
Citation: Robert Carlson. Spectral theory for nonconservative transmission line networks[J]. Networks and Heterogeneous Media, 2011, 6(2): 257-277. doi: 10.3934/nhm.2011.6.257
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In this study, we consider a mathematical model for a particular component of the nephron, the functional unit of kidney [1]. It describes the ionic exchanges through the nephron tubules in the Henle's loop. The main function of kidneys is to filtrate the blood. Through filtration, secretion and excretion of filtered metabolic wastes and toxins, the kidneys are able to maintain a certain homeostatic balance within cells. The first models of nephrons and kidneys were developed in the 1950s with the purpose of explaining the concentration gradient, [10] (or [16] in Chap.3). In their attempt, the authors describe for the first time the urinary concentration mechanism as a consequence of countercurrent transport in the tubules. At a later time the point of view starts to change and a simple mathematical model for a single nephron has been introduced and described, [17]. Furthermore, thanks to the work of J.L. Stephenson [26] it turns out that the formation of a large axial concentration gradient in the kidney depends mainly on the different permeabilities in the tubules and on their counterflow arrangement. Most mathematical models were stationary systems of coupled differential equations. It is possible to find an historical excursus and how these types of models have been developed, referring to [26]. Sophisticated models have been developed to describe the transport of water and electrolyte in the kidney in the recent literature. At the microscopic level of description, cell-based models have been proposed in [23,30,14,31,4] to model small populations of nephrons at equilibrium. Macroscopic models have been also considered in [28,27,21,5,2,9] without accounting for cell-specific transport mechanisms. Despite the development of such sophisticated models, some aspects of the fundamental functions of the kidney remain yet to be fully explained [15]. For example, how a concentrated urine can be produced by the mammalian kidney when the animal is deprived of water remains not entirely clear.
The loop of Henle and its architecture play an important role in the concentrated or diluted urine formation. In order to explain the regulation of urine's concentration, we analyse the counter-current transport in the 'ascending' and 'descending' tubules. There the ionic exchanges between the cell membrane and the environment where tubules are immersed, take place.
We consider a simplified model for sodium exchange in the kidney nephron : the nephron is modelled by an ascending (resp. descending) tubule, of length denoted
$ {a1∂tu1+α∂xu1=J1=2πr1P1(q1−u1)a2∂tu2−α∂xu2=J2=2πr2P2(q2−u2)a3∂tq1=J1,e=2πr1P1(u1−q1)+2πr1,eP1,e(u0−q1)a4∂tq2=J2,e=2πr2P2(u2−q2)+2πr2,eP2,e(u0−q2)−G(q2)a0∂tu0=J0=2πr1,eP1,e(q1−u0)+2πr2,eP2,e(q2−u0)+G(q2), $
|
(1) |
complemented with the boundary and initial conditions
$ u1(t,0)=ub(t),u1(t,L)=u2(t,L),∀t>0u1(0,x)=u01(x),u2(0,x)=u02(x),u0(0,x)=u00(x),q1(0,x)=q01(x),q2(0,x)=q02(x),∀x∈(0,L). $
|
(2) |
In this model, we have used the following notations :
●
●
● Sodium's concentrations (
● Permeabilities
● Section areas
$ a_1 = \pi r_1^2, \ a_2 = \pi r_2^2, \ a_3 = \pi (r_{1, e}^2 - r_1^2), \ a_4 = \pi (r_{2, e}^2 - r_2^2), \ a_0 = \pi \left(\frac{r_{1, e}^2+r_{2, e}^2}{2}\right). $ |
In this work we indicate as lumen the limb under consideration and as tubule the segment together with its epithelial layer. In physiological common language, the term 'tubule' refers to the cavity of lumen together with its related epithelial layer (membrane) as part of it, [16].
In the ascending tubule, the transport of solutes both by passive diffusion and active re-absorption uses
$ G(q2)=Vm,2(q2kM,2+q2)3 $
|
(3) |
where
In a recent paper [19], the authors have studied, from the modelling and biological perspective, the role of the epithelial layer in the ionic transport. The aim of this work is to clarify from the mathematical point of view the link between model (1) taking into account the epithelial layer and models neglecting it. In particular, when the permeability between the epithelium and the lumen is large it is expected that these two regions merge, allowing to reduce system (1) to a model with no epithelial layer. More precisely, as the permeabilities
The outline of the paper is the following. In the next section, we provide the mathematical model and state our main result. Section 3 is devoted to the proof of existence of solution of our model. Then, we focus on the convergence of this solution when permeabilies go to infinity. To this aim, we first establish in section 4 a priori estimates. Using compactness results, we prove convergence in section 5. Numerical illustration of this convergence result is proposed in section 6. Finally an appendix is devoted to a formal computation of the convergence rate at stationary state.
Before presenting our main result, we list some assumptions which will be used throughout this paper.
Assumption 2.1. We assume that the initial solute concentrations are non-negative and uniformly bounded in
$ 0≤u01,u02,q01,q02,u00∈BV(0,L)∩L∞(0,L). $
|
(4) |
For detailed definitions of the
Assumption 2.2. Boundary conditions are such that
$ 0≤ub∈BV(0,T)∩L∞(0,T). $
|
(5) |
Assumption 2.3. Regularity and boundedness of
We assume that the non-linear function modelling active transport in the ascending limb is an odd and
$ ∀x≥0,G(−x)=−G(x),0≤G(x)≤‖G‖∞,0≤G′(x)≤‖G′‖∞. $
|
(6) |
We notice that the function
To simplify our notations in (1), we set
$ a1∂tuε1+α∂xuε1=1ε(qε1−uε1) $
|
(7a) |
$ a2∂tuε2−α∂xuε2=1ε(qε2−uε2) $
|
(7b) |
$ a3∂tqε1=1ε(uε1−qε1)+K1(uε0−qε1) $
|
(7c) |
$ a4∂tqε2=1ε(uε2−qε2)+K2(uε0−qε2)−G(qε2) $
|
(7d) |
$ a0∂tuε0=K1(qε1−uε0)+K2(qε2−uε0)+G(qε2) $
|
(7e) |
Formally, when
$ a1∂tuε1+a3∂tqε1+α∂xuε1= K1(uε0−qε1)a2∂tuε2+a4∂tqε2−α∂xuε2= K2(uε0−qε2)−G(qε2). $
|
Passing formally to the limit when
$ (a1+a3)∂tu1+α∂xu1= K1(u0−u1) $
|
(8) |
$ (a2+a4)∂tu2−α∂xu2= K2(u0−u2)−G(u2), $
|
(9) |
coupled to the equation for the concentration in the interstitium obtained by passing into the limit in equation (7e)
$ a0∂tu0=K1(u1−u0)+K2(u2−u0)+G(u2). $
|
(10) |
This system is complemented with the initial and boundary conditions
$ u1(0,x)=(a1u01(x)+a3q01(x))(a1+a3),u2(0,x)=(a2u02(x)+a4q02(x))(a2+a4), $
|
(11) |
$ u0(0,x)=u00(x), $
|
(12) |
$ u1(t,0)=ub(t),u2(t,L)=u1(t,L). $
|
(13) |
Finally, we recover a simplified system for only three unknowns. From a physical point of view this means fusing the epithelial layer with the lumen. It turns out to merge the lumen and the epithelium into a single domain when we consider the limit of infinite permeability. For
The aim of this paper is to make these formal computations rigorous. For this sake, we define weak solutions associated to the limit system (8)-(10) :
Definition 2.4. Let
$ \mathcal{S}_3: = \left\{ {\boldsymbol{\phi} } \in C^{1}([0, T]\times[0, L])^3, \quad {\boldsymbol{\phi} }(T, x) = 0 , \; \phi_1(t, L) = \phi_2(t, L) , {\rm{ and }} \phi_2(t, 0) = 0\right\}, $ |
we have
$ ∫T0∫L0u1((a1+a3)∂tϕ1+α∂xϕ1)dxdt+α∫T0ub(t)ϕ1(t,0)dt+∫L0(a1+a3)u1(0,x)ϕ1(0,x)dx+∫T0∫L0u2((a2+a3)∂tϕ2−α∂xϕ2)dxdt+∫L0(a2+a4)u2(0,x)ϕ2(0,x)dx+∫T0∫L0{a0u0∂tϕ3+K1(u1−u0)(ϕ3−ϕ1)+K2(u2−u0)(ϕ3−ϕ2)+G(u2)(ϕ3−ϕ2)}dxdt+∫L0a0u00(x)ϕ3(0,x) dx=0. $
|
(14) |
More precisely, the main result reads
Theorem 2.5. Let
$ uεi→ε→0uii=0,1,2,stronglyinL1([0,T]×[0,L]),qεj→ε→0ujj=1,2,stronglyinL1([0,T]×[0,L]), $
|
where
The system (7) can be seen as a particular case of the model without epithelial layer introduced and studied in [28] and [27].
