On the infiltration of rain water through the soil with runoff of the excess water

Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800 www.elsevier.com/locate/na

On the in%ltration of rain water through the soil with runo) of the excess water I. Borsi, A. Farina, A. Fasano∗ Dipartimento di Matematica “U. Dini”, Universita degli Studi di Firenze, Viale Morgagni 67/A, Firenze 50134, Italy Received 31 March 2003; received in revised form 1 May 2003; accepted 23 June 2003

Abstract This paper deals with the modelling of the rain water in%ltration through the soil above the aquifer in case of runo) of the excess water. The main feature of the model lies on the correct de%nition of the boundary condition on the ground surface. The latter allows to estimate, after saturation, the real amount of the water that penetrates the soil and the one which runs o). The quantity playing a key role is the so-called rain pressure, de%ned as the pressure exerted by the rain on the soil. Although its importance is basically theoretical and it can be neglected for practical purposes, it helps understanding the real evolution of the physical problem, providing a theoretical justi%cation of the empirical procedures. c 2004 Published by Elsevier Ltd.  Keywords: Rain water percolation; Constrained boundary conditions; Free boundary problems; Classical solutions

1. Introduction This paper focuses on the mathematical modelling of a well-known classical problem, namely the rain water in%ltration through the soil layer above the aquifer, i.e. through the so-called vadose zone. At %rst sight this may look like a simple incompressible Darcyan >ow in a partially saturated medium. The %rst attempt to describe rain water in%ltration is the famous Green-Ampt model (1911) [17] with no capillarity. Such a model is however too crude to be realistic and more accurate models have been proposed over the years. Even though thumb rules are e)ectively used by soil engineers it seems to us that a satisfactory answer to the main question, concerning the appropriate ∗

Corresponding author. Fax: +39-055-4573898. E-mail address: [email protected]%.it (A. Fasano).

c 2004 Published by Elsevier Ltd. 1468-1218/$ - see front matter
doi:10.1016/j.nonrwa.2003.06.002

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formulation of the boundary condition at in>ow surface, has never been given. This is the issue we want to address in this paper. The study of the penetration of water through ground surface is of fundamental importance when, for instance, we want to reconstruct the history of contamination processes on the basis of recorded rainfall data over a speci%c area or when we want to predict the concentration of pollutants within the aquifer and in the vadose zone. In fact, contaminants are carried to aquifers with the water in%ltrating from the ground surface. Modelling this process (sometimes referred to as percolation) is important also because it allows to evaluate the real amount of water that in%ltrates through the ground surface. Such an information is fundamental for calculating peak >ows, thus achieving an optimal design of drainage networks or other storm water control facilities, especially in urban areas. A complete 3D percolation model is necessarily very complicated (the reader is referred to [1–5,7,11,14,20]). However, it makes sense to study a one-dimensional model because the basic features are more easily illustrated in a simple geometry. Moreover, there are physical situations in which assuming planar geometry does not introduce signi%cant errors in the computation of the relevant quantities. We consider the case in which at time t = 0 (the rain starting time) the vadose zone is nowhere saturated. The moisture distribution may correspond, for instance, to the equilibrium with no >ux through the ground surface (disregarding evaporation). We consider a rainfall event of particularly high intensity occurring in the time interval (0; T ): In these conditions saturation of the ground surface takes place at the so-called saturation time tJ; 0 ¡ tJ ¡ T . At that time a saturation front starts moving downward, becoming the (unknown) interface between two di)erent >ow regimes. After the onset of saturation, if the incoming rain >ux is large enough, the ground may not be able to absorb it all. The excess water may form a pond or it can run away (of course mixed situations are possible). Here, we take the extreme case of total runo), so that the ground surface is always exposed to the direct action of rain. The case of ponding has been studied in [16]. A peculiar feature of the model we are going to present is the fact that, after the appearance of the saturation front, the water pressure at the in>ow surface cannot exceed a threshold that we called rain pressure (Prain ). This is, so to speak, the pressure at which rain water is supplied, which can be de%ned as the linear momentum transfer rate to the ground from the population of falling rain drops. Such a quantity is not only easily computable but is also measurable. The pressure controlled regime is also subject to a constraint, since the corresponding surface >ux cannot exceed the one which is really available, i.e. the rain water >ux. Thus two possible conditions can be imposed at the in>ow surface (i.e. the ground surface): • Flux condition (all rainwater impinging on the unit surface per unit time penetrates the soil) with the constraint P ¡ Prain . • If the above constraint is violated, the boundary condition becomes P = Prain , as long as the >ux penetrating into soil is less than the rainwater >ux.

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We will not deal with the fully general case, in which there can be in%nitely many switches from one boundary condition to the other, but we shall analyze the particular case of an isolated switching point, selecting the data so that such a situation does occur. Existence and uniqueness of a classical solution will be proved following the techniques presented in [12,13]. Although the introduction of the concept of rain pressure turns out to be substantially theoretical (Prain is in fact very small even for intense rainfall), the model we are going to discuss highlights some essential qualitative features, clarifying the mathematical nature of the only admissible boundary conditions in the physical situation considered. After the problem has been formulated in a correct physical and mathematical way, one can realize that the procedures followed in practice usually introduce small errors and are therefore justi%ed. 2. Modelling rain water ow through unsaturated soil In Section 2.1 we give the general de%nitions, state the physical assumptions and write the mathematical model for percolation through unsaturated soil (the so-called Richards’s equation [4, p. 373]). A critical analysis of the boundary condition which is usually imposed on the ground surface is presented. 2.1. General de7nitions, physical assumptions and Richards’ equation We consider a gravity driven one-dimensional >ow in an unsaturated porous layer. The vertical coordinate z varies between z = 0, the so-called water table or phreatic surface, to z=Z, the ground surface (see Fig. 1). Since we are considering a suLciently short-time interval during which the rainfall occurs, it is reasonable to assume that the aquifer level is practically constant. We introduce the porosity n (volume fraction occupied by the pores), the moisture content (0 6 6 n) and we denote by  the water density. Saturation is de%ned as the ratio between the moisture content and the porosity n. If porosity and water density are constant, one can write the usual mass conservation equation @ @q + = 0; @t @z

(2.1)

where q represents the discharge. Assuming that Darcy’s law applies, we write
 k @P q=− + (2.2)  @z with k permeability, P water pressure,  speci7c weight and  viscosity of water. Dealing with an unsaturated porous medium, at every point a capillary pressure Pc is de%ned, so that Pc = Pair − P;

(2.3)

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Fig. 1.

where Pair is the air pressure and represents the pressure of the gas in the pores. Introducing hc = Pc =; hair = Pair = and =

P ; 

(2.4)

usually called suction, Eq. (2.3) becomes hc = hair − :

(2.5)

We now stipulate that Pair is constantly equal to the atmospheric pressure. Rescaling the latter to 0, (2.5) entails hc = − :

(2.6)

In general, hc is an experimentally determined function of the moisture content

(see [4,9,10,24] for a detailed description). More generally hysteresis in wetting and drying cycles produces a more complex relationship between hc and . Nevertheless, in the present study we consider hc as a single-valued decreasing function of (see [19] for instance) hc = −f( ):

(2.7)

From (2.6) and (2.7) we obtain = f( );

= g( );

with g = f−1 . We assume that: G1: g : (−∞; 0] → (0; n]. G2: g(0) = n and lim →−∞ g( ) = 0. G3: g is two times continuously di)erentiable in (−∞; 0].

(2.8)

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G4: g ( ) ¿ 0 ∀ ∈ (−∞; 0]. G5: g (0) = 1= ¿ 0. G6: g (0) = 0. Conditions G5 and G6 have only the purpose of simplifying the mathematical treatment. The graph of g( ) is called retention or saturation curve. Because of (2.8), f : (0; n] → [0; −∞), with f(n) = 0 and lim →0 f( ) = −∞. In particular, G4 –G6 entail f ( ) ¿ 0, ∀ ∈ (0; n], f (n) =  ¿ 0 and f (n) = 0. Assuming as usual k = k( ), we de%ne the hydraulic conductivity as  (2.9) K( ) = k( ):  We suppose that K satis%es the following conditions: K1: K : (−∞; 0] → (0; Ksat ]; lim →−∞ K( ) = 0 and K(0) = Ksat . For is constantly equal to Ksat , where  Ksat = k(0): 

¿ 0, K( ) (2.10)

K2: K( ) is twice continuously di)erentiable. K3: K  ( ) ¿ 0 ∀ ∈ (−∞; 0); K  (0) = 0. Substituting (2.2), (2.8) and (2.9) into Eq. (2.1), we obtain the so-called Richards’ equation for the dependent variable 
 @ @ @  − K( ) + 1 = 0; (2.11) g( ) @t @z @z which has to be coupled with suitable initial and boundary conditions. As initial condition we take (z; 0) =

0 (z) 6 0;

0 6 z 6 Z;

(2.12)

with: I1: I2:

0 (z) ¡ 0 0 (z)

∀z ∈ (0; Z) and 0 (0) = 0. is twice continuously di)erentiable and we set 1

1 =  0 0 ; 2 = (

 0 )−  0 :

(2.13) (2.14)

1 If f : I → R, · denotes the sup norm, i.e.  f  =sup 0 0 ∈I |f()|. Further ( f )− denotes the negative part of f.

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∀z ∈ (0; Z) and 0 (Z) = −1. 0 (z) satis%es the following inequality:  0 (z) ¡ 0

