Global weak solutions to a sixth order Cahn-Hilliard type equation

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1 Global weak solutions to a sixth order Cahn-Hilliard type equation M. D. Korzec, P. Nayar, P. Rybka December 8, 21 Abstract In this paper we study a sixth order Cahn-Hilliard type equation that arises as a model for the faceting of a growing surface. We show global in time existence of weak solutions and uniform in time a priori estimates in the H 3 norm. These bounds enable us to show the uniqueness of weak solutions. Key words: Cahn-Hilliard equation, self-assembly, global weak solution 21 Mathematics Subject Classification. Primary: 35K55 35Q56, 35G2) 1 Introduction During the last two or three decades it has become popular to model the evolution of thin solid films in terms of continuum theory. One example for a thin film approximation of a surface diffusion based process that describes the faceting of a growing surface has been given by Savina et al. [3]. It can be extended to more complex self-assembly systems such as quantum dots [5-8]. However here, we stick to the one-material model established before. Additional information on self-arranging nano-surfaces, quantum dots and faceting of growing surfaces can be found in the references mentioned above. Mathematically, the problem is interesting and challenging, since the regularizing Wilmore term in the surface energy results, when applying a long wave approximation, in a sixth order term that dominates the semilinear partial differential equation. More precisely, the model describes an evolving surface, a graph of function h : Ω R 2 [, T] R. The surface is governed in Ω by h t = D 2 h h+ 3 h [βh 2 yh xx +h 2 xh yy +4h x h y h xy )+αh 2 xh xx +h 2 yh yy )]. 1) Here, α, β > are anisotropy coefficients, D > is a parameter related to the deposition rate, is the standard Laplacian and subscripts indicate differentiation with respect to the noted variables. Furthermore, as described in the derivation of 1

2 this equation see Savina et al. [3] or Korzec [6]), the overall surface is in a moving frame. As usually, an initial condition supplements the problem, hx, y,) = h x, y), for x, y) Ω 2) and also boundary conditions have to be imposed. There are various possibilities, but the two most common ones are given by defining the domain as Ω = R 2 or Ω = T 2, where T 2 is the flat torus. The latter one yields a periodic surface, it seems as realistic as an infinite domain. Hence we choose the bounded version to gain additional technical advantages in the analysis. We establish the existence of global weak solutions, i.e. we show that there exists a function h C[, T], H 3 ) with h t L, T), H 3 ), such that h satisfies 1) in the distributional sense. The main result is stated below, it will be proved in Section 3. Theorem 1 Let us assume that h H 3 T 2 ), then there exists a unique weak solution 1), which is well-defined on [, ). Before we proceed with the proof, we want to record the structure of the problem, which has also been found in the originating paper [3]. Basically, equation 1) is a perturbed gradient system h t = D 2 h 2 + H. 3) For a proper definition of H, see 4) below. It turns out that getting an a priori estimate in H 3 is the crucial part of the work, this is the content of Theorem 4. We achieve that by a bootstrapping argument, where we use the constant variation formula representation of the solution. On the other hand the H 2 estimates are much easier to establish. We take advantage of the boundedness of the domain and availability of the Sobolev inequalities. It turns out that we cannot repeat this part of the argument on an unbounded domain, e.g. R 2. Once we set the objectives, we describe the methods to achieve that goal. We use the notation and the guidance of the semigroup theory, see [3]. From our perspective, problem 1) does not justify the full-fledged theory. We choose an easier approach that bases on Fourier series. Here, we are content with establishing global in time existence. We do not study here the asymptotic behavior of the system. We postpone it for a future work. We should also mention, that [6], [7] and [8] are the only closely related papers we are aware of. In [6] the authors are concerned with the one-dimensional version of the same problem. However, the approach applied there is completely different, for the authors use the Galerkin method. This general tool is not best suited for the regularity study, so that they have to overcome additional technical difficulties which are absent here, in their uniqueness result. Moreover, [6] presents also numerical results on coarsening and stationary states. The other papers are [7] and [8]. The authors study a similar sixth order problem, which also belongs to a class of Cahn-Hilliard equations. The motivation to study that problem comes from a different physical phenomenon, namely the phase transitions in ternary oil-water-surfactant systems considered in a bounded domain. 2