A priori estimates uniform with respect to the parameter
Definition 2.6. Let
$ {\mathcal S}_5: = \left\{ {\boldsymbol{\phi} } \in C^{1}([0, T]\times[0, L])^5, \quad {\boldsymbol{\phi} }(T, x) = 0 , \; \phi_1(t, L) = \phi_2(t, L) , {\rm{ and }} \phi_2(t, 0) = 0\right\} $ |
we have
$ ∫T0∫L0(uε1(a1∂tϕ1+α∂xϕ1)+1ε(qε1−uε1)ϕ1)dxdt+α∫T0uεb(t)ϕ1(t,0)dt+∫L0a1u01(x)ϕ1(0,x)dx+∫T0∫L0(uε2(a2∂tϕ2−α∂xϕ2)+1ε(qε2−uε2)ϕ2)dxdt+∫L0a1u02(x)ϕ2(0,x)dx+∫T0∫L0(a3qε1(∂tϕ3)+K1(uε0−qε1)ϕ3−1ε(qε1−uε1)ϕ3)dxdt+∫L0a3q01(x)ϕ3(0,x) dx+∫T0∫L0(a4qε2(∂tϕ4)+K2(uε0−qε2)ϕ4−1ε(qε2−uε2)ϕ4−G(qε2)ϕ4)dxdt+∫L0a4q02(x)ϕ4(0,x)dx+∫T0∫L0(a0uε0(∂tϕ5)+K1(qε1−uε0)ϕ5+K2(qε2−uε0)ϕ5+G(qε2)ϕ5)dxdt+∫L0a0u00(x)ϕ5(0,x) dx=0. $
|
(15) |
Theorem 2.7. (Existence). Under assumptions (4), (5), (6) and for every fixed
We define the Banach space
$ {a1∂tu1+α∂xu1=(˜q1−u1)ε,a2∂tu2−α∂xu2=(˜q2−u2)ε,a3∂tq1=(˜u1−q1)ε+K1(˜u0−q1),a4∂tq2=(˜u2−q2)ε+K2(˜u0−q2)−G(˜q2),a0∂tu0=K1(˜q1−u0)+K2(˜q2−u0)+G(˜q2), $
|
(16) |
with initial data
$ u_{1}(t, 0) = u_b(t)\geq 0\ , \quad u_{2}(t, L) = u_{1}(t, L), \quad {\rm{for }} t > 0, $ |
where
First we define the solutions of (16) using Duhamel's formula. Under these hypothesis, we may compute
$ u1(t,x)={u01(x−αa1t)e−ta1ε+1a1ε∫t0e−t−sa1ε˜q1(x−αa1(t−s),s) ds,ifx>αa1t,ub(t−a1xα)e−xεα+1αε∫x0e−1αε(x−y)˜q1(t−a1(x−y)α,y) dy,ifx<αa1t, $
|
(17) |
with
$ u2(t,x)={u02(x+αa2t)e−ta2ε+1a2ε∫t0e−t−sa2ε˜q2(s,x+αa2(t−s)) ds,ifx<L−αa2t,u1(t+a2(x−L)α,L)ex−Lαε+1αε∫Lxex−yεα˜q2(t+a2(x−y)α,y)dyifx>L−αa2t. $
|
(18) |
Then, for the other unknowns, one simply solves a system of uncoupled ordinary differential equations leading to :
$ {q1(t,x)=q01(x)e−(1ε+K1)ta3+1a3∫t0e−(1ε+K1)(t−s)a3(1ε˜u1+K1˜u0)(s,x) ds,q2(t,x)=q02(x)e−(1ε+K2)ta4+1a4∫t0e−(1ε+K2)(t−s)a4(1ε˜u2+K1˜u0−G(˜q2))(s,x) ds,u0(t,x)=u00(x)e−(K1+K2)ta0+1a0∫t0e−(K1+K2)(t−s)a0(K1˜q1+K2˜q2+G(˜q2))(s,x) ds. $
|
(19) |
Using Theorem B.1, the previous unknowns solve the weak formulation reading :
$ {∫T0∫L0(−u1(a1∂t+α∂x)φ1(t,x)+1ε(u1−˜q1)φ1(t,x))dxdt+a1[∫L0u1(t,x)φ1(t,x)dt]t=Tt=0+α[∫T0u1(t,x)φ1(t,x)]x=Lx=0=0,∫T0∫L0(−u2(a2∂t−α∂x)φ2(t,x)+1ε(u2−˜q2)φ2(t,x))dxdt+a2[∫L0u2(t,x)φ2(t,x)dt]t=Tt=0+α[∫T0u2(t,x)φ2(t,x)]x=Lx=0=0, $
|
(20) |
for any
$ ∫T0∫L0−|u1|(a1∂t+α∂x)φ1(t,x)+1ε(|u1|−sgn(u1)˜q1)φ1(t,x)dxdt+a1[∫L0|u1|(t,x)φ1(t,x)dt]t=Tt=0+α[∫T0|u1|(t,x)φ1(t,x)]x=Lx=0=0. $
|
(21) |
The same holds also for the other unknowns
$ {a1∂t|u1|+α∂x|u1|=1ε(sgn(u1)˜q1−|u1|)a2∂t|u2|−α∂x|u2|=1ε(sgn(u2)˜q2−|u2|), $
|
we actually mean that these inequalities hold in the previous sense, i.e. in the sense of (21). The reader should notice that the stronger regularity of the integrated forms (17), (18) and (19) allows to define these solutions on the boundaries of the domain
$ {a1∂t|u1|+α∂x|u1|≤1ε(|˜q1|−|u1|)a2∂t|u2|−α∂x|u2|≤1ε(|˜q2|−|u2|)a3∂t|q1|≤1ε(|˜u1|−|q1|)+K1(|˜u0|−|q1|)a4∂t|q2|≤1ε(|˜u2|−|q2|)+K2(|˜u0|−|q2|)−|G(˜q2)|a0∂t|u0|≤K1(|˜q1|−|u0|)+K2(|˜q2|−|u0|)+|G(˜q2)|. $
|
(22) |
We have used the fact that
Adding all equations and integrating on
$ ddt∫L0(a1|u1|+a2|u2|+a0|u0|+a3|q1|+a4|q2|)≤α|u1(t,0)|+1ε∫L0(|˜u1|+|˜u2|+|˜q1|+|˜q2|) dx+(K1+K2)∫L0|˜u0| dx, $
|
where we use the boundary condition
$ ‖U(t,x)‖L1(0,L)5≤ ‖U(0,x)‖L1(0,L)5+αminiai∫T0|ub(s)| ds+η∫T0‖˜U(t,x)‖L1(0,L)5 dt, $
|
(23) |
with
On the other hand, using (17), (18) and (19), one quickly checks that
$ ‖U‖L∞((0,T)×(0,L))5≤max(‖U0‖L∞(0,L)5,‖ub‖L∞(0,T))+CTε‖˜U‖L∞((0,T)×(0,L))5 $
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(24) |
where the generic constant
Let us now prove that
$ ‖T(˜U)−T(˜W)‖L1(0,L)5=‖U−W‖L1(0,L)5≤η∫T0‖˜U−˜W‖L1(0,L)≤ηT‖˜U−˜W‖XT. $
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Again similar computations as in (24), show that
$ \left\lVert{U-W}\right\rVert_{{L^\infty((0, T)\times(0, L))^5}} \leq \frac{C T}{ \varepsilon } \left\lVert{\tilde{U}-\tilde{W}}\right\rVert_{{L^\infty((0, T)\times(0, L))^5}}. $ |
Therefore, as soon as
As a result of above computations, we have also that if
$ ‖U1−U2‖L1((0,T)×(0,L))≤‖U0,1−U0,2‖L1(0,L)+αminiai‖u1b−u2b‖L1(0,T). $
|
(25) |
which shows and implies uniqueness as well.
