1. Introduction In this paper we consider the following parabolic problem with dynamic boundary conditions:

Transcription

1 Differential and Integral Euations Volume 14, Number 12, December 2001, Pages PARABOLIC PROBLEMS WITH NONLINEAR DYNAMICAL BOUNDARY CONDITIONS AND SINGULAR INITIAL DATA José M. Arrieta 1 Departamento de Matemática Aplicada, Facultad de Matemáticas Universidad Complutense de Madrid, Madrid, Spain Pavol Quittner 2 Institute of Applied Mathematics, Comenius University Mlynská dolina, Bratislava, Slovakia Aníbal Rodríguez-Bernal 1 Departamento de Matemática Aplicada, Facultad de Matemáticas Universidad Complutense de Madrid, Madrid, Spain (Submitted by: Herbert Amann) 1. Introduction In this paper we consider the following parabolic problem with dynamic boundary conditions: u t + Au = f(x, t, u, u), x Ω, t > 0, (γu) t + Bu = g(x, t, γu), x Γ, t > 0, (1.1) u(x, 0) = u 0 (x), x Ω, (γu)(x, 0) = v 0 (x), x Γ, where Ω is a bounded domain in R n of class C 2, Γ = Ω, ν denotes the outer normal on Γ, γ is the trace operator, and Au = u + ωu, Bu = u ν + ωu. Although we will consider this particular case, the techniues we use can also be applied to the case of systems in which, as in [10], Au = j (a jk k u) + a j j u+a 0, Bu = a jk ν j γ k u+b 0 γu, with smooth-enough coefficients. On the nonlinear terms, f and g, we assume that they are smooth functions with 1 Partially supported by DGES PB Partially supported by Swiss National Science Foundation and VEGA Grant 1/7677/20. Accepted for publication: September 2000 AMS Subject Classifications: 35K

2 1488 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal certain growth assumptions that will be specified later (see conditions (Hf) and (Hg) below). Observe that the setting for (1.1) consists in writing an evolution euation for a two-component vector containing the function u in Ω and the trace of u on Γ. Therefore, (1.1) is considered as an evolution euation in a product of function spaces in Ω and Γ, and the condition that the latter component is the trace of the former has to be imposed. The well-posedness of (1.1) was considered in [10] for the case of smooth and compatible initial data, that is, when v 0 is the trace of the smooth function u 0. In this paper, we are interested in obtaining the largest possible growth on the nonlinear terms such that (1.1) can be solved with initial data in Lebesgue spaces, L 1 (Ω) L 2 (Γ), or even measures M(Ω) M(Γ), where M(Ω) and M(Γ) denote the space of bounded Radon measures in Ω and on Γ, respectively. That is, we are interested in obtaining the critical exponents for f and g for these classes of initial data. A general approach for this type of problem was introduced in [5], where it was used for the problem of critical exponents for the heat euation. This approach was then refined in [6], and it was used to determine the critical exponents for the heat euation with nonlinear boundary conditions. The approach mentioned above can be summarized as follows. One first writes the euation in an abstract form Z t + AZ = F(t, Z) where A represents the linear part of the euation and F includes all nonlinear terms. Then one shows that in a suitable Banach space F, A with domain E = D(A) generates an analytic semigroup e At. One then constructs a suitable scale of interpolation spaces X α for α [0, 2], such that X 2 = E, X 1 = F and the restriction of e At to X α is also an analytic semigroup. Then the properties of the mapping F on different spaces of the scale plus the variation-of-constants formula Z(t) = e At Z(0) + t 0 e A(t s) F(s, Z(s)) ds allow one to show that for Z(0) F the euation has a uniue solution, even when F grows critically. Now we describe the contents of the paper. After Section 2, where we introduce preliminaries, we study the linear semigroup associated to euation (1.1) in Section 3. In this section we include an analysis of different families of scales of Banach spaces associated to the linear operator. These families

3 parabolic problems with nonlinear dynamical boundary conditions 1489 are related to different products of spaces of functions defined in Ω and on Γ that can be chosen to obtain a well-posed problem. In Section 4 we refine again the abstract result given in [5, 6]. This is needed since the results developed in these papers do not apply directly to our situation. We do not include a detailed proof of this new abstract result since it can be obtained by minor modifications of the one given in [5]. The main results of the paper are contained in Section 5. Here we obtain that for initial data in certain spaces, M(Ω) M(Γ) among others, there is a well-defined solution of (1.1). We need to analyze how the nonlinearity of the problem acts between the different scales of spaces defined in Section 3 in order to be able to apply the abstract results from Section 4. We describe this property of the nonlinearity by controlling the growth of f and g at infinity. As a matter of fact, the main hypotheses on f and g are expressed as the following: (Hf) The function f is a C 1 -function, and there exist ρ 0, ρ 0, ρ 1, ρ 1 1 and a constant C > 0 such that for x Ω, t R +, ξ R, η R n, ξ f(x, t, ξ, η) C(1 + ξ ρ0 1 + η ρ 0 1 ), η f(x, t, ξ, η) C(1 + ξ ρ η ρ1 1 ). (1.2) (Hg) The function g can be extended to a C 2 -function in Ω, which we denote also by g, and there exists ρ 2 1 such that ξ g(x, t, ξ) C(1 + ξ ρ 2 1 ), x Ω, t R +, ξ R. (1.3) The critical exponents for a given space of initial data are the least upper bounds for the numbers ρ 0, ρ 1, ρ 2, ρ 0, ρ 1 for which we have a well-posed problem in that space. Denote ρ c 2 0 () = 1 + n, ρc 1 () = 1 + n+, ρc 2 0 () = 1 + n+, ρ c 1 () = 1 + n, ρc 2 () = 1 + n 1. (1.4) It is well known that the exponents ρ c 0 () or ρc 1 () are critical for the wellposedness of the problem u t + Au = f(x, t, u, u), x Ω, t > 0, γu = 0, x Γ, t > 0, u(x, 0) = u 0 (x), x Ω (1.5) in L (Ω) if f(x, t, u, u) = u ρ 0 or f(x, t, u, u) = u ρ 1, respectively (see [8] and [7], for example). Moreover, replacing the homogeneous Dirichlet boundary condition in (1.5) by the nonlinear boundary condition Bu = g(x, t, u) (where g satisfies (1.3)) one obtains well-posedness in L (Ω) for ρ 2 < ρ c 2 () := 1 + n (see [6] and notice that ρc 2 () < ρc 2 ()). Our results in

