p 8 - Semantic Scholar

A Parabolic Quasilinear Problem for Linear Growth Functionals F. Andreu  V. Caselles y and J. M. Mazon z ,

Dedicated to Ph. Benilan on the Occasion of His 60th Birthday

Abstract

We prove existence and uniqueness of solutions for the Dirichlet problem for quasilinear parabolic equations in divergent form for which the energy functional has linear growth. A tipical example of p energy functional we consider is the one given by the nonparametric area integrand 1 + k k2 , which corresponds with the time-dependent minimal surface equation. We f (x;  ) = also study the asimptotic behavoiur of the solutions.

Key words: Linear growth functionals, nonlinear parabolic equations, accretive operators, semigroups.

AMS (MOS) subject classi cation: 35K65, 35K55, 47H06, 47h20.

1 Introduction and preliminaries Let be an open bounded set in IRN with Lipschitz continuous boundary @ . We are interested in the problem 8 > > > > > > < > > > > > > :

@u = div a(x; Du) @t u(t; x) = '(x)

in

Q = (0; 1) 

on

S = (0; 1)  @

(1.1)

in x 2

being a function with linear growth as k k ! 1. A tipical example of pa function f (x;  ) satisfying the conditions we need is the nonparametric area integrand f (x;  ) = 1 + k k2 . Problem (1.1) for this particular f , that is, the time-dependent minimal surface equation, has been studied in [12] and [17]. Another examples of problems of type (1.1) included in our case are the following: The evolution problem for plastic antiplanar shear, studied in [21], which corresponds to the plasticity functional f given by 8 1 2 > if k k  1 < 2 k k f () = > : kk ? 21 if kk  1;

u(0; x) = u0 (x) 2 where u0 2 L ( ) and a(x;  ) = r f (x;  ), f

 Dept. de Analisis Matematico, Universitat de Valencia, 46100 Burjassot (Valencia), Spain, [email protected] y Dept. de Tecnologia, Universitat Pompeu-Fabra, La Rambla 30-32, 08002 Barcelona, Spain,

[email protected] z Dept. de Analisis Matematico, Universitat de Valencia, 46100 Burjassot (Valencia), Spain, [email protected]

1

and the evolution problems associated with the Lagrangians: q

f (x; ) = 1 + aij (x)i j ; where the functions aij are continuous and satisfy aij (x) = aji(x), k k2  aij (x)i j  C k k2 for all  2 IRN ; and the Lagrangian q f (x; ) = 1 + x2 + kk2 ; which was considered by S. Bernstein ([8]) On the other hand, problem (1.1) is studied in [14] for some Lagrangians f , which do not include the nonparametric area integrand, but instead include the plasticity functional and the total variation ow, that is, the case f ( ) = k k. Now, the concept of solution given in [14] is the one obtained by considering the abstract Cauchy problem in L2 ( ) associated to the relaxed energy, but the subdierential of the energy functional is not characterized . For the particular case of the total variacion ow, we give in [4] a dierent approach to the Dirichlet problem. There, we studied the problem in the framework of the L1 - theory, and we characterized the subdierential in L2 ( ) of its relaxed energy (we refer also to [3] where we treated the L1 -theory for the Neuman problem for the total variational ow). In general, the problem (1.1) does not have a classical solution. The aim of this paper is to introduce a concept of solution of the Dirichlet problem (1.1), for which existence and uniqueness for initial data in L2 ( ) is proved. To do that we characterize the subdierential of the energy associated with the problem and we use the nonlinear semigroup theory. In a forthcoming paper we will study the same problem in the framework of the L1 -theory, as we did with the Dirichlet problem for the total variational ow ([4]). In order to consider the relaxed energy we recall the de nition of function of measure (see for instance, [6] or [12]). Let g :  IRN ! IR be a Caratheodory function such that jg(x; )j  M (1 + kk) 8 (x; ) 2  IRN ; (1.2) for some constant M  0. Furthermore, we assume that g possesses an asymptotic function, i.e. for almost all x 2 there exists the nite limit   lim tg x; t = g0 (x;  ): t!0+ 

(1.3)

It is clear that the function g0 (x;  ) is positively homogeneous of degree one in  , i.e. g0 (x; s) = sg0 (x; ) for all x;  and s > 0: We denote by M( ; IRN ) the set of all IRN -valued bounded Radon measures on . Given  2 M( ; IRN ), we consider its Lebesgue decomposition  =  a + s ; where a is the absolutely continuous part of  with respect to de Lebesgue measure N of IRN , and s is singular with respect to N . We denote by a(x) the density of the measure a with respect to N and by (ds =djjs )(x) the density of s with respect to jjs. Given  2 M( ; IRN ), we de ne ~ 2 M( ; IRN +1 ) by ~(B ) := ((B ); N (B )); 2

for every Borel set B  IRN . Then, we have

~ = ~a + ~s = ~a(x)N + ~s = (a (x); 1)N + (s; 0): Hence, we have

d~s =  ds ; 0 jsj ? a:e: dj~sj djsj

j~ j = j j; s

s

For  2 M( ; IRN ) and g satisfying the above conditions, we de ne the measure g(x; ) on as   s g0 x; dd ( x ) djjs (1.4) s j  j B B B for all Borel set B  . In formula (1.4) we may write (d=djj)(x) instead of (ds =djjs )(x), because the two functions are equal jjs -a.e. Another way of writing the measure g(x; ) is the following. Let us consider the function g~ :

 IRN  [0; +1[! IR de ned as 8  > > g (x; )t if t > 0 < t (1.5) g~(x; ; t) := > > : g0 (x; ) if t = 0 As it is proved in [6], if g is a Caratheodory function satisfying (1.2), then  Z Z d (x); dN (x) d; (1.6) g(x; ) = g~ x; d d B B where  is any positive Borel measure such that jj + N  . Due to the linear growth condition on the Lagrangian g, the natural energy space to study (1.1) is Z

g(x; ) :=

Z

g(x; a (x)) dx +

Z

the space of functions of bounded variation. Let us recall several facts concerning functions of bounded variation (for further information concerning functions of bounded variation we refer to [13], [22] or [2]). A function u 2 L1 ( ) whose partial derivatives in the sense of distributions are measures with nite total variation in is called a function of bounded variation. The class of such functions will be denoted by BV ( ). Thus u 2 BV ( ) if and only if there are Radon measures 1 ; : : : ; N de ned in with nite total mass in and Z



Z

uDi 'dx = ? 'di

(1.7)

for all ' 2 C01( ), i = 1; : : : ; N . Thus, the gradient of u is a vector valued measure with nite total variation Z k Du k= supf u div ' dx : ' 2 C01( ; IRN ); j'(x)j  1 for x 2 g: (1.8)

The space BV ( ) is endowed with the norm

k u kBV =k u kL1 ( ) + k Du k : 3

(1.9)

For u 2 BV ( ), the gradient Du is a Radon measure that decomposes into its absolutely continuos and singular parts Du = Da u + Dsu. Then Da u = ru N where ru is the Radon-Nikodym derivative of the measure Du with respect to the Lebesgue measure N . There is also the polar decomposition s Dsu = ? D?! ujDsuj where jDsuj is the total variation measure of Dsu. We shall need several results from [5] (see also [16]). Following [5], let

X ( ) = fz 2 L1( ; IRN ) : div(z) 2 L1 ( )g: (1.10) If z 2 X ( ) and w 2 BV ( ) \ L1 ( ) we de ne the functional (z; Dw) : C01( ) ! IR by the formula Z

< (z; Dw); ' >= ? w ' div(z) dx ?

