The obstacle problem for the porous medium equation

The obstacle problem for the porous medium

equation

V. Bogelein, T. Lukkari and C. Scheven

REPORT No. 33, 2013/2014, fall

ISSN 1103-467XISRN IML-R- -33-13/14- -SE+fall

THE OBSTACLE PROBLEM FOR THE POROUS MEDIUM EQUATION

VERENA BOGELEIN, TEEMU LUKKARI, AND CHRISTOPH SCHEVEN

ABSTRACT. We prove existence results for the obstacle problem related to the porousmedium equation. For sufficiently regular obstacles, we find continuous solutions whosetime derivative belongs to the dual of a parabolic Sobolev space. We also employ the notionof weak solutions and show that for more general obstacles, such a weak solution exists.The latter result is a consequence of a stability property of weak solutions with respect tothe obstacle.

1. INTRODUCTION

The porous medium equation (PME for short)

(1.1) ∂tu−∆um = 0

is an important prototype example of a nonlinear parabolic equation. The name stems frommodeling the flow of a gas in a porous medium: a combination of the continuity equation,Darcy’s law, and an equation of state lead to equation (1.1) for the density of the gas, afterscaling out various physical constants. There is an extensive literature concerned with thisequation, and we refer to the monographs [5, 7, 21, 22] for the basic theory and numerousfurther references in a variety of directions.

In this work, we are concerned with the so-called obstacle problem related to this equa-tion. Roughly speaking, in the obstacle problem we want to find solutions to the porousmedium equation subject to the constraint that they lie above a given function ψ, the ob-stacle. This leads to a variational inequality; formally, a function u solves the obstacleproblem for the porous medium equation if u ≥ ψ and

(1.2)∫

ΩT

∂tu(vm − um) +∇um · (∇vm −∇um) dz ≥ 0

for all comparison maps v such that v ≥ ψ, and with the same boundary values as u; seeDefinitions 2.1 and 2.2 below for the rigorous interpretation of this inequality. The classicreferences for the obstacle problem to parabolic equations include [1] and the monograph[18], and some of the more recent ones are [2, 3, 20]. Alternatively, the obstacle problemcan be thought of as finding the smallest supersolution to an equation staying above theobstacle function. This approach is analogous to the balayage concept in classical potentialtheory, and it is used in a nonlinear parabolic context in [15, 17]. One of our motivations isthe need for a method of constructing supersolutions with favorable properties in nonlinearpotential theory, see e.g. [11, 12, 16]. A recent example of this for the PME can be foundin [14]. Our main interest is in the degenerate case, meaning that one has m > 1 in (1.1).Since the arguments also work – with occasional minor modifications – in the supercriticalsingular range (n− 2)+/n < m < 1, we included this case as well.

Our main results concern the existence of solutions to the obstacle problem for theporous medium equation. The first of them is the existence of strong solutions. Herestrong refers to the fact that the time derivative ∂tu of a solution u belongs to the dual ofa parabolic Sobolev space. This makes (1.2) meaningful, since we can then use the usual

Date: March 31, 2014.1991 Mathematics Subject Classification. Primary 35K65, Secondary 35K86, 35D05, 47J20.Key words and phrases. obstacle problem, porous medium equation, existence.

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2 V. BOGELEIN, T. LUKKARI, AND C. SCHEVEN

duality pairing. Apart from the time regularity, the strong solutions we find are locallyHolder continous. The continuity comes out as a byproduct of the existence proof. As faras we know, no previous existence result for the obstacle problem related to (1.1) providessolutions with these regularity properties. To make the inequality (1.2) meaningful in moregeneral situations, we use a similar notion of a weak solution to the obstacle problem asin Alt & Luckhaus [1]. Our second main result is the existence of weak solutions. Herethe regularity assumptions on the obstacles can be considerably relaxed, but the controlon the time derivative and the continuity of the solution are lost. In both cases, only mildregularity assumptions are made on the obstacles; in particular, contrary to [1, 18], noassumption about boundedness and monotonicity in time is needed.

We use a penalization method to prove the existence of strong solutions. The heuristicidea of our choice of penalization is as in [2]: roughly speaking, we solve the equation

(1.3) ∂tu−∆um = (∂tψ −∆ψm)+

for a given obstacle function ψ, where the positive part (f)+ := maxf, 0 is taken onthe right-hand side. In order to properly define the right-hand side, we certainly needsome regularity assumptions on the obstacle ψ. We merge the penalization method withthe usual way of constructing solutions to the porous medium equation via approximationby uniformly parabolic equations. The latter is described for instance in Chapter 5 of [21].Comparison results using ideas from [4] play a key role in proving that the solutions indeedstay above the obstacle function. Note that there are several ways to choose a penalization;for instance, one could use projection operators as in [1, 18]. However, our choice has theadvantage that we can obtain uniform estimates for the penalized equations.

The generalization of the existence result to weak solutions and irregular obstacles is aconsequence of two facts. The first of them is the natural observation that strong solutionsare also weak solutions. The second step is the more difficult one: weak solutions to theobstacle problem turn out to be stable with respect to convergence of the obstacles in aparabolic Sobolev space. In other words, given a sequence of obstacles ψi converging to alimit obstacle ψ, we show that one has convergence of the corresponding solutions ui, upto a subsequence, to a limit function. Further, the limit function turns out to be a solution tothe limiting obstacle problem. Thereby, the difficulty relies in identifying the weak limits ofui and umi . Now the existence of weak solutions follows by choosing a sufficiently smoothapproximation of the given obstacle, ensuring that strong solutions exist, and then applyingthe stability property. Finally, we note that our proof produces local weak solutions, i.e. thevariational inequality can be localized on smaller domains. Such a property can be usefulwhen proving regularity of weak solutions, which we postpone to a subsequent work.

The paper is organized as follows. In § 2, we give the rigorous definitions of strongand weak solutions to the obstacle problem and the statements of our existence results.§ 3 contains various auxiliary results needed for our existence proofs. In § 4, we startworking on the obstacle problem, and establish some basic estimates for later use. § 5deals with continuity in time, and we prove that the weak solutions to the obstacle problemare continuous in time with values in Lm+1. We establish suitable comparison principlesfor the uniformly parabolic approximating equations in § 6, and these are then employed in§ 7 to show the existence of solutions to a penalized porous medium equation. We finallyprove the existence of strong and weak solutions in § 8 and § 9, respectively.

Acknowledgements. We acknowledge the warm hospitality of the Institut Mittag-Leffler inthe Fall 2013 during the program “Evolutionary problems”, during which we initiated ourcollaboration on this paper.





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THE OBSTACLE PROBLEM FOR THE POROUS MEDIUM EQUATION 3

2. WEAK AND STRONG SOLUTIONS TO THE OBSTACLE PROBLEM

In this section, we define our notions of solution to the obstacle problem and state themain results. We begin by introducing some notation, and recalling several other defini-tions.

Throughout the paper we always assume that m > mc := (n−2)+

n and that Ω is abounded open subset of Rn such that Rn \ Ω is uniformly 2-thick; see Definition 2.4below. We use the notation ΩT = Ω × (0, T ) and Ut1,t2 = U × (t1, t2), where U ⊂ Ωis open and 0 < t1 < t2 < T . The parabolic boundary ∂pUt1,t2 of a space-time cylinderUt1,t2 consists of the initial and lateral boundaries, i.e.

∂pUt1,t2 = (U × t1) ∪ (∂U × [t1, t2]).

The notation Ut1,t2 b ΩT means that the closure Ut1,t2 is compact and Ut1,t2 ⊂ ΩT .We useH1(Ω) to denote the usual Sobolev space, i.e. the space of functions u in L2(Ω)

such that the weak gradient exists and also belongs to L2(Ω). The norm of H1(Ω) is

‖u‖H1(Ω) = ‖u‖L2(Ω) + ‖∇u‖L2(Ω).

The Sobolev space with zero boundary values, denoted by H10 (Ω), is the completion of

C∞0 (Ω) with respect to the norm of H1(Ω). The dual of H10 (Ω) is denoted by H−1(Ω)

and 〈·, ·〉 will indicate the related duality pairing between H−1(Ω) and H10 (Ω). We note

that Lm+1(Ω) → H−1(Ω), since m > mc.The parabolic Sobolev space L2(0, T ;H1(Ω)) consists of all measurable functions

u : [0, T ]→ H1(Ω) such that∫

ΩT

|u|2 + |∇u|2 dz <∞.

The definition ofL2(0, T ;H10 (Ω)) is identical, apart from the requirement that u : [0, T ]→

H10 (Ω). We say that u belongs to L2

loc(0, T ;H1loc(Ω)) if u ∈ L2(t1, t2;H1(U)) for all

Ut1,t2 b ΩT . Let u ∈ L∞(0, T ;Lm+1(Ω)). By u ∈ H1(0, T ;H−1(Ω)) we mean thatthere exists ut ∈ L2(0, T ;H−1(Ω)) such that

∫ T

0

u(·, t)ϕt(t) dt = −∫ T

0

ut(·, t)ϕ(t) dt, for all ϕ ∈ C∞0 (0, T ).

The previous equality makes sense due to the inclusion Lm+1(Ω) → H−1(Ω) whichallows us to identify u as an element of L2(0, T ;H−1(Ω)). Since ut ∈ L2(0, T ;H−1(Ω))implies that u ∈ C0([0, T ];H−1(Ω)), it is clear what we mean by saying u(·, 0) = uo ∈H−1(Ω) in the H−1-sense.

We consider obstacle functions ψ : ΩT → R≥0 with

(2.1) ψm ∈ L2(0, T ;H1(Ω)

), ∂t(ψ

m) ∈ Lm+1m (ΩT ) and ψm(·, 0) ∈ H1(Ω).

Note that in particular, this implies ψ ∈ C0([0, T ];Lm+1(Ω)). The class of admissiblefunctions is defined by

Kψ(ΩT ) :=v ∈ C0

([0, T ];Lm+1(Ω)

): vm ∈ L2

(0, T ;H1(Ω)

), v ≥ ψ a.e. on ΩT

.

Furthermore, the class of admissible comparison functions will be denoted by

K ′ψ(ΩT ) :=v ∈ Kψ(ΩT ) : ∂t(v

m) ∈ Lm+1m (ΩT )

.

Note that ψ ∈ Kψ and ψ ∈ K ′ψ and therefore Kψ,K′ψ 6= ∅. As the initial data, we take a

function uo so that

(2.2) umo ∈ H1(Ω) with uo ≥ ψ(·, 0) a.e. on Ω.

In order to introduce the notion of a weak solution to the obstacle problem, wehave to attribute a meaning to the term containing the time derivative even if we donot know that ∂tu exists in some sense. Therefore, following Alt & Luckhaus [1], for



u ∈ L2m(ΩT ) ∩ C0([0, T ];Lm+1(Ω)) and v ∈ L2m(ΩT ) with ∂t(vm) ∈ Lm+1m (ΩT ) and

α ∈W 1,∞([0, T ]) with α(T ) = 0 and η ∈ L∞(Ω) we define

〈〈∂tu, αη(vm − um)〉〉uo :=

∫

ΩT

η[α′[

1m+1u

m+1 − uvm]− αu∂tvm

]dz

+ α(0)

∫

Ω

η[

1m+1u

m+1o − uovm(·, 0)

]dx,

where uo is the given initial datum. Here, we note that the assumptions on v imply thatv ∈ C0([0, T ];Lm+1(Ω)), so that all integrals are well defined. Moreover, since m > mc,assumption (2.2) implies that uo ∈ Lm+1(Ω). Now, we can define what we mean by localweak and strong solution to the obstacle problem.

Definition 2.1. A nonnegative function u ∈ Kψ(ΩT ) is a local strong solution to theobstacle problem for the porous medium equation if and only if ∂tu ∈ L2(0, T ;H−1(Ω))and

(2.3)∫ T

0

〈∂tu, αη(vm − um)〉dt+

∫

ΩT

α∇um · ∇(η(vm − um)

)dz ≥ 0

holds for all comparison maps v ∈ Kψ(ΩT ), every cut-off function in time α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0 and and every cut-off function in space η ∈C1

0 (Ω,R≥0).A nonnegative function u ∈ Kψ(ΩT ) is a local weak solution to the obstacle problem

for the porous medium equation if and only if

(2.4) 〈〈∂tu, αη(vm − um)〉〉uo +

∫

ΩT

α∇um · ∇(η(vm − um)

)dz ≥ 0

holds true for all comparison maps v ∈ K ′ψ(ΩT ), every cut-off function in time α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0 and every cut-off function in space η ∈ C1

0 (Ω,R≥0).

We note that if u is a local strong solution with u(·, 0) = uo in the H−1(Ω)-sense, thenit is also a local weak solution; see Lemma 3.2. Further, the variational inequality (2.4)implies that u(·, 0) = uo; see Lemma 5.2.

The preceding definition is local in the sense that there are no prescribed Dirichletboundary data on the lateral boundary ∂Ω × (0, T ). However, our aim in this paper isto prove existence of weak solutions to the obstacle problem with prescribed initial andDirichlet boundary data. Therefore, we consider Dirichlet boundary data

(2.5) g ∈ K ′ψ(ΩT ) and g(·, 0) = uo.