4 1490 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal Section 5 indicate that the exponents (1.4) are critical for the well-posedness of (1.1) in L (Ω) L (Γ). More precisely, we obtain the following assertions: For initial conditions (u 0, v 0 ) M(Ω) M(Γ), the following conditions are sufficient for the well-posedness (see Subsection 5.2, Theorem 5.11): ρ 0 < ρ c 0(1), ρ 1 < ρ c 1(1), ρ 0 < ρ c 0(1), ρ 1 < ρ c 1(1), ρ 2 < ρ c 2(1). For initial conditions (u 0, v 0 ) L #(Ω) L (Γ), with 1 < < and # = n/(n + 1), the following conditions are sufficient (see Subsection 5.3, Theorem 5.13): ρ 0 < ρ c 0 (# ), ρ 1 < ρ c 1 (# ), ρ 0 < ρ c 0 (# ), ρ 1 < ρ c 1 (# ), ρ 2 < ρ c 2 (). We also study well-posedness in the space L (Ω) B 1 1/ (Γ) where 1 < < and B s denotes the Besov space. In this case, we have to assume some additional technical assumption on and/or g, and then we obtain the following sufficient conditions (see Subsection 5.1, Theorem 5.7): ρ 0 < ρ c 0(), ρ 1 < ρ c 1(), ρ 0 < ρ c 0(), ρ 1 < ρ c 1(), (n )ρ 2 n, where the last condition is empty if n. If < n then this condition can be written in the form ρ 2 ρ c 2 ( ), where := (n 1)/(n ) is the biggest exponent such that B 1 1/ (Γ) is embedded into L (Γ). Finally, in Section 6 we include an analysis of critical exponents for an elliptic euation in the presence of dynamical boundary conditions. 2. Preliminaries In this section we introduce some notation, known results and preliminaries for (1.1). For s R and 1 < < we consider the Bessel potential spaces H s(ω) with norm s, and Besov spaces B s (Γ) := Bs 1/ (Γ) with norm s,. The norm in the Lebesgue space H 0 (Ω) = L (Ω) will be denoted simply by. and the norm in L (Γ) by,γ. By we denote the dual exponent to ; i.e., 1/ + 1/ = 1. By a + (respectively a ) we mean a number b > a (respectively b < a) which is sufficiently close to a. We put also a b := max{a, b} for any a, b R. By D(A) we denote the domain of definition of a linear operator A. If X and Y are Banach spaces then we denote by L(X, Y ) the space of continuous linear operators A : X Y and we set Isom(X, Y ) := {A L(X, Y ) : A is an isomorphism}. We write X =. Y if X, Y are isomorphic and X Y if X is continuously imbedded into Y.

5 parabolic problems with nonlinear dynamical boundary conditions 1491 It is known (see [16]) that if s > 1/, then the trace operator γ L ( H s (Ω), B(Γ) s ). It is also known that there exists a continuous right inverse T L ( B(Γ), s H s (Ω) ) of γ; i.e., γ(t g) = g for g B(Γ). s Moreover, T g can (and will) be defined as the uniue (generalized) solution of the problem Au = 0 in Ω, u = g on Γ. Similarly, let Sf denote the solution of the problem Au = f in Ω, u = 0 on Γ. Then S L ( H s (Ω), Hs+2 (Ω) ) for any s 0. Following [12], we identify the dual space ( L (Ω) ) of L (Ω) with L (Ω), the dual of H s (Ω) with a subspace of H s (Ω) for s > 0 and the dual space of B s (Γ) with B1 s (Γ). If Z := (u, v) L (Ω) L (Γ) and Υ := (ϕ, ψ) L (Ω) L (Γ), we set Z, Υ = u, ϕ Ω + v, ψ Γ = Ω uϕ dx + vψ ds. Γ Throughout the paper we shall use standard embedding, interpolation and duality properties for Bessel potential and Besov spaces (see [16], for example). 3. The linear semigroup In this section we study the linear operators associated to problem (1.1). For this, following [10], we set E 1 := {(u, v) H 2(Ω) B2 (Γ) : γu = v}, F 1 := L (Ω) B 1 (Γ), A 1 : E 1 F 1 (u, v) (Au, Bu). It is shown in [10] that choosing ω sufficiently large, the triplet (A, E, F ) := (A 1, E 1, F 1 ) has the following properties: (λ + A) Isom(E, F ) for Re λ 0, (3.1) ( A) generates a strongly continuous analytic semigroup e ta in F. This implies that one can solve the linear euation Z t + A 1 Z = 0 with Z = (u, v) and initial data Z 0 F 1. Since the solution lies in E 1, observe

6 1492 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal that this implies that v = γ(u) for t > 0 and that Z solves } } u t + Au = 0 v in Ω, t + Bu = 0 u(0) = u 0 v(0) = v 0 on Γ, (3.2) with (u 0, v 0 ) = Z 0 F 1. Now, following [1], a completely abstract argument can be carried out so one can solve (3.2) for a larger class of initial data. The sketch of such a procedure is as follows. If a triplet (A, E, F ) satisfies (3.1) then one can construct the extrapolated space G defined as the completion of F under the norm A 1. The realization of A in G, which we still denote by A, is an extension of A and A : F G. Moreover the triplet (A, F, G) satisfies that (λ + A) Isom(F, G) for Re λ 0, (3.3) ( A) generates a strongly continuous analytic semigroup e ta in G. At this point we can construct now the scale of interpolation-extrapolation spaces as follows. Let [, ] η, 0 < η < 1, be the complex interpolation functor. If we denote X 1+η = [E, F ] η, X η = [F, G] η, 0 η 1, then we have that the realization of the operator A in X η, which we still denote by A, is such that the triplet (A, X 1+η, X η ) satisfies (3.1) for every 0 η 1. Moreover we have the estimates e At x X η Ct α η x X α, 0 α η 2. (3.4) We can apply this abstract procedure starting with the triplet (A 1, E 1, F 1 ) introduced above, and then we construct the extrapolated space G 1 and the scale of interpolation-extrapolation spaces X 1+η 1 = [E 1, F 1 ] η, X η 1 = [F 1, G 1 ] η, 0 η 1, which satisfy all the above. In particular, we can now solve (3.2) with initial data (u 0, v 0 ) = Z 0 G 1, and by the smoothing effect of the semigroup, the solution enters E 1 after positive time. Some description of G 1 and the interpolated spaces we just constructed will be given further below. On the other hand, by [10], A 1 also admits a uniue continuous extension to A 0 Isom(E 0, F 0 ), where E 0 = L (Ω) B 0 (Γ), and F 0 is the completion of E 0 with respect to an appropriate norm and the triplet (A, E, F ) := (A 0, E 0, F 0 ) satisfies again (3.1). Using the abstract procedure above we can construct the space G 0 and the scale X β 0 for 0 β 2 as X1+η 0 = [E 0, F 0 ] η, X η 0 = [F 0, G 0 ] η, 0 η 1, which satisfy all the above. In particular, we can now solve the linear euation Z t + A 0 Z = 0 with Z = (u, v) and initial data Z 0 G 0, and the solution lies in E 0 for positive

7 parabolic problems with nonlinear dynamical boundary conditions 1493 times. After giving some description of the spaces obtained by the abstract procedure above, it will be shown below that in fact Z solves also (3.2). Now we can interpolate the results above as follows. Observe first that E 1 E 0, and by construction F 1 F 0 which gives in turn G 1 G 0. Denote as before by [, ] θ, 0 < θ < 1, the complex interpolation functor. Set E θ = [E 1, E 0 ] θ, F θ = [F 1, F 0 ] θ, and let A θ be the realization of the operator A 0 in F θ ; i.e., A θ Z = A 0 Z for Z D(A θ ) := {W D(A 0 ) : A 0 W F θ }. From the results in [10], it turns out that the triplet (A θ, E θ, F θ ) satisfies (3.1), and the abstract techniues of interpolation and extrapolation give us the space G θ and the scale of spaces X 1+η θ = [E θ, F θ ] η, X η θ = [F θ, G θ ] η, 0 η 1, which satisfy all the above. When we need to stress the dependence of the scales on the parameter we will write X η θ (), X1+η θ (). It can also. be shown that, as expected, G θ = [G1, G 0 ] θ. In particular, we can now solve the linear euation Z t + A θ Z = 0 with Z = (u, v) and initial data Z 0 G θ, and the solution lies in E θ for positive times. The structure of these spaces and operators is described in the following diagram: A 1 F1 E 1 A 1 E θ A θ G1 A θ Fθ A E 0 A 0 0 F0 Gθ G0 and X 1+η 1 X 1+η θ X 1+η 0 A 1 A θ A 0 X η 1 X η θ X η 0 0 η 1, where the arrows represent the natural embeddings. It is clear that to make the techniues above useful for practical purposes, some description and properties of the spaces we just constructed must be given. One of the technical difficulties of (3.2) is that only an incomplete description of the scales X η θ is available. This will translate into some results for (1.1) which may probably be improved with a better knowledge of the scales. We start by observing that it is a straightforward generalization of some results in [10] that E θ. = {(u, v) H 2θ (Ω) B2θ (Γ) : γu = v} if 1/ < 2θ 2, E θ H 2θ (Ω) B 2θ (Γ) if 2θ = 1/,. E θ = H 2θ (Ω) B2θ (Γ) if 0 2θ < 1/, (3.5). [ F θ = {(ϕ, ψ) H 2 2θ (Ω) B 2 2θ (Γ) : γϕ = ψ} ] if 0 2θ < 1 + 1/,

