 2. It is this corollary that enables us to solve a problem leftopen in [22].We can easily foresee other potential applications for the functional inequalities developed in thispaper. An immediate example is the study of epitaxial growth of thin films ( see ([1], [9]) and thereferences therein). A family of continuum models has been established, one of which has the form(1.12) ∂ t u + u ∆ u = 0 in Ω T . OURTH ORDER PARABOLIC EQUATIONS 3
Using u β as a test function yields(1.13) 1 β + 1 ddt Z Ω u β +1 dx + Z Ω ∆ u ∆ u β +2 dx = 0 , and Lemma 2.5 in Section 2 becomes applicable. Of course, the resulting inequality is far fromenough to obtain an existence assertion for (1.12). However, the idea behind the derivation of theinequality can lead to the discovery of additional estimates. Since our inequalities do not depend onthe space dimension N , their applications will inevitably lead to the relaxation of the restrictionson N in previous studies [9]. Theorem 1.1.
Let Ω be a bounded convex domain in R N . Assume: (H1) α ∈ [1 , ) , n ∈ [1 , σ ) , where (1.14) σ = if N < , N if N > ,any number in (0 , if N = 4 ; (H2) u ∈ L ∞ (Ω) with inf Ω u > .Then there is a weak solution to (1.1)-(1.3) in the following sense: (C1) u ∈ L α + σ (Ω T ) with u ≥ on Ω T , u α ∈ L ( , T ; W , (Ω)) ; (C2) ∇ u α · ν = 0 a.e. on Σ T ; (C3) for each ξ ∈ C ∞ (Ω T ) with ξ ( x, T ) = 0 and ∇ ξ · ν = 0 on Σ T there holds − Z Ω T u∂ t ξdxdt − Z Ω u ( x ) ξ ( x, dx + Z Ω T (cid:18) nα u n + α − ∇ u α ∆ u α ∇ ξ + u α + n − ∆ u α ∆ ξ (cid:19) dxdt = 0 . (1.15)We would like to make some remarks about Theorem 1.1. We can conclude from Lemma 2.2below that ∇ u α ∈ (cid:0) L (Ω T ) (cid:1) N . Thus each integral in (1.15) makes sense. Assumption (H1) islargely due to the restrictions for (1.4) to hold. Theorem 1.2.
Let Ω be a bounded convex domain in R N and (H2) hold. Assume: (H3) α = 1 , n ∈ ( , σ ) , where σ is given as in (1.14).Then there is a weak solution to (1.1)-(1.3) in the sense of (C3). In comparison with previous results on the thin-film equation (see the reviews in [24], [5] and[13]), this theorem has removed all the restrictions on the space dimension. Thus this is truly amulti-dimensional result. As we all know, most of the existing results on non-linear fourth-orderparabolic equations involve restrictions on the space dimensions with the one-dimensional problemsattracting the most attention. See ,e.g., ([2], [3], [7], [25] and [6]), where various properties ofsolutions are investigated.Our approach to the question of existence is to construct a sequence of smooth, positive approx-imate solutions such that the calculations similar to (1.8)-(1.10) can be employed. A well-knowndifficulty in the study of fourth-order equations is that the maximum principle is no longer true.In fact, the heat kernel for the heat biharmonic equation changes signs. Thus arguments basedupon the maximum principle for second order equations do not work here. We must rely on thenonlinear structure of our equation to obtain non-negative solutions. It turns out that the term u α − = u − α in (1.1) plays a key role in the existence of non-negative solutions. The case where n = 1 , α ≤ α > JIAN-GUO LIU AND XIANGSHENG XU the gradient flow theory fails [22]. The key to our success seems to be that we have found a rightway to approximate the term u − α with the exponent being negative.The optimal transport theory has been successfully employed to treat many different types ofparabolic equations as gradient flows of various entropy functionals for various transportation met-rics, the canonical example being the regular scalar heat equation viewed by Jordan, Kinderlehrerand Otto [14] as the gradient flow of the Boltzmann entropy for the quadratic Monge-KantorovichMK2 (frequently named Wasserstein) metric. We have seen a very large body of work done onthis subject in the last 20 years ( in the study of the heat equation in a very general framework,porous-medium equations, thin-film flow equations, chemotaxis models, etc.. See ([10], [22], [18])and the references therein as examples.). However, in the generality considered in Theorems 1.1and 1.2, the transport theory is no longer applicable [22]. We discretize the time derivative in (1.1)and transform it into a system of two second order elliptic equations. Our approximation schemeseems to be standard. However, the genius is in the details, and we have to overcome numeroustechnical difficulties for it to work here. On the one hand, we need to introduce new terms in ourapproximate problems in order to ensure high regularity and positivity of our approximate solu-tions. On the other hand, we have to make sure that these new terms do not destroy the essential aprior estimates that hold for positive, smooth solutions of the original equations. Striking a suitablebalance between the two constitutes the core of our development.This paper is organized as follows. In section 2 we develop a class of functional inequalities.Section 3 is devoted to the fabrication of our approximation schemes. Here the key is how tohandle the term u α − . Then we proceed to obtain discretized versions of the a priori estimatesthat hold for positive, smooth solutions of the original equations, which eventually leads to theestablishment of Theorems 1.1 and 1.2 in the two subsequent sections.2. Functional Inequalities
In this section we study the functional inequality (1.4). We will focus on the case where Ω is abounded convex domain in R N . Our method is algebraic in nature. In this regard, it is similar to[17].The key to our development is the following lemma, which is a substantial improvement overLemma 2.1 in [28]. Lemma 2.1.
Let Ω be a bounded domain in R N with Lipschitz boundary ∂ Ω . Assume that (2.1) α = 0 . Then we have Z Ω u α − β |∇ u β | dx ≥ β (2 + N ) α Z Ω |∇ u α | dx + β (2 + N ) α Z Ω (∆ u α ) dx + 16 β ( α − β )( α − β )(2 + N ) α Z Ω |∇ u α | dx (2.2) for all u ∈ W α .Proof. If β = 0, then the lemma is trivially true. Thus assume that β = 0. Note that(2.3) (∆ u β ) ≤ N |∇ u β | . Thus if α = β , then (2.2) is still true. Without loss of generality, we suppose β = α. For convenience, we also assume that u ∈ W α is bounded above and bounded away from 0 below.As a result, u ∈ W s for each s ∈ R . (Otherwise, use a suitable approximation [10]. The sameis understood in the subsequent calculations in this section. In this connection, we would like to OURTH ORDER PARABOLIC EQUATIONS 5 mention that in our applications to partial differential equations our approximate solutions satisfythis condition.) We compute, for i, j = 1 , · · · , N , that ∂ i u β = ∂ i ( u α ) βα = βα u β − α ∂ i u α , (2.4) ∂ ij u β = β ( β − α ) α u β − α ∂ i u α ∂ j u α + βα u β − α ∂ ij u α . (2.5)First, we let i = j in the above equation and then sum up over i to derive(2.6) ∆ u β = β ( β − α ) α u β − α |∇ u α | + βα u β − α ∆ u α . Square both sides of this equation and multiply through the resulting equation by α β u α − β toarrive at(2.7) α β u α − β | ∆ u β | = | ∆ u α | + 2 β − αα u α |∇ u α | ∆ u α + (cid:18) β − αα (cid:19) u α |∇ u α | . Square both sides of (2.5), multiply through the resulting equation by α β u α − β , and then sum up i, j to obtain(2.8) α β u α − β |∇ u β | = |∇ u α | + 2 β − αα u α ∇ u α · ∇ u α ∇ u α + (cid:18) β − αα (cid:19) u α |∇ u α | . Note that ∇ u α = 2 u α ∇ u α . Keeping this in mind, we can rewrite (2.8) and (2.7) as2 ∇ u α · ∇ u α ∇ u α = α β − α ) β u α − β |∇ u β | − α β − α ) |∇ u α | − β − α ) α |∇ u α | , (2.9) |∇ u α | ∆ u α = α β − α ) β u α − β | ∆ u β | − α β − α ) | ∆ u α | − β − α ) α |∇ u α | . (2.10)Note that u − α |∇ u α | = u − α |∇ u α | ∇ u α · ∇ u α = div (cid:0) u − α |∇ u α | ∇ u α u α (cid:1) − div (cid:0) u − α |∇ u α | ∇ u α (cid:1) u α = div (cid:0) u − α |∇ u α | ∇ u α u α (cid:1) − u − α |∇ u α | ∆ u α − u − α ∇ u α ∇ u α · ∇ u α + 2 u − α |∇ u α | . (2.11)Integrating this equation over Ω, we obtain, with the aid of the fact that ∇ u α · ν = 0 on ∂ Ω, that(2.12) 4 Z Ω |∇ u α | dx = 2 Z Ω ∇ u α · ( ∇ u α ∇ u α ) dx + Z Ω |∇ u α | ∆ u α dx. Integrate (2.9) and (2.10) over Ω, add the two resulting equations, then make use of (2.12), therebyderive α β − α ) β Z Ω u α − β |∇ u β | dx + α β − α ) β Z Ω u α − β | ∆ u β | dx = α β − α ) Z Ω |∇ u α | dx + α β − α ) Z Ω | ∆ u α | dx − α − β ) α Z Ω |∇ u α | dx. (2.13) JIAN-GUO LIU AND XIANGSHENG XU
Multiplying through this equation by β − α ) β α , we can conclude the lemma from the inequality(2.3). The proof is complete. (cid:3) Notice that the only inequality we have used in the proof of the above lemma is (2.3). Thus (2.2)is just as sharp an inequality as (2.3). Obviously, the lemma has been obtained by sharpening theproof of Lemma 2.1 in [28].