Z

w z
r' dx:

(1.11)

8 w 2 W 1;1( ) \ L1( )

(1.12)





Then (z; Dw) is a Radon measure in , Z



and

(z; Dw) = Z

B

Z



z
rw dx

(z; Dw) 

Z

B

Z

j(z; Dw)j  kzk1 kDwk B

(1.13)

for any Borel set B  . Moreover, (z; Dw) is absolutely continuous with respect to kDwk with Radon-Nikodym derivative (z; Dw; x), which is a kDwk measurable function from to IR such that Z

B

(z; Dw) =

Z

B

(z; Dw; x)kDwk

(1.14)

for any Borel set B  . We also have that

k(z; Dw; :)kL1 ( ;kDwk)  kzkL1 ( ;IR ): N

By writing

(1.15)

z
Dsu := (z; Du) ? (z
ru) dN ;

we see that z
Dsu is a bounded measure. Furthermore, in [16] it is proved that z
Dsu is absolutely continuous with respect to jDs uj (and, thus, it is a singular measure with respect to N ), and

jz
Dsuj  kzk1 jDsuj:

(1.16)

As a consequence of Theorem 2.4 of [5], we have:

s If z 2 X ( ) \ C ( ; IRN ); then z
Ds u = (z
? D?! u) djDs uj

(1.17)

In [5], a weak trace on @ of the normal component of z 2 X ( ) is de ned. Concretely, it is proved that there exists a linear operator : X ( ) ! L1(@ ) such that

k (z)k1  kzk1

(z)(x) = z(x)
 (x) for all x 2 @ if z 2 C 1( ; IRN ): 4

We shall denote (z )(x) by [z;  ](x). Moreover, the following Green's formula, relating the function [z;  ] and the measure (z; Dw), for z 2 X ( ) and w 2 BV ( ) \ L1( ), is established: Z



Z

w div(z) dx + (z; Dw) =

Z

@



[z;  ]w dH N ?1 :

(1.18)

Let g be a function satisfying (1.2). Then for every u 2 BV ( ) we have the measure g(x; Du) de ned by Z Z Z ??! s g(x; Du) = g(x; ru(x)) dx + g0 (x; D u(x)) djDsuj B

B

B

for all Borel set B  . If we assume that has a Lipschitz boundary, and that g(x;  ) is de ned also for x 2 @ , we may consider the functional G in BV ( ) de ned by

G(u) :=

Z



g(x; Du) +

Z

@



g0 (x;  (x)['(x) ? u(x)]) dH N ?1;

(1.19)

where ' 2 L1 (@ ) is a given function and  is the outer unit normal to @ . It is proved in [6] that, if g~(x; ; t) is continuous on  IRN  [0; +1[ and convex in (; t) for each xed x 2 , then G is the greatest functional onZ BV ( ) which is lower-semicontinuous with respect to the L1 ( )-convergence and satis es G(u)  g(x; ru(x)) dx for all functions u 2 C 1 ( ) \ W 1;1 ( ) with u = ' on @ .

The paper is organized as follows. In Section 2 we give the de nition of solution for the Dirichlet problem and we state the existence and uniqueness result for this type of solutions. Section 3 is devoted to prove the existence and uniqueness result. To do that, we study the problem from the point of view of nonlinear semigroup theory. We characterize the subdierential in L2 ( ) of the relaxed energy functional associated with the problem. In Section 4 we give a weakened form of the maximum principle and we study the asymptotic behaviour of solutions proving that they stabilize as t ! 1 by converging to a solution of the steady-state problem. Finally, the Appendix contains the proof of the approximation Lemma stated in Section 3.

2 The existence and uniqueness result. In this section we de ne the concept of solution for the Dirichlet problem (1.1) and we state the existence and uniqueness result for this type of solutions when the initial data are in L2 ( ). Here we assume that is an open bounded set in IRN , N  2, with boundary @ of class C 1 , and the Lagrangian f :  IRN ! IR satis es the following assumptions, which we shall refer collectively as (H): (H1 ) f is continuous on  IRN , convex and dientiable with continuous gradient in  for each xed x 2 and satis es the linear growth condition

C0 kk ? C1  f (x; )  M (kk + C2 ):

(2.1)

for some positive constants C0 , C1 , C2 . Moreover, f 0 exists and f 0 (x; ? ) = f 0 (x;  ) for all  2 IRN and all x 2 . (H2 ) f~(x; ; t) is continuous on  IRN  [0; +1[ and convex in (; t) for each xed x 2 . 5

f

We consider the function a(x;  ) = r f (x;  ) associated to the Lagrangian f . By the convexity of

a(x; )
( ? )  f () ? f ();

and the following monotonicity condition is satis ed (a(x; ) ? a(x;  ))
( ?  )  0: Moreover, it is easy to see that ja(x; )j  M 8 (x; ) 2  IRN : We consider the function h :  IRN ! IR de ned by h(x; ) := a(x; )
: From (2.2) and (2.1), it follows that C0kk ? D1  h(x; )  M kk for some positive constants D1 . We assume that (H3 ) h0 exists and the function h~ is continuous on  IRN  [0; +1[. We need to consider the mapping a1 de ned by a1(x; ) := t!lim a(x; t): +1 Observe that

(2.2) (2.3) (2.4)

(2.5)

h0 (x; ) = a1(x; )
 and C0kk  h0 (x; )  M kk:

(H4 ) a1(x;  ) = r f 0 (x;  ) for all  6= 0 and all x 2 . In particular, as a consequence of Euler's Theorem, we have f 0(x; ) = a1(x; )
 = h0 (x; ); for all  2 IRN and all x 2 . (H5 ) a(x;  )
  h0 (x; ) for all ;  2 IRN , and all x 2 . Either from (H4 ) or (H5 ) it follows that a1 (x;  )
  h0 (x; ) for all ;  2 IRN ,  6= 0, and all x 2 . Indeed, it suces to replace  by t in (H5 ) and let t ! +1.

De nition 1 Let ' 2 L1(@ ) and u0 2 L2( ). A measurable function u : (0; T )  ! IR is a solution of (1.1) in QT = (0; T )  if u 2 C ([0; T ]; L2 ( )), u(0) = u0 , u0 (t) 2 L2 ( ), u(t) 2 BV ( ) \ L2 ( ), a(x; ru(t)) 2 X ( ) a.e. t 2 [0; T ], and for almost all t 2 [0; T ] u(t) satis es: u0 (t) = div(a(x; ru(t)) in D0( )

(2.6)

a(x; ru(t))
Dsu(t) = f 0(x; Dsu(t))

(2.7)

[a(x; ru(t));  ] 2 sign(' ? u(t))f 0 (x;  (x)) H N ?1 ? a:e: on @ :

(2.8)

6

Our main result is the following:

Theorem 1 Let ' 2 L1 (@ ) and assume we are under assumptions (H). Given u0 2 L2( ), there

exists a unique solution u of (1.1) in QT for every T > 0 such that u(0) = u0 .

