Simulation of reactive contaminant transport with
non-equilibrium sorption by mixed finite elements and Newton
method
Citation for published version (APA):
Radu, F. A., & Pop, I. S. (2009). Simulation of reactive contaminant transport with non-equilibrium sorption by
mixed finite elements and Newton method. (CASA-report; Vol. 0934). Technische Universiteit Eindhoven.
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EINDHOVEN UNIVERSITY OF TECHNOLOGY
Department of Mathematics and Computer Science
CASA-Report 09-34
October 2009
Simulation of reactive contaminant transport with
non-equilibrium sorption by mixed finite
elements and Newton method
by
F.A. Radu, I.S. Pop
Centre for Analysis, Scientific computing and Applications
Department of Mathematics and Computer Science
Eindhoven University of Technology
P.O. Box 513
5600 MB Eindhoven, The Netherlands
ISSN: 0926-4507
Simulation of rea tive ontaminant transport with
non-equilibrium sorption by mixed nite elements and
Newton method
Florin A.Radu
1;2
Iuliu Sorin Pop
3
Re eived:date/A epted:date
Abstra t We present a omputational numeri al s heme for rea tive ontaminant
transportwithnon-equilibriumsorptioninporousmedia.Themass onservatives heme
isbasedonEulerimpli it,mixed niteelements(MFE)andNewtonmethod.We
on-siderthe aseof aFreundli htypesorption.Inthis asethe sorptionisothermis not
Lips hitz,butjustHolder ontinuous. Todeal withthis,we performaregularization
step. The onvergen e of the s heme is analysed. An expli it order of onvergen e
dependingonlyontheregularizationparameter,thetimestepandthemeshsizeis
de-rived.WegivealsoasuÆ ient onditionforthequadrati onvergen eoftheNewton
method.Finallyrelevantnumeri alresultsarepresented.
1Introdu tion
Waterpollutionisagraveproblemnowadays.Questionslikehowdangerousa
ontam-inatedsiteis,whether naturalattenutation ano urr, whi hpro essesaredominant
andwhetherana tiveremediationisne essarybe omerelevantforde issionsonwhat
wehavetodoinordertoproperlyprote tourselvesandthenextgenerations,aswell
as the environment. To answer these questions, predi tions based on mathemati al
modelling and numeri al simulationsare needed, requiring eÆ ient and reliable
nu-meri al odes. Besides modern dis retization tools and eÆ ient linear andnonlinear
solvers,itis veryimportantthat theimplementedmodelallows tosimulaterealisti ,
ompli ateds enarios.Themodelmustin lude: owinsaturated/unsaturated,
hetero-geneoussoilsandmulti omponentadve tive-di usive-rea tivetransportwithsorption
F.A.Radu
UFZ-HelmholtzCenter forEnvironmentalResear h,Permoserstr.15, D-04318Leipzig,
Ger-many
UniversityofJena,Wollnitzerstr.7,D-07749,Jena,Germany
E-mail: orin.raduufz.de
I.S.Pop
inporousmedia.Weespe iallyemphasizethatareliablesimulationtoolmust
ompre-hendthee e tsofsorptiononthetransportof ontaminantsinsoil,seee.g.[36,37℄.In
thispaperwepresentandanalyseamass onservativenumeri als hemetoeÆ iently
simulaterea tivesolute transportwithnon-equilibriumsorptioninaquifers.Herewe
ontinuetheworkin[32,33℄,wherethe aseofequilibriumsorptionwastreated.
Ageneralmathemati almodelfor rea tive solutetransportwithnon-equilibrium
sorptionis S t + b t s r(D S r Q )= S r( ) in J; (1) ts=ks( ( ) s) in J; (2)
with (t;x)denotingthe on entrationofthedissolvedspe ies,s(t;x)the on entration
oftheadsorbedspe ies,Dthe onstantdi usion oeÆ ient(wetakeforsimpli ityD=
1),r()area tionrateandkstheDamkohlernumberassumedalsoofmoderateorder.
Wefurtherassumethatthewater ontent
S
is onstant,asen ounteredforexamplein
thesaturated owregime,where
S
=1.IR
d
denotesthe omputationaldomain,
dthespa edimensionandJ=(0;T)thetimeinterval,T beingsome niteendtime.
Further, ()isasorptionisotherm,withapossibleunboundedderivative.This hoi e
overstherealisti , pra ti al aseofaFreundli htypeisotherm,e.g. (x)=x
,with 2(0;1℄.Initial (0;x)= I (x);s(0;x)=s I
(x)andhomogeneousDiri hletboundary
onditions ompletethemodel.ThehomogeneousDiri hletboundary onditionswere
hosen justfor the sake of simpli ity,all the results presented inthis paper anbe
extendedtomoregeneralboundary onditions.Thewater uxQ(x)solves
rQ=0; (3)
Q= K
S
r( +z); (4)
in,with (x)beingthepressurehead,K
S
thehydrauli ondu tivityandztheheight
againstthe gravitationaldire tion.We onsiderhere onlythe ase offully saturated
ow,butthes hemeisimplementedfor thegeneralsaturated/unsaturated ow ase.
Furthermore,theanalysispresentedinthenextse tion anbeextendedalsotostri tly
unsaturated ow(seealso[32℄).
Inthis paperwe present an Euler-impli it, mixed nite element s heme for the
system(1){(2).ThesorptionisothermisassumedmonotoneandHolder ontinuous.
Choosingforamixed niteelementdis retizationensuresthelo almass onservation.
Be ause themodelis degenerate,itssolutionla ksregularity,therefore weonly
on-siderthelowestorderRaviart-Thomas niteelements.Thenonlinearproblemsarising
atea htimesteparesolvedbyaNewton method.Tothis aim,we dis retizea
regu-larizedapproximation ofthe original model. We provideerror estimates forthe fully
dis retesolutionandderiveanexpli it onditionforthe onvergen eofthes heme,in
termsofthedis retizationandtheregularizationparameters.This onditionallowsto
ontrolapriorithetimestep,theregularizationparameterandthemeshdiameter,so
thatanoptimal onvergen eisa hieved.Weusethesolutionoftheprevioustimestep
n 1asthestarting hoi efortheNewtoniterationatthe urrentstepn.Inthisway
theinitialerror anbequanti edduetoastabilityestimate,andanexpli it onstraint
onthedis retizationparamaters anbederivedtoensurethe ertainquadrati
time. In this sense we mention [14℄, proposing an adaptive methodthat takes into
a ounttheerrormadeinea hNewtonstepwithrespe ttothetotalerror,resulting
intoaveryfastmethod.
Inspiteoftheri hliteratureonnumeri almethodsfor solutetransportinporous
media (like [2,5{8,13,15{20,23,24,30,32 {3 5℄), only a few are onsidering nonlinear
sorption.Wereferto[5,6℄and[13℄ for onforming niteelementdis retization.There
boththe asesofnon-equilibriumandequilibriumsorptionare onsidered.Thewater
ow is assumedsaturated, with the ux Q being given analyti ally. A similar
situ-ation is also onsidered in [11℄. We also refer to [16℄ for a ombined nite
volume- nite element s heme for transport withequilibrium sorption. The mixed nite
ele-ment dis retizationfor equilibriumsorption is onsidered in[12℄and [32℄.Thelatter
also onsideredsaturated/unsaturated ow,takingexpli itlyintoa ountthelow
reg-ularity of the solutionto the Ri hards equation. Theresulting estimates dependon
thea ura yofthes hemefor thewater ow.Inall thisworksthenonlinearsystems
arising at ea h timestep are supposed to be solvedexa tly.The degenera y related
withaFreundli htypeisothermmakessolvingthesesystemsa ompli atedtask.The
resultingfullydis retenonlinearproblemsare ommonlysolvedbydi erentmethods:
theNewtons heme,whi hislo allyquadrati onvergent,somerobust rst-order
lin-earizations hemes(see[26,38℄),or theJager-Ka ur s heme[23,19℄. The onvergen e
of the Newton method applied to the system provided by a MFE dis retization of
anellipti problem is studied in[25℄. Con erning thesystems providedby theMFE
dis retization of degenerate paraboli equations we mention [26℄ for a robust linear
s hemeand[29,33℄for theNewtonmethod.