Lemma 2.2.
Assume that Ω is bounded and convex. Then we have (2.14) Z Ω |∇ u α | dx ≤ Z Ω (∆ u α ) dx for all u ∈ W α .Proof. This lemma is taken from [28]. The proof is rather simple. Thus we repeat it here.Remember that in this case (1.6) holds. Taking note of this, we calculate from (2.12) that4 Z Ω |∇ u α | dx ≤ (cid:18)Z Ω |∇ u α | dx (cid:19) (cid:18)Z Ω |∇ u α | dx (cid:19) + (cid:18)Z Ω |∇ u α | x (cid:19) (cid:18)Z Ω | ∆ u α | dx (cid:19) (2.15) ≤ (cid:18)Z Ω | ∆ u α | dx (cid:19) (cid:18)Z Ω |∇ u α | dx (cid:19) from whence the lemma follows. (cid:3) Now we are ready to study the functional(2.16) I ( u ) = Z Ω u γ − α − β ∆ u α ∆ u β dx. At this point, we only assume(2.17) αβ > , γ = 0 . Recall from (2.6) that ∆ u α = α ( α − γ ) γ u α − γ |∇ u γ | + αγ u α − γ ∆ u γ , (2.18) ∆ u β = β ( β − γ ) γ u β − γ |∇ u γ | + βγ u β − γ ∆ u γ . (2.19)Plugging these two into (2.16) yields γ αβ I ( u ) = Z Ω (∆ u γ ) dx + 16( α − γ )( β − γ ) γ Z Ω |∇ u γ | dx + 4( α + β − γ ) γ Z Ω |∇ u γ | ∆ u γ dx. (2.20)Let us first consider the special case where N = 1. In this case, we have Z Ω ∇ u γ ∇ u γ ∇ u γ dx = Z Ω |∇ u γ | ∆ u γ dx. Thus by (2.12), we obtain Z Ω |∇ u γ | ∆ u γ dx = 43 Z Ω |∇ u γ | dx. OURTH ORDER PARABOLIC EQUATIONS 7
Use this in (2.20) to derive(2.21) γ αβ I ( u ) = Z Ω (∆ u γ ) dx + 16( γ − α + β ) γ + 3 αβ )3 γ Z Ω |∇ u γ | dx. If γ − α + β ) γ + 3 αβ ≥
0, we are done. If γ − α + β ) γ + 3 αβ <
0, i.e.,(2.22) α + β − p α + β − αβ < γ < α + β + p α + β − αβ, then we apply (2.14) to (2.21) to get(2.23) γ αβ I ( u ) ≥ γ − α + β ) γ + 9 αβγ Z Ω (∆ u γ ) dx. For the coefficient of the integral in the preceding inequality to be positive, we must impose theconditions γ > α or γ < β in the case where α ≥ β , or(2.24) γ > β or γ < α in the case where α < β .(2.25)In summary, we have Lemma 2.3. If N = 1 and α ≥ β , then (1.4) holds whenever γ > min { α, α + β + p α + β − αβ, } or (2.26) γ < max { β, α + β − p α + β − αβ } . (2.27)Now we deal with the more general case N >
1. It turns out that the sign of the term 2 γ − α − β plays a significant role. Lemma 2.4.
Let Ω be a bounded convex domain in R N and γ a number satisfying (2.28) 2 γ − α − β > . Without loss of any generality, we assume (2.29) β ≤ α. If either
35 ( α + β ) > γ ≥ α, or (2.30) γ < min { α, β } , (2.31) then there is a positive number c such that (1.4) holds.Proof. Under (2.28)-(2.30), the coefficient of the second integral in (2.20) is non-negative, while thecoefficient of the third integral is negative. Thus we can deduce from (2.20) and (2.14) that γ αβ I ( u ) ≥ Z Ω (∆ u γ ) dx + 4( α + β − γ ) γ Z Ω |∇ u γ | ∆ u γ dx ≥ Z Ω (∆ u γ ) dx + 4( α + β − γ ) γ (cid:18)Z Ω |∇ u γ | dx (cid:19) (cid:18)Z Ω (∆ u γ ) dx (cid:19) ≥ Z Ω (∆ u γ ) dx + 3( α + β − γ ) γ Z Ω (∆ u γ ) dx = 3( α + β ) − γγ Z Ω (∆ u γ ) dx. (2.32) JIAN-GUO LIU AND XIANGSHENG XU
The coefficient of the last integral in the above inequality is positive by (2.30). This completes theproof of the first part of the lemma.If γ < α , then the coefficient of the second integral in (2.20) is negative. Then it follows from(2.20) and (2.14) that γ αβ I ( u ) ≥ Z Ω (∆ u γ ) dx + 9( α − γ )( β − γ ) γ Z Ω (∆ u γ ) dx + 3( α + β − γ ) γ Z Ω (∆ u γ ) dx = 4 γ − α + β ) γ + 9 αβγ Z Ω (∆ u γ ) dx. (2.33)Note that 4 γ − α + β ) γ + 9 αβ = 4( γ − β )( γ − α ). Thus it is positive if (2.31) holds. Theproof is complete. (cid:3) Next we analyze the case where γ = α + β . In this direction, we have the following result. Lemma 2.5.
Let Ω be a bounded convex domain in R N . Then for each α ∈ ( β , β ) there is apositive number c = c ( α, β ) such that (2.34) Z Ω ∆ u α ∆ u β dx ≥ c Z Ω (cid:16) ∆ u α + β (cid:17) dx for all u ∈ W α + β .Proof. . Let γ = α + β in (2.20) to obtain(2.35) ( α + β ) αβ I ( u ) = Z Ω (∆ u α + β ) dx − α − β ) ( α + β ) Z Ω |∇ u α + β | dx In view of (2.14), we have(2.36) ( α + β ) αβ I ( u ) ≥ (cid:18) − α − β ) ( α + β ) (cid:19) Z Ω (∆ u α + β ) dx. If α ∈ ( β , β ), then the coefficient on the right-hand side of the preceding inequality is positive.The proof is complete. (cid:3) For the case where(2.37) 2 γ − α − β < , we deduce from (2.12) that γ αβ I ( u ) = Z Ω (∆ u γ ) dx + 16( αβ − γ ) γ Z Ω |∇ u γ | dx − α + β − γ ) γ Z Ω ∇ u γ ∇ u γ ∇ u γ dx. (2.38)Hence the key is how to handle the term R Ω ∇ u γ ∇ u γ ∇ u γ dx . To this end, we infer from (2.9) that ∇ u γ · ∇ u γ ∇ u γ = γ η − γ ) η u γ − η |∇ u η | − γ η − γ ) |∇ u γ | − η − γ ) γ |∇ u γ | , (2.39) OURTH ORDER PARABOLIC EQUATIONS 9 where η is a number to be determined later. Substituting this into (2.38) , we arrive at γ αβ I ( u ) = Z Ω (∆ u γ ) dx + α + β − γη − γ Z Ω |∇ u γ | dx + 16[( α + β − γ )( η − γ ) + αβ − γ ] γ Z Ω |∇ u γ | dx + ( α + β − γ ) γ ( γ − η ) η Z Ω u γ − η |∇ u η | dx. (2.40)This puts us in position to apply (2.2). To do this, we need to suppose(2.41) γ − η > Z Ω u γ − η |∇ u η | dx ≥ η (2 + N ) γ Z Ω |∇ u γ | dx + η (2 + N ) γ Z Ω (∆ u γ ) dx + 16 η ( γ − η )( γ − η )(2 + N ) γ Z Ω |∇ u γ | dx. (2.42)Use this in (2.40) to derive γ αβ I ( u ) ≥ α + β + N γ − (2 + N ) η (2 + N )( γ − η ) Z Ω (∆ u γ ) dx − ( α + β − γ ) N ( γ − η )( N + 2) Z Ω |∇ u γ | dx + 16[(( N − η − ( N + 1) γ )( α + β − γ ) + (2 + N )( αβ − γ )](2 + N ) γ Z Ω |∇ u γ | dx. (2.43)We choose η so that the coefficient of the last integral in the above equation is 0. This leads to(2.44) η = ( N + 1) γ ( α + β − γ ) − (2 + N )( αβ − γ )( N − α + β − γ ) . The number η chosen above must satisfy (2.41). Plug the value of η into (2.43) and take a note of(1.6) and the fact that the coefficient of the second integral in (2.43) is negative to arrive at(2 + N ) γ αβ I ( u ) ≥ (1 − N )( α + β ) + 3 N γ − (2 + N ) ηγ − η Z Ω (∆ u γ ) dx. Thus our last hypothesis is that the coefficient of the above integral is positive, i.e.,(2.45) (1 − N )( α + β ) + 3 N γ − (2 + N ) η > . To summarize our results, we have
Lemma 2.6.
Let Ω be a bounded convex domain in R N . Assume that (2.17) and (2.37) hold. If η given by (2.44) satisfies (2.41), and (2.45), then there is a positive number c = c ( α, β, γ, N ) suchthat (1.4) holds. Corollary 2.1.
Let Ω be a bounded convex domain in R N . Then for each α ∈ ( ( N − N +1 , ) there isa positive number c = c ( α, N ) such that (2.46) Z Ω u α − ∆ u ∆ u α dx ≥ c Z Ω (∆ u α ) dx for all u ∈ W α .Proof. This corollary is largely contained in [28]. We also refer the reader to (1.38) in [22] for adifferent version of (2.46). It is also an easy consequence of our preceding development. To seethis, note that in this case we have(2.47) β = 1 , γ = α, and 2 γ − α − β = α − . If α = 1, then (2.46) is trivially true. If α >
1, we apply Lemma 2.4. The conditions (2.28), (2.29),and (2.30) are equivalent to 1 < α < . If α <
1, we substitute (2.47) into (2.44) to obtain(2.48) η = − αN − . Obviously, (2.17) is true. Since η <
0, we see that (2.41) is also satisfied. Plugging (2.47) and(2.48) into (2.45), we arrive at(2.49) (2 N + 1) α > ( N − . Thus (2.45) holds under our assumptions on α . We conclude (2.46) from Lemma 2.6. (cid:3) Corollary 2.2.
