Analysis
of
a
model
for
precipitation
and
dissolution
coupled
with
a
Darcy
flux
A. Agosti ∗,L. Formaggia,A. Scotti
DepartmentofMathematics,PolitecnicodiMilano,PiazzaLeonardodaVinci32,Milano,20133,Italy
Received10December2014 Availableonline6June2015 SubmittedbyW.Layton
1. Introduction
Thestudyofreactiveflowinporousmediaisofparticularrelevanceinseveralapplicationsrangingfrom geochemical reactions insedimentarybasins [3,4],kerogen degradationand expulsion inoilreservoirs [23], groundwater contamination and remediation processes [21], biomedical applications such as drug release fromdrugelutingdevices[17,10,2].
These processes can be characterized by the presence of phenomena such as adsorption that, at the macroscale, can be effectively modeled by discontinuous reaction terms. The process of crystal precipita-tion/dissolution,whichisthesubjectofthiswork,isusuallymodeledwithadiscontinuousdissolutionrate
* Correspondingauthor.
E-mailaddresses: abramo.agosti@polimi.it (A. Agosti),luca.formaggia@polimi.it (L. Formaggia), anna.scotti@polimi.it (A. Scotti).
totakeintoaccountthatthedissolutiontakesplaceonlyifthecrystalconcentrationisaboveagivenvalue. Thus,thediscontinuitydependson thesolutionitself. Therefore, to provethe existenceof asolution, and todetermineitsbehavioratthediscontinuity,wehavetoresorttoFilippovtheory[9].Whenthetrajectory ofthesolutionreachesthesurfaceofdiscontinuity,inthephasespaceitcancrossthesurfaceorslideontoit untilasuitableexitconditionismet.This classofproblemsmaybe interpretedas differentialinclusions:a recentreviewontheirnumericaltreatment,inthecontextofODEs,isfoundin[8].
In this workwe consider thesimplified, yetrealistic model of precipitation and dissolution proposed in [25] whose numerical treatment has been considered in [15,14]. The model is defined at Darcy scale and describesthe dissolution and precipitation processleading to the formation/degradation of crystalsinside a porous matrix, where the reactants are advected by a given velocity field. We consider here a variable velocity field described by the coupling with a Darcy model where porosity and permeability are linked to the crystal concentration byan empirical law.The model involvesa non-linear discontinuous reaction termthatdescribesthefactthatthedissolutionprocessstartsonlywhentheconcentrationofthereactants reachesacertaincriticalvalue,andiscastasadifferentialinclusion.
In [26] and [16] rigorous models for crystal dissolution and precipitation ina porous medium, in which the amount of precipitate does affect the pore structure, are developed, starting from two-dimensional micro-scale models with free boundaries and using upscaling techniques. These models provide the pre-cise dependency of the porosity and permeability on the amount of precipitate. The simplified coupled model studied here retains the main mathematical difficulties of the coupling laws introduced in [26] and [16].
In [20] a model inwhich reactive transport of asolute coupledto unsaturatedDarcy flow, obtainedby solvingtheRichardsequationforwaterfluxandwatersaturation,isconsidered,andanEulerimplicit-mixed finite elementschemeisanalyzed. Inthatstudy,thefocus isonthesolute transportand reaction,andthe convergence of the scheme is shown assuming that at each time level the solute concentration does not influencethewaterflowquantities.
In[19] anEulerimplicit-mixedfiniteelementschemeformulticomponenttransportandflowisproposed, with a dependence of the porosity and the permeability on the concentrations, deduced from the models introducedin[26].Inthisstudythenecessaryestimatestoshowtheconvergenceoftheschemearededuced onlyforthecaseofconstantporosity.
Wefocushereontheanalysisofthefullycoupledproblem,showingforthefirsttimeitswellposedness,as wellas theconvergence to itsweaksolutionof aparticular finite elementapproximation.This result isan important extension of thatprovided inthe quotedreferences, where the velocityfieldwas agiven datum orwasnotaffectedbythesolutetransportandreaction,andwherearegularizationapproachwasfollowed.
Themaindifficultieswehadto facearethefollowing:
• to show regularity and non-degeneracy of the Darcy field, which is described by a Darcy law with discontinuouspermeabilityandforcingterm;
• toboundthenonlinearcouplingtermsbetweentheDarcy fieldand thecationconcentration;
• to obtain energy estimates for the Darcy velocity and pressure that guarantee the convergence of a sequence ofapproximationstothesolution.
We show some numerical results that underline the fact that by using a method that captures the discontinuity accurately we get sharper dissolution fronts than regularization methods. For the sake of brevity,wehaveomittedthedetailsofthenumericaltechniquesbasedonevent-drivenmethods[6] thatwe haveemployed,whicharethesubjectofaforthcomingwork.
The paper is organized as follows: in Section 2 we outline the mathematical model for dissolution– precipitation, which is taken from [15], and the coupling with a Darcy velocity field. We first give the main result of uniqueness and regularity of the solution of the coupled problem. The existence of
solu-tionsisprovedinSection 2.5,viaaFaedo–Galerkinapproach.Weconstructafiniteelementapproximation and we prove that the limit solution exists and coincides with that of the original differential problem. The final section is devoted to the illustration of a numerical result that shows the effectiveness of the model.
2. A simplifiedmodelofdissolutionandprecipitation
Weintroduceamodelthatdescribestheflowinaporousmediumwithionsdissolvedinwaterthatmove under the action of transport and diffusionand precipitate in acrystal form [25]. Wemodel the problem at the Darcy scale: the medium is a continuum, and the pores and solid particles are homogenized on a reference volumeelement.Weareinterestedinstudyingtheinteractionoftransportanddiffusion,with the chemicalreactionsthatdeterminetheprocessofion(anionandcation)precipitationanddissolution.These processes transform the dissolved ions into immobile solid species, with the consequent formation/ dissolutionofcrystals.
2.1. Nomenclature
With Ω ⊂ R2 we indicate the domain of the problem, occupied by the porous medium. The domain is open and polygonal, with boundary ∂Ω= Γ = ΓD ∪ ΓN . Toobtain uniquenessand regularity resultswe may need to impose further restrictions later on. With T > 0 we denote a givenfinal time. Wefurther define
ΩT = (0, T ] × Ω, ΓTD,N = (0, T ] × Γ
and weindicate with Lp(Ω),Hm(Ω) and Lp((0,T );V ) the usual Banach and Sobolev spaces and spaces with valuesinSobolev spaces [1], for ap ∈ [1,∞] and m ∈ N. While, Hdiv(Ω) = {v ∈ [L2(Ω)]2, div v ∈
L2(Ω)}. With· weindicatetheL2(Ω) norm.
Forafunctionu: ΩT → R: u= u(t,x) we setu(t): Ω → R: u(t)(x)= u(t,x),andanalogouslyfor vectorfunctions.
Furthermore,C,D,E,. . . denotethroughoutgenericpositiveconstantsindependentoftheunknown vari-ablesorthediscretizationparameters,thevalueofwhichmight changefromlineto line.
2.2. Themodel
Weusethesimplifiedmodelconsidered in[15,14],herebrieflyrecalledfortheconvenienceofthereader. While reactionsusually takeplace between variouscationsandanions, thissimplifiedmodel considers only one mobile species, whose mass concentration is denoted byu. Themass concentration of the (immobile) precipitateisdenotedbyv.
The following adimensionalized problem for the evolution of the reactant u is derived from the mass conservation principleand thechemicalbalancedynamics [15,14]:
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ ∂t(u + v)− div ∇u − qu= 0 in ΩT, u = g in ΓTD, ∇u · n = h in ΓT N, u = u0 in Ω for t = 0, (1)
where g and h are givendata, and q∈ Hdiv(Ω) isavelocity fieldgoverned byDarcy’s law. Thecoupling
System(1) iscoupledwith theadimensionalizedequation fortheprecipitatethatreads ∂ ∂tv = r(u)− H(v) in Ω T, v = v0 in Ω for t = 0, (2)
where r(u) and H(v) are theproduction and dissolution rates, respectively,so thattherate of changein theprecipitateconcentrationisthenetresultoftheprocessofprecipitationand dissolution.Itisassumed thatr :R→ [0,∞) is locallyLipschitzcontinuouswith thefollowing properties:
⎧ ⎨ ⎩
r(u) = 0 for u≤ 0,
r(u) monotonically increasing for u≥ 0, r(1) = 1.