8 1494 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal F θ [ H 2 2θ (Ω) B 2 2θ (Γ) ] if 2θ = 1 + 1/,. [ F θ = H 2 2θ (Ω) B 2 2θ (Γ) ]. = H 2θ 2 (Ω) B 2θ 1 (Γ) if 1 + 1/ < 2θ 2. The following lemmata give us information about some of the spaces of the scales that we have just constructed. They give us a description of X η θ for some values of η and θ. Lemma 3.1. Let 1 + 1/ < 2θ 2. Then X 1+η θ Proof. We have [E θ, F θ ] η H 2θ 2(1 η) (Ω) B 2θ+η 1 (Γ), for η [0, 1]. (3.6) E θ H 2θ (Ω) B2θ and the result follows by interpolation. (Γ), F. θ = H 2θ 2 (Ω) B 2θ 1 (Γ), Lemma 3.2. For θ 1/2 with 2θ 1 + 1/ and η [0, 1], we have F θ (1 η)/2 [F θ, F θ 1/2 ] η [F θ, G θ ] η X η θ. Proof. A direct application of (3.5) implies that E θ 1/2 F θ for any θ [1/2, 1]. Moreover since A θ 1/2 : E θ 1/2 F θ 1/2 and A θ : F θ G θ are isomorphisms, and both the operators A θ and A θ 1/2 are realizations of the same operator, we will also have that F θ 1/2 G θ, for any θ [1/2, 1]. This last embedding and the properties of the interpolation give us the result. Observe that the arguments in the proof of Lemma 3.2 guarantee that E 0 F 1/2 G 1. This implies, in particular, that the solution of the linear euation Z t + A 0 Z = 0 with Z = (u, v) and initial data Z 0 G 0 lies in E 1 for positive times and solves (3.2). 4. Abstract results In this section we give an abstract result on existence and uniueness of solutions for problems of the form ẋ + Ax = F(t, x), t > 0, (4.1) x(0) = x 0. The results presented here are extensions of the abstract results presented in [5, 6], and they are needed to deal with parabolic problems with dynamic boundary conditions. With this in mind we will indicate only the differences in the proofs and refer to [5, 6] for details. We assume that we are given an interval I [0, 2], a scale of Banach spaces X α, α I, and sectorial operators A α : D(A α ) X α X α, α I, with

9 parabolic problems with nonlinear dynamical boundary conditions 1495 Re σ(a α ) > ω > 0, where σ(a α ) denotes the spectrum of A α. We assume that X β X α and that A β is the realization of A α in X β if α, β I, α < β. Moreover, we assume that the linear semigroup generated by the operator A α satisfies the ineuality t β α e Aα t x β M x α, α β, α, β I, t > 0, where θ denotes the norm in X θ. If no confusion seems likely, we write A instead of A α. The main differences in comparison with the results from [5, 6] rely on the hypotheses for the nonlinearities. In our case we will assume the following hypothesis: (H) There exist numbers 0 < ε i < 1, ρ ij 1 and γ i < 1, i, j = 0, 1,..., k, such that 1 + ε i, γ i I, 1 > γ i max j [ε i + (ρ ij 1)ε j ] =: γ i, min i γ i =: γ > ε := max i ε i, (4.2) and the nonlinearity F : [0, ) X 1+ ε X γ is such that F(t, 0) X γ i, F(t, 0) γi C, i = 1,..., k, and F(t, x) F(t, y) can be written as F(t, x) F(t, y) = k i=0 f i(t, x, y) with k ( ρ f i (t, x, y) γi C x y 1+εi (1 + x ij 1 1+ε j + y ρ ij 1) ) 1+ε j, (4.3) i = 1,..., k. Then, we look for functions that satisfy (4.1) in the sense that x(t) = e A t x 0 + t 0 j=0 e A(t s) F(s, x(s)) ds (4.4) with x 0 X 1. As in [5, 6] we look for ε-regular solutions of (4.4) in the sense of [6, Definition 1.1], i.e., a function x C([0, τ], X 1 ) C ( (0, τ], X 1+ε) which satisfies (4.4), and we have the following result. Theorem 4.1. Assume that the nonlinearity F satisfies (H). Let y 0 X 1 and denote B r (y 0 ) = {x X 1 : x y 0 < r}. Then there exist r = r(y 0 ) > 0 and τ 0 = τ 0 (y 0 ) > 0 with the following properties: for any x 0 B r (y 0 ) there exists a continuous function x : [0, τ 0 ] X 1, with x(0) = x 0, which is the uniue ε regular mild solution to (4.1) starting at x 0. In addition, this solution satisfies x C((0, τ 0 ], X 1+θ ), t θ x(t, x 0 ) 1+θ t 0 + 0, 0 < θ < γ sup I. Moreover, if x 0, z 0 B r (y 0 ), then t θ x(t, x 0 ) x(t, z 0 ) 1+θ C(θ 0, τ 0 ) x 0 z 0 1, (4.5)

10 1496 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal t [0, τ 0 ], 0 θ θ 0 < γ sup I. If γ, 1 + γ I and t F(t, x), as a map from (0, τ) to X γ, is locally Hölder continuous, uniformly on bounded sets of x X 1+γ, then x C 1 ((0, τ 0 ], X γ ) C((0, τ 0 ], X 1+γ ), and x(, x 0 ) satisfies the euation ẋ + Ax = F(t, x) in X γ for t (0, τ 0 ). Finally, assume that all nonlinearities are subcritical; that is, γ i > γ i for all 1 i k. Then r can be chosen arbitrarily large. That is, the time of existence can be chosen uniformly on bounded sets of X 1. Proof. Under the assumptions above, one can repeat the considerations in [5] or [6] in order to prove the uniue solvability of (4.4) in the ball B µ := {x K : x K µ} of the Banach space K := {x C ( (0, τ], X 1+ ε) : x K := sup i sup 0<t τ t ε i x(t) 1+εi < }, (4.6) where µ and τ are sufficiently small. More precisely, one obtains an ε-regular solution. Moreover, this solution is uniue in the class of ε-regular solutions. The only significant change concerns the proofs of Lemmata 2 and 3 in [5] (see also [6]), where one has to use estimates of the following type: t t θ e (t s)a( F ( s, x(s) ) F ( s, y(s) )) ds 0 = t θ t 0 Ct θ t 0 Ct θ t 0 e (t s)a k ( ) 1+θ f i s, x(s), y(s) ds i=0 1+θ k (t s) γi 1 θ ( ) f i s, x(s), y(s) γi ds i=0 k [ (t s) γi 1 θ x(s) y(s) 1+εi i=0 k ( ρ 1 + x(s) ij 1 1+ε j + y(s) ρ ij 1) ] 1+ε j ds j=0 Ct θ t 0 k (t s) γi 1 θ s ε i ε j (ρ ij 1) ( s ε ) i x(s) y(s) 1+εi i,j=0 ( s ε j(ρ ij 1) + ( s ε ) j ρij 1 ( ) ) x(s) 1+εj + s ε j ρij 1 y(s) 1+εj ds c(µ) x y K.