Thefollowingsarethenovelaspe tsinthispaper:
{ We analyze amixed nite element s heme for solute transportin porous media
withnon-equilibriumsorption.
{ We study the onvergen e of the Newton s heme solving the nonlinear systems
arizingatea htimestep.
{ Wederiveexpli it,suÆ ient onditionsonthethreeparameters(thetimestep,the
meshdiameterandtheregularizationparameter)ensuringtheoptimal onvergen e
ofthes heme.
Thepaperisstru turedasfollows:inthenextse tionthedis retizations hemeis
presented.The mainresults on erningthe onvergen e ofthe s hemeare givenand
dis ussed. InSe tion3 we presentstability estimatesand theproofs of thetheorems
statedinthepre edingse tion.Twonumeri alstudies:anexamplewithananalyti al
solutionandarealisti in ltrationproblemare giveninSe tion4.Inthelast se tion
wepresent on luding remarks.
2Dis retization and mainresults
Throughout thispaperwe use ommonnotations inthefun tional analysis. Byh;i
wemeantheinnerprodu tonL
2
().Further,kkandkk
L p
()
standforthenorms
inL
2
()andL
p
()respe tively.Thefun tionsinH(div;)areve torvalued,having
aL
2
divergen e.kk
1
standsforthe norminH
1
().C denotesapositive onstant,
not depending on the unknowns or the dis retization parameters. L
f
Forthedis retizationintimeweletN 2Nbestri tlypositive,andde nethetime step =T=N, as well as t n =n (n2f1;2;:::;Ng). Furthermore,T h is aregular de ompositionofR d
into losedd-simpli es;hstandsforthemeshdiameter.Here
weassume=[
T2T
h
T,hen eispolygonal.Correspondingly,wede nethedis rete
subspa esW h L 2 ()andV h H(div;): W h :=fp2L 2
()jpis onstantonea helementT 2T
h g; V h :=fq2H(div;)jq jT =a+bx forallT 2T h g: (5) Inotherwords,W h
denotesthespa eofpie ewise onstantfun tions,whileV
h
isthe
R T 0
spa e(see[10℄).
Inwhatfollows wemakeuseoftheusualL
2 proje tor: P h :L 2 ()!W h ; hP h w w;w h i=0; (6) forallw h 2W h .Furthermore,aproje tor h anbede nedon(H 1 ()) d (see[10,p. 131℄)su hthat h :(H 1 ()) d !V h ; hr( h v v );w h i=0; (7) for all w h 2W h
.Following [27℄, p.237,this operator an beextendedto H(div;).
Fortheaboveoperatorsthereholds
kw P h wkChkwk 1 ; (8) kv h v kChkv k 1 (9) foranyw2H 1 ()andv2(H 1 ()) d .
Mixed niteelementsareappliedforsolvingthe owproblem(3)-(4)numeri ally.
Onehas(see[28,31℄ fordetails)
k h k+kQ Q h k+kr(Q Q h )kCh; (10) where ( ;Q)L 2 ()H(div;) and( h ;Q h ) W h V h
denotethe ontinuous
and dis retesolutions respe tively.For errorestimates for MFE s hemesfor
unsatu-rated/saturated owwereferto[3,28,31℄ andforstri tlysaturated owto[3℄;abrief
review anbefoundin[32℄.Wementionthattheideasintheanalysis arriedoutin
thispaper analsobeappliedforthe onformal,mixedor nitevolumedis retization
ofthegroundwater ow,aslongastheyprovideestimateslike(10),andtheassumption
(A3)madebelowholds.
TheNewtoniteration onsideredhererequiresthatthederivativesofthenonlinear
fun tionsareboundedawayfrom0andin nity.Thereforeweperformaregularization
step.Forsimpli ityonlytheregularizationisothermfortheFreundli hisotherm (x)=
[x℄ +
isgiven,where 2(0;1℄and[x℄
+
standsforthepositive utofx.Ageneralization
tootherisothermsisstraightforward.
(x)= (x) if x62[0;℄; ( 1) 2 x 2 +(2 ) 1 x if x2[0;℄: (11)
Lemma1 Theregularizedsorptionisothermisnonde reasing.Further, ()and 0
()
are Lips hitz ontinuous on[0;1)with theLips hitz onstants L
= 1 , respe -tivelyL 0 =(2 2 ) 2 .Finally,wehave 0 (x) (x)(1 ) (12) ifx2(0;),whereas (x)= (x)whenever x2=(0;).
Towrite(1){(2)inamixedformwede nethe ontaminant uxq= r + Q.
Inthiswaytherea tivesolutetransportproblemisformulatedas
Problem1 ( ontinuousmixedvariationalformulation)
Find( ;q;s)2H 1 (J;L 2 ())L 2 (J;H(div;))H 1 (J;L 2 ()),with jt=0 = I and s jt=0 =s I
sothatforallt2Jwehave
h t ;wi+ b h t s;wi+hrq;wi=hr( );wi (13) hq;v i h ;rv i h Q;v i=0 (14) h t s;wi=k s (h ( );wi hs;wi); (15) forallw2L 2 ()andv2H(div;).
Throughoutthispaperwemakeuseofthefollowingassumptions:
(A1) The rate fun tion r : IR ! IR is di erentiable with r
0
bounded and Lips hitz
ontinuous.Furthermore,r( )=0forall 0.
(A2) Theinitial
I ;s
I
areessentiallyboundedandpositive.
(A3) Thewater uxanditsnumeri alapproximationareessentiallybounded,Q;Q
h 2
L 1
().
(A4) Thesorption isotherm ()is nonde reasing, nonnegative and Holder ontinuous
withanexponent 2(0;1℄,i.e.j (a) (b)jCja bj
foralla;b2IR.Moreover,
( )=0if 0.
(A5) ForthesolutionofProblem1wehave 2L
1
(J),whilet andtsareHolder
ontinuousintwithexponent =2 and respe tively.Furthermore,thereholds
N X n=1 kq(tn)k 2 C : (16)
Remark1 Theregularity assumedin(A5)for
t
and
t
s isthe maximalregularity
one anexpe tfor transportproblemswithnon-equilibriumsorption,whenthe
sorp-tionisotherm isof Freundli htype.A ording to[21℄ Chapter II.4, ifthe initialand
boundarydataare ompatibleandsuÆ ientlysmoothwehave 2C
2+ ;1+ =2
(J)
and s 2 C
;1+
(J). Furthermore,Proposition 1below justi es(16) inthe one
dimensional ase, when H(div;) = H
1
(). For the essential bounds of one only
needstoassumethatthedataareessentiallybounded.Furthermore,toavoidnegative
valuesfor ands-whi harenonrealisti -theratesrand areextendedby0inthe
Problem2 (fullydis retevariationalformulation) Letn 2f1;:::;Ngand ( n 1 h ;s n 1 h )2W h V h be given.Find ( n h ;q n h ;s n h ) 2W h V h W h
sothatforallt2J wehave
h n h n 1 h ;w h i+ b hs n h s n 1 h ;w h i+ hrq n h ;w h i= hr( n h );w h i (17) hq n h ;v h i h n h ;rv h i h n h Q h ;v h i=0 (18) hs n h s n 1 h ;w h i= k s (h ( n h );w h i hs n h ;w h i); (19) forallw h 2W h andv h 2V h .Wetakeattimet=0: 0 h =P h I ands 0 h =P h s I .
