The Cauchy problem for ut=Δu+|∇u|q

Introduction

Surfaces, such as those of a crystal, chemical deposit, flame, or tumor, may grow due to a variety of mechanisms. One such mechanism is ballistic deposition. A simple-minded picture of this mechanism is that of particles each moving along a straight path approaching a surface and randomly attaching themselves. This point of view is considered appropriate for vapor deposition and the sputter deposition of thin films of aluminium and rare earth metals. The accepted starting point for a continuum approach to this mechanism is the partial differential equation

With this model, surface growth relative to a reference plane, which may move with a constant velocity, is simulated. The unknown h denotes the height of the surface above the plane and t denotes time. The first term on the right-hand side of Eq. (1) describes a diffusive relaxation in which ν may be thought of as an effective surface tension. This can be effectuated by evaporation-condensation processes or gravity-induced restructuring and exhibits a smoothing effect on protrusions on the surface. The second term on the righthand side of (1) arises from the growth process. It is nonlinear and can be derived as the lowest-order nonvanishing component in a gradient expansion. The constant λ in this term is a measure of the net rate of deposition. The last term on the right-hand side of Eq. 1is produced by a stochastic force with zero mean and short-range correlations. In its simplest form η signifies a white noise with a Gaussian distribution dependent on the coordinates of the reference plane and time. The above model was first proposed by Kardar, Parisi, and Zhang [23], and has since been referred to as the KPZ equation [21,25,34].

To account for further surface growth effects, Krug and Spohn [24] advocated extending the model (1) to

where q 1 is a constant. Some properties of solutions of this equation have been studied numerically in [1,29]. With any value q > 0, Eq. (2) has been designated the generalized KPZ equation. In particular, without the noise term η, it is known as the generalized deterministic KPZ equation [21,25,34]. In this paper we study the Cauchy problem for the generalized deterministic KPZ equation. We denote the unknown by u, and, by scaling the variables, normalize the parameters ν and λ to unity. Also, we do not restrict the number of spatial dimensions. Thus we consider the following problem. Let S T denote the strip S T := R n × (0, T ] for some 0 < T < ∞.

Then solve the equation

subject to the condition

where the operators on the right-hand side of (3) denote the standard differential operators with respect to the spatial variable x = (x 1 , x 2 , . . . , x n ) ∈ R n , q is a positive real number, and u 0 is a given real function on R n . The hypothesis on the initial data function u 0 which we shall assume throughout this paper is that u 0 ∈ C(R n ) ∩ L ∞ (R n ). Furthermore, by a solution of problem (3)- (4) we understand a bounded classical solution. In other words, we mean a function u ∈ C 2,1 (S T ) ∩ C(S T ) ∩ L ∞ (S T ) which satisfies Eq. (3) at every point (x, t) ∈ S T and (4) for every x ∈ R n . Under these premises, we shall prove that problem (3)-(4) admits a unique solution for all T > 0 and determine some regularity properties of this solution. In passing, we document a number of cases in which the problem may be solved explicitly.

An equation of the type (3) is often referred to as a viscous Hamilton-Jacobi equation and problem (3)-(4) has attracted much interest [2,[4][5][6][8][9][10][11][12][13][14][15][16][17][18]27,28,30] in recent years. In the earliest of these papers [12], the existence and uniqueness of a classical solution was proven under the assumption that q = 1 and u 0 ∈ C 3 0 (R n ). These results formed the basis for the derivation of a number of explicit and implicit formulae for the solution of the problem when q = 1 and u 0 is a bounded measure with particular monotonicity and symmetry properties [10,11]. In [4] the existence of a suitably-defined weak solution when 1 q < 2 and u 0 is a Radon measure was investigated. In [2], the existence of a unique classical solution was proven under the hypothesis that q > 1 and u 0 ∈

. More recently, attention has focused on the questions of existence and uniqueness of suitablydefined weak solutions when q 1, and u 0 ∈ L p (R n ) for some 1 p < ∞ or u 0 is a bounded measure [5,14,16,17], and on determining the large-time behavior of such solutions [2,5,[8][9][10][11][12]14,15,18,28,30].