(3)
Here, u = 0 and u = 1 are two limiting values. The former sets the minimal concentration required to activatethereaction.Thelatterlimitsthemaximal(adimensional)reactionrateto one.
Remark1.Amoregeneralassumptionfor(3) isgivenby ⎧
⎨ ⎩
r(u) = 0 for u≤ u∗ with 0≤ u∗ < 1, r monotonically increasing for u≥ u∗, r(u∗) = 1 for u∗< u∗.
Notethat,ifr(u)= un,n∈ N+ (in[15] r(u)= u2,asgivenbythemassbalancelaw),thenu
∗ = 0, i.e.r(u)
= ProjR+ (un),andu∗ = 1.
ThedissolutionrateisdescribedbytheHeavisidegraph
H(v) = ⎧ ⎨ ⎩ 0 for v < 0, [0, 1] for v = 0, 1 for v > 0,
which isaset-valuedfunction. Thus,theequalsymbolinequations(1) and(2) shouldinfactbe replacedbyaninclusionsymbol.Following[14,15],we choosetheselection
H(v) = min{1, r(u)} if v = 0. (4)
Withthischoicethedissolution ratebecomesadiscontinuousfunction,dependingonu andv.
Remark 2. With reference to Remark 1, since here u∗ = 1 and all the results below will be obtained forboundaryandinitialconcentrationsu0 andv0 withvaluesbelow1,asitwillbeclearlateronlyaneteffect ofdissolutionwillbe encounteredandnoprecipitation willbe possibleinequation(2).Inthemoregeneral casewhentheinitialdatau0 > u∗ alsoprecipitationisencountered.Nevertheless,thedifficultyinthemodel remainsthesame.
Ifu= 1 everywhereatagiventimet∗,thenthesystemisinequilibriumfort> t∗,i.e.noprecipitationor dissolutionoccur,sincetheprecipitationrateisbalancedbythedissolutionrateregardlessofthevalueofv. Analogously,ifv = 0 everywhereatagiventimet∗,thenthesystemisinequilibriumfort> t∗.Thismodel describes the fact that there is a threshold value for the concentration of the reactant above which dissolutionstarts.
2.3.Analysisofthedifferentialinclusionproblem
Contrary to what has been done in [15] we do not regularize the dissolution term. Since it is a jump discontinuousfunction,thesolutionmaynotbeeverywheredifferentiablein(0,T ).Weneedthentoexploit some results on finite dimensional Ordinary Differential Equations with Discontinuous Right Hand Side (ODEwithDRH)andadoptspecialtechniquesforthenumericalsolutionoftheproblem.
Let us introduce afamily of finite dimensional subspaces Vh of H1(Ω) such that H1(Ω) is a Hilbertian
sum of the Vh. Let us denote by Ph : L2(Ω) → Vh the projection operator, and by uh = Phu, vh = Phv.
Problem(2) becomes:atanytimet,givenuh,findvh suchthat
∂
∂tvh∈ r(uh)− H(vh) in Ω
T,
(5) vh = Ph[v(0)] in Ω for t = 0,
wherewehavesupposedthatv(0)∈ L2(Ω).
Lemma 1.System(5)has aunique solution.
Proof. The proof is based on the fact that (5) represents a family of non-autonomous finite dimensional inclusion problems on the parameter h. Forany x ∈ Ω this problem is isomorphic to ˙z ∈ F (t,z(t)), t ∈ [0,T ], z(0)= z0 with z : R→ Rn, F: R × Rn → 2Rn, n ∈ N. An existence and uniqueness result for this
inclusionproblemisexpressedinthefollowingtheorem:
Theorem 1.Letthesetvalued mapF = F (t,z) satisfythefollowingconditions: 1. thesets F (t,z) are closedandconvex,
2. themapF (·,z) isLipschitz continuous(measurabilityisonlyrequiredforexistence)andthemapF (t,·) isUpper Semi-Continuous (USC),i.e.the closureof theset {F (t,z)|z− z0< δ}, δ > 0,iscompact
∀z0∈ Rn,
3. F (t,z) satisfies a growth condition: ∀y ∈ F (t,z) ∃k(t) > 0 and a(t) | y ≤ k(t)z+ a(t), for all
z∈ Rn.
Thenthere isan absolutelycontinuoussolution tothedifferentialinclusion,forevery z0∈ Rn.Ifmoreover
themapF (t,z) isOne-SidedLipschitzContinuous(OSLC),i.e.ifthereexistsaconstantL(t) suchthatfor every z1,z2,andforevery y1∈ F (t,z1) andy2∈ F (t,z2),
(z1− z2, y1− y2)≤ L(t)z1− z22,
thenthesolutionisunique.Analogousresultsexistatthecontinuouslevel,givenbytheHille–Yosidatheory formaximal monotonenonlinearoperators.
Theproofmaybefoundin[9].Theset-valuedmapH(vk) hasthefollowingproperties:
1. itisconvex,compactandmaximalmonotone(indeed itcanbecharacterizedasthesubdifferentialofa convexfunction);
2. itsatisfiesthegrowthcondition,withk = a= 1 (inparticularitsatisfiesaboundednessconditionwith k = 0);
Therefore,accordingto Theorem 1,andusingthepropertiesofthemapr(·) andthetimeregularityofthe solutionsuh whichwill begivenbelow,problem (5) admitsauniqueabsolutely continuoussolution. 2
Tointegratesystem(5) atdiscretelevel,wehavetoselectanelementoftheset H(0) whenvh = 0.This selectionshouldcoincidewiththeprescriptionintroducedinequation (4) atthecontinuouslevel,i.e.
H(vh) = min{1, r(uh)} if vh= 0. (6)
WeusetheresultscomingfromthetheoryofFilippov [9],and exploitedforthenumericalsolutionin [6,7], inthecontextofevent-driven methods.Thereadermayrefertothequotedreferencesfordetails,whichwe donotreporthereforthesakeofbrevity.
2.4. Couplingwith aDarcy model
Weconsider the couplingof the precipitation–dissolutionmodel (1)–(2) withthe Darcy equationsfor a singlephasefluidwithconstantdensity(water,inthecaseofourinterest).Namely,theadvectionvelocity fieldforthecationtransportprocessin(2) isthesolutionoftheproblemforq andp givenby
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂φ ∂t + div q = 0 in Ω T, q =−k(φ) μ ∇p in Ω T, p = pD on ΓTD, q· n = η on ΓTN, (7)
wherewehaveindicatedtheessentialandthenaturalboundaryconditionswithdataη andpD,respectively.
Here φ istheporosity,q isthemacroscopicvelocity, p is thefluid pressure,μ> 0 is thedynamicviscosity andk isascalarpermeability.
The Darcy modelis coupled to the cation dynamics by the factthatthe precipitation and dissolution processes influence the porosity and thus also the permeability of the medium. In particular, with the increase of the precipitate concentration there is a consequent reduction of porosity; an empirical law forthevariationofporositywithvaryingprecipitateconcentrationisgivenby [12]
dφ dt =−
∂v
∂t. (8)
Ateachtime wehave φ= φ0 − v, withφ0 aconstantwhich wetake,forthe sakeofsimplicity,equalto 1. The permeability coefficient is modeled as a positive Lipschitz continuous function of porosity. A possibleempiricallawis [12]
k(φ) = (φ)2→ k(φ(v)) = (φ0− v)2= (1− v)2, (9)
forv ∈ [0,1].
Remark3.In[26] and[16] rigorousmodelsforcrystaldissolutionandprecipitationinaporousmedium,in which the amount of precipitate does affect the pore structure, are developed, starting from two-dimensional micro-scale models with free boundaries and using upscaling techniques. These models provide theprecisedependency oftheporosityand permeabilityontheamountofprecipitate,intermsof the thickness of thelayer of precipitated crystal around the grain boundaries, where the porosity enters alsointhemassbalanceequationsforthesoluteandtheprecipitate.
