UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS TO DISCRETE MORSE FLOWS

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1 Internationa Journa of Pure and Appied Mathematics Voume 67 No., 93-3 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS TO DISCRETE MORSE FLOWS Yoshihiko Yamaura Department of Mathematics Coege of Humanities and Sciences Nihon University 3-5-4, Sakurajosui, Setagaya-ku, Tokyo, , JAPAN e-mai: Abstract: The present paper shows Lipschitz continuity of discrete Morse (variationa) fows to the reguarized At-Caffarei variationa functiona generating free boundaries; the continuity is uniform with respect to the discrete Morse fows and the reguarization. Together with the uniform Höder continuity shown in another paper, it is possibe to construct Morse fows, which wi be deat with a forthcoming paper. AMS Subject Cassification: 35J Key Words: Bernstein method, Rothe s scheme, discrete Morse fows. Introduction Our main objective is to construct Morse fows of the functiona introduced by At and Caffarei () which induces a free boundary probem. For this purpose, the author has tried to appy the method of discrete Morse fows deivered in the paper Kikuchi 5, the argument of which reies ony on the variationa structure. However, because of the difficuty caused by the singuarity contained in the functiona, the tria is not competed. We adopt in this paper the reguarized method used in Caffarei et a 3, where we are abe to make use of some properties of differentia equations. We describe in this paper a proof of uniform Lipschitz continuity for the space variabe of soutions to discrete Morse fows of a reguarized functiona. The assertion and the outine of the proof has been stated in Yamaura 7 whose main resut is the uniform Höder continuity Received: December 7, c Academic Pubications

2 94 Y. Yamaura for the time variabe. Since these two uniform continuities are not dependent on approximate soutions, our main objective stated above is estabished, the proof of which sha be stated in another paper. For a positive integer m, R m is a m-dimensiona Eucidean space. For a = (a,a,...,a m ), b = (b,b,...,b m ) R m, a,b is inner product of a and m b defined by a,b = a j b j. L m is an m-dimensiona Lebesgue measure. For j= a function f defined on R m, we denote by spt f the support of f. We use the standard notation for the Lebesgue and Soboev spaces L p (R m ), p and W k,p (R m ), k N, p, equipped with the norms L p (R m ), W k,p (R m ), respectivey. We aso use the notation H k (R m ) instead of W k, (R m ). L oc(r m ) is the metric space of functions whose square is ocay integrabe in R m. We sha state the At-Caffarei functiona and the discrete Morse fow method to this functiona. The At-Caffarei functiona AC : H (R m ) R is defined by ( ) AC(u) := u + χ(u) dx, R m where χ : R R is the characteristic function of open interva,, that is, for r, χ(r) := for r >. In the construction of Morse fows to functiona AC, the paper Caffarei et a 3, introduced a reguarized functiona: ( ) AC ε (u) := u + χ ε (u) dx, R m where ε is a positive number and χ ε is a reguarization of function χ. The precise definition of the reguarization wi be given in Section. Let u be a given function satisfying the foowing conditions: (i) u W, (R m ) W, (R m ) C,α (R m ) for some α,, (ii) L m (spt u ) < +, () (iii) u in R m, where symbo spt u denotes the support set of function u. The scheme of the discrete Morse fow method to the construction of Morse fows to variationa functiona AC ε is formuated in the foowing: give an initia data u, we inductivey introduce two sets u ε n} n= and ACε n} n= of functions

3 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS and functionas as foows: for each u ε n (n =,,... ), we introduce a variationa functiona AC ε n given by AC ε n(u) = AC ε (u) + u u ε h n dx R m and define u ε n (n =,,... ) as a minimizer of AC ε n in H (R m ). Here h is an arbitrariy fixed positive number. The minimizer u ε n satisfies Euer-Lagrange equations of AC ε n in the form: u ε n u ε n h = AC ε (u ε n ) in H (R m ). By use of the famiy u ε n} n= of minimizers together with the given u, we generate an h-step function u ε : h, H (R m ) by the scheme: u ε u for h t, (t) = u ε n for t n < t t n (n =,,... ). The function u ε is constructed with resort to the minimaity of introduced functionas AC ε n} n=, and named a discrete Morse fow to the functiona ACε ( ) and is sometimes represented in the form: u ε := u ε n } n=. Remark. (Existence of Discrete Morse Fow) The existence of minimizer to variationa probem AC ε n in H (R m ) is shown by the ower semi-continuity of AC ε n with respect to the weak convergence in H (R m ). We remark that AC ε n (uε n ) < + for each n =,,... is verified by taking the -function as a comparison function with the minimizer u ε n. The Lipschitz estimate asserted in this paper is stated in the foowing Theorem, in the proof of which we use the Bernstein method. We note the estimates asserted in the foowing theorem is vaid with no dependence on parameters h nor ε. Theorem. (Uniform Gradient Bound) Let (u ε n) be a discrete Morse fow to AC ε. Then there exists a universa positive constant L depending on M = sup(χ + χ ) and u L (R m ) such that R u ε n L (R m ) L for n =,,,.... ()

4 96 Y. Yamaura. Preiminaries We need to introduce a reguarization of the characteristic function χ so as to be usefu in the proof of Theorem based on the Bernstein method. Lemma. (Approximation of χ) There exists a function χ : R R such that the foowing conditions hod. χ C (R) (3) in, χ = (4) in,+ χ in R (5) χ > in, (6) χ χ in, (7) χ = χ in,+ (8) Proof. We define a C (R)-function γ as foows: for r, γ(r) := e r for r >. In addition, we define for r 3 6, ϕ(r) := 8r + 5 for 3 6 r 5 6, for r 5 6. Let ϕ be a one-dimensiona 6-moified function of ϕ, namey, we define ϕ(τ) := η σ (τ ξ) ϕ(ξ)dξ with σ = 6, R where η σ is the standard moifier (see for instance Evans et a 4, p. ). We set β(r) = γ(r)ϕ(r) (r R). Finay, we define χ (r) := r + β(τ)dτ βdl for r R (9)

