Cauchy problem for multiscale conservation laws: Application to structured cell populations
aa r X i v : . [ m a t h . A P ] O c t Cauchy problem for multiscale conservation laws:Application to structured cell populations ∗ Peipei Shang † October 11, 2010
Abstract
In this paper, we study a vector conservation law that models the growth and selectionof ovarian follicles. During each ovarian cycle, only a definite number of follicles ovulate,while the others undergo a degeneration process called atresia. This work is motivated bya multiscale mathematical model starting on the cellular scale, where ovulation or atresiaresult from a hormonally controlled selection process. A two-dimensional conservationlaw describes the age and maturity structuration of the follicular cell populations. Thedensities intersect through a coupled hyperbolic system between different follicles andcell phases, which results in a vector conservation law and coupling boundary conditions.The maturity velocity functions possess both a local and nonlocal character. We provethe existence and uniqueness of the weak solution to the Cauchy problem with boundedinitial and boundary data.
Keywords: conservation laws, nonlocal velocity, multiscale, biomathematics
In this paper, we study the following vector conservation law ~φ ft ( t, x, y ) + ( A f ~φ f ( t, x, y )) x + ( B f ~φ f ( t, x, y )) y = C ~φ f ( t, x, y ) , (1.1) t ∈ [0 , T ] , ( x, y ) ∈ [0 , , ∗ This work was supported by the large scale INRIA project REGATE (REgulation of the GonAdoTropEaxis). † INRIA Paris-Rocquencourt Centre. Universit´e Pierre et Marie Curie-Paris 6. UMR 7598 LaboratoireJacques-Louis Lions, 75005 Paris, France. E-mail:
[email protected] . ~φ f = ( φ f , · · · , φ fN , b φ f , · · · , b φ fN , e φ f , · · · , e φ fN ) T , f = 1 , · · · , n , and A f := diag { N z }| { g f , · · · , g f , N z }| {b g f , · · · , b g f , N z }| {e g f , · · · , e g f } ,B f := diag { N z }| { h f , · · · , h f , N z }| { , · · · , , N z }| {e h f , · · · , e h f } ,C := − diag { N z }| { λ, · · · , λ, N z }| { , · · · , , N z }| {e λ, · · · , e λ } . Here b g f and e g f are positive constants, g f = g f ( u f ) ∈ C ([0 , ∞ )), h f = h f ( y, u f ), e h f = e h f ( y, u f ), λ = λ ( y, U ) and e λ = e λ ( y, U ) all belong to C ([0 , × [0 , ∞ )), with u f = u f ( M f ( t ) , M ( t ) , t ) the local control and U = U ( M ( t ) , t ) the global control, where u f > U > u f ∈ C ([0 , ∞ ) × [0 , T ]) and U ∈ C ([0 , ∞ ) × [0 , T ]).For instance, the specific case that motivated our work is presented in Appendix 6.1. Thefunction M f is the maturity on the follicular scale given by M f ( t ) := N X k =1 Z Z a γ s yφ fk ( t, x, y ) dx dy + N X k =1 Z Z ( a − a ) γ s y b φ fk ( t, x, y ) dx dy + N X k =1 Z Z a γ ( γ y + γ s ) e φ fk ( t, x, y ) dx dy and M ( t ) := n X f =1 M f ( t )is the global maturity on the ovarian scale. The parameters a , a , γ s and γ are positiveconstants with a > a .This work is motivated by problems of cell dynamics arising in the control of the develop-ment of ovarian follicles. Ovarian follicles are spheroidal structures sheltering the maturingoocyte. During each ovarian cycle, numerous follicles are in competition for their survival.However, only very few follicles reach an ovulatory size, most of them undergo a degenera-tion process, known as atresia (see for instance [12]). A mathematical model, proposed byF. Cl´ement [5] using both multi-scale modeling and control theory concepts, describes thefollicle selection process. For each follicle, the cell population dynamics is ruled by a con-servation law, which describes the changes