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Advances in Mathematical Physics
Volume 2011 (2011), Article ID 606757, 26 pages
http://dx.doi.org/10.1155/2011/606757
Research Article

On the Solution of a Hyperbolic One-Dimensional Free Boundary Problem for a Maxwell Fluid

Dipartimento di Matematica “Ulisse Dini”, Università degli Studi di Firenze, Viale Morgagni 67/A, 50134 Firenze, Italy

Received 11 March 2011; Revised 12 May 2011; Accepted 14 June 2011

Academic Editor: Luigi Berselli

Copyright © 2011 Lorenzo Fusi and Angiolo Farina. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study a hyperbolic (telegrapher's equation) free boundary problem describing the pressure-driven channel flow of a Bingham-type fluid whose constitutive model was derived in the work of Fusi and Farina (2011). The free boundary is the surface that separates the inner core (where the velocity is uniform) from the external layer where the fluid behaves as an upper convected Maxwell fluid. We present a procedure to obtain an explicit representation formula for the solution. We then exploit such a representation to write the free boundary equation in terms of the initial and boundary data only. We also perform an asymptotic expansion in terms of a parameter tied to the rheological properties of the Maxwell fluid. Explicit formulas of the solutions for the various order of approximation are provided.

1. Introduction

In this paper we study the well posedness of a hyperbolic free boundary problem arisen from a one-dimensional model for the channel flow of a rate-type fluid with stress threshold presented in [1]. The model describes the one-dimensional flow of a fluid which behaves as a nonlinear viscoelastic fluid if the stress is above a certain threshold 𝜏𝑜 and like a rate type fluid if the stress is below that threshold. The problem investigated here belongs to a series of extensions of the classical Bingham model we have proposed in recent years (see [25]).

In particular, in [1] we describe the one-dimensional flow of such a fluid in an infinite channel, assuming that in the outer part of the channel the material behaves as a viscoelastic upper convected Maxwell fluid, while in the inner core as a rate-type Oldroyd-B fluid. The general mathematical model is derived within the framework of the theory of natural configurations developed by Rajagopal and Srinivasa (see [6]). The constitutive equations are obtained imposing how the system stores and dissipates energy and exploiting the criterion of the maximization of the dissipation rate.

The main practical motivation behind this study comes from the analysis of materials like asphalt or bitumen which exhibit a stress threshold beyond which they change its rheological properties. Indeed from the papers [79], it is clear that such materials have a viscoelastic behaviour (for instance, upper convected Maxwell fluid) which is observed if the applied stress is greater than a certain threshold (see, in particular, [7]).

The mathematical formulation for the channel flow driven by a constant pressure gradient consists in a free boundary problem involving a hyperbolic telegrapher's equation (Maxwell fluid) and a third-order equation (Oldroyd-B fluid). The free boundary is the surface dividing the two domains: the inner channel core and the external layer. Due to the high complexity of the general problem, here we have considered a simplified version which arises when the order of magnitude of some physical parameters involved in the general model ranges around particular values. In such a case we have that the velocity of the inner core is constant in space and time, while the outer part behaves as a viscoelastic upper convected Maxwell fluid (see [1] for more details). The mathematical formulation turns out to be a hyperbolic free boundary problem which, in the authors knowledge, is new since it involves a telegrapher's equation coupled with an ODE describing the evolution of the interface.

The paper is structured as follows. In Section 2 we formulate the problem, namely, problem (2.1), and specify the basic assumptions. In Section 3 we give an equivalent formulation of the problem which leads to a nonlinear integrodifferential equation for the free boundary. We prove local existence and uniqueness for such an equation (see Theorem 3.3), under specific assumptions on the data.

The interesting aspect of the mathematical analysis lies on the technique we employ to reduce the complete problem to a single integrodifferential equation from which some mathematical properties can be derived (the free boundary equation can be solved autonomously from the governing equation of the velocity field). Such a methodology is a generalization of a technique already introduced in [2].

In Section 4 we perform an asymptotic expansion in terms of a coefficient 𝜔 (representing the ration between the elastic characteristic time and the relaxation time of the viscoelastic material), which typically is of the order 𝑂(101). This procedure allows to obtain approximations of the actual solution up to any order through an iterative procedure. We do not prove the convergence of the asymptotic approximations to the actual solution (whose existence is proved in Theorem 3.3), limiting ourselves to develop only the formal procedure. Indeed, the main advantage of this procedure is that, for each order of approximation, the governing equation for the velocity field is the “standard” wave equation, which is by far easier to handle than the telegrapher's equation. We end the paper with few conclusive remarks.

2. Mathematical Formulation

In this section we state the mathematical problem. We refer the reader to [1] for all the details describing how this simplified model was derived from the general one. In the general case, in the region [0,𝑠], the fluid behaves as an Oldroyd-B type fluid. The problem we are studying here is a particular case of such a model, which stems from some specific assumptions on the physical parameters (fulfilled by materials like asphalt and bitumen). Under such assumptions, the inner core [0,𝑠] moves with uniform constant velocity 𝑉𝑜.

We consider an orthogonal coordinate system 𝑥𝑜𝑦 and assume that the fluid is confined between two parallel plates placed at distance 2𝐿. We assume that the motion takes place along the 𝑥-direction and that the velocity field has the form 𝑖𝑣(𝑦,t)=𝑣(𝑦,𝑡). We rescale the problem (in a nondimensional form) and, because of symmetry, we study the upper part of the layer 𝑦[0,1] (the space variable is rescaled by 𝐿). The geometry of the system we investigate is depicted in Figure 1.

606757.fig.001
Figure 1: Upper part of the channel.

The mathematical model is written for the velocity field 𝑣(𝑦,𝑡) in the viscoelastic region which is separated from the region with zero strain rate (uniform velocity) by the moving interface 𝑦=𝑠(𝑡).

The nondimensional formulation is the following:𝑣𝑡𝑡+2𝜔𝑣𝑡=𝑣𝑦𝑦+𝛽2𝑣𝑦(𝑠,1),𝑡>0,(𝑦,0)=𝑣𝑜𝑠(𝑦)𝑦𝑜,𝑣,1𝑡(𝑠𝑦,0)=0𝑦𝑜,𝑣,1𝑣(1,𝑡)=0𝑡>0,(𝑠,𝑡)=𝑉𝑜𝑣𝑡>0,𝑦(𝑠,𝑡)+̇𝑠𝑣𝑡(𝑠,𝑡)=𝛽2𝑠𝙱𝚗𝑡>0,(0)=𝑠𝑜,𝑠𝑜(0,1).(2.1) where (i)𝜌 is the material density, (ii)𝜂 is the viscosity of the fluid, (iii)𝜇 is the elastic modulus, (iv)𝛽2 is a positive parameter depending on the viscosity 𝜂 (see [1]), (v)𝙱𝚗 is the Bingham number, (vi)𝑉𝑜 is the velocity of the inner core, (vii)2𝜔=𝑡𝑒/𝑡𝑟, (viii)𝑡𝑒=𝐿𝜌/𝜇 is the characteristic elastic time, (ix)𝑡𝑟=𝜂/2𝜇 is the relaxation time.

