Existence and uniqueness of solutions for a degenerate quasilinear parabolic problem

Publicacions Matemàtiques, Vol 38 (1994), 327-352 . EXISTENCE AND UNIQUENES S OF SOLUTIONS FOR A DEGENERAT E QUASILINEAR PARABOLIC PROBLE M MAURIZIO BADII Abstract We consider the following quasilinear parabolic equation of degenerate type with convection term u t = cp(u)xx + b(u)x in (—L, O) x (Q, T) . We solve the associate initial-boundary data problem, wit h nonlinear flux conditions . This problem, describes the evaporatio n of an incompressible fluid from a homogeneous porous media . Th e nonlinear condition in x = O, means that the flow of fluid leavin g the porous media depends on variable meteorological condition s and in a nonlinear manner on u . In x = —L, we have an impervious boundary . Far a sufficiently smooth initial data, one prove s the existence and uniqueness of the global strong solution in th e class of bounded variation functions . 1 . IntroduGtio n In this paper we study the existence and uniqueness of solutions fo r the following degenerate quasilinear parabolic proble m (1) u t = cp(u) xx +b(u) m ,  in (—L, 0) x (O,T ) (2) cp(u(0, t)) x + b(u(O,t)) _ —v(u(0, t))q(t), for t E (O,T ) (3) cp(u(—L, t)) x + b(u(—L, t)) = O,  for t E (O, T ) (4) u(x, O) = uo(x),  in (—L, O) . Partially supported by G .N .A .F .A .-C .N .R . and M .U .R .S .T . 40% . 328  M . BADI I Throughout the remainder of the paper we shall assume that the following hypothesis are satisfie d ~pECI([0,1]}, {p(0) =cp'(0) =0andcp ' (s) > 0fors ~0an d (H9)1 _ 1 cp is Holder continuous of order 0 E (0,1 ) ; (Ha) b E C°' 1 ([0, 1]), b(0) = 0 and IÇ pise ; (Hv) v : [0,1] --~ [0,1] is continuous, increasing wit h v(0) = 0 and v(1) = 1 ; (H q ) d t E [0, T] , q(t) ~ 0 is continuous and nonincreasing ; (Ho) uo E H 1 ( —L, 0), 0 ç uo(x) Ç 1, Vx E [ —L, 0] . Problem (1)-(4) describes the evaporation of a homogeneous, incompressible fluid from a homogeneous, isotropic and rigid soil, with variabl e meteorological conditions . In x = 0 the nonlinear condition of Fourier - Robin means that the flow of water leaving the soil, vanishes for u = 0 while assume its maximal value when u is maximal . Between these values, the flow of water depends in a nonlinear manner on u and with a q(t) which representsvariablemeteorological conditions . Assumption (3 ) means an impervious boundary . Equation (1) is a useful model in many different applications as, fo r instance, the flow of groundwater in a homogeneous, isotropic, rigid, an d unsaturated porous medium . If we choose the coordinate x to measur e the vertical height from ground level and pointing upward, the soil i s represented by the vertical column (—L, O) . If 9(x, t) denotesthe moisture content, defined as the volume of wate r present per unit volume of soil and v (x, t) is the seepage velocity o f the water, the law by which fluid flows through porous media can b e described, was found by Darcy experimentally and is given b y ( 5 )  v = —k(B) . D s and the continuity equatio n (s) 0 t +v s -0 . ° In (5), k(0) is the hydraulic conductivity of soil and ~ is the