11 parabolic problems with nonlinear dynamical boundary conditions Problems with dynamical boundary conditions In this section, we apply the abstract results of Section 4 to problem (1.1) and obtain results on existence and uniueness of solutions when the initial data lie in different spaces. If we denote by A the linear operator (and its different realizations) introduced in Section 3, Z = (u, v) and F(t, Z) = F(t, u, v) = ( f (, t, u( ), u( ) ), g (, t, v( ) )), then problem (1.1) can be written as Ż + AZ = F(t, Z), t > 0, Z(0) = Z 0. (5.1) First, for the case of smooth and compatible initial data, assume 1 + 1/ > 2η > 2θ > 1/, u 0 H 2θ (Ω) and v 0 = γu 0 ; i.e., Z 0 := (u 0, v 0 ) E θ. It follows from [12] and the properties of the Nemytskii mapping F that under the corresponding assumptions on ρ 0, ρ 1, ρ 2, ρ 0 and ρ 1 in (1.2) and (1.3), the problem (1.1) has a uniue maximal weak solution u defined on its maximal existence interval [0, T ). Moreover, u is the first component of Z, where Z C([0, T ), E θ ) satisfies the variation-of-constants formula Z(t) = e ta θ Z 0 + t 0 e (t s)aη F ( s, Z(s) ) ds and v = γ(u). Below we shall prove the existence of a uniue solution of (5.1) in an appropriate Banach space for nonsmooth and noncompatible initial data. We will make use of the scales constructed in Section 3 and we will take Z 0 X η ξ for different values of η and ξ. Depending on η and ξ we will derive different conditions on the growth of the nonlinearities in terms of the exponents ρ 0, ρ 1, ρ 2, ρ 0 and ρ 1. Since the corresponding solution Z(t) will belong to E θ for some 2θ > 1/ and any t > 0, one can easily check that this solution will be a solution in the sense of [10] for any t > 0. Moreover, one can use the methods from [14] to get further regularity properties of Z. In view of the abstract results of Section 4 we need to study how the nonlinearity F acts between the different spaces of the scale of Banach spaces constructed for A. Our goal is to prove that for the scale of Banach spaces X η ξ, for some fixed ξ, the nonlinearity F fulfils hypothesis (H). Once this is done, Theorem 4.1 will guarantee that problem (5.1) is well posed. Notice that condition (H) will impose restrictions on the possible growth of the nonlinearities f and g

12 1498 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal that will be translated into restrictions on the exponents ρ 0, ρ 1, ρ 2, ρ 0 and ρ 1. We will need to distinguish different cases according to where we choose our initial data. Although it would be possible to perform the forthcoming analysis for initial data in X η ξ for a large variety of η and ξ, we will restrict ourselves to some specific choices of η and ξ, as we now explain. One of our motivations for the study of (1.1) with nonsmooth and noncompatible initial data was to consider initial values Z 0 in L 1 (Ω) L 2 (Γ), for 1 < 1, 2 <, or in M(Ω) M(Γ) (which is close to the case 1 = 2 = 1, in some sense). However, these spaces are not included in the scales of spaces described in Section 3. Therefore, we have to restrict ourselves to some spaces from these scales which are as close as possible to L 1 (Ω) L 2 (Γ) and to M(Ω) M(Γ). This will lead us to the following three particular cases: i) Initial data in L (Ω) B 1 (Γ), 1 < <. Observe that in this case F 1 = L (Ω) B 1 (Γ), and therefore we will study the euation in the scale of spaces given by X η = X η 1, η [0, 1]. This case is treated in Subsection 5.1. ii) Initial data in M(Ω) M(Γ). If we choose close to 1 then there exists a small δ > 0 such that M(Ω) M(Γ) F 1 δ/2 X1 1 δ (it is enough to choose δ > n/ ; then these embeddings follow from (3.5) and Lemma 3.2). Therefore, we will use the scale X η = X η δ 1. This case is treated in Subsection 5.2. iii) Initial data in L #(Ω) L (Γ), with # = n/(n + 1), 1 < <. If δ > 0 is small enough and 2ξ = 1 + 1/ + δ, then (3.5) and Lemma 3.2 imply L #(Ω) L (Γ) H 1/ 1 (Ω) B 1/ (Γ) F ξ δ + /2 X 1 δ+ ξ. Therefore we will use the scale X η = X η δ+ ξ. This case is treated in Subsection 5.3. Recall from the Introduction that in all the cases mentioned above we plan to prove well-posedness under the conditions ρ i < ρ c i (), ρ i < ρ c i (), i = 0, 1, (5.2) where should be replaced by, 1 or # if u 0 belongs to L (Ω), M(Ω) or L #(Ω), respectively. The critical exponents ρ c i () and ρc i (), i = 0, 1, fulfill ρ c 1 () = ρc 0 () n n+, ρc 0 () 1 = (ρc 0 () 1) n n+, and ρc 1 () 1 = (ρc n+ 1 () 1) n. Hence, if ρ 0, ρ 1, ρ 0, and ρ 1 satisfy (5.2), then we can always increase their values in such a way that the new values will still satisfy (5.2) and, moreover,

13 parabolic problems with nonlinear dynamical boundary conditions 1499 n ρ 1 = ρ 0 n +, ρ n 0 1 = (ρ 0 1) n +, ρ 1 1 = (ρ 1 1) n + (5.3) n (it is sufficient to choose ρ 0 sufficiently close to ρ c 0 () and define ρ 1, ρ 0, and ρ 1 by (5.3). By doing this we greatly simplify some proofs below). We decompose F as F(t, Z) = F 1 (t, Z) + F 2 (t, Z) = ( F 1 (t, u), 0 ) + ( 0, F2 (t, v) ), where F 1 (t, u) = f (, t, u( ), u( ) ) and F 2 (t, v) = g (, t, v( ) ). If no confusion seems likely, we shall identify the operator F i with F i for i = 1, 2 and, for example, we shall write F 1 (t, u) instead of F1 (t, u). We shall prove that F fulfills the assumption (4.3) in the following form: where F 1 (t, u 1 ) F 1 (t, u 2 ) = f 0 (t, u 1, u 2 ) + f 1 (t, u 1, u 2 ), F 2 (t, v 1 ) F 2 (t, v 2 ) = f 2 (t, v 1, v 2 ), f 0 (t, u 1, u 2 ) = f (, t, u 1 ( ), u 1 ( ) ) f (, t, u 2 ( ), u 1 ( ) ), f 1 (t, u 1, u 2 ) = f (, t, u 2 ( ), u 1 ( ) ) f (, t, u 2 ( ), u 2 ( ) ), f 2 (t, v 1, v 2 ) = g (, t, v 2 ( ) ) g (, t, v 1 ( ) ). The following results describe how the Nemytskii operators f 0, f 1 and f 2 act on Bessel potential spaces on Ω and on Besov spaces on Γ, respectively. Lemma 5.1. Let f fulfill (Hf) with ρ 0, ρ 1, ρ 0, and ρ 1 satisfying (5.3). Let α 0 and θ fulfill n n + α 0 < n, n < θ 0 and ρ 0 = n θ n α 0. Put α 1 := α 0 (n + )/n. Then f 0 (t, u 1, u 2 ) θ, C u 1 u 2 α0, f 1 (t, u 1, u 2 ) θ, C u 1 u 2 α1, ( 1 + ( ( ui ρ 0 1 α 0, i=1 2 ( ui ρ 1 1 α 1, i=1 + ui ρ ) ) 0 1 α 1,, + ui ρ ) ) 1 1 α 0,. (5.4) Proof. Put r i = n n θ ρ i, i = 0, 1, and notice that our assumptions guarantee r i > ρ i, i = 0, 1. The growth assumption (Hf), (5.3) and Hölder s ineuality imply 2 ( f 0 (t, u 1, u 2 ) r0 /ρ 0 C u 1 u 2 r0 (1 + ui ρ 0 1 r 0 + u ρ ) ) 0 1 i r 1, i=1 f 1 (t, u 1, u 2 ) r1 /ρ 1 C (u 1 u 2 ) r1 (1 + 2 ( ui ρ 1 1 r 1 i=1 + u i ρ 1 1 r 0 ) ). (5.5)