The existence, uniqueness, and regularity results in the present paper distinguish themselves from their predecessors in that they apply to all q > 0 and that they do not assume that the initial-data function is in any sense differentiable nor that it lies in an L p (R n )space with p < ∞. They are obtained via a priori estimates on the gradient of a solution which extend and improve similar estimates in the papers already mentioned.

The plan of the remainder of this paper is as follows. In the next section we treat the case q = 2 which is exceptional in that problem (3)-(4) can be solved explicitly. Thereafter, in Section 3, we establish a priori estimates on the gradient of a solution of problem (3)-(4) which are the key to all our results, and prove our main existence theorem and a corollary. In the subsequent section, we prove that the solution whose existence we have established is necessarily the only solution of the problem. This we do by means of a comparison principle whose proof makes use of the a priori estimates on the gradient of the solution derived in the previous section. In Section 4, we also state and prove four useful corollaries of the comparison principle. These place us in a position to analyze another case, n = q = 1, in which problem (3)-(4) can be solved explicitly, in Section 5. In the final section we prove our main regularity result.

Ironically, many of the results of the present paper have already been publicized elsewhere [3,[7][8][9]13,15,17,18,28,32,33] citing the preprint: "The Cauchy problem for u t = ∆u + |∇u| q " or "The Cauchy problem for the KPZ equation," Bulletin Laboratoire Amiénois de Mathématiques Fondamentales et Appliquées 23, Université de Picardie 'Jules Verne,' Amiens, France, 1998. The present paper is a revision of part of this preprint. The remainder is currently being prepared for publication as [20]. Results contained in the present paper which have been mentioned explicitly in one or more of the references [8,9,17] include Theorems 2, 4, 5, 7, and 8, and Lemmas 7 and 8 of the preprint. These results may be found in the present paper as: the primary conclusion of Theorem 2, Theorem 6, the remaining conclusions of Theorem 2, Theorem 3, Corollary 4.2, Theorem 4, and Corollary 4.4, respectively. A few of them have been sharpened.

The case q = 2

When q = 2, corresponding to the deterministic KPZ equation as originally proposed by Kardar et al. [23], problem (3)-(4) may be solved explicitly. This stems from the observation [23] that if u is a solution of Eq. (3) with q = 2 then w := e u satisfies the linear heat equation

In the case of one spatial dimension, noting that when u solves (3) then v := ∇u satisfies the Burgers equation, this transformation is equivalent to the renowned Cole-Hopf transformation. By the well-known Poisson formula for the solution of the Cauchy problem for the heat equation, this leads to the following result.

Theorem 1. Suppose that q

Then problem (3)-(4) has a unique bounded solution which is given by

Existence

To establish the existence of a solution of problem (3)-(4) in the general case, we shall use a regularization process which allows us to fall back on the established theory of quasilinear parabolic equations [26]. This process comprises two elements. One is the regularization of the initial data. The other is the regularization of the nonlinear term in (3) in two steps. This regularization is designed to cope with two primary hurdles in applying the standard theory to Eq. (3). The first is the growth of the nonlinear term as |∇u| → ∞ in the case q > 2. The second is the lack of smoothness in the case q < 2.

Before proceeding with our existence proof, we clarify some notation which will be employed frequently throughout the remainder of this paper. We set

We let

where

We denote by C p,p/2 (Q) the space of functions f defined in Q • , such that the partial derivatives

for p > 1, where

for 0 < p 1, is finite. Note that in the name and norm of these spaces p appears as a subscript. This is to avoid confusion with C p (Ω), where p is a positive integer and Ω may be any subset of R n with interior Ω • such that Ω ⊆ Ω • , and with C p,p/2 (Q), where Q may be any set points (x, t) ∈ R n+1 with interior Q • such that Q ⊆ Q • , respectively. By C p (Ω) we mean the space of real functions defined in Ω • for which every partial derivative of order less than or equal to p exists in Ω • and is continuously extendible to Ω; while, by C p,p/2 (Q) we denote the space of functions f defined in Q • such that

