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

Authors: Lorenzo Fusi; Angiolo Farina
Journal: Advances in Mathematical Physics
 (2011)

---

## 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.


---

## Body

## 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 [2–5]).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 [7–9], 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 Section2 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 Section4 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(10-1). 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 systemxoy 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 v⃗(y,t)=v(y,t)i⃗. 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.Figure 1
Upper part of the channel.The mathematical model is written for the velocity fieldv(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:(2.1)vtt+2ωvt=vyy+β2y∈(s,1),t>0,v(y,0)=vo(y)y∈(so,1),vt(y,0)=0y∈(so,1),v(1,t)=0t>0,v(s,t)=Vot>0,vy(s,t)+ṡvt(s,t)=-β2Bnt>0,s(0)=so,so∈(0,1).
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])(2.2)μ=1MPa,ρ=1.5×103Kg/m3,η=102MPa⋅s.TakingL=500m we get(2.3)te=15s,tr=50s,⟹ω=0.15.Remark 2.1.
In [1] we have proved that problem (2.1) admits a stationary solution provided
(2.4)Vo⩽β2(12+Bn)
and that the stationary solution is given by
(2.5)v∞(y)=-β22(s-Bn-y)2+β2Bn2+Vo,s∞=1+Bn-Bn2+2Voβ2.