In order to prove our convergence result, we first establish some uniform a priori estimates. The strategy of the proof of Theorem 2.5 relies on a compactness argument. In this Section we will omit the superscript
The following lemma establishes that all concentrations of system are non-negative and this is consistent with the biological framework.
Lemma 4.1. (Non-negativity). Let
Proof. We prove that the negative part of our functions vanishes. Using Stampacchia's method, we formally multiply each equation of system (7) by corresponding indicator function as follows:
$ {(a1∂tu1+α∂xu1)1{u1<0}=1ε(q1−u1)1{u1<0}(a2∂tu2−α∂xu2)1{u2<0}=1ε(q2−u2)1{u2<0}(a3∂tq1)1{q1<0}=1ε(u1−q1)1{q1<0}+K1(u0−q1)1{q1<0}(a4∂tq2)1{q2<0}=1ε(u2−q2)1{q2<0}+K2(u0−q2)1{q2<0}−G(q2)1{q2<0}(a0∂tu0)1{u0<0}=K1(q1−u0)1{u0<0}+K2(q2−u0)1{u0<0}+G(q2)1{u0<0}. $
|
Again as in the proof of existence in Section 3, these computations can be made rigorously using the extra regularity provided along characteristics in the spirit of Lemma 3.1, [20] summarized in Theorem B.1.
Using again Theorem B.1, one writes formally :
$ {a1∂tu−1+α∂xu−1≤1ε(q−1−u−1)a2∂tu−2−α∂xu−2≤1ε(q−2−u−2)a3∂tq−1≤1ε(u−1−q−1)+K1(u−0−q−1)a4∂tq−2≤1ε(u−2−q−2)+K2(u−0−q−2)+G(q2)1{q2<0}a0∂tu−0≤K1(q−1−u−0)+K2(q−2−u−0)−G(q2)1{u0<0}. $
|
Adding the previous expressions, one recovers a single inequality reading
$ {{\partial}}_t (a_1 u_1^{-} + a_2 u_2^{-} + a_3 q_{1}^{-}+ a_4 q_{2}^{-} + a_0 u_0^{-}) +\alpha {{\partial}}_x (u_1^{-}- u_2^{-}) \leq G(q_2)({\mathbf{1}_{\{q_{2} < 0\}}} - {\mathbf{1}_{\{u_{0} < 0\} }}). $ |
By Assumption 2.3, we have that
$ ddt∫L0(a1u−1+a2u−2+a3q−1+a4q−2+a0u−0)(t,x) dx≤α(u−2(t,L)−u−2(t,0)−u−1(t,L)+u−1(t,0)). $
|
Since
$ \frac{d}{dt}\int_{0}^{L}(a_1 u_1^{-} + a_2 u_2^{-} + a_3 q_{1}^{-}+ a_4 q_{2}^{-} + a_0 u_0^{-})(t, x)\ dx \leq \alpha u_1^{-}(t, 0) = \alpha u_b^{-}(t). $ |
From Assumptions 2.2 and 2.1, the initial and boundary data are all non-negative. Thus
Lemma 4.2. (
$ 0 \leq u_{0}(t, x)\leq \kappa(1+t), \quad 0 \leq u_{i} (t, x)\leq \kappa(1+t), \quad0 \leq q_{i} (t, x)\leq \kappa(1+t), \quad i = 1, 2, $ |
$ 0 \leq u_2(t, 0) \leq \kappa(1+t), \quad 0\leq u_1(t, L) \leq \kappa(1+t), $ |
where the constant
Proof. We use the same method as in the previous lemma for the functions
$ w_i = (u_i- \kappa(1+t)), \quad i = 0, 1, 2, \quad z_j = (q_j- \kappa(1+t)), \quad j = 1, 2. $ |
From system (7) and using the fact that
$ z_j{\mathbf{1}_{\{w_{i}\geq 0\}}} = z_j^{+}{\mathbf{1}_{\{w_{i}\geq 0\}}}-z_j^{-}{\mathbf{1}_{\{w_{i}\geq 0\}}} \leq z_j^{+}, \quad\quad w_i{\mathbf{1}_{\{z_{j}\geq 0\}}}\leq w_i^{+}, $ |
we get
$ {a1∂tw+1+a1κ1{w1≥0}+α∂xw+1≤1ε(z+1−w+1)a2∂tw+2+a2κ1{w2≥0}−α∂xw+2≤1ε(z+2−w+2)a3∂tz+1+a3κ1{z1≥0}≤1ε(w+1−z+1)+K1(w+0−z+1)a4∂tz+2+a4κ1{z2≥0}≤1ε(w+2−z+2)+K2(w+0−z+2)−G(q2)1{z2≥0}a0∂tw+0+a0κ1{w0≥0}≤K1(z+1−w+0)+K2(z+2−w+0)+G(q2)1{w0≥0}. $
|
(26) |
Adding expressions above gives
$ ∂t(a1w+1+a2w+2+a3z+1+a4z+2+a0w+0)+α∂x(w+1−w+2)≤ −κ1{w0≥0} +G(q2)(1{w0≥0}−1{z2≥0}). $
|
Integrating with respect to
$ ddt∫L0(a1w+1+a2w+2+a3z+1+a4z+2+a0w+0)(t,x)dx≤α(w+2(t,L)−w+2(t,0)−w+1(t,L)+w+1(t,0))+∫L0(G(q2)−κ)1{w0≥0}dx, $
|
where we use the fact that
$ ddt∫L0(a1w+1+a2w+2+a3z+1+a4z+2+a0w+0)(t,x)dx+αw+2(t,0)≤α(ub(t)−κ(1+t))++(‖G‖∞−κ)∫L01{w0≥0}dx. $
|
If we adjust the constant
$ \frac{d}{dt}\int_{0}^{L} (a_1 w_1^{+} + a_2 w_2^{+} + a_3 z_1^{+} + a_4 z_2^{+} + a_0 w_{0}^{+})(t, x)\, dx + \alpha w_2^+(t, 0) \leq 0, $ |
which shows the claim.
