Viorel Barbu

Michael Röckner

Francesco Russo

Probabilistic representation for solutions of an irregular porous media type

Keywords: singular degenerate porous media type equation, probabilistic representation. 2000 AMS-classification: 60H30, 60H10, 60G46, 35C99, 58J65

equation: the degenerate case.

Introduction

We are interested in the probabilistic representation of the solution to a porous media type equation given by

∂ t u = 1 2 ∂ 2 xx (β(u)), t ∈ [0, ∞[ u(0, x) = u 0 (x), x ∈ R, (1.1)
in the sense of distributions, where u 0 is an initial bounded probability density. We look for a solution of (1.1) with time evolution in L 1 (R).

We make the following assumption.

Assumption 1.1

• β : R → R is monotone increasing.

• |β(u)| ≤ const|u|, u ≥ 0.

In particular, β is right-continuous at zero and β(0) = 0.

• There is λ > 0 such that (β + λid)(x) → ∓∞ when x → ∓∞.

Remark 1.2 (i) By one of the consequences of our main result, see Remark 1.6 below, the solution to (1.1) is non-negative, since u 0 ≥ 0.

Therefore, it is enough to assume that only the restriction of β to R + is increasing such that |β(u)| ≤ const|u| for u ≥ 0, and (β +λid)(x) → ∞ when x → +∞. Otherwise, we can just replace β by an extension of the restriction of β to R + which satisfies Assumption 1.1, e.g. take its odd symmetric extension.

(ii) In the main body of the paper, we shall in fact replace β with the "filled" associated graph, see remarks after Definition 2.2 for details; in this way, we consider β as a multivalued function and Assumption 1.1 will be replaced by Hypothesis 3.1.

Since β is monotone, (1.1) implies β(u) = Φ 2 (u)u, u ≥ 0, Φ being a nonnegative bounded Borel function. We recall that when β(u) = |u|u m-1 , m > 1, (1.1) is nothing else but the classical porous media equation.

One of our targets is to consider Φ as continuous except for a possible jump at one positive point, say e c > 0. A typical example is

Φ(u) = H(u -e c ), (1.2)
H being the Heaviside function.

The analysis of (1.1) and its probabilistic representation can be done in the framework of monotone partial differential equations (PDE) allowing multivalued coefficients and will be discussed in detail in the main body of the paper. In this introduction, for simplicity, we restrict our presentation to the single-valued case.

Definition 1.3

• We will say that equation (1.1) or β is non-degenerate if on each compact, there is a constant c 0 > 0 such that Φ ≥ c 0 .

• We will say that equation (1.1) Of course, Φ in (1.2) is degenerate. In order to have Φ non-degenerate, one could add a positive constant to it.

There are several contributions to the analytical study of (1.1), starting from [START_REF] Ph | A semilinear equation in L 1 (R N )[END_REF] for existence, [START_REF] Brezis | Uniqueness of solutions of the initial-value problem for u t -∆ϕ(u) = 0[END_REF] for uniqueness in the case of bounded solutions and [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF] for continuous dependence on the coefficients. The authors consider the case where β is continuous, even if their arguments allow some extensions for the discontinuous case.

As mentioned in the abstract, the first motivation of this paper was to discuss continuous time models of self-organized criticality (SOC), which are described by equations of type (1.1) with β(u) = uΦ 2 (u) and Φ as in (1.2), see e.g. [START_REF] Bak | How Nature Works: The Science of Self-Organized Criticality[END_REF] for a significant monography on the subject and the interesting physical papers [START_REF] Banta | Avalanche dynamics from anomalous diffusion[END_REF] and [START_REF] Cafiero | Local rigidity and self-organized criticality for avalanches[END_REF]. For other comments related to SOC, one can read the introduction of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF]. The recent papers, [START_REF] Barbu | Stochastic porous media equations and Self-organized criticality[END_REF][START_REF] Barbu | Selforganized criticality via stochastic partial differential equations[END_REF], discuss (1.1) in the case (1.2), perturbed by a multiplicative noise.

The singular non-linear diffusion equation (1.1) models the macroscopic phenomenon for which we try to give a microscopic probabilistic representation, via a non-linear stochastic differential equation (NLSDE) modelling the evolution of a single point.

The most important contribution of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF] was to establish a probabilistic representation of (1.1) in the non-degenerate case. For the latter we established both existence and uniqueness. In the degenerate case, even if the irregular diffusion equation (1.1) is well-posed, at that time, we could not prove existence of solutions to the corresponding NLSDE. This is now done in the present paper.

To the best of our knowledge the first author who considered a probabilistic representation (of the type studied in this paper) for the solutions of a nonlinear deterministic PDE was McKean [START_REF] Jr | Propagation of chaos for a class of non-linear parabolic equations[END_REF], particularly in relation with the so called propagation of chaos. In his case, however, the coefficients were smooth. From then on the literature has steadily grown and nowadays there is a vast amount of contributions to the subject, especially when the nonlinearity is in the first order part, as e.g. in Burgers equation. We refer the reader to the excellent survey papers [START_REF] Sznitman | Topics in propagation of chaos[END_REF] and [START_REF] Graham | Probabilistic models for nonlinear partial differential equations[END_REF].

A probabilistic interpretation of (1.1) when β(u) = |u|u m-1 , m > 1, was provided for instance in [START_REF] Benachour | Processu associés à l' équation des milieux poreux[END_REF]. For the same β, though the method could be adapted to the case where β is Lipschitz, in [START_REF] Jourdain | Probabilistic approximation for a porous medium equation[END_REF] the author has studied the evolution equation (1.1) when the initial condition and the evolution takes values in the set of all probability distribution functions on R. Therefore, instead of an evolution equation in L 1 (R), he considers a state space of functions vanishing at -∞ and with value 1 at +∞. He studies both the probabilistic representation and propagation of chaos.

Let us now describe the principle of the mentioned probabilistic representation. The stochastic differential equation (in the weak sense) rendering the probabilistic representation is given by the following (random) non-linear diffusion:

Y t = Y 0 + t 0 Φ(u(s, Y s ))dW s Law density(Y t ) = u(t, •), (1.3)
where W is a classical Brownian motion. The solution of that equation may be visualised as a continuous process Y on some filtered probability space (Ω, F, (F t ) t≥0 , P ) equipped with a Brownian motion W . By looking at a properly chosen version, we can and shall assume that Y :

[0, T ] × Ω → R + is B([0, T ]) ⊗ F-measurable.
Of course, we can only have (weak) uniqueness for (1.3) fixing the initial distribution, i.e. we have to fix the distribution (density) u 0 of Y 0 .

The connection with (1.1) is then given by the following result, see also [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF].

Theorem 1.5 Let us assume the existence of a solution Y for (1.3). Then u : [0, T ] × R → R + provides a solution in the sense of distributions of (1.1) with u 0 := u(0, •).

Remark 1.6 An immediate consequence for the associated solution of (1.1) is its positivity at any time if it starts with an initial value u 0 which is positive. Also the mass 1 of the initial condition is conserved in this case.

However this property follows already by approximation from Corollary 4.5 of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF], which in turn is based on the probabilistic representation in the nondegenerate case, see Corollary 4.2 below for details.

The main purpose of this paper is to show existence of the probabilistic representation equation (1.3), in the case where β is degenerate and not necessarily continuous. The uniqueness is only known if β is non-degenerate and in some very special cases in the degenerate case.

Let us now briefly and consecutively explain the points that we are able to treat and the difficulties which naturally appear in the probabilistic representation.

For simplicity we do this for β being single-valued (and) continuous. However, with some technical complications this generalizes to the multi-valued case, as spelt out in the subsequent sections.

1. Monotonicity methods allow us to show existence and uniqueness of solutions to (1.1) in the sense of distributions under the assumption that β is monotone, that there exists λ > 0 with (β + λid)(R) = R and that β is continuous at zero, see Proposition 3.2 of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF] and the references therein.

2. If β is non-degenerate, Theorem 4.3 of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF], allows to construct a unique (weak) solution Y to the non-linear SDE in the first line of (1.3), for any intial bounded probability density u 0 on R.

3. Suppose β to be degenerate. We fix a bounded probability density

u 0 . We set β ε (u) = β(u) + εu, Φ ε = √ Φ 2 + ε and consider the weak solution Y ε of Y ε t = Y ε 0 + t 0 Φ ε (u ε (s, Y ε s ))dW s , (1.4)
where u ε (t, •) is the law of Y ε t , t ≥ 0 and Y ε 0 is distributed according to u 0 (x)dx. The sequence of laws of the processes (Y ε ) are tight, but the limiting process of a convergent subsequence a priori may not necessarily solve the SDE

Y t = Y 0 + t 0 Φ(u(s, Y s ))dW s .
(1.5)

However, this will be shown to be the case in the following two general situations.

(a) The case when the initial condition u 0 is locally of bounded variation, without any further restriction on the coefficient β.

(b) The case when β is strictly increasing after some zero, see Definition 4.20, and without any further restriction on the initial condition.

In this paper, we proceed as follows. Section 2 is devoted to preliminaries and notations. In Section 3, we analyze an elliptic non-linear equation with monotone coefficients which constitutes the basis for the existence of a solution to (1.1). We recall some basic properties and we establish some other which will be useful later. In Section 4, we recall the notion of C 0 -solution to (1.1) coming from an implicite scheme of non-linear elliptic equations presented in Section 3. Moreover, we prove three significant properties. The

first is that β(u(t, •)) is in H 1 , therefore continuous, for almost all t ∈ [0, T ].
The second is that the solution u(t, •) is locally of bounded variation if u 0 is.

The third is that if β is strictly increasing after some zero, then Φ(u(t, •)) is continuous for almost all t. Section 5 is devoted to the study of the probabilistic representation of (1.1).

Finally, we would like to mention that, in order to keep this paper selfcontained and make it accessible to a larger audience, we include the analytic background material and necessary (through standard) definitions.

Likewise, we tried to explain all details on the analytic delicate and quite technical parts of the paper which form the back bone of the proofs for our main result.

Preliminaries

We start with some basic analytical framework.

If f : R → R is a bounded function we will set f ∞ = sup x∈R |f (x)|.
By C b (R) we denote the space of bounded continuous real functions and by C ∞ (R) the space of all continuous functions on R vanishing at infinity. D (R) will be the space of all infinitely differentiable functions with compact support ϕ : R → R, and D ′ (R) will be its dual (the space of Schwartz distributions). S (R) is the space of all rapidly decreasing infinitely differentiable functions ϕ : R → R, and S ′ (R) will be its dual (the space of tempered distributions).