K( 0 )

 0

+ K  ( 0 ) 0 (

 0

+ 1) ¿ 0

∀z ∈ [0; Z]:

(2.15)

The meaning of (2.15) is clear if we look at Eq. (2.11): the initial moisture distribution is such that it does not produce a decrease of . In other words @ =@t at t = 0 is nonnegative. Such a condition, e.g. compatible with an equilibrium situation, has an important role in the sequel since, as we shall see in Section 6, it will imply the time monotonicity of . In particular, de%ning WM = Wm =

K ( ) ; 0 0 ;0] K( )

(2.16)

K ( ) ; 0 0 ;0] K( )

(2.17)

max

∈[−

min

∈[−

inequality (2.15) holds true if     0 + WM 0 ( 0 + 1) ¿ 0  0

+

Wm 0 ( 0

+ 1) ¿ 0

for

 0

∈ [ − 1; 0];

for

 0

∈ [ − 1; 0]:

We remark that assumption K3 guarantees WM ¿ 0 and Wm ¿ 0. As we said a possible 0 (z) is the steady state solution corresponding to zero >ux at the top surface (no rain, no evaporation), namely 0 (z)

= −z:

(2.18)

Let us now consider the boundary conditions on z = 0 and Z. The surface z = 0 represents the phreatic surface and so we take (0; t) = n. Thus, because of G2, (0; t) = 0;

0 ¡ t:

(2.19)

On z=Z, i.e. on the ground surface, we have, at this stage, a prescribed >ux boundary ˜ denotes the rainfall rate, we impose condition. If the vector N ˜ · ˜e z = ˜q · ˜e z ; N

(2.20)

where ˜e z is the outward normal to the ground surface and ˜q is the speci%c discharge. ˜ = −N (t)˜e z , (N (t) is usually an interpolant of the In the case we are considering, N meteorological data and is generally expressed in mm/h) and ˜q = q˜e z , so 
 @ = N (t); 0 ¡ t: (2.21) K( ) +1 @z z=Z As announced we shall see that we may be forced to change the boundary condition at a later stage of the process.

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For the moment we have the following initial–boundary value problem, denoted as (P ):  
 @ @ @    K( ) = ( ) g + 1 ; 0 6 z 6 Z; t ¿ 0;    @z @z @t      (z; 0) = 0 (z); 0 6 z 6 Z; (P ) (2.22)  (0; t) = 0; 0 ¡ t;     
  @   +1 = N (t); 0 ¡ t:  K( ) @z z=Z Remark 2.1. It will be shown that, under suitable assumptions on N (t), there exists tJ ∈ (0; T ) such that (Z; tJ) = n. At that time (saturation time) a moving saturation front z = s(t) appears and, for t ¿ tJ, a saturated ( = n) and an unsaturated ( ¡ n) region coexist within the vadose zone. Remark 2.2. When (Z; t) reaches saturation there is a limit to the capacity of the soil to take water in. Thus, it may not be possible to impose the >ux condition at the ground surface. Of course any replacement of the >ux boundary condition has to be compatible with (Z; t) = n. If at t = tJ ground surface reaches saturation, i.e. (Z; tJ) = n, some authors (see [10] and the references therein) assume N (tJ) = Ksat ;

(2.23)

writing the alternative boundary conditions in the form 
 @ = N (t) if N (t) 6 Ksat ; K( ) +1 @z z=Z (Z; t) = 0

if N (t) ¿ Ksat :

(2.24) (2.25)

We have to remark however that (2.23) is possibly compatible only with the degenerate case, i.e. f  (n) = 0. Indeed, if one assumes G5 (namely f (n) = ), when saturation occurs, i.e. when t = tJ, (Z; tJ) = 0:

(2.26)

On the other hand, (2.23) together with (2.21) and (2.26) would imply @ (Z; tJ) = 0: @z

(2.27)

But (0; t) = 0; (z; 0) = 0 (z) 6 0 and (Z; tJ) = 0, so, by Hopf’s Lemma (see [22], Chapter 14), we would have @ =@z(Z; tJ) ¿ 0 which contradicts (2.27). So at saturation N (tJ) ¿ Ksat ;

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namely N (tJ) − Ksat = – ¿ 0:

(2.28)

We thus conclude that, in general, (2.23) and the alternative conditions (2.24), (2.25) are not correct under assumption G5. As a matter of fact, the saturation time is singled out by (2.26) and not by (2.23). Further, while (2.21) remains correct if (Z; t) ¡ 0, after the onset of saturation (2.25) has to be modi%ed (we will discuss in detail such an issue in Section 4.1).

3. Unsaturated soil: analysis of the mathematical problem In this section, we analyze problem (P ) (i.e. problem (2.22)) showing also the existence of a %nite saturation time. In Section 3.1 we list the assumptions on the rainfall rate and in Section 3.2 we prove that there exists a time tJ such that ground surface reaches saturation, i.e. (Z; tJ) = n. Finally, in Section 3.3 we show that (P ) is well posed and illustrate some qualitative properties of the solution. 3.1. Assumptions on the rainfall rate N (t) We consider the time interval (0; T ) in which a rainfall event takes place. It is important to recall that the purpose of the present model is studying water in%ltration in the soil during heavy rains. This reason motivates the following assumptions on N (t): N1: N (t) is continuously di)erentiable in [0; T ] and dN (t) ¿0 dt

∀t ¿ 0:

N2: N (t) is such that

 
 t 2 N (t) ¿ WM Ksat ; 1 + 1 + Z+ WM JM Z

(3.1)

with 1 , 2 , WM given by (2.13), (2.14) and (2.16), respectively, and with 2 JM =

g ( ) : 0 0 ;0] K( )

max

∈[−

We assume also N (0) = WM Ksat 2

JM ¿ 0 because of G4.




2 1 + 1 + WM

(3.2)  Z ¿ K0 ;

(3.3)

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where 3 K0 = K(− 0 0 ):

(3.4)

As we shall see, assumption N2 has an important physical consequence: saturation is reached at the top surface in a %nite time. Of course in (3.1) the parameters characterizing the hydrological behavior of the ground come into play. 3.2. Reaching saturation The scope of this section is to show that, if condition N2 holds saturation is reached at z = Z in a %nite time tJ, i.e. (Z; t) vanishes for the %rst time at t = tJ. We also provide an upper estimate of tJ. First, we prove the following: Lemma 3.1. Let

(z; t) be the solution of problem (P ) in the domain

DT = (0; Z) × (0; T );

(3.5)

for some T ¿ 0. If assumptions K1–K3 and (3.3) are ful7lled then −  0 0 6 (z; t)

∀(z; t) ∈ DJ T :

(3.6)

Proof. Suppose that there exists a point (z ∗ ; t ∗ ) ∈ DJ T such that (z ∗ ; t ∗ ) ¡ −  0 0 . From the maximum principle and Hopf ’s Lemma [22, Chapter 14] z∗ = Z

and

z (Z; t

∗

) ¡ 0:

(3.7)

Recalling (3.3), (3.4) and the assumptions on K and N (t), (2.21) yields z (Z; t

∗

)+1=

z (Z; t

∗

) ¿ 0;

N (t ∗ ) N (t ∗ ) N (0) ¿ ¿ ¿ 1; ∗ K( (Z; t )) K0 K0

thus

which contradicts (3.7). Hence (3.6) follows. Proposition 3.1. Assuming that (z; t) is the classical solution of problem (P ) in DT given by (3.5), if N1 and N2 are satis7ed then the ground surface becomes saturated in a time tJ such that tJ 6

JM | 0 (Z)|; WM

where JM and WM are given by (3.2) and (2.16), respectively. 3

Assumption K1 implies K0 ¿ 0.

(3.8)

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Proof. The di)erence % = − 0 satis%es the problem   g ( )%t = K( )(%zz + 0 ) + K  ( )(%z + 0 )2       +K  ( )(%z + 0 );    %(z; 0) = 0;     %(0; t) = 0;      [K( )%z ]z=Z = N (t);

(z; t) ∈ DT ; 0 6 z 6 Z;

(3.9)

0 ¡ t ¡ T; 0 ¡ t ¡ T;

where g ( ), K( ) and K  ( ) have to be considered as functions of (z; t). We remark that assumption I3 has been exploited to obtain the boundary condition on Z. The di)erential equation of problem (3.9) can be rewritten as K 2 g K % + &; %t = %zz + %z (1 + 2 0 ) + K K z K

(3.10)

with &(z; t) =

 0

+

K K

  0( 0

+ 1):

Thanks to assumption I4 we have & ¿ 0 and consequently % ¿ 0. Let us now introduce the function v(z; t) =

A 2 z + Czt − B; 2

and look for the constants A, B, and C so that % ¿ v. Setting % = F + v, the function F satis%es the following problem:   g K   = F + F (Fz + vz + 0 )2  t zz   K K     K   + (Fz + vz + 0 ) + vzz    K     g + 0 − vt ; (z; t) ∈ DT ; K     1   0 ¡ z ¡ Z; F(z; 0) = − Az 2 + B;   2      F(0; t) = B; 0 ¡ t ¡ T;      K( )Fz (Z; t) = N (t) − K( )(AZ + Ct); 0 ¡ t ¡ T: Now we impose that K (vz + K

 0)