## 3. An Equivalent Formulation

Before proceeding in proving analytical results of problem (2.1) we introduce the new coordinate system(3.1)x=1-y,⟺y=1-x,
and the new variable(3.2)U(x,t)=exp⁡(ωt)v(1-x,t),⟺v(y,t)=U(1-y)exp⁡(-ωt).
With transformations (3.1)-(3.2), problem (2.1) becomes(3.3)Uxx-Utt+ω2U=-β2exp⁡(ωt),x∈(0,ξ),t>0,U(x,0)=Uo(x),x∈(0,ξo),Ut(x,0)=U1(x),x∈(0,ξo),U(0,t)=0,t>0,U(ξ,t)=exp⁡(ωt)Vo,t>0,Ux(ξ,t)+ξ̇Ut(ξ,t)-ξ̇ωU(ξ,t)=exp⁡(ωt)β2Bn,t>0,ξ(0)=ξo,ξo∈(0,1),where(3.4)ξ(t)=1-s(t),ξo=1-so,Uo(x)=vo(1-x),U1(x)=ωUo(x).
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])(3.5)U(x,t)=12[Uo(x-t)+Uo(x+t)]+12∫x-tx+t[R(x,t;ζ,0)U1(ζ)-Rθ(x,t;ζ,0)Uo(ζ)]dζ+β22∫0texp⁡(ωθ)dθ∫x-t+θx+t-θR(x,t;ζ,θ)dζ,
where R(x,t;ζ,θ) is the Riemann's function that solves the problem (see again Figure 3)(3.6)Rζζ-Rθθ+ω2R=0(ζ,θ)∈Ω(x,t),R(x,t;x+t-θ,θ)=1θ∈[0,t],R(x,t;x-t+θ,θ)=1θ∈[0,t],Ω(x,t)={(x,t):x-t+θ⩽ζ⩽x+t-θ,0⩽θ⩽t}.Figure 2
Sketch of the domain.Figure 3
The domainDI.To determine the solution of problem (3.6) we set(3.7)z=(t-θ)2-(ζ-x)2,
where(3.8)zθ2-zζ2=1,zθθ-zζζ=1z.By means of (3.7) problem (3.6) becomes(3.9)R′′(z)+R′(z)z-ω2R(z)=0R(0)=1,
where (3.9)(1) is the modified Bessel equation of zero order. The solution of (3.9) is given by(3.10)R(z)=R(x,t;ζ,θ)=Io(ω(t-θ)2-(ζ-x)2),
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](3.11)Io′(x)x=12[Io(x)-I2(x)],one can prove that(3.12)Rθ(x,t;ζ,θ)=ω2(θ-t)2[Io(ω(t-θ)2-(ζ-x)2)-I2(ω(t-θ)2-(ζ-x)2)],
where I2 is the modified Bessel function of second order. Recalling that U(x,0)=Uo(x) and, by (3.3)(3), (3.4), that(3.13)Ut(x,0)=ωUo(x),we see that(3.14)[R(x,t;ζ,0)ω-Rθ(x,t;ζ,0)]Uo(ζ)=[Io(ωt2-(ζ-x)2)(ω+ω2t2)-ω2t2I2(ωt2-(ζ-x)2)]Uo(ζ),and representation formula (3.5) can be rewritten as
(3.15)U(x,t)=12[Uo(x-t)+Uo(x+t)]+12∫x-tx+t[Io(ωt2-(ζ-x)2)(ω+ω2t2)-ω2t2I2(ωt2-(ζ-x)2)]Uo(ζ)dζ+β22∫0texp⁡(kθ2)dθ⋅∫x-t+θx+t-θIo(ω(t-θ)2-(ζ-x)2)dζ.Let us now write a representation formula forU(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 get(3.16)0=12[Uo(-t*)+Uo(t*)]+12∫-t*t*[R(0,t*;ζ,0)ω-Rθ(0,t*;ζ,0)]Uo(ζ)dζ+β22∫0t*exp⁡(ωθ)dθ∫-t*+θt*-θR(0,t*;ζ,θ)dζ,
where
(3.17)t*=t-xis the coordinate of the intersection of the characteristic ζ=x-t+θ with ζ=0. Relation (3.16) can be rewritten as(3.18)0=12[Uo(x-t)+Uo(t-x)]+12∫x-tt-x[R(0,t-x;ζ,0)ω-Rθ(0,t-x;ζ,0)]Uo(ζ)dζ+β22∫0t-xexp⁡(ωθ)dθ∫x-t+θt-x-θR(0,t-x;ζ,θ)dζ.
From (3.18), the extended function Uosx(x), defined in [-ξo,0], fulfills the following Volterra integral equation of second type:
(3.19)Uosx(x-t)-∫0x-t[R(0,t-x;ζ,0)ω-Rθ(0,t-x;ζ,0)]Uosx(ζ)dζ=-Uo(t-x)+∫t-x0[R(0,t-x;ζ,0)ω-Rθ(0,t-x;ζ,0)]Uo(ζ)dζ-β22∫0t-xexp⁡(ωθ)dθ∫x-t+θt-x-θR(0,t-x;ζ,θ)dζ.