If p ≥ 1 by L p (R) (resp. L p loc (R)), we denote the space of all real Borel functions f such that |f | p is integrable (resp. integrable on each compact interval). We denote the space of all Borel essentialy bounded real functions by L ∞ (R). In several situations we will even omit R.

We will use the classical notation W s,p (R) for Sobolev spaces, see e.g. [START_REF] Adams | Sobolev spaces[END_REF].

• s,p denotes the corresponding norm. We will use the notation H s (R) instead of W s,2 (R). If s ≥ 1, this space is a subspace of the space C(R) of real continuous functions. We recall that, by Sobolev embedding, W 1,1 (R) ⊂ C ∞ (R) and that each u ∈ W 1,1 (R) has an absolutely continuous version. Let δ > 0. We will denote by < •, • > -1,δ the inner product

< u, v > -1,δ =< (δ - 1 2 ∆) -1/2 u, (δ - 1 2 ∆) -1/2 v > L 2 (R) ,
and by • -1,δ the corresponding norm. For details about (δ -1 2 ∆) -s , see [START_REF] Stein | Singular integrals and differentiability properties of functions[END_REF][START_REF] Triebel | Interpolation Theory, Function Spaces, Differential Operators[END_REF] and also [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF], section 2. In particular, given s ∈ R, (δ

-1 2 ∆) s maps S ′ (R) (resp. S(R)) onto itself. If u ∈ L 2 (R). (δ - 1 2 ∆) -1 u(x) = R K δ (x -y)v(y)dy, with K δ (x) = 1 √ 2δ e - √ 2δ|x| . (2.6)
Moreover the map (δ -1 2 ∆) -1 continuously maps H -1 onto H 1 and a tempered distribution u belongs to H -1 if and only if (δ

-1 2 ∆) -1/2 u ∈ L 2 . Remark 2.1 L 1 ⊂ H -1 continuously. Moreover for u ∈ L 1 , u -1,δ ≤ K δ 1 2 ∞ u L 1 = (2δ) -1 4 u L 1 .
Let T > 0 be fixed. For functions (t, x) → u(t, x), the notation u ′ (resp. u ′′ ) will denote the first (resp. second) derivative with respect to x.

Let E be a Banach space. One of the most basic notions of this paper is the one of a multivalued function (graph). A multivalued function (graph) β on E will be a subset of E × E. It can be seen, either as a family of couples (e, f ), e, f ∈ E and we will write f ∈ β(e) or as a function β : E → P(E).

We start with the definition in the case E = R.

Definition 2.2 A multivalued function β defined on R with values in subsets of R is said to be monotone if given

x 1 , x 2 ∈ R, (x 1 -x 2 )(β(x 1 )-β(x 2 )) ≥ 0.
We say that β is maximal monotone (or a maximal monotone graph)

if it is monotone and if for one (hence all) λ > 0, β + λid is surjective, i.e.

R(β + λid) := x∈R (β(x) + λx) = R.
For a maximal monotone graph β : R → 2 R , we define a function j : R → R by

j(u) = u 0 β • (y)dy, u ∈ R, (2.7)
where β • is the minimal section of β. It fullfills the property that ∂j = β in the sense of convex analysis see e.g. [START_REF] Barbu | Analysis and control of nonlinear infinite dimensional systems[END_REF]. In other words β is the subdifferential of j. j is convex, continuous and if 0 ∈ β(0), then j ≥ 0.

We recall that one motivation of this paper is the case where

β(u) = H(u - e c )u
g i ∈ Af i , i = 1, 2, we have f 1 -f 2 ≤ f 1 -f 2 + λ(g 1 -g 2 ) ,
for any λ > 0.

This is equivalent to saying the following: for any λ > 0, (I + λA) -1 is a contraction on Rg(I + λA). We remark that a contraction is necessarily single-valued.

Proposition 2.4 Suppose that E is a Hilbert space equipped with the scalar product ( , ) H . Then A is accretive if and only if A is monotone i.e.

(f 1 -f 2 , g 1 -g 2 ) H ≥ 0 for any f 1 , f 2 , g 1 , g 2 ∈ E such that g i ∈ Af i , i = 1,
E = L 1 (R), so E * = L ∞ (R).
The following is taken from [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF], Section 1.

Theorem 2.7 Let β : R → R be a monotone (possibly multi-valued) function such that the corresponding graph is maximal monotone. Suppose that

0 ∈ β(0). Let f ∈ E = L 1 (R). 1. There is a unique u ∈ L 1 (R) for which there is w ∈ L 1 loc (R) such that u -∆w = f in D ′ (R), w(x) ∈ β(u(x)), for a.e. x ∈ R, (2.8)
see Proposition 2 of [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF].

Then, a (possibly multivalued) operator

A := A β : D(A) ⊂ E → E
is defined with D(A) being the set of u ∈ L 1 (R) for which there is

w ∈ L 1 loc (R) such that w(x) ∈ β(u(x)) for a.e. x ∈ R and ∆w ∈ L 1 (R) and for u ∈ D(A) Au = {- 1 2 ∆w|w as in definition of D(A)}.
This is a consequence of the remarks following Theorem 1 in [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF].

In particular, if β is single-valued, then Au = -1 2 ∆β(u). (We will adopt this notation also if β is multi-valued). [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF]. Moreover D(A) = E. [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF]. In particular, for every positive integer n

The operator A defined in 2. above is m-accretive on

E = L 1 (R), see Proposition 2 of

We set

J λ = (I + λA) -1 , which is a single-valued operator. If f ∈ L ∞ (R), then |J λ f ∞ ≤ f ∞ , see Proposition 2 (iii) of
, J n λ f ∞ ≤ f ∞ .
Let us summarize some important results of the theory of non-linear semigroups, see for instance [START_REF] Evans | Nonlinear evolution equations in an arbitrary Banach space[END_REF][START_REF] Barbu | Nonlinear semigroups and differential equations in Banach spaces[END_REF][START_REF] Barbu | Analysis and control of nonlinear infinite dimensional systems[END_REF][START_REF] Ph | A semilinear equation in L 1 (R N )[END_REF] or the more recent monograph [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF],

which we shall use below. Let A : E → E be a (possibly multivalued) accretive operator. We consider the equation

0 ∈ u ′ (t) + A(u(t)), 0 ≤ t ≤ T.
(2.9)

A function u : [0, T ] → E which is absolutely continuous such that for a.e.

t, u(t, •) ∈ D(A) and fulfills (2.9) in the following sense is called strong solution.

There exists η : [0, T ] → E, Bochner integrable, such that η(t) ∈ A(u(t)) for a.e. t ∈ [0, T ] and

u(t) = u 0 - t 0 η(s)ds, 0 < t ≤ T.
A weaker notion for (2.9) is the so-called C 0 -solution, see chapter IV.8 of [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF], or mild solution, see [START_REF] Barbu | Analysis and control of nonlinear infinite dimensional systems[END_REF]. In order to introduce it, one first defines the notion of ε-solution related to (2.9).

An ε-solution is a discretization

D = {0 = t 0 < t 1 < . . . < t N = T }
and an E-valued step function

u ε (t) = u 0 : t = t 0 u j ∈ D(A) : t ∈]t j-1 , t j ],
for which t jt j-1 ≤ ε for 1 ≤ j ≤ N , and

0 ∈ u j -u j-1 t j -t j-1 + Au j , 1 ≤ j ≤ N.
We remark that, since A is maximal monotone, u ε is determined by D and u 0 , see Theorem 2.7 3.

Definition 2.8 A C 0 -solution of (2.9) is an u ∈ C([0, T ]; E) such that
for every ε > 0, there is an ε-solution u ε of (2.9) with

u(t) -u ε (t) ≤ ε, 0 ≤ t ≤ T.
Proposition 2.9 Let A be a maximal monotone (multivalued) operator on a

Banach space E. We set again J λ := (I +λA) -1 , λ > 0. Suppose u 0 ∈ D(A).

Then:

1. There is a unique C 0 -solution u : [0, T ] → E of (2.9) 2. u(t) = lim n→∞ J n t n u 0 uniformly in t ∈ [0, T ].
Proof.

1) is stated in Corollary IV.8.4. of [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF] and 2) is contained in Theorem IV 8.2 of [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF].

The complications coming from the definition of C 0 -solution arise because

the dual

E * of E = L 1 (R) is not uniformly convex.
In general a C 0 -solution is not absolutely continuous and not a.e. differentiable, so it is not a strong solution. For uniformly convex Banach spaces, the situation is much easier.

Indeed, according to Theorem IV 7.1 of [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF], for a given u 0 ∈ D(A), there would exist a (strong) solution u : [0, T ] → E to (2.9). Moreover, Theorem 1.2 of [START_REF] Crandall | On the relation of the operator ∂ ∂s + ∂ ∂τ to evolution governed by accretive operators[END_REF] says the following. Given u 0 ∈ D(A) and given a sequence (u n 0 ) in D(A) converging to u 0 , then the sequence of the corresponding strong solutions (u n ) would converge to the unique C 0 -solution of the same equation.

Elliptic equations with monotone coefficients

Let us fix our assumptions on β which we assume to be in force in this entire section.

Hypothesis 3.1 Let β : R → 2 R be a maximal monotone graph with the property that there exists c > 0 such that

w ∈ β(u) ⇒ |w| ≤ c|u|. (3.1)
We note that (3.1) implies that β(0) = 0, hence j(u) ≥ 0, for any u ∈ R, where j is defined in (2.7). Furthermore, by Hypothesis 3.1,

j(u) ≤ |u| 0 |β • (y)|dy ≤ c|u| 2 . (3.2)
We recall from [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF] that the first ingredient to study well-posedness of equation (1.1) is the following elliptic equation

u -λ∆β(u) ∋ f (3.3)
where f ∈ L 1 (R) and u is the unknown function in L 1 (R).

Definition 3.2 Let f ∈ L 1 (R). Then u ∈ L 1 (R) is called a solution of (3.3) if there is w ∈ L 1 loc with w ∈ β(u) a.e. and u -λ∆w = f (3.4)
in the sense of distributions.