Then, the class of admissible functions subject to the Dirichlet boundary data is given by

Kψ,g(ΩT ) :=v ∈ Kψ(ΩT ) : vm − gm ∈ L2

(0, T ;H1

0 (Ω))

and the class of admissible comparison functions subject to the Dirichlet boundary datawill be denoted by

K ′ψ,g(ΩT ) :=v ∈ K ′ψ(ΩT ) : vm − gm ∈ L2

(0, T ;H1

0 (Ω)).

Note that g ∈ Kψ,g and g ∈ K ′ψ,g(ΩT ) and therefore Kψ,g(ΩT ),K ′ψ,g(ΩT ) 6= ∅. Now,we can define what we mean by a weak solution to the obstacle problem for the porousmedium equation with prescribed boundary data.

Definition 2.2. A nonnegative function u ∈ Kψ,g(ΩT ) is a strong solution to the obstacleproblem for the porous medium equation if and only if ∂tu ∈ L2(0, T ;H−1(Ω)) and

(2.6)∫ T

0

〈∂tu, α(vm − um)〉dt+

∫

ΩT

α∇um · ∇(vm − um) dz ≥ 0

holds for all comparison maps v ∈ Kψ,g(ΩT ) and every cut-off function α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0.


A nonnegative function u ∈ Kψ,g(ΩT ) is a weak solution to the obstacle problem forthe porous medium equation if and only if

(2.7) 〈〈∂tu, α(vm − um)〉〉uo +

∫

ΩT

α∇um · ∇(vm − um) dz ≥ 0

holds for all comparison maps v ∈ K ′ψ,g(ΩT ) and every cut-off function α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0.

Remark 2.3. In the case m > 1, the assumptions ∂tψm, ∂tgm ∈ Lm+1m (ΩT ) could be

replaced by the weaker assumptions ∂tψm, ∂tgm ∈ L2m

2m−1 (ΩT ). This can be achievedby slight modifications in the proof, in particular by different applications of Holder’sinequality. However, in order to keep the exposition as simple as possible and to havea unified proof for both cases m > 1 and m < 1, we will work with the assumptions∂tψ

m, ∂tgm ∈ Lm+1

m (ΩT ) in both cases.

We note that under a mild regularity assumption on the domain Ω, any local weaksolution u ∈ Kψ(ΩT ) in the sense of Definition 2.1 which additionally satisfies um−gm ∈L2(0, T ;H1

0 (Ω)) is a weak solution in the sense of Definition 2.2; see Lemma 3.5. Theregularity assumption on the domain is made precise in the following

Definition 2.4. A set E ⊂ Rn is uniformly p-thick if there exist constants γ, %o > 0 suchthat

capp(E ∩B%(x), B2%(x)

)≥ γ capp

(B%(x), B2%(x)

),

for all x ∈ E and for all 0 < % < %o. 2

Here, capp denotes the usual variational p-capacity, cf. [11]. To require that Rn \ Ω isuniformly p-thick is not very restrictive, since all Lipschitz domains or domains satisfyingan exterior cone condition are uniformly p-thick for every p ∈ (1,∞). If p > n, then thecomplement of every open set Ω ( Rn is uniformly p-thick.

To show the existence of strong solutions, we need the following stronger assumptionson the data:

(2.8)

ψ, g ∈ L∞(ΩT ,R≥0) satisfy (2.1), (2.5), uo ∈ L∞(Ω,R≥0) satisfies (2.2),

Ψ := ∂tψ −∆ψm ∈ L∞(ΩT ).

Note that these assumptions imply that the obstacle ψ is locally Holder continuous; indeed,the assumption Ψ ∈ L∞(ΩT ) contains the fact that ψ is a solution to the porous mediumequation with a right-hand side given by a bounded function. Now, the Holder continuityof ψ follows from [8].

Solutions to the obstacle problem have a relationship to weak solutions and supersolu-tions. Therefore, we recall what we mean by weak solutions and weak supersolutions tothe porous medium equation.

Definition 2.5. A nonnegative function u ∈ C0([0, T ];Lm+1(Ω)) is a local weak solutionof the porous medium equation

(2.9) ∂tu−∆um = 0 in ΩT ,

if um ∈ L2(0, T ;H1(Ω)) and

(2.10)∫

ΩT

[− u∂tϕ+∇um · ∇ϕ

]dz = −

∫

Ω×tuϕdx

∣∣∣∣T

t=0

for all test functions ϕ ∈ C∞(ΩT ) with ϕ = 0 on ∂Ω × [0, T ]. For weak supersolutions,the requirement is that

(2.11)∫

ΩT

[− u∂tϕ+∇um · ∇ϕ

]dz ≥ 0

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for all positive test functions ϕ ∈ C∞0 (ΩT ). For weak subsolutions, the inequality in (2.11)is reversed.

A nonnegative function u is a weak solution in an open set V ⊂ Rn+1 if it is a weaksolution in the sense defined above in all space-time cylinders Ut1,t2 such that Ut1,t2 ⊂ V .

Now we are in a position to state the main results of the present article. We start withthe existence result for strong solutions.

Theorem 2.6. Let Ω be a bounded open subset of Rn such that Rn \ Ω is uniformly 2-thick and m > mc := (n−2)+

n . Assume that the data ψ, uo, g satisfy the regularity andcompatibility conditions (2.8). Then there exists a local strong solution u to the obstacleproblem for the porous medium equation in the sense of Definition 2.1 satisfying um−gm ∈L2(0, T ;H1

0 (Ω)) and u(·, 0) = uo.The function u is also locally Holder continuous, and satisfies u ≥ ψ everywhere in

ΩT . Further, u is a weak supersolution to the porous medium equation in ΩT , and a weaksolution in the open set z ∈ ΩT : u(z) > ψ(z).

Our second result ensures the existence of weak solutions under the presence of anirregular obstacle.

Theorem 2.7. Let Ω be a bounded open subset of Rn such that Rn \ Ω is uniformly 2-thick and m > mc := (n−2)+

n . Assume that the data ψ, uo, g satisfy the regularity andcompatibility conditions (2.1), (2.2) and (2.5). Then there exists a local weak solutionto the obstacle problem for the porous medium equation in the sense of Definition 2.1satisfying um − gm ∈ L2(0, T ;H1

0 (Ω)). Again, u is also a weak supersolution to theporous medium equation in ΩT .

Thanks to Lemma 3.5 this immediately implies the existence of strong and weak solu-tions to the obstacle problem.

Corollary 2.8. Under the assumptions of Theorems 2.6 and 2.7 there exists a strong and aweak solution, respectively, to the obstacle problem for the porous medium equation in thesense of Definition 2.2.

Remark 2.9. We conjecture that (at least) the strong solutions obtained in Theorem 2.6are unique in a certain sense. To simplify matters, assume for a moment that the obstacleand the boundary values are given by the same function, i.e. g = ψ. Define

U := v : ΩT → R≥0 : v is a lower semicontinuous weak supersolution to the PME,Uψ := v ∈ U : v ≥ ψ.

A function w ∈ Uψ is the smallest supersolution above the obstacle ψ, if

(2.12) v ∈ Uψ implies that w ≤ v in ΩT .

The smallest supersolution is unique, if it exists: if u1, u2 ∈ Uψ both have the property(2.12), then two applications of (2.12) yield that u1 ≤ u2 and u2 ≤ u1, implying thatu1 = u2.

A strong solution u to the obstacle problem with the properties given in Theorem 2.6clearly belongs to the class Uψ . Showing that (2.12) also holds would proceed as follows:in the contact set u = ψ one clearly has u ≤ v for any v ∈ Uψ . To conclude that u ≤ valso outside the contact set, i.e. in V := u > ψ, one would like to apply the comparisonprinciple, since u is a weak solution in V and u ≤ v on the boundary ∂V . However, V isa general open set in Rn+1, not a space-time cylinder. As far as we know, the comparisonprinciple for general open sets in Rn+1 remains open for the porous medium equation.


3. PRELIMINARIES

3.1. Mollification in time. The above definitions of weak solution to the obstacle problemdo not include a time derivative for u. Nevertheless, we would like to use test functionsinvolving the solution u itself, and the quantity ∂u

∂t inevitably appears. This situation isusually resolved by employing a mollification procedure in the time direction. The molli-fication

(3.1) [[v]]h(x, t) := e−th vo +

1

h

∫ t

0

es−th v(x, s) ds

with some vo ∈ L1(Ω) has turned out to be convenient for dealing with the porous mediumequation. The aim is to obtain estimates independent of the time derivative of [[u]]h, andthen pass to the limit h ↓ 0. The basic properties of the mollification (3.1) are given in thefollowing lemma, see [19].

Lemma 3.1. Let p ≥ 1.(i) If v ∈ Lp(ΩT ) and vo ∈ Lp(Ω), then

‖[[v]]h‖Lp(ΩT ) ≤ ‖v‖Lp(ΩT ) + c(p)h1p ‖vo‖Lp(Ω),

[[v]]h → v in Lp(ΩT ) as h ↓ 0 and

∂t[[v]]h = 1h (v − [[v]]h).

(ii) If∇v ∈ Lp(ΩT ) and∇vo ∈ Lp(Ω), then∇[[v]]h = [[∇v]]h,

‖∇[[v]]h‖Lp(ΩT ) ≤ ‖∇v‖Lp(ΩT ) + c(p)h1p ‖∇vo‖Lp(Ω),

and∇[[v]]h → ∇v in Lp(ΩT ) as h ↓ 0.(iii) If v ∈ L∞(0, T ;Lp(Ω)) and vo ∈ Lp(Ω), then [[v]]h ∈ C0([0, T ];Lp(Ω)) with

‖[[v]]h‖L∞(0,T ;Lp(Ω)) ≤ ‖v‖L∞(0,T ;Lp(Ω)) + ‖vo‖Lp(Ω).

Moreover, there holds [[v]]h → v in Lp(ΩT ) as h ↓ 0 and [[v]]h(·, 0) = vo.(iv) If vk → v in Lp(ΩT ), then also

[[vk]]h → [[v]]h and ∂t[[vk]]h → ∂t[[v]]h in Lp(ΩT ).

(v) If ∇vk → ∇v in Lp(ΩT ), then∇[[vk]]h → ∇[[v]]h in Lp(ΩT ).(vi) If vk v, (or ∇vk ∇v) weakly in Lp(ΩT ), then [[vk]]h [[v]]h, (or∇[[vk]]h ∇[[v]]h) weakly in Lp(ΩT ).

(vii) If v, ∂tv ∈ L2(0, T ;H−1(Ω)) and v(·, 0) = vo in the H−1(Ω)-sense, then∂t[[v]]h ∂tv weakly in L2(0, T ;H−1(Ω)).

(viii) If ϕ ∈ C(ΩT ), then[[ϕ]]h(x, t)→ ϕ(x, t)

uniformly in ΩT as h ↓ 0.

The next lemma ensures that the weak formulation of the variational inequality in (2.4),respectively (2.7) coincides with the strong form (see Theorem 2.6), if the time derivative∂tu of the solution belongs to L2(0, T ;H−1(Ω)).

Lemma 3.2. Let u ∈ L∞(0, T ;Lm+1(Ω)) be a non-negative function with um ∈L2(0, T ;H1(Ω)), ∂tu ∈ L2(0, T ;H−1(Ω)) and u(·, 0) = uo in the H−1(Ω)-sense anduo ∈ Lm+1(Ω) ⊂ H−1(Ω). Then, there holds

(3.2)∫ T

0

〈∂tu, αη(vm − um)〉dt = 〈〈∂tu, αη(vm − um)〉〉uo

for any v : ΩT → R≥0 satisfying vm ∈ L2(0, T ;H1(Ω)) and ∂t(vm) ∈ Lm+1m (ΩT ) and

α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0 and η ∈ C10 (Ω,R≥0).

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Proof. We define [[u]]h according to (3.1) with the choice vo = uo and similarly, for thedefinition of [[um]]h, we choose vo = umo . We first re-write

∫ T

0

〈∂t[[u]]h, αη(vm − [[um]]h)〉dt

=

∫

ΩT

αη∂t[[u]]h(vm − [[u]]mh ) dz +

∫

ΩT

αη∂t[[u]]h([[u]]mh − [[um]]h) dz.(3.3)

For the first integral on the right-hand side we compute∫

ΩT

αη∂t[[u]]h(vm − [[u]]mh ) dz

= −∫

ΩT

[α′η[[u]]h(vm − [[u]]mh ) + αη[[u]]h∂t(v

m − [[u]]mh )]

dz

− α(0)

∫

Ω

ηuo(vm(·, 0)− umo ) dx

=

∫

ΩT

[α′η[[[u]]m+1

h − [[u]]hvm]

+ mm+1αη∂t[[u]]m+1

h − αη[[u]]h∂tvm]

dz

+ α(0)

∫

Ω

η[um+1o − uovm(·, 0)

]dx

=

∫

ΩT

[α′η[

1m+1 [[u]]m+1

h − [[u]]hvm]− αη[[u]]h∂tv

m]

dz

+ α(0)

∫

Ω

η[

1m+1u

m+1o − uovm(·, 0)

]dx.

For the integrand of the second integral in (3.3) we have due to Lemma 3.1 (i) that

∂t[[u]]h([[u]]mh − [[um]]h) = ∂t[[u]]h([[u]]mh − um) + ∂t[[u]]h(um − [[um]]h)

= − 1h ([[u]]h − u)([[u]]mh − um) + ∂t[[u]]h(um − [[um]]h)

≤ ∂t[[u]]h(um − [[um]]h).