in cell age and maturity. Two acting controls, u f and U (see [1] and also Appendix 6.1), are distinguished. The global control U resultsfrom the ovarian feedback onto the pituitary gland and impacts the secretion of the folliclestimulating hormone (FSH). The feedback is responsible for reducing FSH release, leading tothe degeneration of all but those follicles selected for ovulation. The local control u f , spe-cific to each follicle, accounts for the modulation in FSH bioavailability related to follicularvascularization. In addition, the status of cells are characterized by three phases. Phase 12nd 2 correspond to the proliferation phases, and Phase 3 corresponds to the differentiationphase (see Fig 1). In Appendix 6.2, we have reformulated the original model to a new system(1.1), where the unknowns are all defined on the same domain [0 , T ] × [0 , , so that it canbe treated as a general model for multiscale structured cell populations (see Appendix 6.2for the relation between the original notations and the new notations).The initial conditions are given by ~φ f : = ~φ f (0 , x, y ) (1.2)=( φ f ( x, y ) , · · · , φ fN ( x, y ) , b φ f ( x, y ) , · · · , b φ fN ( x, y ) , e φ f ( x, y ) , · · · , e φ fN ( x, y )) T . We use the simplified notations u f ( t ) := u f ( M f ( t ) , M ( t ) , t ), U ( t ) := U ( M ( t ) , t ), u f ( t ) := u f ( M f ( t ) , M ( t ) , t ) and U ( t ) := U ( M ( t ) , t ) in the whole paper.The boundary conditions at x = 0 are given by φ f ( t, , y ) = 0 , ( t, y ) ∈ [0 , T ] × [0 , ,φ fk ( t, , y ) = 2 τ gf a g f ( u f ( t )) b φ fk − ( t, , y ) , ( t, y ) ∈ [0 , T ] × [0 , , k = 2 , · · · , N. b φ fk ( t, , y ) = a g f ( u f ( t )) τ gf φ fk ( t, , y ) , ( t, y ) ∈ [0 , T ] × [0 , , k = 1 , · · · , N. e φ f ( t, , y ) = 0 , ( t, y ) ∈ [0 , T ] × [0 , , e φ fk ( t, , y ) = e φ fk − ( t, , y ) , ( t, y ) ∈ [0 , T ] × [0 , , k = 2 , · · · , N. (1.3)The boundary conditions at y = 0 are given by φ fk ( t, x,
0) = b φ fk ( t, x,
0) = 0 , ( t, x ) ∈ [0 , T ] × [0 , , k = 1 , · · · , N. e φ fk ( t, x,
0) = φ fk ( t, a a x, , ( t, x ) ∈ [0 , T ] × [0 , a a ] , , ( t, x ) ∈ [0 , T ] × [ a a , , k = 1 , · · · , N. (1.4)The boundary conditions at y = 1 are given by e φ fk ( t, x,
1) = 0 , ( t, x ) ∈ [0 , T ] × [0 , , k = 1 , · · · , N. (1.5)The well-posedness problems of the hyperbolic conservation laws have been widely studiedfor a long time. We refer to the works [7, 8, 9, 15, 16] (and the references therein) in thecontent of weak solutions to systems in conservation laws, and [13, 14] in the content ofclassical solutions to general quasi-linear hyperbolic systems. In this paper, we perform themathematical analysis of this system: we prove the existence, uniqueness and regularity ofthe weak solution to the Cauchy problem defined by (1.1)-(1.5) with initial and boundarydata in L ∞ . The main difficulty tackled with in this paper comes from the nonlocal velocity,the coupling between boundary conditions and the coupling between different follicles in themodel. Additionally, we have to deal with loss terms, and the problem is a 2 - space dimensionone. 3 ✻ · · ·· · · y x