According to Theorem 4.1 of [START_REF] Ph | A semilinear equation in L 1 (R N )[END_REF], and Theorem 1, Ch.1, of [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF], equation

(3.
3) admits a unique solution. Moreover, w is also uniquely determined by u. Sometimes, we will also call the couple (u, w) the solution to (3.4).

We recall some basic properties of the couple (u, w).

Lemma 3.3 Let (u, w) be the unique solution of (3.3). Let J λ β : L 1 (R) → L 1 (R) be the map which associates the solution u of (3.3) to f ∈ L 1 (R).
We have the following:

1. J λ β 0 = 0.
2. J λ β is a contraction in the sense that

J λ β (f 1 ) -J λ β (f 2 ) L 1 ≤ f 1 -f 2 L 1 for every f 1 , f 2 ∈ L 1 . 3. If f ∈ L 1 L ∞ , then u ∞ ≤ f ∞ 4. If f ∈ L 1 L 2 , then u ∈ L 2 and R u 2 (x)dx ≤ R f 2 (x)dx. and R j(u)(x)dx ≤ R j(f )(x)dx ≤ const f L 2 . 5. Let f ∈ L 1 . Then w, w ′ ∈ W 1,1 ⊂ C ∞ (R). Hence, in particular w ∈ W 1,p for any p ∈ [0, ∞].
Proof.

1. is obvious and comes from uniqueness of (3.3).

2. See Proposition 2.i) of [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF].

3. See Proposition 2.iii) of [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF].

4. This follows from [START_REF] Ph | A semilinear equation in L 1 (R N )[END_REF], Point III, Ch. 1 and (3.2).

5. We define g :

= 1 2λ (w + f -u). Since f ∈ L 1 , also u ∈ L 1 , hence w ∈ L 1 by (3.1). Altogether it follows that g ∈ L 1 . (3.4) and (2.6) imply that w = K δ ⋆ g with δ = 1
2λ and hence

w ′ = K ′ δ ⋆ g = sign(y -x)e -λ -1 2 |x-y| g(y)dy,
where

sign x =        -1 : x < 0 0 : x = 0 1 : x > 0.
This implies w, w ′ ∈ L 1 L ∞ . By (3.4) we know that also

w ′′ ∈ L 1 , hence w, w ′ ∈ W 1,1 (⊂ C ∞ ).
u + λδβ(u) -λ∆(β(u)) ∋ f. (3.5)
In fact, [START_REF] Ph | A semilinear equation in L 1 (R N )[END_REF] treats the equation ∆v + γ(v) ∋ f , with γ : R → 2 R a maximal monotone graph. We reduce equation (3.3) and (3.5) to this equation, by

setting v = λβ(u), γ(v) = -β -1 ( v λ )
, where β -1 is the inverse graph of β, and setting v = λβ(u), γ(v) = -β -1 ( v λ )δv, respectively. In both cases γ is a maximal monotone graph.

Since w ′′ ∈ L 1 , (3.4) can be written as

R u(x)ϕ(x)dx -λ R w ′′ (x)ϕ(x)dx = R f (x)ϕ(x)dx ∀ϕ ∈ L ∞ (R). (3.6)
Since w ∈ L ∞ , we may replace ϕ by w in (3.6). In addition, w ′ ∈ L 2 , so by a simple approximation argument, it follows that

R (u(x) -f (x))w(x)dx + λ R w ′ 2 (x)dx = 0. (3.7)
Now, we are ready to prove the following.

Lemma 3.5 Let f ∈ L 1 L 2 and (u, w) be a solution to (3.3). Then R (j(u) -j(f ))(x)dx ≤ -λ R w ′ 2 (x)dx.
Proof. By definition of the subdifferential and since w(x) ∈ β(x) for a.e.

x ∈ R, we have

(j(u) -j(f ))(x) ≤ w(x)(u -f )(x) a.e. x ∈ R. (3.8)
Again (3.7) implies the result after integrating (3.8).

We go on analysing the local bounded variation character of the solution u of (3.3).

If f : R → R, for h ∈ R, we define

f h (x) = f (x + h) -f (x). (3.9)
Writing w h ′′ := (w h ) ′′ we observe that

u h -λw h ′′ = f h , (3.10)
where w(x) ∈ β(u(x)), and w(x

+ h) ∈ β(u(x + h)) a.e.
Let ζ ≥ 0 be a smooth function with compact support.

Lemma 3.6 Assume β is strictly monotone, i.e.

β(x) β(y) = ∅ if x = y. (3.11)
Let u be a solution of (3.3). Then, for each

h ∈ R R ζ(x)|u h (x)|dx ≤ R ζ(x)f h (x)sign(w h (x))dx + c ζ ′′′ ∞ λ|h| u L 1 . (3.12)
where c is the constant from (3.1).

Proof.

(3.10) gives R u h (x)ϕ(x)dx = R (λw h ′′ (x) + f h (x))ϕ(x)dx, ∀ϕ ∈ L ∞ (R). (3.13) We set ϕ(x) = sign(u h (x))ζ(x
= 0} = {λw h ′′ + f h = 0}. Hence (3.13) implies R ζ(x)|u h (x)|dx = {u h =0} (λw h ′′ (x) + f h (x))sign(w h (x))ζ(x)dx = λ R w h ′′ (x)sign(w h )(x)ζ(x)dx + R f h (x)sign(w h (x))ζ(x)dx. It remains to control λ R w h ′′ (x)sign(w h (x))ζ(x)dx. (3.14)
Let ̺ = ̺ L : R → R, be an odd smooth function such that ̺ ≤ 1 and

̺(x) = 1 on [ 1 L , ∞[. (3.14) is the limit when L goes to infinity of λ R w h ′′ (x)̺(w h (x))ζ(x)dx = -λ R w h ′ (x) 2 ̺ ′ (w h (x))ζ(x)dx - λ R w h ′ (x)̺(w h (x))ζ ′ (x)dx.
Since the first integral of the right-hand side of the previous expression is positive, (3.14) is upper bounded by the limsup when L goes to infinity of

-λ R w h ′ (x)̺(w h (x))ζ ′ (x)dx = -λ R (̺(w h (x))) ′ ζ ′ (x)dx,
where ̺(x) =

x 0 ̺(y)dy. But the previous expression is equal to

λ R ̺(w h (x))ζ ′′ (x)dx = λ R ̺(w(x))(ζ ′′ (x -h) -ζ ′′ (x))dx.
Since ̺(x) ≤ |x|, pointwise and w ∈ β(u) a.e., u ∈ L 1 , the previous integral is bounded by

λ ζ ′′′ ∞ |h| R |w(x)|dx ≤ cλ ζ ′′′ ∞ |h| R |u(x)|dx, (3.15)
with c coming from (3.1).

Remark 3.7 Using similar arguments as in Section 4 below, we can show that u is locally of bounded variation whenever f is. We have not emphasized this result since we will not directly use it.

4 Some properties of the porous media equation

Let β : R → 2 R . Throughout this section, we assume that β satisfies Hypothesis 3.1. Our first aim is to prove Theorem 4.15 below, for which we need some preparations. Let

u 0 ∈ (L 1 ∩ L ∞ )(R).
We recall some results stated in [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF] as Propositions 2.11 and 3.2.

Proposition 4.1 1. Let u 0 ∈ (L 1 L ∞ )(R).
Then, there is a unique solution to (1.1) in the sense of distributions. This means that there exists a unique couple (u,

η u ) ∈ L 1 ∩ L ∞ ([0, T ] × R) 2 with R u(t, x)ϕ(x)dx = R u 0 (x)ϕ(x)dx + 1 2 t 0 R η u (s, x)ϕ ′′ (x)dx ds, (4.1
)

where ϕ ∈ C ∞ • (R). Furthermore, t → u(t, •) is in C([0, T ], L 1 ) and η u (t, x) ∈ β(u(t, x)) for dt ⊗ dx-a.e. (t, x) ∈ [0, T ] × R.

We define the multivalued map

A = A β : E → E, E = L 1 (R), where D(A) is the set of all u ∈ L 1 for which there is w ∈ L 1 loc (R) such that w(x) ∈ β(u(x)) a.e. x ∈ R and ∆w ∈ L 1 (R). For w ∈ D(A) we set Au = {- 1 2
∆w|w as in the definition of D(A)}.

Then A is m-accretive on L 1 (R). Therefore there is a unique C 0solution of the evolution problem

   0 ∈ u ′ (t) + Au(t), u(0) = u 0 .
3. The C 0 -solution under 2. coincides with the solution in the sense of distributions under 1. [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF].

4. u ∞ ≤ u 0 ∞ . 5. Let β ε (u) = β(u) + εu, ε > 0 and consider the solution u (ε) to    ∂ t u (ε) = 1 2 β ε (u (ε) ) ′′ u (ε) (0, •) = u 0 . Then u (ε) → u in C([0, T ], L 1 (R)) when ε → 0, see
Corollary 4.2 We have u(t, •) ≥ 0 a.e. for any t ∈ [0, T ]. Moreover R u(t, x)dx = R u 0 (x)dx = 1, for any t ≥ 0.
Proof. In fact the functions u (ε) introduced in point 4. of Proposition 4.1 have the desired property. Taking the limit when ε goes to zero, the assertion follows.

Remark 4.3 Uniqueness to (4.1) holds even only with the assumptions β monotone, continuous at zero and β(0) = 0, see [START_REF] Brezis | Uniqueness of solutions of the initial-value problem for u t -∆ϕ(u) = 0[END_REF].