3 Strong solution for data in L2( )

To prove Theorem 1 we shall use the nonlinear semigroup theory ([9]). For ' 2 L1 (@ ) we de ne the energy functional ' : L2 ( ) ! [0; +1] by 8 Z > >

0. It follows that there exists u 2 BV ( ) \ L2 ( ), such that

uj * u weakly in L2 ( ); Hence,

Z



uj ! u in Lq ( ) for all 1  q < NN?1 : Z

u2 dx  lim sup u2j dx: j

!1



(3.78) (3.79)

After passing to a subsequence, if necessary, we may assume that

a(x; ruj ) * z as j ! 1, weakly in L1( ; IRN ) 21

(3.80)

and

in D0 ( ): (3.81) According to [6], Fact 3.3, there exists a sequence fwk g  C 1 ( ) \ BV ( ) such that wk j@ = ',

?div(z) = v ? u

wk ! u in L2( );

and ' (wk ) ! ' (u):

(3.82)

Now, by the convexity of f we have Z



Z

f (x; ruj ) dx 

Z



Z

a(x; ruj )
ruj dx ? a(x; ruj )
rwk dx + f (x; rwk ) dx:



Thus, having in mind (3.74), (3.75) and (3.76), we get 'j (uj ) =

Z



Z

f (x; ruj ) dx +

Z



Z

Z

f 0(x; Dsuj ) +

@

Z



juj ? 'j jf 0(x;  (x)) dH N ?1 

 f (x; rwk ) dx + a(x; ruj )
ruj dx + f 0(x; Ds uj ) +

Z





Z

Z

Z

@



juj ? 'j jf 0(x;  (x)) dH N ?1 ?

? a(x; ruj )
rwk dx  f (x; rwk ) dx + (v ? uj )uj dx+

+



Z

@



[a(x; ruj );  ]'j dH N ?1 ?



Z



a(x; ruj )
rwk dx:

Using (3.79) and (3.80), it follows that lim sup ' (uj ) = lim sup 'j (uj )  j

Z

!1

j

Z

Z

!1

 f (x; rwk ) dx + uv dx ? u2 dx +



Z



Z

Z

@



[z;  ]' dH N ?1 ?

Z

Z



z
rwk dx 

 f (x; rwk ) dx + (v ? u)u dx + div(z)wk dx:



Hence, by (3.81), letting k ! 1, we arrive to lim sup ' (uj )  klim  (w ) = ' (u): !1 ' k j

!1

Thus, by the lower-semicontinuity of ' , we get ' (u) = jlim  (u ): !1 ' j

(3.83)

Applying Theorem 3 of [18] as in the step 1, it follows that lim j !1 =

Z

Z





h(x; Duj ) +

h(x; Du) +

Z

@

Z

@





juj ? 'j jf 0(x;  (x)) dH N ?1 =

ju ? 'jf 0(x;  (x))

dH

N

?1 :

On the other hand, by Green's formula, (3.74), (3.75) and (3.76), we have Z



h(x; Duj ) +

Z

@



juj ? 'j jf 0(x;  (x)) dH N ?1 = 22

(3.84)

=

Z



(a(x; ruj ); Duj ) +

Z

Z

@



[a(x; ruj );  ]('j ? uj ) dH N ?1 =

Z

uj (v ? uj ) dx + [a(x; ruj );  ]'j dH N ?1 :

@

 1 Since [a(x; ruj );  ] ! [z;  ] weakly in L (@ ), letting j ! +1, and using (3.84), it follows that =

Z



h(x; Du) +

Z

Z

@



ju ? 'jf 0 (x;  (x)) dH N ?1  u(v ? u) dx +

Z



Z

= (z; Du) + Now, by Lemma 4 (i), we have

Z



@



@



[z;  ]' dH N ?1 =

[z;  ](' ? u) dH N ?1 :

H N ?1 ? a:e in @ :

j[z;  ](x)j  f 0(x;  (x)) Moreover, as in step 1, we get

jz
Dsuj  f 0(x; Dsu) as measures in : With this and using Lemma 3, we obtain Z



Z

z
ru dx =



h(x; ru) dx =

Z



a(x; ru)
ru dx

z
Dsu = f 0(x; Dsu) [z;  ] 2 sign (' ? u)f 0 (x;  (x)) H N ?1 ? a:e: As in Step 1, to get that (u; v ? u) 2 B' , we only need to prove that div z = div a(x; ru) in D0 ( )

(3.85) (3.86) (3.87) (3.88)

and

[z;  ] = [a(x; ru);  ] H N ?1 ? a:e: on @ : (3.89) This is a consequence of Lemma 5, if we consider the IRN -valued measures n,  on which are de ned as Z Z n(B ) := Dun + ('n ? un ) dH N ?1 B

(B ) :=

Z

\

B

\

\

B @

Du +

Z

\

B @

(' ? u) dH N ?1

for all Borel sets B  . 2 Step 3. To prove the density of D(B' ) in L2 ( ), we prove that C01 ( )  D(A' )L ( ) . Let v 2 C01( ). By the above, v 2 R(I + n1 B') for all n 2 IN . Thus, for each n 2 IN , there exists un 2 D(B' ) such that (un ; n(v ? un )) 2 B' . Then there is a(x; run ) 2 X ( ) such that n(v ? un ) = ?div(a(x; run )) in D0( ) and Z Z (w ? un )n(v ? un ) dx = (a(x; run ); Dw)?

?



Z

@



[a(x; run );  ](w ? ') dH

N

?1 ?

Z





h(x; Dun ) ? 23

Z

@



j' ? unjf 0(x;  (x)) dH N ?1 :

for every w 2 BV ( ) \ L2 ( ). Taking w = v, we get Z Z Z a(x; run)
rv dx ? [a(x; run);  ](v ? ') dH N ?1? (v ? un )2 dx = n1



@

Z

? h(x; Dun ) ?

Z





j' ? unjf 0(x;  (x)) dH N ?1  @

Z  Z 1 N ?1  n a(x; run)
rv dx ? [a(x; run);  ](v ? ') dH 



@



 Z M N ?1 :  n jrvj dx + @ jv ? 'j dH 2 Letting n ! 1, it follows that un ! v in L2 ( ). Therefore v 2 D(B' )L ( ) and the proof is complete. 2 Z

Proof of Theorem 2. First, we prove that B'  @ '. Let (u; v) 2 B', and w 2 W'1;2( ). Then,

by (2.2), and applying Green's formula, we get Z

Z



(w ? u)v dx = ? (w ? u) div a(x; ru) dx =

Z

= (a(x; ru); Dw ? Du) ? =

Z





Z

@



[a(x; ru);  ](' ? u) dH N ?1 = Z

a(x; ru)
rw dx ? a(x; ru)
ru dx ? a(x; ru)
Dsu? ?

Z

Z



Z

@

Z





[a(x; ru);  ](' ? u) dH N ?1  Z

 f (x; rw) dx ? f (x; Du) dx ? @ j' ? ujf 0(x;  (x)) dH N ?1 = '(w) ? '(u): Suppose that w 2 BV ( ) \ L2( ). According to [6], Fact 3.3, there exists a sequence wn 2 W'1;2 ( ), with wn ! w in L2 ( ), and ' (wn ) ! ' (w). Then, by the above inequality, we have Z

Now, letting n ! 1, we get and, therefore, (u; v) 2 @ ' .



(wn ? u)v dx  '(wn ) ? ' (u):

Z



(w ? u)v dx  '(w) ? ' (u); 2

Since B'  @ ' , and, by Proposition 1, L1 ( )  R(I + B' ), we have @ ' = B' L ( ) . To nish the proof we only need to prove that the operator B' is closed. Let (un ; vn ) 2 B' , and assume that (un ; vn ) ! (u; v) in L2 ( )  L2 ( ). Let us prove that (u; v) 2 B' . Since (un ; vn ) 2 B' , we know that a(x; run) 2 X ( ) is such that

?vn = div a(x; run) in D0( ) a(x; run)
Dsun = f 0(x; Dsun) 24

(3.90) (3.91)

[a(x; run );  ] 2 sign (' ? un )f 0 (x;  (x)) H N ?1 ? a:e: Multiplying (3.90) by un and applying Green's formula we obtain Z

Z

? unvn dx = Hence,



@



[a(x; run );  ]' dH

Z

h(x; Dun ) 



Z

?1 ?