3 They obtain similar results by different methods, i.e. the typical tools of the theory of parabolic equations due to Solonnikov [9]. Notation We will clarify the notation we use. We identify the flat torus T 2 with [, 2π) 2, x, y) is a generic point of T 2. By dv = dxdy we denote the Lebesgue measure. For h : T 2 R, we will write ) 1/2 h = h L2 T 2 ), h = h x ) 2 + h y ) 2 ) dv. T 2 Since we work on the torus, in place of the Fourier transform we consider the Fourier series, which may be written formally as hx, y) = e ixk+yl) ĥk, l) = e ixk+yl) ĥk, l) dµk, l), k,l) Z 2 R 2 where µ is the standard counting measure supported on Z 2. In this formula we use ĥk, l) = 1 2π) 2 hx, y)e ixk+yl) dv x, y). T 2 For the sake of consistency we also recall the inverse Fourier transform for f : Z 2 R. Namely, we define ˇfx, y) = k,l) Z 2 e ixk+yl) fk, l). Moreover, we notice that for any s R, the norm in the Sobolev space H s T 2 ) is equivalent to f H s T 2 ) = ) s/2 ˆf ) L2 µ). 2 Local in time existence We want to discover as much structure of 1) as possible. For this purpose we define a vector field F = α 3 h3 x, h 3 y) + βh 2 yh x, h 2 xh y ) and the functions Ψ = βh 2 yh xx + h 2 xh yy + 4h x h y h xy ) + αh 2 xh xx + h 2 yh yy ), Φ = 1 2 h xx + h yy ) h2 x + h 2 y) + α 12 Note that div F = Ψ. Subsequently we shall write h 4 x + h 4 y) + β 2 h2 xh 2 y. H = h + 2 h Ψ = div h + h F). 4) Thus, indeed 1) takes the form 3). We notice that due to the periodic boundary condition the average of H vanishes, T 2 H dv =. 3

4 Finally, we define the functional L = Φ dv. 5) T 2 The first stage of our analysis of 1) is a study of the following linear equation h t = 3 h + f, h, ) = h ), 6) where f : T 2 R is a given function whose regularity has to be specified yet. Although we first treat 6), we keep in mind that we finally want to consider fh) = D 2 h h Ψh). 7) We proceed formally by applying the Fourier transform to both sides, this yields, d dtĥt, ξ) = ξ 6 ĥt, ξ) + ˆf, ĥ, ξ) = ĥξ). After solving this ODE we obtain an explicit formula for the Fourier transform of the solution, Thus, we can write ht, x, y)) = ĥt, ξ) = e ξ 6 t ĥ ξ) + e ξ 6 t ĥ ξ) + After introducing the following shorthand e 3t f = we can write a solution of 6) in the form: ht) = e 3t h + e ξ 6 t s) ˆfs, ξ) ds. e ξ 6 t s) ˆfs, ξ) ds ) x, y). e 6 t ˆf ), ) 8) e 3 t s) fs) ds. Once we derived the above constant variation formula for solutions to 6), we introduce the operator Fh)t, ) = e 3t h ) ) + e 3 t s) fhs, )) ds 9) with f given by 7). We notice that the above F is well-defined on the following space X T = C [, T];H 3 T 2 ) ). The ball centered at zero with radius M will be denoted by X M T, X M T = X T {v : sup vt) H 3 M}. t [,T] 4