Altogether, we have shown that∫ T

0


≤∫

ΩT

[α′η[

1m+1 [[u]]m+1


m]

dz

+ α(0)

∫

Ω

η[

1m+1u

m+1o − uovm(·, 0)

]dx

+

∫ T

0

〈∂t[[u]]h, αη(um − [[um]]h)〉dt.

Now, Lemma 3.1 ensures that all appearing terms converge in the limit h ↓ 0. In particular,the last term on the right-hand side disappears in the limit h ↓ 0. This proves that

∫ T

0

〈∂tu, αη(vm − um)〉dt ≤ 〈〈∂tu, αη(vm − um)〉〉uo .

In order to prove the reversed inequality, we assume first that u ≥ ε for ε > 0 if m > 1,and u ≤ k < ∞ if m < 1. We note from (3.1) that [[um]]h ≥ εm and [[um]]h ≤ km in thecases m > 1 and m < 1, respectively. Then we write

∫ T

0


=

∫

ΩT

[αη∂t[[u]]hv

m − αη∂t[[um]]1m

h [[um]]h

]dz


+

∫

ΩT

αη∂t([[um]]

1m

h − [[u]]h)[[um]]h dz.(3.4)

Note that the quantity ∂t[[um]]1m

h does not cause any problems, since

|∂t[[um]]1m

h | = 1m [[um]]

1m−1

h |∂t[[um]]h| ≤

1mε

1−m|∂t[[um]]h|, if m > 1,1mk

1−m|∂t[[um]]h|, if m < 1.

For the first integral on the right-hand side we compute∫

ΩT

[αη∂t[[u]]hv

m − αη∂t[[um]]1m

h [[um]]h

]dz

= −∫

ΩT

[α′η[[u]]hv

m + αη[[u]]h∂tvm + 1

m+1αη∂t[[um]]

m+1m

h

]dz

− α(0)

∫

Ω

ηuovm(·, 0) dx

=

∫

ΩT

[α′η[

1m+1 [[um]]

m+1m


m]

dz

+ α(0)

∫

Ω

η[

1m+1u

m+1o − uovm(·, 0)

]dx.

Moreover, by Lemma 3.1 (i) we have∫

ΩT

αη∂t([[um]]

1m

h − [[u]]h)[[um]]h dz

= −∫

ΩT

[αη([[um]]

1m

h − [[u]]h)∂t[[u

m]]h + α′η([[um]]

1m

h − [[u]]h)[[um]]h

]dz

=1

h

∫

ΩT

αη([[um]]

1m

h − u)(

[[um]]h − um)

+ αη(u− [[u]]h

)([[um]]h − um

)dz

+

∫

ΩT

α′η([[um]]

1m

h − [[u]]h)[[um]]h dz

≥ 1

h

∫

ΩT

αη(u− [[u]]h

)([[um]]h − um

)dz +

∫

ΩT

α′η([[um]]

1m

h − [[u]]h)[[um]]h dz

=

∫

ΩT

αη∂t[[u]]h([[um]]h − um

)dz +

∫

ΩT

α′η([[um]]

1m

h − [[u]]h)[[um]]h dz.

Inserting this above, we conclude that∫ T

0


≥∫

ΩT

[α′η[

1m+1 [[um]]

m+1m


m]

dz

+ α(0)

∫

Ω

η[

1m+1u

m+1o − uovm(·, 0)

]dx

+

∫ T

0

⟨∂t[[u]]h, αη

([[um]]h − um

)⟩dt+

∫

ΩT

α′η([[um]]

1m

h − [[u]]h)[[um]]h dz.

Recall that this holds with the assumptions u ≥ ε > 0 if m > 1, and u ≤ k if m < 1. Toestablish the inequality also for a general function u, we apply the estimate proved so farto the functions uε = (um + εm)

1m and uk = min(u, k), and then appeal to Lemma 3.1

for letting ε ↓ 0 and k →∞, respectively. Here, we note that ∂t[[uε]]h = 1h (uε− [[uε]]h)→

1h (u − [[u]]h) = ∂t[[u]]h in Lm+1(ΩT ) and [[um]]h → um in L

m+1m (ΩT ) in the limit ε ↓ 0

holds, respectively the analogous convergences for k → ∞ if m < 1. Finally, another


application of Lemma 3.1 shows that all integrals converge as h ↓ 0 and in particular thelast two integrals disappear. This proves the asserted identity.

3.2. Hardy’s inequality. Another tool we need is the Hardy inequality; see [9, 10, 13] forthe proof.

Theorem 3.3. Let Ω be an open bounded subset of Rn such that Rn \ Ω is uniformly 2-thick, and u ∈ H1

0 (Ω). Then there is a constant c depending only on n, γ, where γ is theconstant from Definition 2.4, such that

∫

Ω

|u|2dΩ(x)2

dx ≤ c∫

Ω

|∇u|2 dx,

wheredΩ(x) := dist(x, ∂Ω).

The next lemma will be a helpful tool to localize certain arguments. The main tool inthe proof will be Hardy’s inequality.

Lemma 3.4. Let Ω be an open bounded subset of Rn such that Rn \Ω is uniformly 2-thick,and let ηλ ∈ C∞0 (Ω, [0, 1]) be a cutoff function such that ηλ = 1 in x ∈ Ω : dΩ(x) ≥ λ,and |∇ηλ| ≤ c

λ . Then for any ϕ ∈ L2(0, T ;H10 (Ω)) we have

ϕηλ ϕ weakly in L2(0, T ;H10 (Ω)) as λ ↓ 0.

Proof. Pick any functional f ∈ L2(0, T ;H−1(Ω)). Then f = g − div h for some g ∈L2(ΩT ) and h ∈ L2(ΩT ,Rn). This means that

∫ T

0

〈f, ϕ(1− ηλ)〉dt =

∫

ΩT

(1− ηλ)ϕg dz +

∫

ΩT

∇(ϕ(1− ηλ)) · hdz,

and the claim follows by showing that the right-hand side tends to zero as λ→ 0.The first term tends to zero by the dominated convergence theorem. For the second

term, we have∫

ΩT

∇(ϕ(1− ηλ)) · hdz =

∫

ΩT

(1− ηλ)∇ϕ · hdz −∫

dΩ(x)<λϕ∇ηλ · hdz,

where we used the fact that ∇ηλ = 0 when dΩ(x) ≥ λ. The first term on the right-handside again tends to zero by the dominated convergence theorem. For the second one, weuse Holder’s inequality and the Hardy inequality from Theorem 3.3 to get

∣∣∣∣∫

dΩ(x)<λϕ∇ηλ · hdz

∣∣∣∣ ≤ c(∫

dΩ(x)<λ

|ϕ|2λ2

dz

) 12(∫

dΩ(x)<λ|h|2 dz

) 12

≤ c(∫

ΩT

|ϕ|2dΩ(x)2

dz

) 12(∫

dΩ(x)<λ|h|2 dz

) 12

≤ c(∫

ΩT

|∇ϕ|2 dz

) 12(∫

dΩ(x)<λ|h|2 dz

) 12

→ 0

since|(x, t) ∈ ΩT : dΩ(x) < λ| → 0

as λ→ 0.

As an immediate consequence of Lemma 3.4 we can show that local weak or strongsolutions possessing the right lateral boundary data are indeed weak or strong solutions.

Lemma 3.5. Any local weak solution u ∈ Kψ(ΩT ) in the sense of Definition 2.1 whichadditionally satisfies um − gm ∈ L2(0, T ;H1

0 (Ω)) is a weak solution in the sense ofDefinition 2.2. The same holds for strong solutions.

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Proof. In the variational inequalities (2.3), respectively (2.4) we simply choose η = ηλwith ηλ as in Lemma 3.4 and then pass to the limit λ ↓ 0.

3.3. Some elementary inequalities. For u, v ≥ 0 we define

I(u, v) := 1m+1

(um+1 − vm+1

)− vm(u− v).

Lemma 3.6. For any m ≥ 1 and u, v ≥ 0 we have1

m+1 |u− v|m+1 ≤ I(u, v).

Proof. The estimate is trivial for v = 0, and in the other case we can divide by vm+1 andlet x = u

v to see that it suffices to show that the inequality

h(x) := 1m+1

(xm+1 − 1)− (x− 1)− 1

m+1 |x− 1|m+1 ≥ 0

holds for every x ≥ 0. But one-dimensional calculus shows that h has critical pointsin 0 and 1, is strictly concave on [0, 1

2 ] and strictly convex on [ 12 ,∞). This implies that

h(1) = 0 is the minimum value of h on the interval [0,∞).

Lemma 3.7. For any 0 < m < 1 and any u, v ≥ 0 we have

I(u, v) ≥ m2m−2

vm−1|u− v|2 if |u− v| < v,

|u− v|m+1 if |u− v| ≥ v.

Proof. In the special case v = 0, the estimate holds true since m2m−2 ≤ m2 ≤ 1

m+1 . Ifv 6= 0, we divide by vm+1 and let x = u

v . It thus remains to prove

1m+1

(xm+1 − 1)− (x− 1) ≥ m2m−2

(x− 1)2 if 0 ≤ x < 2,

(x− 1)m+1 if x ≥ 2.

For the case 0 ≤ x < 2 this is satisfied because the function

g(x) := 1m+1

(xm+1 − 1)− (x− 1)−m2m−2(x− 1)2

is strictly convex on [0, 2] and possesses its only critical point in x = 1 with minimumvalue g(1) = 0. For the case x > 2, we introduce the auxiliary function

h(x) := 1m+1

(xm+1 − 1)− (x− 1)−m2m−2(x− 1)m+1.

The concavity of x 7→ xm and the inequality x < 2(x − 1) for x > 2 imply xm − 1 >mxm−1(x− 1) > m2m−1(x− 1)m and therefore

h′(x) >[m2m−1 −m(m+ 1)2m−2

](x− 1)m > 0

for x > 2. We infer that h is strictly increasing on [2,∞) and consequently, h(x) >h(2) = g(2) ≥ 0 for any x > 2. This completes the proof.

Corollary 3.8. For any 0 < m < 1 and any two functions u, v ∈ Lm+1(Ω) there holds∫

Ω

|u− v|m+1 dx ≤ c∫

Ω

I(u, v) dx+ c

(∫

Ω

I(u, v) dx

)m+12(∫

Ω

vm+1 dx

) 1−m2

for a constant c = c(m).

Proof. We begin by using Holder’s inequality with exponents 2m+1 and 2

1−m in order toestimate∫

|u−v|<v|u− v|m+1 dx =

∫

|u−v|<vv

(m−1)(m+1)2 |u− v|m+1 v

(1−m)(m+1)2 dx

≤(∫

|u−v|<vvm−1|u− v|2 dx

)m+12(∫

Ω

vm+1 dx

) 1−m2

.



We thereby get∫

Ω

|u− v|m+1 dx ≤∫

|u−v|≥v|u− v|m+1 dx

+

(∫

|u−v|<vvm−1|u− v|2 dx

)m+12(∫

Ω

vm+1 dx

) 1−m2

.

Now the claim follows by estimating the integrands with the help of Lemma 3.7. Next, we prove yet another elementary inequality:

Lemma 3.9. For any u, v ≥ 0 and m > 1 we have |v − u|m ≤ |vm − um|.Proof. Without loss of generality, we may assume 0 ≤ u ≤ v. Then we can estimate

vm − um =

∫ 1

0

m(u+ t(v − u))m−1 dt (v − u)

≥∫ 1

0

mtm−1(v − u)m−1 dt (v − u) = (v − u)m,

where we used the assumption m > 1. This implies the asserted inequality. As a first consequence, we get the following Lebesgue space estimate.

Corollary 3.10. For any two maps u, v ∈ L2m(ΩT ) we have

‖v − u‖mL2m(ΩT ) ≤ ‖vm − um‖L2(ΩT ) in the case m > 1

and‖vm − um‖L2(ΩT ) ≤ ‖v − u‖mL2m(ΩT ) in the case 0 < m < 1.

As a second consequence of Lemma 3.9 we have

Corollary 3.11. For any u, v ≥ 0 there holds

|v − u|m+1 ≤ (vm − um)(v − u) in the case m > 1

and|vm − um|m+1

m ≤ (vm − um)(v − u) in the case 0 < m < 1.

Proof. For the proof of the first inequality, we assume without loss of generality that 0 ≤u ≤ v and apply Lemma 3.9 to bound the first factor. The second inequality follows fromthe first by replacing m by 1

m and v, u by vm, um in the first inequality.

4. COMPARISON ESTIMATES

In this chapter we will prove certain comparison estimates for functions u satisfying thevariational inequality (2.4). Throughout this Section we assume that ψ, uo, g satisfy theregularity and compatibility conditions (2.1), (2.2), and (2.5). Note that we do not assumethat u belongs to C0([0, T ];Lm+1(Ω)) here.

Lemma 4.1. Assume that u : ΩT → R≥0 with um − gm ∈ L2(0, T ;H10 (Ω)) and u ≥ ψ

a.e. in ΩT satisfies the variational inequality (2.4). Then, for all comparison maps v ∈K ′ψ,g(ΩT ) and every cut-off function α ∈ W 1,∞([0, T ],R≥0) with α(T ) = 0 and α′ ≤ 0,we have the estimate∫

ΩT

|α′|I(u, v) dz ≤∫

ΩT

α[∂tv

m(v − u) +∇um · ∇(vm − um)]

dz

+ α(0)

∫

Ω

I(uo, vo) dx,

where we abbreviated vo := v(·, 0) and

I(u, v) := 1m+1 (um+1 − vm+1)− vm(u− v).