5 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS The function χ defined in such a way satisfies the properties stated in Lemma. We state ony the proof of (7) which is non-trivia. For the purpose, owing to (3), it is enough to show the desired inequaity ony on the interva,. For r,, the equaity β(r) = e r hods, and so χ + (r) = β (r) e r + βdl = βdl = r e r + β(r) + βdl = βdl = χ (r), where we used the inequaity r. In the foowing of this paper, we et ε be a positive constant ess than, and we define ( r ) χ ε (r) := χ for r R. () ε Then, χ ε have the foowing properties : Lemma. (Properties of χ ε )Let χ ε ( ) be as in (). Then χ ε ( ) C (R). () χ ε in R. () χ ε in R, χ ε > in,ε. (3) for r,, χ ε (r) = (4) for r ε,+. There exists a positive number K such that (5) χ ε K ε, χ ε K ε in R. (6) Proof. The assertions () (4) are shown directy from the properties of χ isted in Lemma. The assertion (6) hod by setting K := max χ L (R), χ L (R)}. Here we notice that K is finite because both χ and χ is continuous and their supports are compact in R. In the proof based on the Bernstein method, a differentiabiity of soutions is needed.

6 98 Y. Yamaura Remark. (Euer-Lagrange Equation) Let (u ε n ) be a discrete Morse fow to AC ε. Then for each n, the function u ε n W, (R m ) is a weak soution of the Partia Differentia Equation u χ ε (u) u uε n (x) h = in R m. In addition, it hods for each n that u ε n C 3 (R m ). Proof. We show ony the reguarity. The case where m =. By Remark, u = u ε satisfies ( u ζ dx = χ ε(u) + u u ) ζ dx for ζ Cc (R). h R R Therefore, u ε is twice weak differentiabe, and the second weak derivative is represented as (u ε ) = χ ε(u ε ) + uε u. (7) h Since from Soboev imbedding theorem u,u ε C, (R). Hence, the right-side of (7) is a continuous function, and hence we have (u ε ) C (R). We thus obtain (u ε ) C (R), and so u ε C (R). Combining this fact with the initia condition u C,α (R), the right-side of (7) is turned out to be a C (R)- function, and as a resut we obtain u ε C3 (R). Having showed u ε C3 (R) C,α (R), we can repeat the above argument by repacing u with u ε in order to prove the assertion with n =. Finay, by induction we accompish the proof in the case where m =. The case where m. In this case, the assertion is shown by directy invoking the reguarity resut Ladyzhenskaya et a 6, p. 84, Theorem 6.4 in the domain B R () for any R >. 3. The Construction of C -diffeomorphism φ The proof of Theorem is based on the Bernstein method, which is estabished by the carefu choice of transformation φ as introduced in this section. Proposition. (The Construction of φ) There exist a function φ :,+,+, and positive numbers r > og 3 and b > such that the foowing properties hod: φ C,+, (8)

7 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS φ in,+, (9) ( φ() =, φ og 3 ) =, φ(r ) =, () φ,og 3 (r) = er for r,og 3, () φ r,+ b in r,+, () φ,og 3 C,og 3, (3) φ r,+ C r,+, (4) φ og 3,r C og 3,r, (5) φ in og 3,r, (6) Set Φ := φ φ in,. Then Φ 3 in og 3,r. (7) Here, we remark that the function Φ is we-defined since φ > in,+ (see Coroary beow). Before showing Proposition, we show properties of φ which are ed directy from Proposition. Coroary. (The Properties of φ) The function φ as in Proposition satisfies the properties isted beow. φ is nondecreasing in,+, (8) φ b in,+, (9) φ 3 in og 3,+, (3) φ ( ) is non increasing in og 3 ),+, (3) φ 3 in og 3,,+ (3) φ is bounded in,+. (33) There exists an inverse function φ :,+,+, so that φ C,+, φ, C,, φ, C,, φ,+ C,+. (34)

8 Y. Yamaura The function Φ defined as in Proposition satisfies Φ is differentiabe in,+ \ og 3 },r, and Φ, (35) Φ cannot be differentiabe at r = og 3,r, (36) Φ is non increasing in,+, (37) Φ in,+, (38) Φ(A) φ(a) φ(b) φ (A) φ (B) (39) for A,B,+ Proof. We show ony the inequaity (39), for other assertions are easiy checked by the eementary argument based on the properties stated in Proposition. In order to show (39), we set for A,B,+ I(A,B) := Φ(A) φ(a) φ(b) φ (A) φ (B). In case A = B, it hods that I(A,B) =, and so the inequaity surey hods. Hence, in the foowing, we assume that A B. We proceed our proof by dividing the argument according to the vaue of A and B as foows: (case) A,B,og 3. Since φ (r) = φ (r) = e r for r,og 3, and so I(A,B) = ea e A e A e B e A e B =. (case) A,B og 3,+. By appying the mean vaue theorem in the cosed interva with the endpoints A and B, there exist θ and θ, such that I(A,B) = φ (A) φ (A) φ ( θa + ( θ)b ) A B φ ( θa + ( θ)b ) A B. (a) The case where B < A. The inequaity θa + ( θ)b A impies that φ (θa + ( θ)b) φ (A), because φ is nondecreasing by (8). Since from (38) it hods that φ (A) φ (A), we have φ (A) φ (A) (A B). Hence we can estimate I(A,B) φ (A) φ ( θa + ( θ)b) } (A B). The first factor of the ast-side is nonpositive because θa + ( θ)b A and that φ is noninceasing in og 3,+ by (3). We thus obtain I(A,B).