Here, startingfrom the modelof precipitation and dissolution (1) and(2), proposedin[25],coupled to theDarcyequations(7) throughtheempiricallaws(8) and(9),weobtainasimplifiedcoupledmodel,where theporosity doesnotenterexplicitlyintothereactivetransportequations.Howeverthemodelretainsthe mainmathematicaldifficultiesofthecouplinglawsintroducedin[26] and[16].
Indeed,using(8) andintroducingtheporosityweightfactorinthemassbalanceequationsforthesolute andtheprecipitate(1) and(2),wewouldobtainthefollowingadimensionalizedreactivetransportequation:
(1− v)∂tu− div
∇u − qu= (1− v)(u − 1)[r(u) − H(v)], (10) which,coupledtotheDarcyequations(7),givesasimilarmodeltothatintroducedin[20].Notethat,for0≤ v < 1, 0 ≤ u ≤ 1, the reactive transport model (10) retains the same mathematical properties as the simplifiedmodeldescribedby(1) and(2).
In order toavoidnumericaldifficulties linkedwiththe degeneracy oftheDarcy equations at thediscrete level, as will be explained in Section 2.5, we introduce the following regularization of the permeability coefficient:
k(v) := k(min{1 − , v}), > 0.
Note that k(v) is a Lipschitz continuous function for v > 0. Forthe analysis we consider η = 0, g = 0
and h = 0. We also assume throughout that |ΓD| > 0. Let us define the following functional spaces:
U := {u ∈ L2((0,T );H
01(Ω)) : ∂tu ∈ L2((0,T );H−1(Ω))}, V := {v ∈ H1([0,T ];L2(Ω))}, Q := L2(ΩT ), Z :=
L2((0,T );Hdiv(Ω)).
Theweakformulationofthecoupledproblem reads:
ProblemP2.Foragiven> 0,find(q,p)∈ Z × Q and(u,v)∈ U × V,with(u(0),v(0))= (u0,v0),such
thatforall(τ ,ψ)∈ Hdiv(Ω)× L2(Ω) and(ω,θ)∈ H01 × L2(Ω),
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ μ k(v)q(t), τ − (p(t), div τ ) = −(pD, τ · n)ΓD, (div q(t), ψ) = (−∂φ(v)∂t , ψ) = (∂tv(t), ψ),
(∂tu(t), ω) + (∇u(t), ∇ω) − (q(t)u(t), ∇ω) ∈ (H(v(t)) − r(u(t)), ω),
(∂tv(t), θ)∈ (r(u(t)) − H(v(t)), θ),
(11)
fort∈ (0,T ),withH(v) satisfyingtheselection(4),u(0)= u0 ∈ L2(Ω) andv(0)= v0 ∈ L2(Ω). Ourmain resultsarestatedinthefollowingtwotheorems.
Theorem2.LetΩ beaconvexpolygonaldomain,andletpD ∈ H1/2(ΓD).Assumemoreoverthat0≤ u0 ≤ 1
and0≤ v0 < 1.Thenthereexistsasolutionof(11).
This theorem will be proven in the next section through the convergence result of the fully discrete problem employingtheEulermethodfortimeintegration.
Theorem 3. Under the same hypothesis of Theorem 2 and the additional assumptions that Ω is a convex polygonal domain with convex anglesgiven by a rational fraction of π (for example Ω= [−L,L]× [−l,l]), and that the Dirichlet and Neumann data are imposed on whole polygon edges, if q ∈ [L∞(Ω)]2, then the
solution of (11) is unique.
Remark 4. The hypothesis of q ∈ [L∞(Ω)]2 is perfectly reasonable due to the additional assumption of Theorem 3. Indeed, it is possible to prove, byusing some results on the regularity of elliptic equations
with rough coefficients in polygonal domains, that p ∈ C0,β ( ¯Ω), for a β ∈ (0,1) (see [13], Chapter 14, in particular Theorem 14.2.1, for the regularity on compact subsets of Ω; see [11], Chapter 6, in particular result (6,4,2,1),for the regularity near the corner of the domain, supposing thatv0 hasno discontinuity near thecorners;seealso [22]).Thismeans thatp isalmost everywhere aLipschitz function,and,together withtheboundednessofk(v),thisprovidesthewantedregularityforthevelocity.Forthesakeofbrevity we
omittheproof.
Remark5.Bycompactnessargumentsitmaybeshownthatif,inadditiontothestatedhypotheses,u0 ∈
H01(Ω),thenu∈ C0([0,T ];H
01(Ω))∩ L2(0,T ;H2(Ω)) isaglobalstrongsolutionof(11).
Inordertoproceedwith theproofofTheorem 3 weneedthefollowing
Lemma2. Under thestated hypotheseson the initialdata and on k,0≤ v(t)< 1 and 0≤ u(t)≤ 1 a.e. in
L2(Ω) forallt∈ (0,T ],independentlyof ,andμ/k(v(t))∈ L∞(Ω) andispositive.Thesolutionu belongs to
L∞(0,T ;L2(Ω)) ∩ L2(0,T ;H01(Ω)), v ∈ L∞(0,T ;L2(Ω)), q ∈ Hdiv(Ω), p∈ L2(Ω), ∂tv(t) ∈ L∞(Ω) and
∂tu(t)∈ L2(0,T ;H−1(Ω)).
Proof. LetussetR(u,v)= r(u)− H(v).Wemaynotethat • R(u,v)≥ 0 ifv≤ 0,forallu;
• R(u,v) iseither0 or−1 ifu≤ 0; • R(u,v)≤ 0 ifv≥ 0 and0≤ u≤ 1;
• R(u,v)≥ 0 ifu≥ u∗,so inparticularwhen u≥ 1.
Toprovethatv isnotnegativeitissufficienttotakeθ = v−(t)= 12 (v(t)− |v(t)|) as testfunctionin(11) to get v−(t)2≤ v− 02+ 2 t 0 (R(u(τ ), v−(τ )), v−(τ ))dτ ≤ 0,
by whichv−(t) is azeroelementofL2(Ω).
Wenowtakeω = u−(t) and ψ = (u−(t))2 in(11),andweexploitthenon-negativityofv provenabove andthepropertiesofR(u,v) to get,afterintegrationbypartsofthedivergenceterm,
1 2 d dtu −(t)2+∇u−(t)2+1 2(∂tv, u −(t)2)∈ −(R(u−, v), u−(t))≤ 0, bywhich,sinceu−0= 0, u−(t)2+ 2 t 0 ∇u−(τ )2dτ ≤ u− 02+ 1 2 t 0 (R(u−(τ ), v(τ )), u−(τ )2)dτ ≤ 1 2 t 0 u−(τ )2dτ.
ByGronwall’s inequalitywegetu−(t)= 0.Thefactthatu cannotexceed1 isobtainedinasimilarway, bylookingatthe negativepartof1− u.Usingthese bounds foru wemayprovethatv≤ 1 bylookingat thenegativepartof1− v.Note that0≤ v(t)< 1 and0≤ u(t)≤ 1 foreach> 0,uniformly in.
Remark 6. Since the reaction rate R(u,v) is zero (for v = 0) or negative (for v = 0) for 0 ≤ u ≤ 1, v canonlydecrease from v0< 1 to zero, sothatv < 1 at alltimes. Thisis aconsequenceofthe factthat
only dissolution is considered, as expressed in Remark 2. Moreover, since 0 ≤ v(t) < 1 and 0 ≤ u(t) ≤ 1 independentlyof ,thereexistsaconstant1 > 0 such that,foreach0≤ < 1,wehavethat
k(v) := k(min{1 − , v}) ≡ k(v).
Hence weobtain, byusing amaximum principleargument,thattheregularized ProblemP2 isequivalent to theunregularizedproblem.
Bythedefinitionofk(v) we obtainthatμ/k(v(t))∈ L∞(Ω) andispositive.
Forwhatconcernstheregularityofthesolutions,wetakeω = u(t) in (11),toobtain 1 2 d dtu(t) 2+∇u(t)2∈ −1 2 ∂tv(t), u2 − ∂tv(t), u =−1 2 r(u)− H(v), u2 − r(u)− H(v), u , afterintegratingbypartsandusing(11)2 withψ = u2.