10 10 10 1 0 10 1 0 10 1 · · · · · · φ f b φ f e φ f φ f b φ f e φ f φ fk b φ fk e φ fk φ fN b φ fN e φ fN · · ·· · · ✒❘ ✒❘ ✒❘ ✒❘ ✒❘✒ ✒ ✲ ✒ ✒ ✲ ✒ ✒ ✲ ✒ ✒ Figure 1: The illustration of N cell cycles for follicle f ; x denotes the age velocity and y denotes the maturity velocity. The top of the domain corresponds to the differentiationphase and the bottom to the the proliferation phase. In the related works that have considered nonlocal velocity problems [3, 4], the problemsare only 1 - space dimension and do not have source terms. The velocities there are alwayspositive, while in our case, the velocities e h f ( f = 1 , · · · , n ) we considered in this paper changesign in Phase 3. In another related work [10], which was also motivated by [5], the author hasperformed a mathematical analysis on this kind of model. He has reduced the model to a 1-space dimension mass-maturity dynamical system of coupled Ordinary Differential Equations,basing on the asymptotic properties of the original law. Once reduced, the model is amenableto analysis by bifurcation theory, that allows one to predict the issue of the selection for onespecific follicle amongst the whole population. And he gave some assumptions on velocitiesaccording to biological observations. In another work [11], the author also studied the well-poseness of the model, he proved the existence of a measure valued solution without provingthe uniqueness of the solution, and he gave the behavior of the solution to the main equationalong its characteristics. An associated reachability problem has been tackled in [6], whichdescribed the set of microscopic initial conditions leading to the macroscopic phenomenon ofeither ovulation or atresia in the framework of backwards reachable sets theory. The authorsalso performed some mathematical analysis on well-poseness of this model in a simplified openloop like case, there the authors assumed that the local control u f and the global control U are given functions of time t . While in this paper, we consider the close/open loop case.This paper is organized as follows. In Section 2 we first give the main results. In Section3 after giving some basic notations, we prove that the vector maturity ~M := ( M , · · · , M n )exists as a fixed point of a map from continuous function space, and then we construct alocal weak solution to the Cauchy problem defined by (1.1)-(1.5). In Section 4 we provethe uniqueness of the weak solution. Finally in Section 5 we prove the existence of a globalsolution. In complement, we introduce in Appendix 6.1 the original mathematical modelproposed by F. Cl´ement [5]. In Appendix 6.2 we reformulate this model to the new system(1.1) and we give the equivalence between the original notations and the new notations. InAppendix 6.3 we give a basic lemma that is used to prove the existence and uniqueness ofthe weak solution. 4 Main results
We recall from [2, Section 2.1], the usual definition of a weak solution to the Cauchyproblem defined by (1.1)-(1.5).
Definition 2.1.
Let
T > ~φ f ∈ L ∞ ((0 , ) be given. A weak solution of Cauchy problem(1.1)-(1.5) is a vector function ~φ f ∈ C ([0 , T ]; L ((0 , )) such that for every τ ∈ [0 , T ] andevery vector function ~ϕ :=( ϕ , · · · , ϕ N , b ϕ , · · · , b ϕ N , e ϕ , · · · , e ϕ N ) T ∈ C ([0 , τ ] × [0 , ) with ~ϕ ( τ, x, y ) = 0 , ∀ ( x, y ) ∈ [0 , , (2.1) ~ϕ ( t, , y ) = 0 , ∀ ( t, y ) ∈ [0 , τ ] × [0 , , (2.2) ϕ ( t, , y ) = e ϕ ( t, , y ) = 0 , ∀ ( t, y ) ∈ [0 , τ ] × [0 , , (2.3) ϕ k ( t, x,
0) = b ϕ k ( t, x,
0) = ϕ k ( t, x,
1) = b ϕ k ( t, x,
1) = 0 , ∀ ( t, x ) ∈ [0 , τ ] × [0 , , (2.4)one has Z τ Z Z ~φ f ( t, x, y ) · ( ~ϕ t ( t, x, y )+ A f ~ϕ x ( t, x, y )+ B f ~ϕ y ( t, x, y )+ C ~ϕ ( t, x, y )) dxdydt (2.5)+ Z Z ~φ f ( x, y ) · ~ϕ (0 , x, y ) dxdy + N X k =1 Z τ Z a a e h f (0 , u f ( t )) φ fk ( t, a a x, e ϕ k ( t, x, dxdt + N X k =2 Z τ Z τ gf a b φ fk − ( t, , y ) ϕ k ( t, , y ) dydt + N X k =2 Z τ Z e g f e φ fk − ( t, , y ) e ϕ k ( t, , y ) dydt + N X k =1 Z τ Z a b g f g f ( u f ( t )) τ gf φ fk ( t, , y ) b ϕ k ( t, , y ) dydt = 0 . With the definition, we have the main result
Theorem 2.1.