5 Theorem 2 Let us assume that h H 3 and let us fix M/2 > h H 3. Then, there exists T > such that F : XT M XM T and F is a contraction on XM T. In particular, there exists a unique solution of the integral equation Fh) = h in XT M. Remark. The solution constructed in the above theorem will be called a mild solution to 1). Proof. We shall write L 2 for L 2 µ), where µ is the counting measure. For any s R we will use H s = H s T 2 ). We shall first check that the operator defined by 8) is continuous on H s for any s and all t >. Indeed, e 3t h H s = 1 + ξ 2 ) s/2 e 3t h ) ξ) L2 = 1 + ξ 2 ) s/2 e ξ 6 t ĥ ξ) L2 1 + ξ 2 ) s/2 ĥ ξ) L2 = h H s. We also want to use continuity of the function, t e 3t h C[, T];H s ). It follows from the Lebesgue s dominated convergence theorem, namely lim t t e 3t e 3 t )h H s = lim t t 1 + ξ 2 ) s/2 e ξ 6t e ξ 6 t )ĥξ) L2 =. We shall establish a regularizing property of F which is a crucial point in our theory. We claim that for any p R, < ε, t t and a function v C[t, t], H p 61 ε) ) we have e 3 t s) vs, ) ds H p Cε) e t t t ) ε v C[,t];H p 61 ε) ). 1) t Indeed, let us notice ) e 3 t s) vs, ) ds H p = 1 + ξ 2 ) p/2 e 3 t s) vs, ) ds ξ) L2 t t = 1 + ξ 2 ) p/2 t e ξ 6 t s)ˆvs, ξ) ds L2 t 1 + ξ 2 ) p/2 e ξ 6 t s)ˆvs, ξ) L2 ds. At this point we make a simple observation, for t > s > ξ 6 t s) t 1 + ξ 6 )t s) t ξ 2 ) 3 t s). As a result, for any ε, 1] we have e 3 t s) vs, ) ds H p t e t e ξ 2 ) 3 t s) t s) 1 ε 1 + ξ 2 ) 31 ε) 1 t t s) 1 ε1 + ξ 2 ) p 31 ε)ˆvs, 2 ξ) L2 ds. If y = 1 + ξ 2 ) 3 t s), then e ξ 2 ) 3 t s) t s) 1 ε 1 + ξ 2 ) 31 ε) = e 1 4 y y 1 ε Cε), 5

6 where Cε) is a constant that may vary during the proof. Therefore, e 3 t s) vs, ) ds H p e t Cε) t e tcε) ε t 1 t s) 1 ε1 + ξ 2 ) p 31 ε)ˆvs, 2 ξ) L2 ds t t ) ε sup vs, ) H p 61 ε). s [,t] Thus, we have derived 1). Subsequently, we take p = 3 and we consider 1) with t =. In order to prove that F maps X T into X T one has to verify that for any h XT M, the following bound holds sup fht, )) H 3 61 ε) CM) <, 11) t [,T] where CM) is independent of h. We select < ε < 1/3. Obviously, by the definition of the norm and our choice of ε, we see that Since the embedding 2 h H 3 61 ε) C h H 7 61 ε) C h H 3. H 2 T 2 ) CT 2 ) L T 2 ) 12) is valid see [1]), then for any element h XT M we have h 2 x 2 h 2 H x ε) L 2 = h 4 x dv h x 2 T 2 h 2 x dv C h 2 H h 3 x 2 L 2 T 2 C h 4 H CM 4. 3 We conclude that sup ht) 2 C[,t];H 3 61 ε) ) CM2. t [,T] Finally, if we restrict ε even further by requiring that ε < 1/6, then we have the following estimate for the nonlinearity, h x h y h xy ) H 3 61 ε) C h x h y h xy H 5 61 ε) C h x h y h xy L2 C h x h y h xy L2 C h 3 H 3. After combining these observation, we conclude that sup Ψh) C[,t];H 3 61 ε) ) CM2 + M 3 ). t [,T] This implies that F : XT M XM T, where T is so chosen, that for given M we have Cε)e T T ε M + M 2 + M 3 ) < M/2. Our next goal is to prove that F : XT M XM T is a contraction for sufficiently small T >. For this purpose, because of 1) it is enough to show that f is Lipschitz continuous in XT M, fv) fu) C[,t];H 3 61 ε) ) CM) u v C[,t];H 3 ) 13) 6