. . ∂x k n n ∂t l exists in Q • and is continuously extendible to Q for all k 1 , k 2 , . . . , k n 0, l 0, and

Also, since we shall frequently use comparison principle arguments, at this stage it will be convenient to introduce a lemma which covers our needs in this respect. The lemma below is an abridged version of a result of Kaplan [22] which has the particular merit that it permits the comparison of subsolutions and supersolutions of nonlinear parabolic equations on unbounded spatial domains.

Lemma 1. Let N be the nonlinear parabolic operator

where f is uniformly continuous and uniformly Lipschitz continuous with respect to u in S T × (−r, r) × B r for every r > 0. Suppose that u (1) and u (2) (1) (x, 0) u (2) (x, 0) for all x ∈ R n , there holds u (1) (x, t) u (2)

Using a standard mollifying process, we can construct a sequence of functions

and

for all x ∈ R n and k 1, and u 0,k → u 0 as k → ∞ uniformly on compact subsets of R n . Set

where 0 < ε < 1. Note that there exist a number 0 < δ < 1 which depends only on q such that F ε ∈ C 2+δ (B ρ ) and

for all ξ 1 , ξ 2 ∈ B ρ , ρ > 0, and 0 < ε < 1. Furthermore, setting

there holds

and

Consider the problem of solving the equation

subject to the initial condition

This problem has a solution u k,ε ∈ C 2+α,1+α/2 (S T ) for some 0 < α < 1 by Theorem V.8.1 of the monograph [26]. Furthermore, writing (15) as

where f (x, t) := (F k,ε (∇u k,ε ))(x, t), and noting that f ∈ C 1+α,(1+α)/2 (S T ), by Theorem IV.5.2 of [26] it follows that u k,ε ∈ C 3+α,(3+α)/2 (S T ).

We shall now obtain estimates of u k,ε which are independent of the behavior of F k,ε (ξ ) for large |ξ |. The first is

where

Because any function of the form κ + F k,ε (0)t with κ a real number is a smooth solution of (15) and 8holds, using Lemma 1 we deduce that m

Since 14holds, this readily yields (18). Our second estimate is

To obtain this, set w := |∇u k,ε | 2 (x, t), and compute that w satisfies

where L is the linear parabolic operator

with (19). From (19) and (14) it follows that u k,ε actually satisfies the equation

This leads to the following deduction.

Lemma 2. There exists 0 < γ < 1, which depends only on n, q, and u 0,k , such that problem (21), (16) has a solution u k,ε ∈ C 4+γ ,2+γ /2 (S T ) satisfying (18) and (19). Furthermore, the norm of u k,ε in C 2+γ ,1+γ /2 (S T ) is bounded above by a number which depends only on n, q, and u 0,k .

Proof. Excepting the regularity, the assertions of the lemma have been established above.

To confirm the regularity, we note that from (10) it can be deduced that |F ε (ξ )| (1 + |ξ | 2 ) q/2 for all ξ ∈ R n and 0 < ε < 1. So, if we look at Eq. (21) in the form (17) with