Let
T > , ~φ f ∈ L ∞ ((0 , ) be given. Let us further assume that g f ( u f ) > , ∀ u f ∈ [0 , ∞ ) ,h f ( y, u f ) > , ∀ ( y, u f ) ∈ [0 , × [0 , ∞ ) , e h f (0 , u f ) > , e h f (1 , u f ) < , ∀ u f ∈ [0 , ∞ ) . Then the Cauchy problem defined by (1.1)-(1.5) admits a unique weak solution ~φ f =( φ f , · · · , φ fN , b φ f , · · · , b φ fN , e φ f , · · · , e φ fN ) T . Moreover, the weak solution ~φ f even belongs to C ([0 , T ]; L p ((0 , )) for all p ∈ [1 , ∞ ) . The sketch of the proof of Theorem 2.1 consists in first proving that the maturity ~M ( t ) =( M ( t ) , · · · , M n ( t )) exists as a fixed point of the map ~M ~G ( ~M ) (see Section 3.1), andthen in constructing a (unique) local solution (see Section 3.2 and Section 4), before finallyproving the existence of a global solution to the Cauchy problem defined by (1.1)-(1.5) (seeSection 5). 5 Fixed point argument and construction of a local solutionto the Cauchy problem
In this section, we first derive the contraction mapping function ~G . Given the existenceof fixed point to this contraction mapping function, we can then construct a local solution tothe Cauchy problem defined by (1.1)-(1.5). First we introduce some notations. Let us define: k φ fk k := k φ fk k L ∞ ((0 , ) := ess sup ( x,y ) ∈ [0 , | φ fk ( x, y ) | , k b φ fk k := k b φ fk k L ∞ ((0 , ) := ess sup ( x,y ) ∈ [0 , | b φ fk ( x, y ) | , k e φ fk k := k e φ fk k L ∞ ((0 , ) := ess sup ( x,y ) ∈ [0 , | e φ fk ( x, y ) | .K :=2 N ( γ + γ s ) n X f =1 N X k =1 (cid:16) a k φ fk k L ((0 , ) +( a − a ) k b φ fk k L ((0 , ) + a k e φ fk k L ((0 , ) (cid:17) , (3.1) K := max nb g f , k g f ( u f ( M f , M, t )) k C ( Q ) , k h f ( y, u f ( M f , M, t )) k C ( Q ) , (3.2) k e h f ( y, u f ( M f , M, t )) k C ( Q ) , k λ ( y, U ( M, t )) k C ( Q ) , k e λ ( y, U ( M, t )) k C ( Q ) o ,K := min n inf ( M f ,M,t ) ∈ Q g f ( u f ( M f , M, t )) , inf ( y,M f ,M,t ) ∈ Q h f ( y, u f ( M f , M, t )) o > , (3.3) Q := [0 , K ] × [0 , T ] , Q := [0 , × [0 , K ] × [0 , T ] , Q := [0 , × [0 , K ] × [0 , T ] . For any δ >
0, letΩ δ,K := n ~M ( t )=( M ( t ) , · · · , M n ( t )) ∈ C ([0 , δ ]) : k ~M k C ([0 ,δ ]) :=max f k M f k C ([0 ,δ ]) ≤ K o , where the constant K is given by (3.1).In order to derive the expression of the contraction mapping function ~G , we solve thecorresponding linear Cauchy problem (1.1)-(1.5) with given ~M ∈ Ω δ,K . For any fixed t ∈ [0 , δ ]and f ∈ { , · · · , n } , we trace back the density function ~φ f at time t along the characteristicsto the initial and boundary data, hence we divide the plane time t into several parts. ForPhase 1, the velocities h f are always positive, we introduce three subsets ω f,t , ω f,t and ω f,t of [0 , (see Fig 2 (a)) ω f,t := n ( x, y ) | Z t g f ( u f ( σ )) dσ ≤ x ≤ , η ( t, Z t g f ( u f ( σ )) dσ ) ≤ y ≤ o ,ω f,t := n ( x, y ) | ≤ x ≤ Z t g f ( u f ( σ )) dσ, η ( t, x ) ≤ y ≤ o ,ω f,t := [0 , \ ( ω f,t ∪ ω f,t ) . y = η ( t, x ) (see Fig 2 (a))satisfies dηds = h f ( η, u f )( s ) , η ( θ ) = 0 , θ ≤ s ≤ t, (3.4)with θ defined by x = Z tθ g f ( u f ( σ )) dσ .If ( x, y ) ∈ ω f,t , we trace back the density function φ fk at time t along the characteristicsto the initial data. Otherwise, we trace back the density function at time t along the char-acteristics to the boundary data. For any fixed t ∈ [0 , δ ] and ( x, y ) ∈ [0 , , let us definecharacteristics ξ i ( s ) := ( x i ( s ) , y i ( s )) , i = 1 , · · · , , x , y ) the points on the bottom face ( t = 0),( t , α ,