Z



unvn dx +



>From (2.5) and (3.93), we have

N

h(x; Dun ) ?

Z

@



Z

Z



Z

Z



Hence,

Z

@



@



j' ? unjf 0(x;  (x)) dH N ?1 :

[a(x; run );  ]' dH N ?1 :

C0 jDunj dx ? D1 N ( ) 

 unvn dx +

Z



(3.92)

(3.93)

h(x; Dun ) dx 

[a(x; run );  ]' dH N ?1 :

8 n 2 IN: (3.94) Therefore, u 2 BV ( ) \ L2 ( ). On the other hand, since ka(x; run )k1  M , we may assume that a(x; run) * z in the weak topology of L1( ; IRN ); (3.95) with kz k1  M . Moreover, since vn ! v in L2 ( ), we have that v = ?div(z ) in D0 ( ). By the

jDunj dx  M1

de nition of the weak trace on @ of the normal component of z , it is easy to see that weakly in L1(@ ):

[a(x; run );  ] * [z;  ]

(3.96)

Now, we prove the convergence of the energies. According to [6], Fact 3.3, there exists a sequence

wj 2 C 1( ) \ BV ( ), with wj j@ = ', wj ! u in L1( ), and '(wj ) ! '(u). Moreover, looking at the proof of Fact 3.3 in [6], we have that, wj = wj1 + wj2 with wj1 j@ = uj@ and wj1 ! u in L1 ( ), wj2 j@ = ' ? uj@ , wj2 ! 0 in L1( ), and, using [5], Lemma 1.8, we have that Z



(z; Dwj1 ) !

Z



(z; Du):

By the convexity of f and taking (3.91) and (3.92) into account we have Z

' (un ) =

f (x; run) dx +



Z

Z



f 0(x; Dsun) + Z

Z

@



jun ? 'jf 0(x;  (x)) dH N ?1  Z

 a(x; run)
run dx ? a(x; run)
rwj dx + f (x; rwj ) dx+

+



Z



a(x; run)
Dsun +

Z

= (a(x; run ); Dun ) ?

+

Z

@





Z

@



Z



[a(x; run );  ](' ? un ) dH N ?1 =

a(x; run)
rwj dx + '(wj )+

[a(x; run );  ](' ? un ) dH N ?1 = 25

= ' (wj ) ?

Z



Z

a(x; run)
rwj dx ? div(a(x; run ))un dx +

= '(wj ) ?



Z

Z

Z

a(x; run)
rwj dx + vnun dx +

@

Hence, by (3.95) and (3.96), it follows that Z

Z

lim sup ' (un )  '(wj ) ? (z; Dwj ) + !1



n

Z

Z

= ' (wj ) ? (z; Dwj1 ) ? (z; Dwj2 ) ? = '(wj ) ?

Z





(z; Dw1 ) +



Z

j



Letting j ! 1, we have that

div(z)w2

Z

lim sup ' (un )  '(u) ? (z; Du) + !1

Z



n



dx +

j

@





Z



Z

@

Z

@

div(z)u dx + @



[z;  ]u dH



[z;  ]' dH N ?1 =

Z

[z;  ]' dH N ?1 =

Z N ?1 dH ?

[z;  ]u N

[a(x; run );  ]' dH N ?1 =

[a(x; run );  ]' dH N ?1 :

uv dx +

Z



?1 ?

@

Z





div(z)u dx:

div(z)u dx = '(u):

Finally, by the lower-semicontinuity of ' , we obtain ' (u) = nlim !1 ' (un ):

(3.97)

If we consider the IRN -valued measures n ,  on which are de ned as

n(B ) := (B ) :=

Z

B

\

Dun +

Z

B

for all Borel sets B  , we have

\

j !  Moreover,

' (u) =

Z

Hence, (3.97) yields

Du +

lim n!1

Z

\

B @



Since

Z

(' ? u) dH N ?1

Z



f~(x; ~n) = Z

Z



Z



h~ (x; ~) =



Z



h(x; Du) +

Z

@



26



f~(x; ~n ):

f~(x; ~):

~ h~ (x; ~) = nlim !1 h(x; ~n ) = nlim !1 h(x; Dun ) +

Z

' (un ) =

and

Then, applying [18], Theorem 3, it follows that Z

(' ? un ) dH N ?1

\

B @

weakly as measures in :

f~(x; ~)



Z

Z

@



jun ? 'jf 0 (x;  (x)) dH N ?1:

j' ? ujf 0(x;  (x)) dH N ?1 ;

we have

Z



h(x; Du) +

= nlim !1

Now, since

lim !1

Z





Z

n

= nlim !1

Z



Z

@



j' ? ujf 0(x;  (x)) dH N ?1 =

h(x; Dun ) +

h(x; Dun ) + Z

Z

@

Z

@





jun ? 'jf 0 (x;  (x)) dH N ?1 =



Z

Z



we nally obtain Z



= nlim !1

Z

@

Z

a(x; run )
run dx + f 0(x; Dsun) + = nlim !1 (a(x; run ); Dun ) +

=

jun ? 'jf 0(x;  (x)) dH N ?1:



Z

@



[a(x; run );  ]' dH

[z;  ]' dH N ?1 ?

h(x; Du) +

Z

@



Z



(3.98)

@ N



@



[a(x; run );  ](' ? un ) dH N ?1 =

[a(x; run );  ](' ? un ) dH N ?1

?1 ?

Z

Z

div(a(x; run ))un dx =



div(z)u dx = (z; Du) +

Z



@

Z

Z

j' ? ujf 0(x;  (x)) dH N ?1 = (z; Du) +



@

[z;  ](' ? u) dH N ?1 ;



[z;  ](' ? u) dH N ?1 :

(3.99)

Again, by (3.95) and (3.96) we can apply Lemma 4 to get the inequality

j[z(x);  (x)]j  f 0(x;  (x)) a.e. in @ :

(3.100)

Moreover, acting as in the proof of Proposition 1, we get that

jz
Dsuj  f 0(x; Dsu) as measures in :

(3.101)

Hence by (3.97), (3.98), (3.99), (3.95) (3.101) and (3.100), we may apply Lemma 3, to obtain Z



z
ru dx =

Z



h(x; ru) dx =

Z



a(x; ru)
ru dx

z
Dsu = f 0(x; Dsu) [z;  ] 2 sign (' ? u)f 0 (x;  (x)) H N ?1 ? a:e:: Now, using Lemma 5 with 'n = ', we have div z = div a(x; ru)

in D0 ( )

and

[z;  ] = [a(x; ru);  ] H N ?1 ? a:e: on @ : Since v = ?div(z ) in D0 ( ), taking (3.105) into account, we get

v = ?div(a(x; ru)) 27

in D0 ( );

(3.102) (3.103) (3.104) (3.105) (3.106)

and, using (3.102), (3.103) and (3.105), we get

a(x; ru)
Dsu = f 0(x; Dsu): Finally, by (3.104) and (3.106) we get [a(x; ru);  ] 2 sing (' ? u)f 0 (x;  (x))

H N ?1 ? a:e:

Therefore, (u; v) 2 B' .

2

Proof of Theorem 1. Let (S (t))t0 be the semigroup in L2( ) generated by the subdiferential of ' . Then by the nonlinear semigroup theory ([9]), given u0 2 L2 ( ) = D(@ ' ), u(t) = S (t)u0 is the only strong solution of problem (3.1). Thus, by Theorem 2, we have that for almost all t 2 [0; +1[, u(t) 2 D(B') and ?u0(t) 2 B'(u(t)). This concludes the proof. 2

4 Behaviour of the solution We have the following weak form of the maximum principle.

Theorem 3 Suppose u1 and u2 are two solutions of (1.1) corresponding to unitial data u1;0 and u2;0 in L2 ( ) and boundary data '1 and '2 in L1 (@ ), respectively. If

u1;0  u2;0 and '1  '2; then u1  u2 .