using (19) we have the estimate |f (x, t)| (1 + ρ 2 k ) q/2 for all (x, t) ∈ S T . By Theorem V.3.1 of [26] applied to problem (17), (16) this implies that every partial derivative ∂u k,ε /∂x i ∈ C β,β/2 (S T ) for i = 1, 2, . . . , n with a constant 0 < β < 1 which depends only on n, q, and u 0,k . Furthermore, the norm of these functions in this space have a bound which depends only on the same factors as β itself. Thus, recalling (11), the function f given by (22) is an element of C γ ,γ /2 (S T ), where γ = min{q, 1}β and the norm of f in this space is bounded above by a number which depends only on n, q, and u 0,k . Consequently, applying Theorem IV.4.2 of [26] to problem (17), (16) we have u k,ε ∈ C 2+γ ,1+γ /2 (S T ) with an upper bound for the norm of u k,ε in this space which does not depend on ε or T . We now use a standard bootstrap argument. The last conclusion means that the function f defined by (22) is a member of C 1+γ ,(1+γ )/2 (S T ). Therefore, u k,ε ∈ C 3+γ ,(3+γ )/2 (S T ) by Theorem IV.5.1 of [26] applied to problem (17), (16). This in turn implies that f given by (22) is a member of C 2+γ ,1+γ /2 (S T ). Whence, by a second application of the lastmentioned theorem to problem (17), (16), we conclude that u k,ε ∈ C 4+γ ,2+γ /2 (S T ). ✷ Our next step is to let ε → 0. Before we do this though we need to obtain estimates on the solution of problem (21), (16) which will allow us to take the final action of letting k → ∞. We use a variant of the Bernstein technique. We define the function φ :

and let

and

and

if q = 1, where, for i = 1, 2,

Proof. Define

where θ is a positive

function. Let N denote the semilinear parabolic operator

where

and H is given by (12). By computation it can be verified that

Take now

Recalling (13), this means that

Hence, for any number η > 0 there holds

Moreover, if we choose η small enough, we have

Thus, we obtain (26). To prove (27) we take

Hence, for any η > 0, we have N {(t + η) −2/q exp(2ε q Υ 2 t)} 0 in S T . So, choosing η small enough once more, and arguing as above, it can be ascertained that w(x, t) (t + η) −2/q exp(2ε q Υ 2 t) for all (x, t) ∈ S T . This yields (27). ✷

With the above lemma in hand, we may carry out the step of letting ε ↓ 0. For fixed ξ ∈ R n observe that the expression F ε defined by (10) is a nondecreasing function of ε.

for every (x, t) ∈ S T . Furthermore, by the bounds in Lemma 2, there holds u k ∈ C 2+γ ,1+γ /2 (S T ) for some 0 < γ < 1 and u k,ε → u k in C 2+ν,1+ν/2 (B r × [0, T ]) for all r > 0 and 0 < ν < γ . We may subsequently also let ε ↓ 0 in (21) to deduce that u k satisfies Eq. (3) in S T . Moreover, by Lemma 1 this is the only solution in C 2,1 (S T ) ∩ L ∞ (S T ) satisfying (16). We capture these conclusions in the following

|∇u

|∇u

and

for every (x, t) ∈ S T . Furthermore, given any 0 < τ < T there exists 0 < δ < min{q, 1} which depends only on n, q, m, M, and τ such that u k ∈ C 2+δ,1+δ/2 (S τ,T ) and the norm of u k in this space is bounded above by a number which depends only on the same factors as δ.

Proof. The existence and uniqueness of u k was established above. While the estimates (30), (31), (32), and (33) follow immediately from (18), (19), (26), and (27), respectively. To verify the remaining assertions of the lemma, we use the fact that u k satisfies (17) with

By (32) we know that f ∈ L ∞ (S T \ S τ ) with a bound for the norm which depends only on q, M − m, and τ . Hence by Theorem V.3.1 of [26] there is a number 0 < β < 1 which depends only on n, q, m, M, and τ such that every partial derivative ∂u k /∂x i for i = 1, 2, . . ., n is an element of C β,β/2 (S τ,T ) with a bound for the norm in this space which depends only on the same factors as β. This implies that f ∈ C δ,δ/2 (S τ,T ), where δ = min{q, 1}β with a similar bound for the norm. The open assertions finally follow from Theorem IV.10.1 of [26]. ✷

We are now in a position to carry out the final step of the construction of a solution of problem (3)-(4). This is our existence result.