Below we fix on an initial condition u 0 ∈ L 1 L ∞ . Lemma 4.4 Let ε > 0. We consider an ε-solution given by

D = {0 = t 0 < t 1 < . . . < t N = T } and u ε (t) =    u 0 , t = 0 u j , t ∈]t j-1 , t j ] , for which for 1 ≤ j ≤ N    u j - t j -t j-1 2
w ′′ j = u j-1 , w j ∈ β(u j ) a.e.. We set

η ε (t, •) =    β(u 0 ) : t = 0, w j : t ∈]t j-1 , t j ].
When ε → 0, η ε converges weakly in L 1 ([0, T ]×R) to η u , where (u, η u ) solves equation (1.1). Furthermore, for p = 1 or p = ∞,

sup t≤T u ε (t, •) L p ≤ u 0 L p and (4.2) sup t≤T η ε (t, •) L p ≤ c u 0 L p ,
where c is as in Hypothesis 3.1. Hence,

sup t≤T u(t, •) L p ≤ u 0 L p . (4.3)
and (4.4) follows by an elementary interpolation argument. Indeed, for r ∈

η L r ([0,T ]×R) ≤ cT 1 r u 0 r-! r L ∞ u 0 r L 1
[1, ∞[, r ′ := r r-1 , ε n → 0 and ϕ ∈ (L r L ∞ )([0, T ] R) we have T 0 R ϕ(t, x)η u (t, x)dxdt = lim n→∞ T 0 R ϕ(t, x)η εn (t, x)dxdt ≤ ϕ L r ′ ([0,T ]×R) lim inf n→∞ T 0 R |η εn (t, x)| r dxdt 1 r ≤ ϕ L r ′ ([0,T ]×R) lim inf n→∞ sup t≤T η εn (t, •) r-1 r L ∞ T 1 r sup t≤T η εn (t, •) 1 r L 1 ≤ ϕ L r ′ ([0,T ]×R) T 1 r c u 0 r-1 r L ∞ u 1 r L 1 , 19
where we used the second part of (4.2) in the last step.

If not mentioned otherwise, in the sequel for N > 0 and ε = T N , we will consider the subdivision

D = {t i = εi, 0 ≤ i ≤ N }. (4.5)
We now discuss some properties of the solution exploiting the fact that the initial condition is square integrable.

Proposition 4.5 Let u 0 ∈ (L 1 L ∞ )(R). Then the solution (u, η u ) ∈ (L 1 ∩ L ∞ )([0, T ] × R) 2 of (1.1) has the following properties. a) η u (t, •) is absolutely continuous for a.e. t ∈ [0, T ] and η u ∈ L 2 ([0, T ]; H 1 (R)). b) R j(u(t, x))dx+ 1 2 t r R (η ′ u ) 2 (s, x)dxds ≤ R j(u(r, x))dx ∀ 0 ≤ r ≤ t ≤ T.
In particular t → R j(u(t, x))dx is decreasing and

[0,T ) R (η ′ u ) 2 (s, x)dxds ≤ 2 R j(u 0 (x))dx := C ≤ const. u 0 2 L 2 < ∞ . (4.6) c) t → j(u(t, x))dx is continuous on [0, T ].
Proof. We consider the scheme considered in Lemma 4.4 corresponding to ε = T N . By Lemma 3.3 5., we have w j ∈ H 1 (R), 1 ≤ j ≤ N , and by Lemma 3.5

R j(u i (x))dx + ε 2 R w ′ i (x) 2 dx ≤ R j(u i-1 (x))dx ∀i = 1, . . . , N. (4.7)
Hence for any 0

≤ l ≤ m ≤ N R j(u m (x))dx + ε 2 m i=l+1 R (w ′ i )(x) 2 dx ≤ R j(u l (x))dx. (4.8)
Using the notation introduced in Lemma 4.4, for all 0 ≤ r ≤ t ≤ T , we obtain

R j(u ε (t, x))dx + 1 2 t r R (η ε′ ) 2 (s, x)dsdx ≤ R j(u ε (r, x))dx. (4.9)
On the other hand, Lemma 4.4 and Lemma 3.3 4. imply that (4.12) and (4.9) say that η ε , ε > 0, are bounded in L 2 ([0, T ]; H 1 (R)). There is then a subsequence (ε n ) with η εn converging weakly in L 2 ([0, T ]; H 1 (R)) and therefore also weakly in L 2 ([0, T ] × R) to some ξ. According to Lemma 4.4 and the uniqueness of the limit, it follows η u = ξ and so

u ε (t) L 2 ≤ u 0 L 2 ∀t ∈ [0, T ]. (4.10) Therefore, T 0 ds R u ε (s, x) 2 dx ≤ T u 0 2 L 2 . ( 4
η u ∈ L 2 ([0, T ]; H 1 (R)),
which implies a). We recall that

T 0 ds R (η ′ u ) 2 (s, x)dx ≤ lim inf ε→0 T 0 ds R (η ε′ ) 2 (s, x)dx. (4.13)
In fact the sequence (η ε′ ) is weakly relatively compact in L 2 ([0, T ] × R). It follows by (3.2) and (4.10) that j(u ε (t)), ε > 0, are uniformly integrable for each t ∈ [0, T ]. Since j is continuous, and

u ε (t, •) → u(t, •) in L 1 (R) for each t ∈ [0, T ], it follows that j(u ε (t, •)) → j(u(t, •)) as ε → 0 in L 1 (R) for each t ∈ [0, T ]. (4.
12) and (4.9) imply that

R j(u(t, x))dx + 1 2 t r R (η ′ u ) 2 (s, x)dxds ≤ R j(u(r, x))dx, (4.14)
for every 0 ≤ r ≤ t ≤ T , which is inequality b).

To prove c), by (4.3) and Lebesgue dominated convergence theorem, us-

ing again that u 0 ∈ (L 1 L ∞ )(R) ⊂ L 2 , we deduce that t → u(t, •) is in C([0, T ], L 2 ) since it is in C([0, T ], L 1 ) by Proposition 4.1, 1. Now let t n → t in [0, T ] as n → ∞, then u(t n , •) → u(t, •) in L 2 as n → ∞, in particu- lar {u 2 (t n , •)|n ∈ N} is equiintegrable, hence by (3.2) {j(u(t n , •))|n ∈ N} is equiintegrable. Since j is continuous, assertion c) follows. Corollary 4.6 R u 2 (t, x)dx ≤ u 0 2 L 2 , ∀t ∈ [0, T ]. (4.15)
Proof. The result follows by Fatou's lemma, from (4.10).

Inequality (4.14) will be shown in Theorem 4.15 below to be indeed an equality.

Remark 4.7 According to Proposition 4.1 1., for every ϕ ∈ C ∞ 0 (R), we have

R u(t, x)ϕ(x)dx - R u 0 (x)ϕ(x)dx = 1 2 t 0 R η u (s, x)ϕ ′′ (x)dxds. (4.16) Since s → η u (s, •) belongs to L 2 ([0, T ]; H 1 (R)), by Proposition 4.5 a), we have s → η ′′ u (s, •) ∈ L 2 ([0, T ]; H -1 (R))
. This, together with (4.16), imply that t → u(t, •) is absolutely continuous from [0, T ] → H -1 (R). So, in This proposition will be important to prove that the real function t → R j(u(t, x))dx is absolutely continuous.

H -1 (R) we have d dt u(t, •) = 1 2 η ′′ u (t, •) t ∈ [0, T ] a.e.. ( 4
Proof. We equip H = H -1 (R) with the inner product •, • -1,δ where δ ∈]0, 1] and

u, v -1,δ = (δ - 1 2 ∆) -1 2 u, (δ - 1 2 ∆) -1 2 v L 2 (R) ,
and corresponding norm • -1,δ . We define Γ :

H → [0, ∞] by Γ(u) =    R j(u(x))dx, if u ∈ L 1 loc +∞, otherwise.
and D(Γ) = {u ∈ H| Γ(u) < ∞}. We also consider

D(A δ ) = {u ∈ D(Γ)| ∃ η u ∈ H 1 , η u ∈ β(u) a.e.}.
For u ∈ D(A δ ), we set A δ u = {(δ -1 2 ∆)η u |η u as in the definition of D(A δ )}. Obviously, Γ is convex since j is convex, and Γ is proper since D(Γ) is nonempty and even dense in H -1 , because L 2 (R) ⊂ D(Γ). The rest of the proof will be done in a series of lemmas. Lemma 4.9 The function Γ is lower semicontinuous.

Proof. First of all we observe that Γ is lower semicontinuous on L 1 loc (R). In fact, defining Γ N , N ∈ N, analogously to Γ, with j ∧ N replacing j, by the continuity of j and Lebesgue's dominated converegence theorem, Γ N is continuous in L 1 loc . Since Γ = sup N ∈N Γ N , it follows that Γ is lower continuous on L 1 loc . Let us suppose now that u n → u in H -1 (R). We have to prove that for each K > 0. Hence, by Dunford-Pettis theorem, the sequence (u n ) is weakly relatively compact in L 1 loc . Therefore, there is a subsequence (n l ) such that (u n l ) converges weakly in L 1 loc , necessarily to u, since u n → u strongly, hence also weakly in H -1 (R). Since Γ is convex and lower semicontinuous on L 1 loc , it is also weakly lower semicontinuous on L 1 loc , sse [START_REF] Choquet | Representation theory[END_REF] p.62, 22.1. This implies that

j(u(x))dx ≤ lim inf l→∞ j(u n k (x))dx = C.
Finally, (4.20)and thus the assertion of Lemma 4.9 is proved.

An important intermediate step is the following.

Lemma 4.10 D(Γ) = D(A δ ) and ∂ H Γ(u) = A δ u, ∀u ∈ D(Γ). In particular D(A δ ) is dense in H -1 .
We observe, that ∂ H depends in fact on δ since the inner product on H -1 depends on δ.

Proof. Let u ∈ D(Γ), h ∈ L 2 (⊂ D(Γ)). For z ∈ ∂ H Γ(u) we have Γ(u + h) -Γ(u) ≥ z, h -1,δ = R v(x)h(x)dx, (4.21)
where [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF], this yields that v ∈ β(u) a.e. Consequently, D(Γ) = D(A δ ) and

v = (δ -1 2 ∆) -1 z. Clearly v ∈ H 1 . By (4.21) it follows that v ∈ ∂ L 2 Γ(u) where Γ is the restriction of Γ to L 2 (R). By Example 2B of Chapter IV.2 in
∂ H Γ(u) ⊂ A δ u, ∀u ∈ D(Ψ).

It remains to prove that

A δ (u) ⊂ ∂ H Γ(u), ∀u ∈ D(Γ). Let u ∈ D(Γ), h ∈ L 2 , η u ∈ β(u) a.e. with η u ∈ H 1 . Since j(u + h) -j(u) ≥ η u h a.e., it follows Γ(u + h) -Γ(u) ≥ R η u (x)h(x)dx = (δ - 1 2 ∆)η u , h -1,δ . (4.22)
It remains to show that (4.22) holds for any h ∈ H -1 such that u+h ∈ D(Γ).

