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Convergence of conforming approximations for inviscid incompressible Bingham fluid flows and related problems
François Bouchut, Robert Eymard, Alain Prignet
To cite this version:
François Bouchut, Robert Eymard, Alain Prignet. Convergence of conforming approximations for inviscid incompressible Bingham fluid flows and related problems. Journal of Evolution Equations, Springer Verlag, 2014, pp.14(3):635-669, 2014. �10.1007/s00028-014-0231-9�. �hal-00859700�
Convergence of conforming approximations for inviscid incompressible Bingham fluid flows and related problems
F. Bouchut, R. Eymard and A. Prignet∗ September 9, 2013
Abstract
We study approximations by conforming methods of the solution to the variational in- equality h∂tu, v−ui+ψ(v)−ψ(u) ≥ hf, v−ui, which arises in the context of inviscid incompressible Bingham fluid flows and of the total variation flow problem. We propose a general framework involving total variation functionals, that enables to prove convergence of space, or time-space approximations, for steady or transient problems. We consider time implicit, or time implicit regularized (linearized or not) algorithms, and prove their conver- gence for general total variation functionals. Comparison with analytical solutions show the accuracy of the methods.
Keywords. Total variation flow, Bingham fluids, conforming approximations, regularization method, convergence
1 Introduction
A series of practical physical and engineering problems involve the flows of the so-called incom- pressible “Bingham fluids”. For such flows within a domain Ω ⊂RN, N = 2 or 3, the relation between the stress tensorσ(t, x) (seen as aN×N matrix), the pressurep(t, x) and the velocity u(t, x)∈RN is given by (see for example [12] and references therein)
σ=−pIN + κ
|Du|+ 2ν
Du, (1.1)
where κ >0 andν >0 are given physical coefficients (ν is called the viscosity of the fluid), IN is the N×N identity matrix, and where
(Du)ij = 1
2(∂iuj+∂jui), i, j = 1, . . . , N, (1.2) denoting by∂i the partial derivative with respect to the i-th coordinate of a point x∈Ω, and
|Du|2 =
N
X
i,j=1
(Du)2ij. (1.3)
∗Universit´e Paris-Est, Laboratoire d’Analyse et de Math´ematiques Appliqu´ees (UMR 8050), CNRS, UPEMLV, UPEC, F-77454, Marne-la-Vall´ee, France(francois.bouchut, robert.eymard, [email protected])
Since the case of Bingham fluids with negligible viscosity arises in practice [23], our motiva- tion is to provide here numerical methods which remain available in the case when ν is small.
Therefore, this paper is focused on approximate methods allowing for the limit caseν = 0. The incompressibility condition for the fluid reads
divu=
N
X
i=1
∂iui = 0. (1.4)
Assuming a constant density ρ = 1 for the fluid and neglecting the nonlinear convection term, the momentum conservation equation is given by
∂tui−
N
X
j=1
∂jσij =fi, i= 1, . . . , N, (1.5) wheref has values inRN. An initial condition
u(0,·) =u0 (1.6)
is considered, as well as boundary conditions. For simplicity, we assume here homogeneous Neumann boundary conditions, which can be written
N
X
j=1
σijnj = 0 on ∂Ω, i= 1, . . . , N, (1.7) wheren is the outward normal unit vector on the boundary∂Ω of Ω.
In order to provide a formal variational formulation, let us define a set E of regular functions v : Ω → RN such that divv = 0. We then multiply (1.5) by vi, sum over i= 1, . . . , N and integrate on Ω. We get, after an integration by parts in space accounting for the homogeneous Neumann boundary conditions and for the relation divv= 0,
∀v∈E, Z
Ω
∂tu·v+ κ
|Du|+ 2ν
Du:Dv
dx= Z
Ω
f ·vdx, (1.8) where we denote by
Du:Dv=
N
X
i,j=1
(Du)ij(Dv)ij. (1.9)
In (1.8) and (1.5), the ratio Du/|Du|is not always defined. The physical understanding of this term indicates that it should be interpreted whenDu = 0 as “any trace-free symmetric matrix with norm less or equal to one”. In the mathematical language such quantity is called “multi- valued”. This fundamental difficulty makes the approximation of the problem (1.8) already a complex challenge forν >0, and has motivated a large literature, see [12, 15, 16] and references therein.
Since our motivation is to provide discretization methods also available in the caseν = 0, a first step is to rewrite (1.8) under a form which provides a well-posed continuous formulation. We then follow [12], writing a variational inequality giving a rigorous sense to this problem. Let us define, for u, v∈E andx∈Ω,
A(u, v)(x) =Du(x) :Dv(x) and a(u)(x) = A(u, u)(x)1/2
. (1.10)
Then from the Cauchy-Schwarz inequality one has
A(u, v) +A(u, u) =A(u, u+v)≤a(u)a(u+v), which gives
1
a(u)A(u, v)≤a(u+v)−a(u).
Let us also note that
2A(u, v)≤A(u+v, u+v)−A(u, u).
Then, denoting
ψ(v) = Z
Ω
(κ a(v) +ν A(v, v)) dx, (1.11)
the formulation (1.8) implies
∀v∈E, Z
Ω
∂tu·vdx+ψ(u+v)−ψ(u)≥ Z
Ω
f·vdx. (1.12)
Reciprocally, letting v=θw in the previous inequality, we note that A(u+θw, u+θw)−A(u, u) =θ 2A(u, w) +θA(w, w)
, and
a(u+θw)−a(u) =θ2A(u, w) +θA(w, w) a(u+θw) +a(u) .
We then formally recover (1.8) from (1.12) by dividing by θ, lettingθ >0 tend to 0 and letting θ <0 tend to 0. We conclude that (1.12) is more general than (1.8) since it can be written in cases whena(u)(x) = 0 occurs for somex∈Ω. The formulation (1.12) can indeed be understood as saying that the linear formv7→R
(f−∂tu)·v dx belongs to the subdifferential of the convex functional ψ atu. The term in (1.8) appears in fact as the formal differential of the functional ψ.