Theorem 2. Suppose that q >

and classically. To confirm the theorem, it subsequently remains to show that u is continuous down to t = 0 and satisfies the initial condition (4). For this purpose, we fix x 0 ∈ R n and observe that for any k 1 there holds lim sup

Hence in the limit k → ∞,

On the other hand, by Lemma 1, there holds u k (x, t) u * k (x, t) for all (x, t) ∈ S T and k 1, where u * k denotes the unique bounded solution of the linear heat equation, i.e., (5), satisfying (16). Concurrently, letting u * denote the unique bounded solution of (5) satisfying (4), using the Poisson formula it can be verified that

In view of the arbitrariness of x 0 , combining (37) and (38) demonstrates that u ∈ C(S T ) and (4) holds. ✷

Remark. In the case q = 2, estimates (35) and (36) can be sharpened to This result may be proven along the lines of (35) and (36), by noting that in the case q = 2 the terms c and d in the proof of Lemma 3 can be amalgamated, and, providentially, the function H defined by (12) is zero. Subsequently using (40) one obtains c + d = −1 and h = 0 in lieu of (28) and (29).

For q > 1, estimates of the type (35) and the specific estimate (36) have been obtained independently by almost the same approach in [5]. Related results assuming q 1 and u 0 ∈ L 1 (R n ) or that u 0 (x) has an explicit decay rate as |x| → ∞ can be found in [5,6,15,30]. The present a priori estimates (35), (36), and (39) extend these results to every q > 0 and sharpen those of the type (35) for q > 1.

In the cases q = 2 and q = 1 the exponent of t in inequalities (35), (36), and (39) can be seen to be sharp. To be specific, when q = 2 the function u(x, t) := ln e M + e m 2 + e M − e m 2 erf

where erf denotes the error function, can be verified to be an explicit solution of Eq. 3with appropriate initial data satisfying (7) for any τ > 0. Similarly,

represents an explicit solution of (3) with q = 1. From both examples it can be determined that for large t, the exponent of t in (35), (36), and (39), i.e., 1/2, cannot be improved upon. In particular, the second example demonstrates that when q = 1 an estimate of the type (36), i.e., |∇u|(x, t) θ(u(x, t); m, M)t −1 for some bounded function θ , cannot hold in general. Note that by employing (35) and (36), or, by integrating Eq. (3) over B r × (τ, T ), applying the divergence theorem, and, subjecting the resulting boundary integrals to (34)-(36), one can obtain diverse upper bounds for the integral of |∇u| q in B r × (τ, T ) for every r > 0 and 0 < τ < T . Similarly, employing (35) and (36), or, by first multiplying Eq. 3by u − m, applying Green's theorem, treating the boundary integrals as described above, and, using the integral estimates already obtained, one can deduce upper bounds for the integral of |∇u| 2 in B r × (τ, T ). The pick of these bounds yields the following

In particular, if q > 2 the integral of |∇u| q in B r × (0, T ) is bounded above by a number which does not depend on T , if 2 > q > 1 the same holds for |∇u| 2 , and if q < 1 it holds for |∇u| q + |∇u| 2 .

Uniqueness

We shall use the estimates on the gradient of the constructed solution of problem (3)-(4) to establish the uniqueness of the solution of this problem. Lemma 1 is unfortunately inadequate for our purposes because in general the regularity required for this lemma is absent.

Theorem 3. Suppose that q > 0 and u

This theorem can easily seen to be a consequence of Theorem 2 and the following comparison principle.

and

there holds

Proof. We trace the proof of the comparison principle in [22] and modify it appropriately. The strategy is the following. For ε > 0, consider the function Moreover, if the dependence of A on ε is such that

we may let ε ↓ 0 in (47) to obtain (44). Our goal is thus to show that for any ε > 0 we can find A > 0 such that (46) holds at any point (x, t) ∈ Q, where (∇z)(x, t) = 0, and such that (48) holds. Direct computation shows that

where

To proceed, we distinguish between the cases q > 1 and q 1. When q > 1, by the mean value theorem,

Hence, by (36),

where

Subsequently, substituting (45) and (52) in (51), we obtain

It follows that the desired inequality (46) holds when A := 2qε{nT (q−1)/q + q(K + εT 1/q ) q−1 }. This also fulfills (48). On the other hand, when q 1,