Equation (3.19) can be put in the more compact form(3.20)Uosx(χ)-∫0χKsx(χ,ζ)Uosx(ζ)dζ=Fsx(χ),where χ=x-t∈[-ξo,0] and
(3.21)Ksx(χ,ζ)=[Io(ωχ2-ζ2)(ω-ω2χ2)+ω2χ2I2(ωχ2-ζ2)],Fsx(χ)=-Uo(-χ)+∫-χ0[R(0,-χ;ζ,0)ω-Rθ(0,-χ;ζ,0)]Uo(ζ)dζ-β22∫0-xexp⁡(ωθ)dθ∫x+θ-x-θR(0,-x;ζ,θ)dζ.Figure 4
The domainDII.Due to the regularity of the kernelKsx(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
(3.22)Uo(x)={Uosx(x),x∈[-ξo,0],Uo(x),x∈[0,ξo],and the solution U(x,t) in the domain DII is given by (3.23)U(x,t)=12[Uo(x+t)-Uo(t-x)]+12∫t-xx-t[R(0,t-x;ζ,0)ω-Rθ(0,t-x;ζ,0)]Uo(ζ)dζ+12∫x-tx+t[R(x,t;ζ,0)ω-Rθ(t,x;ζ,0)]Uo(ζ)dζ+β22∫0t-xeωθdθ∫t-x-θx-t+θR(0,t-x;ζ,θ)dζ+β22∫0teωθdθ∫x-t+θx+t-θR(x,t;ζ,θ)dζ.Remark 3.1.
We notice that, considering the representation formulae (3.5) and (3.23),
(3.24)lim⁡x→t+U(x,t)=lim⁡x→t-U(x,t).
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,
(3.25)lim⁡x→t+Ux(x,t)=lim⁡x→t-Ux(x,t),t∈[0,ξo2],lim⁡x→t+Ut(x,t)=lim⁡x→t-Ut(x,t),t∈[0,ξo2],
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 forU(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., |ξ̇|<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
(3.26)ξ*+t*=x+t,⟹t*=t*(x,t),(3.27)∂t*∂t=1ξ̇(t*)+1,∂t*∂x=1ξ̇(t*)+1.We consider once again the representation formula (3.5) and impose condition (3.3)(5), getting(3.28)2eωt*Vo=Uo(ξ*-t*)+Uo(ξ*+t*)+∫ξ*-t*ξ*+t*[R(ξ*,t*;ζ,0)ω-Rθ(ξ*,t*;ζ,0)]Uo(ζ)dζ+β2∫0t*eωθdθ∫ξ*-t*+θξ*+t*-θR(ξ*,t*;ζ,θ)dζ.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: (3.29)Uodx(ξ*+t*)+∫ξoξ*+t*[R(ξ*,t*;ζ,0)ω-Rθ(ξ*,t*;ζ,0)]Uodx(ζ)dζ=2eωt*Vo-Uo(ξ*-t*)-∫ξ*-t*ξo[R(ξ*,t*;ζ,0)ω-Rθ(ξ*,t*;ζ,0)]Uo(ζ)dζ-β2∫0t*eωθdθ∫ξ*-t*+θξ*+t*-θR(ξ*,t*;ζ,θ)dζ.Recalling (3.26) and proceeding as for the domain DII, the above can be rewritten as(3.30)Uodx(χ)+∫ξoχKdx(χ,ζ)Uodx(ζ)dζ=Fdx(χ),where χ=x+t and(3.31)Kdx(χ,ζ)=[Io(ω(t*(χ))2-(ζ-ξ*(χ))2)(ω+ω2t*(χ)2)-ω2t*(χ)2I2(ω(t*(χ))2-(ξ*(χ)-ζ)2)],Fdx(χ)=[2eωtVo-Uo(x-t)-∫x-tξo[R(x,t;ζ,0)ω-Rθ(x,t;ζ,0)]Uo(ζ)dζ-β2∫0teωθdθ∫x-t+θx+t-θR(x,t;ζ,θ)dζ]|(x=ξ*(χ),t=t*(χ)).
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] as(3.32)Uo(x)={Uosx(x),x∈[-ξo,0],Uo(x),x∈[0,ξo],Uodx(x),x∈[ξo,ξ(ξo)+ξo].
The solution in the domain DIII is thus given by(3.33)U(x,t)=eωt*Vo+12[Uo(x-t)-Uo(ξ*-t*)]+12∫x-tx+tP(x,t;ζ,0)Uo(ζ)dζ-12∫ξ*-t*x+tP(ξ*,t*;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θx+t-θR(x,t;ζ,θ)dζ-β22∫0t*eωθdθ∫ξ*-t*+θx+t-θR(ξ*,t*;ζ,θ)dζ,