Let Ω be a bounded convex domain in R N . Then there is an ε ∈ [0 , ) such thatto each α ∈ ( , there corresponds a positive number c = c ( ε, α ) with the property (2.50) Z Ω u ε − ∆ u α ∆ udx ≥ c Z Ω (∆ u α + ε ) dx for all u ∈ W α + ε .Proof. In this case, we have(2.51) β = 1 , γ = α + ε . Thus α + β − γ = 1 − ε . Hence we need to show that there exists an ε ∈ [0 , ) such that γ > η, (2.52) − ( N − α + 1) + 3 N ( α + ε )2 − (2 + N ) η > , (2.53)where η is defined by (2.44). Plugging (2.51) into (2.44), we derive(2.54) η = − N ε + 2( N + 1 + α ) ε + ( N + 2) α − N + 3) α N − − ε ) . Using this value of η in (2.53), after some elementary calculations we arrive at − N − + 4( N + 1)( N + 2) α − ( N + 2) α > N (5 N − ε + [2 N ( N + 2) α − N (2 N − ε ≡ h ( ε ) . (2.55)The right-hand side is a quadratic function in ε , which achieves its minimum value at(2.56) ε = − ( N + 2) α + 2(2 N − N − . But this number is not always non-negative. It becomes negative only when α > N − N +2 . Thus wetake ε = ( α > N − N +2 , − ( N +2) α +2(2 N − N − otherwise. OURTH ORDER PARABOLIC EQUATIONS 11
Obviously, we have ε ∈ [0 , − α ). Next we will show that ε selected above satisfies (2.52)-(2.53). If ε = 0, then(2.57) η = [( N + 2) α − N + 3)] α N − < α <
2. Thus (2.52) is trivially true. Set ε = 0 in (2.55) to obtain(2.58) − N − + 4( N + 1)( N + 2) α − ( N + 2) α > . Solutions to this inequality form the interval N + 1) − √ NN + 2 , N + 1) + 4 √ NN + 2 ! , which contains the interval ( ,
2) if N ≤
4. That is to say, if the space dimension does not exceed4, we can simply take ε = 0. We will have to do a little bit more work if we want (2.50) to hold forall the space dimensions. To this end, we substitute (2.56) into (2.55) to deduce − ( N − α + (3 N − α + 2 − N > . Solutions to this inequality are the interval N − − p N (5 N − N − , N − p N (5 N − N − ! , which contains the interval ( ,
2) if
N >
2. To see (2.52), we substitute (2.51) and (2.54) into(2.52) to obtain ( N − ε + (2 N α + 4) ε + ( N + 2) α − N + 1) α < . Remember that ε lies in the interval (0 , − α ) and the function on the left-hand side of the aboveinequality is an increasing function of ε over the interval. Thus it is sufficient for us to prove H ( α ) ≡ ( N −
2) (4 − α )
25 + (2
N α + 4) 4 − α N + 2) α − N + 1) α < . It is easy to see that H ( α ) is a convex quadratic function of α . An elementary calculation showsthat H ( 12 ) < , H (2) < . Thus H ( α ) < α ∈ ( , (cid:3) From our proof we see that this lemma can hold for more general α .Similarly, we can investigate the functional J ( u ) = Z Ω u γ − α ∆ ln u ∆ u α dx. A simple calculation shows(2.59) ∆ ln u = − γ u − γ |∇ u γ | + 1 γ u − γ ∆ u γ . Plug this and (2.18) into J ( u ) to obtain(2.60) γ α J ( u ) = Z Ω (∆ u γ ) dx + 4( α − γ ) γ Z Ω |∇ u γ | ∆ u γ dx − α − γ ) γ Z Ω |∇ u γ | dx. It is interesting to note that the arguments of Lemmas 2.4 and 2.5 do not work here. If(2.61) α − γ > , we can still mimic the proof of Lemma 2.6 to obtain the following lemma. Lemma 2.7.
Let Ω be a bounded convex domain in R N and (2.61) be satisfied. Set (2.62) η = (2 + N )( γ − α ) γ + ( α − γ ) α ( N − γ − α ) . If η satisfies the inequalities η − γ < and (2.63) (2 + N )( η − γ ) + ( N − α − γ ) < , (2.64) then there is a positive number c = c ( α, γ, N ) such that (2.65) J ( u ) ≥ c Z Ω (∆ u γ ) dx. Finally, we remark that it is possible to extend the inequality (1.4) to other types of domains Ω.For example, if the boundary of Ω is C , β = 1, and α = γ = , a result of [27] asserts that(2.66) Z Ω ∆ u ∆ √ u √ u dx ≥ c (cid:18)Z Ω |∇ √ u | dx + Z Ω u |∇√ u | dx (cid:19) − c Z Ω udx for u ∈ W . Here the complication is largely due to the fact that (1.6) is no longer true in thiscase. In its place, we have(2.67) Z Ω (∆ u ) dx + Z Ω |∇ u | dx ≥ c Z Ω |∇ u | dx. It is also interesting to pursue the case where the Neumann boundary condition is replaced withthe Dirichlet boundary condition.3.
The Approximate Problem
In this section we will show how to construct a sequence of positive, smooth approximate solu-tions. Then we proceed to derive a priori estimates for the sequence that hold under more generalconditions than these in Theorems 1.1 and 1.2. Our approximation scheme is based upon thefollowing lemma.
Lemma 3.1.
Let Ω be a bounded domain in R N with Lipschitz boundary ∂ Ω . Assume that α ≥ , ε ∈ [0 , , n ∈ R , and (3.1) p > max { N , } . Then for each > τ > and each f ∈ L ∞ (Ω) there is a solution ( ρ, F ) with ρ ≥ in the space (cid:0) W , (Ω) ∩ L ∞ (Ω) (cid:1) to the problem − div [( ρ + τ ) n ∇ F ] + τ F = ρ − fτ in Ω , (3.2) − ∆ ρ α + τ ρ p = − ρ − ε ρ α − ε + τ F + τ in Ω , (3.3) ∇ ρ α · ν = ∇ F · ν = 0 on ∂ Ω . (3.4) Furthermore, we have that ρ, F ∈ C ,β (Ω) for some β ∈ (0 , and ρ ≥ c in Ω for some c > ,where β, c depend on the given data. Of course, the equations (3.2)-(3.4) are satisfied in the sense of distributions. The last term τ in(3.3) has been added to ensure that ρ cannot be identically 0. As we shall see, it is also the mainreason why ρ has a positive lower bound. This idea was first employed in [26]. The real tricky part,though, is that we have used the term ρ − ε ρ α − ε + τ to approximate ρ − α . That is, a term with a negativeexponent is being approximated by a term with two positive exponents. It serves two purposes: OURTH ORDER PARABOLIC EQUATIONS 13 one is that we avoid having to seek solutions in a function space whose functions must have positivelower bounds; the other is that it ensures that solutions to (3.3) is non-negative. If our solution isnon-negative then the term τ in (3.3) guarantees that it is bounded away from zero below. If wefurther assume that f is H¨older continuous on Ω, then the classical Schauder theory [11] indicatesthat the pair ( ρ, F ) is a classical solution. This, together with the fact that ρ is bounded awayfrom 0 below, enables us to achieve higher regularity, thereby justifying all our calculations in thederivation of a prior estimates for the sequence of approximate solutions to be constructed later. Proof.