Since r(u) isnon-negative andH(v) takesvalues between 0 and 1,the applicationofCauchy–Schwarz and Younginequalitiesgives
1 2 d dtu(t) 2+∇u(t)2≤1 2u(t) 2+|Ω|1/2u(t) ≤|Ω| 2 +u(t) 2.
Then, thankstoGronwall’s inequality,
u(t)2+ 2 t 0 ∇u(s)2ds≤ (u 02+ T|Ω|)(1 + T e2T) for t∈ (0, T ]. (12)
Asforv,wetakeθ = v in(11); we usethelocalLipschitzcontinuityofr(u) and (12) toobtainthatv(t) isboundeduniformlyin(0,T ].
Theboundednessof∂tv,andconsequentlythatofdiv q,derivesfromthatofr(u)− H(v).Ifwechoose τ
= ∇η in(11),whereη satisfies
−Δη = p,
η|ΓD = 0, ∂η∂n = 0 on ΓN,
weobtainp2 ≤ C + Dq2,thankstotheboundednessofμ/k(v) and totheLax–Milgramestimate ∇η
≤ Cp.ThenotationD meansthatthevalueofthepositiveconstantD dependson.
Wenowtakeτ = q andψ = p in(11),exploitthefactsthat0≤ H(v)≤ 1,r(u) is boundedandk(v) is
positivelyboundedawayfromzero,andusetheestimateforp,toobtainq≤ C.
Since u ∈ L2(0,T ;H
01(Ω)) and qu ∈ L∞(0,T ;L2(Ω)), we have that ∂tu ∈ L2(0,T ;H−1(Ω)). Note
that, if u0 ∈ H01(Ω), since q ∈ [L∞(Ω)]2, by choosing ω = ∂tu, we obtain that u ∈ L∞(0,T ;H01(Ω))∩ H1(0,
T ;L2(Ω)). 2
Wearenowinthepositiontoprove Theorem 3:
ProofofTheorem 3. Assumethereexist twosolutionsof(11),indicatedby(u1,v1,q1,p1) and (u2,v2,q2,p2),
q
¯(0)= 0 andp¯(0)= 0. Takingθ = ¯v(t,x) from thefourthequationofsystem(11) weobtain 1 2 d dt¯v 2∈ (r(u 1)− r(u2), ¯v)− (H(v1)− H(v2), ¯v).
Using themonotonicity property of theset valuedmap H(v) (note thatthis property is still validwith the prescription (4)), the Lipschitz continuity of r(u) (with constant Lr) and Schwarz and Young
inequalities, weobtain d dt¯v(t, x) 2≤ L2 r¯u(t, x)2+¯v(t, x)2. Wethenhave ¯v(t, x)2≤ C eT t 0 ¯u(s, x)2ds. (13)
Takingnowω = ¯u(t,x) in thedifferencebetweenequations(11)3 for(u1,v1,q1,p1) and (u2,v2,q2,p2)
wehave
(∂tu(t), ¯¯ u(t)) +∇¯u(t)2− (¯qu1(t),∇¯u(t)) − (q2u(t),¯ ∇¯u(t)) ∈ (∂t¯v(t), ¯u(t)), (14)
whichwerewrite,after integrationbyparts ofthefourthtermand integrationintimefrom 0 tot< T ,as thefollowingdifferentialinclusion:
1 2¯u(t) 2+ t 0 ∇¯u(s)2ds∈ t 0 (¯qu1(s),∇¯u(s))ds − 1 2 t 0 ∂v2(s) ∂t , ¯u 2(s) ds + t 0 (H(v1(s))− H(v2(s)), ¯u(s))ds− t 0
(r(u1(s))− r(u2(s)), ¯u(t)). (15)
Wenowintroducethefollowinglemma.
Lemma3.Giventwosolutionsof(11),indicatedby(u1,v1,q1,p1) and(u2,v2,q2,p2),anddefining:u¯ = u1
− u2,v¯ = v1 − v2,wehave
H(v1)− H(v2) ≤ C¯u + D¯v. (16)
Proof. Indeed, if v1 = v2 = 0, from (4), Lemma 2 and from the Lipschitz continuity of r(u), we get the estimate H(v1)− H(v2)≤ Cu¯; if v1,v2 = 0, we get the estimate H(v1)− H(v2)≤ Dv¯; if v1 = 0
and v2 = 0 (or vice versa), from estimate(13) wehave thatu1 = u2 (almost everywhere inΩ), andfrom (4) wehaveH(v1)− H(v2)≤ C1− u2≤ Cu¯. 2
Thanks to Lemma 2, the monotonicity and the Lipschitz continuity of r(u), we can then write the followinginequality: ¯u(t)2+ t 0 ∇¯u(s)2ds≤ C t 0 ¯q(s)2ds + D t 0 ¯u(s)2ds + E t 0 s 0 ¯u(r)2drds,
which implies(seeforinstance [18]) ¯u(t)2≤ C t ¯q(s)2ds. (17) 0
Takingthedifferencebetweenthefirstandsecondequationsofsystem(11)for(u2,v2,q2,p2) and (u1,v1,
q1,p1),weobtain ⎧ ⎨ ⎩ μ k(v1)− μ k(v2) q1, τ + μ k(v2)q, τ¯ − (¯p, div τ ) = 0, (div ¯q, ψ) = (∂tv, ψ).¯ (18)
If wechooseτ =∇η,whereη satisfies
−Δη = ¯p,
η|ΓD = 0, ∂η∂n = 0 on ΓN,
¯p2+
thefirstequationof(18)provides μ k(v1)− μ k(v2) q1,∇η + μ k(v2) ¯ q,∇η = 0,
by which,thankstoLemma2,theLax–Milgramestimate∇η≤ p¯,theLipschitzcontinuitypropertyof k(v) and (13),wefinallyobtain
¯p2≤ C¯q ¯p + D¯p ¯v → ¯p2≤ C¯q2+ D t
0
¯u(s)2ds. (19)
Wenowtakeτ = q¯ andψ = ¯q in(18),obtaining μ k(v1)− μ k(v2) q1, ¯q + μ k(v2) ¯ q, ¯q ∈r(u1)− r(u2), ¯p −H(v1)− H(v2), ¯p . (20) Since r(u) andk(v) areLipschitz continuous, q is bounded, 1/k(v) ispositively bounded away from
zero,and given(16),wehavethat
(21) q¯2 ≤ Cu¯ p¯ + Dv¯ p¯ + Ev¯ q¯.
Combinedwith(19),itallowsusto obtain
¯q2≤ C¯u2+ D t
0
¯u(s)2ds, (22)
which,substitutedin(17),provides
¯u(t)2≤ C t 0 ¯u(s)2ds + D t 0 s 0 ¯u(r)2drds, (23)
and,consequently,u¯(t)2 = 0 [18].Asaconsequenceof(13),(22) and(19),wethenhave:v¯(t)= q¯(t)= p¯(t)= 0. 2
2.5. Existence ofsolutions of thecoupledproblem
TheexistenceofsolutionsisprovedthroughaFaedo–Galerkinapproachusingadiscretizedproblem.We writeadiscreteapproximationofProblemP2 byafinitedifferenceschemeintime,adualmixedhybridized finite element discretization in space for the Darcy equation and a primal hybrid finite element discretization inspace for the species transport equations. Since for the numericalsolution of the discrete ODE DRH we use event-driven methods [8] that employ the Filippov prescriptions, we have chosen an explicit Euler time discretization. The Euler semi-implicit method used in [14] (in the case of a given velocity field) is notfeasible in ourcase, since it updates the v variable at a given time step using a value ofthe u variableatthenexttimestep,whichmakesevent-driventechniquesimpracticable.