where for simplicity of notation we have introduced(3.34)P(x,t;ζ,θ)=R(x,t;ζ,θ)ω-Rθ(x,t;ζ,θ),and where Uo(x) is given by (3.32). Therefore for any fixed C1 function ξ(t) with |ξ̇|<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 get(3.35)Ux(x,t)=12[Uo′(x-t)-Uo′(ξ*-t*)ξ̇(t*)-1ξ̇(t*)+1]+(ωVoeωt*ξ̇*+1)+12[P(x,t;x+t,0)Uo(x+t)-P(x,t;x-t,0)Uo(x-t)]-12[P(ξ*,t*;x+t,0)Uo(x+t)-P(ξ*,t*;ξ*-t*,0)Uo(ξ*-t*)ξ̇(t*)-1ξ̇(t*)+1]-12∫ξ*-t*x+t[Px(ξ*,t*;ζ,0)ξ̇(t*)+Pt(ξ*,t*;ζ,0)]Uo(ζ)ξ̇(t*)+1dζ+12∫x-tx+tPx(x,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θx+t-θRx(x,t;ζθ)dζ-β22∫0t*eωθ[R(ξ*,t*;x+t-θ,θ)-R(ξ*,t*;ξ*-t*+θ,θ)ξ̇(t*)-1ξ̇(t*)+1]dθ-β22∫0t*eωθ∫ξ*-t*+θx+t-θ[Rx(ξ*,t*;ζ,θ)ξ̇(t*)-Rt(ξ*,t*;ζ,θ)]dζξ̇(t*)+1,while, differentiating (3.33) with respect to t, we obtain
(3.36)Ut(x,t)=12[-Uo′(x-t)-Uo′(ξ*-t*)ξ̇(t*)-1ξ̇(t*)+1]+(ωVoeωt*ξ̇*+1)+12[P(x,t;x+t,0)Uo(x+t)+P(x,t;x-t,0)Uo(x-t)]-12[P(ξ*,t*;x+t,0)Uo(x+t)-P(ξ*,t*;ξ*-t*,0)Uo(ξ*-t*)ξ̇(t*)-1ξ̇(t*)+1]-12∫ξ*-t*x+t[Px(ξ*,t*;ζ,0)ξ̇(t*)+Pt(ξ*,t*;ζ,0)]Uo(ζ)ξ̇(t*)+1dζ+β2∫0teωθdθ+12∫x-tx+tPt(x,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θx+t-θRt(x,t;ζθ)dζ-β22∫0t*eωθ[R(ξ*,t*;x+t-θ,θ)-R(ξ*,t*;ξ*-t*+θ,θ)ξ̇(t*)-1ξ̇(t*)+1]dθ-β22∫0t*eωθ∫ξ*-t*+θx+t-θ[Rx(ξ*,t*;ζ,θ)ξ̇(t*)-Rt(ξ*,t*;ζ,θ)]dζξ̇(t*)+1.
Notice that
(3.37)P(x,t;x+t,0)=P(x,t;x-t,0)=ω+ω2t2,R(x,t;x+t-θ,θ)-R(x,t;x-t+θ,θ)=0.
Now we evaluate (3.35) and (3.36) on the free boundary x=ξ(t), that is,(3.38)Ux(ξ,t)=Uo′(ξ-t)ξ̇+1-(ω+ω2t2)Uo(ξ-t)1ξ̇+1+(ωVoeωtξ̇+1)+12∫ξ-tξ+t[Px(ξ,t;ζ,0)-Pt(ξ,t;ζ,0)]Uo(ζ)dζξ̇+1-β2∫0teωθξ̇+1dθ+β22∫0teωθ∫ξ-t+θξ+t-θ[Rx(ξ,t;ζ,θ)-Rt(ξ,t;ζ,θ)]dζξ̇+1,Ut(ξ,t)=-Uo′(ξ-t)ξ̇ξ̇+1+(ω+ω2t2)Uo(ξ-t)ξ̇ξ̇+1+(ωVoeωtξ̇+1)+12∫ξ-tξ+t[Pt(ξ,t;ζ,0)-Px(ξ,t;ζ,0)]ξ̇Uo(ζ)dζξ̇+1+β2∫0teωθξ̇ξ̇+1dθ+β22∫0teωθ∫ξ-t+θξ+t-θ[Rt(ξ,t;ζ,θ)-Rx(ξ,t;ζ,θ)]ξ̇dζξ̇+1.
At this point we insert (3.38), (3.3)(5) in (3.3)(6), obtaining (3.39)(ξ̇-1)[(ω+ω2t2)Uo(ξ-t)-Uo′(ξ-t)+β2ω(eωt-1)+12∫ξ-tξ+t[Pt(ξ,t;ζ,0)-Px(ξ,t;ζ,0)]Uo(ζ)dζ+β22∫0teωθdθ∫ξ-t+θξ+t-θ[Rt(ξ,t;ζ,θ)-Rx(ξ,t;ζ,θ)]dζ]=eωtβ2Bn,
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).Figure 5
The domainDIII.Next we remark that (3.39) can be further simplified. Indeed, recalling (3.10) and (3.12),
(3.40)Rx(x,t;ζ,θ)=-Rζ(x,t;ζ,θ),Rθx(x,t;ζ,θ)=-Rθζ(x,t;ζ,θ),so that, on (ξ(t),t;ζ,θ), we have
(3.41)∫ξ-t+θξ+t-θRxdζ=-∫ξ-t+θξ+t-θRζdζ=R(ξ,t;ξ+t-θ,θ)-R(ξ,t;ξ-t+θ,θ)=0,while, on (ξ(t),t;ζ,θ),(3.42)∫ξ-tξ+tPxdζ=-∫ξ-tξ+tRζωdζ+∫ξ-tξ+tRθζdζ=ω[R(ξ,t;ξ-t,0)-R(ξ,t;ξ+t,0)]+[Rθ(ξ,t;ξ+t,0)-Rθ(ξ,t;ξ-t,0)]=0.Hence (3.39) reduces to
(3.43)(ξ̇-1)[(ω+ω2t2)Uo(ξ-t)-Uo′(ξ-t)+β2ω(eωt-1)+12∫ξ-tξ+tPt(ξ,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫ξ-t+θξ+t-θRt(ξ,t;ζ,θ)dζ]=eωtβ2Bn.Remark 3.2.
The functionU(x,t) is continuous across the characteristic x+t=ξo. Indeed
(3.44)lim⁡x+t→ξo+⁡U(x,t)=lim⁡x+t→ξo-⁡U(x,t),where the limit limx+t→ξo+ is evaluated using (3.33) and the limit limx+t→ξo- 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