Note that all the theoretical background of [12] relies on the viscous term, and strongly depends on the assumption ν >0, while here we want to also handle the case ν = 0. We would like to extend the definition of ψ(v) to all functionsv∈L2(Ω)N, thus we write forv∈E
Z
Ω
|Dv|dx= sup
ϕ∈(Cc1(Ω))N2,kϕkL∞(Ω)≤1
Z
Ω
Dv:ϕdx, (1.13)
Z
Ω
|Dv|2dx 1/2
= sup
ϕ∈(Cc1(Ω))N2,kϕkL2(Ω)≤1
Z
Ω
Dv:ϕdx, (1.14)
whereCc1(Ω) is the set ofC1(Ω) functions with compact support in Ω. Integrating by parts, we thus extend the definition of ψ(v) to all functionsv∈L2(Ω)N by defining
ψ(v) =κ sup
w∈VN,N
hv, wi+ν sup
w∈WN,N
hv, wi
!2
∈[0,∞], (1.15)
whereh·,·idenotes theL2(Ω)N scalar product, and where VN,N =
w∈L2(Ω)N,∃ϕ∈ Cc1(Ω)N2
, wi = 1 2
N
X
j=1
∂j(ϕij +ϕji),
N
X
i,j=1
ϕ2ij ≤1 in Ω
, (1.16) WN,N =
w∈L2(Ω)N,∃ϕ∈ Cc1(Ω)N2
, wi = 1 2
N
X
j=1
∂j(ϕij +ϕji), Z
Ω N
X
i,j=1
ϕ2ijdx≤1
. (1.17) We consider the space H of all functions v∈L2(Ω)N such that divv= 0. Then the functional ψ may be defined by (1.15) as a mapping fromH to [0,∞], and we may define the set B of all functions v ∈ H such that ψ(v) < ∞. Indeed, the finiteness of the first term in (1.15) means that Dv is a finite measure on Ω, while the finiteness of the second term (ifν >0) means that Dv∈L2(Ω).
The problem is then to find, for a given T >0, a function u such that u∈L2(0, T;H),
Z T 0
ψ(u(t))dt <∞, ∂tu∈L2(0, T;H), u(0) =u0 and
∀v∈L2(0, T;H), Z T
0
h∂tu(t), v(t)−u(t)i+ψ(v(t))−ψ(u(t))
dt
≥ Z T
0
hf(t), v(t)−u(t)idt.
(1.18)
For such nonlinear monotone variational inequalities in a Hilbert space, involving a convex lower semi-continuous functional (which is precisely the case for the function ψ defined above and valued in ]− ∞,∞]), the theory of [6] applies, giving the existence and uniqueness of the solution.
This theory has been used in several works, applied to the total variation flow, and we refer to [22] for a general exposition on the subject. The total variation flow problem is a scalar problem which consists in looking for the solution u : Ω→Rto the problem
∂tu−div ∇u
|∇u|
=f, (1.19)
with Neumann boundary condition (∇u/|∇u|)·n= 0 on∂Ω. Indeed, this problem may also be written under the form of inequality (1.18), denoting by H the space L2(Ω), and introducing the BV seminorm defined by
ψ(v) = sup
w∈VN
hv, wi, (1.20)
whereh·,·idenotes theL2(Ω) scalar product and where VN =
(
w∈L2(Ω),∃ϕ∈ Cc1(Ω)N
, w= divϕ,
N
X
i=1
ϕ2i ≤1 in Ω )
. (1.21)
The total variation problem is however more “regular” than the inviscid Bingham problem, since it is also monotone inL1(Ω), as proved in [1]. The theory of monotone problems in Banach spaces is provided in [9, 2, 3]. This L1 monotonicity (implying the almost everywhere monotonicity,
via the well-known Crandall-Tartar lemma [10]) enables to use the Kruzkov entropies, as for hyperbolic scalar conservation laws, and therefore to include a transport term. This is done in [4], including convergence results for numerical approximations.
The Bingham problem does not have this L1 structure. Nevertheless, there is a very strong result in [20], that states that in two-dimensions, the Bingham evolution problem has a smooth solution, without viscosity, and including advection.
It is worth noticing that the form of inequality (1.18) includes both linear problems and nonlinear problems such as the p−Laplacian, which shows that it is quite general. Therefore it presents some interest to give general lines for its approximation.
This paper is devoted to the approximation by conforming methods of the solution to (1.18) in the general case of a convex lower semi-continuous (l.s.c. for short) functional on a Hilbert space.
It is then applied to the framework of general total variation functionals, which includes the particular cases of the inviscid Bingham problem and of the total variation flow problem. In this situation we use a regularization procedure and time implicit or linearized implicit integration.
Our results generalize the ones of [13, 14], obtained for the total variation flow, and for more regular data. Our analysis of the linearized implicit scheme, which is the one applied in practice for Bingham flows (see [19, 21]), seems to be new.
The paper is organized as follows. In Section 2, we first recall some properties of the steady problem, following [6], and we propose sufficient conditions for a convergent approximation in a finite dimensional subspace, provided that some interpolation conditions be satisfied (Subsection 2.1). We then provide in Subsection 2.2 the analysis of the regularization method, applied to general total variation functionals. We turn to the transient problem in Section 3: we first state some basic properties in Subsection 3.1, and we then prove the convergence of a time-space conforming approximation of fully implicit type in Subsection 3.2. We provide in Subsection 3.3 the proof of convergence for the implicit regularization method for general total variation functionals, and treat the linearized algorithm in Subsection 3.4. We finally propose in Section 4 the study of numerical convergence in the particular case of the total variation flow. In a first subsection, we show that the problem to be solved in a finite dimensional space is itself approximated in the steady case. In a second subsection we consider the transient case, for which we show the convergence of the implicit method, where the use of a linearization is studied in the case of the regularized problem. Our results extend the ones of [13, 14], since they apply without further hypotheses on the regularity of the continuous solution. A short conclusion is finally given in Section 5.
2 Approximation of the steady problem
2.1 General framework
As stated in the introduction of this paper, we focus on the approximation of a steady version of the problem (1.18), using the framework of [6].
LetH be a Hilbert space, with scalar producth·,·iand normk · k. Letψ : H→]− ∞,∞] be a convex, lower semi-continuous function such that the set B ={v∈H, ψ(v) <∞}(the domain of ψ) is not empty. Classical results in this situation can be found for example in [7].
We first recall the following standard lemma in convex analysis.
Lemma 2.1 A functional ψ is convex, lower semi-continuous with non-empty domain B, and first order positively homogeneous (i.e. for all λ∈ R, v ∈ B, ψ(λv) = |λ| ψ(v)) if and only if there exists a non empty setV ⊂H such that
∀w∈V, −w∈V, (2.1)
and
∀v∈H, ψ(v) = sup
w∈V
hw, vi. (2.2)
In this case, B is a subspace of H, andψ satisfies ψ(u+v)≤ψ(u) +ψ(v).