7 for a positive ε, 1/3). Once we establish 13), taking e T T ε < 1 2CM)Cε) will finish the proof. Now we show 13). Here the linear term 2 v does not cause any problems, while some more work has to be invested for the nonlinearities. In order to deal with the term v 2, we observe that for ε < 1/2 the number s = 3 61 ε) is negative. Therefore, u 2 x v 2 x H s u 2 x v 2 x L2 C u x v x L u x + v x L CM u v H 3. In the above estimates we used the embedding 12). In order to finish the proof we consider the nonlinear term v x v y v xy ). We have u x u y u xy v x v y v xy H s u x u y u xy v x v y v xy H s+2 u x v x )u y u xy H s+2 + v x u xy u y v y ) H s+2 + v x v y u xy v xy ) H s+2. Note that for ε, 1/6) we have s + 2 <, hence H s+2 C L2. Therefore u x v x )u y u xy H s+2 C u x v x )u y u xy L2 C u x v x u y u xy L2 CM 2 u v H 3 and similarly Finally, we have v x u xy u y v y ) H s+2 CM 2 u v H 3. v x v y u xy v xy ) H s+2 C v x v y u xy v xy ) L2 C v x v y u xy v xy L2 CM 2 u v H 3. The same technique may be used to estimate the other two terms. We have derived 13). Once we have established existence of a unique fixed point of F, we will prove that the solution of the equation Fh) = h enjoys some additional regularity. Namely, any fixed point is locally Hölder continuous in the norm H 3 T 2 ) with respect to time. Lemma 1 Let us take any p R. For every < a 1 there exists a constant C a > such that for δ > e 3δ Id)g H p C a a δa g H 6a+p. Proof. We begin with an observation about the exponential function. Namely, there exists a constant C a such that for x we have 1 e x C a a xa. Indeed, for x = both sides are equal, hence it is enough to show the inequality for the derivatives e x C a x a 1 for some C a >. But this is obvious, since for a = 1 we have e x 1 and for a, 1) the function, ) x e x x a 1 has infinite limits when x + and x. 7

8 We use this observation in the following estimate, e 3δ I)g H p = e ξ 6δ 1)ĝ1 + ξ 2 ) p/2 L2 C a a δa ĝ ξ 6a 1 + ξ 2 ) p/2 L2 C a a δa g H 6a+p. Now we can show better regularity of the fixed point constructed in the previous theorem. Here is the first step in this direction. Lemma 2 The unique solution of the equation Fh) = h, where F is given by formula 9), is locally Hölder continuous in the norm H 3 T 2 ) with respect to time. More precisely, there exist constants a, ε 1 > such that for a constant C = Cε 1, M, a). Proof. We have the following estimate ht + δ) ht) H 3 Cδt 1 + δ a t ε 1 + δ ε 1 ) ht + δ, ) ht, ) H 3 e 3δ I)e 3t h H δ t e 3δ I)e 3 t s) fhs, )) ds H 3 e 3 t+δ s) fhs, )) ds H 3. We observe that the first term on the RHS can be bounded as follows, e 3 t+δ e 3t )h H 3 = 1 + ξ 2 ) 3/2 e ξ 6t 1 e ξ 6δ )ĥ L2 C 1 + ξ 2 ) 3/2 e ξ 6t ξ 6 tδ 1 t ĥ L2 C δ t 1 + ξ 2 ) 3/2 ĥ L2 = C δ t h H 3 CM δ t. This means that the first term is even locally Lipschitz continuous. From 1) and 11) we deduce +δ e 3 t+δ s) fhs, )) ds H 3 Cε)Mδ ε. t Finally, using Lemma 1 for any positive a and formula 1) with t = and any ε 1 >, we obtain e 3δ I)e 3 t s) fhs, )) ds H 3 C a δ a a e 3 t s) fhs, )) H 3+6a ds C a δ a a tε 1 fh) C[,t];H 3+6a 61 ε 1 ) ). Once we apply 11) with a + ε 1 < ε < 1/6 to the above term, we will come to the desired conclusion, i.e. e 3δ I)e 3 t s) fhs, )) ds H 3 C a δ a a tε 1 CM). Next is our regularity theorem, which explains that h, the mild solution to 1), is in fact a weak solution to 1), in the sense that h C[, T];H 3 ) and h t C, T); H 3 ) and the equation is satisfied in the distributional sense. 8