14 1500 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal Since our assumptions and the choice of r i imply H α 0 (Ω) L r0 (Ω), H α 1 1 (Ω) L r1 (Ω), and L ri /ρ i (Ω) H θ (Ω), i = 0, 1, the estimates (5.5) guarantee (5.4). In certain situations we will need to impose some extra restrictions on the nonlinearity g. We express it as the following: (Hg+) The function g satisfies hypothesis (Hg) and ξ x g(x, t, ξ) C(1 + ξ ρ2 1 } ) ξξ 2 g(x, t, ξ) C(1 + ξ ρ2 2 x Ω, t R ) +, ξ R (5.6) (if ρ 2 2, then the last ineuality should be read as ξξ 2 g(x, t, ξ) C). Lemma 5.2. Let g satisfy (Hg) for some ρ 2 1. Let 1 < < n and let ρ 2 = (n θ)/(n α), where one of the following conditions holds: i) 0 (1 n + n/) θ < 1/ and 1/ < α < n/, ii) 1/ < θ 1, 1 α < n/, ρ 2 2 and g satisfies (Hg+). Then the Nemytskii operator G = G(t) : B α (Γ) Bθ (Γ) : v g(, t, v( ) ), satisfies G(v 1 ) G(v 2 ) θ, C v 1 v 2 α, (1 + v 1 ρ 2 1 α, + v 2 ρ 2 1 α, ). (5.7) Proof. i) Put p := (n 1)/(n θ). Then 1 p < and L p (Γ) B θ (Γ), so that (1.3) and Hölder s ineuality imply G(v 1 ) G(v 2 ) θ, C G(v 1 ) G(v 2 ) p,γ ( c v 1 v 2 ps,γ 1 + v1 ρ 2 1 ps (ρ 2 1),Γ + v 2 ρ ) 2 1 (5.8) ps (ρ 2 1),Γ for any s > 1. The choice s = ρ 2 = (n θ)/(n α) guarantees α n = n 1 ps = n 1 ps (ρ 2 1) ; hence, B α (Γ) L ps (Γ) and B α (Γ) L ps (ρ 2 1)(Γ). These embeddings and (5.8) imply (5.7). ii) Since the trace operator γ : H β (Ω) B β (Γ), β > 1/, and its right inverse T are continuous linear operators and G(v) = γ G(T v), where G(u) = g (, t, u( ) ) is the Nemytskii operator defined for functions u : Ω R, the estimate (5.7) will follow from G(u 1 ) G(u 2 ) θ, C u 1 u 2 α, ( 1 + u1 ρ 2 1 α, + u 2 ρ 2 1 α, ). (5.9) Let p be defined by 1 n/p = θ n/. Then 1 < p and H 1 p (Ω) H θ (Ω); hence, G(u 1 ) G(u 2 ) θ, C G(u 1 ) G(u 2 ) 1,p

15 parabolic problems with nonlinear dynamical boundary conditions 1501 C ( G(u 1 ) G(u 2 ) p + G(u 1 ) G(u 2 ) p ) C ( g(, u 1 ( )) g(, u 2 ( ))) p + x g(, u 1 ( )) x g(, u 2 ( ) p + ξ g(, u 1 ( )) u 1 ξ g(, u 2 ( )) u 2 p ) = C(I1 + I 2 + I 3 ), where we suppressed the dependence of g on t in our notation. From (1.3), (5.6) and Hölder s ineuality, we have I 1 + I 2 c (u 1 u 2 )(1 + u 1 ρ u 2 ρ 2 1 ) p c u 1 u 2 ps ( 1 + u1 ρ 2 1 ps (ρ 2 1) + u 2 ρ 2 1 ps (ρ 2 1)), I 3 ( ξ g(, u 1 ( )) ξ g(, u 2 ( ))) u 1 p + ξ g(, u 2 ( ))( u 1 u 2 ) p ( ξ g(, u 1 ( )) ξ g(, u 2 ( ))) u 1 p + c ( 1 + u 2 ρ 2 1 pt(ρ 2 1)) u1 u 2 pt, for s, t > 1. If ρ 2 > 2, then we have ( ξ g(, u 1 ( )) ξ g(, u 2 ( ))) u 1 p c ( 1 + u 1 ρ 2 2 pr 1 (ρ 2 2) + u 2 ρ 2 2 pr 1 (ρ 2 2)) u1 u 2 pr2 u 1 pr3, where 1 = 1/r 1 + 1/r 2 + 1/r 3. The estimates above imply (5.9) provided we find s, r 1, r 2, r 3, and t in such a way that the following embeddings hold: H α L ps, L ps (ρ 2 1), L pr1 (ρ 2 2), L pr2, H 1 pr 3, L pt(ρ2 1), H 1 pt. These embeddings are true as long as α n max{ n ps, n ps (ρ 2 1), n pr 1 (ρ 2 2), n, 1 n, pr 2 pr 3 which can be achieved by the following choice: n + (1 θ) n + (1 α) n + (1 θ) s, t = n α n + (1 θ) r 1 = (ρ 2 2)(n α), r n + (1 θ) 2 =, r 3 = n α This concludes the proof for the case ρ 2 > 2. If ρ 2 = 2 then we have n pt(ρ 2 1), 1 n }, pt n + (1 θ), (α θ) ( ξ g(u 1 ) ξ g(u 2 )) u 1 p c u 1 u 2 pr u 1 pr. n + (1 θ) n + (1 α). Similar to the case ρ 2 > 2, it is sufficient to find s, r, and t in such a way that the following embeddings hold: H α L ps, L ps (ρ 2 1), L pr, H 1 pr, L pt(ρ 2 1), H 1 pt,