Proof. For λ > 0 we let ηλ ∈ C∞0 (Ω, [0, 1]) be a cutoff function in space satisfying ηλ = 1in x ∈ Ω : dΩ(x) ≥ λ, and |∇ηλ| ≤ c

λ . The variational inequality (2.4) implies∫

ΩT

α∇um · ∇[ηλ(vm − um)] dz

≥∫

ΩT

[(−α′)ηλ

[1

m+1um+1 − uvm

]+ αηλu∂tv

m]

dz

− α(0)

∫

Ω

ηλ[

1m+1u

m+1o − uovmo

]dx.(4.1)

For the the right-hand side of (4.1) we have

limλ↓0

∫

ΩT

[(−α′)ηλ

[1

m+1um+1 − uvm

]+ αηλu∂tv

m]

dz

− α(0)

∫

Ω

ηλ[

1m+1u

m+1o − uovmo

]dx

=

∫

ΩT

[(−α′)

[1

m+1um+1 − uvm

]+ αu∂tv

m]

dz

− α(0)

∫

Ω

[1

m+1um+1o − uovmo

]dx

=

∫

ΩT

[|α′|[

1m+1 (um+1 − vm+1)− vm(u− v)

]+ αu∂tv

m]

dz

+

∫

ΩT

α′ mm+1v

m+1 dz − α(0)

∫

Ω

[1

m+1um+1o − uovmo

]dx

=

∫

ΩT

[|α′|[

1m+1 (um+1 − vm+1)− vm(u− v)

]− α∂tvm(v − u)

]dz

− α(0)

∫

Ω

[1

m+1 (um+1o − vm+1

o )− vmo (uo − vo)]

dx,

where we also used the assumption α′ ≤ 0 and an integration by parts. For the left-handside of (4.1) we find by Lemma 3.4 that

limλ↓0

∫

ΩT

α∇um · ∇[ηλ(vm − um)] dz =

∫

ΩT

α∇um · ∇(vm − um) dz.

Therefore, using the preceding computations in (4.1), passing to the limit λ ↓ 0 and re-arranging terms we arrive at∫

ΩT

|α′|[

1m+1 (um+1 − vm+1)− vm(u− v)

]dz

≤∫

ΩT

α∂tvm(v − u) dz +

∫

ΩT

α∇um · ∇(vm − um) dz

+ α(0)

∫

Ω

[1

m+1 (um+1o − vm+1

o )− vmo (uo − vo)]

dx.

This finishes the proof of the Lemma. Later on, we need a local version of Lemma 4.1 for comparison maps v, which do not

admit the right boundary values on the lateral boundary ∂Ω× (0, T ).

Lemma 4.2. Assume that u : ΩT → R≥0 with um − gm ∈ L2(0, T ;H10 (Ω)) and u ≥ ψ

a.e. in ΩT satisfies the variational inequality (2.4). Then, for all comparison maps v ∈K ′ψ(ΩT ), and every cut-off function in time α ∈ W 1,∞([0, T ],R≥0) with α(T ) = 0 andα′ ≤ 0 we have the estimate∫

Br(xo)×(0,T )

|α′|I(u, v) dz ≤∫

Br(xo)×(0,T )

α[∂tv

m(v − u) +∇um · ∇(vm − um)]

dz



+ 2

∫

∂Br(xo)×(0,T )

α|∇um||vm − um|dz

+ α(0)

∫

Br(xo)

I(uo, vo) dx,

for a.e. r ∈ (0,dist(xo, ∂Ω)), where we abbreviated vo := v(·, 0) and

I(u, v) := 1m+1 (um+1 − vm+1)− vm(u− v).

Proof. By ηλ ∈ C10 (Br, [0, 1]) we denote a cut-off function in space satisfying ηλ = 1 on

Br−λ and |Dηλ| ≤ 2/λ. Using this in the variational inequality (2.4), we infer

0 ≤ 〈〈∂tu, αηλ(vm − um)〉〉uo +

∫

Br×(0,T )

α∇um · ∇[ηλ(vm − um)] dz.(4.2)

For the first term, we have

〈〈∂tu, αηλ(vm − um)〉〉uo =

∫

ΩT

[α′ηλ

[1

m+1um+1 − uvm

]− αηλu∂tvm

]dz

+ α(0)

∫

Ω

ηλ[

1m+1u

m+1o − uovmo

]dx

−→λ↓0

∫

Br×(0,T )

[α′[

1m+1u

m+1 − uvm]− αu∂tvm

]dz

+ α(0)

∫

Br

[1

m+1um+1o − uovmo

]dx

= 〈〈∂tu, αχBr (vm − um)〉〉uo .For the second term in (4.2), we have∫

Br×(0,T )

α∇um · ∇[ηλ(vm − um)] dz

≤∫

Br×(0,T )

αηλ∇um · ∇(vm − um) dz + 2λ

∫

(Br\Br−λ)×(0,T )


−→λ↓0

∫

Br×(0,T )

α∇um · ∇(vm − um) dz + 2

∫

∂Br×(0,T )


for a.e. r > 0. Joining the two preceeding formulae with (4.2), we arrive at

0 ≤ 〈〈∂tu, αχBr (vm − um)〉〉uo +

∫

Br×(0,T )

α∇um · ∇(vm − um) dz

+ 2

∫

∂Br×(0,T )

α|∇um||vm − um|dz.

At this stage, we can proceed in the same way as in Lemma 4.1 on Br instead of Ω, withthe last integral as additional term. This yields the claim.

5. CONTINUITY IN TIME

Here, we prove that the variational inequality (2.4) already implies u ∈C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo. The following lemma is a version of [2, Lemma2.5]; see also [3, Lemma 2.4].

Lemma 5.1. Assume that u, ψ, g ∈ L∞(0, T ;Lm+1(Ω)) satisfy u ≥ ψ a.e. on ΩT and∂tψ

m, ∂tgm ∈ Lm+1

m (ΩT ). Then the maps wh defined by

(wh)m := maxψm, [[um − gm]]h + gm

,



where

[[um − gm]]h(x, t) := e−th

(umo (x)− gm(x, 0)

)+

1

h

∫ t

0

es−th

(um − gm

)(x, s) ds

satisfy

lim suph↓0

supto∈(0,T )

∫

Ωto

∂t(wmh )(wh − u) dz ≤ 0.

Proof. We define mollifications of ψm − gm by

[[ψm − gm]]h(x, t) := e−th

(ψm − gm

)(x, 0) +

1

h

∫ t

0

es−th

(ψm − gm

)(x, s) ds.

Due to the assumption u ≥ ψ, we have

(5.1) wmh ≥ [[ψm − gm]]h + gm

and moreover, the assumption ∂tψm ∈ Lm+1m (ΩT ) implies by Lemma 3.1 that

(5.2) ∂t[[ψm − gm]]h → ∂tψ

m + ∂tgm strongly in L

m+1m (ΩT ) as h ↓ 0.

For a fixed h > 0 we split the term under consideration into

supto∈(0,T )

∫

Ωto

∂t(wmh )(wh − u) dz

≤ supto∈(0,T )

∫

Ωto∩wh>ψ∂t[[u

m − gm]]h(wh − u) dz

+ supto∈(0,T )

∫

Ωto

∂tgm(wh − u) dz

+ supto∈(0,T )

∫

Ωto∩wh=ψ(∂tψ

m − ∂tgm)(ψ − u) dz

=: Ih + IIh + IIIh.

The integrand in the first integral is nonpositive since

∂t[[um − gm]]h(wh − u) = − 1

h

([[um − gm]]h − (um − gm)

)(wh − u)

= − 1h (wmh − um)(wh − u) ≤ 0

by Lemma 3.1 (i) and the monotonicity of the function r 7→ rm. This implies Ih ≤ 0. Next,we consider the second term IIh. Here, we note that wh → u in Lm+1(ΩT ) by Lemma 3.1and therefore limh↓0 IIh = 0. Finally, we turn our attention to the third term IIIh, for whichwe only have to consider points z ∈ Ωto with ψ(z) = wh(z). In such points we have

∂t[[ψm − gm]]h = 1

h

(ψm − gm − [[ψm − gm]]h

)= 1

h

(wmh − gm − [[ψm − gm]]h

)≥ 0,

where we used (5.1) for the last estimate. Since u ≥ ψ a.e. by assumption, this implies∂t[[ψ

m − gm]]h(ψ − u) ≤ 0 on the domain of integration of IIIh. We thereby obtain thebound

IIIh ≤ supto∈(0,T )

∫

Ωto∩wh=ψ∂t[[ψ

m − gm]]h(ψ − u) dz

+ supto∈(0,T )

∫

Ωto∩wh=ψ

(∂t(ψ

m − gm)− ∂t[[ψm − gm]]h)(ψ − u) dz

≤∥∥∂t(ψm − gm)− ∂t[[ψm − gm]]h

∥∥Lm+1m (ΩT )

‖ψ − u‖Lm+1(ΩT ) −→h↓0

0

by the convergence (5.2). This proves that lim suph↓0 IIIh ≤ 0. Since we have shownalready that Ih ≤ 0 for every h > 0 and IIh → 0 as h ↓ 0, the proof is complete.


Lemma 5.2. Let m > 0 and assume that ψ, uo, g satisfy (2.1), (2.2), (2.5) and that u ∈L∞(0, T ;Lm+1(Ω)) with um−gm ∈ L2(0, T ;H1

0 (Ω)) and u ≥ ψ a.e. in ΩT satisfies thevariational inequality (2.4). Then, we have u ∈ C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo.

Proof. For h > 0 we define the maps wh by

(wh)m := maxψm, [[um − gm]]h + gm

,

where [[um − gm]]h is defined as in Lemma 5.1. In Lemma 4.1 we use v = wh as compar-ison map. Since wh(·, 0) = uo, this implies the estimate∫

ΩT

|α′|I(u,wh) dz ≤∫

ΩT

α∂t(wmh )(wh − u) dz +

∫

ΩT

α∇um · ∇(wmh − um) dz,

for any cut-off function α ∈ W 1,∞(0, T,R≥0) with α(T ) = 0 and α′ ≤ 0. For a timeto ∈ (0, T ) and 0 < ε < to we define the cut-off function by α ≡ 1 on [0, to − ε],α(t) := 1

ε (to − t) on [to − ε, to] and α ≡ 0 on [to, T ]. Plugging this function into thepreceding estimate and letting ε ↓ 0, we infer

ess supto∈(0,T )

∫

Ω×toI(u,wh) dx(5.3)

≤ supto∈(0,T )

∫

Ωto

∂t(wmh )(wh − u) dz +

∫

ΩT

|∇um| |∇wmh −∇um|dz.

The last integral vanishes in the limit h ↓ 0 by Lemma 3.1 (ii), while the first integral onthe right-hand side can be estimated in the limit h ↓ 0 using Lemma 5.1. We thereby arriveat

limh↓0

ess supto∈(0,T )

∫

Ω×toI(u,wh) dx = 0.

By Lemma 3.6 if m ≥ 1, respectively Corollary 3.8 and the fact that wh is boundedin L∞(0, T ;Lm+1(Ω)) independently of h if 0 < m < 1, this implies that wh → ustrongly in L∞(0, T ;Lm+1(Ω)) as h ↓ 0. Moreover, we have wh ∈ C0([0, T ];Lm+1(Ω))by Lemma 3.1 (iii). We have thus shown C0([0, T ];Lm+1(Ω)) 3 wh → u inL∞(0, T ;Lm+1(Ω)) as h ↓ 0. This implies the claim u ∈ C0([0, T ];Lm+1(Ω)). Sincewh(·, 0) = uo, we also have u(·, 0) = uo.

6. COMPARISON LEMMAS FOR PSEUDOMONOTONE OPERATORS

Our aim in this section is to derive certain comparison principles for porous mediumtype equations. We will also use some other, truncated porous medium equations as partof our proofs. Let a : R≥0 → R≥0 be a function satisfying

0 < α ≤ a(s) ≤ 1α ,

for some 0 < α < 1. More precisely, we assume that the mapping R≥0 × Rn 3 (s, ξ) 7→a(s)ξ has the following properties:

|a(s)ξ| ≤ 1α |ξ|(

a(s)ξ1 − a(s)ξ2)· (ξ1 − ξ2) ≥ α|ξ1 − ξ2|2

|a(s1)ξ − a(s2)ξ| ≤ L|ξ||s1 − s2|,(6.1)

for any s, s1, s2 ∈ R≥0 and ξ, ξ1, ξ2 ∈ Rn. The above properties of the map (s, ξ) 7→a(s)ξ imply that we can define weak solutions as follows.

Definition 6.1. A nonnegative function u is a local weak solution to the equation

(6.2) ∂tu− div(a(u)∇u) = 0 in ΩT


if u ∈ C0([0, T ];L2(Ω)) ∩ L2loc(0, T ;H1

loc(Ω)), and

(6.3)∫

ΩT

[− u∂tϕ+ a(u)∇u · ∇ϕ

]dz = −

∫

Ω×tuϕdx

∣∣∣∣T

t=0

for all test functions ϕ ∈ C∞(ΩT ) with ϕ = 0 on ∂Ω× [0, T ]. The requirement for a weaksupersolution is that

(6.4)∫

ΩT

[− u∂tϕ+ a(u)∇u · ∇ϕ

]dz ≥ 0,

for all positive test functions ϕ ∈ C∞0 (ΩT ). For weak subsolutions, the inequality in (6.4)is reversed.