Proof. It is given in [8] or [18, theorem 3.1.1], but for completeness we give it shortly. Since the “if” part is obvious, considerψ convex, lower semi-continuous with non-empty domain, and first order positively homogeneous. Then applying the homogeneity property to some v0 ∈ B andλ= 0 yields thatψ(0) = 0. We deduce thatψ(λv) =|λ|ψ(v) for allλ∈Randv ∈H, with the convention that 0× ∞ = 0. In particular, ψ(−v) = ψ(v). Applying the Fenchel-Moreau theorem we have thatψis the supremum of all affine functions upper dominated byψ. Consider such an affine functionv7→µ+hw, viwith µ∈R and w∈H. We have
ψ(v)≥µ+hw, vi for all v∈H. (2.3) Applying this inequality to λv, using the homogeneity and lettingλ→ ∞ yields that
ψ(v)≥ hw, vi for all v∈H. (2.4) Applying this to −v gives then that ψ(v) ≥ |hw, vi| ≥ 0. Since (2.3) applied to v = 0 gives µ≤0, we deduce that the linear function v7→ hw, vi, which is upper dominated byψ by (2.4), is greater than the affine functionv7→µ+hw, vi. Therefore,ψis also the supremum of all linear functions upper dominated byψ. In other words, (2.2) holds with
V ={w∈H, ∀v∈H hw, vi ≤ψ(v)}. (2.5)
It is easy to check finally that 0∈V and that (2.1) holds.
Let us notice that the examples provided in the introduction of this paper (without viscosity) fall into the class given by Lemma 2.1.
We recall that, according to the convexity ofψ, the property of lower semi-continuity also holds for the weak topology ofH, which implies that, for any sequence (vn)n∈Nof elements of B that weakly converges to v ∈ H, and such that there exists C ∈ R with ψ(vn) ≤ C for all n ∈ N, thenψ(v)≤lim inf
n ψ(vn), which implies thatv∈B.
Let α > 0 andf ∈H be given. The aim of this section is to approximate the solution to the following problem: find u such that
u∈B,
∀v∈B, αhu, v−ui+ψ(v)−ψ(u)≥ hf, v−ui. (2.6) Note that (2.6) may also be written as
αu+∂ψ(u)3f, (2.7)
with introducing the subdifferential of the function ψatu,
∂ψ(u) ={w∈H, ∀v∈H, ψ(v)≥ψ(u) +hw, v−ui}. (2.8) Then according to [6] we have the following result.
Lemma 2.2 There is existence and uniqueness of the solution u to the problem (2.6), which moreover satisfies
u= argmin
v∈B
J(v), (2.9)
where J : B →R is defined by J(v) = α
2kvk2+ψ(v)− hf, vi, ∀v∈B. (2.10) Corollary 2.3 There exists u0 ∈B such that ∂ψ(u0) is not empty.
Proof. Take α = 1 and f = 0. Then the solution u obtained by Lemma 2.2 satisfies (2.7),
which implies that ∂ψ(u) is not empty.
Remark 2.4 In the case whenψ is first order positively homogeneous (the case of Lemma 2.1), thenu∈B is solution to the problem (2.6) if and only if
u∈H
∀v∈H, αhu, vi+ψ(v)≥ hf, vi, (2.11) and
αkuk2+ψ(u) =hf, ui. (2.12)
It indeed suffices to let v = 0 and v = 2u in (2.6) for obtaining (2.12), which also shows that u∈B. This characterization is used in the examples for exhibiting an analytical solution.
Remark 2.5 If we have two solutions u1 andu2 to (2.6) associated to two different right-hand sides f1 and f2, then one hasku2−u1k ≤ kf2−f1k/α. This is obtained by taking v=u2 in the formulation (2.6) for u1, taking v=u1 in the formulation (2.6) for u2, and adding the results.
This contraction property is in the heart of the theory of monotone operators (here ∂ψ is the monotone operator in (2.7)).
Let us now introduce a reduction argument. Take u0 ∈B such that ∂ψ(u0) is not empty, and pick some w ∈ ∂ψ(u0). Then one has ψ(v) ≥ ψ(u0) +hw, v−u0i for all v ∈ H. Therefore, setting for v∈H
ψ(v) =e ψ(v+u0)−ψ(u0)− hw, vi, (2.13) the functional ψe : H →]− ∞,∞] is convex and lower semi-continuous, and satisfies ψe ≥ 0, and ψ(0) = 0 (implying 0e ∈Be={v∈H,ψ(v)e <∞}). For a∈R(chosen later) and for v∈H, define then
J(v) =e J(v+u0)−a= α
2kv+u0k2+ψ(v+u0)− hf, v+u0i −a.
Then the minimum of Jeis obtained at the point ue = u−u0, where u is the solution to the problem (2.6). Settingfe=f−w−αu0, we have
Je(v) = α
2kvk2+ψ(v)e − hf , vie +ψ(u0) + Dα
2u0−f, u0
E
−a, thus choosinga=ψ(u0) +hα2u0−f, u0i, the problem is to find the minimum of
J(v) =e α
2kvk2+ψ(v)e − hf , vi.e Therefore, eu is the solution to the problem
ue∈B,e
∀v∈B,e αhu, ve −eui+ψ(v)e −ψ(eu)e ≥ hf , ve −ui.e (2.14) Indeed, (2.14) can also be deduced directly from (2.6). We conclude that we may assume, without loss of generality, that ψ≥0 andψ(0) = 0. This is done in the following Hypothesis.
Hypothesis 2.6 We consider the following assumptions:
1. H is a Hilbert space,
2. ψ : H→]− ∞,∞]is convex and lower semi-continuous, 3. ψ(v)≥0 for all v∈H, and ψ(0) = 0.
Note that the third assumption ensures that if f = 0, the solution to (2.6) isu= 0.
Assuming Hypothesis 2.6 (that implies 0∈B={v∈H, ψ(v)<∞}), letHb be a closed subspace of H and let Bb =Hb∩B. Note that 0 ∈Bb (in the examples, B is a subspace ofH and Hb is a finite dimensional subspace of H). We define the approximate problem as: findubsuch that
bu∈B,b
∀v∈B,b αhu, vb −uib +ψ(v)−ψ(u)b ≥ hf, v−ui.b (2.15) It is then immediate to get the existence and uniqueness ofubsolution to (2.15).
Lemma 2.7 Under Hypothesis 2.6, let α >0 and f ∈H be given. Let Hb be a closed subspace of H and let Bb =Hb ∩B. Then there exists one and only one solution to the problem (2.15), which moreover satisfies
ub= argmin
v∈Bb
J(v), (2.16)
where J is defined by (2.10).
Proof. It suffices to consider f, the orthogonal projection ofb f on H. Then (2.15) is identicalb to (2.6) replacing ψ by its restriction to Hb and f by fb, since hf , vib = hf, vi for all v ∈ H.b Therefore, Lemma 2.2 gives the existence and uniqueness of the solution to (2.15).
The scheme (2.15) leads to the resolution of a convex minimization problem in a finite dimen- sional space. There are many numerical methods well-suited to that (gradient or conjugate gra- dient methods for example). We analyze a different type of method, the regularization method, in the particular case of total variation functionals in Subsection 2.2, based on the particular form of the function ψ. Let us however give the following error estimate result for the scheme (2.15) in the general case.