For the last estimate on
$ ddt∫L0(a1w+1+a3z+1)dx+αw+1(t,L)≤ αw+1(t,0)+K1∫L0(w+0−z+1)dx−κ∫L0(1{w1≥0}+1{z1≥0})dx. $
|
Integrating on
$ \alpha \int_0^T w_1^+(t, L)\, dt \leq \alpha \int_0^T w_1^+(t, 0)\, dt = 0, $ |
for
Lemma 4.3. (
$ \mathcal{H}(t) = \int_{0}^{L} (a_1|u_1|+a_2|u_2|+a_0|u_0|+a_3|q_1|+a_4|q_2|)(t, x)\, dx. $ |
Then, under hypothesis (4), (5), (6) the following a priori estimate, uniform in
$ \mathcal{H}(t) \leq \alpha \|u_b\|_{L^{1}(0, T)} +\mathcal{H}(0), \quad \forall t > 0. $ |
Moreover the following inequalities hold:
$ ∫T0|u2(t,0)|dt≤‖ub‖L1(0,T)+1αH(0), $
|
and
$ ∫T0|u1(t,L)|dt≤∫L0(|u01(x)|+|q01(x)|)dx+CT $
|
with
Proof. Since from Lemma 4.1 all concentrations are non-negative, we may write from system (7)
$ {a1∂t|u1|+α∂x|u1|=1ε(|q1|−|u1|)a2∂t|u2|−α∂x|u2|=1ε(|q2|−|u2|)a3∂t|q1|=1ε(|u1|−|q1|)+K1(|u0|−|q1|)a4∂t|q2|=1ε(|u2|−|q2|)+K2(|u0|−|q2|)−|G(q2)|a0∂t|u0|=K1(|q1|−|u0|)+K2(|q2|−|u0|)+|G(q2)|. $
|
(27) |
Adding all equations and integrating on
$ ddtH(t)+α|u2(t,0)|=α|u1(t,0)|=α|ub(t)|. $
|
(28) |
Integrating now with respect to time, we obtain:
$ H(t)+α∫t0|u2(s,0)|ds≤α∫t0|ub(s)| ds+H(0). $
|
(29) |
with
$ \frac{d}{dt}\int_{0}^{L}(a_1 |u_1|+ a_3 |q_{1}|)\ dx +\alpha|u_1(t, L)| \leq \alpha|u_{b}(t)| + K_{1}\int_{0}^{L}|u_{0}|\ dx. $ |
Since we have shown that
Here we detail the notion of regularization for
$ f_\delta (x) = \sum\limits_{k = 0}^\infty \eta_{\delta_k} * (f \zeta_k) $ |
where
$ \left\{ ζk∈C∞c(Vk),0≤ζk≤1,(k=1,2,…)∞∑k=1ζk≡1on(0,L) \right. $
|
where
$ \left\{ Θ0=∅,Θk:={x∈(0,L);dist(x,{0,L})>1k},(k=1,2,…)Vk:=Θk+1∖¯Θk−1,(k=1,2,…) \right. $
|
and there exists
$ \left\{ Supp(ηδk∗(ζkf))⊆Vk,∫(0,L)|ηδk∗(ζkf)−ζkf|dx≤δ2−k,∫(0,L)|ηδk∗(ζ′kf)−ζ′kf|dx≤δ2−k, \right. $
|
Although
$ \lim\limits_{\delta\to0} \left\lVert{ {\partial_x} f_\delta}\right\rVert_{{L^1(0, L)}} = \left\lVert{\lambda_f((0, L))}\right\rVert_{{}}, $ |
where the right hand side is the total variation of the Radon measure associated to the derivative of
$ ‖∂xfδ‖L1(0,L)≤‖f‖BV(0,L). $
|
(30) |
Then we set
$ fδc(x):=(1−χδ(x))fδ(x)+cχδ(x),∀x∈(0,L) $
|
(31) |
where
$ \chi_{{\delta}}(x) : = \chi\left( \frac{x}{\delta} \right) , \quad \chi(x) : = {1if|x|<10if|x|>2 , $
|
Lemma 4.4. If
$ \left\lVert{ {\partial_x} f^\delta_c}\right\rVert_{{L^1(0, L)}} \leq C \left( 1+\left\lVert{f}\right\rVert_{{ {{{\rm{BV}}}}(0, L)}} \right) $ |
where the generic constant
Proof. Differentiating (31) and integrating in space gives :
$ ‖∂xfδc‖L1(0,L)≤2‖∂xfδ‖L1(0,L)+1δ∫δ0|fδ(x)−c||χ′(x/δ)|dx≤2‖∂xfδ‖L1(0,L)+‖χ′‖L∞(R)(|fδ(0+)−c|+1δ∫δ0|fδ(x)−fδ(0+)|dx)≤C(‖f‖BV(0,L)+|fδ(0+)−c|)≤C(‖f‖BV(0,L)+c+|fδ(0+)|)≤C(‖f‖BV(0,L)+c+‖fδ‖W1,1(0,L))≤C(1+‖f‖BV(0,L)) $
|
where we used (30).
Definition 4.5. If
$ \left\{ u0,δ1(x):=(1−χδ(x)−χδ(L−x))u01,δ+c1χδ(x)+c2χδ(L−x),∀x∈[0,L]uδb(t):=(1−χδ(t))ub,δ+c1χδ(t),∀t∈[0,T]u0,δ2(x):=(1−χδ(L−x))u02,δ+c2χδ(L−x),∀x∈[0,L]q0,δ1(x):=q01,δ,∀x∈[0,L]q0,δ2(x):=q02,δ,∀x∈[0,L]u0,δ0(x):=u00,δ,∀x∈[0,L] \right. $
|
where
The matching is
This regularization procedure allows then to obtain
Lemma 4.6. Assume hypotheses 2.3, and let
$ {a1∂t+α∂x)uδ1,t=1ε(qδ1,t−uδ1,t)a2∂t−α∂x)uδ2,t=1ε(qδ2,t−uδ2,t)a3∂tqδ1,t=−1ε(qδ1,t−uδ1,t)+K1(uδ0,t−qδ1,t)a4∂tqδ2,t=−1ε(qδ2,t−uδ2,t)+K2(uδ0,t−qδ2,t)−G′(qδ2)qδ2,ta0∂tuδ0,1=K1(qδ1,t−uδ0,t)+K2(qδ2,t−uδ0,t)−G′(qδ2)qδ2,t $
|
(32) |
where
$ {uδ1,t(t,0)=∂tuδb(t),a1uδ1,t(0,x)=−α∂xu0,δ1+1ε(q0,δ1−u0,δ1)a2uδ2,t(0,x)=α∂xu0,δ2+1ε(q0,δ2−u0,δ2)a3qδ1,t(0,x)=−1ε(q0,δ1−u0,δ1)+K1(u0,δ0−q0,δ1)a4qδ2,t(0,x)=−1ε(q0,δ2−u0,δ2)+K2(u0,δ0−q0,δ2)−G(q0,δ2)a0uδ0,t(0,x)=K1(q0,δ1−u0,δ0)+K2(q0,δ2−u0,δ0)+G(q0,δ2) $
|
(33) |
Proof. The Duhamel's formula obtained by the fixed point method in the proof of Theorem 2.7 provides a solution
Remark 1. A priori estimates from previous sections, when applied to the problem (32) complemented with initial-boundary data (33), do not provide a control of
This remark motivates next paragraphs.