In the case of asphalt typical values are (see [8, 9])𝜇=1MPa,𝜌=1.5×103K𝑔/m3,𝜂=102MPas.(2.2)

Taking 𝐿=500m we get𝑡𝑒=15s,𝑡𝑟=50s,𝜔=0.15.(2.3)

Remark 2.1. In [1] we have proved that problem (2.1) admits a stationary solution provided 𝑉𝑜𝛽212+𝙱𝚗(2.4) and that the stationary solution is given by 𝑣𝛽(𝑦)=22(𝑠𝙱𝚗𝑦)2+𝛽2𝙱𝚗2+𝑉𝑜,𝑠=1+𝙱𝚗𝙱𝚗2+2𝑉𝑜𝛽2.(2.5)

3. An Equivalent Formulation

Before proceeding in proving analytical results of problem (2.1) we introduce the new coordinate system𝑥=1𝑦,𝑦=1𝑥,(3.1) and the new variable𝑈(𝑥,𝑡)=exp(𝜔𝑡)𝑣(1𝑥,𝑡),𝑣(𝑦,𝑡)=𝑈(1𝑦)exp(𝜔𝑡).(3.2) With transformations (3.1)-(3.2), problem (2.1) becomes𝑈𝑥𝑥𝑈𝑡𝑡+𝜔2𝑈=𝛽2exp(𝜔𝑡),𝑥(0,𝜉),𝑡>0,𝑈(𝑥,0)=𝑈𝑜(𝑥),𝑥0,𝜉𝑜,𝑈𝑡(𝑥,0)=𝑈1(𝑥),𝑥0,𝜉𝑜,𝑈𝑈(0,𝑡)=0,𝑡>0,(𝜉,𝑡)=exp(𝜔𝑡)𝑉𝑜𝑈,𝑡>0,𝑥̇(𝜉,𝑡)+𝜉𝑈𝑡̇(𝜉,𝑡)𝜉𝜔𝑈(𝜉,𝑡)=exp(𝜔𝑡)𝛽2𝙱𝚗,𝑡>0,𝜉(0)=𝜉𝑜,𝜉𝑜(0,1),(3.3)

where𝜉(𝑡)=1𝑠(𝑡),𝜉𝑜=1𝑠𝑜,𝑈𝑜(𝑥)=𝑣𝑜(1𝑥),𝑈1(𝑥)=𝜔𝑈𝑜(𝑥).(3.4) Notice that, by means of (3.2), the evolution equation for the new variable 𝑈(𝑥,𝑡) has become a nonhomogeneous Klein-Gordon equation [10].

The domain of problem (3.3) is depicted in Figure 2. We begin by considering the domain 𝐷𝐼 (see Figure 3). Here the solution has the representation formula (see [11])1𝑈(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)+𝑈𝑜(+1𝑥+𝑡)2𝑥+𝑡𝑥𝑡𝑅(𝑥,𝑡;𝜁,0)𝑈1(𝜁)𝑅𝜃(𝑥,𝑡;𝜁,0)𝑈𝑜(+𝛽𝜁)𝑑𝜁22𝑡0exp(𝜔𝜃)𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝑅(𝑥,𝑡;𝜁,𝜃)𝑑𝜁,(3.5) where 𝑅(𝑥,𝑡;𝜁,𝜃) is the Riemann's function that solves the problem (see again Figure 3)𝑅𝜁𝜁𝑅𝜃𝜃+𝜔2𝑅[],[],𝑅=0(𝜁,𝜃)Ω(𝑥,𝑡),(𝑥,𝑡;𝑥+𝑡𝜃,𝜃)=1𝜃0,𝑡𝑅(𝑥,𝑡;𝑥𝑡+𝜃,𝜃)=1𝜃0,𝑡Ω(𝑥,𝑡)={(𝑥,𝑡)𝑥𝑡+𝜃𝜁𝑥+𝑡𝜃,0𝜃𝑡}.(3.6)

606757.fig.002
Figure 2: Sketch of the domain.
606757.fig.003
Figure 3: The domain 𝐷𝐼.

To determine the solution of problem (3.6) we set𝑧=(𝑡𝜃)2(𝜁𝑥)2,(3.7) where𝑧2𝜃𝑧2𝜁𝑧=1,𝜃𝜃𝑧𝜁𝜁=1𝑧.(3.8)

By means of (3.7) problem (3.6) becomes𝑅𝑅(𝑧)+(𝑧)𝑧𝜔2𝑅𝑅(𝑧)=0(0)=1,(3.9) where (3.9)(1) is the modified Bessel equation of zero order. The solution of (3.9) is given by𝑅(𝑧)=𝑅(𝑥,𝑡;𝜁,𝜃)=𝐼𝑜𝜔(𝑡𝜃)2(𝜁𝑥)2,(3.10) where 𝐼𝑜 is the modified Bessel function of zero order. It is easy to prove that the function defined by (3.10) satisfies problem (3.6). Moreover, since [12]𝐼𝑜(𝑥)𝑥=12𝐼𝑜(𝑥)𝐼2(𝑥),(3.11)

one can prove that𝑅𝜃𝜔(𝑥,𝑡;𝜁,𝜃)=2(𝜃𝑡)2𝐼𝑜𝜔(𝑡𝜃)2(𝜁𝑥)2𝐼2𝜔(𝑡𝜃)2(𝜁𝑥)2,(3.12) where 𝐼2 is the modified Bessel function of second order. Recalling that 𝑈(𝑥,0)=𝑈𝑜(𝑥) and, by (3.3)(3), (3.4), that𝑈𝑡(𝑥,0)=𝜔𝑈𝑜(𝑥),(3.13)we see that𝑅(𝑥,𝑡;𝜁,0)𝜔𝑅𝜃𝑈(𝑥,𝑡;𝜁,0)𝑜=𝐼(𝜁)𝑜𝜔𝑡2(𝜁𝑥)2𝜔𝜔+2𝑡2𝜔2𝑡2𝐼2𝜔𝑡2(𝜁𝑥)2𝑈𝑜(𝜁),(3.14)and representation formula (3.5) can be rewritten as 1𝑈(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)+𝑈𝑜+1(𝑥+𝑡)2𝑥+𝑡𝑥𝑡𝐼𝑜𝜔𝑡2(𝜁𝑥)2𝜔𝜔+2𝑡2𝜔2𝑡2𝐼2𝜔𝑡2(𝜁𝑥)2𝑈𝑜+𝛽(𝜁)𝑑𝜁22𝑡0exp𝑘𝜃2𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝐼𝑜𝜔(𝑡𝜃)2(𝜁𝑥)2𝑑𝜁.(3.15)

Let us now write a representation formula for 𝑈(𝑥,𝑡) in the domain 𝐷𝐼𝐼 (see Figure 4). We once again make use of (3.6), where now 𝑈𝑜 has to be extended to the domain [𝜉𝑜,0]. Following [2], we extend 𝑈𝑜 imposing condition (3.3)(4), that is, 𝑈(0,𝜃)=0. From the representation formula we get10=2𝑈𝑜𝑡+𝑈𝑜𝑡+12𝑡𝑡𝑅0,𝑡;𝜁,0𝜔𝑅𝜃0,𝑡𝑈;𝜁,0𝑜+𝛽(𝜁)𝑑𝜁22𝑡0exp(𝜔𝜃)𝑑𝜃𝑡𝜃𝑡+𝜃𝑅0,𝑡;𝜁,𝜃𝑑𝜁,(3.16) where 𝑡=𝑡𝑥(3.17)is the coordinate of the intersection of the characteristic 𝜁=𝑥𝑡+𝜃 with 𝜁=0. Relation (3.16) can be rewritten as10=2𝑈𝑜(𝑥𝑡)+𝑈𝑜(+1𝑡𝑥)2𝑡𝑥𝑥𝑡𝑅(0,𝑡𝑥;𝜁,0)𝜔𝑅𝜃(𝑈0,𝑡𝑥;𝜁,0)𝑜(+𝛽𝜁)𝑑𝜁220𝑡𝑥exp(𝜔𝜃)𝑑𝜃𝑡𝑥𝜃𝑥𝑡+𝜃𝑅(0,𝑡𝑥;𝜁,𝜃)𝑑𝜁.(3.18) From (3.18), the extended function 𝑈𝑠x𝑜(𝑥), defined in [𝜉𝑜,0], fulfills the following Volterra integral equation of second type: 𝑈𝑜𝑠𝑥(𝑥𝑡)0𝑥𝑡𝑅(0,𝑡𝑥;𝜁,0)𝜔𝑅𝜃(𝑈0,𝑡𝑥;𝜁,0)𝑜𝑠𝑥(𝜁)𝑑𝜁=𝑈𝑜(𝑡𝑥)+0𝑡𝑥𝑅(0,𝑡𝑥;𝜁,0)𝜔𝑅𝜃(𝑈0,𝑡𝑥;𝜁,0)𝑜(𝛽𝜁)𝑑𝜁220𝑡𝑥exp(𝜔𝜃)𝑑𝜃𝑡𝑥𝜃𝑥𝑡+𝜃𝑅(0,𝑡𝑥;𝜁,𝜃)𝑑𝜁.(3.19) Equation (3.19) can be put in the more compact form𝑈𝑜𝑠𝑥(𝜒)𝜒0𝐾𝑠𝑥(𝜒,𝜁)𝑈𝑜𝑠𝑥(𝜁)𝑑𝜁=𝐹𝑠𝑥(𝜒),(3.20)where 𝜒=𝑥𝑡[𝜉𝑜,0] and 𝐾𝑠𝑥(=𝐼𝜒,𝜁)𝑜𝜔𝜒2𝜁2𝜔𝜔2𝜒2+𝜔2𝜒2𝐼2𝜔𝜒2𝜁2,𝐹𝑠𝑥(𝜒)=𝑈𝑜(𝜒)+0𝜒𝑅(0,𝜒;𝜁,0)𝜔𝑅𝜃𝑈(0,𝜒;𝜁,0)𝑜𝛽(𝜁)𝑑𝜁220𝑥exp(𝜔𝜃)𝑑𝜃𝑥𝜃𝑥+𝜃𝑅(0,𝑥;𝜁,𝜃)𝑑𝜁.(3.21)

606757.fig.004
Figure 4: The domain 𝐷𝐼𝐼.