0) the points on the left face ( y = 0), ( t , α ,
1) the points on the right face ( y = 1),( τ , , β ) the points on the back face ( x = 0) and ( τ , , β ) the points on the front face( x = 1). (a) (b) Figure 2: For Phase 1, time t plane is divided into three parts ω f,t , ω f,t and ω f,t . (a) Case( t, x, y ) ∈ ω f,t , characteristic ξ connects ( t, x, y ) with (0 , x , y ); (b) Case ( t, x, y ) ∈ ω f,t ,characteristic ξ connects ( t, x, y ) with ( τ , , β ); According to the coupled boundary betweenPhase 1 at x = 1 and Phase 2 at x = 0, we define characteristic ξ which connects ( τ , , β )with (0 , x , y ).For any fixed x ∈ [0 , ξ = ( x , y ) by (see Fig 2 (a)) dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( t ) = ( x, . If ( x, y ) ∈ ω f,t (see Fig 2 (a)), we define characteristic ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . ξ ( s ) ∈ [0 , , ∀ s ∈ [0 , t ]. Let us define( x , y ) := ( x (0) , y (0)) . (3.5)If ( x, y ) ∈ ω f,t (see Fig 2 (b)), we define characteristic ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique τ ∈ [0 , t ] such that x ( τ ) = 0, so that we can define β := y ( τ ) . (3.6)According to the coupled boundary between Phase 1 at x = 1 and Phase 2 at x = 0, for anyfixed ( x , y ) ∈ [0 , , we define characteristic ξ =( x , y ) by (see Fig 2 (b)) dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ (0) = ( x , y ) . For Phase 3, due to the fact that velocities e h f change sign in Phase 3, we divide the planeat time t into four subsets e ω f,t , e ω f,t , e ω f,t and e ω f,t of [0 , (see Fig 3 (a)) e ω f,t := n ( x, y ) | e g f t ≤ x ≤ , e η ( t, e g f t ) ≤ y ≤ e η ( t, e g f t ) o , e ω f,t := n ( x, y ) | ≤ x ≤ e g f t, ≤ y ≤ e η ( t, x ) o ∪ n ( x, y ) | e g f t ≤ x ≤ , ≤ y ≤ e η ( t, e g f t ) o , e ω f,t := n ( x, y ) | ≤ x ≤ e g f t, e η ( t, x ) ≤ y ≤ e η ( t, x ) o , e ω f,t := [0 , \ ( e ω f,t ∪ e ω f,t ∪ e ω f,t ) . Here y = e η ( t, x ) and y = e η ( t, x ) (see Fig 3 (a)) satisfy d e η ds = e h f ( e η , u f )( s ) , e η ( t − x e g f ) = 0 , t − x e g f ≤ s ≤ t,d e η ds = e h f ( e η , u f )( s ) , e η ( t − x e g f ) = 1 , t − x e g f ≤ s ≤ t. If ( x, y ) ∈ e ω f,t , we trace back the density function e φ fk at time t along the characteristics to theinitial data. Otherwise, we trace back the density function at time t along the characteristicsto the boundary data. For any fixed t ∈ [0 , δ ] and ( x, y ) ∈ [0 , , let us define characteristics ξ i ( s ) := ( x i ( s ) , y i ( s )) , i = 5 , · · · ,