Subsequently, substituting (45) and (53) in (51), in this case we can achieve our goal if we choose A := 2(nε + ε q ). ✷

As corollaries of Theorem 4 we have the following

Therefore, if (54) holds with equality at some point (x 0 , t 0 ) ∈ S T , then (x 0 , t 0 ) must be a maximum for z with z(x 0 , t 0 ) = 0. The maximum principle (see, for instance, Theorem 5 of Chapter 3 of [31]) subsequently implies that z ≡ 0 in S t 0 . Hence, by (55), |∇u| ≡ 0 in S t 0 . It follows that if (54) holds with equality at some point (x 0 , t 0 ) ∈ S T , necessarily u 0 is constant. ✷ (iv) u 0 (x) = u 0 (y) for all x, y ∈ R n with x 1 = y 1 + a 1 , and σ u 0 (x) σ u 0 (y) for all x, y ∈ R n with |x

Proof. (i) The hypothesis means that u 0 is periodic with period a. The uniqueness of the solution of problem (3)-(4) implies that u(•, t) must also be periodic with the same period. (ii) Let z ∈ R n be such that z 1 0. Defineũ 0 (x) := u 0 (x + z) for all x ∈ R n . Let u denote the solution of problem (3)-(4) with initial dataũ 0 . By the uniqueness of the solution of this problem established in Theorem 3,ũ(x, t) = u(x + z, t) for all (x, t) ∈ S T . Simultaneously, since σ u 0 (x) σũ 0 (x) for all x ∈ R n by the hypothesis on u 0 , Theorem 4 implies that σ u(x, t) σũ(x, t) for all (x, t) ∈ S T . This gives the result.

(iii) The hypothesis is such that u 0 is radially symmetric. Consequently, the uniqueness of solutions of problem (3)-(4) implies that u(•, t) must also be radially symmetric for every 0 < t T . It therefore remains to show that σ u(x, t) increases as |x| increases for every 0 < t T . For this purpose fix 0 < τ T and inf{u(x, τ ): x ∈ R n } < µ < sup{u(x, τ ): x ∈ R n }. For 0 < t τ define ζ(t) := sup{|x|: σ u(x, t) < σ µ} and Q := {(x, t) ∈ S τ : |x| < ζ(t)}. By construction, Q fits the hypothesis on the domain in Theorem 4. Setting Γ := Q \ Q, in the light of the continuity of u there holds u(x, t) = µ for all (x, t) ∈ Γ ∩ S τ . Whereas, σ u 0 (x) σ µ for all (x, 0) ∈ Γ . Theorem 4 subsequently implies that σ u(x, t) σ µ for all (x, t) ∈ Q. Hence, σ u(x, τ ) σ µ for all |x| ζ(τ ). This gives the result in view of the arbitrariness of µ and τ .

(iv) By the uniqueness of the solution of problem (3)-(4) it can be verified that u(x, t) = u(y, t) for all (x, t), (y, t) ∈ S T such that |x 1 | and |y 1 | differ by an integer multiple of a 1 . Consequently, it remains to prove that σ u(x, t) increases as x 1 increases for 0 x 1 a 1 /2 for every 0 < t T . Picking τ and µ as in the proof of part (iii), defining ζ(t) := sup{x 1 ∈ (0, a 1 /2): σ u(x, t) < σ µ}, Q := {(x, t) ∈ S τ : |x 1 | < ζ(t)}, and Γ := Q \ Q, by design there holds σ u(x, t) σ µ for all (x, t) ∈ Γ ∩ S τ , while σ u 0 (x) σ µ for all (x, 0) ∈ Γ . The proof is subsequently similar to that of the previous part. ✷

The case n = q = 1

In the case n = q = 1, identifying x 1 with x, Eq. (3) may be viewed as the equation

It follows that if for a solution u the sign of the derivative ∂u/∂x is known a priori, the equation is effectively a linear equation with piecewise constant coefficients. In this section, we shall capitalize on this observation to document a number of situations in which Eq. (3) with n = q = 1 can be solved explicitly, whereby the requisite a priori information is provided by Corollary 4.4.