Lemma 2.8 Under Hypothesis 2.6, letα >0 and f ∈H be given and letu∈B be the solution to the problem (2.6). Let Hb be a closed subspace of H, let Bb = Hb ∩B and let ub ∈ Bb be the solution to (2.15). Then
ku−buk ≤2 Rbu,f α
!1/2
, (2.17)
and
ψ(u)−ψ(u)b
≤6kfk Rbu,f α
!1/2
, (2.18)
with
Rbu,f = inf
v∈Bb
kfk kv−uk+ ψ(v)−ψ(u)+
, (2.19)
where we denote for all x∈R, x+= max(0, x).
Proof. Let us first observe that ubsatisfies, setting v= 0 in (2.15) and using thatψ(0) = 0, αkukb 2+ψ(u)b ≤ hf,bui,
which gives, according to the Young inequality, α
2kukb 2+ψ(bu)≤ 1
2αkfk2. (2.20)
Since ψ≥0 this shows that
kuk ≤b 1
αkfk. (2.21)
We similarly write, from taking v= 0 in (2.6), kuk ≤ 1
αkfk. (2.22)
From (2.15), we get for anyv∈Bb
αhbu, u−bui+ψ(u)−ψ(u) +b R(v)≥ hf, u−ui,b with
R(v) =αhu, vb −ui+ψ(v)−ψ(u) +hf, u−vi. (2.23) Hence, taking the infimum, we get
αhbu, u−bui+ψ(u)−ψ(u) + infb
v∈Bb
R(v)≥ hf, u−ui,b (2.24) while from (2.23) and (2.19) we have
inf
v∈Bb
R(v)≤2Rbu,f. (2.25)
We have also, lettingv= 0 in (2.23) and using that ψ(0) = 0, ψ≥0 and (2.21)-(2.22), inf
v∈Bb
R(v)≤R(0)≤ 2
αkfk2. (2.26)
Taking v=bu in (2.6), we get
αhu,ub−ui+ψ(bu)−ψ(u)≥ hf,ub−ui. (2.27) Adding the inequalities (2.24) and (2.27) yields
αku−buk2 ≤ inf
v∈Bb
R(v), (2.28)
which provides (2.17), using (2.25). We deduce also that the right-hand side of (2.28) is non- negative. We then write, again using (2.24) and (2.21),
ψ(bu)−ψ(u)≤2kfkku−ukb + inf
v∈Bb
R(v),
and similarly from (2.27)
ψ(u)−ψ(bu)≤αhu,ub−ui+hf, u−bui ≤2kfkku−uk.b We deduce, with (2.28) and (2.26), that
ψ(u)−ψ(u)b
≤2kfkku−buk+ inf
v∈Bb
R(v)
≤r inf
v∈Bb
R(v) 2kfk
√α +r inf
v∈Bb
R(v)
!
≤ 4kfk
√α r
inf
v∈Bb
R(v),
which leads to (2.18) using (2.25).
We deduce the following convergence result for the approximation method.
Corollary 2.9 Under Hypothesis 2.6, let (Hbn)n∈N be a sequence of closed subspaces of H, and let, for all n∈N, Bbn=Hbn∩B. We assume that
n→∞lim inf
w∈Bbn
kw−vk+ ψ(w)−ψ(v)+
= 0, ∀v∈B. (2.29)
Let, for alln∈N,ubn∈Bbnbe the unique solution ubto (2.15) withBb=Bbn. Then, ubn converges in H to the unique solution u∈B to the problem (2.6) asntends to ∞ and ψ(ubn) converges to ψ(u).
2.2 Total variation functionals
In this subsection we apply the framework of the previous section to the case of functionalsψof total variation type, generalizing (1.20), or (1.15) withν = 0, in the introduction of this paper.
Hypothesis 2.10 We assume Hypothesis 2.6. Moreover, we assume that there exists a subspace H1 ⊂ H, an open set Ω ⊂ RN with N ≥ 1, a nonnegative Borel measure µ on Ω such that
µ(Ω)<∞, and a symmetric bilinear mappingA : H1×H1 →L1µ(Ω)such thatA(u, u)(x)≥0 for µ a.e. x∈Ω, for all u∈H1, and
∀u∈H1, ψ(u) = Z
Ω
a(u)(x)dµ, (2.30)
denoting by a(u)(x) = (A(u, u)(x))1/2 for µ a.e. x ∈ Ω. In particular, ψ is finite on H1, i.e.
H1⊂B.
The two examples we are interested in obviously satisfy these conditions, with the following choices.
Example 2.11 (Total variation flow) The Hilbert space is taken H = L2(Ω), with Ω a bounded open set inRN,ψis as in (1.20), (1.21),B =L2(Ω)∩BV(Ω),µis the Lebesgue measure onΩ, H1 is any space such that C∞(Ω)⊂H1 ⊂H1(Ω), andA(u, v)(x) =∇u(x)· ∇v(x).
Example 2.12 (Inviscid Bingham flow) The Hilbert space is takenH ={u∈L2(Ω)N, divu= 0} with Ω a bounded open set in RN, ψ is as in (1.15), (1.16) with ν = 0, B = {u ∈ L2(Ω)N, divu = 0, Du ∈ M(Ω)}, where Du = (∇u + (∇u)t)/2, M(Ω) is the space of fi- nite measures over Ω, µ is κ times the Lebesgue measure on Ω, H1 is any space such that {u∈C∞(Ω)N,divu= 0} ⊂H1⊂ {u∈H1(Ω)N,divu= 0}, andA(u, v)(x) =Du(x) :Dv(x).
We next analyze the algorithm byregularizationfor computing an approximate solution to (2.16) (which converges to the continuous solution according to Corollary 2.9).
Lemma 2.13 Under Hypothesis 2.10, let α > 0 and f ∈ H. Let Hb be a finite dimensional subspace of H1. Then, forε >0, there exists one and only one function buε∈Hb solution to
ubε∈H,b
∀v∈H, αhb buε, vi+ Z
Ω
A(ubε, v)(x)
ε+a(ubε)(x)dµ=hf, vi. (2.31) Moreover, denoting by bu∈Hb the unique solution to (2.16), we have
kubε−uk ≤b
rµ(Ω)ε
α , (2.32)
and
|ψ(buε)−ψ(bu)| ≤2kfk
rµ(Ω)ε
α + 3µ(Ω)ε. (2.33)
The estimates (2.32), (2.33) provide, with Corollary 2.9, the convergence of (ubε)n to u under the conditions (2.29) andε→0. Note that sinceHbn⊂H1, it is necessary for having (2.29) that the following condition holds:
∀v ∈B, inf
w∈H1
kw−vk+ ψ(w)−ψ(v)+
= 0. (2.34)
This condition is proved to hold true for the total variation flow and for the Bingham flow in the appendix. Then to recover (2.29) from (2.34), it suffices to require that
n→∞lim inf
w∈Hbn
kw−vk+ ψ(w)−ψ(v)+
= 0, ∀v∈H1, (2.35)
which is easily fulfilled.