10 (see Fig 3), which will be used to construct the contractionmapping function.If ( x, y ) ∈ e ω f,t (see Fig 3 (a)), we define characteristic ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . One has ξ ( s ) ∈ [0 , , ∀ s ∈ [0 , t ]. Let us define( x , y ) := ( x (0) , y (0)) . (3.7)8f ( x, y ) ∈ e ω f,t (see Fig 3 (a)), we define characteristic ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique t ∈ [0 , t ] such that y ( t ) = 0, so that we can define α := x ( t ) . (3.8)For any fixed cell cycle k = 1 , · · · , N , according to the coupled boundary between Phase1 and Phase 3 (corresponding to φ fk ( t, x, y ) at y = 1 and e φ fk ( t, x, y ) at y = 0), we separatethe right face of Phase 1 into two parts. For k = 1 , · · · , N , we define characteristic ξ passingthrough the right face intersects with the bottom face at (0 , x , y ) (see Fig 3 (b)). For k = 2 , · · · , N , we define characteristic ξ passing through the right face intersects with theback face, and then back to the bottom face at (0 , x , y ) of the k − x , y ) ∈ [0 , , we define ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ (0) = ( x , y ) , and ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( 1 − x b g f ) = (0 , y ) . (a) (b) (c) Figure 3: For Phase 3, time t plane is divided into four parts e ω f,t , e ω f,t , e ω f,t and e ω f,t . (a)Case ( t, x, y ) ∈ e ω f,t , characteristic ξ connects ( t, x, y ) with (0 , x , y ); Case ( t, x, y ) ∈ e ω f,t ,characteristic ξ connects ( t, x, y ) with ( t , α , y = 1 and Phase 3 at y = 0, we separate the right face into two parts,and we define characteristic ξ and characteristic ξ ; (c) Case ( t, x, y ) ∈ e ω f,t , characteristic ξ connects ( t, x, y ) with ( τ , , β ); According to the coupled boundary between two consecutivecell cycles, we define characteristic ξ which connects ( τ , , β ) with (0 , x , y ).9f ( x, y ) ∈ e ω f,t (see Fig 3 (c)), we define characteristic ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique τ ∈ [0 , t ] such that x ( τ ) = 0, so that we can define β := y ( τ ) . (3.9)According to the coupled boundary between two consecutive cell cycles (corresponding to e φ fk − ( t, x, y ) at x = 1 and e φ fk ( t, x, y ) at x = 0) ( k = 2 , · · · , N ), for any fixed ( x , y ) ∈ [0 , ,we define ξ = ( x , y ) by (see Fig 3 (c)) dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ (0) = ( x , y ) . With all of the above defined characteristics and noting Lemma 6.1 in Appendix 6.3, weare now able to define a map ~G ( ~M )( t ) := (cid:16) G ( M , M )( t ) , · · · , G n ( M n , M )( t ) (cid:17) for all t ∈ [0 , δ ]with G f ( M f , M )( t ):= Z y (0)0 Z − R t g f ( u f ( σ )) dσ a γ s N X k =1 φ fk ( x , y ) y ( t ) e − R t λ ( y ,U )( s ) ds dx dy (3.10)+ Z − b g f t Z y ( − x b gf )0 a − a ) γ s N X k =2 b φ fk − , ( x , y ) y ( t ) e − R t − x b gf λ ( y ,U )( s ) ds dy dx + Z Z − b g f t ( a − a ) γ s N X k =1 y b φ fk ( x , y ) dx dy + Z − R t g f ( u f ( σ )) dσ Z ζ ( x )0 a γ s N X k =1 ( y + Z τ h f ( y , u f )( σ ) dσ ) φ fk ( x , y ) e − R τ λ ( y ,U )( s ) ds dy dx + Z Z − e g f t a γ N X k =1 ( γ y ( t )+ γ s ) e φ fk ( x , y ) e − R t e λ ( y ,U )( s ) ds dx dy + Z y (0) Z ζ − ( y )0 a γ N X k =1 ( γ y ( t )+ γ s ) φ fk ( x , y ) e − ( R t λ ( y ,U )( s ) ds + R tt e λ ( y ,U )( s ) ds ) dx dy + Z − b g f t Z y ( − x b gf ) a − a ) γ N X k =2 ( γ y ( t )+ γ s ) b φ fk − , ( x , y ) e − ( R t − x b gf λ ( y ,U )( s ) ds + R tt e λ ( y ,U )( s ) ds ) dy dx + Z Z − e g f t a γ N − X k =1 ( γ y ( t )+ γ s ) e φ fk ( x , y ) e − ( R − x e gf e λ ( y ,U )( s ) ds + R t − x e gf e λ ( y ,U )( s ) ds ) dx dy . Here y i ( t ) ( i = 2 , · · · ,