Theorem 5. Suppose that n = q = 1 and u 0 ∈ C(R) ∩ L ∞ (R). Let G be defined by (6).

When complemented with the initial condition

this constitutes a problem for the linear heat equation in one spatial dimension which can be solved using standard techniques [19]. Explicitly computing the solution of problem (61)-(63) with Laplace transformation [19], and, retracing the above argument, the requisite formulae can be obtained after some manipulation. ✷ Part (ii) has been previously proven using probabilistic arguments in [10]. Moreover, in [11] these arguments have been extended to obtain an implicit probabilistic representation of the solution when n 2 and u 0 (x) u 0 (y) for all x, y ∈ R n with |x| |y|.

Regularity

In this final section we determine some additional smoothness properties of the solution constructed in Section 3. We set Moreover, if u 0 is uniformly continuous in R n then u is uniformly continuous in S T independently of the magnitude of T . In particular, if u 0 ∈ C α (R n ) for some α ∈ P \{2, 3, 4, . . .} then u ∈ C β,β/2 (S T ), where

and the norm of u in this space has an upper bound which is independent of T . Finally, if u 0 ∈ C 1 (R n ) there holds

with strict inequality if u 0 is not identically constant.

For the specific case q = 1, conclusion (64) has been noted in [14]. Inequality (66) without any mention of strictness has been proven assuming q = 1 and u 0 ∈ C 3 0 (R n ) in [12], and assuming q > 1 and u 0 ∈ C 2 (R n ) ∩ W 2,∞ (R n ) in [2].

We shall prove Theorem 6 in stages. First we prove the assertions which are independent of the regularity of u 0 . Next we confirm that (66) holds when u 0 ∈ C 1 (R n ) and that u ∈ C α,α/2 (S T ) as stated when u 0 ∈ C α (R n ) for a noninteger α > 1 in P. Thirdly, armed with (66), we show that u ∈ C 1,1/2 (S T ) with an upper bound on the norm which does not depend on T . Thereafter, we prove the remaining assertions regarding the continuity of u. Finally, we demonstrate that (66) is strict if |∇u 0 | ≡ 0.

For the first step in the proof of Theorem 6, we employ a standard boot-strap argument, considering Eq. (3) in the form (17) with f (x, t) := |∇u| q (x, t).

Recalling the construction of u as the limit of the sequence {u k } ∞ k=1 in Section 3, the last statement of Lemma 4 implies that given any 0 < τ < T there holds u ∈ C 2+δ,1+δ/2 (S τ,T ) for some 0 < δ < min{q, 1} which depends only on n, q, m, M, and τ . Furthermore, the norm of u in this space has an upper bound which depends only on the same factors as δ.

However, supposing that u ∈ C i+δ,(i+δ)/2 (S τ,T ) for some 0 < δ < 1 and integer i 1, every component of ∇u will be a member of C i−1+δ,(i−1+δ)/2 (S τ,T ). This means that f ∈ C ν,ν/2 (S τ,T ) for all 0 < ν i − 1 + δ if q is an even integer, and 0 < ν min{i − 1 + δ, q} otherwise. Whereupon, by Theorem IV.10.1 of [26], u ∈ C ν+2,(ν+2)/2 (S τ,T ) for all such noninteger ν. Thus, if u ∈ C i+δ,(i+δ)/2 (S τ,T ) for some 0 < δ < 1 and integer i 1 we deduce that u ∈ C p,p/2 (S τ,T ) for all p ∈ P with 0 < p i + 1 + δ. Moreover, throughout this induction process, it can be checked that the norm of u in C p,p/2 (S τ,T ) can be bounded above by a number which depends only on n, q, m, M, p, and τ . This provides (64). The proof of (65) is similar. Since s → s q is smooth for s > 0, given any set of the form Q := {(x, t) ∈ S t 0 ,t 1 : |x − x 0 | < r} with x 0 ∈ R n , r > 0, and 0 < t 0 < t 1 T , such that Q ⊂ Q, one deduces u ∈ C p,p/2 (Q ) for every p > 0.