LetThbearegularconformingdecompositionofΩ into triangles,andletusintroducethefollowingfinite
elementspaces: Zh:={qh∈ K∈Th Hdiv(K)|qh|K∈ RT0(K)∀K ∈ Th}, Vh:={ph∈ L2(Ω)|ph|K∈ P0(K)∀K ∈ Th}, Ph:={λh∈ K∈Th H1/2(∂K)|λh|∂K ∈ P0(∂K)∀K ∈ Th, λh|∂Ω= 0}, Wh:={vh∈ K∈Th H1(K)|vh|K∈ P1(K)∀K ∈ Th}, Qh:={μh∈ K∈Th H−1/2(∂K)|μh|∂K∈ P0(∂K)∀K ∈ Th},
wherePi(K) indicatesthespaceofpolynomialsoforderi onK,RT0(K) isthezero-thorderRaviart–Thomas
space, and P0(∂K) is the zero-th order polynomialon each edge of ∂K. Introducing thelocal projection
operatorsΠK : Hdiv(K)→ RT0(K);PK0 : L2(K)→ P0(K);PK1 : H1(K)→ P1(K);ρK :
K∈ThL2(∂K)→
R0(∂K),werecallsomewellknownresultsfrom interpolationtheory[5].
τ − ΠKτ0,K ≤ ChKτ 1,K, for τ ∈
H1(K)2, divτ− ΠKτ
0,K ≤ ChK div τ 1,K, for div τ ∈ H1(K),
v − P0
Kv0,K≤ ChKv1,K, for v∈ H1(K),
v − P1
Kv0,K≤ ChKv1,K, for v∈ H1(K),
λ − ρKλ1/2,eh ≤ ChKw2,K, for w∈ H2(K)|w|∂K= λ,
λ − ρKλ−1/2,eh≤ ChK div q1,K, for q∈
H2(K)2| q · n|∂K = λ, (24) whereeh istheset ofedgesofK.Wesetτ = T /N foran N ∈ N,andtn = nτ ,n= 1,. . . ,N .Startingwith
u0
h= Phu0 and vh0 = Phv0, where Ph is aglobalinterpolation operator,withu0∈ H01(Ω) andv0∈ L2(Ω),
wedefine ProblemPh
2.Given(un−1h ,vhn−1)∈ Uh×Uh,find(qhn,pnh,λhn)∈ Zh×Vh×Phand(unh,vnh,μnh)∈ Wh×Wh×Qh,
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ K∈Th K μ k(vnh−1) qhn· τh− K pnhdiv τh+ ∂K λnhτh· n + ∂K∩ΓD pDτh· n = 0, K∈Th K ψhdiv qhn = Ω r(un−1h )− H(vn−1h )ψh, K∈Th ∂K ρhqnh· n = 0, K∈Th K (unh− unh−1)ωh+ τ K ∇un h∇ωh− τ K qnhunh∇ωh− τ ∂K μnhωh =−τ Ω (r(un−1h )− H(vn−1h ))ωh, Ω (vnh− vhn−1)θh∈ τ Ω (r(unh−1)− H(vnh−1))θh, K∈Th ∂K unhνh= 0. (25)
This particular choice of finite elements is useful for the analysis, since we can easily treat the terms element-wise. However, we wish to point out that numerical experiments show the suitability of more standardfinite elementdiscretizations.
Defining the operator B := (Bq· n,ρ) = K∂Kρq· n ∀ρ ∈ K∈ThH1/2(∂K),ρ∂Ω = 0, we can
identify Hdiv(Ω)= ker B. Having introducedafinite element space Ph of functionsinP0(∂K),which are
discontinuousattheedgevertices of∂K,theformulationenforces thelocal reciprocityconstraint[5]. Analogously,bydefiningtheoperatorC := (Cv,μ)=KKvμ∀μ∈K∈ThH−1/2(∂K) wecanidentify H1
0(Ω)= ker C.SinceQh|K=P0(∂K),thespace ker Ch isgiven byfunctionsinWh which arecontinuous
at the middle point of each edge of ∂K; the formulation is thus equivalent to a non-conforming primal formulationonaCrouzier–Raviartfiniteelementspace[5].
Remark 7.Note that we cannot exploitamaximum principle, since, forour kindof problem and forthe finite element spaceswe are using, there are nostandard discrete maximumprinciples. Hence,we cannot show thatifunh−1≤ 1 thenunh≤ 1 forasufficientlysmalltimestep,northatvhn< 1 ifvnh−1< 1.However, the introduction of the regularization of permeability k(vh) avoids the problem of low regularity of the
Darcy fieldduetothedegeneracyat vh= 1.Moreover,discretemaximumprinciplewhichguaranteesthat
ifun−1h ≤ 1 thenun
h≤ 1 isavailableinthecasevh|K ∈ P0(K).
Thepossibilityofnegativeconcentrationsofvn
handunh,whichcanbeobtainedwhenvn−1h 1,vhn−1= 0,
r(un−1h ) 1 andthetimestepτ isgreaterthanagivenτ∗ 1,canbeavoidedbylocalizingthethreshold v = 0 with aneventdrivenstrategy,like theonesin[7,8].
From the fourthand fifth equations of system (11), byapplying elliptic regularity on each element K and noting that the forcing term r(unh−1)− H(vhn−1) ∈ L2(K), we obtain that uh,vh ∈ L∞(K), and
k(vh)∈ L∞(K).
Existence anduniquenessofthesolutionto ProblemP2h derivefrom thefollowingfacts: • Thetermμ/k(vhn−1) isboundedandalwayspositive,thequadraticform (
μ k(vn−1h )
qnh,qnh)K is
contin-uous andcoercive over ker Bh, thefinite dimensionalspaces satisfythediscrete inf–sup conditionand
theforcingtermsarewell defined.Thereforethemixedhybridized formulationoftheDarcy equations hasauniquesolution[5].
• Thebilinearform associated tothetransport equationforuh isweaklycoerciveoverker Ch, thefinite
dimensionalspacessatisfythediscreteinf–supconditionandtheforcingtermiswelldefined.Therefore, theprimalhybridformulationforun
• The application of the Filippov selection procedure ensures that there exists a unique sequence of solutions vhn,which converges, for N → ∞, to the uniquesolutionof thecontinuous intime inclusion problem uniformlyinC0([0,T ],R2)[9].
We proceed now to obtain energy estimates, which will be used later to show the convergence of a time continuousapproximationofthediscretesolutionstotheweaksolutionofthecontinuousproblem.Toease notation, from now on, where not otherwise indicated, the sum over all the elements K of the mesh is understood.
Lemma 4(Energy estimates). Thereexistconstants C > 0 independentof τ andh such that thefollowing estimateshold: sup k=1,...,Nv k h− vhk−1K+ N n=1 un h− unh−1 2 K ≤ Cτ, (26) N n=1 ∇(un h− un−1h ) 2 K+ τ N n=1 ∇un h2K ≤ C, (27) sup k=1,...,Nq k hK ≤ C, (28) sup k=1,...,Np k hK+ sup k=1,...,Nλ k h∂K≤ C, (29) N n=1 qn h− qnh−12K ≤ Cτ, (30) N n=1 pn h− pnh−1 2 K+ N n=1 λn h− λnh−1 2 ∂K ≤ Cτ, (31) τ N n=1 div qn h2K ≤ C, (32) N n=1 div(qn h− qn−1h ) 2 K ≤ Cτ. (33)
Proof. Wetakeωh= unh and νh= μhn inthefourthandinthesixthequationsofsystem(25),respectively,
tohave 1 2 un h2K− un−1h 2K+unh− un−1h 2K + τ∇unh2K+ τ (r(un−1h )− H(vhn−1), (unh)2)K =−τ(r(unh−1)− H(vnh−1), unh)K.
Since r(·) is positive, the term (r(un−1h ),(un
h)2) is positive. Using the boundedness of H(·), theLipschitz
continuityofr(·),Cauchy–Schwarzand Younginequalities,weobtain 1 2 un h2K− un−1h 2 K+unh− un−1h 2 K + τ∇unh2K ≤ Cτun h2K+ τ Lrunh−1KunhK+ CτunhK≤ Cτunh2K+ Cτu n−1 h 2K+ Cτ + 1 2τu n h2K,
and,bysummingover n= 1,. . . ,k, foranarbitraryk≤ N, 1 2u k h2K+ 1 2 k n=1 un h− unh−12K+ τ k n=1 ∇un h2K ≤ 1 2u I h2K+ C.