total potential . When absorption and chimical osmotic and thermal effect ar e negligible, the total potential may be expressed as (I) = 0( B ) + x, wher e 0(0) is the hydrostatic potential due to capillary suction . Combinin g both equations (5), (S), we obtai n ( 7 )  B t = (k(0)'¿Po(8)Bs + k(0)} s = (D(0)0 5 + k(0)} s DEGENERATE PARABDLIC PROBLEM  32 9 where D(9) := k(0)0 0 (9) denotes the soil moisture diffusivity . By defining cp ( s ) := j ó D(r) dr and b(s) :_ k(s), (7) yields (1) . In problem (1)-(4), u denotes the saturation of soil, for this we requir e the condition u > O . In the present paper, we prove the existence and uniqueness of solution for (1)-(4), considering at first a quasilinear parabolic problem o f nondegenerate type, approximating problem (1)-(4) . This nondegenerate problem is obtained adding a so called "artificial viscosity " term, substituting Sp with cp + 1/k, k E N in (1) . Th e nondegenerate problem, is studied using a semi-discretization scheme i n the time . One proves the existence and uniqueness of solution fo r the approximate problem . Existence of solution u for (1)-(4) is the n proved, going to the limit for k ~ oo . For this reason, we look fo r estimates which are independent of k . With the assumption (u 0 ) x + b(uo) E BV( —L, O), we prove the existence of solution u far (1)-(4) in th e B V (0, Ti L l ( —L, 0} } spaces . Finally, we prove the uniqueness of solutio n for (1)-(4), with the further assumption that ó((,p W l } is Holder continuous of order 1/2 draw our inspiration from [6] where is proved th e uniqueness of a bounded variational solution for a nonlinear degenerat e diffusion-convection variational inequality connected to an oil engineering problem . In [2], [11] is studied a nonlinear parabolic problem, wit h a nonlinear integro-differential term and with nonlinear boundary candi - tions . This authors, prove the uniqueness of solution in the class of B V functions . We remember also [1] and [7] . Related work, although rather different, can be found in [5] and [10 ] and references given therein . 2 . Existence of solutions far an approximate proble m In the following, we denote with BV(O, T) the space of functions u such that are locally integrable on (0, T) and whose generalized derivativa is a n integrable measure of Radon on (0, T) . Far more details on BV spaces , see [10] . Let V := H 1 (—L, O) and V' its dual space, we denote with ( ., .) bot h the pairing of duality V', V and the usual inner product in L 2 (-L, 0) . The inner product in V is defined by (u, v) 1 = (u, v) + (u, v x ) . By Sobolev's embedding Theorem, V c C( [--L, 0] ), with continuou s injection . Definition 1 . For a strong solution of (1)-(4) on (0, T), we mean a function u E BV(O, T ; L x (—L, o)} f - 1 L°°(Q T ) , 0 Ç u(x,t) Ç 1 a .a . on 330  M . BADI I Q T := (—L,0) x (O, T), such that u t E L Z (O, T ; V ' ), cp(u) E H 1 (Q T ) n L O °(Q T ), u(x, 0) = uo(x), a .a . on (—L, 0) an d fo t ( u s ,v) ds + J t v(u(O,$))q(s)v(O,$) d s 0 t o o J - L (t .p(u) x + b(u))v x dx ds = O , (8 ) for any v E L 2 (0, T ; V) and for all t E (0,T) . To