16 1502 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal and this is possible with the same choice of t and s as above and with r = n (θ 1) n α. Remark 5.3. Notice that if n = 1 and g satisfies (Hg) for some ρ 2 1 then (5.7) is trivially true for any α, θ,. We also have the following related result: Lemma 5.4. Let g satisfy (Hg+) for some ρ 2 2, let n and ε > 0. Then the Nemytskii operator G : B 1+ε (Γ) B 1 (Γ) : v g (, t, v( ) ) satisfies G(v 1 ) G(v 2 ) 1, C(ε) v 1 v 2 1+ε, ( 1 + v1 ρ ε, + v 2 ρ ε,). (5.10) Proof. The proof is very similar to that of Lemma 5.2 ii). We estimate G(u 1 ) G(u 2 ) 1, C ( g(, u 1 ( )) g(, u 2 ( )) + x g(, u 1 ( )) x g(, u 2 ( )) + ξ g(, u 1 ( )) u 1 ξ g(, u 2 ( )) u 2 ) = C(I 1 + I 2 + I 3 ). Using (1.3), (5.6) and Hölder s ineuality, we obtain I 1 + I 2 c (u 1 u 2 )(1 + u 1 ρ2 1 + u 2 ρ2 1 ) ( c u 1 u u1 ρ u 2 ρ ) 2 1, I 3 ( ξ g(, u 1 ( )) ξ g(, u 2 ( ))) u 1 + ξ g(, u 2 ( ))( u 1 u 2 ) c ( 1 + u 1 ρ u 2 ρ ) 2 2 u1 u 2 u 1 + c ( 1 + u 2 ρ 2 1 ) u1 u 2. Since B 1+ε (Γ) L (Γ) for all ε > 0, we obtain the result from the previous estimates on I 1, I 2 and I Case of initial data in F 1 = L (Ω) B 1 (Γ). In this case we use the scale X 1+η := X 1+η 1 = [E 1, F 1 ] η, X η := X η 1 = [F 1, G 1 ] η, 0 η 1. In Propositions 5.5 and 5.6 below, we shall verify (4.3) for the operators F 2 and F 1, respectively. Proposition 5.5. Let 1 < <, and let g satisfy (Hg) with ρ 2 ρ m 2 := n/(n ) if < n (no restriction on ρ 2 if n). Then the operator F 2 fulfills (4.3) with ρ 2i = 1 for i = 0, 1, and with ρ 22,γ 2,ε 2 as follows: i) If n and g satisfies (Hg+), then we can choose ρ 22 ρ 2, ρ 22 > 1, ε 2 = 0 +, γ 2 = 1 > ρ 22 ε 2.

17 parabolic problems with nonlinear dynamical boundary conditions 1503 ii) If n > > n/2 and g satisfies (Hg+), then we can choose [ 1 ], ρ 22 = ρ m 2, ε 2 = γ2 = ρ 22 ε 2 = 1. iii) If n/2 > 1, then we can choose [ 1 ], [ 1 ]. ρ 22 = ρ m 2, ε 2 = γ2 = ρ 22 ε 2 = Moreover, if < n and ρ 2 < ρ m 2 above and γ 2 = (ρ 22 ε 2 ) +. ρ m 2 ρ m 2 Proof. Due to Lemmas 3.1 and 3.2, we have then one can choose ρ 22 = (ρ m 2 ), ε 2 as X 1+ε 2 H 2ε 2 (Ω) B 1+ε 2 (Γ) and X γ 2 F (1+γ2 )/2 {0} B γ 2 (Γ) for any ε 2, γ 2 [0, 1]. Therefore, we just need to check the corresponding properties of the Nemytskii mapping G : B 1+ε 2 (Γ) B γ 2 (Γ) : v g (, t, v( ) ). i) If n, then Lemma 5.4 guarantees the reuired properties of G for any ε 2 < 1/ρ 22 and γ 2 (ε 2 ρ 22, 1). ii) iii) Let < n. Notice that ρ m 2 > 2 if > n/2. Choose ε 2 = [1/ρ m 2 ] if > n/2, ε 2 = [1/(ρ m 2 )] if n/2 and define γ 2 := ρ 22 ε 2 + n ( 1 ρ 22 ρ m 2 ). (5.11) We shall use Lemma 5.2 with θ = γ 2 and α = 1 + ε 2. This choice and (5.11) imply ρ 22 = n θ n α. Moreover, fixing ρ 22 = ρ m 2 or ρ 22 = (ρ m 2 ), we obtain γ 2 = ρ 22 ε 2 or γ 2 = (ρ 22 ε 2 ) +, respectively, and 0 (1 n + n/) θ < 1/ and 1/ < α < n/ if n/2, 1/ < θ 1 and 1 α < n/ if > n/2, so that Lemma 5.2 concludes the proof. Proposition 5.6. Let 1 < < and let f fulfill (Hf) with ρ 0, ρ 1, ρ 0, and ρ 1 satisfying (5.3) and ρ 0 = [ρ c 0 ()]. Then F 1 fulfills (4.3) with ρ 00 = ρ 0, ρ 01 = ρ 0, ρ 10 = ρ 1, ρ 11 = ρ 1, ρ 02 = ρ 12 = 1 and ε 0 = (1/ρ 0 ), ε 1 = (1/ρ 1 ), 1 > γ 0 = γ 1 > max i,j=0,1 {ε i + (ρ ij 1)ε j } = 1. More precisely, if η := ρ c 0 () ρ 0 is small and a > 0 is such that aρ c 0 () + 1/ρc 0 () (n/2, n/), then we may choose ε 0 = 1 ρ c aη, 0 () ε 1 = n + n ε 0 and γ 0 = γ 1 = 1 + 2ρ 0 ε 0 (ρ 0 1) n.

18 1504 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal Proof. First let us show that the choice of ε i, γ i in the proposition guarantees 1 > γ 0 > ε 0 ρ 0 = 1 if η > 0 is small enough. (5.12) Consider ρ 0, ε 0, and γ 0 as functions of η 0. Then ρ 0 = 1, ε 0 = a, γ 0 (0) = ρ 0 (0)ε 0 (0) = 1, γ 0 (0) = 2( aρ c 0 () + 1/ρc 0 ()) + n/ < 0, and (γ 0 ρ 0 ε 0 ) (0) = ( aρ c 0 () + 1/ρc 0 ()) + n/ > 0, which implies (5.12). This estimate and (5.3) imply also γ 0 > max i,j {ε i + (ρ ij 1)ε j } = 1. According to Lemmas 3.1 and 3.2, X 1+ε H 2ε (Ω) B 1+ε (Γ) and X γ F (1+γ)/2 [H 1 γ (Ω)] {0} =. H γ 1 (Ω) {0} if γ > 1/, so that (4.3) will follow for i = 0, 1 if we show ( 1 ) f i (t, u 1, u 2 ) γi 1, C u 1 u 2 2εi, 1 + u k ρ ij 1 2ε j,, i = 0, 1. j,k=0 These estimates follow from Lemma 5.1 with θ = γ 0 1 and α i = 2ε i. The assumptions in Lemma 5.1 are satisfied due to n/(n + ) < 2/ρ c 0 () < n/ and γ 0 = 1. Theorem 5.7. Let 1 < <, u 0 L (Ω), v 0 B 1 (Γ) and assume that conditions (Hf) and (Hg) are satisfied with ρ 0 < ρ c 0 (), ρ 1 < ρ c 1 (), ρ 0 < ρ c 0 (), ρ 1 < ρ c 1 (), (n )ρ 2 n. Assume that one of the following two conditions holds: i) > n/2 and g satisfies (Hg+). ii) 1 < ( n n). Then for any τ small enough there exists a uniue solution (u, v) C ( [0, τ], L (Ω) B 1 (Γ) ) C ( (0, τ], H 2θ (Ω) B 1+θ (Γ) ) of (4.4), where θ = 1 in case i) and θ = [1/] in case ii). γu(t) = v(t) for t > 0. Moreover, Proof. Increasing ρ 0, ρ 1, ρ 0, and ρ 1, if necessary, we may assume that (5.3) is true and ρ 0 = [ρ c 0 ()]. We apply Theorem 4.1 to prove the assertion. For this we need to check that hypothesis (H) holds. Notice that Propositions 5.5 and 5.6 determine the values of ε j and ρ ij in (4.2). In all the possible cases we have the following values: ε 0 = [1/ρ c 0 ()], ε 1 = [1/ρ c 1 ()], ρ 00 = [ρ c 0 ()], ρ 01 = [ ρ c 0 ()], ρ 10 = [ ρ c 1 ()], ρ 11 = [ρ c 1 ()], and ρ 02 = ρ 12 = ρ 20 = ρ 21 = 1. Propositions 5.5 and 5.6 guarantee that the first condition of (4.2) is satisfied. Therefore, we just need to check γ > ε. Since γ 0 and γ 1 are always close to 1, it is sufficient to fulfill γ 2 > ε.