9 Theorem 3 The solution h XT M of the integral equation Fh) = h is differentiable with respect to time in the H 3 norm and h t t, ) = 3 ht, ) + fht, )) in the distributional sense, with initial condition h, ) = h ). As a result, it is a weak solution of 1). Proof. We shall show that h is a limit in the C 1 [a, T a];h 3 ) norm) of functions with the desired property. This approach was used in the proof of [3, Lemma 3.2.1]. For t > δ > we define Then, h δ t, ) = e 3t h ) + δ dh δ dt t, ) = 3 e 3t h ) + e 3δ fht δ, )) + e 3 t s) fhs, )) ds. δ 3 e 3 t s) fhs, )) ds, where we treat the above functions like elements of H 3 T 2 ). Indeed, using our standard arguments we notice 3 e 3t h ) H 3 < CM, e 3δ fht δ, )) H 3 < CM. Moreover, for any s R 3 e 3t g ) H s ξ 6 e ξ 6t 1+ ξ 2 ) s/2 ĝ ) L2 C t 1+ ξ 2 ) s 2ĝ ) L2 = C t g H s. Hence the norm of 3 e 3t in LH s, H s ) may be bounded by C/t. As a result we arrive at δ δ 3 e 3 t s) 1 fhs, )) ds H 3 sup fhs, )) H 3 s [,t δ] t s ds sup fhs, )) H 3 61 ε) ln δ/t s [,t δ] CM)ln δ/t <. We have dh δ dt t, ) = e 3δ fht δ, )) + 3 h δ s, ). In order to finish the proof we have to show that h δ t, ) ht, ) H 3 δ, e 3δ fht δ, )) H 3 fht, )), δ and use the limit differentiation theorem. Our first observation is ht, ) h δ t, ) H 3 = t δ e 3 t s) fhs, )) ds H 3 3 h δ s, ) H 3 3 hs, ) δ CT, ε)δ ε sup fhs, )) H 3 61 ε) CT, M)δ ε. s [,t] δ 9

10 Secondly, we note e 3δ fht δ, )) fht, )) H 3 e 3δ Id)fht δ, )) H 3 Due to 13), we arrive at + fht δ, )) fht, )) H 3. fht δ, )) fht, )) H 3 CM) ht δ, ) ht, ) H3 δ, because h C[, T];H 3 ). Moreover, using Lemma 1 we have e 3δ I)fht δ, )) H 3 C b b δb fht δ, )) H 6b 3 C C b b δb δ, because 6b ε) for sufficiently small b >. Finally, Theorem 2 implies that, 3 h δ t, ) 3 ht, ) H 3 = 3 e 3 t s) fhs, )) ds H 3 = t δ t δ + t δ t δ t δ 3 e 3 t s) fhs, )) fht, ))) ds + t δ 3 e 3 t s) H 3 H 3 ) fhs, )) fht, )) H 3 ds d ) e 3 t s) fht, ))) ds t δ ds H 3 C t s ht, ) hs, ) H 3 ds + e 3δ I)fht, )) H 3 ) CT) t s = C θ CT) δ ε 1 ε 1 + t s) θ + t s t δ ) δ C a t δ a δa ds + C C a a δa. δ Moreover, the convergence is uniform for t in compact subsets of, T). 3 A priori estimates, global existence 3 e 3 t s) fht, ))) ds H 3 In this Section we derive an a priori estimate in the space L 2 [, T];H 3 T 2 )). Before we present this main result, let us prove a useful bound Lemma 3 For ρ, τ we have sup1 + y) τ e ρy3 Cτ)max{1, ρ τ/3 }. 14) y Proof. If y 1 then and if y 1 then 1 + y) τ e ρy3 2 τ 1 + y) τ e ρy3 2y) τ e ρy3 = 2 τ ρ τ/3 ρy 3 ) τ/3 e ρy3 Cτ)ρ τ/3. 1

11 Theorem 4 Let us assume that h is a weak solution to 1) and 2), which was constructed in Theorem 2. In addition, we assume that h H 3. Then, h L, T; H 3 ) and h L,T;H 3 ) C 3 h, T), where the constant Ch, T) depends only of T and the initial data h. Proof. Step 1. Differentiating L with respect to time see 5)) and integrating by parts we obtain dl dt = h h t h x h xt + h y h yt ) + α h 3 T 2 3 x h xt + h 3 ) yh yt + β h 2 y h x h xt + h 2 ) ) xh y h yt dv = 2 h + h xx + h yy ) α h 2 xh xx + h 2 ) yh yy β 4hx h y h xy + h 2 yh xx + h 2 )) xh yy ht dv T 2 = Hh t dv. T 2 Thus, since h is a weak solution of 1), then ) dl D dt = H T 2 2 h 2 + H dv = H 2 dv + D T 2 2 Since T 2 H dv =, we have the Sobolev inequality Moreover, As a result, H 2 dv 2π H 2 dv. T 2 T 2 D 2 H h 2 1 2π H2 + πd2 8 h 4. T 2 H h 2 dv. dl dt πd2 h 4 dv C 1 + C 2 L, 15) 8 T 2 where C i = C i D, α, β) >, i = 1, 2, because we can find D i = D i α, β) >, i = 1, 2 such that h 4 D 1 + D h2 x + h 2 y) + α ) h 4 12 x + h 4 β y) + 2 h2 xh 2 y D 1 + D 2 Φ. Due to the Gronwall inequality we deduce from 15) that ) C1 Lt) + L) e C2t C 1, C 2 C 2 so h is bounded in L [, T];H 2 T 2 )) for a fixed T <. Let us notice that this bound is not uniform with respect to T >. We keep the following observation in mind, It will be used below. K 1 u H 2α Id ) α u L2 K u H 2α. 16) 11