Proof. For almost all t 2 [0; +1[, we have u0i(t) 2 L2( ), ui (t) 2 BV ( ) \ L2( ), a(x; rui (t)) 2 X ( ), and

u02 (t) ? u01 (t) = div[a(x; ru2 (t)) ? a(x; ru1(t))] in D0( )

(4.1)

a(x; rui(t))
Dsui(t) = f 0(x; Ds ui(t))

(4.2)

[a(x; rui (t));  ] 2 sign('i ? ui (t))f 0 (x;  (x)) H N ?1 ? a:e: on @ : Multiplying in (4.1) by (u2 (t) ? u1 (t))+ , integrating in , and using Green's formula, we get 1 Z d [(u (t) ? u (t))+ ]2 = 1 2 dt 2 =

Z



=? +

div[a(x; ru2 (t)) ? a(x; ru1 (t))](u2 (t) ? u1(t))+ =

Z



Z

@



(a(x; ru2 (t)) ? a(x; ru1 (t)); D((u2 (t) ? u1

(t))+ ))+

[a(x; ru2 (t)) ? a(x; ru1 (t));  ](u2 (t) ? u1 (t))+ : 28

(4.3)

(4.4)

Now, by the chain rule for BV-functions ([2],[14], Lemma 1.2), there exists a scalar function (t), with 0  (t)  1, such that Z

(a(x; ru2 (t)) ? a(x; ru1 (t)); D((u2 (t) ? u1 (t))+ )) =



=

Z

fu2 u1 g Z

+



(a(x; ru2 (t)) ? a(x; ru1 (t))
(ru2 (t) ? ru1 (t)) dx+

(t)(a(x; ru2(t)) ? a(x; ru1 (t))
Ds(u2(t) ? u1(t)):

Observe that, by the monotonicity of a, (H5 ) and (4.2), we have that Z



(a(x; ru2 (t)) ? a(x; ru1 (t)); D((u2 (t) ? u1 (t))+ ))  0:

(4.5)

On the other hand, since '1  '2 , from (4.3), it is easy to see that Z

@



[a(x; ru2 (t)) ? a(x; ru1 (t));  ](u2 (t) ? u1 (t))+  0:

(4.6)

From (4.4), (4.5) and (4.6), it follows that 1 Z d [(u (t) ? u (t))+ ]2  0: 1 2 dt 2 Since u1;0  u2;0 , we have u1  u2 , and the proof is concluded. 2 We shall now prove that the solution u(t) stabilizes as t ! +1 by converging to a solution of the steady-state problem. To do that, we follow the proof of Theorem 4.2 in [17].

Theorem 4 Suppose u0 2 L2 ( ) \ BV ( ) and ' 2 L1(@ ). Then the solution u(t) of (1.1) converges as t ! +1 to some limit w 2 B'?1 (0) in the following sense: u(t) ! w strongly in L1( ) and weakly in L2( ): Proof. Since B' is the subdierential of ', by a classical result of Bruck ([10], Theorem 4), to

prove the weak convergence in L2 ( ), it is sucient to prove that ' attains its minimun in L2 ( ). In fact, let fun g be a minimizing sequence for '. Without loss of generality, we may assume that un 2 BV ( ) \ L2 ( ). Now, by approximation we may assume that un 2 W 1;1( ) \ L2( ). Denote by J : IR ! IR the truncature function 8 ?k'k1 if r < ?k'k1 > > > > >
r > > > > :

if

x > k'k1 : If we take wn := J  un , wn 2 W 1;1 ( ) \ L1 ( ), and, using that jJ 0 j  1, we have ' (wn ) =

Z



k'k1

jrj  k'k1

f (x; rwn) +

if

Z

@



jwn ? 'jf 0(x;  (x)) dH N ?1 =

29

=

Z

fjun jk'k1 g Z

f (x; run) +

 f (x; run) +

Z

Z

@

@





jJ  un ? J  'jf 0(x;  (x)) dH N ?1  jun ? 'jf 0(x;  (x)) dH N ?1 :

Thus, fwn g is still a minimizing sequence for ' . Moreover, this sequence is bounded in W 1;1( ) \ L1( ), hence, relatively compact in L1( ). We may extract a subsequence converging in L1( ) to some u 2 L1 ( ) \ BV ( ). Therefore, ' (u) = inf2 ' (u): 2 ( )

u L

Then, by Bruck's result ([10], Theorem 4), there exists w 2 B'?1 (0), such that u(t) ! w weakly in L2( ). Finally, we prove the strong convergence in L1( ). Since (u(t); ?u0 (t)) 2 @ ', using [9], Lemma 3.3, we have d  (u(s)) = ? Z u0 (s)2  0; ds '

hence,

'(u(t))  ' (u0 ) 8 t > 0: Thus, fu(t) : t  0g is bounded in BV ( ), and, therefore relatively compact in L1 ( ). The result follows. 2

5 Appendix In this appendix we prove the approximation Lemma. Before giving the proof, let us construct a substitute for the distance function to the boundary d(:; @ ). That construction would be unnecessary if @ would be of class W 2;1 ([7]). We follow the proof of Lemma 5.1 in [7] for C 2 domains. If @ is a manifold of class C 1 , then there is some  > 0 such that for all points y 2 such that d(y; ) <  there is z 2 @ and t 2 (0; ) such that y = z ? t (z),  (z) being the outer unit normal to @ at z ([11]). In other words,  := fx 2 : x = y ? t (y); y 2 @ ; t 2 (0; )g is open. Then there is a function D 2 C 1 ( ) such that D = 0 on @ , D > 0 on and rD(x) = ? (x) for all x 2 @ . This is a consequence of Withney's extension Theorem ([15], p.48, [13], p. 245). Indeed, since

y  (y) jxx ? ? yj ! 0 as x; y ! p, x 6= y, x; y 2 @ ,

by Withney's Theorem, we know that there exists a function D~ 2 C 1 ( ) such that D~ = 0 on @ and rD~ (x) = ? (x) for all x 2 @ . Now, let y 2 @ and t 2 (0; ). Using the mean value theorem, we know that D~ (y ? t (y)) = D~ (y) ? trD~ (y ? s (y))
 (y) = ?trD~ (y ? s (y))
 (y): Since D~ 2 C 1 ( ), we have D~ (y ? t (y)) = t(1 + !(t)) where !(t) = o(1) as t ! 0+ and is a modulus of continuity for rD~ . Without loss of generality we may assume that  > 0 is such that !(t) < 12 for all t 2 (0; ). In particular, we have that D~ (x) > 0 for all x 2 : (5.1) 30

We shall modify D~ so that the modi ed function is > 0 in . Let  2 C ([0; 1)), (t) > 0, for all t 2 (0; 1), (t) = o(t) as t ! 0+. Let 1 be an open set, 1  , with smooth boundary @ 1   such that 0 <  ? () < D~ (x) <  + () for all x 2 @ 1 for some  > 0. Let 02 be an open set with smooth boundary such that 02  1 and () < d(@ 1 ; @ 02 ) < 2(), where d(@ 1 ; @ 02 ) = inf fjx ? yj : x 2 @ 1 ; y 2 @ 02 g. Let d@ 02 be the distance function to @ 02, > 0 in