For the existence of solutions in the above sense we refer to [18]. In order to makethe following computations rigorous concerning the use of the time derivative, we need toreformulate the equation in terms of the mollified solution [[u]]h. Here, we choose vo = 0in (3.1). If u is a weak solution to the porous medium equation (2.9), then [[u]]h satisfies

(6.5)∫

ΩT

[∂t[[u]]hϕ+∇[[um]]h · ∇ϕ

]dxdt =

1

h

∫

Ω

u(·, 0)

∫ T

0

ϕ(·, s)e− sh dsdx,

for all test functions ϕ ∈ L2(0, T ;H10 (Ω)). The equation (6.5) follows from (2.10) by

straightforward manipulations involving a change of variables and Fubini’s theorem. Sim-ilarly, the mollification [[u]]h of a solution to (6.2) satisfies

(6.6)∫

ΩT

[∂t[[u]]hϕ+ [[a(u)∇u]]h · ∇ϕ

]dx dt =

1

h

∫

Ω

u(·, 0)

∫ T

0

ϕ(·, s)e− sh dsdx

for all ϕ ∈ L2(0, T ;H10 (Ω)). For inhomogeneous equations involving a zero order term

f(x, t, u), one has to add the integral∫

ΩT[[f ]]h(x, t, u)ϕdxdt on the right-hand side.

Our first aim is to establish some comparison principles for weak solutions, respectivelysub- and supersolutions to (6.2). We adapt the arguments from [4] for the proof.

Lemma 6.2. Let u1 and u2 be weak solutions to (6.2) such that (u1 − u2)+ ∈L2(0, T ;H1

0 (Ω)) and u1(·, 0) ≤ u2(·, 0) a.e. on Ω. Then

u1 ≤ u2 almost everywhere in ΩT .

The proof is based on the following estimate. In the proof, we make use of the usualtwo-sided truncation operator Tε with ε > 0, given by

Tε(s) :=

s, if |s| ≤ εε sgn(s), if |s| > ε.

Lemma 6.3. Let u1 and u2 be as in Lemma 6.2. Then

supt∈(0,T )

∫

0<u1−u2<ε(u1 − u2)2(·, t) dx+ ε sup

t∈(0,T )

∫

u1−u2≥ε(u1 − u2)(·, t) dx

+ α

∫

0<u1−u2<ε|∇(u1 − u2)|2 dz

≤ L2ε2

α

∫

0<u1−u2<ε|∇u1|2 dz.

Proof. For τ ∈ (0, T ] we let χ[0,τ ] : [0, T ] → [0, 1] be the characteristic function of theinterval [0, τ ]. We take ϕ = χ[0,τ ]Tε(u1−u2)+ with ε > 0 as test-function in (6.6) writtenfor u1 and u2, subtract the results and use the assumption u1(·, 0) ≤ u2(·, 0). This leads to∫

Ωτ

[∂t[[u1−u2]]hTε(u1−u2)+ + [[a(u1)∇u1−a(u2)∇u2]]h∇Tε(u1−u2)+

]dz ≤ 0.



We want to get rid of the time derivative in the first term. To achieve this, we write

∂t[[u1 − u2]]hTε(u1 − u2)+ = ∂t[[u1 − u2]]hTε([[u1 − u2]]h

)+

+ ∂t[[u1 − u2]]h[Tε(u1 − u2)+ − Tε

([[u1 − u2]]h

)+

].

With the abbreviation v = u1 − u2, we can write the second term as

∂t[[v]]h(Tε(v)+ − Tε([[v]]h)+

)=v − [[v]]h

h

[Tε(v)+ − Tε([[v]]h)+

]≥ 0,

where the positivity follows from the fact that v 7→ Tε(v)+ is increasing. Thus, we have∫

Ωτ

∂t[[u1 − u2]]hTε(u1 − u2)+ dz ≥∫

Ωτ

∂t[[u1 − u2]]hTε([[u1 − u2]]h

)+

dz

=1

2

∫

0<[[u1−u2]]h<ε[[u1 − u2]]2h(·, τ) dx

+ ε

∫

[[u1−u2]]h≥ε[[u1 − u2]]h(·, τ) dx.

The time derivative has been eliminated, and we may let h ↓ 0 to get1

2

∫

0<u1−u2<ε(u1 − u2)2(·, τ) dx+ ε

∫

u1−u2≥ε(u1 − u2)(·, τ) dx

+

∫

Ωτ∩0<u1−u2<ε

(a(u1)∇u1 − a(u2)∇u2

)· (∇u1 −∇u2) dz ≤ 0.

We note that(a(u1)∇u1 − a(u2)∇u2

)· (∇u1 −∇u2)

=(a(u1)− a(u2)

)∇u1 · (∇u1 −∇u2) + a(u2)|∇u1 −∇u2|2

≥ −L|u1 − u2||∇u1||∇u1 −∇u2|+ α|∇u1 −∇u2|2,where in the last line we used the structure conditions (6.1). Therefore, the precedinginequality yields that

1

2

∫

0<u1−u2<ε(u1 − u2)2(·, τ) dx+ ε

∫

u1−u2≥ε(u1 − u2)(·, τ) dx

+ α

∫

Ωτ∩0<u1−u2<ε|∇u1 −∇u2|2 dz

≤ L∫

Ωτ∩0<u1−u2<ε|u1 − u2||∇u1||∇u1 −∇u2|dz

≤ Lε∫

Ωτ∩0<u1−u2<ε|∇u1||∇(u1 − u2)|dz

≤ α

2

∫

Ωτ∩0<u1−u2<ε|∇(u1 − u2)|2 dz +

L2ε2

2α

∫

Ωτ∩0<u1−u2<ε|∇u1|2 dz.

This leads to the estimate∫

0<u1−u2<ε(u1 − u2)2(·, τ) dx+ ε

∫

u1−u2≥ε(u1 − u2)(·, τ) dx

+ α

∫

Ωτ∩0<u1−u2<ε|∇(u1 − u2)|2 dz

≤ L2ε2

α

∫

0<u1−u2<ε|∇u1|2 dz.

The desired estimate follows from this inequality by taking the supremum over τ ∈ (0, T ]in the first two terms and τ = T in the third one.


Proof of Lemma 6.2. From Lemma 6.3 we have

supt∈(0,T )

∫

u1−u2≥ε(u1 − u2)(·, t) dx ≤ L2ε

α

∫

0<u1−u2<ε|∇u1|2 dz → 0

in the limit ε ↓ 0. Thus, we have proved

supt∈(0,T )

∫

u1−u2≥0(u1 − u2)(·, t) dx ≤ 0

which implies that u1 ≤ u2 almost everywhere in ΩT .

The following variants of Lemma 6.2 follow by essentially the same arguments asabove.

Lemma 6.4. Let u1 be a weak subsolution, and u2 a weak supersolution to (6.2) such that(u1 − u2)+ ∈ L2(0, T ;H1

0 (Ω)) and u1(·, 0) ≤ u2(·, 0) on Ω. Then


We will use the following version when proving that solutions to penalized equationsstay above the obstacle function.

Lemma 6.5. Let u, ψ ∈ L2(0, T ;H1(Ω)) be such that (ψ − u)+ ∈ L2(0, T ;H10 (Ω)) and

1

2

∫

0<ψ−u<ε(ψ − u)2(·, τ) dx+ ε

∫

ψ−u≥ε(ψ − u)(·, τ) dx

+

∫

Ωτ∩0<ψ−u<ε

(a(ψ)∇ψ − a(u)∇u

)· ∇(ψ − u) dz ≤ 0

for all τ ∈ [0, T ] and all ε > 0. Then

ψ ≤ u almost everywhere in ΩT .

7. SOLUTIONS OF THE PENALIZED PME

In this section, we construct solutions to a penalized porous medium equation by ap-proximating the PME with pseudomonotone equations. In the next two sections, we im-pose the stronger regularity assumptions (2.8) on the functions ψ, g and uo.

We fix a (small) number δ > 0. For the penalty term, we pick an increasing functionζδ ∈ C∞(R, [0, 1]) such that ζδ ≡ 1 on [0,∞), ζδ ≡ 0 on (−∞,−δ] and |∇ζδ| ≤ c/δ.The aim in this section is to construct a weak solution to the Cauchy-Dirichlet problem forthe penalized PME

(7.1)

∂tu−∆um = Ψ+ζδ(ψm − um) in ΩT ,

u = g on ∂Ω× (0, T ),

u(·, 0) = uo in Ω,

by an approximation process. Therefore, we will first consider the following equation

(7.2) ∂tu− div(a(u)∇u) = Ψ+ζδ(ψm − um) in ΩT ,

where a satisfies the assumptions (6.1). The definition of weak solutions to (7.1) and (7.2)are analogous to Definitions 2.5 and 6.1. The first result concerning equation (7.2) is acomparison lemma.

Lemma 7.1. Let u1 and u2 be weak solutions of (7.2) under the assumptions (6.1) and(2.8) and such that (u1 − u2)+ ∈ L2(0, T ;H1

0 (Ω)) and u1(x, 0) ≤ u2(x, 0). Then



Proof. The right-hand side in (7.2) is decreasing in u, so it is straightforward to incorporatethis term into the proof of Lemma 6.2. More specifically, testing with Tε(u1 − u2)+ andletting h ↓ 0 yields

∫

ΩT

Ψ+

[ζδ(ψ

m − um1 )− ζδ(ψm − um2 )]Tε(u1 − u2)+ dz ≤ 0.

To see that the sign is indeed negative, we note that Tε(u1 − u2)+ is nonzero only whenu2 < u1, and the function t 7→ ζδ(ψ

m − t) is decreasing.

To pass to the limit in the approximation scheme, we need an energy estimate for equa-tion (7.1).

Lemma 7.2. Let u be a weak solution to the Cauchy-Dirichlet problem (7.1) under theassumptions (6.1) and (2.8). Then, we have

supt∈[0,T ]

∫

Ω×tum+1 dx+

∫

ΩT

|∇um|2 dz ≤ cM

and‖∂tu‖2L2(0,T ;H−1(Ω)) ≤ cM,

where c = c(n,m,diam(Ω), T ) and

M := supt∈[0,T ]

∫

Ω×tgm+1 dx+

∫

Ω

um+1o dx+

∫

ΩT

[|Ψ+|2 + |∇gm|2 + |∂tgm|

m+1m

]dz.

Proof. For τ ∈ (0, T ] we let χ[0,τ ] : [0, T ] → [0, 1] be the characteristic function of theinterval [0, τ ]. We test the regularized equation (6.5) with ϕ = χ[0,τ ](u

m − gm) to get∫

Ωτ

[∂t[[u]]h(um − gm) +∇[[um]]h · ∇(um − gm)

]dz

=

∫

Ωτ

[[ζδ(ψm − um)Ψ+]]h(um − gm) dz

+1

h

∫

Ω

uo

∫ τ

0

(um − gm)e−sh dsdx.

For the first term on the left-hand side, we find, using Lemma 3.1 (i) that∫

Ωτ

∂t[[u]]hum dz =

∫

Ωτ

∂t[[u]]h[[u]]mh dz +

∫

Ωτ

∂t[[u]]h(um − [[u]]mh

)dz

≥∫

Ωτ

∂t[[u]]h[[u]]mh dz =1

m+ 1

∫

Ωτ

∂t[[u]]m+1h dz

=1

m+ 1

∫

Ω×τ[[u]]m+1

h dx.

Moreover, integrating by parts, we find that∫

Ωτ

∂t[[u]]hgm dz = −

∫

Ωτ

[[u]]h∂tgm dz +

∫

Ω×τ[[u]]hg

m dx.

Inserting this above, letting h ↓ 0 with the help of Lemma 3.1 and rearranging terms wefind that

1

m+ 1

∫

Ω×τum+1 dx+

∫

Ωτ

|∇um|2 dz

≤∫

Ω×τugm dx−

∫

Ωτ

u∂tgm dz +

∫

Ωτ

∇um · ∇gm dz

+

∫

Ωτ

ζδ(ψm − um)Ψ+(um − gm) dz +

∫

Ω×0

(um+1o − uogm

)dx.


We proceed by taking absolute values, applying Young’s inequality, and taking the supre-mum over t ∈ (0, T ] to get

1

m+ 1

∫

Ω×τum+1 dx+

∫

Ωτ

|∇um|2 dz

≤ 1

2(m+ 1)sup

t∈(0,T ]

∫

Ω×tum+1 dx+

1

4

∫

Ωτ

|∇um|2 dz

+ c supt∈(0,T ]

∫

Ω×tgm+1 dx+

∫

Ωτ

|∇gm|2 dz + c(T )

∫

Ωτ

|∂tgm|m+1m dz

+

∫

Ωτ

ζδ(ψm − um)Ψ+(um − gm) dz +

∫

Ω

um+1o dx,

where we have also taken into account that uogm(·, 0) ≥ 0. For the second last term onthe right, we have for ε > 0 that

∣∣∣∣∫

Ωτ

ζδ(ψm − um)Ψ+(um − gm) dz

∣∣∣∣

≤ ε∫

Ωτ

|um − gm|2 dz + cε

∫

Ωτ

Ψ2+ dz

≤ cε∫

Ωτ

|∇(um − gm)|2 dz + cε

∫

Ωτ

Ψ2+ dz

≤ cε∫

Ωτ

|∇um|2 dz + c

∫

Ωτ

|∇gm|2 dz + cε

∫

Ωτ

Ψ2+ dz,

where we used Young’s and Poincare’s inequalities and the fact that |ζδ| ≤ 1. Choosingε > 0 small enough, we can absorb the terms containing ∇um from the right-hand sideinto the left. We thus have arrived at

1

m+ 1

∫

Ω×τum+1 dx+

1

2

∫

Ωτ

|∇um|2 dz

≤ 1

2(m+ 1)sup

t∈(0,T ]

∫

Ω×tum+1 dx+ cM,

with a constant c depending on n,m,diam(Ω), T and where M is defined in the statementof the lemma. We now take the supremum over τ ∈ (0, T ] in the first term on the left-handside and τ = T in the second one. Absorbing the matching terms from the right-hand sideto the left, we deduce the first estimate claimed in the lemma.