+ vzz +

 0

−

g vt ¿ 0; K

i.e. K (Az + Ct + K

 0)

+A+

 0

−

g Cz ¿ 0; K

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

which is satis%ed if A−




sup

∈[−

0 0 ;0]

K K



 0 0 − (

 0 )−  0


−

sup

∈[−

0 0 ;0]

g K

773

 CZ ¿ 0;

namely, recalling (2.13), (2.14), (2.16) and (3.2), if A ¿ 1 WM + 2 + JM CZ: Our %nal choice is A = WM (1 + 1) + 2 ; C=

WM : JM Z

In order to have F(z; 0) ¿ 0 it is enough to take B = 12 AZ 2 ; and keeping Fz (Z; t) ¿ 0 for all t requires N (t) ¿ Ksat (AZ + Ct): Thus N2 implies F ¿ 0. We conclude that (Z; t) ¿ 0 (Z) + v(Z; t), implying that (Z; t) has to vanish not later than the time t ∗ such that 0 (Z) + v(Z; t ∗ ) = 0, i.e. (Z; tJ) = 0 for some tJ 6

JM | 0 (Z)|: WM

The estimate of tJ just obtained is in agreement with physical intuition. As a matter of fact, it depends on how far we started from saturation and on how fast a pressure increase improves conductivity and take the soil surface closer to saturation. 3.3. Existence, uniqueness and qualitative properties of the solution of (P ) Problem (P ) presents boundary conditions of mixed type: on z = 0 a Dirichlet condition is de%ned while on z =Z a nonlinear Robin condition is prescribed. Applying the same reasoning of Lemma 3.1 we deduce that the solution is bounded. Extending to the interval [ − Z; Z] as an odd function, the problem is formulated in a way that Theorem 7.4 of [18, p. 491] applies directly. 4 Thus, we conclude that (P ) is solvable up to the onset of saturation. Theorem 3.1. Suppose assumptions G1–G6, K1–K3 and (3.3) are ful7lled, then there exists a unique solution 5 ∈ H 2+; 1+=2 (DJ tJ) to problem (P ), for arbitrary  ∈ (0; 1). 4

Extending g( ) as (2n − g(− )) for

5

¿ 0, assumptions G6 ensures that g is C 2 in (−∞; +∞).

For what the HNolder spaces and HNolder norms is concerned, we refer the reader to [18, Chapter 1].

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Let us now show that during this stage

is monotone in time.

(z; t) be a solution of problem (P ) in DT , T ¡ tJ. Then, for all

Proposition 3.2. Let z ∈ DJ T we have t (z; t) ¿ 0:

(3.11)

Proof. First we note that taking assumptions G4 and (2.15) into account, from problem (P ) we get t (z; 0)

=

K( 0 )

 0

+ K  ( 0 )( g ( 0 )

 0

+ 1)

¿ 0:

(3.12)

We then introduce %(z; t) =

t (z; t);

and di)erentiating the PDE of (2.22) with respect to time we obtain g %t = K%zz + F1 (z; t)%z + F2 (z; t)%; where g and K have to be considered as functions of z and t and where F1 (z; t) = K  + 2K  z ; 

K  g F2 (z; t) = − g K

(3.13)

 t

K + z ( z + 1) K − K 

2

 :

(3.14)

Considering now -(z; t) = %(z; t) e−Ct with C positive constant such that

F2 C ¿ max ; (z; t)∈D g it easy to check that - satis%es the following problem:   g -t = K-zz + F1 (z; t)-z + F3 (z; t)-;         -(z; 0) = t (z; 0) ¿ 0; -(0; t) = 0;      K  ( (Z; t)) dN (t) −Ct   N (t) -(Z; t) + K( (Z; t)) -z (Z; t) = e ;  K( (Z; t)) dt where F3 (z; t) = g





 F2 − C ¡ 0: g

(3.15)

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J t ∗ ¿ 0, is a point such that -(z ∗ ; t ∗ ) ¡ 0 then, by the maximum prinIf (z ∗ ; t ∗ ) ∈ D, ciple (see, for instance, Theorem 14.2.2 of [22, p. 329]), we deduce that - achieves a negative minimum at a point (Z; tˆ), 0 ¡ tˆ 6 t ∗ . On the other hand, the parabolic version of Hopf ’s Lemma and assumption I1 ensure -z (Z; tˆ ) ¡ 0 and dN (t)=dt ¿ 0, respectively. We thus have K dN (tˆ ) −C tˆ ¿ 0; e N (tˆ )-(Z; tˆ) = −K-z (Z; tˆ ) + dt K which contradicts -(Z; tˆ ) ¡ 0. We, therefore, conclude that the point (z ∗ ; t ∗ ) cannot J So -(z; t) ¿ 0, ∀(z; t) ∈ DJ and (3.11) follows. exist in D. We can be slightly more precise. Indeed, in a similar way we can prove also the following: Proposition 3.3. If (z; t) is a solution of (P ) then t (z; t) ¿ 0 for 0 ¡ z 6 Z, 0 ¡ t 6 tJ.

t (Z; t) ¿ 0

∀t ¿ 0, implying

4. Percolation in partially saturated soil This section is devoted to the analysis of the >ow through the vadose zone after the onset of saturation, i.e. for t ¿ tJ. As pointed out in Remark 2.1, we model percolation considering the soil divided into two time-varying domains representing the saturated the unsaturated region, respectively. The general features of such a model are presented in Section 4.1. Here, the rain pressure is de%ned and the analysis of the boundary condition on z = Z is presented. The mathematical model is summarized in Section 4.2. 4.1. Rain pressure For t ¿ tJ we have to study a problem de%ned into two time-varying domains: • s(t) ¡ z 6 Z, saturated domain, = n. • 0 6 z 6 s(t), unsaturated domain, ¡ n. The saturation front z = s(t), where (s(t); t) = n, is a priori unknown. In the saturated region mass conservation and Darcy’s law yield 
 @ @ Ksat +1 =0 @z @z and so the >ow is quasi-steady, i.e. zz

= 0;

(4.1)

which has to be coupled with suitable boundary conditions on z = s(t) and on z = Z. On the front z = s(t), as we said, (s(t); t) = 0

for t ¿ tJ:

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The boundary condition on the ground surface is more involved since we have to take into account both the pressure and the >ux. Assuming no ponding, our model is based on the hypothesis that the maximum pressure experienced by the water saturating the soil surface is the pressure caused by the linear momentum transfer due to the rain droplets anelastic collisions. We call it rain pressure, Prain . To determine Prain in terms of known quantities, we take a simpli%ed picture in which we consider averaged values for the droplets size and velocity. We de%ne: • N, number of rain droplets impinging on the unit area per unit time [N]=[t −1 ‘−2 ]. • m, mass of a droplet. • v, droplets velocity. Supposing that all the droplets momentum is transferred to the water saturating the ground surface, we can write Prain = Nmv: But, since we de%ned N as the volume of rain water reaching the soil per unit area and time, we have N = N × m=. It follows 6 Prain = N (t)v:

(4.2)

Assuming to know the value m (see [23] for the description of the Shifrin’s method allowing to evaluate the droplet radius) and the relationship v = v(m) (e.g. using the Stokes’ formula), we have that Prain = Prain (t) is a known function. So, since the problem is written for = P=, we de%ne rain (t)

=

Prain (t) v(m) = N (t):  

(4.3)

Now, going back to the boundary condition on Z, we have that the >ux condition Ksat ( z (Z; t) + 1) = N (t)

(4.4)

is accompanied by the constraint (Z; t) 6

rain (t);

(4.5)

since, as mentioned, a pressure larger than Prain is not allowed. If at time tJJ ¿ tJ condition (4.5) is violated, then we replace (4.4) with (Z; t) =

rain (t);

t ¿ tJJ:

(4.6)

Obviously (4.6) has to be compatible with the rainwater >ux, i.e. the >ux of water in%ltrating the soil must be less or equal than the rain >ux. Thus, the following 6

For a rather intense rain releasing 10 cm of water per hour (N (t) = 10 cm=3600 s) with drops falling at a speed of 900 cm=s, we obtain Prain = 2:5 g cm−1 s−2 . An appropriate unit for Prain is mmH2 O , in which units we have ∼ 2:5 × 10−2 .

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

777

inequality: Ksat ( z (Z; t) + 1) 6 N (t);

(4.7)

has to be coupled with (4.6). Summarizing, the boundary condition on z = Z has the alternative form  N (t)    z (Z; t) = K − 1 if (Z; t) ¡ rain (t); sat  N (t)   (Z; t) = rain (t) if z (Z; t) 6 − 1: Ksat 4.2. The mathematical model We summarize the free boundary problem describing the partially saturated regime g ( )

t

= [K( )(

(z; tJ) = J (z);

0 ¡ z ¡ Z;

(s(t)− ; t) = 0;

zz

−

; t) =

= 0;

z (Z; t)

(4.10)

+

   (Z; t) =

(4.11) t ¿ tJ;

; t);

s(t) ¡ z ¡ Z;

=

(4.12)

t ¿ tJ;

(4.13)

t ¿ tJ;

N (t) − 1; Ksat rain (t);

(4.8) (4.9)

t ¿ tJ; z (s(t)

(s(t)+ ; t) = 0;    

0 ¡ z ¡ s(t); t ¿ tJ;

+ 1)]z ;

t ¿ tJ;

(0; t) = 0;

z (s(t)

z

(4.14) if

(Z; t) ¡

rain (t);

if

N (t) − 1; z (Z; t) 6 Ksat

t ¿ tJ; (4.15)

where J (z) is the solution of problem (P ) at time t = tJ and where “+”, respectively “−”, refers to the saturated, respectively, unsaturated, part of the soil. Condition (4.12) expresses the continuity of the discharge and can be determined from the general mass balance equation for the soil–water mixture (see [21] for instance). Problem (4.8)–(4.15) presents many diLculties due both to the elliptic–parabolic coupling and above all to the nonstandard boundary condition (4.15). The study of its well-posedness in the fully general case is beyond the scope of the present paper. We will limit our analysis to the case of an isolated switching point for the boundary condition (4.15), which on the other hand is the most interesting from the practical point of view.