prove the existence of solutions far (1)-(4), we consider a semidiscretized scheme . Divide [o, T] in steps of ugual length h = q t = T/N , N N E N (discretization time step) so, [0, T] = U [(n — 1)h, nh] . ?7,= 1 Now, we consider an approximation of u at time nh defining u n (x) : = u(x, nh) . Set g n := ( 1/ h) f1)h q(t) dt and 0 k (r) := (r)+r/k, dk E N . Since  ( • ) is an increasing function, there exists ç 1 ( . ) . It is not a prioriknown that solution u is in [0,1], therefore we conside r the continuations on all IR f), (,p, b, respectively of v, ço and b, definin g Ks) = b(s) . = p(s) o for s Ç o and v ( s ) = v(1), b(s) = b(1), (,«s) = ço( 1 ) for s ~ 1 . Now, we can resolve the followin g Problem (P a ) : Let uo be given such that 0 Ç u 0 (x) Ç 1, a . a . in (- --- L, 0 ) . To fin d uZ E V, d n ~ 1 solution of the nonlinear elliptic equatio n 0 (1/h)(uZ — uk —1 , v) + v ~ 7d k( 0 ))Q'nv ( 0 ) + f  (uv x d x L o  o +(1 /k) f u~ x v x dx~- f b(uZ)v x dx = 0, V v E V , n = 1, 2, . . . , N-- 1 L  L o 7 .L k = uo . The solution of (P a ) is based oil the solution of the following nonlin e equation a r Equation (P a ) : To find ,zk E V, such tha t µ(z k , v) + v(z k (0))g k v(0) + J o cp(z k ) x v x dx -~ (l/k) J ~ z kx v x d x L  L 0 / ' + J b(z k )v x dx = (g, v), bv E V, µ > 0, g E L Z (—L , —L O) . DEGENERATE PARABOLIC PROBLEM  33 1 Proposition 1 . If (H, p ), (H b ), (H„) hold, there exists a solution z k E V of equation (P S ) . Proof : The existence of solution, is proved by Schauder's fixed poin t Theorem . ■ Proposition 2 . l f (H (p ) , (H b ), (H o ) hold, there exists a so l ution u l E V for problem (P a ) . Proo f : Solved equation (P a ), it is possible to resolve problem (P a ) b y recurrence with respect to n . ■ We show some properties of solutions uZ . Proposition 3 . Let uo be given with 0 < uo(x) < 1, a . a . in (—L, 0 ) and (H (p ), (H b ), (H„) hold, then u¡~ are nonnegative on [—L, 0] . Proof : We proceed by recurrence . We conside r ¡ 0 (9) (1/h)('i4 — uo,v) + v(uk(0))g l v(0) + J cí« u l k ) s v x d x L  f 0  ¡ 0 + (1/k) J uk x v x dx + J b(uk)v x dx = 0  L  L and choose v = (ui) — , one ha s (10) (1/h)(4 — u o,(u))) + Kuk(0))41(u '', ( 0 )) 0  0  f 0 + f  (u(u) ; C~x~-(1/k) f ukx(ullc)x C1~x+ J  (u)(u) ; dx = 0 . L  L  L Now, v(uk(0))gl(uk(0))—  0, thus (10) give s (11) — (1/h)f24 2 dx — (i/k)f  lu~ x 1 2 dx u~ <o~uk <o] —(1/h)  uou k dx < o ] that is a contradiction . Hence, u~ > o and by recurrence one proves tha t u~ > o in [—L, O] . ■ In the following Proposition, we use a Lipschitz increasing approxima- - tion of the function of Heaviside, which is well posed with respect to th e following assumptio n (12)  I — 6(kl ( s ))I < cIs — s 1 1 / 2 . 