Then we have u + h, u ∈ L 1 loc and j(u), j(u + h) ∈ L 1 . We first prove that (4.22) holds if h ∈ L 1 (⊂ H -1 ). We truncate h setting

h n = 1 {|h|≤n} h, n ∈ N, so that h n ∈ L 2 (R). Now j(u + h n )(x) =    j(u(x) + h(x)) if |h(x)| ≤ n, j(u(x)) if |h(x)| > n,
and it is dominated by

j(u + h) + j(u) ∈ L 1 .
We have

R j(u + h n )(x)dx ≥ R j(u(x))dx + (δ - 1 2 ∆)η u , h n -1,δ . ( 4

.23)

Since h n → h in L 1 (and so in H -1 ), using Lebesgue's dominated convergence theorem, (4.22) follows for h ∈ L 1 .

Let M > 0 and consider a smooth function χ : R → [0, 1] such that χ(r) = 1 for 0 ≤ |r| ≤ 1, χ(r) = 0 for 2 ≤ |r| < ∞. We define

χ M (x) = χ x M , x ∈ R.
Then

χ M (x) =    1 : |x| ≤ M 0 : |x| ≥ 2M. Since hχ M ∈ L 1 , we have R (j(u + hχ M )(x) -j(u)(x))dx ≥ (δ - 1 2 ∆)η u , hχ M -1,δ . (4.24)
Since j is convex and non-negative, we have

j(u + hχ M ) = j((1 -χ M )u + χ M (u + h)) ≤ (1 -χ M )j(u) + χ M j(u + h) ≤ j(u) + j(u + h).
Hence Lebesgue's domintated convergence theorem allows to take the limit in the left-hand side, when M → ∞ of (4.24) to obtain R (j(u + h)(x))j(u(x))dx.

The right-hand side of (4.24) converges to (δ -1 2 ∆)η u , h H -1 because of the next lemma. Hence, the assertion of Lemma 4.10 follows.

Lemma 4.11 Define h

M := χ M h in H -1 , M > 0. Then lim M →∞ h M = h weakly in H -1 .

Proof (of Lemma 4.11). Let us first show that the sequence (h

M ) is bounded in H -1 . In fact, given ϕ ∈ H 1 , R h M (x)ϕ(x)dx = R h(x)χ M (x)ϕ(x)dx ≤ h H -1 δ 1 2 χ M ϕ L 2 + χ ′ M ϕ + ϕ ′ χ M L 2 ≤ h H -1 δ 1 2 ϕ L 2 + ϕ ′ L 2 + χ ′ ∞ M ϕ L 2 ≤ const h H -1 ϕ H 1
for some positive constant independent of M .

Hence there is a subsequence weakly converging to some

k ∈ H -1 . Since R h M (x)ϕ(x)dx → M →∞ R h(x)ϕ(x)dx
for any ϕ ∈ C ∞ 0 (R), k must be equal to h. Now the assertion of Lemma 4.11 follows.

By Corollary IV 1.2 in [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF], we know that A δ is maximal monotone on H -1 and therefore m-accretive with domain D(A δ ) = D(Γ).

We go on with the proof of Proposition 4.8. Since our initial condition u 0 belongs to L 1 ∩ L ∞ and L 2 ⊂ D(Γ), clearly u 0 ∈ D(A δ ). According to Komura-Kato theorem, see [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF]Proposition IV.3.1], there exists a (strong)

solution u = u δ : [0, T ] → E = H -1 of    du dt + A δ u ∋ 0, t ∈ [0, T ] u(0, •) = u 0 , (4.25) which is Lipschitz. In particular, for almost all t ∈ [0, T ], u δ (t, •) ∈ D(A δ ) and there is ξ δ (t, •) ∈ H 1 such that ξ δ (t, •) ∈ β(u δ (t, •)) a.e., t → (δξ δ - 1 2 ∆ξ δ )(t, •) ∈ H -1 is measurable and u δ (t, •) = u 0 + t 0 δξ δ - 1 2 ∆ξ δ (s, •)ds (4.26) in H -1 .
Furthermore, for the right-derivative D + u δ (t), we have

D + u δ (t, •)) + (A δ ) • u δ (t, •) = 0 in H -1 , ∀t ∈ [0, T ], (4.27)
where (A δ ) • denotes the minimal section of A δ and the map t

→ (A δ ) • u δ (t, •) -1,δ
is decreasing. On the other hand (4.26) implies that

du δ dt (t, •) + δξ δ (t, •) - 1 2 (ξ δ ) ′′ (t, •) = 0 for a.e. t ∈ [0, T ]. (4.28)
Consequently, for almost all t ∈ [0, T ]

δξ δ (t, •) - 1 2 (ξ δ ) ′′ (t, •) -1,δ = (A δ ) • u δ (t, •) -1,δ ≤ (A δ ) • u 0 -1,δ , (4.29)
i.e. setting ξ 0 = (δ -1 2 ∆) -1 (A δ ) • u 0 , we observe that it belongs to H 1 and that

H -1 (δ - 1 2 ∆)ξ δ (t, •), ξ δ (t, •) H 1 ≤ H -1 (δ - 1 2 ∆)ξ 0 , ξ 0 H 1 , for a.e. t ∈ [0, T ].
Consequently, for a.e.

t ∈ [0, T ], R (δξ δ (t, x) 2 + 1 2 ξ ′ δ (t, x) 2 )dx ≤ δ R ξ 2 0 (x)dx + R ξ ′ 0 2 (x)dx (4.30) ≤ ξ 0 H 1 =: C since δ ≤ 1.
We now consider equation (4.25) from an L 1 perspective, similarly as for equation (1.1), see Proposition 4.1 2. Since our initial condition u 0 belongs to (L 1 ∩ L ∞ )(R), equation (4.25) can also be considered as an evolution problem on the Banach space E = L 1 (R). More precisely define

D( Ãδ ) := {u ∈ L 1 (R)|∃w ∈ L 1 loc : w ∈ β(u)
a.e. and (δ -

1 2 ∆)w ∈ L 1 (R)} and for u ∈ D( Ãδ ), Ãδ u := {(δ - 1 2 ∆)w|w as in D( Ãδ )}.
Note that for w as in the definition of D (A δ ), we have (δ

-1 2 ∆)w ∈ H -1 , since L 1 (R) ⊂ H -1 . Therefore, w ∈ H 1 , hence D( Ãδ ) ⊂ D(A δ ) and Ãδ = A δ on D( Ãδ ).
(4.31) Furthermore, as indicated in Section 3, it is possible to show that Ãδ is an m-accretive operator on L 1 .

For λ > 0, the following four points are then a consequence of Remark 3.4 and Lemma 3.3.

1. For each f ∈ L 1 (R) there is u ∈ L 1 , w ∈ L 1 with w ∈ β(u) a.e. and u + λ(δw - 1 2 λw ′′ ) = f. (4.32)
2. The map

f → u := (I + λ Ãδ ) -1 (f ) is a contraction on L 1 . (4.33) 3. D( Ãδ ) = L 1 . 4. We recall that whenever f ∈ L ∞ , then u ∈ L ∞ and u ∞ ≤ f ∞ . (4.34)
Therefore, there is a C 0 -solution ũ : [0, T ] × R → R of (4.25). Since by (4.31), every ε-solution of (4.25) in L 1 (R) is also an ε-solution of (4.25) in

H -1 and L 1 ⊂ H -1 continuously, ũ is also a C 0 -solution of (4.25) in H -1 .
Since, by Proposition IV 8.2 and 8.7 of [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF], the solution above is the unique C 0 -solution of (4.25) in H -1 , we have proved the first part of the following lemma.

Lemma 4.12 The solution ũ coincides with the H -1 -valued solution u δ .

Moreover, for p = 1 or p = ∞ and c as in Hypothesis 3.1

sup t≤T u δ (t, •) L p ≤ u 0 L p and esssup t≤T ξ δ (t, •) L p ≤ c u 0 L p (4.35)
Proof.

It remains to show (4.35). As in the proof of (4.2) by (4.33), (4.34) and induction, we easily obtain that for any ε-solution in L 1 and p = 1 or p = ∞,

sup t≤T u ε (t, •) L p ≤ u 0 L p .
The conclusion follows because for every t ∈ [0, T ], there is a sequence Proof. It will be enough to prove that for δ small enough, we have

(ε n ) such that u εn (t, •) → ũ(t, •) = u δ (t,
R |u δ (t, x) -u(t, x)|dx ≤ cT u 0 L 1 δ. (4.36)
Using point 5. of Proposition 4.1 in a slightly modified form, and approximating β by β ε (u) = β(u) + εu, it is enough to suppose that β is strictly monotone, i.e. (3.11) holds. In the lines below the parameter ε will play however a different role.

We need to go back to the L 1 -ε-solutions related to u δ and u.

For ε > 0 we consider a subdivision 0 = t ε 0 < . . . < t ε j < . . . < t ε N = T such that t ε jt ε j-1 < ε, j = 1, . . . , N . Similarly as in Lemma 4.4

u ε δ (t j , •) = u ε δ (t j-1 , •) + (t j -t j-1 ) (η ε δ ) ′′ 2 (t j , •) -(t j -t j-1 )δη ε δ (t j , •) (4.37)
and

u ε (t j , •) = u ε (t j-1 , •) + (t j -t j-1 ) 1 2 (η ε ) ′′ (t j , •) (4.38) with η ε δ ∈ β(u ε δ ), η ε ∈ β(u ε ) a.e.
. Taking the difference of the previous two equations we obtain

u ε δ (t j , •) -u ε (t j , •) = u ε δ (t j-1 , •) -u ε (t j-1 , •) (4.39) + t j -t j-1 2 (η ε δ -η ε ) ′′ (t j , •) -δ(t j -t j-1 )η ε δ (t j , •).
Let Ψ κ : R → [-1, 1] be an odd smooth increasing function such that Ψ κ (x) → sign x as κ → 0 pointwise, We integrate (4.39) against Ψ κ (η ε δ (t j , •)η ε (t j , •)) and we get

R (u ε δ (t j , x) -u ε (t j , x))Ψ κ (η ε δ (t j , x) -η ε (t j , x))dx = R (u ε δ (t j-1 , x) -u ε (t j-1 , x))Ψ κ (η ε δ (t j , x) -η ε (t j , x))dx - (t j -t j-1 ) 2 R (η ε δ -η ε ) ′ (t j , x) 2 Ψ ′ κ (η ε δ (t j , x) -η ε (t j , x))dx -δ(t j -t j-1 ) R (η ε δ (t j , x)Ψ κ (η ε δ (t j , x) -η ε (t j , x))dx.
Using the fact that

Ψ ′ κ ≥ 0, |Ψ κ | ≤ 1, that, by strict monotonicity of β sign(η ε δ (t j , •) -η ε (t j , •)) = sign(u ε δ (t j , •) -u ε (t j , •)),
a.e. on {u ε δ (t j , •) = u ε (t j , •)}, and letting κ → 0, by (4.2), we obtain

R |u ε δ (t j , x) -u ε (t j , x)|dx ≤ R |u ε δ (t j-1 , x) -u ε (t j-1 , x)|dx + cδ(t j -t j-1 ) u 0 L 1 . (4.40)
Since R |u ε δ (0, x)u ε (0, x)|dx = 0, an induction argument implies that

R |u ε δ (t j , x) -u ε (t j , x)|dx ≤ cT u 0 L 1 • δ for every j ∈ {0, . . . , N }. Consequently, for any t ∈ [0, T ] R |u ε δ (t, x) -u ε (t, x)|dx ≤ cT u 0 L 1 δ.
Letting ε → 0, (4.36) follows and Lemma 4.13 is proved.