Proof of Lemma 2.13. Let us begin with the proof of uniqueness. Consider two elements u1
and u2 ofHb satisfying (2.31). Subtracting (2.31) withu1 and u2, and takingv=u1−u2 leads to
αku1−u2k2+ Z
Ω
A(u1, u1−u2)(x)
ε+a(u1)(x) −A(u2, u1−u2)(x) ε+a(u2)(x)
dµ= 0. (2.36) We have according to the Cauchy-Schwarz inequalityA(u1, u2)(x)≤a(u1)(x)a(u2)(x) forµa.e.
x∈Ω. Sinces→s/(ε+s) is strictly increasing from [0,∞) to [0,1), omitting for simplicity the argumentx we get
0≤
a(u1)
ε+a(u1) − a(u2) ε+a(u2)
a(u1)−a(u2)
≤
A(u1, u1−u2)
ε+a(u1) −A(u2, u1−u2) ε+a(u2)
. Using this information in (2.36) we obtain
αku1−u2k2 ≤0,
which concludes the proof of uniqueness. Turning to the existence proof, we consider the operator T :Hb →Hb defined for u∈Hb by
∀v∈H,b hT(u), vi= Z
Ω
A(u, v)(x)
ε+a(u)(x)dµ. (2.37)
SinceAis bilinear andHb is finite dimensional, the restriction ofAtoHb×Hb is continuous, with values inL1µ(Ω). Therefore, the operatorT is continuous fromHb to itself. Then, we have that any solution to (2.31), in which a factorλ∈[0,1] is introduced in front of the integral, satisfies the estimate
α
2kubεk2+λ Z
Ω
a(ubε)(x)2
ε+a(buε)(x)dµ≤ 1
2αkfk2. (2.38)
Since, for λ = 0, the problem is a finite dimensional invertible linear problem, we get by a standard topological degree argument that there exists at least one solution to the problem for λ= 1. Let us now turn to the proof of (2.32). We remark that, foru, v∈H1 andµ a.e. x∈Ω, omitting for simplicity the argumentx, applying the Cauchy-Schwarz inequality and (5.6),
A(u, v−u)
ε+a(u) ≤ a(u)a(v)−a(u)2
ε+a(u) ≤ε+a(v)−a(u). (2.39) Using (2.31) where we replacev by v−ubε, we get
∀v∈H,b
αhubε, v−ubεi+µ(Ω)ε+ψ(v)−ψ(buε)≥ hf, v−buεi. (2.40) Lettingv =bu (the solution to (2.16)) in (2.40), we obtain
ψ(buε)−ψ(bu)≤ hαubε−f,ub−ubεi+µ(Ω)ε. (2.41) Then, taking v=ubε in (2.15) we get
ψ(bu)−ψ(ubε)≤ hαub−f,buε−bui. (2.42)
Adding the two inequalities, we then get (2.32). Let us now turn to the proof of (2.33). Setting v= 0 in (2.40), we get
αkubεk2+ψ(ubε)≤µ(Ω)ε+hf,buεi, which implies, sincehf,buεi ≤ 2α1 kfk2+α2kubεk2,
α2kbuεk2 ≤2αµ(Ω)ε+kfk2≤(kfk+p
2αµ(Ω)ε)2. This leads with (2.41) to
ψ(ubε)−ψ(u)b ≤(2kfk+p
2αµ(Ω)ε)kbuε−buk+µ(Ω)ε.
Similarly, using (2.21) in (2.42) yields
ψ(u)b −ψ(ubε)≤2kfkkubε−buk.
Using (2.32), we finally get (2.33).
The last step is to approximate the solution ubε to the regularized problem (2.31), since it is nonlinear. For given initialu(0)∈Hb (and fixedε >0), we now define the sequence (u(k))k∈Nby
u(k+1) ∈H,b
∀v∈H,b αhu(k+1), vi+ Z
Ω
A(u(k+1), v)(x)
ε+a(u(k))(x) dµ=hf, vi. (2.43) We have the following result.
Lemma 2.14 Under Hypothesis 2.10, let α > 0 and f ∈ H. Let Hb be a finite dimensional subspace of H1, and let u(0) ∈ Hb and ε > 0 be given. Then there exist a unique sequence (u(k))k∈N defined by (2.43). Moreover, as k→ ∞, u(k) converges to buε, the unique solution to (2.31), and ψ(u(k)) converges to ψ(ubε).
Proof. The estimate α
2ku(k+1)k2+ Z
Ω
a(u(k+1))(x)2
ε+a(u(k))(x)dµ≤ 1
2αkfk2, (2.44)
obtained by letting v =u(k+1) in (2.43) and applying the Young inequality, shows that with a null right-hand side, the square linear system to be solved has the unique solution 0. Hence it is invertible, showing the existence and uniqueness of the sequence (u(k))k∈N.
We then let v=u(k+1)−u(k) in (2.43). Since the Cauchy-Schwarz inequality implies, µa.e. in Ω,
a(u(k+1)) ε+a(u(k))
a(u(k+1))−a(u(k))
≤ A(u(k+1), u(k+1)−u(k)) ε+a(u(k)) , we get from (5.4) proved in Lemma 5.1 that
α 2
ku(k+1)k2+ku(k+1)−u(k)k2− ku(k)k2 +
Z
Ω
Fε
a(u(k+1))(x)
−Fε
a(u(k))(x)
+ a(u(k+1))(x)−a(u(k))(x)2
2(ε+a(u(k))(x))
! dµ
≤ hf, u(k+1)−u(k)i.
(2.45)
Therefore, summing the above inequality for k = 0, . . . , m and applying the Young inequality to the right-hand side, we get
α 2
ku(m+1)k2+
m
X
k=0
ku(k+1)−u(k)k2− ku(0)k2
+ Z
Ω
Fε
a(u(m+1))(x)
−Fε
a(u(0))(x)
+1 2
m
X
k=0
a(u(k+1))(x)−a(u(k))(x)2
ε+a(u(k))(x)
! dµ
≤ 1
αkfk2+α
4ku(m+1)k2− hf, u(0)i.
(2.46)
This proves on one hand that α4ku(m+1)k2+R
ΩFε(a(u(m+1))(x))dµ remains bounded indepen- dently of m, and using (5.3) proved in Lemma 5.1, we get thatψ(u(m+1)) remains bounded.