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5. Study of problem (4.8)–(4.15) in the case of an isolated switching point In this section we show that, if N (t) and P0 (t) satisfy certain conditions (which will be speci%ed in Sections 5.3 and 7.1), there exists tJJ, tJ ¡ tJJ 6 T , such that the %rst of conditions (4.15) holds in [tJ; tJJ) while the second applies in (tJJ; T ). It is useful to de%ne the domain of the (z; t) plane where we operate. For some given T1 , T2 , 0 6 T1 ¡ T2 6 T , and s(t) continuous function de%ned in [T1 ; T2 ] such that 0 6 s(t) 6 Z, RT1 ; T2 denotes the domain RT1 ;T2 = {(z; t) : 0 ¡ z ¡ s(t); T1 ¡ t ¡ T2 }:

(5.1)

In order to have lighter notation, we shall use RT2 when T1 = 0. 5.1. Problems statement Recalling Remark 2.2 (i.e. inequality (2.28)) we have  v(m)  v(m) Ksat ¿ 0 N (tJ) ¿ rain (tJ) =   and so, at least for a short-time interval after tJ, the boundary condition on z = Z is still given by the >ux condition (4.4). Therefore the corresponding mathematical problem consists of Eqs. (4.8)–(4.14) and (4.4). The function J (z), representing, as mentioned, the solution of (2.22) at time tJ, is such that J (0) = J (Z) = 0. In Section 5.3 we will show that there exists tJJ ¿ tJ such that condition (4.5) would be violated if we keep (4.4). Thus, as long as (4.7) holds true, as we shall see in Section 7.1, for t ¿ tJJ we have to switch to the boundary condition (Z; t) =

tJJ ¡ t;

rain (t);

(5.2)

and, as a consequence, problem (4.8)–(4.15) becomes g ( )

t

= [K( )(

(z; tJJ ) = ’ (z); (0; t) = 0;

zz

−

t ¿ tJJ;

(5.3) (5.4) (5.5)

t ¿ tJJ; z (s(t)

+

; t);

(5.6) t ¿ tJJ;

(5.7)

s(t) ¡ z ¡ Z; t ¿ tJJ;

(5.8)

t ¿ tJJ;

(5.9)

(s(t)+ ; t) = 0; (Z; t) =

0 ¡ z ¡ s(t);

t ¿ tJJ;

; t) =

= 0;

+ 1)]z ;

0 ¡ z ¡ b;

(s(t)− ; t) = 0; z (s(t)

z

rain (t);

t ¿ tJJ;

(5.10)

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779

where ’ (z) (solution of (4.8)–(4.14), (4.4) at time t = tJJ ) has the following properties: ’ ∈ C 2 [0; b];

(5.11)

’ (z) ¡ 0

∀z ∈ (0; b);

(5.12)

’ (b) = 0

’ (0) = 0:

(5.13)

The constant b is the thickness of the unsaturated layer at time tJJ and is such that J N (tJJ ) rain (tJ ) = − 1; Z −b Ksat and so ’ (b) =

J

rain (tJ )

Z −b

:

(5.14)

Problems (4.8)–(4.14), (4.4) and (5.3)–(5.10) are amenable, as we shall see in the next section, to Stefan-like problems. 5.2. Reduction of problems (4.8)–(4.14), (4.4) and (5.3)–(5.10) to problems of Stefan type Recalling (2.28), i.e. N (t) ¿ Ksat for t ¿ tJ, the solution of the elliptic problem (4.13), (4.14), (4.4) is
 N (t) (z; t) = − 1 (z − s(t)): (5.15) Ksat Thus, condition (4.12) rewrites N (t) − − 1: z (s(t) ; t) = Ksat Di)erentiating (4.11) with respect to time ˙ z (s(t); t)s(t)

+

t (s(t); t)

= 0;

(5.16)

we get the following di)erential equation for s(t): s(t) ˙ =−

@ Ksat [K( )( g ( (s(t); t))(N (t) − Ksat ) @z

with s(tJ) = Z: We thus formulate Problem 21 . Find a triple (T1 ; s; ) such that: (i) tJ ¡ T1 ¡ T , s ∈ C[tJ; T1 ] ∩ C 1 (tJ; T1 ), s ∈ [0; Z).

z

+ 1)]z=s(t)

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

and z are continuous in RJ t;J T1 , t and zz are continuous in Rt;J T1 with Rt;J T1 given by (5.1) with T1 ; T2 replaced by tJ; T1 , respectively. (iii) The following equations are satis7ed (ii)

g ( )

t

= [K( )(

(z; tJ) = J (z);

+ 1)]z ;

(z; t) ∈ Rt;J T1 ;

(5.17)

0 ¡ z ¡ Z;

(5.18)

tJ ¡ t ¡ T1 ;

(0; t) = 0; z (s(t); t)

z

=

N (t) − 1; Ksat

˙ =− g ( (s(t); t))s(t)

(5.19) tJ ¡ t ¡ T1 ;

@ Ksat [K( )( N (t) − Ksat @z

(5.20) z

+ 1)]z=s(t) ;

s(tJ ) = Z:

tJ ¡ t ¡ T1 ;

(5.21) (5.22)

For t ¿ tJJ the free boundary conditions change, as constraint (4.5) is violated. Solving (5.8)–(5.10) yields (z; t) =

rain (t)

Z − s(t)

(z − s(t));

and so condition (5.7) becomes z (s(t)

−

; t) =

rain (t) ; Z − s(t)

t ¿ tJJ:

(5.23)

Now, di)erentiating (5.6) with respect to time and recalling (5.3), we obtain the Cauchy problem  @ Z − s(t)   s(t) ˙ =− [K( )( z + 1)]z=s(t) ; t ¿ tJJ;  rain (t)g ( (s(t); t)) @z (5.24)   JJ s(t ) = b; 0 ¡ b ¡ Z: Summarizing, the following free boundary problem is de%ned: Problem 22 . Find a triple (T2 ; s; ) such that: (i) tJJ ¡ T2 6 T; s ∈ C[tJJ; T2 ] ∩ C 1 (tJJ; T2 ); s ∈ (0; Z): (ii) ; z are continuous in RJ t;JJ T2 (given by (5.1) with T1 ; T2 replaced by tJJ; T2 , respectively) and t , zz are continuous in Rt;JJ T2 . (iii) The following equations are satis7ed: g ( )

t

= [K( )(

(z; 0) = ’(z);

z

+ 1)]z ;

0 ¡ z ¡ b;

(z; t) ∈ Rt;JJ T2 ;

(5.25) (5.26)

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

tJJ ¡ t ¡ T2 ;

(0; t) = 0; z (s(t); t) rain (t)

Z − s(t)

=

rain (t)

Z − s(t)

s(t) ˙ =−

(5.27)

tJJ ¡ t ¡ T2 ;

;

(5.28)

@ 1 [K( )( (s(t); t)) @z

g (

781

z

+ 1)]z=s(t) ;

s(tJJ) = b ¡ Z:

tJJ ¡ t ¡ T2 ;

(5.29) (5.30)

If (T1 ; s; ), (T2 ; s; ) solve problems 21 , 22 , respectively, one can easily check that conditions (4.11) and (5.6) are satis%ed. Thus, exploiting assumptions G5 and K3, (5.21) and (5.29) imply 2 Ksat zz (s(t); t); N (t) − Ksat Ksat (Z − s(t)) zz (s(t); t); s(t) ˙ =− rain (t)

s(t) ˙ =−

(5.31) (5.32)

respectively. Conversely, let us suppose that (T1 ; s; ) and (T2 ; s; ) solve problems 21 and 22 where Eqs. (5.21) and (5.29) have been replaced by (5.31) and (5.32), respectively. We have the following. Proposition 5.1. If (T1 ; s; ) and (T2 ; s; ) solve problems (5.17)–(5.20), (5.31), (5.22) and (5.25)–(5.28), (5.32), (5.30), respectively, then (T1 ; s; ) and (T2 ; s; ) solve problems (5.17)–(5.22) and (5.25)–(5.30), i.e. problems 21 and 22 . Proof. Due to the similarity of the procedures, we will only show that problem (5.25)– (5.28), (5.32), (5.30) is equivalent to 22 . To this aim it is enough to prove that (s(t); t) = 0 ∀t ¿ tJJ. Setting y(t) = (s(t); t); we have y(t) ˙ =

˙ z (s(t); t)s(t)

= −Ksat  =

+

zz (s(t); t)

K(y) − Ksat g (y)



t (s(t); t)

+

1 g (y)