By (4.26), for every α ∈ C ∞ 0 (R) and all t ∈ [0, T ], R u δ (t, x)α(x)dx = R u 0 (x)α(x)dx -δ t 0 ds R ξ δ (s, x)α(x)dx + 1 2 t 0 ds R dxξ δ (s, x)α ′′ (x),
ξ δ ∈ β(u δ ) a.e. Letting δ → 0, by (4.35) and Lemma 4.13, we obtain that

R u(t, x)α(x)dx = R u 0 (x)α(x)dx + 1 2 lim δ→0 t 0 ds R ξ δ (s, x)α ′′ (x)dx. (4.41) By (4.35) it follows that for each K > 0, u δ → u in L 2 ([0, T ] × [-K, K]) and that (ξ δ ), is bounded in L 2 ([0, T ] × [-K, K]
). Since, by [START_REF] Showalter | Monotone operators in Banach space and nonlinear partial differential equations[END_REF] Example IV.2C,

the map u → β(u) is m-accretive on L 2 ([0, T ] × [-K, K]
), it is weaklystrongly closed, see [START_REF] Barbu | Nonlinear semigroups and differential equations in Banach spaces[END_REF], p.37 Proposition 1.1 (i) and (ii). So, there is a

sequence (δ n ) such that ξ δn → ξ weakly in L 2 ([0, T ] × [-K, K]) for some ξ ∈ β(u) a.e. Hence, (4.41) implies R u(t, x)α(x)dx = R u 0 (x)α(x)dx + 1 2 t 0 ds R ξ(s, x)α ′′ (x)dx. (4.42)
By the uniqueness part of Proposition 4.1 1., we conclude that ξ ≡ η u .

By Proposition 4.5, we already knew that η u (t, •) ∈ H 1 (R) for almost any t.

By (4.30) for a.e. fixed t, there is a sequence (δ n ) such that (ξ δn )(t, •) weakly converges to some ξ(t,

•) in H 1 (R) hence in L 2 (R). Consequently ξ(t, •) = η u (t,
|Γ(u(t, •)) -Γ(u(s, •))| ≤ max r∈{t,s} | < (δ - 1 2 ∆)η u (r, •), u(t, •) -u(s, •) > -1,δ | ≤ max r∈{t,s} (δ - 1 2 ∆)η u (r, •) -1,δ u(t, •) -u(s, •) -1,δ ≤ esssup r∈[0,T ] δ R η u (r, x) 2 dx + 1 2 R η ′ u (r, x) 2 dx u(t, •) -u(s, •) -1,
|Γ(u(t, •)) -Γ(u(s, •))| ≤ const u(t, •) -u(s, •) -1,δ , ∀t, s ∈ [0, T ],
and the assertion follows.

We are now prepared to prove the first main result of this section, which will be used in the next section in a crucial way. for every 0 ≤ r ≤ t ≤ T .

Proof. For a.e. t ∈ [0, T ], (4.17) gives

d dt u(t, •), ϕ = 1 2 η u (t, •), ϕ ′′ , ∀ϕ ∈ C ∞ 0 (R).
By density arguments,

H -1 d dt u(t, •), ψ H 1 = - 1 2 R η ′ u (t, x)ψ ′ (x)dx
for every ψ ∈ H 1 (R). For ψ = η u (t, •), we get

H -1 d dt u(t, •), η u (t, •) H 1 = - 1 2 R η ′ u 2 (t, x)dx. (4.44) Since u ∈ (L 1 L ∞ )([0, T ] × R) and |j(u)| ≤ c|u| 2 ,
u(t + h, •) -u(t, •) h h→0 -→ d dt u(t, •) in H -1 (R).
Let h > 0 such th, t + h are both positive. We have by (4.22)

R j(u(t, x)) -j(u(t -h, x)) h dx ≤ u(t, •) -u(t -h, •) h , η u (t, •) L 2 .
Taking limsup for h → 0, we get

lim sup h→0 R j(u(t, x)) -j(u(t -h, x)) h dx ≤ H -1 d dt u(t, •), η u (t, •) H 1 . (4.46)
On the other hand

u(t + h, •) -u(t, •) h , η u (t, •) L 2 ≤ R j(u(t + h, x)) -j(u(t, x)) h dx. So H -1 d dt u(t, •), η u (t, •) H 1 ≤ lim inf h→0 R j(u(t + h, x)) -j(u(t, x)) h dx.
Consequently for a.e. t ∈ [0, T ], Proof (of Proposition 4.17). For h small real fixed, we set

lim sup h→0 R j(u(t, x)) -j(u(t -h, x)) h dx ≤ H -1 du dt (t, •), η u (t, •) H 1 ≤ lim inf h→0 R j(u(t + h, x)) -j(u(t, x)) h dx. ( 4
u h (t, x) = u(t, x + h) -u(t, x).
Let ζ be a smooth nonnegative function with compact support on some compact interval. We aim at establishing the following intermediate result:

R ζ(x)|u h (t, x)|dx ≤ R ζ(x)|(u 0 ) h (x)|dx + c ζ ′′′ ∞ |h| [0,T ]×R |u(s, x)|dsdx.
(4.50)

Approximating β with β ε as in Proposition 4.1 5., we may suppose that β satisfies (3.11) on β. In the rest of this proof ε will however be the discretization mesh related to an ε-solution. We recall that u is the unique C 0 -solution to (1.1). So for fixed t ∈]0, T ]

u(t, •) = lim ε→0 u ε (t, •) in L 1 , (4.51)
where u ε (t, •) is given in Lemma 4.4.

According to Lemma 3.6 we have, for

i = 1, . . . , N , R ζ(x)|u h i (x)|dx ≤ R ζ(x)u h i-1 (x)sign(w h i (x))dx + c ζ ′′′ ∞ |h|ε R |u i (x)|dx ≤ R ζ(x)|u h i-1 (x)|dx + c ζ ′′′ ∞ |h|ε R |u i (x)|dx,
where

u h i = (u i ) h , w h i = (w i ) h , i ∈ {0, .
. . , N }, and u i is defined as in Lemma 4.4 with partition as in (4.5).

Let t ∈]0, T ] and m be an integer such that

t ∈] (m-1)T N , mT N ]. Summing on i = 0, • • • , m, we get R ζ(x)|u h m (x)|dx ≤ R ζ(x)|u h 0 (x)|dx + c ζ ′′′ ∞ |h|ε m i=1 R |u i (x)|dx. Setting u ε,h := (u ε ) h we obtain R ζ(x)|u ε,h (t, x)|dx ≤ R |u h 0 (x)|ζ(x)dx + c ζ ′′′ ∞ |h| T 0 ds R |u ε (s, x)|dx.
So, letting ε → 0 and using (4.51) we get

R ζ(x)|u h (t, x)|dx ≤ R |(u 0 ) h (x)|ζ(x)dx + c ζ ′′′ ∞ |h| T 0 ds R |u(s, x)|dx
and so (4.50). Therefore, lim sup

h→0 1 |h| R ζ(x)|u h (t, x)|dx ≤ 2 ζ ∞ |h| u 0 var +c ζ ′′′ ∞ [0,T ]×R |u(s, x)|dsdx, (4.52)
where • var denotes the total variation.

We denote the right hand-side of (4.52) by C

(ζ). Let K > 0, ϕ ∈ C ∞ 0 (R) such that suppϕ ⊂] -K, K[, and t ∈ [0, T ]. Taking ζ ≡ 1 on ] -K, K[, we can replace ϕ with ϕζ. Then R u(t, x) ϕ(x) -ϕ(x -h) h dx = R u h (t, x) h ζ(x)ϕ(x)dx ≤ 1 |h| ϕ ∞ R ζ(x)|u h (t, x)|dx.
So taking the limsup and using (4.52) we obtain

R u(t, x)ϕ ′ (x)dx ≤ ϕ ∞ C(ζ)
Hence u(t, •) has locally bounded variation on ] -K, K[ and the assertion follows.

We now show that, without particular assumptions on the initial conditions, in the degenerate case, a suitable "section" of Φ(u(t, •)) has at most countably many discontinuities if so has u(t, •). We again consider equation (1.1) in the sense of distributions

   ∂ t u = 1 2 η ′′ u , η u ∈ β(u) u(0, •) = u 0 ∈ L 1 ∩ L ∞ .
We recall that by Proposition 4.5 a), η u (t, •) ∈ H 1 (R) for a.e. t ∈]0, T ], hence has an absolutely continuous version, which will be still denoted by η u (t, •).

Likewise, since u(t, •) ≥ 0 a.e., for ∀t ∈ [0, T ], we shall take a version which is nonnegative everywhere, which will be still denoted by u(t, •) below.

Define

χ u = η u u 1 {|u|>0} . (4.53)
Here we recall that uη u ≥ 0, hence ηu u ≥ 0 on {|u| > 0}, and that χ u is bounded by Hypothesis 3.1.