This proves on the other hand that the two series in the left-hand side of the above inequality converge, and therefore that
ku(k+1)−u(k)k →0, ask→ ∞. (2.47)
We next observe thatv7→ ka(v)kL2
µ(Ω) is a semi-norm onH1. SinceHb has finite dimension, this implies that there exist a constantMcsuch that
∀v∈H,b ka(v)kL2
µ(Ω)≤Mkvk.c (2.48)
Writing
ka(u(k+1))−a(u(k))kL2
µ(Ω)≤ ka(u(k+1)−u(k))kL2
µ(Ω)≤Mcku(k+1)−u(k)k, (2.49) we get that a(u(k+1))−a(u(k)) → 0 in L2µ(Ω). Using again that the dimension of Hb is finite, we deduce from (2.44) that there exists a subsequence of (u(k))k∈N, again denoted (u(k))k∈N, strongly convergent in the finite dimensional vector space to some elementu∈H. This impliesb that (a(u(k)))k∈N converges in L2µ(Ω) toa(u), and that ψ(u(k)) tends to ψ(u). Using (2.47), we may then pass to the limit in (2.43) for this extracted subsequence, and we get that the limitu satisfies (2.31). Since the solution to (2.31) is unique, we get that the whole sequence (u(k))k∈N
converges to this solution.
Lemma 2.15 With the assumptions of Lemma 2.14, assume further that
∀v∈H,b ka(v)kL∞
µ(Ω)<∞. (2.50)
Then the convergence of u(k) to buε as k→ ∞ is asymptotically geometric with ratio arbitrarily close to
ka(ubε)kL∞ µ(Ω)
ε+ka(ubε)kL∞
µ(Ω)
<1. (2.51)
Proof. With the assumption (2.50), v 7→ ka(v)kL∞
µ(Ω) is a semi-norm on H, which is finiteb dimensional. Thus there exists a constant Gb such that
∀v∈H,b ka(v)kL∞
µ(Ω)≤Gkvk.b (2.52)
It implies thata(u(k))→a(buε) inL∞µ(Ω). Then, take successively v=u(k+2)−u(k+1) in (2.43), and in (2.43) withk incremented of 1. The difference yields
αku(k+2)−u(k+1)k2+ Z
Ω
A(u(k+2), u(k+2)−u(k+1))(x) ε+a(u(k+1))(x) dµ
− Z
Ω
A(u(k+1), u(k+2)−u(k+1))(x)
ε+a(u(k))(x) dµ= 0.
(2.53)
We deduce omitting thex that αku(k+2)−u(k+1)k2+ Z
Ω
a(u(k+2)−u(k+1))2 ε+a(u(k+1)) dµ
= Z
Ω
A(u(k+1), u(k+2)−u(k+1))
1
ε+a(u(k)) − 1 ε+a(u(k+1))
dµ
≤ Z
Ω
a(u(k+1))a(u(k+2)−u(k+1))
a(u(k+1))−a(u(k)) (ε+a(u(k)))(ε+a(u(k+1)))dµ
≤ 1 2
Z
Ω
a(u(k+2)−u(k+1))2
ε+a(u(k+1)) +a(u(k+1))2
a(u(k+1))−a(u(k))
2
(ε+a(u(k)))2(ε+a(u(k+1)))
! dµ,
(2.54)
and therefore that
2αku(k+2)−u(k+1)k2+ Z
Ω
a(u(k+2)−u(k+1))2 ε+a(u(k+1)) dµ
≤ Z
Ω
a(u(k+1))2 a(u(k+1)−u(k))2
(ε+a(u(k)))2(ε+a(u(k+1)))dµ.
(2.55)
Now, since a(u(k))→a(buε) in L∞µ(Ω), for givenη >0 there exist somek0 such that
∀k≥k0, µ a.e.in Ω, a(u(k+1))2
(ε+a(u(k)))(ε+a(u(k+1))) ≤(r+η)2, (2.56) wherer is the left-hand side of (2.51). Then, fork≥k0 one has
2αku(k+2)−u(k+1)k2+ Z
Ω
a(u(k+2)−u(k+1))2 ε+a(u(k+1)) dµ
≤(r+η)2 Z
Ω
a(u(k+1)−u(k))2 ε+a(u(k)) dµ,
(2.57)
which implies that for k≥k0 Z
Ω
a(u(k+1)−u(k))2
ε+a(u(k)) dµ≤(r+η)2(k−k0) Z
Ω
a(u(k0+1)−u(k0))2
ε+a(u(k0)) dµ. (2.58)
Plugging this in (2.57) we deduce that for k≥k0,
ku(k+2)−u(k+1)k ≤C(r+η)k−k0, (2.59)
whereC does not depend onk. Ifr+η <1 we finally write fork≥k0 ku(k)−buεk ≤
∞
X
k0=k
ku(k0+1)−u(k0)k ≤C(r+η)k−k0, (2.60)
which proves the claim.
3 Transient problem
3.1 Continuous framework
Assuming Hypothesis 2.6 (we again notice that the assumptions ψ ≥0 and ψ(0) = 0 bring no restriction to generality in Problem (3.3) below), let T > 0 be given. The space L2(0, T;H) is defined as the Hilbert space of all measurable, almost everywhere defined functions u (in the so-called “Bochner integral” sense, recalled for example in [11]) from (0, T) to H such that kuk2L2(0,T;H) :=RT
0 ku(t)k2dt <∞. The setBT is defined as BT =
u∈L2(0, T;H);
Z T 0
ψ(u(t))dt <∞
. (3.1)
For anyu∈L1loc(0, T;H), we denote by ∂tu∈L1loc(0, T;H) the time weak derivative of u when it exists, that is∂tu∈L1loc(0, T;H) is such that
∀ϕ∈Cc1(]0, T[), Z T
0
ϕ0(t)u(t)dt=− Z T
0
ϕ(t)∂tu(t)dt.
We recall that C∞([0, T];H) is dense in the Hilbert spaces L2(0, T;H) and in H1(0, T;H) :=
{u ∈L2(0, T;H);∂tu ∈L2(0, T;H)}, and that H1(0, T;H) ⊂C0([0, T];H) (see [11]). We also recall that, for allu, v∈H1(0, T;H), we have for anyt1, t2∈[0, T]
Z t2
t1
h∂tu(t), v(t)idt+ Z t2
t1
hu(t), ∂tv(t)idt=hu(t2), v(t2)i − hu(t1), v(t1)i. (3.2) Letf ∈L2(0, T;H) and u0 ∈B be given. We look in this section for a functionu such that
u∈H1(0, T;H)∩BT, u(0) =u0,
Z T
0
h∂tu(t), v(t)−u(t)idt+ Z T
0
ψ(v(t))−ψ(u(t)) dt
≥ Z T
0
hf(t), v(t)−u(t)idt, ∀v ∈BT.