Here wehaveusedthefactthatun
h∈ L∞(K)⊂ L2(K).Thisresultimpliestheestimateinthesecondpart
of (27).
Remark8.Thesameresultcouldbeobtainedbyusing thediscreteGronwall inequality,inthiscasewedo not needun
h ∈ L∞(K).Moreover, wecould obtainthesameresult withouttaking anintegration byparts
of theadvectionterminthefourthequation ofsystem(25),byusingthe factthatqhn∈ [L∞(K)]2,or the estimate ofqn
h givenby(35),andapplying thediscrete Gronwallinequality.
Wetakenowθh= vhn− v n−1
h inthefifthequationofsystem(25),consequentlythefirstpartof(26)gives
vn
h− vnh−1 2
K ≤ τr(1)Kvhn− vnh−1K+ τ Cvnh− vhn−1K.
Toobtain(28)and(29),wetakeτh= ΠK∇ηhinthefirstequationofsystem(25),whereηhisasolution
of
− div ∇ηh= pnh in K,
∇ηh· n|∂K = λnh.
(34) Notethatdiv[ΠK∇ηh]= Pk,0div∇ηh=−Pk,0pnh =−pnh,andthatΠK∇ηh∈ RT0(K).Besides,foreachK,
ΠK∇ηhK ≤ ChpnhK.Then pn h2K+λnh∂K2 =−(pD, λnh)∂K∩ΓD− μ k(vn−1h ) qhn, ΠK∇ηh K =−(˜pD, pnh)K− (∇˜pD, ΠK∇ηh)K− μ k(vhn−1) qnh, ΠK∇ηh K ≤ 1 4p n h2K+ C + C sup K μ k(vhn−1) qn hKpnhK ≤ 1 2p n h2K+ C + Cqnh2K,
where p˜D isa harmonic liftingoftheboundarydata.Hence,wemaywrite
pn
h2K+λnh2∂K ≤ C(1 + qnh2K).
Now, letustakeτh= qnh,ψh= pnh andρh= λnh inthefirst,thesecond andthethirdequationsofsystem
(25),respectively.Weobtain μ k(vhn−1) qnh, qhn K − (r(un−1 h )− H(v n−1 h ), pnh)K =−(˜pD, r(unh−1)− H(v n−1 h ))K− (∇˜pD, qnh)K. (35)
Since 1/k ispositiveandboundedawayfromzero,wecanwrite
qn h2K ≤ 1 4Cpp n h2K+ C + 1 4q n h2K≤ 1 2q n h2K+ C,
whereCpistheconstantC intheinequalitypnhK2 ≤ D +Cqnh2.Wethusobtainestimates(28)and(29).
Ifwe takenowψh= div qnh inthesecondequationof system(25),weget
τ div qhn2K ≤ τr(un−1h )− H(vn−1h )K div qnhK ≤ Cτ +
1
2τ div q
n h2K,
Takingωh= unh− un−1h andνh= μnh inthefourthandinthesixthequationsofsystem(25),respectively,
allowsustowrite that un
h− unh−12K+ τ (∇unh,∇(unh− unh−1))K− τ(qnhunh,∇(unh− unh−1))K =−(vnh− vnh−1, unh− unh−1)K.
Then, byapplyingintegration byparts totheterm (qn
hunh,∇(unh− un−1h ))K, usingtheCauchyand Young
inequalities,equations (26),(28)and (32)and thefactthat∇un
h ∈ L∞(K) (since unh ∈ L∞(K)∩ P1(K)), weget un h− un−1h 2K+ 1 2τ ∇un h2K− ∇un−1h 2K+∇(unh− un−1h )2K
=−τ(unhdiv qhn, (uhn− unh−1))K− τ(qnh∇unh, (unh− unh−1))K− (vhn− vnh−1, u n h− unh−1)K ≤ Cτ2 +1 6u n h− unh−1 2 K+ Cτ2+ 1 6u n h− unh−1 2 K+ Cτ2+ 1 6u n h− unh−1 2 K,
by whichwecanwrite 1 2 k n=1 un h− un−1h 2 K+ 1 2τ∇u k h2K+ 1 2τ k n=1 ∇un h− ∇un−1h 2 K≤ ∇u0h2τ + Cτ,
andobtainthesecondpartof(26),andthefirstpartof(27).
Wenow set ψh = div[qnh− qhn−1] and takethedifference between theequations writtenat time n and
n− 1.Thanksto(16)andYoungandCauchy–Schwarzinequalities,weobtain τ div[qnh− qn−1h ]2K ≤ Cτun−1h − un−2h 2K+1 4τ div[q n h− qn−1h ] 2 K+ Dτvn−1h − vn−2h 2 K +1 4τ div[q n h− qhn−1] 2 K. (36)
Estimate(33)isthusobtainedbyusing(26),summingovern= 1,. . . ,k,forak≤ N,andsettingu−1h = u0 h,
vh−1= v0 h.
Weproceed bytakingτh= qnh− qn−1h andρh= λnh inthefirstand thethirdequationsof system(25),
respectively,andconsideringthedifferencebetweentheequationswrittenattimesn and(n− 1),obtaining that τ μ k(vhn−2) [qhn− qhn−1], [qhn− qhn−1] K =−τ μ k(vn−1h )− μ k(vhn−2) qnh, [qhn− qhn−1] K + τ (pnh− pnh−1, div[qhn− qnh−1])K. (37)
Weestimatethefirsttermontheright handsideusingthefactthatqn
h∈ [L∞(K)]2 andthatthefunction
[k(vh)]−1 isLipschitzcontinuous.TheapplicationofYounginequalityand(26)gives
τ μ k(vnh−1) − μ k(vhn−2) qnh, [qhn− qhn−1] K ≤ 1 2τq n h− qn−1h 2+ Cτ3.
Hence, since [k(vhn−2)]−1 is positiveand bounded awayfrom zero,and thanksto (36), we get from (37)
that 1 2τq n h− qhn−1 2 K ≤ Cτ3+ Dτ div[qhn− qhn−1]Kpnh− pn−1h K≤ Cτ3+ Dτ2pnh− pn−1h K.
Summingovern= 1,. . . ,k, forak≤ N,itisnowpossible toshowthat τ k n=1 qn h− qn−1h 2 K ≤ Cτ2+ Dτ2 k n=1 pn h− pn−1h K. (38)
Now, wesetτh= ΠK∇ηhinthefirstequationofsystem(25), where
− div ∇ηh= pnh− pn−1h in K,
∇ηh· n|∂K = λnh− λn−1h
and takethedifferencebetweentheequationswrittenattimesn and(n− 1), obtainingthat pn h− pnh−1 2 K+λnh− λnh−1 2 K = μ k(vhn−1) − μ k(vhn−2) qhn, ΠK∇ηh K + μ k(vhn−2) [qhn− qhn−1], ΠK∇ηh K . (39)
The firsttermintherighthandsidecanbe boundedusing atrilinearHölder inequality(or usingthefact thatqnh ∈ [L∞(K)]2),notingthat,atthediscretelevel,∇qhnK =div qhnK,sinceqnh∈ RT0(K).Hence
μ k(vhn−1) − μ k(vnh−2) qhn, ΠK∇ηh K ≤ CτΠK∇ηh[H1(K)]2qnh[H1(K)]2 ≤1 4p n h− pnh−12K+ Cτ2+ Dτ2 div qnh2K.