At this step we introduce initial layer correctors. For this sake, on the microscopic scale we define for
$ {a1∂t˜u1=˜q1−˜u1a2∂t˜u2=˜q2−˜u2a3∂t˜q1=˜u1−˜q1a4∂t˜q2=˜u2−˜q2, $
|
(34) |
with initial conditions
$ \tilde{u}_{1}(0, x) = q_{1}^{0, \delta}-u_{1}^{0, \delta}, \quad \tilde{u}_{2}(0, x) = q_{2}^{0, \delta}-u_{2}^{0, \delta}, \quad \tilde{q}_1(0, x) = 0, \quad \tilde{q}_2(0, x) = 0. $ |
Actually, this system may be solved explicitly and we obtain for
$ {˜ui(t,x)=ai+ai+2exp(−t(1ai+1ai+2))ai+ai+2(q0,δi(x)−u0,δi(x))˜qi(t,x)=ai(1−exp(−t(1ai+1ai+2)))ai+ai+2(q0,δi(x)−u0,δi(x)) $
|
(35) |
We introduce the following quantities on the macroscopic time scale
$ {vδ1(t,x)=uδ1(t,x)+˜u1(tε,x)vδ2(t,x)=uδ2(t,x)+˜u2(tε,x)sδ1(t,x)=qδ1(t,x)+˜q1(tε,x)sδ2(t,x)=qδ2(t,x)+˜q2(tε,x) $
|
(36) |
Next, we prove uniform bounds on the time derivatives :
Proposition 1. Let
$ \tilde{\mathcal{H}}_{t}(t) = \int_{0}^{L}(a_1 |{{\partial}}_t v^\delta_{1}|+a_2|{{\partial}}_t v^\delta_{2}|+ a_3|{{\partial}}_t s^\delta_{1}|+a_4|{{\partial}}_t s^\delta_{2}|+a_0|{{\partial}}_t u^\delta_{0}|)(t, x)\ dx, $ |
with functions
$ ˜Ht(t)+α∫t0|∂tvδ2(τ,0)|dτ+α∫t0|∂tvδ1(τ,L)|dτ≤C(‖U0,δ‖W1,1(0,L)+‖uδb‖W1,1(0,T)),fora.e.t∈(0,T), $
|
(37) |
where
Proof. From system (7) we deduce
$ {a1∂tvδ1+α∂xvδ1=1ε(sδ1−vδ1)+α∂x˜u1(tε,x)a2∂tvδ2−α∂xvδ2=1ε(sδ2−vδ2)−α∂x˜u2(tε,x)a3∂tsδ1=1ε(vδ1−sδ1)+K1(uδ0−sδ1)+K1˜q1(tε,x)a4∂tsδ2=1ε(vδ2−sδ2)+K2(uδ0−sδ2)+K2˜q2(tε,x)−G(qδ2)a0∂tuδ0=K1(sδ1−uδ0)+K2(sδ2−uδ0)−K1˜q1(tε,x)−K2˜q2(tε,x)+G(qδ2) $
|
(38) |
with following initial and boundary conditions:
$ vδ1(t,0)=u1(t,0)+˜u1(t,0)=ub(t)+˜u1(tε,0),t∈(0,T),vδ2(t,L)=vδ2(t,L)+˜u2(tε,L),t∈(0,T),vδ1(0,x)=uδ1(0,x)+˜u1(0,x)=q01(x),x∈(0,L),vδ2(0,x)=uδ2(0,x)+˜u2(0,x)=q02(x),sδ1(0,x)=qδ1(0,x)+˜q1(0,x)=q01(x),sδ2(0,x)=qδ2(0,x)+˜q2(0,x)=q02(x). $
|
(39) |
As
$ {a1∂tvδ1,t+α∂xvδ1,t=1ε(sδ1,t−vδ1,t)+1ε∂x˜u1,ta2∂tvδ2,t−α∂xvδ2,t=1ε(sδ2,t−vδ2,t)−1ε∂x˜u2,ta3∂tsδ1,t=1ε(vδ1,t−sδ1,t)+K1(uδ0,t−sδ1,t)+1εK1˜q1,ta4∂tsδ2,t=1ε(vδ2,t−sδ2,t)+K2(uδ0,t−sδ2,t)+1εK2˜q2,t−G′(qδ2)q2,ta0∂tuδ0,t=K1(sδ1,t−uδ0,t)+K2(sδ2,t−uδ0,t)−1εK1˜q1,t−1εK2˜q2,t+G′(qδ2)q2,t $
|
in the sense of Definition (2.6). Again formally, we multiply each equation respectively by
$ {a1∂t|vδ1,t|+α∂x|vδ1,t|≤1ε(|sδ1,t|−|vδ1,t|)+|1ε∂x˜u1,t|a2∂t|vδ2,t|−α∂x|vδ2,t|≤1ε(|sδ2,t|−|vδ2,t|)+|1ε∂x˜u2,t|a3∂t|sδ1,t|≤1ε(|vδ1,t|−|sδ1,t|)+K1(|uδ0,t|−|sδ1,t|)+|1εK1˜q1,t|a4∂t|sδ2,t|≤1ε(|vδ2,t|−|sδ2,t|)+K2(|uδ0,t|−|sδ2,t|)+|1εK2˜q2,t|+|G′(qδ2)1ε˜q2,t|−G′(qδ2)|sδ2,t|a0∂t|uδ0,t|≤K1(|sδ1,t|−|uδ0,t|)+K2(|sδ2,t|−|uδ0,t|)+|1εK1˜q1,t|+|1εK2˜q2,t|+|G′(qδ2)1ε˜q2,t|+G′(qδ2)|sδ2,t|. $
|
(40) |
Indeed, the right hand side of the
$ −G′(qδ2)q2,tsgn(sδ2,t)=−G′(qδ2)(sδ2,t(t,x)−1ε˜q2,t(tε,x))sgn(sδ2,t)≤−G′(qδ2)|sδ2,t|+1ε|G′(qδ2)˜q2,t|. $
|
On the other hand
$ −G′(qδ2)q2,tsgn(uδ0,t)=−G′(qδ2)(sδ2,t(t,x)−1ε˜q2,t(tε,x))sgn(u0,t)≤G′(qδ2)|sδ2,t|+1ε|G′(qδ2)˜q2,t|, $
|
since
$ ddt˜Ht(t)+α|vδ2,t(t,0)|≤F1(t)+F2(t)+F3(t)+F4(t), $
|
(41) |
where
$
F1(t):=α(|vδ2,t(t,L)|−|vδ1,t(t,L)|);F2(t):=α|vδ1,t(t,0)|,F3(t):=1ε∫L0|∂x˜u2,t(tε,x)|dx+1ε∫L0|∂x˜u1,t(tε,x)|dx,F4(t):=2K1ε∫L0|˜q1,t(tε,x)|dx+2ε(‖G′‖∞+K2)∫L0|˜q2,t(tε,x)|dx.