12 Step 2. If α < 3 2, then h L,T;H 2α ) C 2α h, T). In order to show this bound we apply Id u) α to both sides of the constant variation formula ht) = e 3t h ) + where f is given by 7). Taking the L 2 norms yields, h H 2α h H 2α + DK 2 + K + K e 3 t s) fhs, ))s, 17) Id ) α e 3 t s) hs, ) 2 ds Id ) α+1 e 3 t s) h ds + K Id ) α+1 e 3 t s) div F ds + K = h H 2α + I 1 + I 2 + I 3 + I 4 + I 5, Id ) α e 3 t s) h ds Id ) α e 3 t s) div F ds where I k, k = 1,...,5 are ordered abbreviations for the five time integral terms. We have I 3 I 2 and I 5 I 4. We will estimate separately the terms I 1, I 2 and I 4. With 14) it is easy to estimate I 2, I 2 = K ) 1+α e 6 t s) h) s, ) ds Cα) essup t [,T] h H 2t) max{1, t s) 1+α)/3 } ds C 2 h, T) <. Here we use 1 + α)/3 < 1. We shall deal with a representative term h 3 x in I 4, estimates for other three terms h 3 y, h 2 yh x, h 2 xh y in F are similar, K 3 t s) α+3/2 3 h 3 x ds K 3 CK essup t [,T] h H 2t) ) 3 t s) α+3/2 3 h x 3 L 6 ds t s) α+3/2 3 ds C 4 h, T) <. We used here the assumption that α < 3/2 and the two-dimensional Sobolev embedding h LpT 2 ) C 2 h L2 T 2 ), p <. We estimate I 1 as follows, I 1 Cessup t [,T] h 2 t)) t s) α 3 ds Cα)essupt [,T] h L4 t)) 2 Cα)essup t [,T] h H 2t)) 2 C 1 h, T) <. If we combine above results, then we come to the following conclusion, h L,T;H 2α ) C 2α h, T), 12

13 as desired. Step 3. For α < 2 we show h L,T;H 2α ) C 2α h, T) + Ct 3 2α 6 h H 3, with the same method. We continue our calculations h H 2α e 3t h H 2α + DK 2 + K Observe that Moreover, Id ) α e 3 t s) hs, ) 2 ds Id ) α+2 e 3 t s) h ds + K = e 3t h H 2α + I 1 + I 2 + I 4. Id ) α+1 e 3 t s) div F ds e 3t h H 2α ) α 3 2 e 6t ) 3 2 ĥ ) Cα)t α/3+1/2. I 2 K ) α+1 e 6 t s) )ĥs, ) ds Cα)essup t [,T] h H 2t) since α < 2. Fix δ such that α < 2 δ. We then have I 3 K 3 t s) α+1 3 ds C 2α h, T), ) α+1+δ e 6 t s) ) 1/2 δ ĥ 3 xs, ) ds Cessup t [,T] h 3 x 1 2δ t) We estimate I 1 as before. In particular, if α = 3 2 t s) α+1+δ 3 ds Cα, δ)essup t [,T] h 3 2δ t). we obtain the desired result. Summing up, we can give a proof of Theorem 1. Namely, Theorem 3 yields local in time existence of weak solutions while the estimates provided by Theorem 4 imply global existence of solutions. Hence, it only remains to show uniqueness. 4 Uniqueness of the solutions In this section we show that the weak solutions we constructed are indeed unique. Theorem 5 Let us assume that h is a weak solution to 1) with the initial condition 2), where h H 3. Then, this is a unique solution. Proof. By Theorem 4, any weak solution will be in L, T; H 3 ) provided that the initial condition is in H 3. Consider the equation for the difference, h = h 1 h 2, where h 1 and h 2 are two weak solutions with the same initial condition. Testing this equation with h we arrive at the following identity, 1 d 2 dt h 2 + h 2 = h 2 + [ D T 2 2 h 2 2 h 1 2 )h + Fh 1 ) Fh 2 )) h]. 18) 13