332  M . BADI I Proposition 4 . With assumptions (H y ,), (14), (H„) and (12), problem (P n ) has a unique solution . Proof : It is sufficient to prove the uniqueness of 24 . Let uk, 111 b e solutions of problem (P I ) . We defin e 0,  ifw<e< 1 (13)  s E (w) = 1 — log w/ log e, if 0 < e < w where w :_ O k (2G k ) — Cpk(ú k ) • (14) (1/h)(uk — u k~ s e( w )) + (~(u(0)) — v Cú % (0 )))91s,(w(0) ) —(1/loge)f  (w 2 /w) d x w>e ] = (1/loge)f  (((u) —(ú))(wk/w) d x w>E ] (—c/L/loge) f  (I w I 2 / w ) dx w>e ] for (12) and the inequality of Holder . Since, (— ú¡~, s E (w)) > 0 and (~(u(0)) — v(f4(0)))g 1 s E (w(0)) > 0 because of the monotonicity of v( . ), we hav e (15) (_1/Ioge)f  ( I w 2 /)d x w> e] (—clL/ log (Iw s I 2 /w) d x f w>e ] which i mplies (16) (l w .I 2 / w) < c 2 L . f w>e ] Going to the limit as e —÷ o in (14), since s,(w) --} sgn + w and becaus e of (16), one has - 1 )+dx ~ O . ~ L Thus, the claim holds . ■ (17) V o DEGENERATE PARABOLIC PROBLEM  33 3 Now we specialize the choice of uo, assuming tha t uo E V, 0 < uo( x ) 1, t/x E [–L, 0] : cp(up) E V (18) satisfies the conditio n 0  ¡ 0 0 k (uo) x v x dx + J b(uo)v x dx > 0, dv E V L  L with v(x) > 0 in [–L, 0] . Then we obtain : Proposition 5 . I , f (H,), (H b ), (H U ) , (12) and (18) hold and g ' q i , i = 1,2, . . .,1V – 1, the n (19) ul(x) u k n—l (x)  uo(x), vx E [—L, o] . Proof We proceed by recurrente . (20) (1/h)(u) – uo,v) + v~~k(0))91v(0 ) + f O k (u~) x vx dx + J ~ b (u k )v x dx = 0 . L  L Subtracting (18) by (20), we obtain for v = s f (w), s f (w) as in (13), an d w :_ 0k( u 1) – 0k( u 0 ) (21) (1/h)(?4 uo, S E( w )) + v ( u l( 0 ))9 1S E( w ( 0 )) – (1/loge)f  ( iwI 2 /w) d x w>e ] < (–1/ IogE) f  (((U) – (u))(w/w) dx . w>e ) Arguing as in Proposition 4 and going to the límit for s -> 0, one ha s (22) u l k (x) < uo(x), Vx E [–L,0] . Now , ( 23 ) ( l / h )( u k — uk, v) + V ( u k( 0 )) g2v ( 0 ) ¡0  ¡ 0 +J c~ ft (uk) s v y dx + J b(uk)v x dx = O . L  L 334  M . BADI I Subtracting (20) by (23), one ha s (24) (1/h)(u k — 1d k — 21, k + 2Lp, 41) + (vluk(0))g2 — 1/(uk(0)W)v (0 )  ¡0  f 0 + J ( 0 k (u l ) ~ k (uk)) x v x dx +  (b(u 2k ) — b(uk))v x dx = 0 .  L  L By (22), for v = = s e (w), w := 0 k ( u k 2 ) --- 4(u k l -) , q 2 Ç g 1 and going to th e limit as E -> o in (24), we obtai n ¡ 0 (25) (1/h) f L (u 2 k - )+ dx < 0 which implie s (2G)  u 2 k (x) ç ul(x), V x E [-L , By recurrence, (19) follows . ■ From 24, we construct functions N-- 1 (27)  uh,k(x,t) := E u k( x )X n ( t ) n = n where x n ( • ) is the characteristic function of [nh, (rt + 1)h], (nh,k is a ste p approximation of u 1c} an d = (t - nh)((4(x) - uk-1(x))/h) t E [nh, (n + 1)h ] (28) o - n,,k( x , t )  +uk -i (x)  n = 1, . . . , N - 1 = uo(x),  t E [0, h l (Cfh,k is a piecewise linear approximation of Uk, continuous in t) . We can prove the followin g Proposition 6 . With the assumptions of Proposition 5, then Uh,k i s bounded with respect to h in L2 (o, T ; V) n L°° (o, T ; L2 (-L, o)} . Proof : Choosing in (P a ) v = uZ , one ha s (29) (1/h)(u k — 2L k —1 ,tGk) +v(uk(0))4"uk(0) + (1/k)IIukx 2 0  ¡ 0 + f Zdx + J b(u)udx = 0 r  - L  —L DEGENERATE PARABOLIC PROBLEM  33 5 since , (30) (u_u_ 1 , u ) = (1/2)[(u,u ) k — (U k ra --1 ra --11  ¡ rz  r a ~ ~~ 1 + I u ÏC — un—1 ,Uk — u ] E )l 7 by (29) we obtain (31) (1/2h)[Iu k 11z –  +  u k -1  + v ( u k (0))4 nu k (0 ) + (1/k)Iu kx2 + I O  C~x + J o b(uk~ukx C~x = O . - L Hence, ¡ 0 (32) (1/2h){Ilu k I 2— Iluk -1 ~~2~ + (1/k)uII + J b(u r k ' )u Z x dx < O . r By the