19 parabolic problems with nonlinear dynamical boundary conditions 1505 i) If > n/2, then we have γ 2 = 1 > ε. ii) If 1 < n/2, then we have γ 2 = [1/], and therefore we need 1 { 1 max 1 } { n ρ c 0 (), ρ c 1 () = max n + 2, n + }, n + 2 which is euivalent to 1 < < ( n n). It remains to show the trace property γu(t) = v(t). Let z = [1/] +, (ϕ, ψ) H z (Ω) B z (Γ) and G(ϕ, ψ) := γϕ ψ. Then G L ( H z (Ω) Bz (Γ), L (Γ) ) and G(E γ ) = {0} for γ > 1/(2). Since e ta Z 0, e (t s)a F ( Z(s) ) E γ with γ > 1/(2) for 0 < s < t and the integral t 0 e (t s)a F ( Z(s) ) ds converges in H z(ω) Bz (Γ), the integral identity (4.4) implies G ( u(t), v(t) ) = 0 for t > 0. Remark 5.8. If 1 < n/2 and f is independent of u then the additional assumption ( n n) in Theorem 5.7 may be replaced by 1/ 1/ρ c 0 (); that is, (n 2) n. This follows from the proof of Theorem Case of initial data in M(Ω) M(Γ). Recall first that M(Ω) M(Γ) X1 1 δ () whenever δ > n, δ > 0, > 1 (see (3.5) and Lemma 3.2) and that we want to use the scale X η = X η δ 1 with δ > 0 small and close to 1. Let us start with an analogue to Proposition 5.6. In this case, we shall need to choose ( n δ, n 2 ). (5.13) 2(n + ) This choice is possible whenever n/ < n 2 /(2n+2), that is, 2 2 +(n 2) < 2n (which is true for close to 1). Proposition 5.9. Let > 1 be close to 1, let δ satisfy (5.13) and let f fulfill (Hf) with ρ 0, ρ 1, ρ 0, and ρ 1 satisfying (5.3) and ρ 0 = [ρ c 0 ()]. Then F 1 fulfills (4.3) with ρ 00 = ρ 0, ρ 01 = ρ 0, ρ 10 = ρ 1, ρ 11 = ρ 1, ρ 02 = ρ 12 = 1 and ε 0 = (1/ρ 0 ), ε 1 = (1/ρ 1 ), and 1 > γ 0 = γ 1 > max i,j=0,1 {ε i +(ρ ij 1)ε j } = 1. Proof. The proof is similar to that of Proposition 5.6. Exactly as in that proof we obtain γ 0 > max i,j {ε i + (ρ ij 1)ε j } = 1. According to Lemmas 3.1 and 3.2, X 1+ε H 2(ε δ) (Ω) B 1+ε δ (Γ) and X γ F (1+γ δ)/2 [H 1 γ+δ (Ω)] {0} H γ 1 (Ω) {0} provided γ > 1/,

20 1506 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal so that (4.3) will follow for i = 0, 1, if we show f i (t, u 1, u 2 ) γi 1, C u 1 u 2 2(εi δ), ( j,k=0 ) u k ρ ij 1 2(ε j δ),, i = 0, 1. These estimates follow from Lemma 5.1 with θ = γ 0 1, α 0 = 2(ε 0 δ) and α 1 = α 0 (n + )/n, because 2(ε 1 δ) > α 1. The assumptions on α 0 in Lemma 5.1 reuire n/(n + ) 2/ρ c 0 () 2δ < n/, which is guaranteed by (5.13). The other assumptions in Lemma 5.1 are obviously satisfied. Next we prove an analogue to Proposition 5.5. Proposition Let be close to 1 and δ close to n/ ( > 1, δ > n/ ). Let g satisfy (Hg) with ρ 2 < n/(n 1). Then the operator F 2 fulfills (4.3) with ρ 2i = 1 for i = 0, 1, ε 2 = [(n 1)/n] if n > 1, ε 2 = δ + if n = 1, ρ 22 = [1/(ε 2 )] and 1 > γ 2 > ρ 22 ε 2 = (1/). Proof. Due to Lemmata 3.1 and 3.2, we have X 1+ε 2 H 2(ε 2 δ) (Ω) B 1+ε 2 δ (Γ) and X γ 2 F (1+γ2 δ)/2 {0} B γ 2 δ (Γ) for any ε 2, γ 2 [δ, 1]. In order to verify the corresponding properties of the Nemytskii mapping G : B 1+ε 2 δ (Γ) B γ 2 (Γ) : v g (, t, v( ) ), it is sufficient to use Lemma 5.2 i) with θ = γ 2 δ and α = 1 + ε 2 δ (or Remark 5.3 if n = 1). If n > 1 and δ is sufficiently close to n/ then the assumptions of Lemma 5.2 i) are satisfied if ε 2 = [(n 1)/n], ρ 22 = n/((n 1)) and γ 2 := ρ 22 (ε 2 δ) + δ + n ( n ) 1 ρ 22. n Moreover, this choice guarantees also 1 > γ 2 > ε 2 ρ 22. Theorem Assume that the nonlinearities f and g satisfy (Hf) and (Hg) with exponents ρ 0, ρ 1, ρ 2, ρ 0 and ρ 1 satisfying ρ 0 < ρ c 0 (1), ρ 1 < ρ c 1 (1), ρ 0 < ρ c 0 (1), ρ 1 < ρ c 1 (1), and ρ 2 < ρ c 2 (1). Then there exist δ > 0 small and > 1, close to 1, such that M(Ω) M(Γ) X1 1 δ () and F 1, F 2 satisfy (H) with respect to the scale X η = X η δ 1 (). In particular, for every (u 0, v 0 ) M(Ω) M(Γ) there exists a uniue ε-regular solution (u, v) C([0, τ], X 1 ) C((0, τ], X 1+ γ ) for some γ = 1. Proof. The proof follows from Propositions 5.9, 5.10 and 4.1 similarly to the proof of Theorem Case of initial data in L #(Ω) L (Γ). Let 1 < <. As already announced, we shall use the scale X η = X η δ+ ξ, where δ > 0 is small and