Thesystem(17){(19)isnonlinear.Tosolveitwe onsideraNewtonmethod,whi h
islo allyquadrati onvergent:
Problem3 (Newtoniterations)
Letn2f1;:::;Ngand( n 1 h ;s n 1 h )2W h V h
begivenandlet
n;0 h = n 1 h ;s n;0 h = s n 1 h .Fori1 nd( n;i h ;q n;i h ;s n;i h )2W h V h W h su hthat h n;i h n 1 h ;w h i+ b hs n;i h s n 1 h ;w h i+ hrq n;i h ;w h i = hr( n;i 1 h )+r 0 ( n;i 1 h )( n;i h n;i 1 h );w h i (20) forallw h 2W h , hq n;i h ;v h i h n;i h ;rv h i h n;i h Q h ;v h i=0 (21) forallv h 2V h and hs n;i h s n 1 h ;w h i= ks(h ( n;i 1 h )+ 0 ( n;i 1 h )( n;i h n;i 1 h );w h i hs n;i h ;w h i); (22) forallw h 2W h .
To xthe notations:n2 f1;:::;Ng always indexesthe timestep, whileiis usedto
indextheiteration.A ordingly,f
n h ;q n h ;s n h
gdenotesthesolutionofProblem2atthe
n th
timestepandf
n;i h ;q n;i h ;s n;i h
gstandsforthesolutiontripleatiterationi1.The
iterationpro essstartswith
n;0 h = n 1 h ,s n;0 h =s n 1 h
.Inprovingthe onvergen eof
thes hemeitissuÆ ienttoshowthat
k n n h k+kq n q n h k+ks n s n h k ; ;h!0 ! 0; (23) whereas k n h n;i h k+kq n h q n;i h k+ks n h s n;i h k i!1 ! 0 (24)
quadrati ally. This will be a hieved for a suÆ iently smalltime step . A suÆ ient
onditiononthedis retizationparameters; ;hisderivedtoensureboth onvergen es
statedabove.Themainresultsinthispaperare
{ Theorem1showingthe onvergen e(23).
{ Theorem2showingthequadrati onvergen e(24).
Theorem 1 Assuming(A1){(A5),wehave kP h (t N ) N h k 2 + N X n=1 k ( (tn)) ( n h )k 1+ L 1+ + N X n=1 kP h (tn) n h k 2 + N X n=1 kq(tn) q n h k 2 C(h 1+ + + 1+ ): (25) and kP h s(t N ) s N h k 2 + N X n=1 kP h s(tn) s n h k 2 C( 2 r 1 + r (h 1+ + + 1+ )): (26) forany r2IR.
Remark2 Ifthesorption ()anditsderivativeareLips hitz ontinuous(thus =1),
aslightlymodi edproofleadstooptimalerrorestimates
kP h (t N ) N h k 2 + N X n=1 k ( (tn)) ( n h )k 2 + N X n=1 kP h (tn) n h k 2 + N X n=1 kq(t n ) q n h k 2 C(h 2 + 2 ): (27)
Remark3 Similarresults anbeobtainedinthestri tlyunsaturated owregime,or
for asteadyunsaturated ow,where istimeindependent(asassumedfor example
in[6,11℄).
Theorem 2 Assuming (A1){(A4),forsuÆ ientlysmall wehave
k n h n;i h k 2 + kq n h q n;i k 2 C 2 (L 2 r 0+L 2 0 )h d k n h n;i 1 h k 4 (28) and ks n h s n;i h k 2 C 2 (L 2 0 + 2 L 2 (L 2 r 0+L 2 0 ))h d k n h n;i 1 h k 4 : (29)
From(28) we anderivesuÆ ient onditionfor thequadrati onvergen e ofthe
Newtons heme C 2 2 4 h d k n h n 1 h k 2 1: (30)
Withthestabilityestimate
N X n=1 k n h n 1 h k 2 C 2(1 )
provedinProposition2,this onditionbe omes
C 3 4 6 h d 1: (31)
Using (31)and Theorem1,one an hoose apriori thetimestep, the regularization
parameterandthemeshsizeinsu hawaythattheoptimal onvergen eisguaranteed.
Inthissensenumeri alexamplesareprovidedinSe tion4.Referringto[14℄,wherethe
Newton-error is ontrolled adaptively withrespe t to the total error, ondition (31)
Remark4 The ondition(31)isderivedunderpessimisti onditions.Inthissensewe
mentionthethe stability estimatesinTheorem2,bounding tothesum
P N n=1 k n h n 1 h k 2
,wherefrom onlyoneterm, k
n h n 1 h k 2
is usedtoderive (31) from(30). In
aseallthetermshavesimilarorders,thiswouldprovidek
n h n 1 h k 2 C( 2 2 2 ), implying C 4 4 6 h d 1: (32)
In the numeri al al ulations presented at the end, (32) was always enough for the
quadrati onvergen e.
3Stability and onvergen eresults
In this se tion we derive stability estimates for both ontinuous,as well as dis rete
problems1and 2,and provethea priorierror estimatesstatedinTheorem1.Then
weproveTheorem2),showing the onvergen eoftheNewton method.Westart with
preliminaryresultsandnotations:
f n =f(tn) ; f n = 1 Z t n t n 1 f(t)dt e n =P h n n h ; e n s =P h s n s n h e n q =q n q n h ; e n;i = n h n;i h e n;i q =q n h q n;i h ; e n;i s =s n h s n;i h
We re allthat n 2f1;:::;Ngis indexing thetime step,while iis used for the
iter-ation. A ordingly,f n h ;q n h ;s n h
gstandsfor the solutionof Problem2at then
th time step,whereasf n;i h ;q n;i h ;s n;i h
gisthei-thiteration(i1)intheNewton s heme.The
iterationpro essstartswith
n;0 h = n 1 h ands n;0 h =s n 1 h .
Nextweusethefollowing elementarylemmas(see[22℄,p.350for theproofofthe
rstone; forthese ondoneiselementary):
Lemma2 Letf :IR!IR di erentiable with f
0
()Lips hitz ontinuous. Thenthere
holds jf(x) f(y) f 0 (y)(x y)j 2 L f 0 2 jx yj 2 ; 8x;y2IR: (33)
Lemma3 Foranysetofm-dimensionalrealve torsa
k ;b k 2R m (k2f0;:::;Ng;m
1)thefollowingidentitiesarevalid
N X n=1 ha n a n 1 ; N X k =n b k i= N X n=1 ha n ;b n i ha 0 ; N X n=1 b n i; (34) N X n=1 h N X k =n a k ;a n i= 1 2 k N X n=1 a n k 2 + 1 2 N X n=1 ka n k 2 ; (35) N X ha n a n 1 ;a n i= 1 2 ka N k 2 + 1 2 N X ka n a n 1 k 2 1 2 ka 0 k 2 : (36)
Thefollowinginequalitieswillbeusedsevertimesbelow:
theinequalityofmeans
ab 1 2Æ a 2 + Æ 2 b 2
; foranya;b2IRandÆ>0; (37)
andtheYounginequality
ab a p p + b q q
foranya;b>0andp;q2(1;1)su hthat
1 p + 1 q =1: (38)
Wealsoneedthefollowingte hni alLemma
Lemma4 Let 2(0;1)andf
n
2L
1+
();n2f1;:::;Ngsu hthat there holds
N X n=1 kf n k 1+ L 1+ () A: (39)
Forallr2R wehave
N X n=1 kf n k 2 C( 2 r 1 + r A): (40)
Proof. Sin e isbounded,L
1+ ()L 2 () ( <1),and kf n kCkf n k L 1+ ()
for all n2 f1;:::;Ng,where C >0onlydependson.ThentheYounginequality
(38)gives N X n=1 kfnk 2 C 1 r N X n=1 r kfnk 2 L 1+ () C N X n=1 rp+1 r p + N X n=1 1 r q kfnk 2q L 1+ () (41)
forall r2IR,andp;q>0with
1 p + 1 q =1.Takingp= 1+ 1 andq= 1+ 2 ,from(41) and(39)weget (40). 3.1 Stability estimates
WestartwiththestabilityestimatesforProblem1.Thesearesimilartothestandard
energyestimatesforparaboli problems,butrestri tedtothetimest
1 ,:::,t
N .