inequality of Young and multiplying by h, one ha s ( 1 / 2 )ft u kII2 – II u % -l Mz] + ( h i le ) < hk 2 o ( l ) f L I b ( u r iD I 2dx +( h / 2k )Il u k~ 11 1 (33) o (p2hk/2)IL 1  2 (for (H b ) ) ( p2hk / 2 )Il u o li z + ( h / 2k )Il u %xII2 (for (19)) . Thus, I I u II 2 2 — Il u k -l llá + ( h / k )I t 4 .II2 5 hC kll u oll z where C k := p 2 k . Adding up on n, we obtai n n (34 ) ( 35 )  II u Mz + ( h / k ) 1 x ill  (TC k + 1)Iluolil, (hN = T) . Hence , (36)  iiu¡~ 11z  (TC k + 1)iiuoiiz 342  M . BADI I Lemma 9 . For any v E V, n 1 mo f[xE(—L,OI :Iv(x)I<I I v xl dx = 0 (se e  =1,  x> o  =1,  x  sgn(x) = o,  x - = o  sgn n (x) = x/rj, I x I i , > 0  1=—1, xCo  =—1, x C The function sgn n is a Lipschitz function on R . Lemma 10 . For any p, 1 Ç p oo, if v E L p (—L, o), then sgn n (v) — > sgn(v) in L p ( —L, o), when rj - --} o(see [9]) . To obtain an estimate on Ukt uniformly with respect to k, we assum e the following conditio n (?6) 0 Ç uo Ç 1, a .a . in (—L, o), ;o(u 0 ) E V and cp(uo) x + b(uo) E BV( —L, 0) . We can prov e Proposition 11 . If assumptions (H, p ), (H b ), (H„), (H q ) are satisfie d and (76) holds, there exists a constant C > 0 such that Uk verifies (74 ) and (75) . Proof : For o C s C s +7 C T, E ( o, T), d v(t) E V, for a .a . t E ( o, T ) we conside r (77) ( u k( s + T), 27) + (U k (O,5 +T))q(S +T)41(0, s ) + J ~ Ok (u k (s + T)) 5 v x dx + J ~ b(u k (s + T))v x dx = 0 L  L ( 78 ) (u k (s), v) + v(uk(O>S))q(s)v(0, s ) 0  ¡ 0 + ~ ~k(uk(s))xvx dx + J b(u k (s))v x dx = 0 . L  L Subtracting (78) to (77), we obtai n (79) (u k8 (s+T) — 2G ks( s ), v ) + (v(Uk(O, s +T))q(S +T))q(s +T) — v( u k (o, s))q(s))v(0, s ) 0 + J (çb(u(s + T)) — 0k(uk(s)))xvs d x —L f 0 + J (b(u k (s + r)) — b(u k (s)))v x dx = 0 —L DEGENERATE PARABOLIC PROBLEM  34 3 choosing v(s) := sgn, 7 (0 k (u k (s + r)) — g5 k(u /c(s))) E V in (79), one ha s ( 80 ) ( uk(S +T~ — 7dks(s), Sgn,ql0k(uk(s +T)) — 0k(uk(s))) ) + (U(Uk(O, s +T))q(s +T~ — v(u k (o, s))q(s)) sgn, 7 (0 k (u k (o, s + r)) 0 — 0/c( u k( O , s))) + j (çb k (u k (s + T)) — 0k( u k( s )))xl Sgn n( u k( s + T) ) L ¡ 0 — 0k(uk(s))))x dx + J (b(uk(s + r)) — b ( u k( s )))( sgn n ( 0 k( u k( s + T )) L ~kluk(s)ffix dx = O . 0 I,(b(uk`s +r)) — b ( u k( s ))) lSgrin(0klu/c (s + — O k ( u k( s )))) x d x (b(~'(zk(s+T)) ) f z k(S+T)—zk(s)l<r/ 1 — b l0k l ( z k( s ))))( 1 /77)( z k( s + T) — zk(s))xd x (where z k (Q) := ~ k (u k (v)) )  Ç  J+ T} — z k (s)) x i d x f H z k( s + , )—zk(s)l<n ] (since b o  is Lipschitz continuous with constant é) . Since z k ( s + T } — z k (s) E V, Lemma 9 implies tha t (82) 1 m  l+ T) - -- z k (s)} x ~ dx = o n—>0+ [I z k( s +T) — z k(5)Ic r T ] hence, by the Lebesgue'sTheorem we obtai n (83) ~1 ó J t J L (b(u k (s + r) ) — b ( u k( 5 )))( S ó n n lY'k(uk(s +T)) — Y'k(uk(s))))x dx = 0 . Novo, 344  M . BaDI I Moreover , (84) L o (Oke U k( S +T)) — Y'/c(uk(S)))xlSón,(0kluk(S - ~ 0 Ok(uk(s))))a C~x = yi j I, 1l0k( 2 ~k( 3 + r)) — Y'kl u k( S )))xl 2l ~ x ~ O ; by (II,) and (H q ) (85) (V(U k (0, s + T))q(s + r ) — 7/(uk(O, s))q(s)) Sgn n (0k( u k(O, 3 + T )J —  (uk(O, 5) ) )~  C , with C a constant