14 It is sufficient to estimate the nonlinear generic terms on the RHS. Let us look at I = h 3 2,x h 3 1,x) h x = h x h 2 2,x + h 2,x h 1,x + h 2 1,x) h x. T 2 T 2 The term in the parenthesis may be bounded by 3K 2, where K = h L,T;H 3 ). Thus, I 9 4ǫ K4 h x 2 + ǫ h x 2 where ǫ shall be chosen later. We may bound the remaining cubic and the quadratic terms in the same way. This yields the estimates, Fh 2 ) Fh 1 )) h C 3K) h 2 + ǫ α T 2 ǫ 3 + β) h 2, D [ h 2 2 h 1 2 )h 2 C 2K) D T 2 2 h 2 + D 4 h 2. As a result we obtain: 1 d 2 dt h 2 + h 2 h 2 + D 4 h 2 +C 2 K) D 2 h 2 + C 3K) h 2 +ǫ α ǫ 3 +β) h 2. 19) We now choose ǫ so that α 3 + β)ǫ = 1/2. In order to continue, we need the interpolation lemma below. Lemma 4 Let us suppose that u H 3, then for any ǫ > there is a constant C ǫ > so that u C ǫ u + ǫ u. Proof. Let C ε = sup x [, ) x 2 εx 3 <. Then, u = 2 û ) C ε û ) + ε 3 û ) C ε u + ε u. Combining this Lemma with h CT 2 ) u we conclude 1 d 2 dt h h 2 K ǫ h 2 + Mǫ h 2. We choose again ǫ, so that Mǫ = 1 2. We apply Gronwall inequality to the resulting estimate, 1 d 2 dt h 2 K ǫ h 2. Since h) =, we conclude that ht) = for all t [, T]. Uniqueness follows. Acknowledgements MK would like to acknowledge the financial support by the DFG Research Center MATHEON, project C1. PN and PR would like to thank Mrs Agnieszka Bodzenta-Skibińska for stimulating discussions. 14

15 References [1] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, Springer-Verlag, Berlin Heidelberg, 21. [2] T.V. Savina, A. A. Golovin, S.H. Davis, A.A. Nepomnyashchy, P.W. Voorhees, Faceting of a growing crystal surface by surface diffusion, Phys. Rev. E 67 23). [3] D. Henry, Geometric Theory of Semilinear Parabolic Equations, Lecture Notes in Mathematics, Springer-Verlag, Berlin, [4] M.D. Korzec, P.L. Evans, A. Münch, B. Wagner, Stationary solutions of driven fourth- and sixth-order Cahn-Hilliard type equations, SIAM J. Appl. Math., 69 28), no. 2, pp [5] M.D. Korzec, Continuum modeling, analysis and simulation of the selfassembly of thin crystalline films Ph.D. thesis, Technical University in Berlin, 21. [6] M.D. Korzec, P.Rybka, On a higher order convective Cahn-Hilliard type equation, preprint. [7] I. Pawłow, W.M. Zajączkowski, Sixth Order Cahn-Hilliard Type Equation, to appear. [8] I. Pawłow, W.M. Zajączkowski, Initial-boundary-value problem for the sixth order Cahn-Hilliard type equation arising in oil-water-surfactant systems, to appear. [9] V.A.Solonnikov, Boundary value problems for linear parabolic systems of differential equations of general type, Trudy Mat. Inst. Steklov 83, 1965), pp Maciek Korzec Weierstrass Institute for Applied Analysis and Stochastics Markgrafenstraße Berlin korzec@wias-berlin.de Piotr Nayar Institute of Mathematics The University of Warsaw 2-97 Warszawa, Poland pn234428@students.mimuw.edu.pl Piotr Rybka Institute of Applied Mathematics and Mechanics The University of Warsaw 2-97 Warszawa, Poland rybka@mimuw.edu.pl 15