9 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... (b) The case where A B. We are abe to show in the same way as in (a). (case 3) A,og 3 and B og 3,+. We first notice that I(A, B) ( = I A,og 3 ) + φ (A) φ (A) ( φ og 3 ) φ(b) φ ( og 3 ) } φ (B). (4) Owing to (case ), we have known I(A,og 3 ) =. Moreover, for φ (A), we have I(A,B) = ( φ og 3 ) φ(b) φ ( og 3 ) φ (B). By appying the mean vaue theorem, there exist θ, θ, such that φ (A) = ea = e A I(A,B) = = φ ( θ og 3 ) + ( θ)b φ ( θ 3 ) og + ( θ B }(og 3 ) B. Therefore, by recaing that φ 3 φ in og 3,+ from (3) and (3), we obtain I(A,B). (case 4) B,og 3 and A og 3,+. We can start from (4). In this case, using (case ), we have I(A,og 3 ), and hence I(A,B) φ (A) φ (A) = ( φ og 3 ) φ(b) φ ( og 3 ) φ (B) = ( φ (A) φ (A) ) ( 3 eb). The ast-side is nonnegative due to (38). In order to construct a function φ as in Proposition, we have ony to define a function φ on the interva og 3,r such that a conditions in Proposition are fufied. For this purpose, we construct a function ζ as foows: Lemma 3. (The Construction of ζ) There exists a positive number r and a function ζ C (R) satisfying the properties isted beow. ζ() =, ζ () = 3, ζ () = 3, (4) ζ(r ) =, ζ (r ) =, (4) ζ in,r, (43)

10 Y. Yamaura ζ in,r, (44) ( ζ ) 3 ζ in,r. (45) We remark that the function ζ ζ is we-defined in,r, because it hods from (4) that ζ () = 3, and from (43) that ζ,r. Therefore, ζ (r) ζ () = 3 > for r,r. If we use Lemma 3, then Proposition can be shown by defining r := og 3 + r, b := ζ (r ) and e r if r,og 3, φ(r) := ζ(r og 3 ) + if r og 3,r, b(r r ) + if r r,+. Proof of Lemma 3. Let α( ) be a function in C (R). Then we consider the foowing initia probem of ODE: ( ζ ) = α in R, ζ ζ() =, ζ () = 3, ζ () = 3 (46). By integrating two times the given ODE and taking the initia conditions into account, the soution is written as where A α (τ) = ζ α (τ) = 3 τ By differentiating (47), we have τ e s à α (s) ds for τ R (47) α(r)dr and à α (τ) = τ A α (r)dr. ζ α(τ) = 3 eτ Ãα(τ), (48) ζ α (τ) = 3 ( A α(τ))e τ Ãα(τ), (49) ζ α (τ) = 3 } ( A α (τ)) α(τ) e τ à α (τ). (5) Now, if we define α 3 in R, then there exists a unique positive number δ < 3 such that ζ α (δ ) = 4. (5)

11 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 3 This is shown by appying the mean-vaue theorem in the interva, 3 on the basis of the facts (a) ζ α () =, ( ) (b) ζ α >, 3 (5) (c) ζ α > in,. 3 The equaity (a) is nothing but the initia condition, whie (c) is obtained by (48). The inequaity (b) is shown as foows: Since A α (τ) < for τ, 3, we have from (49) that ζ α (τ) > for τ, 3. Thus, ζ α is stricty increasing on, 3. Therefore, ζ α (τ) > ζ α () = 3 for τ, 3. By integrating the ast inequaity from to 3, and recaing ζ α () =, we have (b). We set for λ 3 for τ,δ α λ (τ) λe τ δ + 3 for τ δ,+, where δ is a positive constant as in (5). In case λ =, this function coincides with α 3 defined aready. Then, there exists a unique τ α λ δ, 3 such that A αλ (τ αλ ) = ταλ α λ dτ =. (53) This fact is shown by appying the mean-vaue theorem because A αλ is stricty increasing on,, A αλ (δ ) < and A αλ ( 3 ). We here notice that in case λ =, τ α = 3. (54) Now, et us show that there exists a unique λ,+ such that ζ αλ (τ αλ ) =. (55) To prove this we investigate the continuity and monotonicity of λ τ αλ and λ ζ αλ (τ αλ ). Fact. The function λ,+ τ αλ satisfies the foowing properties: λ τ αλ is stricty decreasing on,+. (56) λ τ αλ is continuous on,+. (57) im τ α λ = δ. (58) λ

12 4 Y. Yamaura Proof. (56): Let < λ < Λ. Then α Λ α λ in R and α Λ α λ in δ,. Hence, ταλ α Λ dτ ταλ α λ dτ =. τ αλ is a positive number such that τ αλ α Λ dτ =, and hence it must hod that τ αλ < τ αλ. (57) : We show ony the continuity at λ =, because the continuity at λ > is shown in the same way. From (i), we have im τ αλ τ α. Moreover λ it hods that im τ αλ τ α. Otherwise, there exists a positive number σ such λ that im τ αλ = τ α σ. Then, for any λ >, λ = ταλ α λ dτ imλ τ αλ α λ dτ = τα σ α λ dτ τα α λ dτ 3 σ where we use the inequaity im λ τ αλ τ αλ (λ > ) which foows from the monotone decreasing property of λ τ αλ, and the inequaity α λ 3 in R. We thus have τα α λ dτ + 3 σ. We here et λ. Then by noting that α λ α uniformy in,τ α, we have = τα α dτ + 3 σ, which is a contradiction. (58) : Since τ αλ δ for λ, we have from (i) that δ := im λ τ α λ δ. Moreover, the reverse inequaity δ δ hods. If not, it must hod δ < δ. Then again by (i) we have τ αλ δ, and so = ταλ δ δ α λ dτ α λ dτ = λ δ δ e δ α λ dτ τ δ dτ + 3 (δ δ ). (59) The ast-side diverges as λ, and we attain a contradiction. We next investigate about the function λ ζ αλ (τ αλ ).