(3.45)lim⁡x+t→ξo-Ux(x,t)=12[Uo′(x-t)+Uo′(ξo)+P(x,t;ξo,0)Uo(ξo)-P(x,t;x-t,0)Uo(x-t)]+12∫x-tξoPx(x,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θξo-θRx(x,t;ζ,θ)dζ,(3.46)lim⁡x+t→ξo+Ux(x,t)=12[Uo′(x-t)-Uo′(ξo)ξ̇o-1ξ̇o+1+P(x,t;ξo,0)Uo(ξo)-P(x,t;x-t,0)Uo(x-t)]+ωVoξ̇o+1+12∫x-tξoPx(x,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θξo-θRx(x,t;ζ,θ)dζ,(3.47)lim⁡x+t→ξo-Ut(x,t)=12[Uo′(ξo)-Uo′(x-t)+P(x,t;ξo,0)Uo(ξo)+P(x,t;x-t,0)Uo(x-t)]+β2∫0teωθdθ+12∫x-tξoPt(x,t;ζ,0)Uo(ζ)dζ+β22∫0teωθdθ∫x-t+θξo-θRt(x,t;ζ,θ)dζ,(3.48)lim⁡x+t→ξo+Ut(x,t)=12[-Uo′(x-t)-Uo′(ξo)ξ̇o-1ξ̇o+1+P(x,t;ξo,0)Uo(ξo)-P(x,t;x-t,0)Uo(x-t)]+ωVoξ̇o+1+12∫x-tξoPt(x,t;ζ,0)Uo(ζ)dζ+β2∫0teωθdθ+β22∫0teωθdθ∫x-t+θx+t-θRx(x,t;ζ,θ)dζ,
where ξ̇o=ξ̇(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:
(3.49)ξ̇oUo′(ξo)=ωUo(ξo)=ωVo,
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 get(3.50)Uo′(ξo)=β2Bn.
Moreover, from (3.43) we have that, when t=0,(3.51)(ξ̇o-1)[ωUo(ξo)-Uo′(ξo)]=β2Bn.
We can therefore prove the following.Theorem 3.3.
If one assumes that compatibility conditionUo(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 ξ̇(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<ξ̇(t)<0 if and only if
(3.52)Vo<β2Bn2ω.Proof.
If we suppose that (3.49), (3.50), (3.51) hold then we have
(3.53)(ξ̇o-1)[Uo′(ξo)ξ̇o-Uo′(ξo)]=β2Bn,⟹ξ̇o=0orξ̇o=2.
The initial velocity ξ̇o=2 is not physically acceptable since existence of a solution requires that |ξ̇|<1. Therefore ξ̇o=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
(3.54)-1<ωVoωVo-β2Bn=1+β2BnωVo-β2Bn=ξ̇o<0,
so that for a sufficiently small time t>0 there exists a unique solution with -1<ξ̇<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 whichω=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
(3.55)U(x,t)={12[Uo(x+t)+Uo(x-t)],inDI,12[Uo(x+t)-Uo(t-x)],inDII,12[Uo(x-t)-Uo(ξ-t)]+Vo,inDIII,
and the free boundary equation is the characteristic with positive slope passing through (ξo,0), that is,
(3.56)(ξ̇-1)Uo′(ξ-t)=0⟹Uo(ξ-t)=Uo(ξo),
namely,
(3.57)ξ(t)=ξo+t⟹ξ̇(t)=1.
So, setting to=1-ξo, for t≥to, the region with uniform velocity (the inner core) has disappeared. For t≥to, the solution U(x,t) is thus found solving
(3.58)Uxx=Utt,0⩽x⩽1,t⩾1-ξoU(x,1-ξo)=Uo*(x),0⩽x⩽1,Ut(x,1-ξo)=U1*(x),0⩽x⩽1,U(0,t)=0t⩾1-ξo,U(1,t)=Vot⩾1-ξo,
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
(3.59)W(x,t)=U(x,t)-xVo
and rescale time with
(3.60)θ=t-to.
Problem (3.58) becomes
(3.61)Wxx=Wθθ,0⩽x⩽1,θ⩾0,W(x,0)=Uo*(x)-Vo(x),0⩽x⩽1,Wθ(x,0)=W1(x)=U1*(x),0⩽x⩽1,W(0,θ)=0,θ⩾0,W(1,θ)=0,θ⩾0,
whose solution is [11]
(3.62)W(x,θ)=∑i=1∞[Ancos⁡(πnθ)+Bnsin(πnθ)]sin(πnx),
where
(3.63)An=2∫0θWo(z)sin(πnz)dz,Bn=2πn∫0θW1(z)sin(πnz)dz,U(x,t)=W(x,t-1+ξo)+xVo.