02 , negative outside. Let d@ 02 ;n = n  d@ 02 , n being a positive regularizing kernel. Observe that krd@ 02 ;nk1  1. We may choose n large enough, and 2 such that 2  02, () < d(@ 1 ; @ 2 ) < 2(), 0 < d@ 02 ;n < () in @ 2 , and d@ 02 ;n > 0 in 2 . Let B1;2 = 1 n 2 . Then, using again Withney's extension Theorem, there is a function R 2 C 1 (B1;2 ) such that R = D~ ?  and rR = rD~ on @ 1 , and R = d@ 02 ;n , rR = rd@ 02 ;n on @ 2 . Moreover, krRk1 is bounded by a constant depending on kD~ k1;@ 1 ; kd@ 02 ;nk1;@ 2 krD~ k1;@ 1 ; krd@ 02 ;nk1;@ 2 and

jD~ (x) ?  ? d@ 02 ;n(y)j 2()  () = 2: jx ? yj x2@ 1 ;y 2@ 2 sup

We de ne D : ! IR by

D = D~ in n 1; D = R +  in B1;2; D = d@ 02 ;n +  in 2: Then D 2 C 1 ( ), D = 0 on @ , D > 0 on and rD(x) = ? (x) for all x 2 @ . Proof of Lemma 2. We may think that u and v are extended as BV functions in IRN in such a way that

Z

@



jDuj =

Z

@

jDvj = 0:



We consider a family of radially symmetric positive molli ers j = j # 0+, and we set zj = j  (v + uj )

(5.2) 1

jN

( ),   0, x j

Z

IRN

(x)dx = 1, (5.3)

Clearly, we have zj 2 C 1 ( ) and obviously we have

zj ! v in LN=(N ?1) ( ):

(5.4)

Also, from (5.2) it follows that Z



q

1 + jDzj (x)j2 dx !

Z



q

1 + jDv(x)j2 dx:

(5.5)

This implies, by the Theorem of convergence of traces for BV functions that

zj j@ ! vj@ in L1 (@ ):

(5.6)

By the Theorem of dierentiation of measures ([7]), we obtain

Dzj (x) ! rv(x) N a.e. in : 31

(5.7)

Indeed, since Dzj = j  Dv + 1j j  Du, this is a consequence of the four following limits lim [j  rv](x) = rv(x) N a.e. in ; j

(5.8)

lim [j  (Dv)s ](x) = (Dv)s (x) = 0 N a.e. in ; j

(5.9)

1 [  ru](x) = ru(x) lim 1 = 0  a.e. in ; lim j N j j j j

(5.10)

1 [  (Du)s ](x) = 0  a.e. in ; lim j N j j

(5.11)

lim [j  rv](x) = 0 jDvjs a.e. in ; j

(5.12)

lim [j  ru](x) = 0 jDvjs a.e. in ; j

(5.13)

since (Du)s , (Dv)s are singular with respect to N and jru(x)j < 1 N a.e. in . In the same way, using the Theorem of dierentiation of measures, we have

lim [j  (Du)ss ](x) = 0 jDvjs a.e. in ; j

1 [  (Du)sa ](x) = (Du)sa (x) lim 1 = 0 jDvjs a.e. in ; lim j j j j

(5.14) (5.15)

where (Du)sa , (Du)ss denote the absolutely continuous and singular part of (Du)s with respect to (Dv)s , and we obtain Dv (x) jDvjs a.e. in : Dzj (x) = lim Dzj (x) = jDv (5.16) lim j j[j  (Dv )s ]j(x) j jDzj (x)j j Similarly lim jDzj (x)j = lim j[j  jDvjs ](x)j = 1 jDvjs a.e. in : (5.17) j j Next, we prove that for a suitable choice of the numbers j one has

Du (x) jDujss a.e.: Dzj (x) = lim ss j (1=j )[  jDuj ](x) jDuj j

(5.18)

Assuming this, it is easy to prove that

Dzj (x) = Du (x) jDujss a.e.: lim j jDz (x)j jDuj j

Indeed,

Dzj = j  Dv(x) + 1j j  Du(x)

= j  rv(x) + j  (Dv)s (x) + 1j j  ru(x) + 1   (Du)sa (x) + 1   (Du)ss (x):

j

j

j

32

j

(5.19)

Since j  rv(x) ! 0, j  (Dv)s (x) ! 0, 1j j  ru(x) ! 0, 1j j  (Du)sa (x) ! 0 jDujss -a.e., we see that (5.19) follows from (5.18). To prove (5.18) we observe that Dzj (x) [j  Dv](x) [j  Du](x) : = + (5.20) ss ss (1=j )[j  jDuj ](x) (1=j )[j  jDuj ](x) [j  jDujss ](x) Since [j  Du](x) ! Du (x) jDujss a.e.; (5.21) [  jDujss ](x) jDuj it is sucient to prove that

j

[j  Dv](x) ! 0 jDujss a.e.:

1 [j  jDujss ](x)

To prove (5.22), we de ne

j

Dv](x) : a (x) = [[ jDu jss](x)

Since Dv and jDujss are mutually singular, then



(5.22) (5.23)

a (x) ! 0 jDujss a.e.: Thus, if we consider the sets for any xed j 2 IN we have

E (; j ) = fx 2 : ja (x)j > j12 g;

lim jDujss(E (; j )) = 0: !0 For each j 2 IN , there is some j such that jDujss(E (j ; j )) < 21j ; that is jDujss(fx 2 : j ja (x)j > 1j g) < 21j : 

This easily implies that

lim ja (x) = 0; jDujss a.e.: !1 j

j

which is exactly (5.22). Moreover, we may choose j such that 1j [j  jDujss ](x) ! 1 jDujss a.e.. Using this, and (5.18), it follows that

jDzj j(x) ! 1 jDujss a.e.:

(5.24)

We observe that up to know we have not used neither the hypothesis on the regularity of @ nor the regularity of g. We observe that the functions zj that we have constructed satisfy some of the requirements of the Lemma but not all of them, in particular, (3.6), (3.8), (3.12), (3.13) have yet to be satis ed. For that, we construct suitable correction functions j and j around the boundary and we shall de ne

vj = z j +  j +  j : 33

Let gj 2 C 1 (@ ) be such that gj ! g in L1 (@ ). We shall construct the sequence of functions j 2 C 1( ) such that j = gj ? zj on @

(5.25) Z



j (x) = 0

Z


Dj !



for all 2 C ( ; IRN ),

jj j

?1

!0

(5.26)

if D(x) > j + 2j

Z

Z

N N

@


 (g ? v)dH N ?1



Z

jDj j ! @ jv ? gjdH N ?1

jD(j + zj )(x)j ! 1 H N ?1 a.e. in T = fx 2 @ : g(x) 6= v(x)g. D(j + zj )(x) ! g(x) ? v(x)  (x) H N ?1 a.e. in T = fx 2 @ : g(x) = 6 v(x)g. jD( + z )(x)j jg(x) ? v(x)j j

j

(5.27) (5.28) (5.29) (5.30) (5.31)

Construction of j . For each number  2 (0; 0 ) we consider a function h (t) : [0; 1) ! [0; 1) such that h 2 C 1([0; 1)); h0(t)  0; h0(0) = ? 1 ; h0(t) is not decreasing; h(0) = 1; h(t) = 0 for t   + 2:

Let fn g1 n=1 be a decreasing sequence of numbers such that

21 < 0 < 1; lim j = 0: j Now, let G 2 W 1;1( ) such that Gj@ = g. Since gj 2 C 1 (@ ), we may consider a functionZ Gj 2 C 1 ( ) Z which is an extension of gj . We may assume that Gj ! G in L1 ( ) and jrGj j ! jrGj. We



de ne j = [Gj (x) ? zj (x)]hj (D(x)): (5.32) Clearly, j 2 C 1 ( ), j = gj ? zj on @ ; and, if D(x) > j + 2j , then hj (D(x)) = 0, and, therefore

j (x) = 0: Now, Z



jj j

( ?1)

N= N

=

Z

2j

jj j

( ?1)

N= N



34

Z

2j

jGj (x) ? zj (x)jN=(N ?1)

where, for any  > 0, we denote

 = fx 2 : D(x) < g: The functions Gj ; zj being independent of j , we may choose j > 0 small enough such that Z jG (x) ? z (x)jN=(N ?1) < 1 : j

2j

Hence

Z



j

j

jj jN=(N ?1) ! 0 as j ! 1:

Let 2 C ( ; IRN ). Since rj (x) = r(Gj ? zj )(x)hj (D(x)) + (Gj ? zj )(x)h0j (D(x))rD(x) we have Z



(x)
rj (x)dx = + = +

Z



Z



(x)
(rGj (x) ? rzj (x))hj (D(x))dx (Gj (x) ? zj (x)) (x)
r(hj (D(x)))dx

Z

j +2

Z

j

j +2

(x)
(rGj (x) ? rzj (x))hj (D(x))dx (Gj (x) ? zj (x)) (x)
h0j (D(x))rD(x)dx:

j

Again, since jh j  1 for all  > 0, a proper choice of j guarantees that Z

j +2

(x)
(rGj (x) ? rzj (x))hj (d(x))dx ! 0

j

as j ! 1. Now, by our choice of Gj , (5.5) and a proper choice of j , we have that Z

j +2

(Gj (x) ? zj (x)) (x)
h0j (D(x))rD(x)dx !

j

Z

@




 (g ? v)dH N ?1 :

(5.33)

Indeed, using the change of variable's formula ([13], p. 118, [20], p. 96), Z

j +2

=

Z  j

0

(Gj (x) ? zj (x)) (x)
h0j (D(x))rD(x)dx

j

+2j Z [D=]

rD(y) h0 (D(y))dH N ?1 (y)d (Gj (y) ? zj (y)) (y)
jr D(y)j j

= (j + 2j )h0j (j )

Z

[D=j ]

rD(y) dH N ?1 (y) (Gj (y) ? zj (y)) (y)
jr D(y)j

for some j 2 (0; j + 2j ) by the intermediate value Theorem. Now, since Gj ; zj do not depend on our choice of j , by choosing j ! 0+ suciently fast, we may gurantee that the integral Z rD(y) dH N ?1 (y) (Gj (y) ? zj (y)) (y)
jr D(y)j [D=j ] 35

is as near as we want to

Z

@



(Gj (y) ? zj (y)) (y)
 (y)dH N ?1 (y):

Hence, by choosing j ! 0+ suciently fast, we obtain(5.33). Hence Z



(x)
rj (x)dx !

Z

@

as j ! 1. In particular, we have Z

limj inf




 (g ? v)dH N ?1

Z

jrj (x)jdx 

@



jg ? vjdH N ?1 :

(5.34)

On the other hand, we have Z



jrj (x)jdx 

Z

j +2

jrGj (x) ? rzj (x)jdx +

Z

j +2

j

jGj (x) ? zj (x)jjh0 (D(x))jjrD(x)jdx: j

j

Again, a suitable choice of j guarantees that Z

j +2

jrGj (x) ? rzj (x)jdx ! 0

j

as j ! 1. Similarly, the properties of Gj , zj and a choice of j imply that Z

j +2

jGj (x) ? zj (x)jjh0j (D(x))jjrD(x)jdx !

j

Hence

lim sup

Z

j



jrj (x)jdx 

Z

This, together with (5.34) proves that lim j

Z



jrj (x)jdx =

@



@



jg(x) ? v(x)jdH N ?1 :

jg(x) ? v(x)jdH N ?1 :

Z

@

Z



jg ? vjdH N ?1 :

(5.35)

Finally, since

Dj + Dzj = rGj hj (D) + rzj (1 ? hj (D)) + (Gj ? zj )h0j (D)rD; we may write on @

Hence, on @ , we have

Dj + Dzj = rGj ? (Gj ? zj )h0j (0) (x): Dj + Dzj = rGj ? (Gj ? zj )h0j (0) jDj + Dzj j jrGj ? (Gj ? zj )h0j (0) j Gj + (Gj ? zj ) = jj r r j Gj + (Gj ? zj ) j 36

(5.36)

Now, choosing j such that j rGj ! 0 as j ! 1, we obtain that Dj + Dzj ! g(x) ? v(x)  (x) jDj + Dzj j jg(x) ? v(x)j

(5.37)

H N ?1 a.e. in T = fx 2 @ : g(x) 6= v(x)g. Next, a proper choice of j in (5.36) guarantees that jDj (x) + Dzj (x)j ! 1 (5.38) H N ?1 a.e. in T . Next, we construct a sequence of functions j 2 C 1 ( ) such that

j = 0 Z



jj j

j = 0

for some j > 0,

Z

on @

(5.39)

!0

(5.40)

if D(x) > j2

(5.41)

N N

?1

jDj j ! 0

there is j0 (x)

For H N ?1 -a.e. x 2 T , such that Dj (x) = 0 for all j  j0 (x) If we set vj = zj + j + j then (3.13) holds:

Construction of j .

For all  > 0 consider a function



(5.42) (5.43) (5.44)

: [0; 1) ! [0; 1) such that  

2 C 1([0; 1));

(0) = 0;



(t) = 0 for t  2 ;

j 0 (t)j  4 ; for t 2 (0; 2 ) 0 (0)  1 ;





Z

0

We de ne  : @ ! IR

1 0 j  (t)jdt  2:

u(x) ? g(x) if g(x) = v(x) and g(x) 6= u(x)  (x) = > ju(x) ? g(x)j 8 > < :

(5.45)

0 elsewhere Let j be a sequence of functions in C 1 (@ ) converging to  in L1 (@ ). Now, we may assume that  is the trace Rof a functionR  2 W 1;1( ) and j are traces of functions j 2 C 1 ( ) such that j !  in L1( ) and jDj j ! jDj. Let j be a decreasing sequence of positive numbers that converges to 0 and consider the functions j (x) = j (x) j (D(x)): (5.46) 37

Clearly, j 2 C 1 ( ), j (x) = 0 if x 2 @ . Also (5.41) holds. Since, by our choice of the functions  , we have j  (t)j  2: (5.47) Now, Z Z N=(N ?1)  2j jj jN=(N ?1) jj j



which tends to 0 as j ! 1, which proves (5.40). Our purpose now is to choose the functions j such that (5.43) holds. For that, we consider the sets N + = fx 2 @ :  (x) = 1g N ? = fx 2 @ :  (x) = ?1g N = N + [ N ?: We consider increasing sequences of compact sets Kj+  N + , Kj?  N ? such that lim H N ?1(N + n Kj+) = lim H N ?1(N ? n Kj?) = 0: j j We consider also decreasing sequences of open sets G+j  N + , G?j  N ? such that lim H N ?1(G+j n N +) = lim H N ?1(G?j n N ?) = 0: j j Now, we take functions j+; j? 2 C 1 (@ ) with values in [0; 1] such that 8 >

:

0 in @ n G?j ,

j+(x) = >  ?(x) = j

and we set The functions j satisfy

(5.49)

j = j+ ? j?: 8 > > > > >
0 > > > > :

Moreover

(5.48)

in Kj+ n G?j in @ n (G+j [ G?j )