Proposition 1 Assuming(A1){(A5)there holds
N X n=1 k n k 2 1 + N X n=1 h ( n ); n i+ N X n=1 kq n k 2 C ; (42) ks N k 2 + N X n=1 ks n s n 1 k 2 + N X n=1 ks n k 2 C ; (43) N X n=1 k n n 1 k 2 C ; (44) N X krq n k 2 C: (45)
Proof.Wetakew= n in(13),v= qn in(14)andw= b n in(15)toobtain ht ; n i+ b hts; n i+hrq n ; n i=hr( n ); n i; (46) kq n k 2 h n ;rq n i h n Q;q n i=0; (47) b hts; n i= b ks(h ( n ); n i hs n ; n i): (48)
Adding the three equalities above, summing up the result from n = 1 to N, and
multiplyingby gives N X n=1 h n n 1 ; n i+ N X n=1 kq n k 2 + N X n=1 b ks h ( n ); n i N X n=1 ks b hs n ; n i = N X n=1 h n n 1 t ; n i+ N X n=1 hr( n ); n i+ N X n=1 h n Q;q n i: (49) We nowtake w=ks 2 P N k =n k
in(13), sumup (14)for n=k toN and takeinthe
resultv=ks 2 q n toobatin ks h( n n 1 ); N X k =n k i+ks h t ( n n 1 ); N X k =n k i+ks 2 hrq n ; N X k =n k i +ks b h(s n s n 1 ); N X k =n k i+ks b h t s (s n s n 1 ); N X k =n k i=ks 2 hr( n ); N X k =n k i; (50) and ks 2 h N X k =n q k ;q n i ks 2 h N X k =n k ;rq n i=ks 2 hQ N X k =n k ;q n i: (51)
Adding(50)and(51),summingupfromn=1toN andusingLemma3yields
N X n=1 k s k n k 2 k s h 0 ; N X n=1 n i+ N X n=1 k s b hs n ; n i k s b hs 0 ; N X n=1 n i + ks 2 2 ( N X n=1 kq n k 2 +k N X n=1 q n k 2 )= N X n=1 h( n n 1 t ; N X k =n k i + N X n=1 h(s n s n 1 t s; N X k =n k i+ N X n=1 k s 2 hr( n ); N X k =n k i + N X n=1 k s 2 hQ N X k =n k ;q n i: (52)
From(49) and(52),byLemma3,theCau hy-S hwarzinequalityand(37)onegets
k N k 2 + N X n=1 k n k 2 + N X n=1 kq n k 2 + N X n=1 h ( n ); n i C(k I k 2 +ks I k 2 + N X n=1 1 k n n 1 t k 2 + N X n=1 kr( n )k 2 ) +C( N X n=1 kQ n k 2 + N X n=1 1 ks n s n 1 t sk 2 + N X n=1 k n k 2 ):
Sin e I
;s I
andQarebounded,theLips hitz ontinuityofr(),theHolder ontinuity
of
t
and
t
swithrespe ttotime,aswellastheGronwall Lemmagive
k N k 2 + N X n=1 k n k 2 + N X n=1 h ( n ); n i+ N X n=1 kq n k 2 C : (54)
Formoredetails onbounding thetermsinvolving thetimederivativewe refertothe
proofofTheorem1,estimates(85) ,(87)and(88) ,dealingwithsimilarterms.Further,
from(14)oneimmediatelyobtains
N X n=1 kr n k 2 C N X n=1 kq n k 2 + N X n=1 kQ n k 2 : (55)
Then(42)isfollows from(A3),thePoin areinequality,(54)and(55).
Toprove(43),wetestin(15)byw= s
n
andsumupfromn=1toN toobtain
N X n=1 hs n s n 1 ;s n i+ N X n=1 ks ks n k 2 = N X n=1 hs n s n 1 ts;s n i+ N X n=1 ks h ( n );s n i: (56)
Using(A2),Lemma3,theHolder ontinuityof
t
sand ,aswellastheboundedness
of
n
(see(A5)),(56)givestheestimate(43).
Toshow(44)wetakew= n n 1 in(13),v= qn in(14)writtenfort=tnas wellasfort=t n 1 ,andw= b ( n n 1 )in(15)toobtain h t ; n n 1 i+ b h t s; n n 1 i+hrq n ; n n 1 i=hr( n ); n n 1 i; hq n q n 1 ;q n i h n n 1 ;rq n i h( n n 1 )Q;q n i=0; b h t s; n n 1 i+ b k s h ( n ); n n 1 i= b k s hs n ; n n 1 i:
Addingtheabove,summingupfromn=1toN,multiplyingby andusingLemma
3leadsto N X n=1 k n n 1 k 2 + 2 kq N k 2 + N X n=1 2 kq n q n 1 k 2 = N X n=1 h n n 1 t ; n n 1 i+ 2 kq 0 k 2 + N X n=1 hr( n ); n n 1 i + N X n=1 h( n n 1 )Q;q n i+ N X n=1 b k s h ( n ); n n 1 i + N X n=1 b ks hs n ; n n 1 i: (57)
UsingtheHolder ontinuityof
t
an ,theLips hitz ontinuityofr,theboundedness
ofQand
n
andthestability estimate(43)weobtain(44).
Forproving(45),by(13) and(15) onegets
N X n=1 krq n k 2 C( N X n=1 1 k n n 1 t k 2 + N X n=1 1 k n n 1 k 2 ) +C( N X k ( n )k 2 + N X kr( n )k 2 + N X ks n k 2 ): (58)
From(58),pro eedingasfor(44)andusingthestabilityestimatesobtainedbeforeone
obtains(45).
We ontinuewiththestabilityestimatesfor thesolutionofProblem2.