21 parabolic problems with nonlinear dynamical boundary conditions ξ = 1 + 1/ + δ. Recall also that denoting # = n/(n + 1), (3.5) and Lemma 3.2 imply L #(Ω) L (Γ) H 1/ 1 (Ω) B 1/ (Γ) F ξ δ + /2 X 1 δ+ ξ. As in the preceding cases, we first verify the assumption (H). To simplify the proof, we shall work with the scale X η := X η δ ξ in the following Proposition Since the assumption (H) for the scale X η (with some γ i > γ i and ε i ) implies the same assumption for the scale X η (with γ i = γ i + and ε i = ε + i ), the conclusions of Proposition 5.12 remain true also for the original scale X η. Proposition Let > 1 and let δ > 0 be small enough. (i) Let f fulfill (Hf) with ρ 0, ρ 1, ρ 0, and ρ 1 satisfying (5.3) and ρ 0 = [ρ c 0 (# )]. Then F 1 fulfills (4.3) with ρ 00 = ρ 0, ρ 01 = ρ 0, ρ 10 = ρ 1, ρ 11 = ρ 1, ρ 02 = ρ 12 = 1 and some ε 0 = (1/ρ 0 ), ε 1 = (1/ρ 1 ) and 1 > γ 0 = γ 1 > max i,j=0,1 {ε i + (ρ ij 1)ε j } = 1. (ii) Let g satisfy (Hg) with ρ 2 = [ρ c 2 ()]. Then the operator F 2 fulfills (4.3) with ρ 2i = 1 for i = 0, 1, ρ 22 = ρ 2, ε 2 = [1/ρ 2 ], and 1 > γ 2 > ρ 22 ε 2 = 1. Proof. In the same way as in the proof of Proposition 5.6, we obtain γ 0 > max i,j {ε i + (ρ ij 1)ε j } = 1. According to Lemmas 3.1 and 3.2, X 1+ε = X 1+ε δ ξ X γ = X γ δ ξ H 1/+2ε 1 δ (Ω) B 1/+ε (Γ) H 2ε δ # (Ω) B 1/+ε (Γ), F ξ (1 γ+δ)/2 = F (1/+γ)/2 [H 2 1/ γ (Ω) B 2 1/ γ (Γ)] [H 1 γ ( # ) (Ω) B 2 1/ γ (Γ)]. = H γ 1 # (Ω) B 1/+γ 1 (Γ), for ε > δ and γ > δ 1/ #. (i) The above imbeddings show that (4.3) is true for i = 0, 1 if f i (t, u 1, u 2 ) γi 1, # C u 1 u 2 2εi δ, # (1 + 1 j,k=0 u k ρ ij 1 2ε j δ, # ), i = 0, 1. These estimates follow from Lemma 5.1 with θ = γ 0 1, α 0 = 2ε 0 δ, α 1 = α 0 (n + # )/n and replaced by # because 2ε 1 δ > α 1. The assumptions in Lemma 5.1 are obviously satisfied if δ is small enough. (ii) In order to verify the corresponding properties of the Nemytskii mapping G : B 1/+ε 2 (Γ) B 1/+γ 2 1 (Γ) : v g (, t, v( ) ), it is sufficient to use Lemma 5.2 i) with θ = 1/ + γ 2 1 and α = 1/ + ε 2, where ε 2 = [1/ρ 22 ],

22 1508 josé m. arrieta, pavol uittner, and aníbal rodríguez-bernal ρ 22 = [ρ c 2 ()] and γ 2 = ρ 22 ε 2 + n 1 ( ) ρ c 2 () ρ 22. Then the assumptions of Lemma 5.2 i) are satisfied and we are done. Theorem Let 1 < <, and let the nonlinearities f and g fulfill (Hf) and (Hg) with exponents ρ 0, ρ 1, ρ 2, ρ 0 and ρ 1 satisfying ρ 0 < ρ c 0 (# ), ρ 1 < ρ c 1 (# ), ρ 0 < ρ c 0 (# ), ρ 1 < ρ c 1 (# ), ρ 2 < ρ c 2 (), where # = n n+ 1. If u 0 L #(Ω) and v 0 L (Γ), then there exists a ) ( 2 (δ C (0, τ], X + ) + ) of uniue ε-regular solution (u, v) C ( [0, τ], X 1 δ+ ξ (4.4), where 2ξ = 1 + 1/ + δ and δ > 0 is small. Proof. The proof is a direct conseuence of Proposition 5.12 and Proposition Elliptic euation with dynamic boundary conditions In this section we consider the following problem with dynamic boundary conditions: Au = 0, x Ω, t > 0, (γu) t + Bu = g(x, t, γu), x Γ, t > 0, (6.1) u(x, 0) = u 0 (x), x Ω, (γu)(x, 0) = v 0 (x), x Γ. In this case one can reduce (6.1) to an evolution problem on Γ as follows. As in Section 2, consider the operator T L ( B(Γ), s H s (Ω) ) for s > 1/. Recall that T g, g B(Γ), s is defined as the uniue (generalized) solution of the problem } Au = 0 in Ω, u = g on Γ. Then, define the operator, which we still denote B, since no confusion seems likely, as (T u) Bu = ν. Note that for s > 1/, B is well defined from B 1+s (Γ) into B(Γ). s Therefore (6.1) can be written as u t + Bu = g(x, t, u). It follows from the results in [9], after a suitable interpolation and extrapolation process, that B generates an analytic semigroup on each space of the scale E α = B α(γ), α R, with domain Eα+1 ; see also [13] and [15] for a ξ

23 parabolic problems with nonlinear dynamical boundary conditions 1509 Hilbertian setting. On the other hand, we assume that the nonlinear term g is as in (1.3). Since in this case we are dealing with just one nonlinear term and a suitable scale of spaces, the techniues in [5] can be applied to obtain the following results. Theorem 6.1. Let 1 < <, and let g satisfy (Hg) with ρ 2 ρ c 2 (). Then (6.1) is well-posed in B 1/ (Γ). Proof. In this case we take the scale of spaces X η = E 1/ 1+η, so that X 1 = B 1/ (Γ). Using Lemma 5.2, ρ 2 ρ c 2 () and some elementary computations we see that the Nemytskii operator G is an ε-regular map relative to the scale X η (see [5] or [6] for the corresponding definition). Applying the results of [5] or [6] we obtain the result. Remark 6.2. If ρ 2 < ρ c 2 (), then for each u 0 L (Ω) there exists a uniue solution of (6.1). This is true since L (Γ) B 1/ δ (Γ) for any small δ > 0 and we can consider the scale X η = E 1/ 1 δ+η for small δ > 0. The condition ρ 2 < ρ c 2 () guarantees that G is an ε-regular map relative to the scale X η, and this permits us to apply the abstract result of [5] or [6]. Theorem 6.3. Assume that the nonlinearity g satisfies (Hg) with ρ 2 < ρ c 2 (1). Then for each initial function u 0 M(Γ), there exists a uniue solution of (6.1). Proof. We follow a similar reasoning as that of [6]. For > 1 and s 0 > n/ we have M(Γ) B 1 s 0 (Γ). Moreover, if we choose close to 1 and s 0 > n/ close to 0, and if we let X 1+η = B 1 s 0+η (Γ), we can prove, using Lemma 5.2, that G is an ε-regular map relative to this scale of spaces as long as ρ 2 < ρ c 2 (1). Applying the results of [5] or [6] we obtain the assertion. References [1] H. Amann, Linear and Quasilinear Parabolic Problems, Volume I: Abstract Linear Theory, Birkhäuser, [2] H. Amann, Semigroups and nonlinear evolution euations, Linear Algebra Appl., 84 (1986), [3] H. Amann, Parabolic evolution euations and nonlinear boundary conditions, J. Differ. Euations, 72 (1988), [4] H. Amann and J. Escher, Strongly continuous dual semigroups,, Ann. Mat. Pura Appl. CLXXI (1996), [5] J.M. Arrieta and A.N. Carvalho, Abstract parabolic problems with critical nonlinearities and applications to Navier-Stokes and heat euations, Trans. Amer. Math. Soc., 352 (2000),

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