[K( )(

zz (s(t); t)

+

z

K  (y) g (y)

+ 1)]z z( z

We, therefore, obtain the following Cauchy problem:       y˙ = K(y) − Ksat a1 (t) + K (y) a2 (t); g (y) g (y)   J y(tJ) = 0;

+ 1):

(5.33)

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

where a1 (t)= zz (s(t); t) and a2 (t)= z (s(t); t) [ z (s(t); t) + 1] are continuous coeLcient. Assumptions G3 and K2 ensure that (5.33) has a unique solution. Moreover, thanks to assumptions G5, K1 and K3 it is easy to check the solution of (5.33) is y = 0. The above result allows to reformulate problems 21 and 22 in the following equivalent way. Problem 21 . Find a triple(T1 ; s; ) with the regularities stated at points (i) and (ii) of the previous de7nition, solving (5.17)–(5.20), (5.31) and (5.22). Problem 22 . Find a triple (T2 ; s; ) with the regularities stated at points (i) and (ii) of the previous de7nition, solving (5.25)–(5.28), (5.32) and (5.30). The existence and uniqueness proofs for problems 21 and 22 are very similar. For this reason we shall deal with only one of them (problem 22 ). Remark 5.1. We noticed that if ( ; s; T1 ) and ( ; s; T2 ) are classical solutions of problems 21 and 22 , respectively, then (s(t); t) = 0. So from the maximum principle we obtain −  J 0 6 (z; t) 6 0

∀(z; t) ∈ RJ t;J T1 ;

(5.34)

− ’0 6 (z; t) 6 0

∀(z; t) ∈ RJ t;JJ T2 ;

(5.35)

respectively, with J de%ned in Section 4.2. 5.3. Existence of tJJ In this section, we take the saturation time tJ as the origin of the time scale and we rescale t to t − tJ, still denoted by t. Proposition 5.2. Assuming that 21 has a unique solution for t ¿ 0 and that  Z – –2 [g( J (z)) − n] d z + T ¿ rain (T ) nKsat 0 nKsat with – given by (2.28), there exists tJJ, 0 ¡ tJJ ¡ T , such that (Z; tJJ) =

J

rain (tJ):

Proof. Let us consider the vector %eld     −K( )( z + 1) E1 ˜= = ; E g( ) E2

(5.36)

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

783

Fig. 2.

where indices 1 and 2 are related to axis z and t, respectively, and write   ˜ d z dt = E ˜ · ˜6 d‘; 0= ∇·E Rt

@Rt

where, referring to Fig. 2, @Rt = OABCO. We have   Z   t =− g( J (z)) d z; =− (N (7) + ns(7)) ˙ d7; OA



BC

 =

0

AB

s(t)

0

 g( (z; t)) d z;

CO

 =

and %nally we obtain the mass balance  Z  n(Z − s(t)) = g( J (z)) d z − 0



+ By Hopf’s Lemma

0

0

 ¿

Z

0

Z

t

0

s(t)

0

t

Ksat ( z (0; 7) + 1) d7

g( (z; t)) d z 

(N (7) − Ksat ) d7 − Ksat

z (0; t) ¡ 0,

 n(Z − s(t)) ¿

0

0

t

z (0; 7) d7:

thus

[g( J (z)) − n] d z +

 0

[g( J (z)) − n] d z + –t;

t

(N (7) − Ksat ) d7 (5.37)

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

where assumption G2 and property (2.28) have been exploited. Recalling now (5.15), by virtue of (5.37) 
N (t) − 1 (Z − s(t)) (Z; t) = Ksat
  Z

1 N (t) ¿ −1 [g( J (z)) − n] d z + –t Ksat n 0  Z – –2 ¿ t: [g( J (z)) − n] d z + nKsat nKsat 0 So, if (5.36) is satis%ed, the existence of tJJ follows. 6. Qualitative properties of the solution to problem
1 In this section, assuming that 21 has a unique classical solution for t ¿ tJ, we analyze some particular properties of it, which are crucial in proving the main results of Section 7. As in Section 5.3 we put tJ = 0. Proposition 6.1. Assuming existence and uniqueness of a classical solution to problem 21 in RT , with RT given by (5.1), we have t (z; t) ¿ 0

and s(t) ˙ ¡ 0;

for t ¿ 0;

provided that assumption N1 is ful7lled. Proof. Proposition 3.3 entails s(0) ˙ ¡ 0, since by (5.16) we have Ksat s(t) ˙ =− t (s(t); t): N (t) − Ksat Thus, because of the continuity of s, ˙ there exists t1 ¿ 0 such that s(t) ˙ ¡0

∀t ∈ [0; t1 ):

As a consequence, t (s(t); t) ¿ 0

∀t ∈ [0; t1 ):

Proposition 3.2 and (5.19) imply t (z; 0) ¿ 0

∀z ∈ [0; Z];

t (0; t)

∀t ¿ 0;

respectively.

=0

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

785

We now show that t (s(t); t) ¿ 0 for t ¿ 0. Suppose that there exists t1 , 0 ¡ t1 ¡ T; such that t (s(t1 ); t1 ) = 0 which in turn implies s(t ˙ 1 ) = 0. Proceeding as in Proposition 3.2, we introduce the function -(z; t) =

t (z; t)e

−Ct

;

(6.1)

which solves the following problem:   g -t = K-zz + F1 (z; t)-z + F3 (z; t)-;       -(z; 0) = %(z; 0) ¿ 0;  -(0; t) = 0;      -(s(t); t) =

t (s(t); t)e

−Ct

(6.2)

¿ 0;

where F1 (z; t) is given by (3.13),    F2 F3 (z; t) = g −C ; g with F2 (z; t) given by (3.14) and C constant such that

F2 : C ¿ max (z; t)∈Rt1 g

(6.3)

Because of (6.3), F3 is negative in Rt1 , and so (s(t1 ); t1 ) is a minimum for the function -. Hopf’s Lemma implies -z (s(t1 ); t1 ) ¡ 0, namely t z (s(t1 ); t1 ) ¡ 0:

(6.4)

Di)erentiating now with respect to time Eq. (5.20) yields ˙ zz (s(t); t)s(t)

+

z t (s(t); t)

=

1 dN (t) Ksat dt

and so, by assumption N1, zt (s(t1 ); t1 )

=

1 dN (t1 ) ¿ 0; Ksat dt

which contradicts (6.4). We thus have proved that applying the maximum principle yields -(z; t) ¿ 0

t (s(t); t) ¿ 0

∀t ¿ 0. Finally,

∀(z; t) ∈ RT :

7. Analysis of problem
2 In this section, we show existence and uniqueness of a solution to problem (5.25)–(5.28), (5.32), (5.30) using a technique based on the Banach %xed point Theorem. Once %xed a function s(t) in a suitable function space S, we solve the parabolic

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

problem (5.25)–(5.28). Then we map s to a new function : obtained solving the Cauchy problem (5.32) and (5.30) in which the function (s(t); t) is the solution of the parabolic problem previously de%ned. Before going through this procedure, in Sections 7.1 and 7.2 we deduce some preliminary results and state few basic de%nitions. Throughout this section we rescale tJJ to 0. The time interval (tJJ; T ) becomes (0; T − tJJ ), which we still denote by (0; T ). 7.1. Some a priori properties of the solution In this section, we show that, under a certain condition on N (t), the constraint (4.7) is not violated during a suLciently short-time interval after t = 0. Throughout this section, we assume existence and uniqueness of a classical solution (T2 ; s; ) to problem 22 . Such an assumption implies (see Proposition 5.1) that (s(t); t) = 0. Proposition 7.1. There exists t˜1 ¿ 0 such that t (z; t) ¿ 0

s(t) ˙ ¡0

∀(z; t) ∈ Rt˜1 ;

∀t ∈ (0; t˜1 )

with Rt˜1 given by (5.1). Proof. Because of Proposition 6.1 we have t (z; 0) ¿ 0 for 0 6 z ¡ b, and t (b; 0) ¿ 0. Using then (5.16) and (5.23) yields Z −b s(0) ˙ =− t (b; 0) ¡ 0: Po (0) Since an estimate of the HNolder norm of t is available, then we can estimate from ˙ ¡ 0 ∀t ∈ [0; t˜1 ). Then, proceeding as in Propobelow the time instant t˜1 such that s(t) sition 6.1, we can also show t (z; t) ¿ 0 ∀(z; t) ∈ Rt˜1 . Likewise we can estimate t˜2 ¿ 0 from below so that  s(t)  b (z; t) d z 6 2 ∀0 6 t 6 t˜2 : t t (z; 0) d z 0

0

We now de%ne t˜ = min{t˜1 ; t˜2 ; T2 }:

(7.1) (7.2)

Further, recalling (5.35), we de%ne also  gM =

max

−’0 6 60

g ( ):

(7.3)

Proposition 7.2. If in the time interval (0; t˜), with t˜ given by (7.2)    b  N (t) ¿ Ksat 1 + 2gM t (z; 0) d z ; 0

 where gM is de7ned by (7.3), then condition (4.7) is ful7lled.