Proposition 2 Assuming(A1){(A4)there holds
k N h k 2 + N X n=1 k n h k 2 + N X n=1 k n h n 1 h k 2 + N X n=1 h ( n h ); n h i+ N X n=1 kq n k 2 C (59) and ks N h k 2 + N X n=1 ks n h k 2 + N X n=1 ks n h s n 1 h k 2 C 2 2 ; (60) N X n=1 k n h n 1 h k 2 + kq N h k 2 + N X n=1 kq n h q n 1 h k 2 C( + 2 2 ): (61) Proof.Wetakew h = n h in(17),v h = q h n in(18)andw h = b n h in(19)toobtain h n h n 1 h ; n h i+ b hs n h s n 1 h ; n h i+ hrq n h ; n h i= hr( n h ); n h i (62) kq n h k 2 h n h ;rq h n i h n h Q h ;q h n i=0 (63) b hs n h s n 1 h ; n h i+ b ksh ( n h ); n h i= b kshs n h ; n h i; (64)
Adding(62),(63)and(64)andsumminguptheresultfromn=1toN weget
N X n=1 h n h n 1 h ; n h i+ N X n=1 kq n h k 2 + N X n=1 b k s h ( n h ); n h i = N X n=1 hr( n h ); n h i+ N X n=1 h n h Q h ;q h n i+ N X n=1 b kshs n h ; n h i: (65) Testingwithv h =ks P N k =n k h
in(17)andsummingupfromn=1toN gives
ks N X n=1 h n h n 1 h ; N X k =n k h i+ N X n=1 b ks hs n h s n 1 h ; N X k =n k h i + N X n=1 b ks 2 hrq n h ; N X k =n k h i= N X n=1 b ks 2 hr( n h ); N X k =n k h i: (66)
Further,wewrite(18) forn=k,sumupfromk=ntoN,testtheresultby
2 k s q n h
and nallysumupfromn=1toN toget
N X n=1 2 ksh N X q k h ;q n h i N X n=1 2 ksh N X k h ;rq n h i= N X n=1 2 kshQ h N X k h ;q n h i (67)
Weadd(65),(66)and(67),anduseLemma3toobtain 1 2 k N h k 2 + 1 2 N X n=1 k n h n 1 h k 2 + N X n=1 kq n h k 2 + N X n=1 b k s h ( n h ); n h i 1 2 k 0 h k 2 + N X n=1 hr( n h ); n h i+ N X n=1 h n h Q h ;q h n i+ N X n=1 b ks 2 hr( n h ); N X k =n k h i + h 0 h ; N X n=1 n h i+ hs 0 h ; N X n=1 n h i+ N X n=1 2 b k s hQ h N X k =n k h ;q n h i: (68)
Using(A1),(A2),(A3),theCau hy-S hwarzinequalityand(37)and nallythe
Gron-wallLemma,(59)follows from(68).
Toprove(60) wetest in(19) withw
h
=s
n
h
,sumtheresult upfromn =1 toN
anduseLemma3andtheCau hy-S hwarzinequality,aswellas(37)toobtain
ks N h k 2 + N X n=1 ks n h s n 1 h k 2 + N X n=1 ks n h k 2 Cks 0 h k 2 +C N X n=1 k ( n h )k 2 : (69)
(60)follows immediatelyby(A2),(59),andtheLips hitz ontinuityof .
For (61) we subtra t (18) at t = t
n 1
from (18) at t = tn, take in the result
v h
= q
n
h
andaddtheresultingto(17)testedbyw
h = n h n 1 h .Thisgives k n h n 1 h k 2 + b hs n h s n 1 h ; n h n 1 h i+ hq n h q n 1 h ;q n h i = hr( n h ); n h n 1 h i+ hQ h ( n h n 1 h );q n h i: (70) Takingw h = b ( n h n 1 h
)in(19), addingtheresultto (70) and summingup for
n=1;:::;N,byLemma3weobtain N X n=1 k n h n 1 h k 2 + 1 2 kq N h k 2 1 2 kq 0 h k 2 + 1 2 N X n=1 kq n h q n 1 h k 2 = N X n=1 b ksh ( n h ); n h n 1 h i+ N X n=1 b kshs n h ; n h n 1 h i + N X n=1 hr( n h ); n h n 1 h i+ hQ h ( n h n 1 h );q n h i: (71)
Fromequation(71),re alling(A3)oneimmediatelygets
N X n=1 k n h n 1 h k 2 + kq N h k 2 + N X n=1 kq n h q n 1 h k 2 Ckq 0 h k 2 +C N X n=1 2 k ( n h )k 2 +C N X n=1 2 ks n h k 2 +C N X n=1 2 kr( n h )k 2 +C N X n=1 kq n h k 2 : (72)
3.2 Apriorierror estimates
Inthisse tionweprovetheapriorierrorestimatesstatedinSe tion2.
Theorem 1 Assuming (A1){(A4),wehave
ke N k 2 + N X n=1 ke n e n 1 k 2 + N X n=1 k ( n ) ( n h )k 1+ L 1+ + N X n=1 ke n k 2 + P N n=1 ke n q k 2 + 2 k N X n=1 e n q k 2 C(h 1+ + + 1+ ): (73)
and,forallr2IR,
kP h s(t N ) s N h k 2 + N X n=1 kP h s(tn) s n h k 2 C( 2 r 1 + r (h 1+ + + 1+ )): (74)
Proof.We ombinetheideasin[6,11℄ withtheonesin[32,33℄.Considering Problem
1att=t
n
,wesubtra t(17)from(13),(18)from(14)and(19)from(15),andusethe
propertiesoftheproje tors P
h and h toobtain h t ( n h n 1 h );w h i+ b h ts (s n h s n 1 h );w h i + hr h e n q ;w h i= hr( n ) r( n h );w h i (75) he n q ;v h i he n ;rv h i h n Q n h Q h ;v h i=0 (76) h t s (s n h s n 1 h );w h i+ks he n s ;w h i=ks h ( n ) ( n h );w h i; (77) forallw h 2W h andv h 2V h .Wetakenoww h =e n 2W h in(75),v h = h q n 2V h in(76)andw h = b e n 2W h in(77)toget h t ( n h n 1 h );e n i+ b h t s (s n h s n 1 h );e n i + hr h e n q ;e n i= hr( n ) r( n h );e n i (78) he n q ; h e n q i he n ;r h e n q i h n Q n h Q h ; h e n q i=0 (79) b h t s (s n h s n 1 h );e n i b k s he n s ;e n i= b k s h ( n ) ( n h );e n i: (80) Furthermore,wetakew h =ks N X k =n e n 2W h in(75)andobtain k s h t ( n h n 1 h ); N X k =n e k i+k s b h t s (s n h s n 1 h ); N X k =n e k i +ks 2 hr h e n q ; N X e k i=ks 2 hr( n ) r( n h ); N X e k i: (81)
Writing(76) forn=k,summingitup fromk=ntoN andtakingv h =ks 2 h q n
intheresultleadsto
ks 2 h N X k =n e k q ; h e n q i ks 2 h N X k =n e k ;r h e n q i ks 2 h N X k =n ( k Q k h Q h ); h e n q i=0: (82)
Wesumup(78)to(82)forn=1toN,adtheresultanduseLemma3,aswellasthe
propertyhe 0 s ;e n
i=0foralln2f1;:::;Ngtoobtain