(3.3)
Let us remark that the inequality (3.3) implies Z t2
t1
h∂tu(t), v(t)−u(t)idt+ Z t2
t1
ψ(v(t))−ψ(u(t)) dt
≥ Z t2
t1
hf(t), v(t)−u(t)idt, ∀v∈Bt1,t2, ∀t1 < t2 ∈[0, T],
(3.4)
where Bt1,t2 denotes the set of all v ∈ L2(t1, t2;H) such that Rt2
t1 ψ(v(t))dt < ∞. It indeed suffices, for a given ˜v ∈Bt1,t2, to define v(t) = ˜v(t) for a.e. t ∈]t1, t2[ and v(t) = u(t) for a.e.
t∈]0, t1[∪]t2, T[, and then use thisv ∈BT as test function in (3.3). Then, since the restriction of v∈BT to any interval ]t1, t2[ belongs toBt1,t2, we conclude that the inequality (3.3) is equivalent to
h∂tu(t), v(t)−u(t)i+ψ(v(t))−ψ(u(t))
≥ hf(t), v(t)−u(t)i, for a.e. t∈]0, T[, ∀v∈BT. (3.5) One can derive even a stronger formulation. Let E ⊂]0, T[ a set such that meas(]0, T[\E) = 0 and for all t∈E, ast1, t2→t with 0< t1 < t < t2 < T,
Z t2
t1
ψ(u(s))ds→ψ(u(t)), Z t2
t1
h∂tu(s)−f(s), u(s)ids→ h∂tu(t)−f(t), u(t)i, Z t2
t1
∂tu(s)−f(s)
ds * ∂tu(t)−f(t) weakly inH,
where the bar integral denotes the average, i.e. the integral normalized by the volume. Dividing (3.4) by t2−t1, passing to the limit ast1, t2 →t for all t∈E and taking v equal to a constant element ofB, we get
h∂tu(t), v−u(t)i+ψ(v)−ψ(u(t))
≥ hf(t), v−u(t)i, ∀v∈B, for a.e. t∈]0, T[. (3.6) This strong formulation is therefore equivalent to (3.5) and to (3.3). It may also be written as
∂tu(t) +∂ψ(u(t))3f(t) for a.e. t∈]0, T[, (3.7) with the subdifferential defined in (2.8). However, in this paper we shall not use this formulation.
The study of existence and uniqueness for Problem (3.3) is given in [22, Theorem 20], following [6], in the case when f = 0, using the semi-group approach and the Yosida regularization.
Since we focus on the approximation of this problem, we indeed recover the existence through the convergence of a semi-discrete (in time) approximation. A convergence result is however established for general time-space approximations.
The following lemma is classical. We recall its short proof.
Lemma 3.1 Under Hypothesis 2.6, let T > 0 be given, and f ∈ L2(0, T;H). If u1 and u2
are two solutions to Problem (3.3), with possibly different initial data, then ku2(t)−u1(t)k is nonincreasing in [0, T]. In particular, there exists at most one solution to Problem (3.3) with a given initial data u0 ∈B.
Proof. Let u1 and u2 be two solutions to Problem (3.3), with possibly different intial data.
Choosing, for given t1 < t2 ∈ [0, T], v = u2 (respectively v = u1) in (3.4) with u = u1
(respectively u=u2), and adding the two obtained inequalities, we get Z t2
t1
h∂tu1(t)−∂tu2(t), u2(t)−u1(t)idt≥0.
Taking into account (3.2), it yields 1
2ku2(t2)−u1(t2)k2 ≤ 1
2ku2(t1)−u1(t1)k2,
which proves the claim.
Lemma 3.2 Under Hypothesis 2.6, let T >0 be given, f ∈L2(0, T;H) and u0 ∈B. Then any solution u to Problem (3.3) verifies the a priori estimates
ku(t)k ≤ ku0k+ 2√
tkfkL2(0,T;H), for all t∈[0, T], (3.8) Z T
0
ψ(u(t))dt≤ 1 2
ku0k+ 2√
TkfkL2(0,T;H)
2
. (3.9)
Proof. We take v= 0 in (3.4), which gives for allt1 < t2 ∈[0, T] 1
2ku(t2)k2+ Z t2
t1
ψ(u(t))dt≤ 1
2ku(t1)k2+ Z t2
t1
hf(t), u(t)idt. (3.10) In particular this implies by taking t1= 0 that for allt∈[0, T],
ku(t)k2 ≤ ku0k2+ 2kfkL2(0,T;H)
Z t 0
ku(s)k2ds 1/2
. (3.11)
Definingϕ(t) =Rt
0ku(s)k2ds, it satisfies the differential inequality ϕ0(t)≤ ku0k2+ 2kfkp ϕ(t), which implies that
d dt
ϕ(t) ku0k2+ 2kfkp
ϕ(t)
!
≤1.
Taking into account that ϕ(0) = 0, we deduce that ϕ(t) ku0k2+ 2kfkp
ϕ(t) ≤t, i.e.
ϕ(t)−2tkfkp
ϕ(t)−tku0k2 ≤0, which yields
pϕ(t)≤tkfk+ t2kfk2+tku0k21/2
.
Plugging this into (3.11) we obtain
ku(t)k2 ≤ ku0k2+ 2kfk
tkfk+ t2kfk2+tku0k21/2
≤ ku0k2+ 4tkfk2+ 2√
tku0kkfk
≤ ku0k+ 2√ tkfk2
,
(3.12)
which proves (3.8). Finally, coming back to (3.10) we easily get (3.9).
Remark 3.3 A generalization of the previous result is as follows. Under Hypothesis 2.6, let T >0 be given, f1, f2 ∈L2(0, T;H) and u01, u02 ∈B. Then two solutions u1, u2 to the Problem (3.3) with respective data verify
ku2(t)−u1(t)k ≤ ku02−u01k+ 2√
tkf2−f1kL2(0,T;H), ∀t∈[0, T]. (3.13) This is obtained by taking as test function v =u2 for the formulation associated to u1, taking v =u1 as test function for the formulation associated to u2, adding the results and arguing as above by a Gronwall lemma.
Remark 3.4 The lemmas 3.1 and 3.2 indeed only use that u0 ∈ H, and not that u0 ∈ B.
However, in order to get an estimate on ∂tu, the propertyψ(u0)<∞ is needed, as we shall see below.
3.2 Time-space implicit approximation
In this subsection we consider the space approximation (2.15), but applied to the transient prob- lem of the previous subsection. We thus consider the following approximate method. Assuming Hypothesis 2.6, let Hb be a closed subspace of H, and let Bb = Hb ∩B. We first approximate u0 ∈B by some
bu0 ∈B,b satisfying ψ(bu0)≤G, ψ(u0)≤G, (3.14) for some constant G≥0. We then take n∈N?, we define the timestepτ =T /n, the sequence (fk)k=1,...,n of elements ofH and the function fn(t) by
fk= 1 τ
Z kτ (k−1)τ
f(t)dt, ∀k= 1, . . . , n,
fn(t) =fk, for a.e. t∈](k−1)τ, kτ[, ∀k= 1, . . . , n.