Lemma 4.19 Suppose β is degenerate, let t ∈ [0, T [ such that η u (t, •) ∈ H 1 (R) and x ∈ R. If u(t, •) is continuous in x, then so is χ u (t, •). In partic- ular, χ u (t, •) has at most countably many discontinuities if so has u(t, •). Proof. It is enough to show that χ 2 u (t, •) is continuous in x. Let x n ∈ R, n ∈ N, converge to x. We have χ 2 u (t, x n ) =    ηu(t,xn) u(t,xn) , if u(t, x n ) > 0 0, if u(t, x n ) = 0. • If u(t, x) > 0, then χ 2 u (t, x n ) → η u (t, x) u(t, x) = χ 2 u (t, x).
• If u(t, x) = 0 then, since β is degenerate,

χ 2 u (t, x n ) n→∞ -→ 0 = χ 2 u (t, x).
We have observed that for a relatively general coefficient β, but with a restriction on the initial condition, u(t, •) (and therefore a suitable section of Φ(u(t, •))) is a.e. continuous, for a.e. t ∈ [0, T ], see Proposition 4.11. We now provide some conditions on β (degenerate) for which a suitable section of Φ(u(t, •)) is continuous for any initial condition in L 2 (R). This will prepare the third main result of this section, crucially to be used in the next section.

Let (u, η u ) be as usual the solution to (1.1) and χ u as in (4.53).

Definition 4.20

We say that β is strictly increasing after some zero

if there is e c ≥ 0 such that i) β| [0,ec[ = 0. ii) β is strictly increasing on [e c , ∞[. iii) If e c = 0, then lim u→0 + Φ(u) = 0.
Remark 4.21 1. Condition iii) guarantees that β is degenerate.

A typical example of a function that is strictly increasing after some

zero is given by

β(u) = uH(u -e c ),
Since β -1 is single-valued, continuous on ]0, ∞[ and η u (x 0 ) ∈ β(u(x 0 )), so η u (x 0 ) > 0, we have

u(x 0 ) = β -1 (η u (x 0 )) = β -1 ( lim n→∞ η u (x n )) = lim n→∞ β -1 (η u (x n )) = lim n→∞ u(x n ). Consequently χ 2 u (x n ) → χ 2 u (x 0 ). 3. u(x 0 ) = e c .
Clearly there are three possibilities.

(a) there is a subsequence (n k ) with u(

x n k ) ∈]e c , ∞[, (b) there is a subsequence (n k ) with u(x n k ) ∈ [0, e c [, (c)
there is a subsequence (n k ) with u(x n k ) = e c ∀k ∈ N.

Case (a). First we suppose e c > 0. We have

η u (x n k ) → η u (x 0 ). If η u (x 0 ) = 0 then χ 2 u (x n k ) = η u (x n k ) u(x n k ) → 0 = η u (x 0 ) u(x 0 ) = χ 2 u (x 0 ). If η u (x 0 ) = 0 then the continuity of β -1 implies u(x n k ) = β -1 (η u (x n k )) → β -1 (η u (x 0 )) = u(x 0 ) = e c , so χ 2 u (x n k ) → χ 2 u (x 0 )
. If e c = 0, the result follows since β is degenerate.

Case (b). In this case e c is again strictly positive. Since η

u (x n k ) ∈ β(u(x n k ) = 0 we have χ u (x n k ) = 0, hence χ u (x n k ) k→∞ -→ 0. But 0 = η u (x n k ) k→∞ -→ η u (x 0
). This implies that η u (x 0 ) = 0, so χ 2 u (x 0 ) = 0. Case (c). We have u(

I 3 (n) = E t s χ n (r, Y n r ) 2 -χ 2 u (r, Y n r ) drΘ(Y n r , r ≤ s) .
We start showing the convergence of I 3 (n). Now Θ(Y n r , r ≤ s) converges a.s. to Θ(Y r , r ≤ s) and it is dominated by a constant. so that it suffices to consider the expectation of By Proposition 5.7 below η u (ε) → η u in L 1 ([0, T ]×R) as ε → 0. Furthermore, Proposition 4. 1 5), see also the theorem in the introduction of [START_REF] Ph | The continuous dependence on ϕ of solutions of u t -∆ϕ(u) = 0[END_REF], implies that u ε (t, •) converges to u(t, •) in L 1 (R), as ε → 0, uniformly in t ∈ [0, T ].

Hence Lebesgue's dominated convergence theorem implies that I(n) → 0, since χ u is bounded.

We go on with the analysis of I 2 (n) and I 1 (n). I 2 (n) equals to zero because Y n is a martingale with quadratic variation given by [Y n ] t = t 0 χ n (r, Y n r ) 2 dr.

We finally treat I 1 (n). We recall that Y n → Y a. s. as random elements in C([0, T ]) and that the sequence E (Y n t ) 4 , is bounded, so (Y n t ) 2 are uniformly integrable. Therefore, for t > s we have

E (Y n t ) 2 -(Y n s ) 2 )Θ(Y n r , r ≤ s) -E (Y 2 t -Y 2

Remark 3 . 4

 34 Let δ > 0. The same results included in Lemma 3.3 are valid for the equation

(4. 4 )

 4 for all r ∈ [1, ∞[. Proof. See point 3. in the proof of the Proposition 3.2 in [9]. (4.2) follows by Lemma 3.3, 1-3. and Hypothesis 3.1 by induction. (4.3) is an immediate consequence of the first part of (4.2) and the fact that u is a C 0 solution.

. 11 )

 11 Since |β(u)| ≤ c|u|, (4.11) implies that sup s, x) 2 dx < ∞.(4.12)

R

  j(u(x))dx ≤ lim inf n→∞ R j(u n (x))dx. (4.20)Let us consider a subsequence such that j(u n (x))dx converges to the righthand side of (4.20) denoted by C. We may suppose C < ∞. the sequence (u n ) n∈N is uniformly integrable on [-K, K]

  •) a.e. as n → ∞. The second part of (4.35) then obviously follows by Hypothesis 3.1, since ξ δ (t, •) ∈ β(u δ (t, •)) a.e. for a.e. t ∈ [0, T ]. Lemma 4.13 We have u δ → u in C([0, T ]; L 1 (R)) as δ → 0, where u is the solution to (1.1).

Theorem 4 .

 4 [START_REF] Choquet | Representation theory[END_REF] Under Assumption(4.18), the unique solution to (1.1) verifies R j(u(t, x))dx = R j(u(r, x))dx -

  x n k ) = e c . If e c = 0 the result follows trivially by definition of χ u . Therefore we can suppose again that e c > 0. Then u(x n k ) = e c k→∞ -→ e c , soχ 2 u (x n k ) = η u (x n k ) e c → η u (x 0 ) e c = η u (x 0 ) u(x 0 ) = χ 2 u (x 0 ).

  r, Y n r ) 2χ u (r, Y n r ) εn ) (r, y)χ u (r, y) 2 u n (r, y) dy.

t s χ 2 u 2 u

 22 s )Θ(Y r , r ≤ s) → 0, when n → ∞. It remains to prove that E (r, Y r )drΘ(Y r , r ≤ s) -(r, Y n r )drΘ(Y n r , r ≤ s) → 0. (5.7) Under the assumptions of the theorem, for fixed r ∈ [0, T ], by the second and third main results of Section 4 (see Propositions 4.17, 4.22 and Remark 4.18), χ u (r, •) has at most a countable number of discontinuities. Moreover, the law of Y r has a density and it is therefore non atomic. So, let N (r) be the null event of ω ∈ Ω such that Y r (ω) is a point of discontinuity of χ u (r, •).For ω / ∈ N (r) we havelim n→∞ χ 2 u (r, Y n r (ω)) = χ 2 u (r, Y r (ω)).Now, Lebesgue's dominated convergence and Fubini's theorem imply (5.7).So equation (5.6) is shown.It remains to prove the following result which is based on our first main result of Section 4, see Theorem 4.15.

  or β is degenerate if lim u→0

+ Φ(u) = 0 in the sense that for any sequence of non-negative reals (x n ) converging to zero, and y n ∈ Φ(x n ) we have lim n→∞ y n = 0. Remark 1.4 1. β may be in fact neither non-degenerate nor degenerate. If β is odd, which according to Remark 1.2 (ii), we may always assume, then β is non-degenerate if and only if lim inf u→0+ Φ(u) > 0. 2. Of course, Φ in (1.2) is degenerate. In order to have Φ non-degenerate, one could add a positive constant to it.

  is said to be accretive if for any f 1 , f 2 , g 1 , g 2 ∈ E such that

	. It can be considered as a multivalued map by filling the gap. More
	generally, let us consider a monotone function ψ. Then all the discontinuities
	are of jump type. At every discontinuity point x of ψ, it is possible to com-
	plete ψ by setting ψ(x) = [ψ(x-), ψ(x+)]. Since ψ is a monotone function,
	the corresponding multivalued function will be, of course, also monotone.
	Now we come back to the case of our general Banach space E with norm
	• . An operator T : E → E is said to be a contraction if it is Lipschitz of norm less or equal to 1 and T (0) = 0.
	Definition 2.3 A map A : E → E, or more generally a multivalued map A : E → P(E)

  ). By(3.11) we have w h = 0 on {u h = 0},

dx a.e. Hence, by strict monotonicity we have ϕ(x) = sign(w h (x))ζ(x) a.e. on {u h = 0}. By (3.10), up to a Lebesgue null set, we have {u h

  •) for almost all t ∈ [0, T ]. Proof. Let 0 ≤ s < t ≤ T . Let Γ and D(Γ) be as defined in the proof of Proposition 4.8. Since u(t, •) ∈ D(Γ) for a.e. t ∈ [0, T ], Lemma 4.10 applies and thus for a.e. t, s ∈ [0, T ] we have that (δ -1 2 ∆)η u (t, •) ∈ A δ u(t, •), and

	Recalling (4.30) for a.e. t ∈ [0, T ] we get
	R	dx η ′ u (t, x) 2 =	R	dx ξ′ (t, x) 2 ≤ lim inf δ→0	R	dx (ξ δ ) ′ (t, x) 2 ≤ C.
	This finally completes the proof of Proposition 4.8.
	At this point, we can state and prove the following important theorem.
	Theorem 4.14 Assume that Hypothesis 3.1 and condition (4.18) hold. Let
	u be the solution of (1.1) (or equivalently of (4.1), from Proposition 4.1).
	Then the function t → R j(u(t, x))dx is absolutely continuous.