The estimate for the time derivative follows from the first estimate. For any ϕ ∈C∞0 (ΩT ) we have by (7.1)1 and Holder’s and Poincare’s inequality that

∣∣∣∣∫

ΩT

u∂tϕdz

∣∣∣∣ ≤∣∣∣∣∫

ΩT

∇um · ∇ϕdz

∣∣∣∣+

∣∣∣∣∫

ΩT

ζδ(ψm − um)Ψ+ϕdz

∣∣∣∣≤ ‖∇um‖L2(ΩT )‖∇ϕ‖L2(ΩT ) + ‖Ψ+‖L2(ΩT )‖ϕ‖L2(ΩT )

≤ c[‖∇um‖L2(ΩT ) + ‖Ψ+‖L2(ΩT )

]‖ϕ‖L2(0,T ;H1(Ω))

≤ cM 12 ‖ϕ‖L2(0,T ;H1(Ω)).

Since C∞0 (ΩT ) is dense in L2(0, T ;H10 (Ω)) this proves the claim and therefore finishes

the proof of the lemma.

Proposition 7.3. Let g, ψ, and uo satisfy (2.8). Then there exists a function u, which is aweak solution of (7.1) satisfying

u ≥ ψ almost everywhere in ΩT .


Proof. Fix ε, γ, δ ∈ (0, 1], and define

ψε := ψ + ε, gε,γ := (gm + γm)1m + ε, uo,ε,γ := uo + ε+ γ.

Then gε,γ ≥ g + ε ≥ ψε, and uo,ε,γ ≥ ψε(·, 0). Let

M := max

supΩT

(ψmε + δ)1m , sup

ΩT

gε,γ , supΩuo,ε,γ

.

Then ψε ≤M , gε,γ ≤M , and uo,ε,γ ≤M . Choose ε and γ smaller if necessary, so that

(7.3) M ≤ 1

γ + ε.

The adjustment can be made independently of δ. Now, we define aε : R≥0 → R≥0 by

(7.4) aε(s) :=

mεm−1 if 0 ≤ s ≤ ε,msm−1 if ε < s < 1

ε ,

mε1−m if s ≥ 1ε .

Note that aε satisfies the structure assumptions (6.1) with α = mminεm−1, ε1−m andL = m|m − 1|maxεm−2, ε2−m and let uε,γ ∈ C0([0, T ];L2(Ω)) ∩ L2(0, T ;H1(Ω))be the weak solution of the Cauchy-Dirichlet problem

(7.5)

∂tuε,γ − div(aε(uε,γ)∇uε,γ

)= (Ψε)+ζδ(ψ

mε − umε,γ) in ΩT

uε,γ = gε,γ on ∂Ω× (0, T )

uε,γ(·, 0) = uo,ε in Ω.

We claim that

(7.6) γ + ε ≤ uε,γ ≤1

γ + ε.

The first inequality follows from Lemma 6.4, since uγ,ε is a weak supersolution and theconstant function ε+ γ is a weak solution of the equation

∂tu− div(aε(u)∇u

)= 0

and moreover, (γ+ ε− uγ,ε)+ ∈ L2(0, T ;H10 (Ω)) and γ+ ε ≤ uε,γ(·, 0). For the second

inequality, we note that ψmε −Mm ≤ −δ. Thus the constant function M is a solution of(7.5)1, and since (uγ,ε −M)+ ∈ L2(0, T ;H1

0 (Ω)) and uε,γ(·, 0) ≤M , the claim followsby combining Lemma 7.1 and (7.3).

The final property of the function uε,γ we need is that uε,γ ≥ ψε. We aim at applyingLemma 6.5. We let τ ∈ (0, T ] and start with the mollified form of (7.5) (as before, we mayreplace ΩT by Ωτ in (7.5) after multiplying the test-function by the characteristic functionof [0, τ ]). This gives

∫

Ωτ

[− ∂t[[uε,γ ]]hϕ− [[aε(uε,γ)∇uε,γ ]]h · ∇ϕ

]dz

= −∫

Ωτ

[[(Ψε)+ζδ(ψmε − umε,γ)]]hϕdz − 1

h

∫

Ω

uo,ε,γ

∫ τ

0

ϕ(·, s)e− sh dsdx.

Next, we observe that

∂t[[ψε]]h = [[∂tψε]]h + 1hψε(·, 0)e−

th

which can be checked by an integration by parts, and

[[aε(ψε)∇ψε]]h = [[∇ψmε ]]h


for ε > 0 small enough so that ψε ≤ 1ε . Taking into account these identities, we add

∂t[[ψε]]hϕ+ [[aε(ψε)∇ψε]]h · ∇ϕ to both sides and get∫

Ωτ

∂t[[ψε − uε,γ ]]hϕ+ [[aε(ψε)∇ψε − aε(uε,γ)∇uε,γ ]]h · ∇ϕdz

=

∫

Ωτ

[[Ψε − (Ψε)+ζδ(ψmε − umε,γ)]]hϕdz

+1

h

∫

Ω

(ψε(·, 0)− uo,γ,ε)∫ τ

0

ϕ(·, s)e− sh dsdx.

Now, we choose ϕ = Tλ(ψε − uε,γ)+ as test-function, discard the initial term which hasa negative sign, integrate in the time term and let h ↓ 0 as in the proof of Lemma 6.3. Thisyields

1

2

∫

0<ψε−uε,γ<λ(ψε − uε,γ)2(·, τ) dx

+ λ

∫

ψε−uε,γ≥λ(ψε − uε,γ)(·, τ) dx

+

∫

Ωτ∩0<ψε−uε,γ<λ

(a(ψε)∇ψε − a(uε,γ)∇uε,γ

)· ∇(ψε − uε,γ) dz

≤∫

Ωτ∩ψε−uε,γ>0

(Ψε − (Ψε)+ζδ(ψ

mε − umε,γ)

)Tλ(ψε − uε,γ) dz

= −∫

Ωτ∩ψε−uε,γ>0(Ψε)−Tλ(ψε − uε,γ) dz ≤ 0,

ensuring that the assumption of Lemma 6.5 is satisfied. Therefore, we conclude that uε,γ ≥ψε, as desired.

Next, we pass to the limit ε ↓ 0 using the energy estimate of Lemma 7.2. We haveaε(uε,γ)∇uε,γ = ∇umε,γ by (7.4) and (7.6). Thus uε,γ is also a weak solution of the penal-ized PME. Since the right-hand side (Ψε)+ζδ(ψ

mε − umε,γ) is bounded independently from

ε, we know from [8, Theorem 1.2] that the functions uε,γ are locally Holder continuouswith a quantitative estimate which is uniform in ε. Hence there is a function uγ such thatuε,γ → uγ as ε ↓ 0 locally uniformly in ΩT . Since uε,γ ≥ ψε, we also have that uγ ≥ ψalmost everywhere. The lateral and initial boundary values of uε,γ are

gε,γ = (gm + γm)1/m + ε and uo,ε,γ = uo + ε+ γ

respectively. We have

∇gmε,γ =((gm + γm)

1m + ε

)m−1(gm + γm)

1m−1∇gm,

and similarly for the time derivative. Thus for a fixed γ, the energy estimate of Lemma 7.2is independent of ε for small values of ε. We recall that Lemma 7.2 is applicable since uε,γis also a weak solution of the penalized PME. It follows that uγ ∈ L∞(0, T ;Lm+1(Ω)),umγ ∈ L2(0, T ;H1(Ω)), ∂tuγ ∈ L2(0, T ;H−1(Ω)), ∇umε,γ ∇umγ weakly inL2(ΩT ,Rn), uε,γ

∗ uγ weakly star in L∞(0, T ;Lm+1(Ω)), ∂tuε,γ ∂tuγ weakly

in L2(0, T ;H−1(Ω)) and umγ − gm − γm ∈ L2(0, T ;H10 (Ω)). The last of these holds

since gmε,γ gm + γm weakly in L2(0, T ;H1(Ω)) as ε ↓ 0. Moreover, uγ(·, 0) = uo,o,γin the H−1(Ω)-sense. We let ε ↓ 0 in the penalized PME by using the above facts. Forϕ ∈ C∞(ΩT ) with ϕ = 0 on ∂Ω× [0, T ], we have

limε↓0

∫ T

0

〈∂tuε,γ , ϕ〉dt =

∫ T

0

〈∂tuγ , ϕ〉dt,

limε↓0

∫

ΩT

∇umε,γ · ∇ϕdz =

∫

ΩT

∇umγ · ∇ϕdz


by the weak convergences, and the fact that

limε↓0

∫

ΩT

ζδ(ψmε − umε,γ)(Ψε)+ϕdz =

∫

ΩT

Ψ+ζδ(ψm − umγ )ϕdz

follows from the dominated convergence theorem. Thus we have∫ T

0

〈∂tuγ , ϕ〉dt+

∫

ΩT

∇umγ · ∇ϕdz =

∫

ΩT

Ψ+ζδ(ψm − umγ )ϕdz

for all test-functions ϕ ∈ C∞(ΩT ) with ϕ = 0 on ∂Ω × [0, T ]. In particular, this im-plies that uγ is a local strong solution to the obstacle problem for the PME with obstacleψ ≡ 0 and additional inhomogeneity Ψ+ζδ(ψ

m − uγ) ∈ L∞(ΩT ); see (2.3). FromLemma 3.2 we then infer that uγ is also a local weak solution to this obstacle problemin the sense of Definition 2.1. Therefore, we can apply Lemma 5.2 (which continues tohold for obstacle problems with an additional right-hand side in L∞) to conclude thatu ∈ C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo.

We want to finish the proof by letting γ ↓ 0. To this end, we note that by (7.6) and thepointwise convergence uε,γ → uγ , we have

γ ≤ uγ ≤1

γ.

This means that uγ is also a weak solution of the equation

∂tuγ − div(aγ(uγ)∇uγ

)= Ψ+ζδ(ψ

m − umγ ).

Thus Lemma 7.1 shows that uγ1≤ uγ2

if γ1 < γ2, since both functions are solutions ofthe equation involving aγ1 . We therefore may define the function u by setting

u(x, t) := limγ↓0

uγ(x, t).

Since uγ ≥ ψ almost everywhere, we also have u ≥ ψ almost everywhere. Showing thatthis function u is the weak solution with the right boundary values proceeds similarly tothe limit process ε ↓ 0 above. However, for the sake of completeness we briefly explainthe argument. The lateral and initial boundary values of uγ are

gγ = (gm + γm)1/m and uo,o,γ = uo + γ

respectively. We have ∇gmγ = ∇gm and ∂tgmγ = ∂tgm. Moreover, since γ ≤ uγ ≤ 1

γ ,uγ is also a weak solution of the penalized PME. Therefore, we can apply Lemma 7.2 touγ and the resulting energy estimate is independent of γ for small values of γ. It followsthat um ∈ L2(0, T ;H1(Ω)), ∇umγ ∇um weakly in L2(ΩT ,Rn), and um − gm ∈L2(0, T ;H1

0 (Ω)) and ∂tu ∈ L2(0, T ;H−1(Ω)), ∂tuγ ∂tu in L2(0, T ;H−1(Ω)) andu(·, 0) = uo in the H−1-sense. We now let γ ↓ 0 in the penalized PME by using the abovefacts similarly as before when passing to the limit ε ↓ 0. In conclusion, we find that

∫ T

0

〈∂tu, ϕ〉dt+

∫

ΩT

∇um · ∇ϕdz =

∫

ΩT

Ψ+ζδ(ψm − um)ϕdz

for all test-functions ϕ ∈ C∞(ΩT ) with ϕ = 0 on ∂Ω× [0, T ]. Arguing as above, we findthat u ∈ C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo. We also have u ≥ ψ almost everywhereand therefore, u is the weak solution of (7.1) we were looking for.

Remark 7.4. The solutions constructed in the previous proposition are the unique limitsolutions to the problem: no other solution may arise as a pointwise limit of strictly positivesolutions. To see this, note that the approximating equations in the proof hold for strictlypositive solutions, so the comparison principle continues to hold for limit solutions.


8. EXISTENCE OF STRONG SOLUTIONS TO THE OBSTACLE PROBLEM

In this section, we prove Theorem 2.6 by passing to the limit δ ↓ 0 in the penalizedequations of the previous section.