N X n=1 h t ( n h n 1 h );e n i+ N X n=1 ks h t ( n h n 1 h ); N X k =n e k i + N X n=1 ks b h ts (s n s n 1 ); N X k =n e k i+ N X n=1 he n q ; h e n q i + N X n=1 ks 2 h N X k =n e k q ; h e n q i+ N X n=1 ks b h ( n ) ( n h );e n i = N X n=1 hr( n ) r( n h );e n i+ N X n=1 k s 2 hr( n ) r( n h ); N X k =n e k i + N X n=1 k s h n Q n h Q h ; h e n q i+ N X n=1 k s 2 h N X k =n ( k Q k h Q h ); h e n q i: (83)
WedenotethetentermsintheabovebyT
1 ;:::;T
10
andpro eedbyestimatingea h
ofthemseparately.UsingLemma3,sin ee
0 =0wehave T 1 = N X n=1 h t ( n h n 1 h );e n i = N X n=1 h t ( n n 1 );e n i+ N X n=1 he n e n 1 ;e n i =T 11 + 1 2 ke N k 2 + 1 2 N X n=1 ke n e n 1 k 2 : (84) Sin e t
is Holder ontinuous with exponent =2 (see (A5)), the Cau hy-S hwarz
inequalityand(37)give
jT 11 j N X n=1 1 2 Z ( Z t n t n 1 ( t (tn) t (s))ds) 2 dx+ N X n=1 2 ke n k 2 C + N X n=1 2 ke n k 2 : (85) ToestimateT 2
weuseLemma3andobtain
T 2 = N X n=1 ks h t ( n n 1 ); N X k =n e k i+ N X n=1 ks he n e n 1 ; N X k =n e k i =T 21 + N X ks ke n k 2 : (86)
Asfor T 11 ,forT 21 wehave jT 21 j 1 2 N X n=1 k t ( n n 1 )k 2 + 3 2 N X n=1 k N X k =n e k k 2 C( + N X n=1 ke n k 2 ): (87) ForT 3 were allthat t
sisHolder ontinouswithexponent .Thisgives
T 3 = N X n=1 k s b h t s (s n s n 1 ); N X k =n e k i N X n=1 k s b 2 k t s (s n s n 1 )k 2 +C N X n=1 ke n k 2 C( 2 + N X n=1 ke n k 2 ): (88) Further, T 4 = N X n=1 he n q ; h e n q i= N X n=1 he n q ; h e n q e n q i+ N X n=1 ke n q k 2 =T 41 + N X n=1 ke n q k 2 : (89)
Usingtheinequalities(9)and(37),byand(A5)-andmorepre isely(16)-onehas
jT 41 j Æ 41 2 N X n=1 ke n q k 2 + 1 2Æ 41 N X n=1 kq n h q n k 2 Æ 41 2 N X n=1 ke n q k 2 +C 1 2Æ 41 N X n=1 h 2 kq n k 2 1 Æ 41 2 N X n=1 ke n q k 2 +Ch 2 ; (90) foranyÆ 41 >0. ByLemma3,forT 5 wehave T 5 = N X n=1 k s 2 h N X k =n e n q ; h e n q i = N X n=1 ks 2 h N X k =n e n q ;e n q i+ N X n=1 ks 2 h N X k =n e n q ; h e n q e n q i = ks 2 2 N X ke n q k 2 + ks 2 2 k N X e n q k 2 +T 51 : (91)
Asfor T 41 weimmediatelyget jT 51 j Æ 51 3 2 N X n=1 k N X k =n e n q k 2 + 2Æ 51 N X n=1 kq n h q n k 2 Æ 51 T 2 2 N X n=1 ke n q k 2 +Ch 2 : (92)
Sin e ()ismonotoneandHolder ontinuous,
T 6 = N X n=1 k s b h ( n ) ( n h );e n i = N X n=1 b k s h ( n ) ( n h );e n i+ N X n=1 k s b h ( n h ) ( n h );e n i = N X n=1 k s b h ( n ) ( n h ); n n h i+ N X n=1 k s b h ( n ) ( n h );P h n n i + N X n=1 ks b h ( n h ) ( n h );e n i N X n=1 C k ( n ) ( n h )k 1+ 1+ +T 61 +T 62 : (93)
The Young inequality, the imbedding L
2
() L
1+
(), the stability estimates in
Proposition1andtheestimate(8)imply
jT 61 j Æ 1+ 61 1+ N X n=1 k ( n ) ( n h )k 1+ 1+ + (1+ )Æ 1+ 61 N X n=1 kP h n n k 1+ 1+ Æ 1+ 61 1+ N X n=1 k ( n ) ( n h )k 1+ 1+ +C(h 1+ + 2 + 4 1+ ) (94) foranyÆ 61 >0.ToestimateT 62
weuseLemma1andthepositivityofthe
on entra-tions: T 62 = N X n=1 k s b h ( n h ) ( n h );P h n i | {z } 0 N X n=1 k s b h ( n h ) ( n h ); n h i 1+
The estimates for the next terms are based on the Lips hitz ontinuity of the
degradationrater().WeuseProposition1and(8),
T 7 = N X n=1 hr( n ) r( n h );e n i N X n=1 Lrk n n h kke n k N X n=1 Lr 2 k n P h n k 2 + N X n=1 3 Lr 2 ke n k 2 Ch 2 + N X n=1 3 Lr 2 ke n k 2 : (96) Similarly, T 8 = N X n=1 k s 2 hr( n ) r( n h ); N X k =n e k i C(h 2 + N X n=1 ke n k 2 ): (97) ToestimateT 9
weuse(A3),theestimatesinProposition1,aswellasin(10),(8)and
(9),theessentialboundednessof ,andtheinequality(37)
T 9 = N X n=1 ks h n Q n h Q h ; h e n q i = N X n=1 ks h n (Q Q h ); h e n q i+ N X n=1 ks h( n n h )Q h ; h e n q i CkQ Q h k 2 +Æ 9 N X n=1 ke n q k 2 +Æ 9 N X n=1 kq n h q n k 2 (98) +C N X n=1 ke n k 2 +C N X n=1 k n P h n k 2 Ch 2 +Æ 9 N X n=1 ke n q k 2 +C N X n=1 ke n k 2 : (99)
Similarly,thelasttermgives
T 10 = N X n=1 ks 2 h N X k =n ( n Q n h Q h ); h e n q i Ch 2 +Æ 10 N X ke n q k 2 +C N X ke n k 2 : (100)
From(83) {(100)weimmediatelyobtain ke N k 2 + N X n=1 ke n e n 1 k 2 + N X n=1 k ( n ) ( n h )k 1+ L 1+ + N X n=1 ke n k 2 + N X n=1 ke n q k 2 + 2 k N X n=1 e n q k 2 C(h 2 +h 1+ + 2 + + 4 1+ + 1+ )+C N X n=1 ke n k 2 : (101)
The rstestimateinTheorem1isobtainedbyapplyingthedis reteGronwallLemma.
Toprovetheinequality(26)wetakew
h =e n s in(77)andobtain h t s (s n h s n 1 h );e n s i+k s ke n s k 2 =k s h ( n ) ( n h );e n s i; (102) giving h t s (s n s n 1 );e n s i+he n s e n 1 s ;e n s i+k s ke n s k 2 ks 2 k ( n ) ( n h )k 2 + ks 2 ke n s k 2 : (103)
Summingup the above for n=1;:::;N, sin ee
0 s
=0weuse Lemma3 and(37) to
obtain 1 2 ke N s k 2 + 1 2 N X n=1 ke n s e n 1 s k 2 + N X n=1 ks 2 ke n s k 2 C N X n=1 1 k t s (s n s n 1 )k 2 + N X n=1 ks 4 ke n s k 2 +C N X n=1 k ( n ) ( n h )k 2 : (104) Sin e t
sisHolder ontinuous,the rsttermontherighthandin(104)isboundedby
C 2
.Forthelasttermweuse(25) andLemma4,yielding
N X n=1 k ( n ) ( n h )k 2 2 N X n=1 k ( n ) ( n h )k 2 +2 N X n=1 k ( n h ) ( n h )k 2 C( 2 r 1 + r (h 2 +h 1+ + 2 + 2 + 4 1+ + 1+ )+ 2 ):
Now(26) followsfrom(104){(105)inastraightforwardmanner.