(3.15)
The sequence (buk)k=1,...,n is then defined by ubk ∈B,b
hDku, wb −buki+ψ(w)−ψ(buk)≥ hfk, w−ubki, ∀w∈B,b ∀k= 1, . . . , n, (3.16) where (Dku)b k=1,...,n is expressed in terms ofubk and ubk−1 by
Dkub= ubk−ubk−1
τ , k = 1, . . . , n. (3.17)
The existence and uniqueness of (ubk)k=1,...,n solution to (3.16) is obvious since this problem has the same form as (2.15) with α= 1/τ andf =fk+buk−1/τ. Let us denote by
u(t) =b ubk andDtu(t) =b Dkbu, ∀t∈](k−1)τ, kτ], ∀k= 1, . . . , n,
u(0) =b ub0, (3.18)
henceu(t) is defined for allb t∈[0, T]. We have the following estimates on buand Dtu.b
Lemma 3.5 Under Hypothesis 2.6, let T >0, f ∈L2(0, T;H) and u0 ∈B be given. Let Hb be a closed subspace of H, and let Bb=Hb ∩B. Let bu0 ∈Bb be such that (3.14) holds. Let n∈N?, τ = T /n, and let bu and Dtbu be defined by (3.15)-(3.18). Then, for C2 given by (3.25), there holds
ψ(bu(t))≤C2/2, ∀t∈[0, T], (3.19) kDtbuk2L2(0,T;H)≤C2, (3.20) kbu(t)k ≤ kub0k+ (T C2)1/2, ∀t∈[0, T], (3.21) and
ku(tb 2)−u(tb 1)k ≤(C2)1/2(|t2−t1|+τ)1/2, ∀t1, t2∈[0, T]. (3.22) Proof. Let us first remark that, according to the Cauchy-Schwarz inequality,
kfkk2 ≤ 1 τ
Z kτ (k−1)τ
kf(t)k2dt, which leads to
n
X
k=1
τkfkk2 ≤ kfk2L2(0,T;H). (3.23) Settingw=ubk−1 in (3.16), we get
τkDkbuk2+ψ(ubk)−ψ(ubk−1)≤ hfk,buk−buk−1i ≤ τ
2kfkk2+τ
2kDkukb 2, (3.24) which gives, summing for k= 1, . . . , mfor a givenm= 1, . . . , n,
ψ(bum) +1
2kDtukb 2L2(0,mτ;H)≤ψ(bu0) + 1
2kfk2L2(0,T;H)
≤G+1
2kfk2L2(0,T;H). This gives (3.19) and (3.20) with
C2 = 2G+kfk2L2(0,T;H). (3.25) We then write
bum =bu0+
m
X
k=1
τ Dku,b ∀m= 1, . . . , n, which implies, using the Cauchy-Schwarz inequality and (3.20)
kubm−ub0k ≤
m
X
k=1
τkDkuk ≤b (mτkDtukb 2L2(0,mτ;H))1/2 ≤(T C2)1/2. (3.26)
This leads to (3.21). Finally, lett1 < t2 ∈[0, T] andk1, k2 = 0, . . . , nsuch that (k1−1)τ < t1 ≤ k1τ and (k2−1)τ < t2 ≤k2τ, which implies that (k2−1−k1)τ < t2−t1. We have
ku(tb 2)−u(tb 1)k ≤
k2
X
k=k1+1
τkDkbuk ≤
(k2−k1)τ
k2
X
k=k1+1
τkDkbuk2
1/2
,
which provides (3.22), using (3.20).
We may now prove the following convergence result.
Theorem 3.6 Under Hypothesis 2.6, let T > 0, f ∈ L2(0, T;H) and u0 ∈ B be given. Let (Hbn)n∈N? be a sequence of closed subspaces of H, and let Bbn=Hbn∩B. We assume that
n→∞lim inf
w∈Bbn
kw−vk+ (ψ(w)−ψ(v))+
= 0, ∀v∈B. (3.27)
For all n∈N?, let bu0n satisfy
ub0n∈Bbn, kbu0n−u0k →0 as n→ ∞, G:= sup
n
ψ(ub0n)<∞, (3.28) which is possible according to (3.27). Let τ =T /n, let ubn andDtubn be defined by (3.15)-(3.18) where Hb has to be replaced by Hbn and ub0 = bu0n. Then ubn(t) weakly converges in H to u(t) as n → ∞, uniformly with respect to t ∈ [0, T], and u is solution to Problem (3.3). Moreover, u satisfies
ψ(u(t))≤ 1
2C2, ∀t∈[0, T], (3.29)
k∂tuk2L2(0,T;H) ≤C2, (3.30) ku(t2)−u(t1)k ≤(C2)1/2|t2−t1|1/2, ∀t1, t2∈[0, T], (3.31) where C2 = 2G+kfk2L2(0,T;H).
Remark 3.7 Letting Hbn = H and ub0n = u0 for all n ∈ N? (semi-discretization in time), Theorem 3.6 provides the existence of a solution to Problem (3.3), and allows for taking G = ψ(u0) and C2= 2ψ(u0) +kfk2L2(0,T;H) in (3.29)-(3.31), since the solution is unique.
Remark 3.8 Theorem 3.11 shows indeed stronger convergence properties.
Proof of Theorem 3.6. Applying Lemma 3.5, we get that the hypotheses of the Ascoli- type Lemma 5.3 (provided in appendix) are fulfilled, from which we deduce that there exists u ∈ C0([0, T];H) and of a subsequence of (ubn)n∈N?, again denoted (ubn)n∈N?, such that bun(t) converges tou(t) weakly inH, uniformly for t∈[0, T]. Note that, according to (3.28), we have u(0) = u0. Using (3.19) and the lower semi-continuity of ψ, we get that (3.29) holds, and in particular thatu(t)∈B andu∈BT. Next, (3.31) comes directly from (5.10). Then, according to (3.20), we have that (Dtubn)n∈N is bounded in the Hilbert space L2(0, T;H). Hence extract- ing a subsequence, Dtbun weakly converges in L2(0, T;H) to some w ∈ L2(0, T;H) satisfying kwk2L2(0,T;H)≤C2. We then have, for a given ϕ∈Cc1(]0, T[),
− Z T
0 ubn(t)ϕ0(t)dt=−
n
X
k=1
ubk Z kτ
(k−1)τ
ϕ0(t)dt=
n
X
k=1
τ Dkbu ϕ((k−1)τ).