  δ .

	By (4.19) and (3.1), this is bounded by
	max(c, C) δ u 0 2 L 2 + 1 u(t, •) -u(s, •) -1,δ ,
	where we recall that by Remark 4.7 the map t → u(t, •) is absolutely contin-uous in H -1 . Since by Proposition 4.5 c), t → Γ(u(t, •)) is continuous, we have

  then, in particular, it belongs to L 2 ([0, T ], L 2 (R)). We need the following lemma.

	Lemma 4.16 For a.e. t ∈ [0, T ]
	H -1	d dt	u(t, •), η u (t, •) H 1 =	d dt R	j(u(t, x))dx.	(4.45)
	Proof. Let t ∈]0, T ] such that

  Proposition 4.17 Let Hypothesis 3.1 hold and let u be the unique solution to (1.1) with initial condition u 0 ∈ L 1 L ∞ being locally of bounded variation. Then, for each t ∈ [0, T ], u(t, •) also has locally bounded variation.

	At this point (4.44) and Lemma 4.16 imply that for a.e. t ∈ [0, T ],
			d dt R	j(u(t, x))dx = -	1 2 R	η ′ u	2 (t, x)dx.	(4.48)
	Theorem 4.14 says that t → R j(u(t, x))dx is absolutely continuous. So, after integrating in time, we get
	R	j(u(t, x))dx =	R	j(u(r, x))dx -	1 2	r	t	ds	R	η ′ u (s, x) 2 dx.	(4.49)
	This completes the proof of Theorem 4.15.
	The second main result of this section, also crucially used in Section 5 below,
	is the following.
	Remark 4.18	1. We note that (4.18) is not needed for the above propo-
	sition.
	2. Since u(t, •) has locally bounded variation, it has at most a countable number of discontinuities. We will see that in the degenerate case, i.e.
	if Φ(0) = 0, a suitable section of Φ(u(t, •)), also has at most countably many discontinuities, see Lemma 4.19 below.
												.47)
	On the other hand we know already by Theorem 4.14 that for a.e. t ∈ [0, T ], the limsup and liminf-terms in (4.47) coincide. Hence the assertion follows.

  ) 2 (s, Y n s )ds, and E([Y n ] T ) is finite, Φ being bounded, the continuous local martingales Y n are indeed martingales.By Skorokhod's theorem there is a new probability space (Ω, F, P ) and processes Ỹ n , with the same distribution as Y n so that Ỹ n converge to some process Ỹ , distributed as Y , as C([0, T ])-random elements P -a.s. In particular, those processes Ỹ n remain martingales with respect to the filtrations generated by them. We denote the sequence Ỹ n (resp. Ỹ ), again by Y n (resp. Y ). We observe that, for each t ∈ [0, T ], u(t, •) is the law density of Y t . In fact, for any t ∈ [0, T ], Y n t converges in probability to Y t ; on the other hand u n (t, •), which is the law of u n t converges to u(t, •) in L 1 (R), by Proposition 4.1 5. Remark 5.6 Let Y n (resp. Y) be the canonical filtration associated with Y n (resp. Y ). Those processes W n are standard (Y n t ) -Wiener processes since [W n ] t = t and because of Lévy's characterization theorem of Brownian motion. Then (s, Y s )dW s . (5.6) where χ u is defined as in (4.53). Once this equation is established for the given u, the statement of Theorem 5.4 would be completely proven because of Remark 5.5. In fact, that remark shows in particular the third line of (5.2). Taking into account, Theorem 4.2 of Ch. 3 of [22], to establish (5.6), it will be enough to prove that Y is a Ymartingale with quadratic variation [Y ] t = t 0 χ 2 u (s, Y s )ds. Let s, t ∈ [0, T ] with t > s and Θ a bounded continuous function from C([0, s]) to R. In order to prove the martingale property for Y , we need to show that E ((Y t -Y s )Θ(Y r , r ≤ s)) = 0. This follows by (5.5) because Y n → Y a.s. as C([0, T ])-valued process and which in turn follows, if for t > s we can verify

	Since (χ n Remark 5.5 We set [Y n ] t = t 0 W n t = t 0 1 χ n (s, Y n s )dY n s . one has Y n t = Y n 0 + t 0 χ n (s, Y n s )dW n s . We aim to prove first that Y t = Y 0 + t t -Y n s )Θ(Y n r , r ≤ s)) = 0. It remains to show that Y 2 t -E (Y 2 t -Y 2 s --E (Y n t ) 2 -(Y n s ) 2 -t s χ 2 u (r, Y n r )dr Θ(Y n r , r ≤ s) , χ u E ((Y n I 2 (n) = E (Y n t ) 2 -(Y n s ) 2 -
	0
	This completes the proof.

t 0 χ 2 u (s, Y s )ds, t ∈ [0, T ], defines a Y-martingale, t s χ 2 u (r, Y r )dr)Θ(Y r , r ≤ s) = 0.

The left-hand side decomposes into I 1 (n) + I 2 (n) + I 3 (n) where

I 1 (n) = E (Y 2 t -Y 2 s -t s χ 2 u (r, Y r )dr)Θ(Y r , r ≤ s) t s χ n (r, Y n r ) 2 dr Θ(Y n r , r ≤ s) ,

and

ACKNOWLEDGEMENTS

Financial support through the SFB 701 at Bielefeld University and NSF-Grant 0606615 is gratefully acknowledged.

where e c > 0 and H is the Heaviside function, i.e. 3. We recall that for almost all t ∈]0, T ], η u (t, •) is continuous. This will constitute the main ingredient in the proof of the proposition below. [START_REF] Banta | Avalanche dynamics from anomalous diffusion[END_REF]. Suppose that β is as in Definition 4.20. Then β -1 is single-valued and continuous on ]0, ∞[. Proposition 4.22 Suppose β strictly increasing after some zero. Then for almost all t ∈]0, T [, χ u (t, •) is continuous.

Proof. We first recall that by Corollary 4.2, u(t, •) ≥ 0 a.e. for all t ∈ [0, T ].

Let e c be as in Definition 4.20. Let t ∈]0, T ] for which η u (t, •) is continuous.

Let (x n ) be a sequence converging to some x 0 ∈ R. The principle is to find

In the sequel of the proof, we will omit t and denote the functions u(t, x) (resp. η u (t, x), χ u (t, x)) by u(x) (resp. η u (x), χ u (x)).

We distinguish several cases 1. u(x 0 ) ∈ [0, e c [. Then e c > 0 and η u (x 0 ) ∈ β(u(x 0 )) = 0. Hence χ u (x 0 ) = 0.

• If u(x n k ) < e c for some subsequence (n k ), then

2. We suppose now u(x 0 ) ∈]e c , ∞[.

The probabilistic representation of the deterministic equation

We again consider the β : R → 2 R satisfying Hypothesis 3.1. We aim at providing a probabilistic representation for solutions to equation (1.1). Let

We consider a multi-valued map Φ : R → 2 R + such that

The degenerate case is much more difficult than the non-degenerate case which was solved in [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF].

Definition 5.1 Let (u, η u ) be the solutions in the sense of Proposition 4.1 to equation (1.1). i.e.

We say that (1.1) has a probabilistic representation, if there is a filtered probability space (Ω, F, P, (F t )), an (F t ))-Wiener process W and, at least one process Y , such there exists

We recall the main result of [START_REF] Ph | Probabilistic representation for solutions of an irregular porous media equation[END_REF], Theorem 4.3.

Theorem 5.2 When β is non-degenerate then (1.1) has a probabilistic representation, with

Remark 5.3 In the non-degenerate case the representation is unique.

We will show that, even in the degenerate case, (1.1) has a probabilistic representation.

Theorem 5.4 Suppose that β is degenerate. Then equation (1.1) admits a probabilistic representation if one of the following conditions are verified.

1. β is strictly increasing after some (non-negative) zero.

2. u 0 has locally bounded variation.

Proof. We will make use of Theorem 5.2. Let ε ∈]0, 1] and set

) the solution to the deterministic PDE (1.1), with β ε replacing β. Define

We note that since Φ ε , ε ∈]0, 1] are uniformly bounded, so are χ ε , ε ∈]0, 1].

By Theorem 5.2, there exists a unique solution Y = Y ε in law of

Since Φ is bounded, using the Burkholder-Davies-Gundy inequality one obtains We set u n := u (εn) , where we recall that u n (t, •) is the law of Y n t , and

Proof. We set

where β • ε is the minimal section of β ε and clearly

In particular,

(5.9) So, by (4.4), the family

Let (ε n ) be a sequence converging to zero. There is a subsequence (n k ) such that η k u := η

Taking the limit when k → ∞, we get

Therefore, [START_REF] Barbu | Analysis and control of nonlinear infinite dimensional systems[END_REF], p.37, Proposition 1.1 (i) and (ii), imply that this map is weakly-strongly closed. Since, by (4.3),

a.e. on [0, T ] × [-K, K] for all K > 0, so, ξ ∈ β(u) a.e. By the uniqueness of (1.1) we get ξ = η u a.e.

Let ε n → 0. The rest of the paper will be devoted to the proof of the existence of a subsequence (η k u ) := η u (εn k ) converging (strongly

Hence {η k u } is equintegrable on [0, T ] × R. Therefore, the existence of such a subsequence completes the proof.

We will need the following well-known lemma.

Lemma 5.8 Let H be a Hilbert space, (f n ) be a sequence in H converging weakly to some f ∈ H. Suppose

Then f n → f strongly in H.

We apply the previous Lemma to establish the existence of a subsequence still denoted by (η k u ) such that (η k u ) ′ converges strongly to η ′ u in L 2 ([0, T ]×R). For this, we will prove that lim sup

We consider (5.8) for ε = ε n k and we let k go to infinity. First, for t ∈ [0, T ] we have

This together with (5.8) gives

Since by Corollary 4.6

the last term in (5.13) converges to zero when k → ∞. Taking the limsup when k → ∞ in (5.13) and using (5.12) we get lim sup

Consequently, by Lemma 5.8

Now let us finally prove that η k u → η u (strongly) in L 2 loc ([0.T ], ×R). Let x ∈ R. We recall that η u , η k u (t, •) vanish at infinity since they belong to H 1 (R) = H 1 0 (R). So we can write, for x ∈ R,

(η k uη u ) 2 (t, y)dy