3.3 Convergen eofthe Newtonmethod
Thequadrati onvergen eoftheNewtons heme(20){(22)isproveninthefollowing
Theorem 2 Assuming(A1){(A4),if issmallenoughwe have
ke n;i k 2 + ke n;i q k 2 C 2 (L 2 r 0+L 2 0 )h d ke n;i 1 k 4 (105) and ke n;i s k 2 C 2 (L 2 0 + 2 L 2 (L 2 r 0+L 2 0))h d ke n;i 1 k 4 (106)
Proof.From(19)wehave hs n h ;w h i= ks 1+ ks h ( n h );w h i+ 1 1+ ks hs n 1 h ;w h i (107) forallw h 2W h
.Usingthisin(17)gives
h n h n 1 h ;w h i+ b ks 1+ ks h ( n h );w h i b ks 1+ ks hs n 1 h ;w h i+ hrq n h ;w h i= hr( n h );w h i (108) forallw h 2W h .Similarly,by(20)and(22), h n;i h n 1 h ;w h i+ b ks 1+ k s h ( n;i 1 h )+ 0 ( n;i 1 h )( n;i h n;i 1 h );w h i b ks 1+ k s hs n 1 h ;w h i+ hrq n;i h ;w h i= hr( n;i 1 h )+r 0 ( n;i 1 h )( n;i h n;i 1 h );w h i (109)
Subtra ting(109)and(21)from(108)and(18)respe tivelyleadsto
he n;i ;w h i+ b ks 1+ k s h ( n h ) ( n;i 1 h ) 0 ( n;i 1 h )( n;i h n;i 1 h );w h i + hre n;i q ;w h i= hr( n h ) r( n;i 1 h ) r 0 ( n;i 1 h )( n;i h n;i 1 h );w h i; (110) forallw h 2W h ,and he n;i q ;v h i he n;i ;rv h i he n;i Q h ;v h i=0; (111) forallv h 2V h .Takingw h =e n;i 2W h in(110)andv h = e n;i q 2V h in(111),adding
theresultingyields
ke n;i k 2 + b k s 1+ k s h 0 ( n;i 1 h )e n;i ;e n;i i+ ke n;i q k 2 = b k s 1+ k s h ( n h ) ( n;i 1 h ) 0 ( n;i 1 h )( n h n;i 1 h );e n;i i+ he n;i Q h ;e n;i q i + hr 0 ( n;i 1 h )e n;i ;e n;i i+ hr( n h ) r( n;i 1 h ) r 0 ( n;i 1 h )( n h n;i 1 h );e n;i i: (112) Sin er 0
() is bounded,whereas and jQ
h
j M
Q
(see (A1) and (A3)), usingthe
in-equality(37)intheabovefurnishes
ke n;i k 2 + b ks 1+ k s h 0 ( n;i 1 h )e n;i ;e n;i i+ 2 ke n;i q k 2 b k 1+ k h ( n h ) ( n;i 1 h ) 0 ( n;i 1 h )e n;i 1 ;e n;i i+ (L r +M 2 Q =2)ke n;i k 2 + hr( n h ) r( n;i 1 h ) r 0 ( n;i 1 h )e n;i 1 ;e n;i i: (113)
Wedenotethe rstand thelast termsontherightbyT
N1
and T
N2
andpro eedby
estimating themseparately. For T
N1
we use Lemmas 1 and 2,the assumption(A1)
andtheinequality(37):
T N1 C Z L 0 je n;i 1 j 2 je n;i jdx C 2 L 2 0 ke n;i 1 k 4 L 4 () + 1 ke n;i k 2 : (114)
Analogously,thereholds T N2 C 2 L 2 r 0ke n;i 1 k 4 L 4 () + 1 4 ke n;i k 2 : (115)
From(113){(115), usingalsothat
0 0weimmediatelyobtain 2 ke n;i k 2 + 2 ke n;i q k 2 (Lr+M 2 Q =2)ke n;i k 2 +C 2 (L 2 0 +L 2 r 0)ke n;i 1 k 4 L 4 () : (116)
Theresult(28)followsnowbyusingtheinverseestimate(seee.g[9℄)
ke i 1 k L 4 () Ch d=4 ke i 1 k:
Finally,forprovingtheinequality(29)weuse(19)and(22)andobtain
he n;i s ;w h i= k s 1+ k s h ( n h ) ( n;i 1 h ) 0 ( n;i 1 h )( n;i h n;i 1 h );w h i (117) forallw h 2W h .Takingw h =e n;i s
andusingLemma2gives
ke n;i s kC( L 0 ke n;i 1 k 2 +L ke n;i k 2 ); (118)
whi h,togetherwith(28), leadsto(29).
4Numeri alResults
Inthisse tionwepresenttwonumeri alteststhatverifythetheoreti al ndingsinthe
pre edingse tions.The rstisaproblemofa ademi nature,admittingananalyti al
solution. These ondis arealisti in ltration problem.Spe i ally,wesolve(1){(2)
withasour etermf()andaFreundli htypesorptionisotherm (x)=x
: S t + b ts r(D S r Q)= S r( )+f in J; (119) t s=k s ( s) in J: (120)
In the rstexample the water uxis onstant, whereas inthe se ond exampleit is
obtainedbysolvingthe owequations(3){(4).
Example1.Wesolvetheproblemaboveintheunitsquare=[0;1℄[0;1℄.We
set
S
=1,
b
=1,D =1, ks=1 andtakea onstant water uxQ=(Q
1 ;Q 2 ) T . Herer( )= 0:1 .With f(t;x;y)= t 1= 1
x(1 x)y(1 y)+[x(1 x)y(1 y)℄
(1 e t ) +2[x(1 x)+y(1 y)℄t 1= +[Q 1 (1 2x)y(1 y)+Q 2 (1 2y)x(1 x)℄t 1= +0:1x(1 x)y(1 y)t 1= ;
andsuitableinitialandboundary onditions,(119){(120)admittheanalyti alsolution
(t;x;y)=t
1=
x(1 x)y(1 y); and (121)
The naltimeisT =1,andthewater uxQ=(0:01;0:01) T
.InspiredbyTheorem1
we omputetheerrors
E ;q= N X n=1 kP h (tn) n h k 2 + N X n=1 kq(tn) q n h k 2 ; and Es= N X n=1 kP h s(tn) s n h k 2 :
Inallsimulationswetake=h.Weperform omputationswith =0:75, =0:5and
=0:25. Inviewof Theorem2,moreexa tly ondition (31),taking =h
2 ensures
the quadrati onvergen e of the Newton methodfor the rst two values of . The
results arepresentedinTables1and 2.Theinitial spatialgridhasameshdiameter
h =0:2, and is re neduniformly by halving h. Theother parameters, and , are
hangeda ordingly.
Thetheoreti alestimatesinTheorem1predi tanorderof onvergen eof =2
forE ;q,sotheerrorredu tionshouldbeofatleast2
2
.Thisisex eededforthe ase
=0:75, where =2isobtained. For =0:5 weobserve anorderof =1:7 for
E ;q
,sin ethis erroris redu edby afa torof3:24=2
1:7
.Again, thisis beyondthe
theoreti allypredi tedorder of2 =1.Oneof thereasonsforthis super onvergen e
isintheassumptionont ,namelyitsHolder ontinuitywithanexponent =2. This
a e tsthetheoreti alestimatesdire tly.Withrespe ttoEs,inall asesweobtainedan
orderof onvergen e s-andthereforearedu tionfa torof2
s
-thatismu hsmaller,
inagreementwiththeestimatesinTheorem1.Wementionthatforallthesimulations
maximalthreeNewtoniterationswereneededpertimestep,andthe onvergen ewas
quadrati (asexpe ted).
For the ase = 0:25 wetook again =h
2
,whi h violatesthe ondition (31).
However,this hoi eisstillinagreementwiththeframeworkofRemark4.Again,the
Newtonmethod onvergesquadrati ally.Sin etheHolderexponent isnowsmaller,a
signi ant de rease in the onvergen e order of the dis retization error is expe ted.
Table3 on rmstheseexpe tations.