$
|
Integrating (41) in time, we get
$ ˜Ht(t)+α∫T0|vδ2,t(t,0)|dt≤∫T0(F1(t)+F2(t)+F3(t)+F4(t))dt+˜Ht(0). $
|
(42) |
Let us consider each term of the right hand side of (42) separately:
●
$ ∫T0(|vδ2,t(t,L)|−|vδ1,t(t,L)|)dt≤1ε∫T0|(˜u2,t−˜u1,t)(tε,L)|dt≤∫Tε0|(˜u2,t−˜u1,t)(τ,L)|dτ≤12{|u0,δ1(0)−q0,δ1(0)|+|u0,δ2(0)−qδ,02(0)|}≤C‖U0,δ‖W1,1(0,L) $
|
where we used trace operator's continuity for
●
$ ∫T0|vδ1,t(t,0)| dt≤∫T0|(uδb)′(s)|ds+∫Tε0|(u0,δ1(0)−q0,δ1(0))e−2τ| dτ≤C(‖uδb‖W1,1(0,T)+‖U0,δ‖W1,1(0,L)) $
|
as above.
●
$ ∫T0∫L0|1ε∂x˜ui,t(tε,x)|dxdt=∫Tε0∫L0|∂x˜ui,t(τ,x)|dxdτ≤C‖U0,δ‖W1,1(0,L), $
|
which is uniformly bounded with respect to
●
$ ∫T0∫L0|1ε˜qi,t(tε,x)|dxdt=∫Tε0∫L0|˜qi,t(τ,x)| dxdτ≤C‖U0,δ‖W1,1(0,L) $
|
thanks to the fact that
It remains to estimate
$ ˜Ht(0)=∫L0(a1|vδ1,t(0,x)|+a2|vδ2,t(0,x)|+a3|sδ1,t(0,x)|+a4|sδ2,t(0,x)|+a0|uδ0,t(0,x)|)dx≤C‖U0,δ‖W1,1(0,L) $
|
So, for instance, for the first term of the sum, we use the first equation in (38) and we write
$ a_1 {{\partial}}_t v_1^\delta(0, x) = \frac{1}{ \varepsilon} (s_1^\delta(0, x)-v_1^\delta(0, x)) + \alpha{{\partial}}_x\tilde{u}_1(0, x) - \alpha {{\partial}}_x v_1^\delta(0, x). $ |
Recalling that
$ \int_{0}^{L}|{{\partial}}_t v_1^\delta(0, x)|\ dx \leq \frac{\alpha}{a_1}\int_{0}^{L}|{{\partial}}_x u_1^{0, \delta}(x)|\ dx < C \left\lVert{U^{0, \delta}}\right\rVert_{{ {\mathbf W}^{1, 1}(0, L)}}, $ |
The rest follows exactly the same way. We conclude from (42) and the above calculations that
$ ˜Ht(t)+α∫T0|vδ2,t(t,0)|dt≤˜Ht(0)+∫t0(F1+F2+F3+F4)(s) ds≤C(‖U0,δ‖W1,1(0,L)+‖uδb‖W1,1(0,T)). $
|
Finally, in order to recover (37), we add the first and third inequalities in (40) and integrate on
$ ∫L0(a1|vδ1,t(T,x)|+a3|sδ1,t(T,x)|)dx+α∫T0|vδ1,t(t,L)|dt≤ α∫T0|vδ1,t(t,0)|dt+∫L0(a1|vδ1,t(0,x)|+a3|sδ1,t(0,x)|)dx+∫T0∫L0(K1|uδ0,t(t,x)|+K1ε|˜q1,t(tε,x)|+1ε|∂x˜u1,t(tε,x)|)dxdt. $
|
We have already proved that the second term of the right hand side is bounded. We have also proved above that
As a consequence, we deduce the following estimates on the time derivatives of the original unknowns
Corollary 1. Let
$ ∫T0∫L0(|∂tuδ1|+|∂tuδ2|+|∂tuδ0|+|∂tqδ1|+|∂tqδ2|)(t,x)dxdt≤C‖U0,δ‖W1,1(0,L),∫T0|∂tuδ2(t,0)| dt≤C‖U0,δ‖W1,1(0,L),∫T0|∂tuδ1(t,L)| dt≤C‖U0,δ‖W1,1(0,L). $
|
(43) |
Proof. We recall the expressions
$ v^\delta_1 = u^\delta_1+\tilde{u}_1, \quad v^\delta_2 = u^\delta_2+\tilde{u}_2, \quad s^\delta_1 = q^\delta_1+\tilde{q}_1, \quad s^\delta_2 = q^\delta_2+\tilde{q}_2. $ |
By the triangle inequality, we have for
$ ‖∂tuδi‖L1([0,T]×[0,L])≤‖∂tvδi‖L1([0,T]×[0,L])+1ε‖∂t˜ui(t/ε,x)‖L1([0,T]×[0,L]),‖∂tqδi‖L1([0,T]×[0,L])≤‖∂tsδi‖L1([0,T]×[0,L])+1ε‖∂t˜qi(t/ε,x)‖L1([0,T]×[0,L]). $
|
The first terms of the latter right hand side are bounded from Proposition 1. For the second terms, we have, as above,
$ ∫T0∫L01ε|∂t˜qi(tε,x)| dxdt= 1ε∫T0∫L0|(q0i(x)−u0i(x))e−2tε| dxdt<C‖U0,δ‖W1,1(0,L),∫T0∫L01ε|∂t˜ui(tε,x)| dxdt= 1ε∫T0∫L0|(u0i(x)−q0i(x))e−2tε| dxdt<C′‖U0,δ‖W1,1(0,L). $
|
Furthermore, from (42), we get
$ \int_{0}^{T} \left|v^\delta_{2, t}(t, 0)\right|\ dt \leq C_T \left\lVert{U^{0, \delta}}\right\rVert_{{ {\mathbf W}^{1, 1}(0, L)}}. $ |
By the triangle inequality, it implies the second estimate in (43) To recover the third claim in (43), we notice that by definition of
$ |u^\delta_{1, t}(t, L)|\leq |v^\delta_{1, t}(t, L)| + \frac 1 \varepsilon \left|\tilde{u}_{1, t}\left(\frac{t}{ \varepsilon}, L\right)\right| \leq |v^\delta_{1, t}(t, L)| + \frac 1 \varepsilon \|q^{0, \delta}_ 1- u^{0, \delta}_1\|_{W^{1, 1}(0, L)} e^{-2\frac{t}{ \varepsilon}}. $ |
where again we use the continuity of the trace operator on
Lemma 4.7. Let
$ \int_0^T\int_{0}^{L} \left(\sum\limits_{i = 0}^2 |{{\partial}}_{x}u^\delta_{i}(t, x)|+\sum\limits_{i = 1}^2 |{{\partial}}_{x}q^\delta_{i}(t, x)|\right) \, dxdt \leq C_T \left\lVert{U^{0, \delta}}\right\rVert_{{W^{1, 1}((0, L))}}, $ |
for some non-negative constant
Proof. Adding equation (7a) with (7c) and also (7b) with (7d) we get
$ α∂xuδ1=K1(uδ0−qδ1)−a1∂tuδ1−a3∂tqδ1,−α∂xuδ2=K2(uδ0−qδ2)−a2∂tuδ2−a4∂tqδ2−G(qδ2). $
|
Using Corollary 1 and (6), the right hand sides are uniformly bounded in
We show here how to use Corollary 1 and Lemma 4.7 in order to obtain
Theorem 4.8. Under hypotheses (2.1)-(2.3), there exists a uniform bound such that the
$ \sum\limits_{i = 0}^2 \left\lVert{u_i}\right\rVert_{{BV((0, T)\times(0, L))}} + \sum\limits_{i = 1}^2 \left\lVert{q_i}\right\rVert_{{BV((0, T)\times(0, L))}} \leq C \left(\left\lVert{U^0}\right\rVert_{{ {{{\rm{}}}{\mathbf B}{\mathbf V}}(0, L)}} + \left\lVert{u_b}\right\rVert_{{ {{{\rm{BV}}}}(0, T)}} \right) $ |
where the generic constant
Proof. Setting