Proof of Theorem 2.6. For δ > 0, let uδ be the weak solution to the Cauchy-Dirichletproblem (7.1) constructed in Proposition 7.3. Since the right-hand side Ψ+ζδ(ψ

m − umδ )is bounded independently from δ, we know from [8, Theorem 1.2] that the functions uδare locally Holder continuous with a quantitative estimate which is uniform in δ. Hencethere is a function u such that uδ → u as δ ↓ 0 locally uniformly in ΩT . We will show thatthis function u is the desired solution of the obstacle problem. First, we note that clearlyu ≥ ψ, since uδ ≥ ψ for all δ > 0. From the a priori estimate of Lemma 7.2, we deducethat for a subsequence still indexed by δ that

(8.1)

∂tuδ ∂tu weakly in L2(0, T ;H−1(Ω)),

∇umδ ∇um weakly in L2(ΩT ,Rn)

as δ ↓ 0. Moreover, we have that um − gm ∈ L2(0, T ;H10 (Ω)) and u(·, 0) = uo in the

H−1(Ω)-sense. By the lower semicontinuity of the L2 norm, we also have

(8.2) ‖∇um‖L2(ΩT ) ≤ lim infδ↓0

‖∇umδ ‖L2(ΩT )

for a subsequence.We let ηδ ∈ C∞0 (Ω, [0, 1]) be a cutoff function such that ηδ = 1 in x ∈ Ω : dΩ(x) ≥

δ, and |∇ηδ| ≤ c/δ. We take ϕ = αη(vm − umδ + δηδ), with α, η, v as in the statementof the theorem, as a test-function in the weak formulation of (7.1)1 to get

(8.3)

∫ T

0

〈∂tuδ, αη(vm − umδ + δηδ)〉dt+

∫

ΩT

∇umδ · ∇(αη(vm − umδ + δηδ)

)dz

=

∫

ΩT

αηΨ+ζδ(ψm − umδ )(vm − umδ + δηδ) dz.

We write the first term in (8.3) as∫ T

0

〈∂tuδ, αη(vm − umδ + δηδ)〉dt =

∫ T

0

〈∂tuδ, αηvm − αηumδ + δαηηδ〉dt.

We have

limδ↓0

∫ T

0

〈∂tuδ, αηvm + δαηηδ〉dt =

∫ T

0

〈∂tu, αηvm〉dt,

by (8.1)1 and the fact that δαηηδ → 0 strongly in H1(Ω) as δ ↓ 0. Two applications ofLemma 3.2 with v ≡ 0 yield in view of the local uniform convergence uδ → u that

limδ↓0

∫ T

0

〈∂tuδ, αηumδ 〉dt = − 1m+1 lim

δ↓0

[ ∫

ΩT

ηα′um+1δ dz + α(0)

∫

Ω

ηum+1o dx

]

= − 1m+1

[ ∫

ΩT

ηα′um+1 dz + α(0)

∫

Ω

ηum+1o dx

]

=

∫ T

0

〈∂tu, αηum〉dt.

Putting all of the above together, we see that the limit as δ ↓ 0 of the first term is

limδ↓0

∫ T

0

〈∂tuδ, αη(vm − umδ + δηδ)〉dt =

∫ T

0

〈∂tu, αη(vm − um)〉dt.



Let us now turn our attention to the second term on the left-hand side of (8.3). By (8.1)2and the fact that uδ → u locally uniformly as δ ↓ 0, we find that

limδ↓0

∫

ΩT

α∇umδ · ∇η(vm − umδ + δηδ) dz =

∫

ΩT

α∇um · ∇η(vm − um) dz.

Again by (8.1)2, we have

limδ↓0

∫

ΩT

αη∇umδ · ∇vm dz =

∫

ΩT

αη∇um · ∇vm dz.

Moreover, from (8.2) and the assumption α, η ≥ 0, we conclude that

− lim infδ↓0

∫

ΩT

αη|∇umδ |2 dz ≤ −∫

ΩT

αη|∇um|2 dz.

Finally, by (8.1)2 and the fact that∇δηδ → 0 in L2(ΩT ), we find that

limδ↓0

∫

ΩT

αη∇umδ · ∇(δηδ) dz = 0.

Combining the above convergences, we conclude for the limit of the second term on theleft-hand side of (8.3) that

lim supδ↓0

∫

ΩT

∇umδ · ∇(αη(vm − umδ + δηδ)

)dz ≤

∫

ΩT

∇um · ∇(αη(vm − um)

)dz.

The next step is to show that the limit as δ ↓ 0 of the right-hand side of (8.3) is nonnegative.Denoting g(s) := ζδ(ψ

m − s), a := vm + δζδ , and b := umδ , and taking into account thatg is decreasing, we have

ζδ(ψm − umδ )(vm − umδ + δηδ)− ζδ(ψm − vm − δηδ)(vm − umδ + δηδ)

= (g(b)− g(a))(a− b) ≥ 0.

Since αηΨ+ is non-negative, this shows for the integrand on the right-hand side of (8.3)that

αηΨ+ζδ(ψm − umδ )(vm − umδ + δηδ) ≥ αηΨ+ζδ(ψ

m − vm − δηδ)(vm − umδ + δηδ).

On the set where ηδ = 1 we have ψm − vm − δηδ ≤ −δ, since v ≥ ψ, and henceζδ(ψ

m − vm − δηδ) = 0. This is the only point where we use the fact that v lies above theobstacle ψ. We combine this with the above inequality to see that for the right-hand sideof (8.3) we have

∫

ΩT

αηΨ+ζδ(ψm − umδ )(vm − umδ + δηδ) dz

≥∫

(x,t)∈ΩT :dΩ(x)<δαηΨ+ζδ(ψ

m − vm − δηδ)(vm − umδ + δηδ) dz

≥ −(∫

(x,t)∈ΩT :dΩ(x)<δ|αηΨ+|2 dz

) 12(∫

ΩT

|vm − umδ + δηδ|2 dz

) 12

,

where in the last line we used Holder’s inequality and the fact that ζδ ≤ 1. The secondfactor is bounded for small values of δ, and the first tends to zero as δ ↓ 0, so that

lim infδ↓0

∫

ΩT

αηΨ+ζδ(ψm − umδ )(vm − umδ + δηδ) dz ≥ 0.

Combining the preceding argumentation, we have so far shown that∫ T

0

〈∂tu, αη(vm − um)〉dt+

∫

ΩT

∇um · ∇(αη(vm − um)

)dz ≥ 0.


This shows that u solves the variational inequality (2.3). Finally, from Lemma 3.2 andLemma 5.2, we conclude that u ∈ C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo. Therefore, u isthe strong solution to the obstacle problem we were looking for.

To show that u is a weak supersolution, we pick an arbitrary nonnegative ϕ ∈ C∞0 (ΩT )and choose

vm = um + ϕ ≥ um ≥ ψm

in (2.3), and cut-off functions α, η with αη ≡ 1 on suppϕ. This gives

(8.4)∫

ΩT

[− u∂tϕ+∇um · ∇ϕ

]dz ≥ 0,

proving that u is a weak supersolution of the PME.Next, we show that u is a weak solution of the PME in z ∈ ΩT : u(z) > ψ(z).

To this aim we pick any function ϕ ∈ C∞0 (ΩT ) which is compactly supported in the setz ∈ ΩT : u(z) > ψ(z) and claim that (8.4) is valid also for this testing function. Withoutloss of generality, we may assume that inf ϕ < 0, for otherwise there is nothing to prove.Now, we choose

vm = um + εϕ

in (2.3), where

0 < ε <infsuppϕ (um − ψm)

− inf ϕ.

Note that v ≥ ψ. As a result, we conclude that (8.4) holds also for sign changing testfunctions with compact support in z ∈ ΩT : u(z) > ψ(z). It follows that u is a weaksolution of the PME in z ∈ ΩT : u(z) > ψ(z).

9. EXISTENCE OF WEAK SOLUTIONS TO THE OBSTACLE PROBLEM

Before we can prove the existence of weak solutions to the obstacle problem for theporous medium equation, we still need the following energy estimate.

Lemma 9.1. Assume that u ∈ Kψ,g is a local weak solution of the obstacle problem forthe porous medium equation in the sense of Definition 2.1. Then, there holds the followingenergy estimate

supt∈[0,T ]

∫

Ω×tum+1 dx+

∫

ΩT

(u2m + |∇um|2

)dz

≤ c[

supt∈[0,T ]

∫

Ω×tgm+1 dx+

∫

Ω

um+1o dx+

∫

ΩT

(g2m + |∇gm|2 + |∂tgm|

m+1m

)dz

]

with a constant c depending on n,m,diam(Ω), and T .

Proof. In Lemma 4.1 we choose the cut-off function α ∈ W 1,∞([0, T ]) by α ≡ 1 on[0, τ − ε] and α(t) := 1

ε (τ − t) on (τ − ε, τ) and α ≡ 0 on [τ, T ], where τ ∈ (0, T ] andε ∈ (0, τ). Letting ε ↓ 0 we obtain

∫

Ω×τI(u, v) dx+

∫

Ωτ

∇um · (∇um −∇vm) dz

≤∫

Ωτ

∂tvm(v − u) dz +

∫

Ω

I(uo, v(·, 0)) dx,(9.1)

for all comparison maps v ∈ K ′ψ,g(ΩT ). Here, we choose v = g, which is admissible,since g ∈ K ′ψ,g(ΩT ). We recall from Lemma 4.1 that I(u, g) ≡ 1

m+1 (um+1 − gm+1) −gm(u− g). By elementary computations, we infer that

12(m+1)u

m+1 − c(m)gm+1 ≤ I(u, g) ≤ um+1 + gm+1.


This yields for any δ ∈ (0, 1) that∫

Ω×τum+1 dx+

∫

Ωτ

|∇um|2 dz

≤∫

Ωτ

∇um · ∇gm + ∂tgm(g − u) dz + c

∫

Ω

(um+1o + gm+1(·, 0) + gm+1(·, τ)

)dx

≤ 1

2

∫

Ωτ

(|∇um|2 + |∇gm|2

)dz +

1

2T

∫

Ωτ

(um+1 + gm+1

)dz

+ c(m,T )

[ ∫

Ωτ

|∂tgm|m+1m dz + sup

t∈[0,T ]

∫

Ω×tgm+1 dx+

∫

Ω

um+1o dx

].

Now, we absorb the term containing |∇um|2 from the right-hand side into the left. We usethe resulting inequality in two directions. In the first term on the left-hand side we take thesupremum over τ ∈ (0, T ], while in the second term we choose τ = T . Finally, we absorbthe term containing um+1 from the right-hand side into the left, taking into account that∫

ΩTum+1 dx ≤ T supτ∈[0,T ]

∫Ω×τ u

m+1 dx. In this way, we find that

supτ∈[0,T ]

∫

Ω×τum+1 dx+

∫

ΩT

|∇um|2 dz

≤ c[ ∫

ΩT

(|∇gm|2 + |∂tgm|

m+1m

)dz + sup

t∈[0,T ]

∫

Ω×tgm+1 dx+

∫

Ω

um+1o dx

],

for a constant c = c(m,T ). Moreover, applying Poincare’s inequality slice-wise to um −gm, we find that∫

ΩT

u2m dz ≤ 2

∫

ΩT

(|um − gm|2 + g2m

)dz

≤ c(n,diam(Ω))

∫

ΩT

|∇um −∇gm|2 dz + 2

∫

ΩT

g2m dz.

Combining the last two inequalities proves the energy estimate.

Now, we have all the prerequisites to prove the main result of the paper.

Proof of Theorem 2.7. The proof will be divided into several steps.

Step 1: Regularization. We begin by approximating the obstacle ψ by functions ψi ∈L∞(ΩT ,R≥0), i ∈ N, satisfying (2.8)2 and

ψmi ∈ L2(0, T ;H1(Ω)

), ∂t(ψ

mi ) ∈ Lm+1

m (ΩT ) and ψmi (·, 0) ∈ H1(Ω)

such that

(9.2) ψmi → ψm in L2(0, T ;H1(Ω)) and ∂tψmi → ∂tψ

m in Lm+1m (ΩT )

and

(9.3) ψmi (·, 0) ψm(·, 0) weakly in H1(Ω),

as i → ∞. Next, we approximate the lateral boundary data g by functions gi ∈L∞(ΩT ,R≥0), i ∈ N, satisfying gi ∈ K ′ψi(ΩT ) and

(9.4) gmi → gm in L2(0, T ;H1(Ω)) and ∂tgmi → ∂tg

m in Lm+1m (ΩT )

and

(9.5) gmi (·, 0) gm(·, 0) weakly in H1(Ω),

as i → ∞. Finally, we approximate the initial values uo by uo,i ∈ L∞(Ω,R≥0) withumo,i ∈ H1(Ω), uo,i ≥ ψi(·, 0) and

(9.6) umo,i → umo strongly in H1(Ω).


By ui ∈ Kψi,gi we denote the local strong solution to the obstacle problem for the PMEwith obstacle ψi, lateral boundary data gi and initial boundary data uo,i constructed inTheorem 2.6. From Lemma 3.2 we know that ui is also a local weak solution in the senseof Definition 2.1, i.e. ui solves the variational inequality

(9.7) 〈〈∂tui, αη(vm − umi )〉〉uo,i +

∫

ΩT

α∇umi · ∇[η(vm − umi )] dz ≥ 0

for every cut-off function in time α ∈W 1,∞([0, T ],R≥0) with α(T ) = 0 and every cut-offfunction in space η ∈ C∞0 (Ω,R≥0) and for every comparison map v ∈ K ′ψi,gi(ΩT ).