Table1 Numeri alresultsforExample1with =h
2 = 2 and =0:75. h E ;q Es s 1 0.2 1.552156e-04 | 1.405687e-05 | 2 0.1 3.975638e-05 1.96 7.307479e-06 0.94 3 0.05 9.941875e-06 1.99 2.755015e-06 1.40 4 0.025 2.478454e-06 2.00 7.970379e-07 1.78 5 0.0125 6.174740e-07 2.00 1.980325e-07 2.02
Example 2. Inthe se ond examplewe onsider the in ltration of benzenein a
saturated,heterogeneoussoil.A sket hofthissituationis displayedinFigure1.The
units are miligram, meter and day, and will be not written expli itly further. The
entire omputationaldomainis=[0;2℄[0;3℄,andin ludestwosubdomains,
1 =
[0:2;1:2℄[2:2;2:7℄and
2
=[1:3;1:7℄[0:8;1:8℄,havingamu hsmallerpermeability.
Table2 Numeri alresultsforExample1with =h 2 = 2 and =0:5. h E ;q Es s 1 0.2 1.789932e-04 | 2.864679e-04 | 2 0.1 5.431520e-05 1.71 1.733271e-04 0.72 3 0.05 1.655999e-05 1.70 8.845142e-05 0.96 4 0.025 5.130814e-06 1.68 3.790681e-05 1.22 5 0.0125 1.566092e-06 1.70 1.443045e-05 1.38
Table3 Numeri alresultsforExample1with =h
2 = 2 and =0:25. h E ;q Es s 1 0.2 5.832042e-04 | 2.864994e-03 | 2 0.1 2.963227e-04 0.97 2.005572e-03 0.50 3 0.05 1.420611e-04 1.05 1.295908e-03 0.62 4 0.025 6.243242e-05 1.18 7.598951e-04 0.76 5 0.0125 2.491041e-05 1.3 4.171508e-04 0.84
aregiveninFigure1.Theotherparametersinvolvedare
S
=0:5,
b
=1andD=1.
Thesour etermf in(119)iszero,andfortherea tiontermwehaver( )= 0:2 .
??? ... 1 2 1 (0,0) (2,0) (0,3) (2,3) 1 =[0:5;1:5℄f3:0g 1 =[2:2;2:7℄[0:2;1:2℄; 2 =[1:3;1:7℄[0:8;1:8℄; K Sj1 =K Sj2 =10 2 ,and K Sjn( 1 [ 2 ) =10. Boundary onditions: jy=3 =10; jy=0 =4,elsewhereQn=0, j 1 =1,elsewherer n=0. Initial onditions: jt=0 =s jt=0 =0.
Fig.1 Computational domainforsimulatingbenzenein ltration inaheterogeneous,
satu-ratedsoil.
The rstsetof omputationsareforh==0:05and =0:5.A ordingto(31),a
timestepoforderh
2
ensurestheoptimal onvergen eoftheNewtons heme.Therefore
wetook =0:0025.Figures2and3present ands,thedissolved,respe tivelyadsorbed
benzene on entration pro les,at di erent times.Asfollows fromTable 4presenting
thetotalresiduum,theiteration onvergesquadrati ally.
These ond set of al ulations is arried out again starting withh = = 0:05,
butnow =0:25. A ording to (31), a timestep of orderh
2:3
is neededto ensure
theoptimal onvergen eoftheNewtonmethod.Thereforewetook =0:0015385. As
before,theNewtonmethod onvergesquadrati ally,seeTable5.Thedissolvedbenzene
Figures 3and4,mu hmore ontaminant isadsorbedinthese ond ase, for alower
valueof .Thentheadsorptionrateismu hhigherforsmallvaluesof .
9.99E-01
7.49E-01
5.00E-01
2.50E-01
2.18E-09
9.99E-01
7.49E-01
5.00E-01
2.50E-01
2.21E-09
1.00E-00
7.50E-01
5.00E-01
2.50E-01
2.35E-09
Fig.2 Benzene on entrationpro lesatT =0:025;T=0:05andT =1(days),for =0:5.
2.45E-02
1.84E-02
1.22E-02
6.12E-03
2.30E-10
4.85E-02
3.64E-02
2.43E-02
1.21E-02
5.89E-10
3.93E-01
2.95E-01
1.96E-01
9.82E-02
5.85E-09
Fig.3 Adsorbedbenzene on entration pro lesatT =0:025;T =0:05 andT =1(days),
for =0:5.
Table4 Convergen eoftheNewtonmethodfortheExample2,with =0:5,=h=0:05,
=0:0025.
T=0.025 T=0.05 T=0.125 T=0.5 T=1
2.4750291e-01 1.5140763e-01 3.1237680e-02 1.5634855e-03 9.7716344e-04
2.3360443e-03 1.4145270e-03 2.3762921e-04 6.1729415e-06 9.6888252e-06
1.5700407e-07 9.6783616e-08 9.1480894e-09 5.5553846e-11 6.7827843e-10
Table5 Convergen eoftheNewtonmethodfortheExample2,with =0:25,=h=0:05,
=0:0015385.
T=0.025 T=0.05 T=0.125 T=0.5 T=1
2.4922387e-01 1.4672717e-01 2.9511781e-02 1.8339098e-03 1.1342203e-03
1.8683306e-03 1.2922026e-03 1.7714809e-04 1.7319646e-05 1.0535939e-05
5.4130407e-08 4.5955634e-08 4.4530926e-09 9.2878094e-10 5.4116583e-10
2.42E-02
1.82E-02
1.21E-02
6.05E-03
4.82E-10
4.79E-02
3.59E-02
2.40E-02
1.20E-02
1.24E-09
6.32E-01
4.74E-01
3.16E-01
1.58E-01
2.20E-08
Fig.4 Adsorbedbenzene on entration pro lesatT =0:025;T =0:05 andT =1(days),
for =0:25.
5Con lusions
We analyzeda mass onservative s heme for solute transport with non-equilibrium
sorption inporousmedia. Thetimedis retization is Euler impli it,and mixed nite
elements are employed for the spatial one. At ea h time step, the emerging
nonlin-ear algebrai systemsare solvedbyaNewton method.A Holder ontinuoussorption
isothermisassumed,whi hmakestheanalysisvery hallenging.Thistypeofsorption
orresponds to thereal, pra ti al situations (like Freundli hisotherms) and weakens
the ommonlyassumedLips hitz ontinuity.Stabilityandapriorierrorestimatesare
shown,guaranteeingthe onvergen eofthes heme.Theorderof onvergen edepends
onlyonthedis retizationparameters.Moreover,asuÆ ient onditionforthequadrati
onvergen eoftheNewtons hemeisobtained.Espe iallythis onditionprovides
use-ful information for the pra ti al al ulations, by indi ating a priori how to ontrol
thedis retization parameters.Thissavesalotof omputationaltime.Tosustainthe
theoreti alresults, twonumeri al exampleshavebeen presented. The rst oneis an
a ademi examplehavingananalyti al solution, and the se ond represents a
realis-ti s enario.Thetheoreti alestimatesarenotne essarysharpbutratherpessimisti ,
thepra ti eshowingabetter onvergen ebehavior.However,the al ulationsreveala
leardependen eofthe onvergen eordersbytheHolderexponent,inthesensethat
values loseto0haveanegativein uen e.
Basedonthetheoreti alresultsand thenumeri alexperiments,we on ludethat
thepresentedmass onservativenumeri als hemeise e tiveand anbeemployedfor
A knowledgments
Partofthe workofISP was supportedbythe GermanResear hFoundation(DFG),
within the Clusterof Ex ellen einSimulationTe hnology (EXC310/1) atthe
Uni-versityofStuttgart.Thissupportisgratefullya knowledged.
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