10) can also be determined by ( x , y ), τ is defined by 1 − x = Z τ g f ( u f ( σ )) dσ , t is defined by y ( t ) = 1, and t is defined by y ( t ) = 1. In (3.10), y = ζ ( x ) := η (0 , x ) (see Fig 3 (b)), where η ( s, x ) satisfies dηds = h f ( η, u f )( s ) , η ( θ ) = 1 , ≤ s ≤ θ, θ defined by 1 − x = Z θ g f ( u f ( σ )) dσ .Next we prove the following fixed point theorem. Lemma 3.1. If δ is small enough, ~G is a contraction mapping on Ω δ,K with respect to the C norm. Proof:
It is easy to check that ~G maps into Ω δ,K itself if0 < δ ≤ min { K , T } , (3.11)where K is defined by (3.2). Let ~M ( t )=( M ( t ) , · · · , M n ( t )), ~M ( t )=( M ( t ) , · · · , M n ( t )) ∈ Ω δ,K ,and M = P nf =1 M f , M = P nf =1 M f . In order to estimate k ~G ( ~M ) − ~G ( ~M ) k C ([0 ,δ ]) , we firstestimate the norms k G f ( M f , M ) − G f ( M f , M ) k C ([0 ,δ ]) separately. Observing the definition(3.10) of G f , it is sufficient to estimate k y i − y i k C ([0 ,δ ]) , where y i = C + Z β i α i h f ( y i , u f )( σ ) dσ,y i = C + Z β i α i h f ( y i , u f )( σ ) dσ. (3.12)Here C denotes various constants, h f represents h f or e h f , and 0 ≤ α i ≤ β i ≤ t ≤ δ ≤ min { K , T } .We have | y i ( s ) − y i ( s ) | ≤ Z β i α i | h f ( y i , u f )( σ ) − h f ( y i , u f )( σ ) | dσ ≤ tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) ) + tK k y i − y i k C ([ α i ,β i ]) , hence k y i − y i k C ([0 ,δ ]) ≤ tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) )1 − tK . (3.13)By (3.13), we have Z β i α i | h f ( y i , u f )( σ ) − h f ( y i , u f )( σ ) | dσ (3.14) ≤ tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) ) + tK k y i − y i k C ([0 ,δ ]) ≤ tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) )+ t K ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) )1 − tK = tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) )1 − tK . Similarly, we have Z β i α i | λ ( y i , U )( σ ) − λ ( y i , U )( σ ) | dσ ≤ tK k M − M k C ([0 ,δ ]) + tK k y i − y i k C ([0 ,δ ]) ≤ t K k M f − M f k C ([0 ,δ ]) + tK k M − M k C ([0 ,δ ]) − tK , (3.15)11here λ represents λ or e λ , and Z β i α i | g f ( u f ( σ )) − g f ( u f ( σ )) | dσ ≤ tK ( k M f − M f k C ([0 ,δ ]) + k M − M k C ([0 ,δ ]) ) . (3.16)By (3.13)-(3.16) and the definition (3.10) of G f , we have k G f ( M f , M ) − G f ( M f , M ) k C ([0 ,δ ]) ≤ tC f k M f − M f k C ([0 ,δ ]) + tC f k M − M k C ([0 ,δ ]) . (3.17)The expressions of C f and C f are given by (6.30) in Appendix 6.3. They depend on k φ fk k , k b φ fk k , k e φ fk k , K and K . Thus we have k ~G ( ~M ) − ~G ( ~M ) k C ([0 ,δ ]) (3.18)= (cid:13)(cid:13)(cid:13)(cid:16) G ( M , M ) − G ( M , M ) , · · · , G n ( M n , M ) − G n ( M n , M ) (cid:17)(cid:13)(cid:13)(cid:13) C ([0 ,δ ]) = max f k G f ( M f , M ) − G f ( M f , M ) k C ([0 ,δ ]) ≤ max f (cid:16) tC f k M f − M f k C ([0 ,δ ]) + tC f k M − M k C ([0 ,δ ]) (cid:17) ≤ max f (cid:16) tC f k M f − M f k C ([0 ,δ ]) + tC f k M − M k C ([0 ,δ ]) + · · · + tC f k M n − M n k C ([0 ,δ ]) (cid:17) ≤ max f t ( C f + C