(7.4)

I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

Proof. One can easily realize that condition (4.7) is veri%ed if rain (t) : Z − s(t) ¿ N (t)=Ksat − 1

787

(7.5)

Integrating Eq. (5.25) between 0 and s(t) yields  s(t) rain (t) − Ksat z (0; t); g ( ) t (z; t) d z = Z − s(t) 0 where the fact that (s; t) = (0; t) = 0 has been taken into account. Since z (0; t) ¡ 0 (parabolic version of Hopf’s Lemma) we have rain (t) Z − s(t) ¿  s(t) : g ( ) t (z; t) d z 0 In particular, recalling (7.1) and (7.3), rain (t) Z − s(t) ¿ b  2gM 0 t (z; 0) d z

(7.6)

for 0 6 t ¡ t˜. Thus (7.5) is satis%ed if rain (t) rain (t) ; ¿ Z − s(t) ¿ b  N (t)=K sat − 1 2gM 0 t (z; 0) d z which, because of (7.4), holds true in the time interval (0; t˜). 7.2. New formulation of problem 22 To study problem 22 it is useful to perform the so-called KirkoB’s transformation, de%ned by  (z; t) K() d: (7.7) !(z; t) = 0

Because of the strictly positiveness of K( ), the above is an invertible mapping. Eq. (5.25) rewrites as A1 (!)!zz + A2 (!)!z − !t = 0

(7.8)

with A1 (!) =

K( (!)) ; g ( (!))

A2 (!) =

K  ( (!)) : g ( (!))

(7.9)

We, therefore, end up with the following system: A1 (!)!zz + A2 (!)!z − !t = 0; !(z; 0) = !’ (z); !(0; t) = 0;

0 ¡ z ¡ b;

0 ¡ t ¡ T;

(z; t) ∈ RT ;

(7.10) (7.11) (7.12)

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800 rain (t)

!z (s(t); t) = Ksat s(t) ˙ =−

 rain (t)

Z − s(t)

;

0 ¡ t ¡ T;

(Z − s(t))!zz (s(t); t);

(7.13) 0 ¡ t ¡ T;

s(0) = b; where

(7.15) 

!’ (z) =

(7.14)

’(z)

0

K() d:

Because of (5.35), we have Aim 6 Ai (!) 6 AiM ;

i = 1; 2;

Aim =

AiM =

where

with

inf

!0 6!60

 !0 =

−’0

0

Ai (!);

sup

!0 6!60

Ai (!);

i = 1; 2;

K() d ¡ 0:

In particular, A1m ¿ 0. De0nition 7.1. We shall denote by S the set of functions s : [0; TJ ] → R, 0 ¡ TJ 6 T , such that: S1. S2. S3. S4.

s ∈ C([0; TJ ]) ∩ C 1 ((0; TJ )). s(0) = b, 0 ¡ b ¡ Z. s(t) ˙ 6 a ∀t ∈√(0; TJ ), with a ¿ 0. s ˙ H = (7; T ) 6 >= 7 ∀7 ∈ (0; TJ ), with > ¿ 0 and = ∈ (0; 1).

The constants =, a and > will be de%ned later. However a, Z, b, TJ are subjected to the following condition:

Z −b b J 0 ¡ T ¡ min ; ;T : (7.16) a a Further, de%ning Z˜ = b + aTJ ;

(7.17)

sm = b − aTJ ;

(7.18)

by virtue of (7.16), we have 0 ¡ Z˜ 6 min {Z − s(t)} t∈[0;TJ ]

and

0 ¡ sm 6 min {s(t)}: t∈[0;TJ ]

(7.19)

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Remark 7.1. In Proposition 7.2 we have shown that there exists t˜ ¿ 0 such that constraint (4.7) is not violated in (0; t˜). Introducing rM

= max

t∈(0;TJ )

rain (t);

(7.20)

Nm = min N (t) ¿ 0;

(7.21)

t∈(0;TJ )

one can easily realize that condition (4.7) is satis%ed in (0; TJ ), i.e. t˜ ¿ TJ , if rM

Z˜

¡

Nm − 1: Ksat

7.3. The reduced problem In this section we prove that problem (7.10)–(7.13), de%ned selecting :(t) ∈ S, has a unique solution. The proof is technical and will be developed through many steps. We %rst transform (7.10)–(7.13) in an equivalent problem (Section 7.3.1). Then we prove existence and uniqueness of a solution (Section 7.3.2). 7.3.1. Formulation of problem (7.10)–(7.13) in a 7xed domain Since classical theorems on nonlinear PDEs of parabolic type are generally formulated for cylindrical domains, we take the transformation z ; 06x61 x= :(t) and @(x; t) = !(x:(t); t): Problem (7.10)–(7.13) acquires the form  @t = Q1 (@; t)@xx + Q2 (@; @x ; x; t);         @(x; 0) = !’ (bx); @(0; t) = 0;      :(t) rain (t)   ;  @x (1; t) = Ksat Z − :(t)

(7.22) (x; t) ∈ ETJ ; 0 6 x 6 1; 0 6 t 6 TJ ;

(7.23)

0 6 t 6 TJ ;

where 1 A1 (@); :(t)2   :(t) ˙ 1 Q2 (@; @x ; x; t) = @x x + A2 (@) :(t) :(t) Q1 (@; t) =

(7.24) (7.25)

and ETJ = (0; 1) × (0; TJ ):

(7.26)

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We recall that, because of (5.14), @x (1; 0) = 0: 7.3.2. Existence and uniqueness of a solution We %rst prove an existence result assuming :(t) ∈ C ∞ (0; TJ ) ∩ S given. Proposition 7.3. Let :(t) be an in7nitely diBerentiable function and : ∈ S. Then there exists a unique solution @ ∈ H 2+; 1+=2 ([0; 1] × [0; TJ ]) to problem (7.23) for arbitrary  ∈ (0; 1). Proof. Proceeding as in Theorem 3.1, we extend the problem to the interval [ − 1; 1] and look for an odd function. Exploiting then Theorem 7.4 of [18, p. 491], the existence of a unique @ ∈ H 2+; 1+=2 ([ − 1; 1] × [0; TJ ]) solving the problem is ensured. Let us consider now a sequence {:k } ∈ C ∞ ∩S and let {@k } be the sequence of the corresponding solutions to problem (7.23). For each k problem (7.23) can be regarded as a linear problem (it is enough to consider Qi , i = 1; 2, as functions of x and t, i.e. Qi = Qi (@k (x; t)), i = 1; 2). Thus, we can apply the results reported in Chapter IV, [18, Section 9] and establish a uniform estimate for @k H 1+B; (1+B)=2 . Namely for arbitrary B ∈ (0; 1) @k H 1+B; (1+B)=2 (ETJ ) 6 M;

(7.27)

where M , here and in the following, denotes any constant depending on data but not on k. Theorem 7.1. For any :(t) ∈ S there exists a constant C ∈ (0; 1) such that problem (7.23) possesses a solution @ ∈ C 1; 0 (EJ TJ ) ∩ H C+2; 1+C=2 (ETJ ). Proof. Consider a sequence {:k } ∈ C ∞ ∩ S such that :k − :1 tends to zero as k → ∞. Let us de%ne Uk (x; t) = @k; x (y; t) − Ksat

:k (t) rain (t) : Z − :k (t)

Di)erentiating the PDE of problem (7.23) with respect to x, we have    :k (t) rain (t) Z rain (t):˙k (t) @Q1 @Q2 Uk; t = Q1 Uk; xx + + : Uk; x + − Ksat @x @x Z − :k (t) (Z − :k (t))2 We de%ne also Q3 (@k ; @k; x ; x; t) = Q4 (t) =

:˙k (t) ; :k (t)

A1 (@k ) :˙k (t) A2 (@k ) @k; x + x + ; 2 :k (t) :k (t) :k (t)

(7.28)

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  :k (t) rain (t) :˙k (t) rain (t) A2 (@k ) 2 − Q5 (@k ; @k; x ; x; t) = @k; x − Ksat :k (t) Z − :k (t) Z − :k (t)  Z rain (t):˙k (t) : + (Z − :k (t))2 Regarding @k and @k; x as functions of x and t and extending problem (7.23) in an antisymmetric way to the interval [ − 1; 0] we get  Uk; t = Q1 (x; t)Uk; xx + Q3 (x; t)Uk; x + Q4 (t)Uk + Q5 (x; t);       Uk (x; 0) = U0 (x); (7.29)  Uk (1; t) = 0;      Uk (−1; t) = 0; where U0 (y) = b’ (bx) − Ksat

b

rain (0)

Z −b

:

By the Ascoli–ArzelVa Theorem it follows that there exists in {@k } and {Uk } two subsequences (which will be indicated again by {@k } and {Uk }) such that @k − @H B; B=2 (ETJ ) → 0

as k → +∞;

Uk − U H B; B=2 (ETJ ) → 0

as k → +∞

with @(y; t) and U (y; t) satisfying (7.28) where indices k are deleted. To show that U (y; t) solves problem (7.29) (where indexes k are deleted) we can use Theorem 15 [15, p. 80] or Ciliberto’s results [8]. Indeed, √ we can show that C ∈ (0; 1) can be determined in terms of = such that Q1 H C; C=2 (ETJ ) ,  tQ3 H C; C=2 (ETJ ) , t Q4 H C; C=2 (ETJ ) and tQ5 H C; C=2 (ETJ ) are bounded independently of k. Thus, the above mentioned theorem of [15] ensures the convergence of {Uk } to the corresponding solution of (7.29). Similarly the function @(y; t) satis%es (7.23). We now show uniqueness for problem (7.23). To this aim, we need some delicate estimates for @(x; t). Lemma 7.1. If @(x; t) is a solution to problem (7.23) then: (i) There exist two positive constants, D1 and D2 , such that |@xx (x; t)| 6 D1 + D2 t B=2