Due to the regularity of the kernel 𝐾𝑠𝑥(𝜒,𝜁) the function 𝑈𝑜𝑠𝑥(𝜒) (which can be determined using the iterated kernels method, [13]) is a smooth function. Thus we extend 𝑈𝑜(𝑥) as 𝑈𝑜(𝑈𝑥)=𝑜𝑠𝑥(𝑥),𝑥𝜉𝑜,𝑈,0𝑜(𝑥),𝑥0,𝜉𝑜,(3.22)and the solution 𝑈(𝑥,𝑡) in the domain 𝐷𝐼𝐼 is given by 1𝑈(𝑥,𝑡)=2𝑈𝑜(𝑥+𝑡)𝑈𝑜(+1𝑡𝑥)2𝑥𝑡𝑡𝑥𝑅(0,𝑡𝑥;𝜁,0)𝜔𝑅𝜃(𝑈0,𝑡𝑥;𝜁,0)𝑜(+1𝜁)𝑑𝜁2𝑥+𝑡𝑥𝑡𝑅(𝑥,𝑡;𝜁,0)𝜔𝑅𝜃𝑈(𝑡,𝑥;𝜁,0)𝑜𝛽(𝜁)𝑑𝜁+220𝑡𝑥𝑒𝜔𝜃𝑑𝜃𝑥𝑡+𝜃𝑡𝑥𝜃+𝛽𝑅(0,𝑡𝑥;𝜁,𝜃)𝑑𝜁22𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝑅(𝑥,𝑡;𝜁,𝜃)𝑑𝜁.(3.23)

Remark 3.1. We notice that, considering the representation formulae (3.5) and (3.23), lim𝑥𝑡+𝑈(𝑥,𝑡)=lim𝑥𝑡𝑈(𝑥,𝑡).(3.24) Moreover, taking the first derivatives (with respect to time 𝑡 and space 𝑥) of 𝑈(𝑥,𝑡) for the domains 𝐷𝐼 and 𝐷𝐼𝐼 it is easy to prove that, assuming the compatibility condition 𝑈𝑜(0)=0, lim𝑥𝑡+𝑈𝑥(𝑥,𝑡)=lim𝑥𝑡𝑈𝑥𝜉(𝑥,𝑡),𝑡0,𝑜2,lim𝑥𝑡+𝑈𝑡(𝑥,𝑡)=lim𝑥𝑡𝑈𝑡𝜉(𝑥,𝑡),𝑡0,𝑜2,(3.25) where the derivatives in limits on the l.h.s. of (3.25) are evaluated using (3.5), while the ones on the r.h.s. using (3.23). This implies that the solution is 𝐶1 across the characteristic 𝑥=𝑡, that is, the line that separates the domains 𝐷𝐼 and 𝐷𝐼𝐼.

We now write the representation formula for 𝑈(𝑥,𝑡) in the domain 𝐷𝐼𝐼𝐼. We proceed as in [2] assuming that the velocity of the free boundary 𝑥=𝜉(𝑡) is less than the velocity of the characteristics (i.e., |̇𝜉|<1) and extending 𝑈𝑜 to the domain [𝜉𝑜,𝜉(𝜉𝑜)+𝜉𝑜] (see Figure 2) in a way such that 𝑈(𝜉,𝑡)=exp(𝜔𝑡)𝑉𝑜 (i.e., imposing the free boundary condition (3.3)(5)).