13 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 5 Fact. The function λ,+ ζ αλ (τ αλ ) satisfies the foowing properties: λ ζ αλ (τ αλ ) is stricty decreasing on,+. (6) λ ζ αλ (τ αλ ) is continuous on,+. (6) im ζ α λ (τ αλ ) = ζ α (δ ) = λ 4. (6) Proof. (6): Let λ < Λ < +. Then from the definition α λ α Λ in,+, and hence A αλ A αλ in,+, à αλ Ãα Λ in,+. By making use of the ast inequaity, we obtain ζ αλ (τ αλ ) 3 ταλ e s ταλ eãα λ (s)ds < 3 e s eãα λ (s)ds = ζ α λ (τ αλ ), where we used the inequaity τ αλ < τ αλ that foows from (56). (6): Let λ. Then by the definition of α λ α λ α λ uniformy in, 3 Moreover from the definition of A αλ and Ãα λ, we have as λ λ. Aαλ A αλ à αλ à αλ uniformy in, 3 as λ λ. This impies e s à α λ (s) e s à α λ (s) uniformy in, 3 as λ λ, (63) where we use the fact that A αλ and Ãα λ are uniformy bounded in, 3 as λ λ. We can estimate ταλ ταλ ταλ e s à α λ (s) ds ταλ e s à α λ (s) ds ταλ e s Ãα λ (s) e s Ãα (s) λ ds + e s à α (s) λ ds τ αλ e s à α λ (s) e s à α λ (s) ds + ταλ τ αλ e 3

14 6 Y. Yamaura By etting λ λ, the first term of the ast-side goes to zero because of (63), and the second goes to zero because τ αλ is continuous with respect to λ by (57). We thus have ζ αλ (τ αλ ) ζ αλ (τ αλ ) as λ λ. (6): Since τ αλ > δ, we have for any λ > that ζ αλ (τ αλ ) = 3 ταλ e s δ eãα λ (s)ds > 3 Here recaing that Ãα λ = Ãα in,δ, we have e s eãα λ (s)ds. Moreover, by (58) ζ αλ (τ αλ ) = 3 3 ταλ ζ αλ (τ αλ ) > 3 ταλ e s eãα δ e s e s eãα λ (s)ds (s)ds λ eãα (s)ds = ζ α (δ ). (64) 3 δ e s eãα (s)ds = ζ α (δ ) Inequaity (64) and the convergence (65) tes us that ζ αλ (τ αλ ) ζ α (δ ) as λ. Summing up the properties of the function λ ζ αλ (τ αλ ), we obtain (6), (6), (6) and ζ α (τ α ) = ζ α ( 3 ) > by (54) and (5)-(b). Thus, we are abe to appy the mean-vaue theorem, and as a resut, we obtain (55). If we choose r := τ αλ and ζ := ζ αλ on R, a the properties of Lemma are fufied. Remark 3. Let φ be a function defined as in the proof of Lemma 3. Then, we can easiy check that φ cannot be differentiabe at r = og 3 and r = r. In addition, on the inverse function φ, (φ ) cannot be differentiabe at s = = φ(og 3 ) and s = = φ(r ). 4. Uniform Gradient Bound By using the notation of piecewise constant function u ε, our main theorem in this paper Theorem is restated as foows: Lemma 4. Let u ε be a discrete Morse fow to AC ε. Then there exists a universa positive constant, independent of h and ε, such that sup u ε L. R m h,+

15 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS We state the foowing simpe emma which is used in the proof of Theorem Lemma 5. Let p. If f satisfies (A) f L p (R m ), (A) f L (R m, R m ), then the maximum vaue of f in R m is attained at some point of R m. Proof. By assumption (A), f turns out to be uniformy Lipschitz continuous in R m, which together with assumption (A) impies the boundedness of f in the whoe of the space R m. Let (x j ) R m be a maximizing sequence of sup R m f. Again by the uniform Lipschitz continuity of f and (A), sequence (x j ) turns out to be a bounded set. As a resut, we can find a desired point as a imit point of convergence subsequence of (x j ). Proof of Lemma 4 The proof is divided into five steps as foows. Step. Rescaing and transformation of u ε. The minimizer u ε n is a soution of the Euer-Lagrange equation of ACε n as foows: u ε n χ ε (uε n ) uε n uε n h = in R m for n =,,.... Therefore, h-step function u ε satisfies the foowing equation: u ε (x,t) χ ε (uε (x,t)) uε (x,t) u ε (x,t h) h for (x,t) R m,+. = (65) We rewrite (65) by using the rescaed function U ε (x,t) = ε uε (εx,ε t), where h ε := h ε. ( U ε )(x,t) χ (U ε (x,t)) Uε (x,t) U ε (x,t h ε ) h ε = for (x,t) R m,. (66) Now, et φ :,+,+ be a C -diffeomorphism as constructed in Proposition of Section 5, and set v ε (x,t) := φ ( U ε (x,t) ). Then (66) is rewritten as v ε φ (v ε ) + v ε φ (v ε ) χ (φ(vε )) φ(vε ) φ(ṽ ε ) = h ε (67) in R m,+,

16 8 Y. Yamaura where ṽ ε (x,t) := v ε (x,t h ε ). Since from (9) it hods that φ > in,+, we can divide the both sides of (67) by φ (v ε ) in order to have v ε + φ (v ε ) φ (v ε ) vε χ (φ(vε )) φ (v ε φ(v ε ) φ(ṽ ε ) ) φ (v ε = ) h ε in R m,+. (68) Step. The existence of the maximum of w ε. Let T be a positive number, and set n T as a positive integer satisfying T t nt,t nt, where t j = jh ε = j h. In the foowing, we sha show ε that for any T >, there exists a point (x,t ) R m,t such that w ε (x,t ) = max R m,t wε. For this purpose, it is sufficient to prove that for each n =,,...,n T, there exists max R m (n )h wε. In the seque, we denote ε,nh ε Un(x) ε := ε uε n(εx), vn(x) ε := φ (Un(x)) ε and wn(x) ε := vn(x) ε for x R m. Then, we have ony to show that for each n N there exists max wε n. Moreover, owing to Lemma 5 with p =, its proof is reduced to make sure the foowing two conditions: (A) : wn ε dx < + R m for n N (A) : wn ε L (R m ) < + The assertion (A) is shown by computing according to the definition of wn ε in the foowing way: wn ε dx = vn ε dx = R m R m = R m φ ( ε uε n(εx) ) dx ε m R m R m u ε n dx < +, where we used the property φ. We next show (A). By simpe computation, we can know Thus we must show w ε n L (R m ) C( u ε n L (R m ), u ε n L (R m )). A- : u ε n L (R m ) < +, A- : u ε n L (R m ) < +.