If now u 0 ∈ C α (R n ) for some α 1, the sequence of functions {u 0,k } ∞ k=1 introduced in Section 3 can be so chosen that u 0,k ∈ C α (R n ) whereby the norm of u 0,k in C α (R n ) is less than or equal to the norm of u 0 in the same space, and ρ k is less than or equal to the L ∞ (R n )-norm of |∇u 0 | for every k 1. Theorem V.3.1 of [26] then implies that u k ∈ C 1+δ,(1+δ)/2 (S T ) for every 0 < δ < min{1, α − 1} with an upper bound for the norm which is independent of k and T . Passage to the limit k → ∞ in the light of (31) yields (66) and that u ∈ C 1+δ,(1+δ)/2 (S T ) for every 0 < δ < min{1, α − 1} with an upper bound on the norm which does not depend on T . The conclusion that when u 0 ∈ C α (R n ) for any noninteger α > 1 in P then u ∈ C α,α/2 (S T ) with an upper bound for the norm which is independent of T may be deduced hereafter by the bootstrap argument described in the previous paragraph.

The above completes the first two stages of the proof of Theorem 6. The third is an obvious corollary of (66) and the lemma below. for all x 0 ∈ R n and 0 t 1 < t 2 T , where c n is a positive number which depends only on n.

Proof. Without loss of generality we may take x 0 = 0 and t 1 > 0. We integrate Eq. (3) over B r × (t 1 , t 2 ) for r > 0 and apply the divergence theorem. This yields

Next, we observe that by a standard mollifying process, given any ι > 0 there exists a functionû 0 ∈ C 2 (R n ) such that |û 0 (x) − u 0 (x)| ω(ι) and |∇û 0 |(x) C n ω(ι)/ι for all x ∈ R n , where C n is a number which depends only on n. Letû denote the solution of problem (3)-(4) with initial dataû 0 . Applying (66) and Lemma 5 toû, and Corollary 4.3 to u andû, there holds

for all x ∈ R n and 0 t 1 < t 2 T . Substituting ι = |t 1 −t 2 | 1/q in the above inequality gives a uniform continuity estimate for u with respect to t when q 2, while ι = |t 1 − t 2 | 1/2 gives a similar estimate when q < 2. Complementing these estimates with (67) proves that u is uniformly continuous in S T . In the event that u 0 is uniformly Hölder continuous with exponent 0 < α < 1 in R n , one may suppose that ω(h) = Kh α for some K > 0, whereupon, by choosing ι = |t 1 − t 2 | ν with ν = 1/ max{2, α + q(1 − α)} in the above argument, one obtains the outstanding continuity result.

To complete the proof of Theorem 6 it remains to show that if |∇u| ρ in S T and |∇u|(x 0 , t 0 ) = ρ at some point (x 0 , t 0 ) ∈ S T then ρ = 0. Suppose therefore that this is not true. In this case, necessarily (x 0 , t 0 ) is a maximum for the function z := |∇u| 2 in the set Q given by (65). On the other hand, cf. Section 3,

where b := q|∇u| q−2 ∇u. The maximum principle (for instance, Theorem 5 of Chapter 3 of [31]) subsequently states that z ≡ ρ 2 in the set Q consisting of all those points (x, t) ∈ Q which are connected to (x 0 , t 0 ) by a path lying entirely in Q and along which the t-component is nondecreasing from (x, t) to (x 0 , t 0 ). However, by the continuity of z, this necessitates Q = S t 0 . Whence, using (68), we deduce that every second-order partial derivative of u with respect to the components of x vanishes in S t 0 . So x → u(x, t) is an affine function in R n for every 0 < t t 0 . However, because u is bounded, this means that x → u(x, t) is constant in R n for every such t. Thus, by a contradiction argument, the proof of Theorem 6 has been completed.