$ \left\lVert{U^\delta}\right\rVert_{{ {\mathbf W}_{t, x}^{1, 1}((0, T)\times(0, L))}} \leq C \left( \left\lVert{U^{0, \delta}}\right\rVert_{{ {\mathbf W}_{x}^{1, 1}(0, L)}} + \left\lVert{u_b^\delta}\right\rVert_{{W^{1, 1}(0, T)}} \right) $ |
Now thanks to Lemma 4.4, one estimates the right hand side with respect to the
$ \left\lVert{U^{0, \delta}}\right\rVert_{{ {\mathbf W}_{x}^{1, 1}(0, L)}} \leq C\left(1+\left\lVert{U^0}\right\rVert_{{{ {{{\rm{}}}{\mathbf B}{\mathbf V}}}(0, L)}} \right), \quad \left\lVert{u_b^\delta}\right\rVert_{{W^{1, 1}(0, T)}} \leq C (1+\left\lVert{u_b}\right\rVert_{{ {{{\rm{BV}}}}(0, T)}} ) $ |
We are in the hypotheses of [32,Theorem 5.2.1. p. 222] : by
$ \left\lVert{U}\right\rVert_{{ {{{\rm{}}}{\mathbf B}{\mathbf V}}_{t, x}(V)}} \leq \liminf\limits_{\delta \to 0} \left\lVert{U^\delta}\right\rVert_{{ {{{\rm{}}}{\mathbf B}{\mathbf V}}_{t, x}((0, T)\times(0, L))}} = \liminf\limits_{\delta \to 0} \left\lVert{U^\delta}\right\rVert_{{ {\mathbf W}^{1, 1}_{t, x}((0, T)\times(0, L))}} $ |
and since the
It is divided into two steps.
1. Convergence : From Lemma 4.2, Lemma 4.3 and Corollary 1, sequences
$ uε1→ε→0u1stronglyinL1([0,T]×[0,L]),uε2→ε→0u2stronglyinL1([0,T]×[0,L]), $
|
with limit function
By equations (7a), one shows by testing with the appropriate
$ \|q_1^{ \varepsilon}- u_1^{ \varepsilon}\|_{L^1((0, T)\times(0, L))}\leq C \varepsilon \left\lVert{u_1^ \varepsilon}\right\rVert_{{BV((0, T)\times(0, L))}} $ |
which tends to zero as
$ \|G(q_2^{ \varepsilon})-G(u_2)\|_{L^{1}((0, T)\times(0, L))} \longrightarrow 0. $ |
For the convergence of
$ a_0 {{\partial}}_{t}u_0 = K_1(u_1-u_0) + K_2(u_2-u_0) + G(u_2). $ |
Then, taking the last equation of system (7), subtracting by this latter equation and multiplying by
$ a0∂t|uε0−u0|≤K1|qε1−u1|+K2|qε2−u2|+(K1+K2)|u0−uε0|+|G(qε2)−G(u2)|≤K1|qε1−u1|+K2|qε2−u2|+(K1+K2)|u0−uε0|+‖G′‖∞|qε2−u2|. $
|
Using Grönwall's Lemma, we get, after an integration on
$ ∫L0|uε0−u0|(t,x)dx≤ ∫L0e(K1+K2)a0t|u0−uε0|(0,x)dx+K1a0∫L0∫T0e(K1+K2)a0(t−s)|qε1−u1|(s,x)dsdx+(‖G′‖∞+K2)a0∫L0∫T0e(K1+K2)a0(t−s)|qε2−u2|(s,x)dsdx. $
|
Thus, one concludes that
$ u_{0}^{ \varepsilon} \xrightarrow[ \varepsilon \to 0]{} u_0 {\rm{ strongly in }} L^1((0, T)\times(0, L)). $ |
2. The limit system :
We pass to the limit in (15), the weak formulation of system (7). Suppose that
$ ∫T0∫L0u1((a1+a3)∂tϕ1+α∂xϕ1)dxdt+α∫T0ub(t)ϕ1(t,0)dt+∫L0(a1u01(x)+a3q01(x))ϕ1(0,x)dx+∫T0∫L0u2((a2+a4)∂tϕ2−α∂xϕ2)dxdt+∫L0(a2u02(x)+a4q02(x))ϕ2(0,x)dx+∫T0∫L0(a0u0∂tϕ3+K1(u1−u0)(ϕ3−ϕ1)+K2(u2−u0)(ϕ3−ϕ2)+G(u2)(ϕ3−ϕ2))dxdt+∫L0a0u0(0,x)ϕ3(0,x)dx=0. $
|
which is exactly (14) with initial data coming from system (7). Finally, since the solution of the limit system is unique, we deduce that the whole sequence converges. This concludes the proof of Theorem 2.5.
For the sake of simplicity, we assume, in what follows that
$ \Delta x : = L/N_x, $ |
In order to avoid an
$ {(1+Δtε)un+1i,1−Δtεqn+1i,1=(1−λ5)uni,1+λ5uni−1,1−λ5(1−λ5)2(ϕni+1/2,1−ϕni−1/2,1)(1+Δtε)un+1i,2−Δtεqn+1i,2=(1−λ5)uni,2+λ5uni+1,2−λ5(1−λ5)2(ϕni+1/2,2−ϕni−1/2,2)−Δtεun+1i,1+(1+Δtε+ΔtK1)qn+1i,1−ΔtK1un+1i,0=qni,1−Δtεun+1i,2+(1+Δtε+ΔtK2)qn+1i,2−ΔtK2un+1i,0=qni,2−ΔtG(qni,2)−ΔtK1qn+1i,1−ΔtK2qn+1i,2+(1+Δt(K1+K2))un+1i,0=uni,0+ΔtG(qni,2) $
|
(44) |
where the time step is chosen s.t.
$ \lambda_5 : = \frac{\Delta t}{\Delta x} \alpha $ |
is less than 1, and the flux limiters are defined as
$ ϕni+1/2,1:=minmod(uni+1,1−uni,1,uni,1−uni−1,1),ϕni+1/2,2:=minmod(uni+2,2−uni+1,2,uni+1,2−uni,2), $
|
and the
$ {\rm minmod}(1, r): = \max\left(0, \min\left(2r, \frac{1+r}{2}, 2\right)\right). $ |
We complement the scheme with initial and boundary conditions :
$ u^{0}_{i, j} : = \frac{1}{\Delta x} \int_{i \Delta x}^{(i+1)\Delta x} u^0_{j}, \quad \forall j\in\{0, 1, 2\}, \quad q^{0}_{i, j} : = \frac{1}{\Delta x} \int_{i \Delta x}^{(i+1)\Delta x} q^0_{j}, \quad \forall j\in\{1, 2\} $ |
and
$ un+10,1:=ubn=1Δt∫(n+1)ΔtnΔtub(t)dt,un+1Nx,2:=un+1Nx−1,1,∀n∈{0,Nt−1}. $
|
(45) |
Proposition 2. Under the CFL condition :
$ \Delta t \sum\limits_{n = 1}^{N_t-1} \sum\limits_{i = 1}^{N_x-1} \left\{ \sum\limits_{j = 0}^2 \left| u^{n+1}_{i, j}- u^n_{i, j}\right| + \sum\limits_{j = 1}^2 \left|q^{n+1}_{i, j}- q^n_{i, j} \right|\right\} < C $ |
For the first part, the proof uses at the discrete level similar ideas as in Lemmas 4.1 and 4.2. For the second part, one adds the initial layer (19) to the numerical solutions for each time step and proceeds as in Proposition 1 and Lemma 1. Since the ideas are very similar and the properties of the transport part are rather standard once it is written in a convex form [8,Harten's Lemma], the proof is left to the reader for the sake of concision.