independent of s . Thus because of (84) and (85), equality (80) gives u s (86) ( u ks( s + r) - u ks( s ), Sgn n (0k(Uk(s-t-T)) - 0k(uk(s)))) 5_ C + ~~ (b(u k(s + T) - b( u k( s )))( Sgn n (0 k ( u ~ ( s + T )) - ~k( n k (s))))s dx L Integrating (86) on [S, t], t E (S, T - T] and going to the límit as r~ -> 0 + , because of (83), one ha s (87) L(Uk( t (Uk(t + r) - Uk ( t )) Sgn (0k( uk ( t + T )) - 0k (Uk(t))))d x 0 < < L (2d k (6 +T~—2Lk(6)) S b ri l0k( U k( 5 +T)) — ~k1 u k( 8 ) JJJ dx + C ( T—6 ) V6 > O, Vt E (6,T -TI . Since O k (•) is increasing, then sgn(0 k (u k(s + r)) - 0k(uk(s))) = sgn(u k (s + T) - u k (s)), thus (87) gives u s (88) Jj, Iuk(t+T) -7b k (t)I dx < J L I2L k (S+T) -76k(b) I d x-I'C(T — S ) dS > 0, Vt E (8, T - 7] . But v, k E C([O,T] ; L 2 (-L,0)), hence whe n S -> 0 + we hav e (89) J O Iuk(t+T) — 26k(t) dx < J O I — 2lpl dx+CT , -L  -L DEGENERATE PARABOLIC PROBLEM  345 We look for an uniform estimate with respect to k for f ° L I26k(T) — uol dx . For this, we take v := sgn n (0k(uk)—Ok(uo)) E V in (78) and s E (5,'r) . Then , ( 90 ) (U k s, sgn(ç5k(uk) — 0k (U0))) + v(uk(0, s))q(s) s b n n (0k(uk(O, s) ) 0 — 0k ( u o( 0 ))) + f (ç5k(uk))x(sgn(q5 k (z k ) — ~ k(uo)))x d x L o + J L b ( u k)( sgn ,lOk( u k) — 0 k ( u 0)))x dx = o , henc e (91) (u k, sgn, i (0k(21k) — 0 k ( u o))) + v ( u k( 0 , s))4(s)Sgn n (0k(uk (O, S) ) 0 — ~k( u 01 0 M + J (((u) — 0 k(u0))x + b ( u k) — b ( u o))( Sgn , 7 (0k( u k ) —L o - 0k( u 0)))xd x=— J L ((Ok( u 0))a+ b ( u 0))( s gn r i (0lc( u k) — Ok( u 0)))xd x . Proceeding as aboye, one proves tha t ( 92 ) L(Ok(uk) — 0k( u 0))x( Sgn ,7l0k( u k) — 0k(u0)))x dx > 0 ; (93) l— 0k( u 0(0) ) )~  C with ( ; independent of s ; (94) ,7 ll á LT f(b(u k ) — b ( u o))( sgn n(0k( u k) — ~k( u o)))x dx = O . Therefor e (95) ( u ks, S bn g(0k( u k) — ~ke u o)) ) ¡ 0 C + J (b(u k ) — b ( u o)) Sgn n(0k( uk) — 0k(u0))s d x - L Integrating (95) from b to r and going to the límit when 77 —> 0 + we hav e because of (76), (92), (93) and (94) tha t f 0 (96) J lf L 346  M . BADI I for S —> 0 + , because u k E C([O,T] ; L 2 (-L, O)) . Thus, (89) and (96) gives u s ( 97 )  uk(t +T) — 2ó k( t ) Il L1(—L,0) < CT . Now, Ukt E L Z (Qó), dS > 0 hence (1/r)lu k (t + T) — u k (t) l —> u kt (t) i n L Z (Q 6 ), and in L Z (—L, 0) for a .a . t E (S, T) . Thus (97) gives us (75) , because Cis independent of S . This ends the proof . ■ 5 . Existence of solutions for degenerats proble m By definition, 0 k ( . ) E C 1 ([0,1] ) and as far as aboye, one ha s (98) III Loo (o,T ;L l (—L,o)} Ç C . By (57) and (98) it follows tha t (99) Iç C . Hence, (100) 0k(uk) is bounded in L°° (QT) n H1,1 (0, T ; L1 (—L, 0)} . The (57) and (98) say that 0 k(uk) E W (0, T), therefor e (101) 0k( u k) —> (PM in L 1( QT ) as k —> ao . Hence, (p(u) E BV (0, T ; L 1 ( ---L, 0)}, because it is the limit of a sequenc e in L°° (Q T ) n H 1,1( 0, T ; L1 (—L, 0)} . By (66), u k -4 u in L 1 ( QT ) , moreover, (69) and (75) imply tha t u k is bounded in L°° (QT) n H1,1 (0, T ; L 1 (—L, 0)}, hence u E BV (0, T ; L I (—L, O)) ; a such u has a trace for t = O . Since 0k(uk) t weakly