17 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 9 Since u ε n L (R m ), we can ead A- from A- with the hep of Lemma 5 with p =. Finay, we have ony to prove A-. The proof is divided into severa steps. (step) We show u H (R m ). Since u is a minimizer of the functiona F ε,h, u satisfies u u = u + χ ε (u ) + u u h in R m (69) We denote by f the right hand side of (69). If we show f L (R m ), we can reach the desired fact u H (R m ) with the hep of Brezis, p. 48, Theorem IX.5. u and u beongs to L (R m ) because these functions are W, (R m )-Soboev functions. Moreover, χ ε (u ) dx = χ ε (u ) χ ε () dx R m R m χ ε L (R) u L (R m ) < +, where we use the fact that χ ε() =. We thus have f L (R m ). (step ) We show u H 3 (R m ). By the preceding step we have shown u H (R m ), and hence we can differentiate (69) to have ( u ) u = u + ( ) u χ ε(u ) + u u. (7) h The right hand side beongs to L (R m ) for u,u W, (R m ). In the same way as in (step ), we have u H (R m ), and hence u H 3 (R m ). (step 3) We show u L (R m ). Rewrite (7) as ( ( u ) + χ ε (u ) ) = h h u. Noting the assumption u L (R m ), we appy Ladyzhenskaya 6, p. 99, Theorem 3. in any bas of radius. As a resut, we obtain esssup u C(ε,h,m, u L (R m ), u L (R m ))). (7) R m (step 4) We show u L (R m ). Since in (step ) we have aready shown u H 3 (R m ), we can differentiate (7) to have ( ( u ) + χ ε(u ) ) u = h ( u u )χ ε (u ) h u.

18 Y. Yamaura By using the assumption u L (R m ), we are abe to appy Ladyzhenskaya 6, p. 99, Theorem 3. again to obtain esssup u R m C(m,ε,h, u L (R m ), u L (R m ), u L (R m )). The conditions for u which is needed in the above argument are u W, (R m ), u L (R m ) and u L (R m ). Since these facts with u repaced by u are shown above, we can obtain the assertion A- with n =. Thus, by induction, we accompish the proof of A-. Step 3. Reduction of the concusion. In order to prove the concusion it is sufficient to verify max R m,t wε max M, sup w ε} for T >. (7) R m } This reduction is shown by noting the fact that w ε (x,t) = w ε (x,) in h ε,, and the inequaities sup w ε u L (R m ), R m } sup R m h, u ε b sup R m h ε, v ε, where b is a constant as in (). Step 4. Inequaities. Let us start the proof of (7). For this purpose, we suppose on the contrary that max R m,t wε > max M, sup w ε}. (73) R m } Let (x,t ) R m,t be a maximum point of w ε in R m,t whose existence was shown in Step. Then we can show the foowings: w ε (x,t ) >, (74) w ε (x,t ) =, (75) m w ε (x,t ) = jj w ε (x,t ), (76) j= i v ε i v ε i v ε i ṽ ε (x, (77),t )

19 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... where ṽ ε (x,t) := v ε (x,t h ε ). The expression (74) is from (73), and (74)-(76) are the eementary facts of cacuus due to the fact that (x,t ) is a maximum point. In order to prove the inequaity (77), we use the Schwarz inequaity. i v ε i v ε i v ε i ṽ ε (x,t ) } w ε (x,t ) w ε (x,t h ε ). (78) The ast inequaity is show in the foowing way. If t > h ε, then (x,t h ε ) R m,t. Therefore, by recaing that (x,t ) is a maximum point, it hods that w ε (x,t ) w ε (x,t h ε ). On the other hand, if < t h ε, from (73), w ε (x,t ) > sup w ε w ε (x,) = w ε (x,t h ε ), R m } where the ast equaity hods because h ε < t h ε. Let us show the foowing three types of inequaity. ( φ ) E(x,t ) := (φ ) (v ε )χ (φ(v ε )) χ (φ(v ε )), (79) (x,t ) ( φ ) } F(x,t ) := (φ ) (v ε )χ (φ(v ε )) χ (φ(v ε )) w ε (8) φ (ṽ ε ) ( h ε φ (V ε j v ε j v ε j v ε j ṽ ε) (x,t). h ε ) In particuar, if v ε (x,t ) og 3,r, then G(x,t ) := 3w ε + χ (φ(vε )) χ (x )) (φ(vε. (8),t ) The proof of (79)-(8): The inequaity (79) is directy foowed from (8) by making use of (77) and the fact that φ > in, ((9)). Thus, in the foowing, we sha prove ony (8) and (8). By using abbreviation Φ(r) := φ (r) φ (r), the reation (68) is rewritten in the foowing. v ε + Φ(v ε ) v ε = χ (φ(vε )) φ (v ε + φ(v ε ) φ(ṽ ε ) ) φ (v ε. (8) ) h ε For convenience of the argument beow, we sha denote by I the eft-side of (8) and by I r the right-side. Namey, we put I := v ε + Φ(v ε ) v ε, I r := χ (φ(vε )) φ (v ε + φ(v ε ) φ(ṽ ε ) ) φ (v ε. ) h ε

20 Y. Yamaura (step ) We aim at the inequaity v ε, I r (x,t ) F(x,t ). (83) Differentiating I r and mutipying the resutant by j v ε, we have ( φ j v ε j I r = j v ε j v ε χ (φ(v ε ) )) (φ ) (v ε ) j v ε j v ε χ (φ(v ε )) ( φ ) (φ ) (v ε ) j v ε j v ε φ(vε ) φ(ṽ ε ) + h ε + φ (v ε ) jv ε j v ε φ (v ε ) j v ε j ṽ ε φ (ṽ ε ) h ε at (x,t ). Let us sum this reation with respect to j from to m. Here we reca w ε = v ε, and remark that the numerator of the ast fraction is deformed as foows: j v ε j v ε φ (v ε ) j v ε j ṽ ε φ (ṽ ε ) = ( ) = j v ε j v ε φ (v ε ) j v ε j v ε φ (ṽ ε ) + (84) ( ) + j v ε j v ε φ (ṽ ε ) j v ε j ṽ ε φ (ṽ ε ) = = w ε( φ (v ε ) φ (ṽ ε ) ) ( +φ (ṽ ε ) j v ε j v ε j v ε j ṽ ε). As a resut, we get v ε, I r ( φ ) } = (φ ) (v ε )χ (φ(vε )) χ (φ(vε )) w ε + + φ (ṽ ε ) ( h ε φ (Vh ε j v ε j v ε j v ε j ṽ ε) ε ) w ε } h ε φ (v ε Φ(v ε )φ(v ε ) φ(ṽ ε ) φ (v ε ) φ (ṽ ε ) at (x,t ) ) Appying (39) with A = v ε (x,t ) and B = ṽ ε (x,t ), the vaue of the ast parenthesis } of the ast reation turns out to be non-positive, we finay arrive at (83) since φ > by (9). (step ) We next show the inequaity v ε, I (x,t ) Φ (v ε )(w ε ). (85) In order to differentiate I, we have to excude the case where the two terms of I are not differentiabe, that is, the case where v ε (x,t ) = og 3 or r (r