We just need to modify the proof of Lemma 3.1 in [28]. We still apply the Leray-SchauderFixed Point Theorem (see Theorem 11.3 in [11]). For this purpose, we define an operator B from L ∞ (Ω) into L ∞ (Ω) as follows. Given that ρ ∈ L ∞ (Ω), we consider the problem − div (cid:2)(cid:0) ρ + + τ (cid:1) n ∇ F (cid:3) + τ F = ρ − fτ in Ω , (3.5) ∇ F · ν = 0 on ∂ Ω . (3.6)Eqn (3.5) is uniformly elliptic, and thus by (3.1) we can appeal to the results in ([11], Chap. 8)and thereby conclude that this linear boundary value problem has a unique solution F in thespace W , (Ω) ∩ L ∞ (Ω). For each q ≥
2, the function | F | q − F ∈ W , (Ω) and ∇ (cid:0) | F | q − F (cid:1) =( q − | F | q − ∇ F . Upon using it as a test function in (3.5), we arrive at(3.7) k F k q ≤ τ k ρ − f k q . Now we use the function F so-obtained to form the problem − ∆ ψ + τ | ψ | pα − ψ = − ( ρ + ) − ε ( ρ + ) α − ε + τ F + τ in Ω , (3.8) ∇ ψ · ν = 0 on ∂ Ω . (3.9)Obviously, this problem has a unique solution ψ in the space W , (Ω) ∩ L ∞ (Ω). We define B ( ρ ) = θ ( ψ ) , where θ ( s ) = | s | α − s .It is easy to see that B : L ∞ (Ω) → L ∞ (Ω) is well-defined. By Theorem 8.22 in [11] and a boundaryflattening argument [29] , we can conclude that there exists a number β ∈ (0 , F, ψ ∈ C ,β (Ω). It is not difficult to show that the H¨older continuityof ψ implies that B is continuous and maps bounded sets into precompact ones.Next, we show that(3.10) k ρ k ∞ ≤ c for all σ ∈ [0 ,
1] and ρ such that σB ( ρ ) = ρ . Here and in the remaining proof, c is a generic positivenumber which depends only on the given data. Without loss of generality, assume σ >
0. Thenthe equation σB ( ρ ) = ρ is equivalent to the problem ρ − fτ = − div (cid:2)(cid:0) ρ + + τ (cid:1) n ∇ F (cid:3) + τ F in Ω , (3.11) − ∆ θ − ( ρσ ) + τ (cid:12)(cid:12)(cid:12) θ − ( ρσ ) (cid:12)(cid:12)(cid:12) pα − θ − ( ρσ ) = − ( ρ + ) − ε ( ρ + ) α − ε + τ F + τ in Ω , (3.12) ∇ θ − ( ρσ ) · ν = ∇ F · ν = 0 on ∂ Ω . (3.13)Remember that ε <
1, and thus (cid:0) θ − ( ρσ ) (cid:1) − ( ρ + ) − ε = 0 on Ω. Upon using (cid:0) θ − ( ρσ ) (cid:1) − as a testfunction in (3.12), we deduce that ρ ≥ θ − ( ρσ ) = ρ α σ α . We can rewrite (3.12) as(3.14) − σ α ∆ ρ α + τσ p ρ p = − ρ − ε ρ α − ε + τ F + τ in Ω . Integrate this equation to obtain τ Z Ω ρ p dx = − σ p Z Ω F ρ − ε ρ α − ε + τ dx + τ | Ω |≤ τ (cid:16) k F k pp + ε − + c (cid:17) k ρ k − εp + c ≤ c k ρ k − εp + c k ρ k − εp + c. (3.15)The last step is due to the fact that pp + ε − ≤ p . A simple application of the interpolation inequality ab ≤ ηa p + c ( η ) b q , p + 1 q = 1gives k ρ k p ≤ c. In the sequel, we will not acknowledge this interpolation inequality again when it is being used.Applying the proof of Theorem 8.15 in ([11], p.189), we can derive from (3.11) and (3.14) that k F k ∞ ≤ c k F k + c k ρ − fτ k p + c ≤ c, (3.16) k ρ α k ∞ ≤ c k ρ α k + c k ρ − α k p + c ≤ c. (3.17)Note that the constant c here depends only the lower bound τ n of the elliptic coefficient ( ρ + τ ) n in (3.11), not the upper bound. This completes the proof of existence.Next, we show(3.18) 1 ρ ∈ L s (Ω) for each s ≥ ρ + δ ) s , where δ >
0, as a test function in (3.3) to obtain τ Z Ω ρ + δ ) s dx ≤ τ Z Ω ( ρ + δ ) p − s dx + c Z Ω ( ρ + δ ) − η − s dx from whence follows Z Ω ρ + δ ) s dx ≤ c Z Ω ( ρ + δ ) p − s dx + c. If s ≤ p , then we take δ → Z Ω ρ s dx ≤ c Z Ω ρ p − s dx + c. It is not difficult to see that this inequalities actually holds for each s >
1, and thus (3.18) follows.Now we let v = ρ α + δ , δ >
0. Then we can easily show that v satisfies the boundary value problem − ∆ v + 2 v |∇ v | = (cid:18) ρ − ε ρ α − ε + τ F − τ + τ ρ p (cid:19) v ≡ G in Ω , ∇ v · ν = 0 on ∂ Ωin the sense of distributions. We can conclude from ([11], [28]) again that k v k ∞ ≤ c k v k + c k G k p ≤ c. The last step is due to (3.18). This completes the proof of Lemma 3.1. (cid:3) If α <
1, then our approximate problem can be made a little simpler. In this regard, we have:
OURTH ORDER PARABOLIC EQUATIONS 15
Lemma 3.2.
Let Ω be a bounded domain in R N with Lipschitz boundary ∂ Ω . Assume that α ∈ (0 , , n ∈ R , and (3.19) p > max { N , } . Then for each > τ > and each f ∈ L ∞ (Ω) there is a solution ( ρ, F ) with ρ ≥ in the space (cid:0) W , (Ω) ∩ L ∞ (Ω) (cid:1) to the problem − div [( ρ + τ ) n ∇ F ] + τ F = ρ − fτ in Ω , (3.20) − ∆ ρ α + τ ρ p = − F ρ − α + τ in Ω , (3.21) ∇ ρ α · ν = ∇ F · ν = 0 on ∂ Ω . (3.22) Furthermore, we have that ρ, F ∈ C ,β (Ω) for some β ∈ (0 , and ρ ≥ c in Ω for some c > ,where β, c depend on the given data. The proof is similar to that of the previous lemma.We are ready to construct our approximate solutions. Let
T > , T ] into j equal subintervals, j ∈ { , , · · · } . Set τ = Tj .
We discretize and regularize the system (1.1)-(1.3) as follows. For k = 1 , · · · , j , solve recursivelythe systems ρ k − ρ k − τ = − div [( ρ k + τ ) n ∇ F k ] + τ F k in Ω , (3.23) − ∆ ρ αk + τ ρ pk = − ( ρ k ) − ε ( ρ k ) α − ε + τ F k + τ in Ω , (3.24) ∇ ρ αk · ν = ∇ F k · ν = 0 on ∂ Ω , (3.25) ρ ( x ) = u ( x ) . (3.26)Define the functions˜ u j ( x, t ) = ( t − t k − ) ρ k ( x ) − ρ k − ( x ) τ + ρ k − ( x ) , x ∈ Ω , t ∈ ( t k − , t k ] ,u j ( x, t ) = ρ k ( x ) , x ∈ Ω , t ∈ ( t k − , t k ] ,F j ( x, t ) = F k ( x ) , x ∈ Ω , t ∈ ( t k − , t k ] , We can rewrite the system (3.23)-(3.26) as ∂ ˜ u j ∂t = − div (cid:2) ( u j + τ ) n ∇ F j (cid:3) + τ F j in Ω T , (3.27) − ∆ u αj + τ u pj = − ( u j ) − ε ( u j ) α − ε + τ F j + τ in Ω T , (3.28) ∇ u αj · ν = ∇ F j · ν = 0 on Σ T , (3.29) u j ( x,
0) = u ( x ) on Ω . (3.30) Lemma 3.3.
Let ε ∈ [0 , ) be given as in Corollary 2.2. Assume that α ∈ [1 , ) , n ∈ (0 , − ε ) , and p > max { N , } . Then there is a τ ∈ (0 , such that Z Ω t (∆ u αj ) dxds + τ Z Ω t | ∆ u α + ε j | dxds + τ Z Ω t u p + α − j |∇ u j | dxds + τ Z Ω t u p + ε − j |∇ u j | dxds + τ Z Ω t u α − j |∇ u j | dxds + τ Z Ω t u ε − j |∇ u j | dxds + max ≤ t ≤ T Z Ω G ( u j ( x, t )) dx ≤ c (3.31) for all τ ∈ (0 , τ ) , where (3.32) G ( s ) = s if n > , s − n if n < , s ln s − s if n = 1 .Here and in what follows c denotes a positive constant independent of j . By the proof of Corollary 2.2, we can take ε = 0 if N ≤