Given a point (𝑥,𝑡) in the domain 𝐷𝐼𝐼𝐼 we define the point (𝜉,𝑡) as the intersection of the characteristic (with negative slope) passing from (𝑥,𝑡) and the free boundary 𝑥=𝜉(𝑡) (see Figure 5). It is easy to check that 𝜉+𝑡=𝑥+𝑡,𝑡=𝑡(𝑥,𝑡),(3.26)𝜕𝑡=1𝜕𝑡̇𝜉(𝑡,)+1𝜕𝑡=1𝜕𝑥̇𝜉(𝑡.)+1(3.27)We consider once again the representation formula (3.5) and impose condition (3.3)(5), getting2𝑒𝜔𝑡𝑉𝑜=𝑈𝑜𝜉𝑡+𝑈𝑜𝜉+𝑡+𝜉+𝑡𝜉𝑡𝑅𝜉,𝑡;𝜁,0𝜔𝑅𝜃𝜉,𝑡𝑈;𝜁,0𝑜(𝜁)𝑑𝜁+𝛽2𝑡0𝑒𝜔𝜃𝑑𝜃𝜉+𝑡𝜉𝜃𝑡+𝜃𝑅𝜉,𝑡;𝜁,𝜃𝑑𝜁.(3.28)From (3.28) we see that the extension 𝑈𝑜𝑑𝑥 to the domain [𝜉𝑜,𝜉(𝜉𝑜)+𝜉𝑜] is the solution of the following Volterra integral equation of second kind: 𝑈𝑜𝑑𝑥𝜉+𝑡+𝜉+𝑡𝜉𝑜𝑅𝜉,𝑡;𝜁,0𝜔𝑅𝜃𝜉,𝑡𝑈;𝜁,0𝑜𝑑𝑥(𝜁)𝑑𝜁=2𝑒𝜔𝑡𝑉𝑜𝑈𝑜𝜉𝑡𝜉𝑜𝜉𝑡𝑅𝜉,𝑡;𝜁,0𝜔𝑅𝜃𝜉,𝑡𝑈;𝜁,0𝑜(𝜁)𝑑𝜁𝛽2𝑡0𝑒𝜔𝜃𝑑𝜃𝜉+𝑡𝜉𝜃𝑡+𝜃𝑅𝜉,𝑡;𝜁,𝜃𝑑𝜁.(3.29)Recalling (3.26) and proceeding as for the domain 𝐷𝐼𝐼, the above can be rewritten as𝑈𝑜𝑑𝑥(𝜒)+𝜒𝜉𝑜𝐾𝑑𝑥(𝜒,𝜁)𝑈𝑜𝑑𝑥(𝜁)𝑑𝜁=𝐹𝑑𝑥(𝜒),(3.30)where 𝜒=𝑥+𝑡 and𝐾𝑑𝑥𝐼(𝜒,𝜁)=𝑜𝜔(𝑡(𝜒))2𝜁𝜉(𝜒)2𝜔𝜔+2𝑡(𝜒)2𝜔2𝑡(𝜒)2𝐼2𝜔(𝑡(𝜒))2𝜉(𝜒)𝜁2,𝐹𝑑𝑥(𝜒)=2𝑒𝜔𝑡𝑉𝑜𝑈𝑜(𝑥𝑡)𝜉𝑜𝑥𝑡𝑅(𝑥,𝑡;𝜁,0)𝜔𝑅𝜃𝑈(𝑥,𝑡;𝜁,0)𝑜(𝜁)𝑑𝜁𝛽2𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃||||𝑅(𝑥,𝑡;𝜁,𝜃)𝑑𝜁(𝑥=𝜉(𝜒),𝑡=𝑡(𝜒)).(3.31) Once again the regularity of the kernel 𝐾𝑑𝑥(𝜒,𝜁) ensures the regularity of the solution 𝑈𝑜𝑑𝑥(𝜒). The function 𝑈𝑜(𝑥) can thus be defined in the interval [𝜉𝑜,𝜉(𝜉𝑜)+𝜉𝑜] as𝑈𝑜𝑈(𝑥)=𝑜𝑠𝑥(𝑥),𝑥𝜉𝑜,𝑈,0𝑜(𝑥),𝑥0,𝜉𝑜,𝑈𝑜𝑑𝑥(𝜉𝑥),𝑥𝑜𝜉,𝜉𝑜+𝜉𝑜.(3.32) The solution in the domain 𝐷𝐼𝐼𝐼 is thus given by𝑈(𝑥,𝑡)=𝑒𝜔𝑡𝑉𝑜+12𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉𝑡+12𝑥+𝑡𝑥𝑡𝑃(𝑥,𝑡;𝜁,0)𝑈𝑜(1𝜁)𝑑𝜁2𝜉𝑥+𝑡𝑡𝑃𝜉,𝑡𝑈;𝜁,0𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝛽𝑅(𝑥,𝑡;𝜁,𝜃)𝑑𝜁22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉𝑥+𝑡𝜃𝑡+𝜃𝑅𝜉,𝑡;𝜁,𝜃𝑑𝜁,(3.33) where for simplicity of notation we have introduced𝑃(𝑥,𝑡;𝜁,𝜃)=𝑅(𝑥,𝑡;𝜁,𝜃)𝜔𝑅𝜃(𝑥,𝑡;𝜁,𝜃),(3.34)and where 𝑈𝑜(𝑥) is given by (3.32). Therefore for any fixed 𝐶1 function 𝜉(𝑡) with |̇𝜉|<1 we have that the solution to problem (3.3)(1–5) is given by (3.5), (3.23), (3.33) with 𝑈𝑜 defined by (3.32). At this point we make use of (3.3)(6) to determine the evolution equation of the free boundary 𝑥=𝜉(𝑡). We begin writing the derivatives 𝑈𝑡(𝑥,𝑡) and 𝑈𝑥(𝑥,𝑡). To this aim we exploit formula (3.33) since 𝑈𝑡 and 𝑈𝑥 have to be evaluated on 𝑥=𝜉(𝑡), which belongs to domain 𝐷𝐼𝐼𝐼. Differentiating (3.33) with respect to 𝑥 we get𝑈𝑥1(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉𝑡̇𝜉𝑡1̇𝜉(𝑡+)+1𝜔𝑉𝑜𝑒𝜔𝑡̇𝜉+1+12𝑃(𝑥,𝑡;𝑥+𝑡,0)𝑈𝑜(𝑥+𝑡)𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜1(𝑥𝑡)2𝑃𝜉,𝑡𝑈;𝑥+𝑡,0𝑜𝜉(𝑥+𝑡)𝑃,𝑡;𝜉𝑡𝑈,0𝑜𝜉𝑡̇𝜉𝑡1̇𝜉(𝑡)1+12𝜉𝑥+𝑡𝑡𝑃𝑥𝜉,𝑡̇𝜉𝑡;𝜁,0+𝑃𝑡𝜉,𝑡𝑈;𝜁,0𝑜(𝜁)̇𝜉(𝑡+1)+1𝑑𝜁2𝑥+𝑡𝑥𝑡𝑃𝑥(𝑥,𝑡;𝜁,0)𝑈𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝑅𝑥𝛽(𝑥,𝑡;𝜁𝜃)𝑑𝜁22𝑡0𝑒𝜔𝜃𝑅𝜉,𝑡𝜉;𝑥+𝑡𝜃,𝜃𝑅,𝑡;𝜉𝑡̇𝜉𝑡+𝜃,𝜃1̇𝜉(𝑡𝛽)+1𝑑𝜃22𝑡0𝑒𝜔𝜃𝜉𝑥+𝑡𝜃𝑡+𝜃𝑅𝑥𝜉,𝑡̇𝜉𝑡;𝜁,𝜃𝑅𝑡𝜉,𝑡;𝜁,𝜃𝑑𝜁̇𝜉(𝑡,)+1(3.35)while, differentiating (3.33) with respect to 𝑡, we obtain 𝑈𝑡1(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉𝑡̇𝜉𝑡1̇𝜉(𝑡+)+1𝜔𝑉𝑜𝑒𝜔𝑡̇𝜉+1+12𝑃(𝑥,𝑡;𝑥+𝑡,0)𝑈𝑜(𝑥+𝑡)+𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜1(𝑥𝑡)2𝑃𝜉,𝑡𝑈;𝑥+𝑡,0𝑜𝜉(𝑥+𝑡)𝑃,𝑡;𝜉𝑡𝑈,0𝑜𝜉𝑡̇𝜉𝑡1̇𝜉(𝑡)1+12𝜉𝑥+𝑡𝑡𝑃𝑥𝜉,𝑡̇𝜉𝑡;𝜁,0+𝑃𝑡𝜉,𝑡𝑈;𝜁,0𝑜(𝜁)̇𝜉(𝑡)+1𝑑𝜁+𝛽2𝑡0𝑒𝜔𝜃+1𝑑𝜃2𝑥+𝑡𝑥𝑡𝑃𝑡(𝑥,𝑡;𝜁,0)𝑈𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝑅𝑡𝛽(𝑥,𝑡;𝜁𝜃)𝑑𝜁22𝑡0𝑒𝜔𝜃𝑅𝜉,𝑡𝜉;𝑥+𝑡𝜃,𝜃𝑅,𝑡;𝜉𝑡̇𝜉𝑡+𝜃,𝜃1̇𝜉(𝑡𝛽)+1𝑑𝜃22𝑡0𝑒𝜔𝜃𝜉𝑥+𝑡𝜃𝑡+𝜃𝑅𝑥𝜉,𝑡̇𝜉𝑡;𝜁,𝜃𝑅𝑡𝜉,𝑡;𝜁,𝜃𝑑𝜁̇𝜉(𝑡.)+1(3.36) Notice that 𝜔𝑃(𝑥,𝑡;𝑥+𝑡,0)=𝑃(𝑥,𝑡;𝑥𝑡,0)=𝜔+2𝑡2,𝑅(𝑥,𝑡;𝑥+𝑡𝜃,𝜃)𝑅(𝑥,𝑡;𝑥𝑡+𝜃,𝜃)=0.(3.37) Now we evaluate (3.35) and (3.36) on the free boundary 𝑥=𝜉(𝑡), that is,𝑈𝑥𝑈(𝜉,𝑡)=𝑜(𝜉𝑡)̇𝜔𝜉+1𝜔+2𝑡2𝑈𝑜1(𝜉𝑡)̇+𝜉+1𝜔𝑉𝑜𝑒𝜔𝑡̇+1𝜉+12𝜉+𝑡𝜉𝑡𝑃𝑥(𝜉,𝑡;𝜁,0)𝑃𝑡𝑈(𝜉,𝑡;𝜁,0)𝑜(𝜁)𝑑𝜁̇𝜉+1𝛽2𝑡0𝑒𝜔𝜃̇+𝛽𝜉+1𝑑𝜃22𝑡0𝑒𝜔𝜃𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝑥(𝜉,𝑡;𝜁,𝜃)𝑅𝑡(𝜉,𝑡;𝜁,𝜃)𝑑𝜁̇,𝑈𝜉+1𝑡𝑈(𝜉,𝑡)=𝑜(̇𝜉𝜉𝑡)̇+𝜔𝜉+1𝜔+2𝑡2𝑈𝑜̇𝜉(𝜉𝑡)̇+𝜉+1𝜔𝑉𝑜𝑒𝜔𝑡̇+1𝜉+12𝜉+𝑡𝜉𝑡𝑃𝑡(𝜉,𝑡;𝜁,0)𝑃𝑥(̇𝜉,𝑡;𝜁,0)𝜉𝑈𝑜(𝜁)𝑑𝜁̇𝜉+1+𝛽2𝑡0𝑒𝜔𝜃̇𝜉̇+𝛽𝜉+1𝑑𝜃22𝑡0𝑒𝜔𝜃𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝑡(𝜉,𝑡;𝜁,𝜃)𝑅𝑥(̇𝜉,𝑡;𝜁,𝜃)𝜉𝑑𝜁̇.𝜉+1(3.38) At this point we insert (3.38), (3.3)(5) in (3.3)(6), obtaining ̇𝜔𝜉1𝜔+2𝑡2𝑈𝑜(𝜉𝑡)𝑈𝑜𝛽(𝜉𝑡)+2𝜔𝑒𝜔𝑡+112𝜉+𝑡𝜉𝑡𝑃𝑡(𝜉,𝑡;𝜁,0)𝑃𝑥𝑈(𝜉,𝑡;𝜁,0)𝑜+𝛽(𝜁)𝑑𝜁22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝑡(𝜉,𝑡;𝜁,𝜃)𝑅𝑥(𝜉,𝑡;𝜁,𝜃)𝑑𝜁=𝑒𝜔𝑡𝛽2𝙱𝚗,(3.39) which is a nonlinear integrodifferential equation of the first order and where 𝑈𝑜 is defined by (3.32). Equation (3.39) is the free boundary equation which, as we mentioned in the introduction, does no longer depend on the velocity field 𝑈(𝑥,𝑡).