21 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 3 is as in Proposition ). The excuded case sha be treated beow as (step 4). We differentiate two terms of I with respect to x-variabe at (x,t ) and mutipying by j v ε to obtain j v ε j I = = j v ε ( j v ε ) +Φ (v ε ) j v ε j v ε v ε + + Φ(v ε ) v ε, j v ε ( j v ε ) at (x,t ). (86) Now et us sum (86) with respect to j from to m. We here notice that the reation j v ε ( j v ε ) = v ε. In addition, the direct computation of differentiation tes us that the first term of the right-side of (86) equas to We thus finay have j v ε ( j v ε) = ( v ε ) m ij v ε. i,j= v ε, I r = w ε v ε + Φ (v ε )(w ε ) + + Φ(v ε ) v ε, w ε at (x,t ). (87) Taking into account (75), (76) and v ε, we obtain (85). (step 3) Let us combine (83) and (85) to deduce F(x,t ) Φ (v ε (x,t ))(w ε (x,t )). (88) Since from (35) the right-side of (88) is non-positive, we arrive at (8). On the other hand, in order to show (8), assume v ε (x,t ) og 3,r. By making use of (7) and (77) in the right-side of (88) and dividing the both sides by w ε (x,t ) which is positive because of (74), we obtain ( φ 3w ε ) } + (φ ) (v ε )χ (φ(vε )) χ (φ(vε )) at (x,t ). (89) Since from (3) and (3) φ 3 and φ 3 in og 3,r, we have φ (φ ) (v ε (x,t )). Appying this to (89), we estabish the desired inequaity (8). (step 4) We treat the case excuded in (step ): v ε (x,t ) = og 3 or r. Our aim is to show (79) and (8). Since the former is directy foowed from the atter, we sha show ony (8). The side I r is differentiabe, and so (83) remains vaid aso in this case. Therefore, we have ony to show v ε, I at (x,t ). (9)

22 4 Y. Yamaura For this purpose, we proceed our argument in the foowing two steps of proving the first inequaity and the second one of v ε (x,t ), I (x,t ) v ε (x,t ), v ε (x,t ). (9) We ony treat the case where v ε (x ) = og 3, whie the treatment of other case is simiar (see the end of this proof). In the foowing, for abbreviation, we omit to write t ike as v ε (x ) := v ε (x,t ) because we fix time variabe t as t throughout this proof. At first we define functions φ and ṽ ε. Let φ(r) e r for r R. Then φ, φ C,, and φ φ on,og 3, and therefore φ φ in,, (9) φ φ in,og 3, (93) φ φ in,og 3. (94) In addition, we define ṽ ε (x,t) := φ (U ε (x,t)) for (x,t) R m h ε,. The proof of the first inequaity of (9) We show for each fixed k =,...,m the inequaity k v ε (x ) k I (x ) k v ε (x ) k ṽ ε (x ). (95) Without oss of generaity we may assume k v ε (x ) since in case k v ε (x ) = the inequaity hods because both sides are equa to zero. Let e k be the unit vector whose k-th component is one. We consider the difference quotient as foows: I (x + δe k ) I (x ) δ = ( vε )(x + δe k ) ( v ε )(x ) + δ + Φ(vε (x + δe k )) v ε (x + δe k ) Φ(v ε (x )) v ε (x ) δ for δ R \ } (96) In the foowing, for abbreviation, we denote for function Ψ : R m,+ ) R the difference quotient Ψ(x + δe k ) Ψ(x ) by Ψ. So we have from (96) δ I = δ vε + v ε (x + δe k ) δ Φ(vε ) + Φ(v ε (x )) δ vε.

23 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 5 In case v ε, we can deform the second term of the right-side as foows: δ vε δ I = δ vε δ vε + + v ε (x + δe k ) ( ) Φ(v ε (x + δe k )) Φ(v ε (x )) δ vε v ε (x + δe k ) v ε + (x ) + Φ(v ε (x )) δ vε δ vε. Since from (37) the function Φ is non-increasing, the fraction containing Φ is non-positive. Therefore, δ vε δ I δ vε δ vε + Φ(v ε (x )) δ vε δ vε for δ R \ }. (97) Here we remark that (97) remains vaid in case v ε =. According to the sign of k v ε (x ), we choose an infinitesima (δ) in the foowing way: In case k v ε (x ) >, δ = δ (k) In case k v ε (x ) <, δ = δ (k) < with δ (k) as, > with δ (k) as. (98) Then, by considering N arge enough, we may assume that in any case the inequaity v ε (x + δ (k) e k ) v ε (x ) = og 3 for N. (99) hods. For this reason, we obtain φ (U ε (ξ)) = v ε (ξ),og 3 at ξ = x, x + δ (k) e k, and therefore, U ε (ξ), We have from (9) (94) φ (φ (U ε (ξ))) = φ ( φ (U ε (ξ))) Hence, φ (φ (U ε (ξ))) = φ ( φ (U ε (ξ))) v ε (ξ) = (φ (U ε (ξ))) = at ξ = x, x + δ (k) e k. at ξ = x, x + δ (k) e k.