converges to (U)t in L 2 (Q 6 ), V(5 > 0 as k ~ oo, we obtain the estimate : Ik o ( u )tML 2 (&T ;L 1 L,ofl  C, where C i s a constant independent of (5 . Thus, I  C an d consequently, since V c L°° (---L, 0) with continuous injection , 11Ç0(u) t II L 2 ( o ,T ; vI )  C . Then , ço(u ) E C([0, T] ; L 2 ( —L, 0)} an d th e Hólde r continuit y o f ç o - 1 implies that u E C([0, T] ; L 2 ( —L, 0)} . Moreover, u E L°° (QT) an d cp(u ) E L°° (Q T ) n H1 (QT) . Finally , th e trace in t = 0 fo r u in L 1 (—L, 0 ) coincides with uo . Thus, we have proved the following existence resul t Theorem 12 . If (H p ), (H b ), (Hp), (H q ) and (76) hold, then ther e exists a strong solution for problem (1)-(4) . DEGENERATE PARABOLIC PROBLEM  34 7 6 h Uniqueness of solution for degenerate proble m To study this problem, we introduce in R2 the one dimensional Haus - dorf f ' s measure 7- ~ 1 (for a definition, see [4]) . Since cp is an homeomor - phism and (p(u) E L°° ( QT ) n Hl (QT), in [6, Lemmas 1 and 2], is prove d that cp(u) and any solution u of (1)-(4) are 'H 1 -a .a . £ 2 -approximatel y continuous on Q T (see [4, pg . 158]) . Now u E BV (0, T ; L'(-L, 0)i hence its distributional time derivativ e u t is an integrable Radon ' s measure on Q T , 7-~ 1 absolutely continuou s (see [6) . This means that u t do not charge the complementary set of r2 - approximate continuity points of u, thus we can utilize the integratio n by parts formula . Let u, fi be solutions for (1)-(4) with initial data u0, respectively, fi o . Then t (u t - ic t , v)v, , v p ( T ) d T J t i v ( u ( O , T )) - L ( u ( O , T )) g ( T ) v ( O , T ) p I T ~ C ~ T 0 +J t f_L — (ú))xvxp(T) dx d T o +J t J o (b(u) - b(ú))v s p(T) dx dT = 0 , o  -- - n -r , L Vv E I ;2(0,T ;V), Vt > 0 and p E D(O,t), p(t) > O . Set w := ço(u) - co(ú) and H, 7 (w) := w +2 /(w 2 +77), 77 > 0 and suppos e that (103)  l b (w -1 ( t )) - b(cP-1(S))i  c ' lt - s1 1 / 2 . It is easy to prove that, d r E IR , (104)  lim  (r) = 0, 0 < H q (r) < 1 and 0 rH ' ~(r) 1/2 . Choosing v = H n (w) E H 1 ( QT ) in (102) one has , t (102) (i~(u(0, r )) — v(ú(0,T)))g(r)Hn(w(O,T))p(r) dr > 0 348  M . BADI I because Sp is increasing, the n ( u t — ut a H n( w ))v , vP( T ) d r (105)  + J t J a  ((u) — (~))xI2H(w)p(r) dx d r 0  L + J t J ~ («u) — b(ú))H~(w)xp(T) dx dr G o — n  —r , L Now, ¡t ¡ o — J J L (b(u)—b(ít))H(w)p(r)dxdr ¡t f L (b(u) — b(ú))(cp(u) — cp(ú))xH ',~(w)p(T) dx d r < J t f L Ib(u) — beú)I ((u) — (~))IH(w)p(r)dxd r fJ 1s — sl 12 (s — s)(s — ) p (T) dx d T L (1/2) J t J ~ ( s—)I2H(s—~)p(r) dx d T 0 - L ¡t ¡ 0 + cl J J-L is  s 1H ' ~(s s)p(T) dx d T 0 by (103) with s := co(u), s := cp(ú) and the inequality of Young . Therefore, (105) gives u s ( ut — ut a Hn(w))v,vP(T) d r (107)  +(1/2) J t L L ~ I— ) 2 H(s — s)p(T) dx d r o w <c 1J t J ~ ~0  L hich implie s (u t — ZLt, 1I 71 (w)}V'Vp(T) dT t Ç c l  ls — .111 4 ' (s — s)p(T) dx dT . o 1 : L t lo t f t f DEGENERATE PARABOLIC PROBLEM  34 9 H, 7 () is an C 1 approximation of the Heaviside ' s function and as far a s aboye claimed in (104) and by the Lebesgue's Theorem, we hav e ¡t f o lim J J  — MH ' ~(s — §')p(r) dx dT = 0 . Now, we treat the' term j(ut — ú t , H n (w)) v, V p(T) dT . We known that u— 'ú E L Z (Q T ) and (u—ú) t E Lz(O,T ;V'), pH n (w) E L2 (0, 7' ; V ) ( H n( w ) E H1 (QT) n L°°(QT)) and ( pH n( w ))e E L 2 (Q T ) . Since V is embedded in L 2 (—L, 0) with dense and continuous injection, then defined the intermediate spaces Y := [V, L 2 ] 1/2 and Y' : = [L 2 ,V'11,2 (see [8, pg . 