(5.50)

?1 in Kj? n G+j .

lim H N ?1 (T \ (G+j [ G?j )) = 0: j

Recall that the functions j are extensions of j to . Now,

rj (x) = rj (x)  (D(x)) + j (x) 0 (D(x))rD(x): j

j

38

(5.51)

If x 2 @ , then

rj (x) = ?j (x) 0 (0) (x): Now, using (5.51), for almost all x 2 T = fx 2 @ : g(x) 6= v(x)g, there exists j0 (x) 2 IN such that j (x) = 0 for all j  j0 (x). Hence, also rj (x) = 0 for all j  j0 (x): j

Next Z

Z



jrj j  2j jrj j + kj k1krDk1

Z

2

 2j jrj j + kj k1krDk1 4 Z



j 0 (D(x))jdx j

j Z

2

j

dxdx:

j

Now, for j large enough Z

2

dx =

j

2

dH N ?1(z) d  2 Z j2 Z dH N ?1(z)d  Cj2 0 [D=] [D=] jrD(z )j

Z  Z j

0

where C depends on Per(@ ). Hence Z

Z



jrj j  2j jrj j + kj k1krDk14Cj :

Now, choosing j we may guarantee that



Z



as j ! 1. Now, since

jrj j ! 0

rvj = rGj hj + rzj (1 ? h ) + (Gj ? zj )h0 (D)rD + rj j

j

j

+ j 0 j rD;

on @ , we have

rvj (x) = rGj (x) ? (Gj (x) ? zj (x))h0 (0) (x) ? j (x) 0 (0) (x): j

j

and we may write on @

rvj = rGj ? (Gj ? zj )h0 (0) ? j 0 (0) jrvj j jrGj ? (Gj ? zj )h0j (0) ? j 0 (0) j j rGj ? j (Gj ? zj )h0 (0) ? j j 0 (0) = j rG ?  (G ? z )h0 (0) ?   0 (0) j : j j j j j j j   Now, we choose j such that j rGj ! 0, j (Gj ? zj )h0 (0) ! 0, as j ! 1, we obtain that g(x) ? u(x) rvj jrvj j ! jg(x) ? u(x)j  (x) j

j

j

j

39

j

j

j

j

(5.52)

H N ?1 a.e. on fx 2 @ : g(x) = v(x); u(x) 6= v(x)g. By choosing j to converge suciently fast to 0, from (5.52), we obtain that

jrvj (x)j ! 1 H N ?1 a.e. on fx 2 @ : g(x) = v(x); u(x) 6= v(x)g.

Let us now check that vj = zj + j + j satis es the required properties. Since vj = gj on @ , (3.6) follows immediately. The property (3.7) follows from (5.4), (5.26) and (5.40). To check (3.8), let 2 C 1( ; IRN ), = ( 1 ; :::; n ) and N +1 2 C 1( ; IR). Using (3.6) and (3.7), we obtain Z

N X

[ lim j

i=1 Z

= ? lim j = ?

Z

Z

=





i

(x)Di vj (x) +

div (x)vj (x)dx +

div (x)v(x)dx +



N

[ (x)
Dv(x) +

N

Z

@



+1 (x)]dx =

Z

@



gj
 +

g
 +

+1 (x)]dx +

Z

N



Z

@

Z





N

+1 (x)dx

+1 (x)dx

(g ? v)
:

Now, because of the lower semicontinuity of the total variation with respect to weak convergence, we have Z q Z Z q N ?1 2 1 + jDvj + jg ? vjdH  limjinf 1 + jrvj j2dx:

@

On the other hand, since Z

q







1 + jrvj j2 dx 

Z

q



1 + jrzj j2 dx +

Z



Z

jrj j + jrj j

using (5.5), (5.29), (5.42) we obtain that lim sup j

Z

q



1 + jrvj j2 dx 

Z

q

1 + jDvj2 +



Z

@



jg ? vjdH N ?1 :

This proves (3.8). Now, using (5.7), (5.27) and (5.41) we obtain (3.9). Next, we observe that (3.10) is a consequence of (5.16), (5.27) and (5.41). In the same way, (3.11) is a consequence of (5.19), (5.27) and (5.41). We observe that (3.12) follows from (5.31) and (5.43). Finally, (3.13) has already been proved. 2

Acknowledgements. The rst and third authors have been partially sypported by the Spanish

DGICYT, Project PB98-1442. The second author acknowledges partial support by the TMR European Project 'Viscosity Solutions and their Applications', reference FMRX-CT98-0234.

References [1] R. A, Adams, Sobolev spaces, Academic Press, Amsterdam, 1975. [2] L. Ambrosio, N. Fusco and D. Pallara, Functions of Bounded Variation and Free Discontinuity Problems, Forthcoming book. 40

[3] F. Andreu, C. Ballester, V. Caselles and J. M. Mazon, Minimizing Total Variational Flow, To appear in Di. and Int. Equt. [4] F. Andreu, C. Ballester, V. Caselles and J. M. Mazon, The Dirichlet problem for the total variational ow, Preprint. [5] G. Anzellotti, Pairings Between Measures and Bounded Functions and Compensated Compactness, Ann. di Matematica Pura ed Appl. IV (135) (1983), 293-318. [6] G. Anzellotti, The Euler equation for functionals with linear growth, Trans. Amer. Math. Soc. 290 (1985), 483-500. [7] G. Anzellotti, BV solutions of quasilinear PDEs in divergent form, Commun. in Partial Dierential Equations 12 (1987), 77{122. [8] S. Bernstein Sur les equations du calcul des variations, Ann. Sci. Ecole Norm. Sup. 29 (1912), 431-485. [9] H. Brezis, Operateurs Maximaux Monotones, North Holland, Amsterdam, 1973. [10] R. E. Bruck, Asymmptotic convergence of nonlinear contraction semi-groups in Hilbert spaces, J. Functional Anal. 18 (1975), 15-26. [11] M. C. Delfour and J. P. Zolesio, Shape Analysis via Oriented Distance Functions, J. Functional Anal. 123 (1994), 129-201. [12] F. Demengel and R. Teman, Convex Functions of a Measure and Applications, Indiana Univ. Math. J. 33 (1984), 673-709. [13] L. C. Evans and R. F. Gariepy, Measure Theory and Fine Properties of Functions, Studies in Advanced Math., CRC Press, 1992. [14] R. Hardt and X. Zhou, An evolution problem for linear growth functionals, Commun. Partial Dierential Equations 19 (1994), 1879-1907. [15] L. Hormander, The Analysis of Linear Partial Dierential Operators I, Springer-Verlag, 1983. [16] R. Kohn and R. Temam, Dual space of stress and strains with application to Hencky plasticity, Appl. Math. Optim. 10 (1983), 1-35. [17] A. Lichnewski and R. Temam, Pseudosolutions of the Time Dependent Minimal Surface Problem, Journal of Dierential Equations 30 (1978), 340-364. [18] Yu. G. Reshetnyak, Weak convergence of completely additive vector functions on a set, Sibirsk. Mat. Z. 9 (1968), 1386-1394. (Translated) [19] R. T. Rockafellar, Duality and stability in extremum problem involvimg convex functions, Paci c J. Math. 21 (1967), 167-187. [20] L. Schwartz, Analyse IV. Applications a la theorie de la mesure, Hermann, 1993. [21] X. Zhou, An Evolution Problem for Plastic Antiplanar Shear, Appl. Math. Optm. 25 (1992), 263-285. 41

[22] W. P. Ziemer, Weakly Dierentiable Functions, GTM 120, Springer Verlag, 1989.

42