The second term in the right hand side can be bounded using the Cauchy–Schwarz inequality and the boundedness ofthepermeability,
μ k(vhn−2) [qnh− qnh−1], ΠK∇ηh K ≤ 1 4p n h− p n−1 h 2K+ Cqhn− q n−1 h 2K. (40)
Thanksto equation(32),wemaywrite 1 2 k n=1 pn h− pn−1h 2 K+ k n=1 λn h− λn−1h 2 ∂K≤ Cτ + D k n=1 qn h− qn−1h 2 K, (41)
and, bysubstituting(38)into(41),weget
k n=1 pn h− pn−1h 2 K ≤ Cτ + Dτ k n=1 pn h− pn−1h K. (42)
This inequalitycanberefinedstartingfrom thefollowingidentity: k n=1 pn h− pn−1h K 2 + k n=1 k m>n ([pnh− pn−1h K− pmh − pm−1h K])2= k k n=1 pn h− pn−1h 2K,
which, substitutedinto (42),gives k n=1 pn h− pnh−1K 2 ≤C τ k n=1 pn h− pnh−12K≤ C + D k n=1 pn h− pnh−1K. (43)
Thisquadraticinequalityimpliesthat k n=1 pn h− pn−1h K ≤ C. (44)
Using(44)in(38),wegetestimate(30),whileusing (30)in(41), wegetestimate(31). 2 Wenow associate to the sequence ofdiscrete solutions(qn
h,pnh,λnh,uhn,vhn) ofProblem P2h the following
timecontinuousapproximation:
Qτh(t) := qnht− tn−1 τ + q n−1 h tn− t τ , P τ h(t) = pnh t− tn−1 τ + p n−1 h tn− t τ , Λτh(t) = λnht− tn−1 τ + λ n−1 h tn− t τ , Uhτ(t) := unht− tn−1 τ + u n−1 h tn− t τ , V τ h(t) := vnh t− tn−1 τ + v n−1 h tn− t τ , (45)
fort∈ [tn−1,tn],n= 1,· · · ,N .Theyareafamilyoflineartimeinterpolantsthatdependontheparameters
h andτ .
To simplify the notation we introduce the following: (f,g)T A =
T
0 (f (t),g(t))Adt for a given L2(A)
product (f,g)A, omitting A if A = Ω. It effectively indicates the L2((0,T ),L2(A)) product. While
(f,g)tnA =tn−1tn (f (t),g(t))Adt isusedto indicatetheL2((tn−1,tn),L2(A)) product.
Weconsider system(25),bymultiplyingitbyaC1([0,T ]) functionwhichiszeroatT andintegratingin
timefrom 0 toT ,to obtainthat(Qτ
h,Phτ,Λτh,Uhτ,Vhτ) satisfiesthefollowing weakformulation:
For any τ ∈ L2((0,T ); K∈ThHdiv(K)), ψ ∈ L2((0,T ); K∈ThL2(K)), ρ ∈ L2((0,T ); K∈ThH1/2(∂K)),ω∈ L2((0,T );H01(Ω)), θ∈ L2((0,T );L2(Ω)),givenτh= ΠKτ , ψh= Pk,0ψ,ρh= ρKρ, ωh= Pk,1ω,θh= Pk,1θ, ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (k(Vμτ h)Q τ h, τ )TK− (Phτ, div τ )TK+ (Λτh, τ · n)T∂K+ (pD, τ · n)T∂K∩ΓD =Nn=1 (k(Vμτ h)Q τ h, [τ− τh])tnK + ([ μ k(Vτ h)− μ k(vn h)]Q τ h, τh)tnK + ( μ k(Vτ h)[Q τ h− qnh], τh)tnK − ([ μ k(Vτ h)− μ k(vn h)][Q τ h− qhn], τh)tnK − (Phτ, div[τ − τh])tnK − ([Phτ− pnh], div τh)tnK +Nn=1(Λτ h, [τ− τh]· n)tn∂K+Nn=1([Λτh− λnh], τh· n)tn∂K+Nn=1(pD, [τ− τh]· n)tn∂K∩ΓD , (div Qτ h, ψ)TK− (∂tVhτ, ψ)TK = N n=1 (div Qτ h, [ψ− ψh])tnK + (div[Qhτ− qhn], ψh)tnK − (∂tVhτ, [ψ− ψh])tnK , (Qτ h· n, ρ)T∂K = N n=1 (Qτ h· n, [ρ − ρh])tn∂K+ tn tn−1([Q τ h− qhn]· n, ρh)tn∂K , (∂tUhτ, ω)T + (∇Uhτ,∇ω)T− (QτhUhτ,∇ω)T + (∂tVhτ, ω)T =Nn=1 (∂tUhτ, [ω− ωh])tn+ tn tn−1(∂tV τ h, [ω− ωh])tn+ tn tn−1(∇U τ h, [∇ω − ∇ωh])tn + ([∇Uτ h− ∇unh],∇ωh)tn− (QτhUhτ, [∇ω − ∇ωh])tn− ([Qτh− qnh]Uhτ,∇ωh)tn − (Qτ h[Uhτ− unh],∇ωh)tn+ ([Qhτ− qnh][Uhτ− unh],∇ωh)tn , (∂tVhτ, θ)T ∈ (r(Uhτ)− H(Vhτ), θ)T + N n=1 (∂tVhτ, [θ− θh])tn− (r(Uhτ), [θ− θh])tn − ([r(Uτ h)− r(u n−1 h )], θh)tn+ (H(Vhτ), [θ− θh])tn+ ([H(Vhτ)− H(v n−1 h )], θh)tn . (46)
Note that, since Wh ∈ ker Ch, the terms in the equations for species transport that correspond to the
hybrid inter-element variables canbe eliminated, and the equations for (Uτ
h,Vhτ) correspond to aprimal
formulationeventuallyconvergingtothecontinuousweakformulation(11)ofProblemP2.Theequationfor
(Qτ
h,Phτ,Λτh) is a hybridformulation oneachelement K, whicheventuallyconverges toacontinuous dual
mixedhybridformulation,whichisequivalentto thedualmixedformulation(11)ofProblemP2, sincethe
continuous and the discrete quadratic forms are continuous and coercive over ker B and ker Bh, and the
continuous andthediscreteinf–supconditionsaresatisfied.
In order to passto thelimit in(46), for h,τ → 0, andidentify thesystemsatisfied bythelimit points we needthefollowing results.
Lemma 5.The continuousinterpolantssatisfy
Qτh∈ L2 (0, T ); K∈Th Hdiv(K) ∩ L∞(0, T ); K∈Th L2(K) , (47) Phτ∈ L∞ (0, T ); K∈Th L2(K) , (48) Λτh∈ L∞ (0, T ); K∈Th L2(∂K) , (49) Uhτ∈ L2((0, T ); H01(Ω))∩ L∞((0, T ); L2(Ω)), (50) ∂tUhτ ∈ L2((0, T ); L2(Ω)), (51) Vhτ ∈ L∞((0, T ); L2(Ω)), (52) ∂tVhτ ∈ L2((0, T ); L2(Ω)). (53)
Proof. Considertheequation ∇Uτ h2=∇un−1h +∇[u n h− un−1h ] t− tn−1 τ 2≤ 2∇un−1 h 2+ 2(t− tn−1)2 τ2 ∇[u n h− un−1h ] 2.
Wehave,byintegratingintimeandusingestimate(27),that
T 0 ∇Uτ h2dt≤ 2 N n=1 tn tn−1 ∇un−1h 2+ 2 N n=1 tn tn−1 (t− tn−1)2 τ2 ∇[u n h− un−1h ] 2 ≤ 2τ N n=1 ∇un−1 h 2+ 2 3τ N n=1 ∇[un h− un−1h ]2≤ C.
From this estimatewe obtain(50) and(52). For whatconcernsthe estimates onthe timederivatives, we havethat T 0 ∂tVhτ2dt = N n=1 tn tn−1 vhn− v n−1 h τ 2dt≤ N n=1 τv n h− v n−1 h τ 2≤ C,
thanksto(26).Anestimatefor∂tUhτΩ2T isprovedsimilarlyandweobtain(51)and(53).Weconsidernow
theexpression div Qτ h2K= div Q n−1 h + div[Qnh− Q n−1 h ] t− tn−1 τ 2 K ≤ 2 div Qn−1h 2 K+ 2 (t− tn−1)2 τ2 div[Q n h− Qn−1h ] 2 K,
whichintegratedintime,thanksto(32)and(33),provides T 0 div Qτ h2Kdt≤ C.
Fromthis estimateand(28)and(29)we obtain(47),(48)and(49). 2 Wearenowintheposition ofderivingthefollowingconvergenceresult.