606757.fig.005
Figure 5: The domain 𝐷𝐼𝐼𝐼.

Next we remark that (3.39) can be further simplified. Indeed, recalling (3.10) and (3.12), 𝑅𝑥(𝑥,𝑡;𝜁,𝜃)=𝑅𝜁(𝑥,𝑡;𝜁,𝜃),𝑅𝜃𝑥(𝑥,𝑡;𝜁,𝜃)=𝑅𝜃𝜁(𝑥,𝑡;𝜁,𝜃),(3.40)so that, on (𝜉(𝑡),𝑡;𝜁,𝜃), we have 𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝑥𝑑𝜁=𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝜁𝑑𝜁=𝑅(𝜉,𝑡;𝜉+𝑡𝜃,𝜃)𝑅(𝜉,𝑡;𝜉𝑡+𝜃,𝜃)=0,(3.41)while, on (𝜉(𝑡),𝑡;𝜁,𝜃),𝜉+𝑡𝜉𝑡𝑃𝑥𝑑𝜁=𝜉+𝑡𝜉𝑡𝑅𝜁𝜔𝑑𝜁+𝜉+𝑡𝜉𝑡𝑅𝜃𝜁[]+𝑅𝑑𝜁=𝜔𝑅(𝜉,𝑡;𝜉𝑡,0)𝑅(𝜉,𝑡;𝜉+𝑡,0)𝜃(𝜉,𝑡;𝜉+𝑡,0)𝑅𝜃(𝜉,𝑡;𝜉𝑡,0)=0.(3.42)Hence (3.39) reduces to ̇𝜔𝜉1𝜔+2𝑡2𝑈𝑜(𝜉𝑡)𝑈𝑜𝛽(𝜉𝑡)+2𝜔𝑒𝜔𝑡+112𝜉+𝑡𝜉𝑡𝑃𝑡(𝜉,𝑡;𝜁,0)𝑈𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉+𝑡𝜃𝜉𝑡+𝜃𝑅𝑡(𝜉,𝑡;𝜁,𝜃)𝑑𝜁=𝑒𝜔𝑡𝛽2𝙱𝚗.(3.43)

Remark 3.2. The function 𝑈(𝑥,𝑡) is continuous across the characteristic 𝑥+𝑡=𝜉𝑜. Indeed lim𝑥+𝑡𝜉+𝑜𝑈(𝑥,𝑡)=lim𝑥+𝑡𝜉𝑜𝑈(𝑥,𝑡),(3.44)where the limit lim𝑥+𝑡𝜉+𝑜 is evaluated using (3.33) and the limit lim𝑥+𝑡𝜉𝑜 using (3.5) or (3.23). If we evaluate the derivatives 𝑈𝑥 and 𝑈𝑡 on the characteristic 𝑥+𝑡=𝜉𝑜 we get two different results depending on whether we are evaluating such derivatives in 𝐷𝐼 or 𝐷𝐼𝐼𝐼. We can prove that lim𝑥+𝑡𝜉𝑜𝑈𝑥1(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)+𝑈𝑜𝜉𝑜+𝑃𝑥,𝑡;𝜉𝑜𝑈,0𝑜𝜉𝑜𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜+1(𝑥𝑡)2𝜉𝑜𝑥𝑡𝑃𝑥(𝑥,𝑡;𝜁,0)𝑈𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉𝑜𝜃𝑥𝑡+𝜃𝑅𝑥(𝑥,𝑡;𝜁,𝜃)𝑑𝜁,(3.45)lim𝑥+𝑡𝜉+𝑜𝑈𝑥1(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉𝑜̇𝜉𝑜1̇𝜉𝑜+1+𝑃𝑥,𝑡;𝜉𝑜𝑈,0𝑜𝜉𝑜𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜+(𝑥𝑡)𝜔𝑉𝑜̇𝜉𝑜+1+12𝜉𝑜𝑥𝑡𝑃𝑥(𝑥,𝑡;𝜁,0)𝑈𝑜(𝛽𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉𝑜𝜃𝑥𝑡+𝜃𝑅𝑥(𝑥,𝑡;𝜁,𝜃)𝑑𝜁,(3.46)lim𝑥+𝑡𝜉𝑜𝑈𝑡1(𝑥,𝑡)=2𝑈𝑜𝜉𝑜𝑈𝑜(𝑥𝑡)+𝑃𝑥,𝑡;𝜉𝑜𝑈,0𝑜𝜉𝑜+𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜(𝑥𝑡)+𝛽2𝑡0𝑒𝜔𝜃1𝑑𝜃+2𝜉𝑜𝑥𝑡𝑃𝑡(𝑥,𝑡;𝜁,0)𝑈𝑜𝛽(𝜁)𝑑𝜁+22𝑡0𝑒𝜔𝜃𝑑𝜃𝜉𝑜𝜃𝑥𝑡+𝜃𝑅𝑡(𝑥,𝑡;𝜁,𝜃)𝑑𝜁,(3.47)lim𝑥+𝑡𝜉+𝑜𝑈𝑡1(𝑥,𝑡)=2𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉𝑜̇𝜉𝑜1̇𝜉𝑜+1+𝑃𝑥,𝑡;𝜉𝑜𝑈,0𝑜𝜉𝑜𝑃(𝑥,𝑡;𝑥𝑡,0)𝑈𝑜+(𝑥𝑡)𝜔𝑉𝑜̇𝜉𝑜+1+12𝜉𝑜𝑥𝑡𝑃𝑡(𝑥,𝑡;𝜁,0)𝑈𝑜(𝜁)𝑑𝜁+𝛽2𝑡0𝑒𝜔𝜃+𝛽𝑑𝜃22𝑡0𝑒𝜔𝜃𝑑𝜃𝑥+𝑡𝜃𝑥𝑡+𝜃𝑅𝑥(𝑥,𝑡;𝜁,𝜃)𝑑𝜁,(3.48) where ̇𝜉𝑜=̇𝜉(0). It is easy to check that, imposing that (3.45) equals (3.46) and that (3.47) equals (3.48), we get the following condition: ̇𝜉𝑜𝑈𝑜𝜉𝑜=𝜔𝑈𝑜𝜉𝑜=𝜔𝑉𝑜,(3.49) which is the condition that must be fulfilled if we want the first derivatives of 𝑈(𝑥,𝑡) to be continuous across the characteristic 𝑥+𝑡=𝜉𝑜.

If we assume that the free boundary equation (3.3)(6) holds up to 𝑡=0 we get𝑈𝑜𝜉𝑜=𝛽2𝙱𝚗.(3.50) Moreover, from (3.43) we have that, when 𝑡=0,̇𝜉𝑜1𝜔𝑈𝑜𝜉𝑜𝑈𝑜𝜉𝑜=𝛽2𝙱𝚗.(3.51) We can therefore prove the following.