∀(x; t) ∈ ETJ :

(7.30)

(ii) @xx is continuous up to x = 1 and D3 @H 2+E; 1+E=2 (E 7J ) 6 √ ; T 7

7 ∈ (0; TJ )

(7.31)

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with ET7J = (0; 1) × (7; TJ ); for any E ∈ (0; 1) with D3 positive constant depending on E. (iii) There exists a positive constant D4 such that D4 |@xxx (x; t)| 6 √ t

∀(x; t) ∈ ETJ :

(7.32)

Proof. Consider (7.29) and recall that, because of (7.27), Q1 H B; B=2 (ETJ ) can be uniformly estimated in terms of a positive constant. Exploiting (4.11) of [12] (see also [12, Appendix 1]) gives |Ux (x; t)| 6 Dˆ 1 + Dˆ 2 t B=2

∀(x; t) ∈ ETJ ;

with ETJ given by (7.26). From (7.28) estimate (7.30) follows trivially. Using now formula (4.22) of [12] (we refer also to [12, Appendix 2]) we obtain D3 U H 2+E; 1+E=2 (E 7J ) 6 √ ; T 7

0 ¡ 7 ¡ TJ :

(7.33)

Thus going back to (7.28) we get both estimate (7.31) and the fact that @xx is continuous up to x = 1 for t ¿ 0. We conclude the proof referring to (4.24) of [12] which provides D4 |Uxx | ¡ √ 7

∀(x; t) ∈ ETJ ;

implying (7.32). Theorem 7.2. For any given : ∈ S the solution @(x; t) of (7.23) is unique. Proof. Consider the di)erence ˜ t) = @1 (x; t) − @2 (x; t): @(x; of two solutions of (7.23). It is easily found that @˜ satis%es the following problem:  @˜ t = Q1 (x; t)@˜ xx + F1 (x; t)@˜ x + F2 (x; t);       @(x; ˜ 0) = 0; (7.34)  ˜ t) = 0;  @(0;     ˜ @x (1; t) = 0; where Q1 is given by (7.24), F1 (x; t) = x

:(t) ˙ 1 + A2 (@1 (x; t)) :(t) :(t)

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and F2 (x; t) = [Q1 (@1 (x; t)) − Q1 (@2 (x; t))]@2xx (x; t) +

1 @2x (x; t)[A2 (@1 (x; t)) − A2 (@2 (x; t))]: :(t)

The source term F2 is bounded by ˜ t)|: |F2 (x; t)| ¡ C1 sup |@2xx @(x; x∈(0;1)

(7.35)

˜ t) Extending problem (7.34) to whole interval [−1; 1] we can represent the solution @(x; in the following way:  t +1 ˜ t) = @(x; G(x; ; t; 7)F2 (; 7) d d7; (7.36) 0

−1

where G(y; ; t; 7) is the Green’s function of problem (7.34) de%ned on (−1; 1) × (0; TJ ) (see [18, Chapter IV, Section 16]). Estimates (7.35) and (7.30) along with (16.16) [18, p. 412] yield  t C ˜ t)| 6 ˜ 7)| d7: √ |@(x; (1 + 7B=2 ) sup |@(z; t − 7 z∈(−1;1) 0 Hence a well-known generalized version of Gronwall’s Lemma (see [6]) gives the conclusion of the proof. 7.4. The operator M As a consequence of previous results, for any : ∈ S, we can de%ne a function s : [0; TJ ] → R as follows:  (Z − :(t))   s(t) ˙ =− @xx (1; t) :(t)2 rain (t) (7.37)   s(0) = b; thus obtaining an operator M acting on S s = M:: Proposition 7.4. For a suitable choice of a, > and TJ 6 T and for any = ∈ (0; 1=2) the operator M maps S into itself. Proof. From point (ii) of Lemma 7.1 and the second of (7.19) we have that s˙ ∈ C(0; TJ ). Estimate (7.30) then implies that the r.h.s. of (7.37) is integrable, namely the function s(t) is de%ned and continuous on [0; TJ ]. Properties S1 and S2 of De%nition 7.1 are thus proved.

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Once %xed B ∈ (0; 1), let us de%ne 
  2=B D1 ˆ T = min T; : D2

(7.38)

Taking 0¡a¡

b ; Tˆ

(7.37) yields |s(t)| ˙ 6

Z (b − at)2

rm

(D1 + D2 t B=2 );

where (7.30) has been taken into account and where rm

= min

t∈[0;T ]

rain (t) ¿ 0:

(7.39)

We thus obtain |s(t)| ˙ 6

2D1  Z (b − aTˆ )2

∀t ∈ [0; Tˆ ]:

rm

Property S3 will be obtained if we select a ∈ (0; b= Tˆ ) such that 2D1 Z (b − aTˆ )2

rm

6a ⇔

2D1 Z rm

6 a(b − aTˆ )2 :

The function f(a) = a(b − aTˆ )2 ; in the interval (0; b= Tˆ ) takes its maximum for a = b=3Tˆ and f(b=3Tˆ ) = 4b3 =27Tˆ . So, if we choose a=

b ; 3Tˆ

(7.40)

with Tˆ given by (7.38) and we set 
2=B  3 Z − b b 2 D1 rm TJ = min ;T ; ; ; a 27D1 Z D2

(7.41)

we have |s(t)| ˙ 6a

∀t ∈ [0; TJ ]:

and condition (7.16) is veri%ed. Recalling now point (ii) of Lemma 7.1 we choose E in (1=2; 1) and %x = = ∈ (0; 12 );

(7.42)

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such that (E=2 − =) ¿ 0. Now, adding and subtracting suitable terms in Eq. (7.37) we get |s(t ˙ 1 ) − s(t ˙ 2 )| 6 F1 |@xx (1; t1 ) − @xx (1; t2 )| + F2 |t1 − t2 |;

(7.43)

where Z

F1 =

2 sm rm

Z 2 rm sm

F2 = with  rM

;


2a + sm

 rM



rm

(D1 + D2 TJ B=2 )

   d rain (t)  :  = max  dt  t∈[0;TJ ]

Setting



>=a

TJ + F1 D3 TJ E=2−= + F2 TJ (3+E)=2−=

(7.44)

and using (7.43) and (7.31) we have s ˙ H = (7; TJ) = sup |s(t)| ˙ + t1 ∈(7;TJ )

|s(t ˙ 1 ) − s(t ˙ 2 )| |t1 − t2 |= t1 ; t2 ∈(7;TJ ) sup

6 a + {F1 @H 2+E; 1+E=2 (E 7J ) + F2 TJ }|t1 − t2 |E=2−= T

 > 1   6 √ a TJ + F1 D3 TJ E=2−= + F2 TJ (3+E)=2−= = √ : 7 7 Thus, if : ∈ S and a, TJ , = and > are given by (7.40), (7.41), (7.42) and (7.44), respectively, then the same is true for s = M:. 7.5. Existence and uniqueness theorem for problem 22 Now we consider S endowed with the C 1 (0; TJ ) norm and show that M is a contraction. Proposition 7.5. There exists a time T2 6 TJ such that M is a contractive mapping of S into itself. Proof. Let us take :1 , :2 ∈ S and the corresponding solutions @1 (y; t), @2 (y; t) to problem (7.23). We introduce the functions :(t) ˜ = :1 (t) − :2 (t);

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˜ t) = @1 (x; t) − @2 (x; t); @(x; U˜ (x; t) = U1 (x; t) − U2 (x; t); Ui (x; t), i = 1; 2, being de%ned by (7.28), where the index k has been replaced by i. Extending problem (7.23) in an antisymmetric way to the interval [ − 1; 0] we have ˜ t) solves that @(x;  @˜ t = Q1 (@1 ; :1 )@˜ xx + F1 (@1 ; :1 )@˜ x + F3 (x; t)@˜ + F4 (x; t):(t); ˜       @(x; ˜ 0) = 0; (7.45)   @˜ x (1; t) = h(t):(t); ˜     ˜ @x (−1; t) = h(t):(t); ˜ where F1 has been introduced in Theorem 7.2 and F3 (x; t) = QJ1@ @2 xx + FJ1@ @2x ; F4 (x; t) = FJ1: @2x ; h(t) = Ksat

rain (t)

Z ; (Z − :1 )(Z − :2 )

where the symbols QJ1@ , FJ1@ , FJ1: are de%ned via the mean value theorem. It is easy to obtain an estimate for the source term F4 :˜ directly from its de%nition, namely 7    d :˜  (7.46) |F4 (x; t):(t)| ˜ 6 C1   ; dt t where C1 is a positive constant and where we have exploited the fact that : ˜ t6 td :=dt ˜ ˜ = 0). Further, since boundary data are dominated by C2 : ˜ t, t (recall :(0) using maximum principle and Gronwall’s Lemma we can write    d :˜  ˜ |@(x; t)| 6 Ct   : (7.47) dt t Let us now consider the problem solved by U˜ (y; t), namely  U˜ t = Q1 (x; t)U˜ xx + Q3 (x; t)U˜ x + Q6 (x; t);       U˜ (x; 0) = 0;  U˜ (1; t) = 0;      ˜ U (−1; t) = 0 7