11]) one has that u — ú E C([O,T] ;Y'), pH n E C([0, T] ; Y) and the following formula hold s (109) J t (u t — ú t , H~n(w))v,vP(T) dr - ( u ( t ) — u(t),H~n(w(t))P(t))Y' Y 0 — (u(0) — ú(0), H n( w (0 ))P (0))Y'Y - J t f 0 (u — ú )(P H n( w ))t dx dr . 0  L But u E C 8 ([0, T] ; L 2 (—L, 0 )} , so the pairing of duality Y', Y is a n integral in L 2 (—L, 0) (see [11, Chapter III] } and f o r p E D(O, t) (109 ) becomes f ( 110)  (u t — ú t , H~(w))v,vP(T)C~T ¡t ¡ o _  J (u—~)(pH 1 (w)) t dxdr . 0  L To the purpose of apply the results of [11, Chapter II], see also [2], w e work vvith the Borelian representative of u in the £ 2 -a .a . class (that i s u -= ú £ 2 -a .a .), this allow to use the Borel measure . We agree to writ e ¡t ¡o  f -- J LL (u—~)(pH,,(w)),dxdr==— J (u—íi)(pH,(w)) t o  Q t because pH, 7 (w) E H1 ( Q T ), belongs to BY(O, T ; L 1 (—L, 0)} and th e Borel measure (pH, 7 (w)) t is £ 2 -Lebesgue absolutely continuous, wit h density (pH, 7 (W))t . 350  M . BADI I Moreover, u — ú E BV (0, T ; L 1 (—L, 0)} n L°° (Q T ) , pH n(w) E L°° (QT ) n H1 ( QT ) ad (u --- ú} t is a Radon's measure 7-i 1 -absolutely continuous, by a result of [12, sections 13 .2 and 14 .4], (u — ú)pH, 7 (w) E B V (0, T ; L 1 ( —L, O)) and we can use in (110) the formula of integratio n by parts (see [2] and [11]) to obtai n f (111)  (u— ú, H(w))v,vP(T) dT — f P(t)H(w)(u — ú) t . Q t Taking in consideration that H n converges pointwise to sgn + as —> 0 + and (pea) — ço(ü) is defined and bounded 7-t 1 -a . a . in Q T by its L °° (QT ) norm, the increasing of (p ( • ) implies tha t li óH T,( w ) = sgn + (c P ( u ) — y( u )) = sgn+ (u—ú), 7-l l -a .a ., (sgn EBV(R)) . By the boundedness of H, I (w) it is possible to apply the Theorem o f Lebesgue, since (u — ic } t is bounded . Then, (112) lim J p(t)H n (w) (u — ú) t = J p(t)(sgn + (u — fi)) (u — ú) t n ~ 0+ Qt  Qt = f p(t)((u — ü ) + ) t , Q t (see Corollary 3 .3 in [11], sgn + is a bounded Borel function) . Going to the limit as r1 ---~ 0 + in (108), we hav e (113) p ( t )(( u — ú) + ) t < 0 , Q t that implies the nonincreasingof (u(x, .) — ú ( x, • )} + as function of t . Hence, 0  0 (114) f (u(x,t)—~(x,t))dz~ f (uo(x)—~o(x))dx, dt > o L  L and the application uo ~ u(•, t) is a T -contraction in L 1 (—L, 0) . Fo r uo = üo in [—L, 0] the uniqueness of solution follows . Thus, the choic e of ú becomes no more important . At Iast, the following result holds DEGENERATE PARABOLIC PROBLEM  35 1 Theorem 13 . If ( .14) — (H„), (76) and (103) hold, the problem (1) - (4) has a unique strong solution. Moreover, (114) holds for any t > 0 and gives an order preserving for the solution . Reference s 1. 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