4. Thus in this case n ∈ (0 , Proof.
For r ∈ [0 , ∞ ) we define(3.33) K ( r ) = Z r s + τ ) n ds = (cid:26) − n (cid:2) ( r + τ ) − n − (1 + τ ) − n (cid:3) if n = 1,ln( s + τ ) − ln(1 + τ ) if n = 1.We use K ( ρ k ) as a test function in (3.23) to obtain(3.34) Z Ω F k ∆ ρ k dx − τ Z Ω F k K ( ρ k ) dx + 1 τ Z Ω ( ρ k − ρ k − ) K ( ρ k ) dx = 0 . We proceed to estimate each integral in the above equation. For this purpose, we solve (3.24) for F k to yield(3.35) F k = ρ α − k ∆ ρ αk + τ ρ ε − k ∆ ρ αk − τ ρ p + α − k − τ ρ p + ε − k + τ ρ α − k + τ ρ ε − k . This can be done because ρ k is bounded away from 0 below. Observe(3.36) 1 τ Z Ω ( ρ k − ρ k − ) K ( ρ k ) dx ≥ τ Z Ω Z ρ k ρ k − K ( r ) drdx. This is due to the fact that K ( r ) is an increasing function on [0 , ∞ ). Substituting (3.35) into thefirst integral in (3.34) gives Z Ω F k ∆ ρ k dx = Z Ω ∆ ρ k ρ α − k ∆ ρ αk dx + τ Z Ω ∆ ρ k ρ ε − k ∆ ρ αk dx +( p + α − τ Z Ω ρ p + α − k |∇ ρ k | dx + ( p + ε − τ Z Ω ρ p + ε − k |∇ ρ k | dx − ( α − τ Z Ω ρ α − k |∇ ρ k | dx − ( ε − τ Z Ω ρ ε − k |∇ ρ k | dx. (3.37)By Corollaries 2.1 and 2.2, we have Z Ω ρ α − k ∆ ρ αk ∆ ρ k ≥ c Z Ω (∆ ρ αk ) dx, (3.38) Z Ω ρ ε − k ∆ ρ αk ∆ ρ k ≥ c Z Ω (∆ ρ α + ε k ) dx. (3.39) OURTH ORDER PARABOLIC EQUATIONS 17 If α >
1, then the coefficient of the sixth integral in (3.37) is negative. To address this issue, wecompute the integral as follows: Z Ω ρ α − k |∇ ρ k | dx = Z Ω | ρ α − k ∇ ρ k | dx = 4 α Z Ω |∇ ρ α k | dx ≤ δτ Z Ω |∇ ρ α k | dx + τ c ( δ ) ≤ δ τ Z Ω | ∆ ρ αk | dx + c ( δ ) , (3.40)where δ is a positive number. Using (3.38)-(3.40) in (3.37) and choosing δ suitably small, we obtain Z Ω F k ∆ ρ k dx ≥ c Z Ω (∆ ρ αk ) dx + cτ Z Ω (∆ ρ α + ε k ) dx +( p + α − τ Z Ω ρ p + α − k |∇ ρ k | dx + ( p + ε − τ Z Ω ρ p + ε − k |∇ ρ k | dx +(1 − ε ) τ Z Ω ρ ε − k |∇ ρ k | dx − c. (3.41)Plugging (3.35) into the second integral in (3.34) yields − τ Z Ω F k K ( ρ k ) dx = − τ Z Ω ( ρ α − k + τ ρ ε − k ) K ( ρ k )∆ ρ αk dx + τ Z Ω ( ρ p + α − k + τ ρ p + ε − k − ρ α − k − τ ρ ε − k ) K ( ρ k ) dx ≡ I ,k + I ,k . (3.42)A simple integration by parts enables us to represent I ,k in the form I ,k = ατ Z Ω (( α − ρ α − εk − (1 − ε ) τ ) K ( ρ k ) ρ α + ε − k |∇ ρ k | dx + ατ Z Ω ρ α − εk + τ ( ρ k + τ ) n ρ α + ε − k |∇ ρ k | dx. (3.43)We first consider the case where(3.44) α > . Set(3.45) b τ ≡ (cid:18) − εα − (cid:19) α − ε τ α − ε . Then we can choose τ ∈ (0 ,
1) so that(3.46) b τ < . From here on, we assume that(3.47) τ ≤ τ . Recall from the definition of K ( r ) that(3.48) K ( r )( r − ≥ , ∞ ) . We can easily deduce that the integrand of the first integral in (3.43) is non-positive only on theset A k = { x ∈ Ω : b τ ≤ ρ k ( x ) ≤ } . On this set, we have − K ( ρ k ) = Z ρ k s + τ ) n ds ≤ Z ρ k s n d ≤ ρ − nk n − if n > , − ln ρ k if n = 1 , − n if n < α, n, ε imply that ρ α − n − εk ≤ A k , and(3.49) τ ρ − ( α − ε ) k ≤ c on A k (3.50)Keeping these in mind, we calculate, for n >
1, that I ,k ≥ ατ Z A k (( α − ρ α − εk − (1 − ε ) τ ) K ( ρ k ) ρ α + ε − k |∇ ρ k | dx ≥ α ( α − τ Z A k K ( ρ k ) ρ α − k |∇ ρ k | dx ≥ − cτ Z A k ρ α − − nk |∇ ρ k | dx ≥ − c Z A k ρ α − − n − εk |∇ ρ k | dx ≥ − c Z A k ρ α − n − εk |∇ ρ α k | dx ≥ − δ Z Ω |∇ ρ α k | dx − c ( δ ) ≥ − δ Z Ω (∆ ρ αk ) dx − c ( δ ) , (3.51)where δ >
0. The above inequality still holds if n ≤
1. Thus if δ is sufficiently small, this term canbe incorporated into the second integral in (3.41).If α = 1, then we can express I ,k in the form(3.52) I ,k = τ Z Ω " − (1 − ε ) τ K ( ρ k ) + ρ k ( ρ − εk + τ )( ρ k + τ ) n ρ ε − k |∇ ρ k | dx. Set B k = { x ∈ Ω : ρ k ( x ) ≥ } . On the set B k , we have K ( ρ k ) ≤ n − if n > ρ k if n = 1, − n ρ − nk if n < ρ ε − n − k ≤ ρ p − k on B k .For n <
1, we estimate I ,k ≥ − (1 − ε ) τ Z Ω K ( ρ k ) ρ ε − k |∇ ρ k | dx ≥ − cτ Z B k ρ ε − − nk |∇ ρ k | dx ≥ − cτ Z Ω ρ p − k |∇ ρ k | dx. (3.53) OURTH ORDER PARABOLIC EQUATIONS 19
In view of the coefficient of the fourth integral in (3.41), we just need to impose a further condition(3.54) cτ < p, where c is the same as the one in the last line of (3.53). Then the fourth term in (3.41) can absorbthe term on the right-hand side of (3.53). The case where n ≥ I ,k in the form(3.55) I ,k = τ Z Ω K ( ρ k ) ρ ε − k ( ρ α − εk + τ )( ρ pk − dx. The integrand in the above integral is always non-negative.Summarizing our preceding estimates, we obtain Z Ω (∆ ρ αk ) dx + τ Z Ω (∆ ρ α k ) dx + τ Z Ω ρ p + α − k |∇ ρ k | dx + τ Z Ω ρ p + ε − k |∇ ρ k | dx + τ Z Ω ρ α − k |∇ ρ k | dx + τ Z Ω ρ ε − k |∇ ρ k | dx + 1 τ Z Ω Z ρ k ρ k − K ( r ) drdx ≤ c (3.56)for τ ∈ (0 , τ ). Multiplying through this inequality by τ and summing up over k , we obtain Z Ω t (∆ u αj ) dxds + τ Z Ω t (∆ u α j ) dxds + τ Z Ω t u p + α − j |∇ u j | dxds + τ Z Ω t u p + ε − j |∇ u j | dxds + τ Z Ω t u α − j |∇ u j | dxds + τ Z Ω t u ε − j |∇ u j | dxds + Z Ω Z u j u K ( r ) drdx ≤ c (3.57)for τ ∈ (0 , τ ). By the definition of K ( r ), we have Z u j u K ( r ) dr = Z u K ( r ) dr + Z u j K ( r ) dr ≥ ( ( u j + τ ) − n (1 − n )(2 − n ) − (1+ τ ) − n u j − n − c if n = 1,( u j + τ ) ln( u j + τ ) − (1 + ln(1 + τ )) u j − c if n = 1.(3.58)Here the fact that the second integral in (3.58) is bounded is due to our assumptions on u . Therest is rather obvious. The proof is complete. (cid:3) Lemma 3.4.
Let the assumptions of Lemma 3.3 hold. Then we have Z Ω t u α − j (∆ u αj ) dxds + τ Z Ω t u ε − j (∆ u αj ) dxds + τ Z Ω t u p +2 α − j |∇ ρ k | dxds + τ Z Ω t u p + ε + α − j |∇ u j | dxds + τ Z Ω t u α + ε − j |∇ u j | dxds + max ≤ t ≤ T Z Ω u α − n ) + j ( x, t ) dx ≤ c. (3.59) Proof.
Here we use a different test function. Let(3.60) L ( r ) = Z r αs α − ( s + τ ) n ds. Then use L ( ρ k ) as a test function in (3.23) to obtain(3.61) − Z Ω ∇ F k · ∇ ρ αk dx − τ Z Ω F k L ( ρ k ) dx + 1 τ Z Ω ( ρ k − ρ k − ) L ( ρ k ) dx = 0 . The first integral in the above equation is equal to Z Ω F k ∆ ρ αk dx = Z Ω ρ α − k (∆ ρ αk ) dx + τ Z Ω ρ ε − k (∆ ρ αk ) dx +( p + α − ατ Z Ω ρ p +2 α − k |∇ ρ k | dx +( p + ε − ατ Z Ω ρ p + ε + α − k |∇ ρ k | dx − ( α − ατ Z Ω ρ α − k |∇ ρ k | dx − ( ε − ατ Z Ω ρ α + ε − k |∇ ρ k | dx. (3.62)Owing to Lemma 2.4, for each α ∈ [1 , ) there is a positive number c with the property(3.63) Z Ω ρ α − k (∆ ρ αk ) dx ≥ c Z Ω (∆ ρ α − k ) dx. If α >