Here we use the symbol  · t with the meaning ft = sup |f(7)|: 7∈(0; t)

(7.48)

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797

with Q6 (x; t) = [Q1 (@1 ; :1 ) − Q1 (@2 ; :2 )]U2 xx +[Q3 (@1 ; @1x ; :1 ) − Q3 (@2 ; @2x ; :2 )]U2x + Q4 (:1 )U1 −Q4 (:2 )U2 + Q5 (@1 ; @1x ; :1 ) − Q5 (@1 ; @1x ; :1 ); the quantities Q3 , Q4 , Q5 being equal to those de%ned in Theorem 7.1. Because of (7.28) we have |@˜ x (x; t)| | 6 |U˜ (x; t)| + C|:(t)|; ˜ and so, exploiting (7.47) along with the above estimates for |U2xx | and |:|, ˜ we get
    d :˜  |Q6 (x; t)| 6 C   + |U˜ (x; t)| : (7.49) dt t We can represent the solution U˜ (x; t) in the following way:  t  +1 U˜ (x; t) = G(x; ; t; 7)Q6 (; 7) d d7; 0

−1

where G(x; ; t; 7) is the Green’s function of problem (7.48) and so
    t  +1  d :˜  1   + |U˜ (; 7)| d d7; √ |U˜ (x; t)| 6 C t − 7  dt 7 0 −1 where estimate (7.49) has been used. Hence, introducing G(t) = sup |U˜ (x; t)|; x∈[−1;1]

we get

   √  d :˜  G(t) 6 C t   + C dt t

0

t

G(7) √ d7: t−7

The generalized version of Gronwall’s Lemma entails   √  d :˜  G(t) 6 tC   : dt t

(7.50)

Then, with the aid of Theorem 16.3 [18, p. 413] and exploiting (7.49) and (7.50) we obtain    t  +1   √  d :˜  1+ 7 (x − )2 ˜ |@xx (x; t)| = |U˜ (x; t)| 6 C   d d7 exp −C dt t 0 −1 t − 7 t−7   √  d :˜  6 C t   : (7.51) dt t

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I. Borsi et al. / Nonlinear Analysis: Real World Applications 5 (2004) 763 – 800

Recalling now (7.37), (7.39) and estimates (7.47) and (7.51) we have   Z ˜ 3Z 2  ˜ |@xx (1; t)| + 4 |@2 xx (1; t)| |:| |s˙1 (t) − s˙2 (t)| 6 2 sm r m sm   √  d :˜  ˜ t) 6 C( t   + : dt t   √  d :˜  6 C t   : dt 0

(7.52)

Since M:1 − M:2 0 = s1 − s2 0 6 ts˙1 − s˙2 0 , using (7.52) we easily obtain   √  d :˜  s1 − s2 C 1 (0; t) 6 (t + 1)s˙1 − s˙2 0 6 C(t + 1) t   dt 0 √ ˜ C 1 (0; TJ) : 6 C(TJ + 1) t: Thus there exists a time interval (0; T2 ), T2 6 TJ , in which M is a contraction with respect to the C 1 norm. The proposition is proved.

8. Conclusions In this paper, we have formulated a 1D model describing percolation when rainfall events of particularly high intensity takes place in the extreme case of total runo) of the excess water. Despite the fact that rainwater percolation is an old and well-known problem, a rigorous analysis is missing and mainly thumb rules have been proposed. We have emphasize that the most important physical feature is the presence of a constraint on the value reached by the pressure at the in>ow surface. Such a fact makes the formulation of the correct boundary condition highly nontrivial. We have modelled the phenomenon considering a >ow through an unsaturated porous medium and determined a condition on the rainfall rate ensuring that the top surface comes to saturation in a %nite time. The crucial question that has been discussed is how to write the in>ow boundary condition on the soil surface after the onset of saturation. We realized that two di)erent types of boundary condition are possible. The switch from one type of boundary condition to the other is controlled by physical constraints. Indeed, assuming no ponding (namely no water accumulation), when all the incoming rain >ux is allowed to enter the ground the pressure at the top surface has to be less than the pressure at which rain is supplied (a quantity physically well de%ned and called rain pressure). On the other hand, when keeping the >ux condition would violate the pressure constraint, we must replace it by the condition that pressure is precisely the rain pressure. At the same time we must check that the new boundary condition does not make the in>ow larger than the one which is actually available, i.e. the rainfall rate.

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799

Once the saturation front has appeared it separates two regions with di)erent >ow regimes (percolation in partially saturated soil). The presence of a constrained boundary condition generates a peculiar free boundary problem, namely a “second order” Stefan problem (see [13]) with self-determined boundary conditions. We have shown that, under particular circumstances, the problem can be analyzed in two distinct time intervals, corresponding to di)erent boundary value problems. In both intervals we have proved existence and uniqueness of the solution using %xed point techniques. Due to the fact that derivatives of order higher than one govern the motion of the free boundary (which is the interface between saturated and unsaturated region) several nontrivial estimates had to be derived. Although for practical purposes the rain pressure can be neglected, we believe that the present study %lls a theoretical gap in the way the rain water percolation problem is approached. In this way not only the scope of providing a satisfactory description of the physical mechanism involved is obtained, but a sound justi%cation is produced for the approximated procedures that are commonly used in practice. References [1] H.W. Alt, A free boundary problem associated with the >ow of ground water, Arch. Rat. Mech. Anal. 64 (1977) 111–126. [2] H.W. Alt, S. Luckhaus, Quasilinear elliptic-parabolic di)erential equations, Math. Z. 183 (1983) 311–341. [3] J. Bear, Dynamics of Fluids in Porous Media, American Elsevier, New York, 1972. [4] J. Bear, Y. Bachmat, Introduction to modeling of transport phenomena, in: J. Bear (Ed.), Porous Media, Theory and Applications of Transport in Porous Media, Vol. 4, Kluwer, Hardbound, 1991. [5] J. Bear, A. Verruijt, Modelling groundwater >ow and pollution with computer programs for sample cases, theory and applications of transport, in: J. Bear (Ed.), Porous Media, Vol. 1, Kluwer, Hardbound, 1987. [6] J.R. Cannon, The one-dimensional heat equation, in: G.C. Rota (Ed.), Encyclopedia of Mathematics and its Applications, Vol. 23, Addison Wesley, Menlo Park, CA, 1984. [7] J.R. Cannon, R.B. Guenther, F.A. Mohamed, On the rainfall in%ltration through a soil medium, SIAM J. Appl. Math. 49 (1989) 720–729. [8] C. Ciliberto, Formule di maggiorazione e teoremi di esistenza per le soluzioni delle equazioni paraboliche in due variabili, Ric. Mat. 3 (1954) 40–75. [9] C. Corradini, F. Melone, R.E. Smith, Modelling in%ltration during complex rainfall sequences, Water Resour. Res. 30 (1994) 2777–2784. [10] C. Corradini, F. Melone, R.E. Smith, A uni%ed model for in%ltration and redistribution during complex rainfall patterns, J. Hydrol. 192 (1997) 104–124. [11] C.J. van Duyn, J.B. McLeod, Nonstationary %ltration in partially saturated porous media, Arch. Rat. Mech. Anal. 78 (1982) 173–198. [12] A. Fasano, M. Primicerio, Free boundary problems for nonlinear parabolic equations with nonlinear free boundary conditions, J. Math. Anal. Appl. 72 (1979) 247–273. [13] A. Fasano, M. Primicerio, Liquid >ow in partially saturated porous media, J. Inst. Math. Appl. 23 (1979) 503–517. [14] J. Filo, S. Luckhaus, Modelling surface runo) and in%ltration of rain by an elliptic–parabolic equation coupled with a %rst-order equation on the boundary, Arch. Rat. Mech. Anal. 146 (1999) 157–182. [15] A. Friedman, Partial Di)erential Equations of Parabolic Type, Prentice-Hall, Engelwood Cli)s, NJ, 1964. [16] R. Gianni, A %ltration problem with ponding, Bollettino U.M.I. 5B (1991) 875–891. [17] W.H. Green, G.A. Ampt, Studies on soil physics—the >ow of air and water through soils, J. Agric. Sci. 4 (1911) 1–24.

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[18] O.A. Ladyzenskaja, V.A. Solonnikov, N.N. Ural’ceva, Linear and Quasi-linear Equations of Parabolic Type, in: Translation of Mathematical Monographs, Vol. 23, American Mathematical Society, Providence, RI, 1968. [19] H.V. Nguyen, D.F. Durso, Absorption of water by %ber webs: an illustration of di)usion transport, TAPPI J. 66 (1983) 76–79. [20] F. Otto, L1 Contraction and uniqueness for unstationary saturated–unsaturated porous media >ow, Adv. Math. Sci. Apl. 7 (1997) 537–553. [21] K.R. Rajagopal, L. Tao, Mechanics of Mixtures, World Scienti%c, River Edge, NJ, 1985. [22] I. Rubinstein, L. Rubinstein, Partial Di)erential Equations in Classical Mathematical Physics, Cambridge University Press, Cambridge, 1998. [23] K.S. Shifrin, Physical Optics of Ocean Water, American Institute of Physics, Providence, RI, 1988. [24] R.E. Smith, F. Melone, Modelling in%ltration for multistorm runo) events, Water Resour. Res. 29 (1993) 133–144.