Theorem 3.3. If one assumes that compatibility condition 𝑈𝑜(𝜉𝑜)=𝑉𝑜 and hypotheses (3.49), (3.50), (3.51) hold, then necessarily either 𝑉𝑜=0 or 𝜔=0 and problem (3.3) admits a unique local 𝐶1 solution (𝑈,𝜉), such that ̇𝜉(0)=0. If one does not assume hypothesis (3.49) (meaning that the first derivatives of 𝑈 are not continuous along the characteristic 𝑥+𝑡=𝜉𝑜), then problem (3.3) admits a unique local solution (𝑈,𝜉), such that ̇1<𝜉(𝑡)<0 if and only if 𝑉𝑜<𝛽2𝙱𝚗2𝜔.(3.52)

Proof. If we suppose that (3.49), (3.50), (3.51) hold then we have ̇𝜉𝑜𝑈1𝑜𝜉𝑜̇𝜉𝑜𝑈𝑜𝜉𝑜=𝛽2̇𝜉𝙱𝚗,𝑜̇𝜉=0or𝑜=2.(3.53) The initial velocity ̇𝜉𝑜=2 is not physically acceptable since existence of a solution requires that |̇𝜉|<1. Therefore ̇𝜉𝑜=0 and, recalling (3.49), we have either 𝑉𝑜=0 or 𝜔=0, since 𝑈𝑜(𝜉𝑜)=𝛽2𝙱𝚗±. If, on the other hand, we suppose that condition (3.49) does not hold, but we assume (3.52), then it is easy to show that 1<𝜔𝑉𝑜𝜔𝑉𝑜𝛽2𝛽𝙱𝚗=1+2𝙱𝚗𝜔𝑉𝑜𝛽2=̇𝜉𝙱𝚗𝑜<0,(3.54) so that for a sufficiently small time 𝑡>0 there exists a unique solution with ̇1<𝜉<0. The existence of such a solution can be proved using classical tools like iterated kernels method (see [13]).

Remark 3.4. Let us consider the limit case in which 𝜔=0 and 𝛽2=0. In this particular situation the Riemann's function 𝑅(𝑥,𝑡;𝜁,𝜃)1 and the solution 𝑈(𝑥,𝑡) is given by 1𝑈(𝑥,𝑡)=2𝑈𝑜(𝑥+𝑡)+𝑈𝑜(𝑥𝑡),in𝐷𝐼,12𝑈𝑜(𝑥+𝑡)𝑈𝑜(𝑡𝑥),in𝐷𝐼𝐼,12𝑈𝑜(𝑥𝑡)𝑈𝑜(𝜉𝑡)+𝑉𝑜,in𝐷𝐼𝐼𝐼,(3.55) and the free boundary equation is the characteristic with positive slope passing through (𝜉𝑜,0), that is, ̇𝑈𝜉1𝑜(𝜉𝑡)=0𝑈𝑜(𝜉𝑡)=𝑈𝑜𝜉𝑜,(3.56) namely, 𝜉(𝑡)=𝜉𝑜̇+𝑡𝜉(𝑡)=1.(3.57) So, setting 𝑡𝑜=1𝜉𝑜, for 𝑡𝑡𝑜, the region with uniform velocity (the inner core) has disappeared. For 𝑡𝑡𝑜, the solution 𝑈(𝑥,𝑡) is thus found solving 𝑈𝑥𝑥=𝑈𝑡𝑡,0𝑥1,𝑡1𝜉𝑜𝑈𝑥,1𝜉𝑜=𝑈𝑜𝑈(𝑥),0𝑥1,𝑡𝑥,1𝜉𝑜=𝑈1(𝑥),0𝑥1,𝑈(0,𝑡)=0𝑡1𝜉𝑜,𝑈(1,𝑡)=𝑉𝑜𝑡1𝜉𝑜,(3.58) where 𝑈𝑜(𝑥) and 𝑈1(𝑥) are determined evaluating (3.55) at time 𝑡=𝑡𝑜. To solve problem (3.58) we introduce the new variable 𝑊(𝑥,𝑡)=𝑈(𝑥,𝑡)𝑥𝑉𝑜(3.59) and rescale time with 𝜃=𝑡𝑡𝑜.(3.60) Problem (3.58) becomes 𝑊𝑥𝑥=𝑊𝜃𝜃,0𝑥1,𝜃0,𝑊(𝑥,0)=𝑈𝑜(𝑥)𝑉𝑜𝑊(𝑥),0𝑥1,𝜃(𝑥,0)=𝑊1(𝑥)=𝑈1(𝑥),0𝑥1,𝑊(0,𝜃)=0,𝜃0,𝑊(1,𝜃)=0,𝜃0,(3.61) whose solution is [11] 𝑊(𝑥,𝜃)=𝑖=1𝐴𝑛cos(𝜋𝑛𝜃)+𝐵𝑛sin(𝜋𝑛𝜃)sin(𝜋𝑛𝑥),(3.62) where 𝐴𝑛=2𝜃0𝑊𝑜(𝑧)sin(𝜋𝑛𝑧)𝑑𝑧,𝐵𝑛=2𝜋𝑛𝜃0𝑊1𝑈(𝑧)sin(𝜋𝑛𝑧)𝑑𝑧,(𝑥,𝑡)=𝑊𝑥,𝑡1+𝜉𝑜+𝑥𝑉𝑜.(3.63)

4. Asymptotic Expansion

In this section we look for a solution to problem (3.3)(1) in the following form: 𝑈(𝑥,𝑡)=𝑖=0𝜔𝑖𝑈(𝑖)(𝑥,𝑡).(4.1) This allows to obtain a sequence of problems for each 𝑖=0,1,2 with the free boundary being given by 𝜉(𝑖)(𝑡). (We remark that the sequence {𝜉(𝑖)(𝑡)} is not, in general, an asymptotic sequence.) Such an analysis is motivated by the fact that, in practical cases (asphalt and bitumen), 𝜔=𝑂(101) (see (2.3)). Hence, it makes sense to look for a “perturbative” approach for the system (3.3).

We do not discuss the issue of the convergence of series (4.1) and of the sequence {𝜉(𝑖)(𝑡)}, which is beyond the scope of the present paper. We limit ourselves to a formal derivation of the free boundary problems that can be obtained plugging (4.1)(1) into (3.3): 𝑖=0𝜔𝑖𝑈(𝑖)𝑥𝑥(𝑥,𝑡)𝜔𝑖𝑈(𝑖)𝑡𝑡(𝑥,𝑡)+𝜔𝑖+2𝑈(𝑖)(𝑥,𝑡)=𝛽2𝑖=0(𝜔𝑡)𝑖.𝑖!(4.2) Hence, for each 𝑖=0,1,2,, we have 𝑖=0,𝑈(𝑜)𝑥𝑥(𝑥,𝑡)𝑈(𝑜)𝑡𝑡(𝑥,𝑡)=𝛽2,𝑖=1,𝑈(1)𝑥𝑥(𝑥,𝑡)𝑈(1)𝑡𝑡(𝑥,𝑡)=𝛽2𝑡,𝑖=2,𝑈(2)𝑥𝑥(𝑥,𝑡)𝑈(2)𝑡𝑡(𝑥,𝑡)=𝛽2𝑡22!𝑈(𝑜)(𝑥,𝑡),𝑖>2,𝑈(𝑖)𝑥𝑥(𝑥,𝑡)𝑈(𝑖)𝑡𝑡(𝑥,𝑡)=𝛽2𝑡𝑖𝑖!𝑈(𝑖2)(𝑥,𝑡)(4.3)