Lemma6(Convergenceresults).Thereexistsasubsequenceof continuousinterpolantsthat,for(h,τ )→ 0, satisfy Uhτ u in L2((0, T ); H01(Ω)), ∂tUhτ ∂tu in L2((0, T ); H−1(Ω)), Vhτ v in L2((0, T ); L2(Ω)), ∂tVhτ ∂tv in L2((0, T ); L2(Ω)), Qτh q in L2 (0, T ); K∈Th Hdiv(K) , Phτ p in L2 (0, T ); K∈Th L2(K) , Λτh λ in L2 (0, T ); K∈Th H1/2(∂K) → L2 (0, T ); K∈Th L2(∂K) . While, Uhτ → u in Lq(0, T ; L2(Ω)),∀q ≥ 1.
Proof. The first set of results derive from (47)–(53), by the application of the Banach–Alaoglu theo-rem[24].Inparticular,theconvergenceresultforthesequenceΛτ
h isgivenbyitsboundednessinthespace
L2((0,T );
K∈ThH1/2(∂K)), which is continuously embedded in L2((0,T );K∈ThL2(∂K)). The
bound-ednessinL2((0,T );
K∈ThH1/2(∂K)) can beobtained byextendingΛτh from theedge toK by meansof
thelifting
− div ∇η
h= Phτ in K,
ηh|∂K = Λτh,
(54) usingmoreover(29)andstandardresultsonellipticregularity, tracetheoremsandinterpolationspaces[1]. The last result is obtained thanks to compactness embedding, from the application of the method of the Hilbertian triad [24] and from the Lebesgue dominated convergence theorem. Indeed, thanks to (50) and (51) the set Uτ
h is relatively compact inL2(0,T ;L2(Ω)) andthere exists asubsequence of Uhτ which
convergesto thelimitpointu in Lq(0,T ;L2(Ω))∀q ≥ 1. 2 Thestrong convergenceofUτ
h inL2(0,T ;L2(Ω)) makesit possible topassto thelimitinthe nonlinear
termr(Uτ
h) ofequation(46).
Note that the family of functions Vτ
h is only weakly convergent to a limit point in L2(0,T ;L2(Ω)).
This is not a problem when passing to the limit interms like 0T(H(Vτ h),θ) →
T
the propertiesof the multivaluedmap H. Namely,the set H(Vτ
h) is bounded and convex, so itis weakly
closed. Hence it is weakly compact, and admits a weakly convergent subsequence: lim(h,τ )(H(Vhτ),θ) →
(H(v),θ). Moreover, sinceVhτ isweaklyconvergent inL2(0,T ;L2(Ω)) andusing theupper semicontinuity and themaximal monotonicitypropertyof themultivaluedmap H,wehavethatlim(h,τ )
T
0 (H(Vhτ),θ)=
T
0 lim(h,τ )(H(V τ
h),θ). This would be sufficient if we were solving the problem with a given Darcy flux:
solvingdirectlytheinclusionproblem withoutregularization avoidsthenecessityofstrong convergenceof Vτ
h inL2(0,T ;L2(Ω)). Since howeverwe areconsidering thecoupling withaDarcy field, inorder to pass
to thelimitintermscontainingthenon-linear permeabilityfactor[k(Vhτ)]−1 we needstrongconvergence.
Weobtainitinthefollowing lemma. Lemma 7.Vhτ → v inL2(0,T ;L2(Ω)).
Proof. The proof is the same as that introduced in [15] in Proposition 4.6 and Lemma 4.7. The only difference here is that, instead of a Lipschitz continuous regularization of the dissolution rate, we have a semicontinuous rate, satisfying the selection (6). Here we only report the generalization of the result proved inProposition 4.6to thisparticular case.Remembering thatvh ∈ Wh and using (6), theresultin Proposition 4.6 becomes
∇PK1 H(vh)K ≤ Lr∇uhK ,
where Lr istheLipschitzconstantofthefunctionr(·).Wereferto[15]forthedetailsoftheproof. 2 Wefinallyinvestigatethelimitequationsofsystem(46) for(τ,h)→ 0.
Theorem4.Thelimitpoint(q,p,λ,u,v) istheweaksolutionofa hybrid formulationoftheweakProblem P2, whichisequivalenttotheweaksolutionofProblemP2.
Proof. Let usstartfrom considering thefirstequationof system(46). Thelefthandsideconverges tothe limit T 0 μ k(v) q, τ K dt− T 0 (p, div τ )Kdt + T 0 (λ, τ · n)∂K+ T 0 (pD, τ · n)∂K∩ΓD.
For all butthe first termthis isa directconsequence of theconvergence results (47), (48)and (49). The firsttermcanberewrittenas
T 0 μ k(Vhτ) Qτh, τ K dt = T 0 μ k(v) Qτh, τ K dt + T 0 μ k(Vhτ)− μ k(v) Qτh, τ K dt. Since Vτ
h → v strongly and is inL∞(Ω), and since Qτh is weaklyconvergent inL2(0,T ;L2(Ω)), the first
termontherighthandsideconvergestothedesiredlimit.Choosingatestfunctionτ ∈ L2(0,T ;[C∞ 0 (Ω)]2),
we canshowthatthesecondtermontherighthandside iszerobyboundingitusing theestimate
T 0 μ k(Vhτ)− μ k(v) Qτh, τ K dt≤ μ k(Vhτ)− μ k(v) Qτh [L1(Ω)]2 ≤ CVτ h − v[L2(Ω)]2QτhL2(Ω)→ 0,
wherewehaveusedtheLipschitzcontinuityofthefunction[k(·)]−1 andthefactthatVτ → v intheL2(Ω) h
norm.Hence,thelefthandsideofthefirstequationofsystem(46)convergesinthedistributionalsenseto thecontinuoushybridformulation,extendedtoΩ,ofthefirstequationofProblemP2.
Wenowshowthatthetermsintherighthandsideofthefirstequation ofsystem(46)convergeto zero for(h,τ )→ 0.LetusdenotethesetermsbyI1,· · · ,I9.Consideringestimate(47)andthefirstinterpolation
estimateof(24),weconcludethat |I1| ≤ C T 0 Qτ h2Kdt 1/2N n=1 tn tn−1 τ − τh2Kdt 1/2 ≤ Chτ L2(0,T ; K∈Th[H1(K)]2)→ 0.
Consideringthefirst estimatein(26),the estimate(47),the Lipschitzcontinuityproperty of thefunction [k(·)]−1,recallingthatdiv QτhK =∇QτhK anddiv τhK =∇τhK,andapplyingatrilinearHölder
inequality,weobtain |I2| ≤ C N n=1 tn tn−1 vn h− vhn−1 2 K div Qτh2Kdt 1/2N n=1 tn tn−1 ∇τh2Kdt 1/2 ≤ C N n=1 τ3 div Qτh2K 1/2N n=1 τ∇τh2K 1/2 → 0. While,byconsideringestimate(30),weget
|I3| ≤ C N n=1 τqhn− qhn−12K 1/2N n=1 ττh2K 1/2 → 0.
Thankstothefirst estimateinequation (26), estimate(33),andbyapplying atrilinearHölder inequality, weareabletowrite
|I4| ≤ C N n=1 tn tn−1 vn h− vhn−1 2 K div[qhn− qhn−1] 2 Kdt 1/2N n=1 tn tn−1 ∇τh2Kdt 1/2 ≤ C N n=1 τ3 div[qhn− qhn−1]2K 1/2N n=1 τ∇τh2K 1/2 → 0. Considering(29)and thesecond interpolationestimatein(24),we get
|I5| ≤ C T 0 Pτ h2Kdt 1/2N n=1 tn tn−1 div[τ − τh]2Kdt 1/2 ≤ Chτ L2(0,T ; K∈Th[H2(K)]2)→ 0.
Note that for this estimate we need to restrict the test functions to be in L2(0,T ;K∈Th[H2(K)]2). Density arguments ensure that the limit points satisfy the continuous weak formulation also for τ ∈
L2(0,T ;K∈Th[H1(K)]2).Thankstothefirstestimatein(31),we have
|I6| ≤ C N n=1 tn tn−1 pn h− pnh−1 2 Kdt 1/2N n=1 tn tn−1 div τh2Kdt 1/2 ≤ C N n=1 τpnh− pnh−12K 1/2N n=1 τ div τh2K 1/2 → 0.