1, then the coefficient of the sixth integral in (3.62) is negative. We will use (3.63) to dealwith the term. To do this, we estimate Z Ω ρ α − k |∇ ρ k | dx = Z Ω ρ α − k ρ α − k |∇ ρ k | dx ≤ δτ Z Ω ρ α − k |∇ ρ k | dx + τ c ( δ ) Z Ω ρ α − k = 4 δ (3 α − τ Z Ω |∇ ρ α − k | dx + τ c ( δ ) Z Ω ρ α − k dx ≤ δ (3 α − τ Z Ω | ∆ ρ α − k | dx + c ( δ ) Z Ω ρ α − k dx ≤ cδτ Z Ω ρ α − k (∆ ρ αk ) dx + c ( δ ) Z Ω ρ α − k dx, (3.64)where δ is a positive number. Using (3.63)-(3.64) in (3.62) and choosing δ suitably small, we obtain Z Ω F k ∆ ρ αk dx ≥ c Z Ω ρ α − k (∆ ρ αk ) dx + cτ Z Ω ρ ε − k (∆ ρ αk ) dx +( p + α − ατ Z Ω ρ p +2 α − k |∇ ρ k | dx +( p + ε − ατ Z Ω ρ p + ε + α − k |∇ ρ k | dx +(1 − ε ) ατ Z Ω ρ α + ε − k |∇ ρ k | dx − c Z Ω ρ α − k . (3.65)Plugging (3.35) into the second integral in (3.61) yields − τ Z Ω F k L ( ρ k ) dx = − τ Z Ω ( ρ α − k + τ ρ ε − k ) L ( ρ k )∆ ρ αk dx + τ Z Ω ( ρ p + α − k + τ ρ p + ε − k − ρ α − k − τ ρ ε − k ) L ( ρ k ) dx ≡ J ,k + J ,k . (3.66) OURTH ORDER PARABOLIC EQUATIONS 21
The term J ,k can be written in the form J ,k = ατ Z Ω (( α − ρ α − εk − (1 − ε ) τ ) L ( ρ k ) ρ α + ε − k |∇ ρ k | dx + α τ Z Ω ρ α − εk + τ ( ρ k + τ ) n ρ α + ε − k |∇ ρ k | dx. (3.67)If α >
1, we can define A k , b τ as before. Note that the integrand of the first integral in (3.67) isnon-positive only on the set A k . For x ∈ A k , we have(3.68) − L ( ρ k ) = Z ρ k αs α − ( s + τ ) n ds ≤ Z ρ k αs α − n − ds ≤ αα − n if α > n , − α ln ρ k if α = n , αα − n ρ α − nk if α < n .If α < n , we have J ,k ≥ ατ Z A k (( α − ρ α − εk − (1 − ε ) τ ) L ( ρ k ) ρ α + ε − k |∇ ρ k | dx ≥ α ( α − τ Z A k L ( ρ k ) ρ α − k |∇ ρ k | dx ≥ − cτ Z A k ρ α − − nk |∇ ρ k | dx ≥ − c Z A k ρ α − − n − εk |∇ ρ k | dx ≥ − c Z A k ρ α − − n − εk |∇ ρ α − k | dx ≥ − δ Z Ω |∇ ρ α − k | dx − c ( δ ) ≥ − δ Z Ω (∆ ρ α − k ) dx − c ( δ ) ≥ − δ Z Ω ρ α − k (∆ ρ αk ) dx − c ( δ ) , (3.69)where δ >
0. Thus J ,k can be absorbed into the second integral in (3.65) if δ is small. If α ≥ n , asimilar argument can be made.If α = 1, then we can express J ,k in the form(3.70) J ,k = τ Z Ω " − (1 − ε ) τ L ( ρ k ) + ρ k ( ρ − εk + τ )( ρ k + τ ) n ρ ε − k |∇ ρ k | dx. Let B k = { x ∈ Ω : ρ k ( x ) ≥ } be given as before. On the set B k , we have L ( ρ k ) ≤ n − if n > ρ k if n = 1, − n ρ − nk if n < ρ ε − n − k ≤ ρ p − k on B k . For n <
1, we estimate J ,k ≥ − (1 − ε ) τ Z Ω L ( ρ k ) ρ ε − k |∇ ρ k | dx ≥ − cτ Z B k ρ ε − − nk |∇ ρ k | dx ≥ − cτ Z Ω ρ p − k |∇ ρ k | dx. (3.71)In view of the coefficient of the fourth integral in (3.65), we just need to impose a further condition(3.72) cτ < p, where c is the same as the one in the last line of (3.71). The case where n ≥ J ,k in the form(3.73) J ,k = τ Z Ω L ( ρ k ) ρ ε − k ( ρ α − εk + τ )( ρ pk − dx. The integrand in the above integral is always non-negative.If n > α = n , we have L ( r ) = Z r αs α − ( s + τ ) n ds = Z r αs α − d ( s + τ ) − n − n = α (1 + τ ) − n n − − αn − r α − ( r + τ ) − n + α ( α − n − Z r s α − ( s + τ ) n − ds ≥ ( αα − n ( r + τ ) α − n + α (1+ τ ) − n ( α − n ) − α ( α − τ ) α − n ( n − α − n ) if r > αα − n r α − n + α (1+ τ ) − n ( α − n ) − α ( α − n − α − n ) if r ≤ n > α = n , we have(3.75) L ( r ) ≥ ( n ln r + τ τ + n ((1+ τ ) − n − n − if r > n ln r + n ((1+ τ ) − n − n − if r ≤ Z Ω Z u j u L ( r ) drdx ≥ c Z Ω u α − n ) + j dx − c, where u j = u j ( x, t ), provided that n >
1. It is not difficult to see the same inequality holds for n ≤
1. Collecting all the previous estimates in (3.61), we arrive at Z Ω (∆ ρ α − k ) dx + τ Z Ω ρ ε − k (∆ ρ αk ) dx + τ Z Ω ρ p +2 α − k |∇ ρ k | dx + τ Z Ω ρ p + ε + α − k |∇ ρ k | dx + τ Z Ω ρ α + ε − k |∇ ρ k | dx + 1 τ Z Ω Z ρ k ρ k − L ( r ) drdx ≤ c Z Ω ρ α − k dx + c. (3.76)Multiply through the inequality by τ , note that 0 ≤ α − <
1, and sum up over k to obtain thedesired result. The proof is complete. (cid:3) OURTH ORDER PARABOLIC EQUATIONS 23
Lemma 3.5.
Let the assumptions of Lemma 3.3 hold. Then the sequence { u αj } is bounded in L (0 , T ; W , (Ω)) .Proof. Note that ∇ u j = 2 α u − α j ∇ u α j . We calculate Z Ω |∇ u j | αNα + N dx = (cid:18) α (cid:19) αNα + N Z Ω u (2 − α ) αN α + N ) j |∇ u α j | αNα + N dx ≤ c (cid:18)Z Ω u − α ) αN α + N ) − αN j (cid:19) − αN α + N ) (cid:18)Z Ω |∇ u α j | dx (cid:19) αN α + N ) ≤ c (cid:18)Z Ω | ∆ u αj | dx (cid:19) αN α + N ) . (3.77)The last step is due to the fact that 2(2 − α ) αN α + N ) − αN ≤ . On account of the Sobolev Embedding Theorem, we have (cid:18)Z Ω u αj dx (cid:19) α ≤ c (cid:18)Z Ω |∇ u j | αNα + N dx (cid:19) α + NαN + c Z Ω u j dx ≤ c (cid:18)Z Ω | ∆ u αj | dx (cid:19) + c. (3.78)Consequently, there holds(3.79) Z T (cid:18)Z Ω u αj dx (cid:19) α dt ≤ c. Recall the interpolation inequality Z Ω |∇ u αj | dx ≤ c Z Ω |∇ u αj | dx + c (cid:18)Z Ω u αj dx (cid:19) . This, together with the fact that α ∈ [1 , ), implies the desired result. (cid:3) Lemma 3.6.
Let the assumptions of Lemma 3.3 hold. Then we have τ Z Ω T u p +2 α − α − n ) + +1) N j dxdt ≤ c. Proof.
By the Sobolev inequality, we estimate, for α > n , that Z Ω T u p +2 α − α − n +1) N j dxdt ≤ c Z T (cid:18)Z Ω u p +2 α −
12 2 NN − j dx (cid:19) N − N (cid:18)Z Ω u α − n +1 j dx (cid:19) N dt ≤ c (cid:18)Z Ω T |∇ u p +2 α − j | dxdt + Z Ω T u p +2 α − j dxdt (cid:19) · sup ≤ t ≤ T Z Ω u α − n +1 j dx ! N ≤ c (cid:18)Z Ω T u p +2 α − j |∇ u j | dxdt + Z Ω T u p +2 α − j dxdt (cid:19) ≤ c Z Ω T u p +2 α − j |∇ u j | dxdt + δ Z Ω T u p +2 α − α − n +1) N j dxdt + c. Choosing δ suitably small yields Z Ω T u p +2 α − α − n +1) N j dxdt ≤ c Z Ω T u p +2 α − j |∇ u j | dxdt + c. If α ≤ n , we have(3.80) Z Ω T u p +2 α − N j dxdt ≤ c Z Ω T u p +2 α − j |∇ u j | dxdt + c. Multiplying through the inequality by τ and taking a note of Lemma 3.3 give the desired result. (cid:3) Proof of Theorem 1.1
The proof is divided into several lemmas.
Lemma 4.1.