and the following free boundary problems 𝑈𝑖=0,(𝑜)𝑥𝑥(𝑥,𝑡)𝑈(𝑜)𝑡𝑡(𝑥,𝑡)=𝛽2𝑈(𝑜)(𝑥,0)=𝑈𝑜(𝑈𝑥),𝑡(𝑜)𝑈(𝑥,0)=0,(𝑜)𝑈(0,𝑡)=0,(𝑜)𝜉(𝑜),𝑡=𝑉𝑜𝑈𝑥(𝑜)𝜉(𝑜)+̇𝜉,𝑡(𝑜)𝑈𝑡(𝑜)𝜉(𝑜),𝑡=𝛽2𝜉𝙱𝚗,(𝑜)(0)=𝜉𝑜,𝑈(4.4)𝑖=1,(1)𝑥𝑥(𝑥,𝑡)𝑈(1)𝑡𝑡(𝑥,𝑡)=𝑡𝛽2𝑈(1)𝑈(𝑥,0)=0,𝑡(1)(𝑥,0)=𝑈𝑜𝑈(𝑥),(1)𝑈(0,𝑡)=0,(1)𝜉(1),𝑡=𝑉𝑜𝑡𝑈𝑥(1)𝜉(1)+̇𝜉,𝑡(1)𝑈𝑡(1)𝜉(1)̇𝜉,𝑡(1)𝑉𝑜=𝑡𝛽2𝜉𝙱𝚗,(1)(0)=𝜉𝑜,𝑈(4.5)𝑖2,(𝑖)𝑥𝑥(𝑥,𝑡)𝑈(𝑖)𝑡𝑡𝑡(𝑥,𝑡)=𝑖𝛽𝑖!2𝑈(𝑖2)𝑈(𝑥,𝑡)(𝑖)𝑈(𝑥,0)=0,𝑡(𝑖)(𝑈𝑥,0)=0,(𝑖)𝑈(0,𝑡)=0,(𝑖)𝜉(𝑖)=𝑡,𝑡𝑖𝑉𝑖!𝑜𝑈𝑥(𝑖)𝜉(𝑖)+̇𝜉,𝑡(𝑖)𝑈𝑡(𝑖)𝜉(𝑖)̇𝜉,𝑡(𝑖)𝑉𝑜𝑡𝑖1=𝑡(𝑖1)!𝑖𝛽𝑖!2𝜉𝙱𝚗,(𝑖)(0)=𝜉𝑜.(4.6) We immediately remark that, in each problem, the governing equation is no longer a telegrapher's equation, but a nonhomogeneous wave equation. Hence, using classical d'Alembert formula, we can write the representation formula for each domain 𝐷𝐼, 𝐷𝐼𝐼, 𝐷𝐼𝐼𝐼 and for each order of approximation 𝑖=0,1,.... In particular, in 𝐷𝐼 we have 𝐷𝐼𝑈(𝑜)1(𝑥,𝑡)=2𝑈𝑜(𝑥+𝑡)+𝑈𝑜+𝛽(𝑥𝑡)2𝑡2,𝑈2!(1)1(𝑥,𝑡)=2𝑥+𝑡𝑥𝑡𝑈𝑜𝛽(𝜁)𝑑𝜁+2𝑡3,𝑈3!(𝑖)𝛽(𝑥,𝑡)=22𝑡0𝑥+𝑡𝜃𝑥𝑡+𝜃𝑈(𝑖2)𝛽(𝜁,𝜃)𝑑𝜁𝑑𝜃+2𝑡𝑖+2(𝑖+2)!,𝑖2,(4.7) while in 𝐷𝐼𝐼𝐷𝐼𝐼𝑈(𝑜)1(𝑥,𝑡)=2𝑈𝑜(𝑥+𝑡)𝑈𝑜+𝛽(𝑡𝑥)2𝑡2𝛽2!2(𝑥𝑡)2,𝑈2!(1)1(𝑥,𝑡)=2𝑥+𝑡𝑡𝑥𝑈𝑜𝛽(𝜁)𝑑𝜁+2𝑡3𝛽3!2(𝑡𝑥)3,𝑈3!(𝑖)𝛽(𝑥,𝑡)=22𝑡0𝑥𝑡+𝜃𝑡𝑥+𝜃𝑈(𝑖2)𝛽(𝜁,𝜃)𝑑𝜁𝑑𝜃+22𝑡0𝑥+𝑡𝜃𝑥𝑡+𝜃𝑈(𝑖2)+𝛽(𝜁,𝜃)𝑑𝜁𝑑𝜃2𝑡𝑖+2𝛽(𝑖+2)!2(𝑡𝑥)𝑖+2(𝑖+2)!,𝑖2,(4.8) and in 𝐷𝐼𝐼𝐼𝐷𝐼𝐼𝐼𝑈(𝑜)(𝑥,𝑡)=𝑉𝑜+12𝑈𝑜(𝑥𝑡)𝑈𝑜𝜉(𝑜)𝑡+𝛽2𝑡2𝛽2!2𝑡2,𝑈2!(1)(𝑥,𝑡)=𝑉𝑜𝑡+12𝜉(1)𝑡𝑥𝑡𝑈𝑜(𝛽𝜁)𝑑𝜁+2𝑡3𝛽3!2𝑡3,𝑈3!(𝑖)(𝛽𝑥,𝑡)=22𝑡0𝑥+𝑡𝜃𝑥𝑡+𝜃𝑈(𝑗2)(𝛽𝜁,𝜃)𝑑𝜁𝑑𝜃22𝑡0𝜉𝑥+𝑡𝜃(𝑖)𝑡+𝜃𝑈(𝑖2)(𝜁,𝜃)𝑑𝜁𝑑𝜃+𝑉𝑜𝑡𝑖+𝛽𝑖!2𝑡𝑖+2𝛽(𝑖+2)!2𝑡𝑖+2(𝑖+2)!.𝑖2.(4.9) Proceeding as in Section 3 we can show that the evolution equations of the free boundary 𝑥=𝜉(𝑖)(𝑡) at each step are given bẏ𝜉𝑖=0,(0)𝛽12𝑡𝑈𝑜𝜉(𝑜)𝑡=𝛽2̇𝜉𝙱𝚗,(4.10)𝑖=1,(1)𝑈1𝑜𝜉(1)𝑡𝑉𝑜+𝛽2𝑡22!=𝑡𝛽2̇𝜉𝙱𝚗,(4.11)𝑖2,(𝑖)𝛽12(𝑖+2)𝑡𝑖+1𝑉(𝑖+1)!𝑜𝑡𝑖1(𝑖1)!+𝛽2𝑡0𝑈(𝑖2)𝜉(𝑖)=𝑡𝑡+𝜃,𝜃𝑑𝜃𝑖𝛽𝑖!2𝙱𝚗.(4.12) At the zero order, assuming the compatibility condition 𝑈(𝜉𝑜)=𝛽2Bn (see problem (4.4)), we have ̇𝜉1𝑜(𝑜)𝑈𝑜𝜉𝑜=𝛽2̇𝜉𝙱𝚗,𝑜(𝑜)=0.(4.13)

At the first order (see problem (4.5)), we assume that the compatibility condition of second order holds in the corner (𝜉𝑜,0). This means that we can differentiate the free boundary equation (4.5)(6) and take the limit for 𝑡0. We have 𝑈(1)𝑥𝑥𝜉(1)̇𝜉,𝑡(1)+𝑈(1)𝑥𝑡𝜉(1)+̈𝜉,𝑡(1)𝑈𝑡(1)𝜉(1)+̇𝜉,𝑡(1)2𝑈(1)𝑥𝑡𝜉(1)̈𝜉,𝑡(1)𝑉𝑜=𝛽2𝙱𝚗,(4.14)

which, when 𝑡0, reduces to𝑈𝑜𝜉𝑜̇𝜉1+(1)2𝑜=𝛽2̇𝜉𝙱𝚗,𝑜(1)=0.(4.15)

For the generic 𝑖th order (see problem (4.6)), we assume that the compatibility conditions in the corner (𝜉𝑜,0) hold up to order (𝑖1). Therefore we can take the (𝑖1)th derivative of (4.6)(6), obtaining 𝑑𝑖1𝑑𝑡𝑖1𝑈𝑥(𝑖)𝜉(𝑖)+𝑑,𝑡𝑖1𝑑𝑡𝑖1̇𝜉(𝑖)𝑈𝑡(𝑖)𝜉(𝑖)+̇𝜉,𝑡(𝑖)𝑑𝑖1𝑑𝑡𝑖1𝑈𝑡(𝑖)𝜉(𝑖)̇𝜉,𝑡(𝑖)𝑉𝑜𝑉𝑜𝑡𝑖1(𝑑𝑖1)!𝑖1𝑑𝑡𝑖1̇𝜉(𝑖)=𝑡𝛽2𝙱𝚗,(4.16)

which, in the limit 𝑡0, reduces tȯ𝜉𝑜(𝑖)𝑉𝑜̇𝜉=0𝑜(𝑖)=0.(4.17) We therefore conclude that, assuming enough regularity for each problem 𝑖=0,1,2,, (4.10)–(4.12) posses a unique local solution with 𝜉(𝑖)(0)=𝜉𝑜 and ̇𝜉(𝑖)(0)=0.

Before proceeding further we suppose that 𝑈(𝑥) has the following properties:

(H1)  𝑈𝑜(𝑥)𝐶([0,𝜉𝑜]),

(H2) 0<𝑈𝑜(𝑥)<𝑉𝑜forall𝑥(0,𝜉𝑜), 𝑈𝑜(0)=0,𝑈𝑜(𝜉𝑜)=𝑉𝑜,

(H3) 𝑈𝑜(𝑥)>0forall𝑥[0,𝜉𝑜] and 𝑈(𝜉𝑜)=𝛽2𝙱𝚗,

(H4) 𝑈𝑜(𝑥) satisfies all the compatibility conditions up to any order in the corner (𝜉𝑜,0).