## 4. Asymptotic Expansion

In this section we look for a solution to problem (3.3)(1) in the following form:
(4.1)U(x,t)=∑i=0∞ωiU(i)(x,t).
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(10-1) (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):
(4.2)∑i=0∞[ωiUxx(i)(x,t)-ωiUtt(i)(x,t)+ωi+2U(i)(x,t)]=-β2∑i=0∞(ωt)ii!.
Hence, for each i=0,1,2,…, we have
(4.3)i=0,Uxx(o)(x,t)-Utt(o)(x,t)=-β2,i=1,Uxx(1)(x,t)-Utt(1)(x,t)=-β2t,i=2,Uxx(2)(x,t)-Utt(2)(x,t)=-β2t22!-U(o)(x,t),⋮i>2,Uxx(i)(x,t)-Utt(i)(x,t)=-β2tii!-U(i-2)(x,t)and the following free boundary problems(4.4)i=0,{Uxx(o)(x,t)-Utt(o)(x,t)=-β2U(o)(x,0)=Uo(x),Ut(o)(x,0)=0,U(o)(0,t)=0,U(o)(ξ(o),t)=VoUx(o)(ξ(o),t)+ξ̇(o)Ut(o)(ξ(o),t)=β2Bn,ξ(o)(0)=ξo,(4.5)i=1,{Uxx(1)(x,t)-Utt(1)(x,t)=-tβ2U(1)(x,0)=0,Ut(1)(x,0)=Uo(x),U(1)(0,t)=0,U(1)(ξ(1),t)=VotUx(1)(ξ(1),t)+ξ̇(1)Ut(1)(ξ(1),t)-ξ̇(1)Vo=tβ2Bn,ξ(1)(0)=ξo,(4.6)i≥2,{Uxx(i)(x,t)-Utt(i)(x,t)=-tii!β2-U(i-2)(x,t)U(i)(x,0)=0,Ut(i)(x,0)=0,U(i)(0,t)=0,U(i)(ξ(i),t)=tii!VoUx(i)(ξ(i),t)+ξ̇(i)Ut(i)(ξ(i),t)-ξ̇(i)Voti-1(i-1)!=tii!β2Bn,ξ(i)(0)=ξo.
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
(4.7)DI{U(o)(x,t)=12[Uo(x+t)+Uo(x-t)]+β2t22!,U(1)(x,t)=12∫x-tx+tUo(ζ)dζ+β2t33!,⋮U(i)(x,t)=β22∫0t∫x-t+θx+t-θU(i-2)(ζ,θ)dζdθ+β2ti+2(i+2)!,i⩾2,
while in DII(4.8)DII{U(o)(x,t)=12[Uo(x+t)-Uo(t-x)]+β2t22!-β2(x-t)22!,U(1)(x,t)=12∫t-xx+tUo(ζ)dζ+β2t33!-β2(t-x)33!,⋮U(i)(x,t)=β22∫0t∫t-x+θx-t+θU(i-2)(ζ,θ)dζdθ+β22∫0t∫x-t+θx+t-θU(i-2)(ζ,θ)dζdθ+β2ti+2(i+2)!-β2(t-x)i+2(i+2)!,i⩾2,
and in DIII(4.9)DIII{U(o)(x,t)=Vo+12[Uo(x-t)-Uo(ξ(o)*-t*)]+β2t22!-β2t*22!,U(1)(x,t)=Vot*+12∫x-tξ(1)*-t*Uo(ζ)dζ+β2t33!-β2t*33!,⋮U(i)(x,t)=β22∫0t∫x-t+θx+t-θU(j-2)(ζ,θ)dζdθ-β22∫0t*∫ξ(i)*-t*+θx+t-θU(i-2)(ζ,θ)dζdθ+Vo(t*)ii!+β2ti+2(i+2)!-β2(t*)i+2(i+2)!.i≥2.
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 by(4.10)i=0,(ξ̇(0)-1)[β2t-Uo′(ξ(o)-t)]=β2Bn,(4.11)i=1,(ξ̇(1)-1)[Uo(ξ(1)-t)-Vo+β2t22!]=tβ2Bn,(4.12)i⩾2,(ξ̇(i)-1)[β2(i+2)ti+1(i+1)!-Voti-1(i-1)!+β2∫0tU(i-2)(ξ(i)-t+θ,θ)dθ]=tii!β2Bn.
At the zero order, assuming the compatibility condition U′(ξo)=β2Bn (see problem (4.4)), we have
(4.13)(1-ξ̇o(o))Uo′(ξo)=β2Bn,⟹ξ̇o(o)=0.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 t→0. We have
(4.14)Uxx(1)(ξ(1),t)ξ̇(1)+Uxt(1)(ξ(1),t)+ξ̈(1)Ut(1)(ξ(1),t)+ξ̇(1)2Uxt(1)(ξ(1),t)-ξ̈(1)Vo=β2Bn,which, whent→0, reduces to(4.15)Uo′(ξo)(1+ξ̇o(1)2)=β2Bn,⟹ξ̇o(1)=0.For the genericith order (see problem (4.6)), we assume that the compatibility conditions in the corner (ξo,0) hold up to order (i-1). Therefore we can take the (i-1)th derivative of (4.6)(6), obtaining (4.16)di-1dti-1[Ux(i)(ξ(i),t)]+di-1dti-1[ξ̇(i)]Ut(i)(ξ(i),t)+ξ̇(i)di-1dti-1[Ut(i)(ξ(i),t)]-ξ̇(i)Vo-Voti-1(i-1)!di-1dti-1[ξ̇(i)]=tβ2Bn,which, in the limitt→0, reduces to(4.17)-ξ̇o(i)Vo=0⟹ξ̇o(i)=0.
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 ξ̇(i)(0)=0.Before proceeding further we suppose thatU(x) has the following properties:(H1)Uo(x)∈C∞([0,ξo]),(H2)0<Uo(x)<Vofor allx∈(0,ξo), Uo(0)=0,Uo(ξo)=Vo,(H3)Uo′(x)>0for allx∈[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 variableϕ(o)=ξ(o)-t, so that (4.10) can be rewritten as(4.18)ϕ̇(o)[β2t-Uo′(ϕ(o))]=β2Bn,withϕ(o)(0)=ξo. Then we look for the solution t=t(ϕ(o)) which fulfills the following Cauchy problem:(4.19)β2Bndtdϕ(o)=[β2t-Uo′(ϕ(o))],t(ξo)=0,