24 6 Y. Yamaura = Uε (ξ) φ ( φ (U ε (ξ))) U ε (ξ) φ ( φ (U ε (ξ))) φ ( φ (U ε (ξ))) 3 = = ( φ (U ε (ξ))) = ṽ ε (ξ) at ξ = x,x + δ (k) e k. Due to (97) with δ = δ (k), δ (k) v ε δ (k) δ (k) v ε I δ (k) Passaging, we get ṽ ε + Φ(v ε (x )) δ (k) v ε δ (k) v ε for N. k v ε (x ) k I (x ) k v ε (x ) k ( ṽ ε )(x ) + Φ(v ε (x )) k v ε (x ) k w ε (x ). () Recaing here w ε (x ) = ((75)), we estabish the concusion (95). The proof of the second inequaity of (9) We aim at the inequaity v ε (x ), jj ṽ ε (x ) for j =,...,m. () For this purpose, we need the foowing emma, which is a non-smooth version of the fact that if smooth function of one variabe ζ = ζ(r) takes its maximum at r = α, then ζ (α) hods. Lemma 6. Let ζ : R R be a C (R)-function such that the maximum is attained at α R. Then there exists both an infinitesima (σ j ) such that σ j > and ζ (α + σ j ) ζ (α) σ j for j N, () and an infinitesima (σ j ) such that σ j < and ζ (α + σ j ) ζ (α) σ j for j N. (3) Proof. Since the proof of (3) are same as (), we show ony (). Suppose on the contrary that there exists a positive number σ such that ζ (α + σ) ζ (α) σ > for < σ < σ. (4)

25 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 7 Since ζ takes the maximum at α, we have ζ (α) =. Therefore, from (4), we have ζ (α + σ) > for < σ < σ. Integrating ζ on the interva α,α+σ, we have < α+σ α ζ (ξ)dξ = ζ(α + σ ) ζ(α), which contradicts a fact that ζ(α) is a maximum vaue. Let us appy Lemma 6 to ζ j (r) := ( v ε )(x + τe j,t ) (r R). We remark that from our assumption that v ε takes its maximum at (x,t ), the function ζ j takes its maximum at r =. As a resut, we can choose for each j =,...,m an negative infinitesima ( ) such that ζ ( ) ζ () for N. Here, in addition, we assume that j U ε (x ) >. Then, for sufficienty arge, it hods that U ε (x + e j ) U ε (x ). Thus, we can choose an infinitesima ( ) such that <, ( j v ε )(x + e j ) ( j v ε )(x ), U ε (x + e j ) U ε (x ). for N (5) In the same way, under the condition j U ε (x ) <, we can choose an infinitesima ( ) such that >, ( j v ε )(x + e j ) ( j v ε )(x ), for N (6) U ε (x + e j ) U ε (x ). On the other hand, ony in case j U ε (x ) =, it is impossibe to compare generay the vaue U ε (x + e j ) and U ε (x ), so we impose an infinitesima ony the conditions >, ( j v ε )(x + e j ) ( j v ε )(x ) for N. (7),

26 8 Y. Yamaura Now we are in a position to prove the desired inequaity (). Let ( ) be as above. Then ( j v ε )(x + e j ) ( j v ε )(x ) = + Therefore, m ( k v ε )(x ) ( kj v ε )(x + e j ) ( kj v ε )(x ) + k= m ( kj v ε )(x + e j ) ( k v ε )(x + e j ) ( k v ε )(x ) k= = m ( k v ε )(x ) ( kj v ε )(x + e j ) ( kj v ε )(x ) k= m ( kj v ε )(x + e j ) ( k v ε )(x + e j ) ( k v ε )(x ) k= (8) In the foowing, we divide our argument according to the sign of j U ε (x ) is zero or not. (case ) The case where j U ε (x ). In this case, by (5) and (6), we have U ε (x + e j ) U ε (x ) =, which impies φ (φ (U ε (ξ))) = φ ( φ (U ε (ξ))) φ (φ (U ε (ξ))) = φ ( φ (U ε (ξ))) at ξ = x and x + e j. (9) Hence, for ξ = x,x + e j it hods that kj v ε (ξ) = ) = kj (φ (U ε (ξ)) = = kju ε (ξ) φ ( φ (U ε (ξ))) j U ε (ξ) k U ε (ξ) φ ( φ (U ε (ξ))) = φ ( φ (U ε (ξ))) 3 ( φ ) = kj (U ε (ξ)) = kj ṽ ε (ξ).

27 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 9 Thus, we can exchange kj v ε of the eft-side of (8) to kj ṽ ε, so that m ( k v ε )(x ) ( kj ṽ ε )(x + e j ) ( kj ṽ ε )(x ) k= m ( kj v ε )(x + e j ) ( k v ε )(x + e j ) ( k v ε )(x ). k= () Here, passaging, we can arrive at the concusion (): v ε (x ), jj ṽ ε (x ) m ( kj v ε )(x )}. (case ) The case where j U ε (x ) =. If we use abbreviation f(x) = kj U ε (x) φ (v ε (x)) and g(x) = ku ε (x), then we have the reation φ (v ε (x)) 3 k= kj v ε (x) = f(x) g(x) j U ε (x)φ (v ε (x)). () If we denote f(x) = kju ε (x) φ and g(x) = ku ε (x), then it hods that (ev ε (x)) 3 φ (ev ε (x)) kj ṽ ε (x) = f(x) g(x) j U ε (x) φ (ṽ ε (x)). Let us consider a difference quotient of both sides of (). Here, notice that it is possibe to drop the term containing j U ε (x ) which is assumed to be zero. So that we have kj v ε (x + e j ) kj v ε (x ) = x + e j = f(ξ) φ (V (x + e j )) g(ξ) j U ε (ξ) x x + e j. x Therefore im = k= m ( k v ε )(x ) kj v ε (x + e j ) kj v ε (x ) = m k v ε (x ) k= im φ (φ (U ε (x ))) im f(ξ) x + e j x g(ξ) j U ε (ξ) x + e j x } ()