Let the assumptions of Lemma 3.3 hold. If n ≥ , then τ F j → strongly in L (Ω T ) .Proof. Recall that(4.1) τ F j = τ u α − j ∆ u αj + τ u ε − j ∆ u αj − τ u p + α − j − τ u p + ε − j + τ u α − j + τ u ε − j . We will show that each term on the right hand side of the above equation tends to 0 strongly in L (Ω T ) as τ →
0. We begin with the last term. For this purpose, assume τ ≤ τ , where τ is givenas in Lemma 3.3. Set I ,j = τ Z Ω T K ( u j ) u ε − j ( u α − εj + τ )( u pj − dxdt. By the proof of Lemma 3.3, we have(4.2) I ,j ≤ c. Let A j = { ( x, t ) ∈ Ω T : u j ( x, t ) ≤ } , B j = Ω T \ A j . OURTH ORDER PARABOLIC EQUATIONS 25
Then we can rewrite (4.2) as τ Z B j K ( u j ) u p + α − j dxdt + τ Z B j K ( u j ) u p + ε − j dxdt − τ Z A j K ( u j ) u α − j dxdt − τ Z A j K ( u j ) u ε − j dxdt ≤ − τ Z A j K ( u j ) u p + α − j dxdt − τ Z A j K ( u j ) u p + ε − j dxdt + τ Z B j K ( u j ) u α − j dxdt + τ Z B j K ( u j ) u ε − j dxdt + c. (4.3)On the set B j , we have(4.4) K ( u j ) ≤ (cid:26) n − if n > u j if n = 1,while on the set A j , there holds(4.5) − K ( u j ) ≤ (cid:26) n − u − nj if n > − ln u j if n = 1.We wish to show that the right-hand side of (4.3) is bounded. If n >
1, we have(4.6) − τ Z A j K ( u j ) u p + α − j dxdt ≤ τ n − Z A j u p + α − nj dxdt ≤ cτ . The last step is due to the fact that p + α − n ≥
0. The second integral on the right-hand side of(4.3) can be handled in an entirely similar way. The third one there can be estimated as follows:(4.7) τ Z B j K ( u j ) u α − j dxdt ≤ cτ Z B j u αj dxdt ≤ cτ . Here we have used Lemma 3.5 and the fact that ln u j ≤ u j on the set B j . As for the last integral,remember that ε − <
0. Hence u ε − j ≤ B j . Subsequently, we have(4.8) τ Z B j K ( u j ) u ε − j dxdt ≤ cτ Z B j u j dxdt ≤ cτ . Now we can conclude that(4.9) − τ Z A j K ( u j ) u ε − j dxdt ≤ c. This implies(4.10) τ Z Ω T u ε − j dxdt → τ → τ Z Ω T u ε − j dxdt = τ Z { u j ≤ τ } u ε − j dxdt + τ Z { u j >τ } u ε − j dxdt (4.11) ≤ | K ( τ ) | τ Z { u j ≤ τ } | K ( ρ k ) | u ε − j dxdt + cτ ε (4.12) ≤ c | K ( τ ) | + cτ ε → τ → n ≥ | K ( τ ) | → ∞ as τ → We can derive from Lemma 3.3 that Z Ω T τ u α − j | ∆ u αj | dxdt ≤ (cid:18)Z Ω T τ u α − j dxdt (cid:19) (cid:18)Z Ω T (∆ u αj ) dxdt (cid:19) ≤ cτ (4.14)because 2 α − <
1. With the aid of Lemma 3.4, we obtain Z Ω T τ u ε − j | ∆ u αj | dxdt ≤ (cid:18)Z Ω T τ u ε − j dxdt (cid:19) (cid:18) τ Z Ω T u ε − j (∆ u αj ) dxdt (cid:19) ≤ c (cid:18)Z Ω T τ u ε − j dxdt (cid:19) → . (4.15)We deduce from Lemma 3.5 that Z Ω T τ u p + α − j dxdt = Z { u j ≤ } τ u p + α − j dxdt + Z { u j > } τ u p + α − j dxdt ≤ cτ + Z { u j > } τ u p +2 α − α − n ) + +1) N j dxdt ≤ cτ + cτ → τ → τ R Ω T u p + ε − j dxdt → ε →
0. This completes the proof. (cid:3)
Lemma 4.2.
Let the assumptions of Lemma 4.1 hold. If n ≤ σ , then the sequence { ∂ t ˜ u j } isbounded in L ((0 , T ); ( W , ∞ (Ω)) ∗ ) , where σ is given as in (1.14).Proof. We first claim that(4.17) Z Ω T u σ +2 αj dxdt ≤ c. We will prove this only in the case where
N <
4. Assuming this, we have Z Ω T u αj dxdt ≤ Z T Z Ω u j dx k u αj k ∞ dt ≤ c Z T (cid:0) k∇ u αj k + k u αj k (cid:1) dt ≤ c. (4.18)The last step is due to Lemma 3.5.Recall that( u j + τ ) n − ∇ u j F j = u α − j ( u j + τ ) n − ∇ u j ∆ u αj + τ u ε − j ( u j + τ ) n − ∇ u j ∆ u αj − τ u p + α − j ( u j + τ ) n − ∇ u j − τ u p + ε − j ( u j + τ ) n − ∇ u j + τ u α − j ( u j + τ ) n − ∇ u j + τ u ε − j ( u j + τ ) n − ∇ u j . (4.19)Our objective here is to show that each term on the right-hand side of the above equation is boundedin ( L (Ω T )) N . To this end, we note( u j + τ ) n − ≤ u n − j + τ n − since n − <
1. By our assumption, 0 < − α + 4 n − ≤ α + σ . We compute Z Ω T u α + n − j |∇ u j ∆ u αj | dxdt = 43 α − Z Ω T u − α + n − j |∇ u α − j | u α − j | ∆ u αj | dxdt ≤ c (cid:18)Z Ω T u − α +4 n − j dxdt (cid:19) (cid:18)Z Ω T |∇ u α − j | dxdt (cid:19) · (cid:18)Z Ω T u α − j | ∆ u αj | dxdt (cid:19) ≤ c. (4.20)There are too many terms on the right-hand side of (4.19), and so we will skip the obvious ones.Now we look at the second term on the right-hand side of (4.19). We have τ Z Ω T u ε + n − j |∇ u j ∆ u αj | dxdt = 4 α + ε Z Ω T τ u ε − α + n − j τ |∇ u α + ε j | τ u ε − j | ∆ u αj | dxdt ≤ c (cid:18)Z Ω T τ u ε − α +4 n − j dxdt (cid:19) (cid:18)Z Ω T τ |∇ u α + ε j | dxdt (cid:19) · (cid:18)Z Ω T τ u ε − j | ∆ u αj | dxdt (cid:19) ≤ cτ . (4.21)Here we have used the fact that 0 < ε − α + 4 n − ≤ α + σ . Next we estimate τ Z Ω T u p + α + n − j |∇ u j | dxdt ≤ (cid:18)Z Ω T τ u p + α − j |∇ u j | dxdt (cid:19) (cid:18)Z Ω T τ u p + α − nj dxdt (cid:19) ≤ c. (4.22)The last step is due to Lemma 3.6 because n ≤ σ . The rest of the terms can be estimatedsimilarly.We still need to consider the term( u j + τ ) n F j = u α − j ( u j + τ ) n ∆ u αj + τ u ε − j ( u j + τ ) n ∆ u αj − τ u p + α − j ( u j + τ ) n − τ u p + ε − j ( u j + τ ) n + τ u α − j ( u j + τ ) n + τ u ε − j ( u j + τ ) n . (4.23)It is easy to see that it is also bounded in L (Ω T ). Let ξ be a C ∞ test function with ∇ ξ · ν = 0 on ∂ Ω. We have( ∂ t ˜ u j , ξ ) = Z Ω ( u j + τ ) n ∇ F j · ∇ ξdx + τ Z Ω F j ξdx = − Z Ω F j (cid:0) n ( u j + τ ) n − ∇ u j · ∇ ξ + ( u j + τ ) n ∆ ξ (cid:1) dx + τ Z Ω F j ξdx, (4.24)where ( · , · ) is the duality pairing between W , ∞ (Ω) and its dual space ( W , ∞ (Ω)) ∗ , from which thelemma follows. (cid:3) Lemma 4.3.
Let the assumptions of Lemma 4.2 hold. Then the sequence { u j } is precompact in L α ((0 , T ); L α (Ω)) . Proof.
Set q = 8 α + 4 σ σ , where σ is given as before. By our assumption on α , we obviously have q > α . We estimate that Z Ω T |∇ u j | q dxdt = 2 q α q Z Ω T u (2 − α ) q j |∇ u α j | q dxdt ≤ c (cid:18)Z Ω T |∇ u α j | dxdt (cid:19) q (cid:18)Z Ω T u − α ) q − q j dxdt (cid:19) − q . Note that − α ) q − q = 2 α + σ . Therefore, we obtain from (4.17)(4.25) Z Ω T |∇ u j | q dxdt ≤ c. We can easily deduce from the definitions of u j , ˜ u j that Z Ω T | ˜ u j | α dxdt ≤ Z Ω T | u j | α dxdt + 12 τ Z Ω | u | α dx, (4.26) Z Ω T |∇ ˜ u j | α dxdt ≤ Z Ω T |∇ u j | α dxdt + 12 τ Z Ω |∇ u | α dx. (4.27)Thus { ˜ u j } is bounded in L α ((0 , T ); W , α (Ω)). Note that for t ∈ ( t k − , t k ] we have˜ u j ( x, t ) − u j ( x, t ) = ( t k − t ) ∂ t ˜ u j ( x, t ) . This together with Lemma 4.2 implies that(4.28) Z T k u j − ˜ u j k ( W , ∞ (Ω)) ∗ dt ≤ cτ. Observe that the embedding W , α (Ω) ֒ → L α (Ω) is compact and L α (Ω) ֒ → (cid:0) W , ∞ (Ω) (cid:1) ∗ iscontinuous. A result of [23] asserts that { ˜ u j } is precompact in both L α ((0 , T ); L α (Ω)) and L ((0 , T ); ( W , ∞ (Ω)) ∗ ). According to (4.28), we also have that { u j } is precompact in L ((0 , T ); ( W , ∞ (Ω)) ∗ ).This puts us in a position to apply the results in [23] again, from which the lemmas follows. Theproof is complete. (cid:3) We are ready to complete the proof of Theorem 1.1. We can extract a subsequence of { j } , stilldenoted by { j } , such that u j → u strongly in L α (Ω T ) and a.e. , (4.29) u αj ⇀ u α weakly in L ((0 , T ); W , (Ω)).(4.30)Equipped with this, we calculate that(4.31) Z Ω T |∇ u αj | dxdt = − Z Ω T ∆ u αj u αj dxdt → − Z Ω T ∆ u α u α dxdt = Z Ω T |∇ u α | dxdt. This implies that(4.32) u αj → u α strongly in L ((0 , T ); W , (Ω)).Without loss of generality, we may also assume(4.33) ∇ u αj → ∇ u α a.e. on Ω T .Note that α ≥ ∇ u j = α u α − j ∇ u αj . This along with (4.29) shows(4.34) ∇ u j → ∇ u a.e. on Ω T . OURTH ORDER PARABOLIC EQUATIONS 29
Next we wish to prove(4.35) ( u j + τ ) n − F j ∇ u j ⇀ α u α + n − ∆ u α ∇ u α weakly in L (Ω T ).This can be derived from the proof of Lemma 4.2. To see this, first observe that(4.36) u α + n − j ∇ u j → u α + n − ∇ u a.e. on Ω T .According to Egoroff’s Theorem, to each δ > E δ ⊂ Ω T with the property(4.37) u α + n − j ∇ u j → u α + n − ∇ u uniformly on Ω T \ E δ and | E δ | < δ .Due to our assumption, we have σ − n + 4 >
0. By a calculation identical to (4.20), we obtain (cid:12)(cid:12)(cid:12)(cid:12)Z E δ u α + n − j ∇ u j ∆ u αj dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ c (cid:18)Z E δ u α +4 n − j dxdt (cid:19) ≤ c (cid:18)Z E δ u α + σj dxdt (cid:19) α +2 n − α + σ ) | E δ | σ − n +44(2 α + σ ) ≤ cδ σ − n +44(2 α + σ ) . (4.38)Consequently, we have lim sup j →∞ (cid:12)(cid:12)(cid:12)(cid:12)Z Ω T u α + n − j ∇ u j ∆ u αj dxdt − Z Ω T u α + n − ∇ u ∆ u α dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ cδ σ − n +44(2 α + σ ) + (cid:12)(cid:12)(cid:12)(cid:12)Z E δ α u α + n − ∆ u α ∇ u α dxdt (cid:12)(cid:12)(cid:12)(cid:12) (4.39)The right-hand side goes to 0 as δ →
0. Therefore,(4.40) u α + n − j ∇ u j ∆ u αj ⇀ u α + n − ∇ u ∆ u α weakly in L (Ω T ).We can also prove(4.41) Z Ω T τ u p + α − nj dxdt → τ → u p + α − nj ≤ δu p + α + N j + c ( δ ) , δ > . Then apply Lemma 3.6 to yield the desired result. The remaining terms on the right-hand side of(4.17) are very easy to handle. Thus (4.35) follows.On account of (4.23), we have( u j + τ ) n F j ⇀ u α + n − ∆ u α weakly in L (Ω T ).We can infer from (4.28) that(4.42) ˜ u j → u strongly in L α (Ω T ).Assume ξ ( x, T ) = 0 in (4.24), integrate it over (0 , T ), then let j → ∞ , and thereby obtain thetheorem. The proof is complete. Proof of Theorem 1.2
The proof of Theorem 1.2 relies on the following lemma
Lemma 5.1.