that is,(4.20)t(ϕ(o))=-1β2Bn∫ξoϕ(o)Uo′(z)exp⁡(ϕ(o)-zBn)dz.
Recalling that |ξ̇(o)|<1, that is,(4.21)|ϕ̇(o)+1|<1,⟹-2<ϕ̇(o)<0,⟹dtdϕ(o)<-12,
from (4.19)(1) we realize that (4.21) is fulfilled if(4.22)ξo+Bn2<1β2inf⁡z∈[0,ξo]Uo′(z).
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 problem(4.23)(ξ̇(1)-1)[Uo(ξ(1)-t)-Vo+β2t22!]=tβ2Bn,
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) becomes(4.24)ϕ̇(1)[Uo(ϕ(1))-Vo+β2t22!]=tβ2Bn,and we have to solve the following Cauchy problem:(4.25)dtdϕ(1)=1tβ2Bn[Uo(ϕ(1))-Vo+β2t22!],t(ξo)=0.
We notice that (4.25)(1) is a Bernoulli equation. Therefore, setting w=t2, problem (4.25) becomes(4.26)dwdϕ(1)=wBn+2[Uo(ϕ(1))-Voβ2Bn],w(ξo)=0.
whose solution is given by(4.27)t2(ϕ(1))=∫ξoϕ(1)2exp⁡{ϕ(1)-sBn}[Uo(s)-Voβ2Bn]ds,
which make sense only if ϕ(1)⩽ξo. Integrating (4.27) by parts we get(4.28)t2(ϕ(1))=2∫ξoϕ(1)exp⁡{ϕ(1)-sBn}[Uo′(s)β2]ds+2β2[Vo-Uo(ϕ(1))].
We recall from the previous section that the condition |ξ̇(1)(t)|<1 is guaranteed if(4.29)dtdϕ(1)<-12,which, by virtue of(4.25)(1), is equivalent to require that(4.30)t2+Bnt+2β2(U(ϕ(1))-Vo)<0.
Hence, under assumption (H2), the discriminant Δ=Bn2+8β-2(Vo-Uo(ϕ(1)))>Bn2>0, and (4.30) is fulfilled when(4.31)-Bn-Δ2<0⩽t<-Bn+Δ2.
Therefore, in order to have a unique local solution, we must require that(4.32)t2<Bn2+Δ-2ΔBn4<Bn22+2β2[Vo-Uo(ϕ(1))],
which, exploiting (4.28), becomes(4.33)2∫ξoϕ(1)exp⁡{ϕ(1)-sBn}[Uo′(s)β2]ds<Bn22.
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 theith-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 as(4.34)ϕ̇(i)[β2(i+2)ti+1(i+1)!-Voti-1(i-1)!+β2∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ]=tii!β2Bn,
with ϕ(i)(0)=ξo. Once again we look for t=t(ϕ(i)), solving this Cauchy problem(4.35)dtdϕ(i)=(i+2)(i+1)tBn-Voiβ2Bnt+i!tiBn∫0tU(i-2)(ϕ(i)-t+θ,θ)dθt(ξo)=0,t(ξo)=0.
Now, hypothesis (H4) and (4.17) entail(4.36)lim⁡t→0+dϕ(i)dt=lim⁡t→0+dtdϕ(i)=-1.Therefore(4.37)Voiβ2=lim⁡t→0+i!ti-1∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ.
So for t sufficiently small, we can approximate the integral on the r.h.s. of (4.37) in the following way:(4.38)∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ=Ci(ϕ(i))ti-1,
where Ci is a smooth function of ϕ(i), determined exploiting (4.7), (4.8), and (4.9). In particular,(4.39)Ci(ξo)=Vo(i-1)!β2Bn.Hence, setting(4.40)Ai=i+2i+11Bn,Bi(ϕ(i))=[i!Ci(ϕ(i))Bn-iVoβ2Bn],problem (4.35) acquires the following structure:(4.41)dtdϕ(i)=Ait+Bi(ϕ(i))1t,t(ξo)=0,t(ξo)=0,
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(i-2)(x,t) are everywhere non negative for every i), Bi(ϕ(i))<0 and Bi(ξo)=0. Equation (4.41)(1) is once again a Bernoulli equation which can be integrated providing(4.42)t2(ϕ(i))=∫ξoϕ(i)2Bi(s)exp⁡{2Ai(ϕ(i)-s)}ds,
where we recall once again that ϕ(i)⩽ξo. Also in this case we can integrate by parts getting(4.43)t2(ϕ(i))=-BiAi+∫ξoϕ(i)2Bi′(s)Aiexp⁡{2Ai(ϕ(i)-s)}ds.Proceeding as before we derive the conditions ensuring|ξ̇(i)|<1. Hence, making use of (4.41) we obtain the following ineqaulity:
(4.44)t2+t2Ai+BiAi<0,
which is satisfied if(4.45)t2<18Ai2-BiAi.Therefore, |ξ̇(i)|<1 when (4.46)∫ξoϕ(i)2Bi′(s)exp⁡{2Ai(ϕ(i)-s)}ds<18Ai.
So, if condition (4.46) is fulfilled, the solution is given by ξ(i)(t)=ϕ(i)(t)+t, where ϕ(i)(t) is obtained inverting (4.42).