28 Y. Yamaura By means of (9) with ξ = x, we can repace φ (φ (U ε (x ))) by φ ( φ (U ε (x ))). In addition, we can repace f and g by f and g respectivey. This is shown as foows: Since U ε (,x ) is a C 3 -function, we can compute the differentiation of f at x and so im = f(ξ) x + e j x = j ( kjj U ε (x ) φ ( φ (U ε (x ))) kj U ε (ξ) φ (φ (U ε (ξ))) ) ξ=x = } kj U ε (x ) j U ε (x ) φ ( φ (U ε (x ))) φ ( φ (U ε (x ))) 3 = im x + e j f(ξ). x In the third equaity, (9) with ξ = x is used. For the same reason, we aso get im g(ξ) j U ε (ξ) x + e j x = im g(ξ) j U ε (ξ) x + e j. x We can thus repace in () f and g by f and g respectivey to have im = im k= k= = im m ( k v ε )(x ) kj v ε (x + e j ) kj v ε (x ) = m ( k v ε )(x ) f(ξ) x + x e j g(ξ) j U ε (ξ) φ (ṽ ε (ξ)) x + e j m ( k v ε )(x ) kj ṽ ε (x + e j ) kj ṽ ε (x ). k= This recovers () in the case where j U ε (x ) =, and therefore we are abe to estabish () in the same way as in (case ). In order to accompish our proof, we comment the case where v ε (x ) = r. In this case the assertion is shown by modifying the foowing two definitions in the proof of the case where v ε (x ) = og 3. We define φ(r) b(r r ) + for r R, where b is a constant as in Proposition. Then, we choose an x } =

29 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... infinitesima ( ) in (5) such that >, and in (6) such that In addition, we choose (δ (k) <. ) in (98) such that δ (k) > in case k v ε (x ) >, and δ (k) < in case k v ε (x ) <. Step 5. Inequaity U ε (x,t ). We show the inequaity U ε (x,t ). (3) Suppose on the contrary that U ε (x,t ) <. In the case where U ε (x,t ) =. Then, by the initia condition () and Theorem??, it turns out that U ε in R m h ε,+, so that U ε (x,t ) =. Hence, v ε (x,t ) = U ε (x,t )(φ ) (U ε (x,t )) = Uε (x,t ) φ (v ε (x,t )) =, and w ε (x,t ) = v ε (x,t ) =. This contradicts (73). In the case where < U ε (x,t ). Since < v ε (x,t ) og 3, φ (v ε (x,t )) = φ (v ε (x,t )) = e vε (x,t ). Therefore, (79) impies that E(x,t ) = e vε (x,t ) χ (φ(v ε (x,t )) χ (φ(v ε (x,t )). It hods that e vε (x,t ) < for v ε (x,t ) >. Furthermore, < φ(v ε (x,t )) = U ε (x,t ) hods, and so from (6) χ (φ(vε (x,t )). Hence, E(x,t ) χ (φ(vε (x,t )) χ (φ(vε (x,t )). On the other hand, by (7), we have χ χ in,, and then finay we have E(x,t ) <, which is a contradiction. In the case where < Uε (x,t ) <. Since v ε (x,t ) (og 3,r ), the assertion (8) hods. G(x,t ) 3w ε (x,t ) + χ (φ(v ε (x,t ))) χ (φ(v ε (x,t ))). By the assumption (73), w ε (x,t ) > M hods. Both χ by M = χ L (R) + χ L (R), and as a resut and χ are estimated which is a contradiction. Step 6. A contradiction. G(x,t ) 3M + M = M,

30 Y. Yamaura Finay, in order to accompish the proof of Lemma 4 we ead a contradiction by using (3) shown above. Let =,,,... be a non-negative integer such that t h ε,( + )h ε. Invoking the resut of Step5, we have φ (v ε (x,t)) = U ε (x,t). Therefore, from (8) of Lemma, χ (φ(vε (x,t )) = and χ (φ(vε (x,t )) =. Appying these facts to (8), we infer F(x,t ) = h ε φ (ṽ ε ) φ (v ε ) Since from (9) φ > in,+, we deduce Noting here (78), we get ( j v ε j v ε j v ε j ṽ ε) (x,t ). j v ε j v ε j v ε j ṽ ε (x,t ). w ε (x,t ) = w ε (x,t h ε ). (4) In case =, that is, t h ε ( h ε,, it hods that sup w ε w ε (x,t h ε ) = w ε (x,t ) = max R m } R m,t wε, (5) which contradicts the assumption (73). In case, that is, t h ε R m,t, we have from (4) that the point (x,t h ε ) is a maximum point of w ε in R m,t, and w ε (x,t h ε ) > max M, sup w ε}. R m } Hence, we can appy our argument from (73) to (4) by repacing (x,t ) with (x,t h ε ) to have w ε (x,t h ε ) = w ε (x,t h ε ). By induction, we have w ε (x,t (+)h ε ) = w ε (x,t h ε ). Finay, we have w ε (x,t (+)h ε ) = w ε (x,t ). Since t ( + )h ε ( h ε,, we have a contradiction sup w ε w ε (x,t ( + )h ε ) = w ε (x,t ) = max R m } R m,t wε. Remark 4 (On the initia condition ()). If we restrict our argument to the case where m 5, we can show the existence of the maximum point or w ε under the foowing initia conditions: (i) u H 3 (R m ) W, (R m ) C,α (R m ) for some α, (ii) L m (spt u ) < + (6) (iii) u in R m

31 UNIFORM LIPSCHITZ CONTINUITY OF SOLUTIONS... 3 Comparing the condition (), the condition above does not need the boundedness of u. References H.W. At, L.A. Caffarei, Existence and reguarity for a minimum probem with free boundary, J. Reine Angew. Math., 35 (98), H. Brezis, Anayse Fonctionnee, Masson, Editeur, Paris (983), Japanese Transated. 3 L.A. Caffarei, J.L. Vázquez, A free-boundary probem for the heat equation arising in fame propagation, Transaction of the AMS, 347 (995), L.C. Evans, R.F. Gariepy, Measure Theory and Fine Properties of Functions, CRC Press (99). 5 N. Kikuchi, A method of constructing Morse fows to variationa functionas, Noninear Word, (994), O.A. Ladyzhenskaya, N.N. Ura tseva, Linear and Quasiinear Eiptic Equations, Academic Press, New York-London-Toronto-Sydney-San Francisco (968). 7 Y. Yamaura, Höder continuity of discrete Morse fows to the At-Caffarei variationa probem generating free boundaries, To be submitted.

32 4