4.1. Zero-Order Approximation

We introduce the new variable 𝜙(𝑜)=𝜉(𝑜)𝑡, so that (4.10) can be rewritten aṡ𝜙(𝑜)𝛽2𝑡𝑈𝑜𝜙(𝑜)=𝛽2𝙱𝚗,(4.18)

with 𝜙(𝑜)(0)=𝜉𝑜. Then we look for the solution 𝑡=𝑡(𝜙(𝑜)) which fulfills the following Cauchy problem:𝛽2𝙱𝚗𝑑𝑡𝑑𝜙(𝑜)=𝛽2𝑡𝑈𝑜𝜙(𝑜),𝑡𝜉𝑜=0,(4.19) that is,𝑡𝜙(𝑜)1=𝛽2𝙱𝚗𝜙(𝑜)𝜉𝑜𝑈𝑜𝜙(𝑧)exp(𝑜)𝑧𝙱𝚗𝑑𝑧.(4.20) Recalling that |̇𝜉(𝑜)|<1, that is,|||̇𝜙(𝑜)|||̇𝜙+1<1,2<(𝑜)<0,𝑑𝑡𝑑𝜙(𝑜)1<2,(4.21) from (4.19)(1) we realize that (4.21) is fulfilled if𝜉𝑜+𝙱𝚗2<1𝛽2inf𝑧0,𝜉𝑜𝑈𝑜(𝑧).(4.22) Therefore, under hypothesis (4.22), local existence of a classical solution is guaranteed. Such a solution is given by 𝜉(𝑜)(𝑡)=𝜙(𝑜)(𝑡)+𝑡, where 𝜙(𝑜)(𝑡) is determined inverting (4.20).

4.2. First-Order Approximation

We now have to solve the probleṁ𝜉(1)𝑈1𝑜𝜉(1)𝑡𝑉𝑜+𝛽2𝑡22!=𝑡𝛽2𝙱𝚗,(4.23) with 𝜉(1)(0)=𝜉𝑜. Proceeding as in Section 4.1 we introduce the new variable 𝜙(1)=𝜉(1)𝑡, so that (4.23) becomeṡ𝜙(1)𝑈𝑜𝜙(1)𝑉𝑜+𝛽2𝑡22!=𝑡𝛽2𝙱𝚗,(4.24)

and we have to solve the following Cauchy problem:𝑑𝑡𝑑𝜙(1)=1𝑡𝛽2𝑈𝙱𝚗𝑜𝜙(1)𝑉𝑜+𝛽2𝑡2,𝑡𝜉2!𝑜=0.(4.25) We notice that (4.25)(1) is a Bernoulli equation. Therefore, setting 𝑤=𝑡2, problem (4.25) becomes𝑑𝑤𝑑𝜙(1)=𝑤𝑈𝙱𝚗+2𝑜𝜙(1)𝑉𝑜𝛽2,𝑤𝜉𝙱𝚗𝑜=0.(4.26) whose solution is given by𝑡2𝜙(1)=𝜙(1)𝜉𝑜𝜙2exp(1)𝑠𝑈𝙱𝚗𝑜(𝑠)𝑉𝑜𝛽2𝙱𝚗𝑑𝑠,(4.27) which make sense only if 𝜙(1)𝜉𝑜. Integrating (4.27) by parts we get𝑡2𝜙(1)=2𝜙(1)𝜉𝑜𝜙exp(1)𝑠𝑈𝙱𝚗𝑜(𝑠)𝛽22𝑑𝑠+𝛽2𝑉𝑜𝑈𝑜𝜙(1).(4.28) We recall from the previous section that the condition |̇𝜉(1)(𝑡)|<1 is guaranteed if𝑑𝑡𝑑𝜙(1)1<2,(4.29)

which, by virtue of(4.25)(1), is equivalent to require that𝑡22+𝙱𝚗𝑡+𝛽2𝑈𝜙(1)𝑉𝑜<0.(4.30) Hence, under assumption (𝐻2), the discriminant Δ=𝙱𝚗2+8𝛽2(𝑉𝑜𝑈𝑜(𝜙(1)))>𝙱𝚗2>0, and (4.30) is fulfilled when𝙱𝚗Δ2<0𝑡<𝙱𝚗+Δ2.(4.31) Therefore, in order to have a unique local solution, we must require that𝑡2<𝙱𝚗2+Δ2Δ𝙱𝚗4<𝙱𝚗22+2𝛽2𝑉𝑜𝑈𝑜𝜙(1),(4.32) which, exploiting (4.28), becomes2𝜙(1)𝜉𝑜𝜙exp(1)𝑠𝑈𝙱𝚗𝑜(𝑠)𝛽2𝑑𝑠<𝙱𝚗22.(4.33) The latter is automatically satisfied, under assumption (𝐻3), recalling that 𝜙(1)𝜉𝑜. So, also for the first order we have local uniqueness and existence of the solution 𝜉(1)(𝑡)=𝜙(1)(𝑡)+𝑡, where 𝜙(1)(𝑡) is obtained inverting (4.28).

4.3. 𝑖th-Order Approximation

We now consider here the 𝑖th-order approximation. The evolution equation of the free boundary is given by (4.12). Proceeding as in the previous sections we set 𝜙(𝑖)=𝜉(𝑖)𝑡, so that (4.12) can be rewritten aṡ𝜙(𝑖)𝛽2(𝑖+2)𝑡𝑖+1𝑉(𝑖+1)!𝑜𝑡𝑖1(𝑖1)!+𝛽2𝑡0𝑈(𝑖2)𝜙(𝑖)=𝑡𝑡+𝜃,𝜃𝑑𝜃𝑖𝛽𝑖!2𝙱𝚗,(4.34) with 𝜙(𝑖)(0)=𝜉𝑜. Once again we look for 𝑡=𝑡(𝜙(𝑖)), solving this Cauchy problem𝑑𝑡𝑑𝜙(𝑖)=(𝑖+2)𝑡(𝑖+1)𝑉𝙱𝚗𝑜𝑖𝛽2+𝙱𝚗𝑡𝑖!𝑡𝑖𝙱𝚗𝑡0𝑈(𝑖2)𝜙(𝑖)𝜉𝑡+𝜃,𝜃𝑑𝜃𝑡𝑜𝑡𝜉=0,𝑜=0.(4.35) Now, hypothesis (𝐻4) and (4.17) entaillim𝑡0+𝑑𝜙(𝑖)𝑑𝑡=lim𝑡0+𝑑𝑡𝑑𝜙(𝑖)=1.(4.36)Therefore𝑉𝑜𝑖𝛽2=lim𝑡0+𝑖!𝑡𝑖1𝑡0𝑈(𝑖2)𝜙(𝑖)𝑡+𝜃,𝜃𝑑𝜃.(4.37) So for 𝑡 sufficiently small, we can approximate the integral on the r.h.s. of (4.37) in the following way:𝑡0𝑈(𝑖2)𝜙(𝑖)𝑡+𝜃,𝜃𝑑𝜃=𝐶𝑖𝜙(𝑖)𝑡𝑖1,(4.38) where 𝐶𝑖 is a smooth function of 𝜙(𝑖), determined exploiting (4.7), (4.8), and (4.9). In particular,𝐶𝑖𝜉𝑜=𝑉𝑜(𝑖1)!𝛽2.𝙱𝚗(4.39)Hence, setting𝐴𝑖=𝑖+21𝑖+1𝙱𝚗,𝐵𝑖𝜙(𝑖)=𝑖!𝐶𝑖𝜙(𝑖)Bn𝑖𝑉𝑜𝛽2𝙱𝚗,(4.40)problem (4.35) acquires the following structure:𝑑𝑡𝑑𝜙(𝑖)=𝐴𝑖𝑡+𝐵𝑖𝜙(𝑖)1𝑡𝜉,𝑡𝑜𝑡𝜉=0,𝑜=0,(4.41) provided 𝑡 sufficiently small. From (4.37) it is easy to check that in a right neighborhood of 𝑡=0 the function (Recall that 𝑈(𝑖2)(𝑥,𝑡) are everywhere non negative for every 𝑖), 𝐵𝑖(𝜙(𝑖))<0 and 𝐵𝑖(𝜉𝑜)