For various values of
$ \left\{ (1+ΔtK12)vn+1i,1−ΔtK12vn+1i,0=(1−λ3)vni,1+λ3vni−1,1−λ3(1−λ3)4(ϕni+1/2,1−ϕni−1/2,1)(1+ΔtK22)vn+1i,2−ΔtK22vn+1i,0=(1−λ3)vni,2+λ3vni+1,2−λ3(1−λ3)4(ϕni+1/2,2−ϕni−1/2,2)−ΔtG(vni,1)−ΔtK1vn+1i,1−ΔtK2vn+1i,2+(1+Δt(K1+K2))vn+1i,0=vni,0+ΔtG(vni,1) \right. $
|
where
$ \lambda_3 : = \frac{\Delta t}{2 \Delta x} \alpha = \frac{\lambda_5}{2} $ |
We initialize the data setting :
$ v^0_{i, 1} = \frac{u^0_{i, 1}+q^0_{i, 1}}{2}, \quad v^0_{i, 2} = \frac{u^0_{i, 2}+q^0_{i, 2}}{2}, \quad v^0_{i, 0} = u^0_{i, 0}. $ |
and boundary conditions are similar to (45). We compare
$ E_{ \varepsilon, \Delta x}: = \Delta t \Delta x \sum\limits_{n = 0}^{N_t} \sum\limits_{i = 0}^{N_x} \left\{\sum\limits_{j = 1}^2 \left| 2 v^n_{i, j} - (u^n_{i, j}+q^n_{i, j}) \right| + \left| v^n_{i, 0} - u^{n}_{i, 0}\right| \right\}, $ |
the results are displayed in Fig. 2.
Although the study of the rate of convergence for the dynamical system is quite complicated, explicit computations may be performed for the stationary system. This system reads, for
$ α∂xˉu1=1ε(ˉq1−ˉu1) $
|
(46a) |
$ −α∂xˉu2=1ε(ˉq2−ˉu2) $
|
(46b) |
$ 0=1ε(ˉu1−ˉq1)+K1(ˉu0−ˉq1) $
|
(46c) |
$ 0=1ε(ˉu2−ˉq2)+K2(ˉu0−ˉq2)−G(ˉq2) $
|
(46d) |
$ 0=K1(ˉq1−ˉu0)+K2(ˉq2−ˉu0)+G(ˉq2), $
|
(46e) |
completed with the boundary conditions
$ \bar{u}_1(0) = u_b, \qquad \bar{u}_2(L) = \bar{u}_1(L). $ |
This system may be reduced by noticing that, after adding the equations in (46), we obtain
$ \alpha {{\partial}}_x (\bar{u}_1-\bar{u}_2) = 0. $ |
From the boundary condition at
$ 2ˉu=ˉq1+ˉq2. $
|
(47) |
From (46e), we deduce
$ ˉu0=K1ˉq1+K2ˉq2K1+K2+G(ˉq2)K1+K2. $
|
(48) |
Injecting this latter expression and (47) into (46c), we obtain
$ \bar{q}_1\left(1+2 \varepsilon \frac{K_1 K_2}{K_1+K_2}\right) = \frac{2 \varepsilon K_1}{K_1+K_2}(K_2 \bar{q}_2 + G(\bar{q}_2)) + \bar{q}_2. $ |
With (46e), we deduce
$ ˉu=ˉq2+εK1K1+K2+2εK1K2G(ˉq2). $
|
(49) |
Using also (46a), we obtain the following Cauchy problem
$ {α(K1+K2+ε(2K1K2+K1G′(ˉq2)))∂xˉq2=K1G(ˉq2),ˉq2(x=0)=ˉq02,with ˉq02+εK1K1+K2+2εK1K2G(ˉq02)=ub. $
|
(50) |
Solving this system, we deduce
When,
$ {α(K1+K2)∂xˉq2,0=K1G(ˉq2,0),ˉq2,0(x=0)=ub. $
|
(51) |
In order to have semi-explicit expression, we introduce an antiderivative of
$ α(K1+K2+2εK1K2)G(ˉq2)+αεlnG(ˉq2)=K1x+α(K1+K2+2εK1K2)G(ˉq02)+αεlnG(ˉq02). $
|
By the same token on (51), we have
$ \alpha (K_1+K_2) \mathcal{G}(\bar{q}_{2, 0}) = K_1 x + \alpha (K_1+K_2) \mathcal{G}(u_b). $ |
Subtracting the last two equalities, we obtain
$ (K1+K2)(G(ˉq2)−G(ˉq2,0))=(K1+K2)(G(ˉq02)−G(ub))+2εK1K2(G(ˉq02)−G(ˉq2))+ε(lnG(ˉq02)−lnG(ˉq2)). $
|
Thanks to a Taylor expansion, there exist
$ ˉq2−ˉq2,0=G(ζ2)G(ζ02)(ˉq02−ub)+2εK1K2(G(ˉq02)−G(ˉq2))+ε(lnG(ˉq02)−lnG(ˉq2))=−εK1G(ζ2)G(ˉq02)(K1+K2+2εK1K2)G(ζ02)+2εK1K2(G(ˉq02)−G(ˉq2))+ε(lnG(ˉq02)−lnG(ˉq2)), $
|
where we use (50) for the last equality. Hence we conclude that, formally, the convergence of
We define the problem :
$ {∂tu+α∂xu=ω(w−u)(t,x)∈(0,T)×(0,L)∂tv−α∂xv=ω(z−v)(t,x)∈(0,T)×(0,L)u(t,0)=ub(t),v(t,L)=vb(t)t∈(0,T)u(0,x)=u0(x),v(0,x)=v0(x)(t,x)∈{0}×(0,L) $
|
(52) |
where the data
$ (w,z)∈L∞((0,T)×(0,L))2,(u0,v0)∈L∞(0,L)2,(ub,vb)∈L∞(0,T)2, $
|
and
$ u(t,x):={ub(t−x/α)exp(−ωx/α)++ω∫0−x/αexp(−ω(s+x/α))w(t+s,x+αs)ds}ift>x/αu0(x−αt)exp(−ωt)+ω∫0−texp(−ω(s+t))w(t+s,x+αs)dsotherwise $
|
(53) |
and
Theorem B.1. Let
$ ∫T0∫L0Φ(u(t,x))(−∂t−∂x+ω)φdxdt+[∫L0Φ(u(t,x))φ(t,x)dx]t=Tt=0+[∫T0Φ(u(t,x))φ(t,x)dt]x=Lx=0=∫T0Φ′(u(t,x))ωw(t,x)φ(t,x)dxdt $
|
for all
The authors acknowledge the unkonwn referees for their comments and remarks which help in improving the paper.
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