## 4.1. Zero-Order Approximation

We introduce the new variableϕ(o)=ξ(o)-t, so that (4.10) can be rewritten as(4.18)ϕ̇(o)[β2t-Uo′(ϕ(o))]=β2Bn,withϕ(o)(0)=ξo. Then we look for the solution t=t(ϕ(o)) which fulfills the following Cauchy problem:(4.19)β2Bndtdϕ(o)=[β2t-Uo′(ϕ(o))],t(ξo)=0,
that is,(4.20)t(ϕ(o))=-1β2Bn∫ξoϕ(o)Uo′(z)exp⁡(ϕ(o)-zBn)dz.
Recalling that |ξ̇(o)|<1, that is,(4.21)|ϕ̇(o)+1|<1,⟹-2<ϕ̇(o)<0,⟹dtdϕ(o)<-12,
from (4.19)(1) we realize that (4.21) is fulfilled if(4.22)ξo+Bn2<1β2inf⁡z∈[0,ξo]Uo′(z).
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 problem(4.23)(ξ̇(1)-1)[Uo(ξ(1)-t)-Vo+β2t22!]=tβ2Bn,
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) becomes(4.24)ϕ̇(1)[Uo(ϕ(1))-Vo+β2t22!]=tβ2Bn,and we have to solve the following Cauchy problem:(4.25)dtdϕ(1)=1tβ2Bn[Uo(ϕ(1))-Vo+β2t22!],t(ξo)=0.
We notice that (4.25)(1) is a Bernoulli equation. Therefore, setting w=t2, problem (4.25) becomes(4.26)dwdϕ(1)=wBn+2[Uo(ϕ(1))-Voβ2Bn],w(ξo)=0.
whose solution is given by(4.27)t2(ϕ(1))=∫ξoϕ(1)2exp⁡{ϕ(1)-sBn}[Uo(s)-Voβ2Bn]ds,
which make sense only if ϕ(1)⩽ξo. Integrating (4.27) by parts we get(4.28)t2(ϕ(1))=2∫ξoϕ(1)exp⁡{ϕ(1)-sBn}[Uo′(s)β2]ds+2β2[Vo-Uo(ϕ(1))].
We recall from the previous section that the condition |ξ̇(1)(t)|<1 is guaranteed if(4.29)dtdϕ(1)<-12,which, by virtue of(4.25)(1), is equivalent to require that(4.30)t2+Bnt+2β2(U(ϕ(1))-Vo)<0.
Hence, under assumption (H2), the discriminant Δ=Bn2+8β-2(Vo-Uo(ϕ(1)))>Bn2>0, and (4.30) is fulfilled when(4.31)-Bn-Δ2<0⩽t<-Bn+Δ2.
Therefore, in order to have a unique local solution, we must require that(4.32)t2<Bn2+Δ-2ΔBn4<Bn22+2β2[Vo-Uo(ϕ(1))],
which, exploiting (4.28), becomes(4.33)2∫ξoϕ(1)exp⁡{ϕ(1)-sBn}[Uo′(s)β2]ds<Bn22.
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 theith-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 as(4.34)ϕ̇(i)[β2(i+2)ti+1(i+1)!-Voti-1(i-1)!+β2∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ]=tii!β2Bn,
with ϕ(i)(0)=ξo. Once again we look for t=t(ϕ(i)), solving this Cauchy problem(4.35)dtdϕ(i)=(i+2)(i+1)tBn-Voiβ2Bnt+i!tiBn∫0tU(i-2)(ϕ(i)-t+θ,θ)dθt(ξo)=0,t(ξo)=0.
Now, hypothesis (H4) and (4.17) entail(4.36)lim⁡t→0+dϕ(i)dt=lim⁡t→0+dtdϕ(i)=-1.Therefore(4.37)Voiβ2=lim⁡t→0+i!ti-1∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ.
So for t sufficiently small, we can approximate the integral on the r.h.s. of (4.37) in the following way:(4.38)∫0tU(i-2)(ϕ(i)-t+θ,θ)dθ=Ci(ϕ(i))ti-1,
where Ci is a smooth function of ϕ(i), determined exploiting (4.7), (4.8), and (4.9). In particular,(4.39)Ci(ξo)=Vo(i-1)!β2Bn.Hence, setting(4.40)Ai=i+2i+11Bn,Bi(ϕ(i))=[i!Ci(ϕ(i))Bn-iVoβ2Bn],problem (4.35) acquires the following structure:(4.41)dtdϕ(i)=Ait+Bi(ϕ(i))1t,t(ξo)=0,t(ξo)=0,
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(i-2)(x,t) are everywhere non negative for every i), Bi(ϕ(i))<0 and Bi(ξo)=0. Equation (4.41)(1) is once again a Bernoulli equation which can be integrated providing(4.42)t2(ϕ(i))=∫ξoϕ(i)2Bi(s)exp⁡{2Ai(ϕ(i)-s)}ds,
where we recall once again that ϕ(i)⩽ξo. Also in this case we can integrate by parts getting(4.43)t2(ϕ(i))=-BiAi+∫ξoϕ(i)2Bi′(s)Aiexp⁡{2Ai(ϕ(i)-s)}ds.Proceeding as before we derive the conditions ensuring|ξ̇(i)|<1. Hence, making use of (4.41) we obtain the following ineqaulity:
(4.44)t2+t2Ai+BiAi<0,
which is satisfied if(4.45)t2<18Ai2-BiAi.Therefore, |ξ̇(i)|<1 when (4.46)∫ξoϕ(i)2Bi′(s)exp⁡{2Ai(ϕ(i)-s)}ds<18Ai.
So, if condition (4.46) is fulfilled, the solution is given by ξ(i)(t)=ϕ(i)(t)+t, where ϕ(i)(t) is obtained inverting (4.42).


## 5. Conclusions

We have studied a hyperbolic (telegrapher's equation) free boundary problem derived from the model for a pressure-driven channel flow of a particular Bingham-like fluid described in [1]. The motivation of this analysis comes from the study of the rheology of materials like asphalt and bitumen. Exploiting the representation formulas, determined by means of modified Bessel functions, we have shown that the free boundary equation (which has turned out to be a nonlinear integrodifferential equation) can be rewritten only in terms of the initial and boundary data of the problem. In other words, the free boundary dynamics can be solved autonomously from the problem for the velocity field.We have shown that local existence and uniqueness is guaranteed under some appropriate assumptions on the initial and boundary data (Theorem3.3). Moreover, when ω<1 (and this is the case of asphalt and bitumen), we approximate the solution performing an asymptotic expansion in which each term can be iteratively evaluated. We did not prove the convergence of the asymptotic series.A further extension of the analysis we have performed in this paper (which is a limit case of the model presented in [1]) would be to study the one-dimensional problem in its general structure, in which the inner part of the layer is treated as an Oldroyd-B fluid (with nonuniform velocity). Of course this problem is by far more complicated than the one described in this paper. Nevertheless the procedure we have employed here seems to be a promising tool.



---

*Source: 606757-2011-07-31.xml*