Step 2: Energy bounds and weak convergence. In view of the energy bound fromLemma 9.1 and our assumptions on gi, we have

supt∈[0,T ]

∫

Ω×tum+1i dx+

∫

ΩT

(u2mi + |Dumi |2

)dz ≤ c,(9.8)

where c = c(n,m,diam(Ω), T, g) is independent of i ∈ N. Therefore, passing to a (notrelabeled) subsequence, we can assume weak convergence

(9.9) umi um weakly in L2(0, T ;H1(Ω))

as i → ∞ for a limit map um satisfying um − gm ∈ L2(0, T ;H10 (Ω)). Applying

Gagliardo-Nirenberg’s inequality to umi , cf. [6, Proposition I.3.1] and (9.8), we infer that∫

ΩT

uqi dz ≤∫

ΩT

(u2mi + |Dumi |2

)dz

(supt∈[0,T ]

∫

Ω×tum+1i dx

) 2n

≤ c,(9.10)

where q = 2(m+ m+1

n

)> 2m.

Step 3: Strong convergence of the time mollifications as i → ∞. We define mollifica-tions of umi in time by

[[umi ]]h(·, t) := e−thumi (·, 0) +

1

h

∫ t

0

es−th umi (·, s) ds

for h ∈ (0, T ], and similarly, we mollify the obstacles by

[[ψmi ]]h := e−thψmi (·, 0) +

1

h

∫ t

0

es−th ψmi (·, s) ds.

Moreover, in order to restore the obstacle condition, we define the maps wi,h by

wmi,h := [[umi ]]h − [[ψmi ]]h + ψmi and wmh := [[um]]h − [[ψm]]h + ψm.

The definition ensures the obstacle condition wi,h ≥ ψi a.e. in ΩT , and the initial val-ues are wi,h(·, 0) = uo,i. The convergences (9.2), (9.3), (9.6), and (9.9), together withLemma 3.1 (vi) yield [[umi ]]h [[um]]h and [[ψmi ]]h [[ψm]]h weakly in L2(0, T ;H1(Ω))as i→∞, and therefore

(9.11) wmi,h wmh weakly in L2(0, T ;H1(Ω)) as i→∞.The maps vmi,h := wmi,h − ψmi = [[umi ]]h − [[ψmi ]]h satisfy

‖∇vmi,h‖L2(ΩT ) ≤ ‖∇umi ‖L2(ΩT ) + ‖∇ψmi ‖L2(ΩT )

and∂tv

mi,h = ∂t[[u

mi ]]h − ∂t[[ψmi ]]h = − 1

h (wmi,h − umi ) ∈ L2(ΩT ).

From the energy bounds in (9.8), the strong convergence (9.2) of the obstacles and theconvergence (9.11), we thus get L2-bounds for both ∇vmi,h and ∂tvmi,h, uniformly in i ∈ Nfor a fixed h > 0. Rellich’s theorem implies strong subconvergence of vmi,h in L2(ΩT ).This implies

(9.12) wmi,h = vmi,h + ψmi → wmh in L2(ΩT ), as i→∞,


for any h > 0, and this convergence even holds without passing to a subsequence becausewe already have identified the limit by (9.11). In view of Corollary 3.10 we get in the casem ≥ 1 that

(9.13) wi,h → wh in L2m(ΩT ), as i→∞,

for every h > 0.

Step 4: Locally uniform convergence of the time mollifications as h ↓ 0. Since v = wi,hdoes not admit the right boundary values on the lateral boundary, it is not admissible ascomparison function in Lemma 4.1, so that we have to apply Lemma 4.2 instead. For asuitable cut-off construction, we consider a fixed ball BR = BR(xo) with B2R(xo) ⊂ Ω.By the mean value theorem, we may choose a radius r ∈ (R, 2R), possibly depending oni, with

(9.14)∫ T

0

∫

∂Br

|∇umi |2 dxdt ≤ 1

R

∫ T

0

∫

B2R

|∇umi |2 dxdt.

We can also choose r in such a way that the application of Lemma 4.2 is possible with thecomparison function v = wi,h. Keeping in mind that I(ui, wi,h) is non-negative and thatwi,h(0) = uo,i, this yields the bound

−∫

Br×(0,T )

α∂twmi,h(wi,h − ui) dz ≤

∫

Br×(0,T )

α∇umi · (∇wmi,h −∇umi ) dz

+ 2

∫

∂Br×(0,T )

α|∇umi | |wmi,h − umi |dz.

Here, we define the cut-off function α ∈ W 1,∞([0, T ]) by α ≡ 1 on [0, T − ε] andα(t) := 1

ε (T − t) on (T − ε, T ]. Letting ε ↓ 0, we infer

−∫

Br×(0,T )

∂twmi,h(wi,h − ui) dz ≤

∫

Br×(0,T )

∇umi · (∇wmi,h −∇umi ) dz(9.15)

+ 2

∫

∂Br×(0,T )

|∇umi | |wmi,h − umi |dz.

Next, we compute

(9.16) − ∂twmi,h = 1h (wmi,h − umi )− ∂tψmi .

We now plug (9.16) into (9.15) to deduce that∫

Br×(0,T )

(wmi,h − umi )(wi,h − ui) dz

≤ h[ ∫

Br×(0,T )

∂tψmi (wi,h − ui) dz +

∫

Br×(0,T )

∇umi · (∇wmi,h −∇umi ) dz

+ 2

∫

∂Br×(0,T )

|∇umi | |wmi,h − umi |dz]

=: h[I + II + III]

with the obvious labeling of the terms I, II, III. Here, we can estimate

I ≤ ‖∂tψmi ‖Lm+1m

(‖wi,h‖Lm+1 + ‖ui‖Lm+1

)

≤ c‖∂tψmi ‖Lm+1m

(‖ui‖Lm+1 + ‖ψi‖Lm+1 + ‖uo,i‖Lm+1 + ‖ψi(·, 0)‖Lm+1

),

and

II ≤ ‖∇umi ‖L2

(‖∇wmi,h‖L2 + ‖∇umi ‖L2

)

≤ c(‖∇umi ‖2L2 + ‖∇ψmi ‖2L2 + ‖∇uo,i‖Lm+1 + ‖∇ψi(·, 0)‖Lm+1

).


Finally, by (9.14) and the embedding H1(Br) → L2(∂Br) we have

III ≤∫

∂Br(xo)×(0,T )

|∇umi |2 + |wmi,h − umi |2 dz

≤ 1

R

∫

B2R×(0,T )

|∇umi |2 + u2mi + |∇wmi,h|2 + w2m

i,h dz.

From the energy bounds (9.8) and the assumptions (9.2) and (9.3) on ψi we thereby con-clude that I + II + III is bounded independently from i ∈ N and h > 0 and hence∫

Br×(0,T )

(wmi,h − umi )(wi,h − ui) dz ≤ C h

with a constant C independent from i ∈ N and h > 0. We now employ Corollary 3.11 todeduce that in the case m ≥ 1 there holds

(9.17) ‖wi,h − ui‖Lm+1(BR×(0,T )) ≤ Ch1

m+1 for all i ∈ N,

while in the case mc < m < 1, we have

(9.18) ‖wmi,h − umi ‖Lm+1m (BR×(0,T ))

≤ Ch mm+1 for all i ∈ N,

with a constant C independent from i ∈ N and h > 0.

Step 5: Pointwise convergence of the solutions. As in the last step, we consider afixed ball BR = BR(xo) with B2R(xo) ⊂ Ω. In the following, we abbreviate BTR :=BR × (0, T ). We start by treating the case m ≥ 1. For any i ∈ N and h > 0, we canestimate

‖ui − u‖Lm+1(BTR)

≤ ‖ui − wi,h‖Lm+1(BTR) + ‖wi,h − wh‖Lm+1(BTR) + ‖wh − u‖Lm+1(BTR).(9.19)

For an arbitrary ε > 0, we can exploit the uniform convergence (9.17) in order to chooseh > 0 so small that

‖ui − wi,h‖Lm+1(BTR) ≤ Ch1

m+1 ≤ 12ε

holds for all i ∈ N. By Lemma 3.1, we moreover can achieve

‖wh − u‖Lm+1(BTR) ≤ ‖wmh − um‖1m

Lm+1m (BTR)

≤ 12ε

by choosing h > 0 smaller if necessary. Plugging the two preceding estimates into (9.19)and letting i→∞, we arrive at

lim supi→∞

‖ui − u‖Lm+1(BTR) ≤ ε+ limi→∞

‖wi,h − wh‖Lm+1(BTR) = ε,

where the last step follows from (9.13). Since ε > 0 and the ball BR were arbitrary, thisimplies ui → u strongly in Lm+1

loc (ΩT ), as i → ∞. From this, we can conclude that

umi → um strongly in Lm+1m

loc (ΩT ), as i → ∞. At this point, we recall from (9.10) that uiis uniformly bounded in Lq(ΩT ) with q > 2m. Therefore, the interpolation property ofLp-spaces ensures that

(9.20) umi → um strongly in L2loc(ΩT ), as i→∞.

Now, we turn our attention to the case m < 1. Here, we estimate for any i ∈ N andh > 0:

‖umi − um‖L2(BTR)

≤ ‖umi − wmi,h‖L2(BTR) + ‖wmi,h − wmh ‖L2(BTR) + ‖wmh − um‖L2(BTR).(9.21)

For an arbitrary ε > 0, since m+1m > 2 we can exploit the uniform convergence (9.18) in

order to choose h > 0 so small that

‖umi − wmi,h‖L2(BTR) ≤ C‖umi − wmi,h‖Lm+1m (BTR)

≤ Ch mm+1 ≤ 1

2ε


holds for all i ∈ N. By Lemma 3.1, we moreover can achieve

‖wmh − um‖L2(BTR) ≤ 12ε

by choosing h > 0 smaller if necessary. Plugging the two preceding estimates into (9.21)and letting i→∞, we arrive at

lim supi→∞

‖umi − um‖L2(BTR) ≤ ε+ limi→∞

‖wmi,h − wmh ‖L2(BTR) = ε,

where the last step follows from (9.12). Since ε > 0 was arbitrary, this implies

(9.22) umi → um strongly in L2loc(ΩT ), as i→∞.

As a consequence of (9.20), respectively (9.22) we conclude that u ≥ ψ a.e. on ΩT .By passing to another subsequence, we can also assume almost everywhere convergence.We note that this is the crucial step to identify the limit map. If we would know only weakconvergence, the weak limit of ui and the weak limit of umi might not be related. Butknowing (9.20), respectively (9.22), we can now conclude

(9.23)

umi um weakly in L2(0, T ;H1(Ω)), as i→∞,ui u weakly in Lm+1(ΩT ), as i→∞.

Step 6: Passage to the limit. Here, we will to pass to the limit i→∞ in the variationalinequality (9.7). Therefore, we consider α ∈ W 1,∞([0, T ],R≥0) with α(T ) = 0, η ∈C∞0 (Ω,R≥0) and v ∈ K ′ψi,gi(ΩT ). By (9.23) and (9.6) we have

limi→∞〈〈∂tui, αη(vm − umi )〉〉uo,i = 〈〈∂tu, αη(vm − um)〉〉uo .

Again by (9.23) we find that

limi→∞

∫

ΩT

α∇umi · ∇(ηvm) dz =

∫

ΩT

α∇um · ∇(ηvm) dz.

and

− lim infi→∞

∫

ΩT

αη∇umi · ∇umi dz ≤ −∫

ΩT

αη∇um · ∇um dz.

holds true. Finally, the strong convergence (9.20), respectively (9.22) and the weak con-vergence (9.23) imply

limi→∞

∫

ΩT

αumi ∇umi · ∇η dz =

∫

ΩT

αum∇um · ∇η dz.

In conclusion, the above considerations allow us to pass to the limit i → ∞ in the varia-tional inequality (9.7) to deduce that

〈〈∂tu, αη(vm − um)〉〉uo +

∫

ΩT

α∇um · ∇[η(vm − um)] dz ≥ 0

holds for any v, α and η as in Definition 2.1. At this point, we can apply Lemma 5.2 todeduce that u ∈ C0([0, T ];Lm+1(Ω)) and u(·, 0) = uo. This shows that u is the localweak solution to the obstacle problem for the PME we were looking for. Finally, to checkthat u is also a weak supersolution, we note that

∫

ΩT

−u∂tϕ+∇um · ∇ϕdz = limi→∞

∫

ΩT

−ui∂tϕ+∇umi · ∇ϕdz ≥ 0

for all nonnegative ϕ ∈ C∞0 (ΩT ) by (9.23) and the fact that the functions ui are weaksupersolutions.


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VERENA BOGELEIN, DEPARTMENT MATHEMATIK, UNIVERSITAT ERLANGEN–NURNBERG, CAUERSTR.11, 91056 ERLANGEN, GERMANY

E-mail address: [email protected]

TEEMU LUKKARI, DEPARTMENT OF MATHEMATICS AND STATISTICS, P.O. BOX 35 (MAD), 40014UNIVERSITY OF JYVASKYLA, JYVASKYLA, FINLAND


CHRISTOPH SCHEVEN, FAKULTAT FUR MATHEMATIK, UNIVERSITAT DUISBURG-ESSEN, THEA-LEYMANN-STR. 9, 45127 ESSEN, GERMANY












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