Let the assumptions of Lemma 3.3 hold. Assume (5.1) α = 1 , < β < n. Then there is τ ∈ (0 , such that for all τ ∈ (0 , τ ) we have Z Ω t (∆ u β j ) dxds + τ Z Ω t (∆ u ε + β j ) dxds + τ Z Ω t u p + β − j |∇ ρ k | dxds + τ Z Ω t u p + ε + β − j |∇ u j | dxds + τ Z Ω t u β + ε − j |∇ u j | dxds + τ Z Ω t (1 + τ u ε − j )( u pj − M ( u j ) dxds ≤ c, (5.2) where (5.3) M ( r ) = Z r βs β − ( s + τ ) n ds. Proof.
Let M ( r ) be given as above. We use M ( ρ k ) as a test function in (3.23) to obtain(5.4) − Z Ω ∇ F k · ∇ ρ βk dx − τ Z Ω F k M ( ρ k ) dx + 1 τ Z Ω ( ρ k − ρ k − ) M ( ρ k ) dx = 0 . The first integral in the above equation is equal to Z Ω F k ∆ ρ βk dx = Z Ω ∆ ρ k ∆ ρ βk dx + τ Z Ω ρ ε − k ∆ ρ k ∆ ρ βk dx + pβτ Z Ω ρ p + β − k |∇ ρ k | dx +( p + ε − βτ Z Ω ρ p + ε + β − k |∇ ρ k | dx − ( ε − βτ Z Ω ρ β + ε − k |∇ ρ k | dx. (5.5)By virtue of Lemma 2.5, we have(5.6) Z Ω ∆ ρ k ∆ ρ βk dx ≥ c Z Ω (∆ ρ β +12 k ) dx, β ∈ ( 12 , , while Corollary 2.2 implies(5.7) Z Ω ρ ε − k ∆ ρ k ∆ ρ βk dx ≥ c Z Ω (∆ ρ β + ε k ) dx, β ∈ ( 12 , . Using (5.6)-(5.7) in (5.5), we obtain Z Ω F k ∆ ρ αk dx ≥ c Z Ω (∆ ρ β +12 k ) dx + cτ Z Ω (∆ ρ β + ε k ) dx + pβτ Z Ω ρ p + β − k |∇ ρ k | dx +( p + ε − βτ Z Ω ρ p + ε + β − k |∇ ρ k | dx − ( ε − βτ Z Ω ρ β + ε − k |∇ ρ k | dx. (5.8) OURTH ORDER PARABOLIC EQUATIONS 31
We calculate the second integral in (5.4) to obtain − τ Z Ω F k M ( ρ k ) dx = − τ Z Ω (1 + τ ρ ε − k ) M ( ρ k )∆ ρ k dx + τ Z Ω ( ρ pk − τ ρ ε − k ) M ( ρ k ) dx ≡ K ,k + K ,k . (5.9)Notice that M ( r ) changes from negative to positive at 1, and thus we always have(5.10) K ,k ≥ . The term K ,k can be written in the form K ,k = − τ Z Ω (1 − ε ) M ( ρ k ) ρ ε − k |∇ ρ k | dx + βτ Z Ω ρ − εk + τ ( ρ k + τ ) n ρ β + ε − k |∇ ρ k | dx. (5.11)Let B k = { x ∈ Ω : ρ k ( x ) ≥ } be given as before. On the set B k , we have M ( ρ k ) ≤ βn − β . Keeping this in mind, we estimate K ,k ≥ − (1 − ε ) τ Z Ω M ( ρ k ) ρ ε − k |∇ ρ k | dx ≥ − cτ Z B k |∇ ρ k | dx ≥ − cτ Z Ω ρ p + β − k |∇ ρ k | dx. (5.12)In view of the coefficient of the fourth integral in (5.8), we just need to select a number τ in (0 , cτ < pβ, where c is the same as the one in the last line of (5.12). Then K ,k can be absorbed into the fourthterm in (5.8).By a calculation similar to (3.74), we have(5.14) M ( r ) ≥ ( ββ − n (cid:2) ( r + τ ) β − n − (1 + τ ) β − n (cid:3) if r > ββ − n r β − n if r ≤ Z Ω Z u j u M ( r ) drdx ≥ − c. The remaining proof is similar to that of Lemma 3.3. The proof is complete. (cid:3)
We are ready to conclude the proof of Theorem 1.2. Since n > , we can pick a number β withthe property 12 < β < min { , n } . Then we apply Lemma 5.1 to obtain(5.15) τ Z Ω T (1 + τ u ε − j )( u pj − M ( u j ) dxds ≤ c. This combined with the fact that(5.16) lim r → + M ( r ) = −∞ implies(5.17) τ F j → L ( Ω T ).We can easily infer this from the proof of Lemma 4.1. That is, if we replace K ( r ) with M ( r ) inthe proof, all the arguments there still work. By examining the rest of the calculations in the proofof Theorem 1.1, we see that all of them are still applicable here except (4.21), for which we makesome adjustments. To this end, we set α = 1 , γ = β , u = u j in (2.18) to obtain∆ u j = 2(1 − β )(1 + β ) u − βj |∇ u β j | + 21 + β u − β j ∆ u β j = 8(1 − β )(1 + β ) u − β j |∇ u β j | + 21 + β u − β j ∆ u β j . (5.18)Substitute this into the left-hand side of (4.21) to obtain τ Z Ω T u ε + n − j |∇ u j ∆ u j | dxdt ≤ − β )(1 + β ) τ Z Ω T u − β + ε + n − j |∇ u j ||∇ u β j | dxdt + 21 + β τ Z Ω T u − β + ε + n − j |∇ u j ∆ u β j | dxdt ≡ A + A . (5.19)We estimate A to yield A ≤ c (cid:18)Z Ω T τ u β + ε − n − β )+ εj |∇ u j | dxdt (cid:19) (cid:18)Z Ω T | ∆ u β j | dxdt (cid:19) ≤ c (cid:18)Z Ω T τ u β + ε − n − β )+ εj |∇ u j | dxdt (cid:19) . (5.20)Thus if the exponent β + ε − n − β ) + ε <
0, then there holds the inequality u β + ε − n − β )+ εj = (cid:18) u j (cid:19) − ( β + ε − − [2( n − β )+ ε ] ≤ δ (cid:18) u j (cid:19) − ( β + ε − + c ( δ )= δu β + ε − j + c ( δ ) . (5.21)Consequently, we can deduce from Lemma 5.1 thatlim sup τ → A ≤ lim sup τ → c (cid:18) δ Z Ω T τ u β + ε − j |∇ u j | dxdt + c ( δ ) τ Z Ω T |∇ u j | dxdt (cid:19) ≤ cδ . (5.22)Since δ is arbitrary, we have lim τ → A = 0. If the exponent β + ε − n − β ) + ε ≥
0, then weuse the inequality u β + ε − n − β )+ εj ≤ δu p − j + c ( δ ) . OURTH ORDER PARABOLIC EQUATIONS 33
This can be done because from our assumptions we always have β + ε − n − β ) + ε < p − τ → A = 0. The term A can be handled in the exactlysame way. This completes the proof. Acknowledgment:
Portion of this work was completed while the second author was visiting DukeUniversity. He would like to express his gratitude for the hospitality of the hosting institution andfinancial support from the KI-Net for his visit. The research of JL was partially supported byKI-Net NSF RNMS grant No. 1107291 and NSF grant DMS 1514826.
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