f ) (cid:16) k M − M k C ([0 ,δ ]) + · · · + k M n − M n k C ([0 ,δ ]) (cid:17) ≤ nt max f ( C f + C f ) max f k M f − M f k C ([0 ,δ ]) . We finally get k ~G ( ~M ) − ~G ( ~M ) k C ([0 ,δ ]) ≤ nt max f ( C f + C f ) k ~M − ~M k C ([0 ,δ ]) . (3.19)Hence we can choose δ small enough (depending on k φ fk k , k b φ fk k , k e φ fk k , K , K , T ) so that k ~G ( ~M ) − ~G ( ~M ) k C ([0 ,δ ]) ≤ k ~M − ~M k C ([0 ,δ ]) . (3.20)By Lemma 3.1 and the contraction mapping principle, there exists a unique fixed point ~M = ~G ( ~M ) in Ω δ,K . Now we show how the fixed point ~M allows to find a solution to Cauchy problem (1.1)-(1.5) for t ∈ [0 , δ ]. Let us recall the definition of three subsets ω f,t , ω f,t and ω f,t of [0 , ,the definition of the characteristics ξ , ξ and also the definition (3.5) of ( x , y ) and (3.6) of( τ , β ). For k = 1, we define φ f ( t, x, y ) by φ f ( t, x, y ) := φ f ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , , else . (3.21)12or k = 2 , · · · , N , we define φ fk ( t, x, y ) by φ fk ( t, x, y ) := φ fk ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , τ gf b φ fk − ( τ , , β ) a g f ( u f ( τ )) e − R tτ [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , , else . (3.22)Since the dynamic in Phase 2 amounts to pure transport equations, we define b φ fk ( t, x, y ), k = 1 , · · · , N by b φ fk ( t, x, y ) := b φ fk ( x − b g f t, y ) , if ( x, y ) ∈ [ b g f t, × [0 , ,a g f ( t − x b g f ) τ gf φ fk ( t − x b g f , , y ) , if ( x, y ) ∈ [0 , b g f t ] × [0 , . (3.23)Let us recall the definition of the four subsets e ω f,t , e ω f,t , e ω f,t and e ω f,t of [0 , , the definitionof the characteristics ξ , ξ , ξ and also the definition (3.7) of ( x , y ), (3.8) of ( t , α ) and(3.9) of ( τ , β ). For k = 1, we define e φ f ( t, x, y ) by e φ f ( t, x, y ) := e φ f ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ f ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (3.24)For k = 2 , · · · , N , we define e φ fk ( t, x, y ) by e φ fk ( t, x, y ) := e φ fk ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ fk ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , e φ fk − ( τ , , β ) e − R tτ [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (3.25)Next we prove that the vector function ~φ f defined by (3.21)-(3.25) is a weak solution toCauchy problem (1.1)-(1.5) for t ∈ [0 , δ ]. To that end, we first prove that ~φ f defined by (3.21)-(3.25) satisfies equality (2.5) of Definition 2.1, then we prove that ~φ f ∈ C ([0 , δ ]; L ((0 , )).Let τ ∈ [0 , δ ]. For any vector function ~ϕ ∈ C ([0 , τ ] × [0 , ), ~ϕ := ( ϕ , · · · , ϕ N , b ϕ , · · · , b ϕ N , e ϕ , · · · , e ϕ N ) T such that (2.1)-(2.4) hold, by definition (3.21)-(3.25) of ~φ f , we have Z τ Z Z ~φ f ( t, x, y ) · ( ~ϕ t + A f ~ϕ x + B f ~ϕ y + C ~ϕ ) dxdydt := X i =1 A i . (3.26)13n (3.26), A := N X k =1 Z τ Z Z ω f,t φ fk ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ ( ϕ kt + g f ϕ kx + h f ϕ ky − λϕ k ) dxdydt,A := N X k =2 Z τ Z Z ω f,t τ gf b φ fk − ( τ , ,β ) a g f ( u f ( τ )) e − R tτ [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ ( ϕ kt + g f ϕ kx + h f ϕ ky − λϕ k ) dxdydt,A := N X k =1 Z τ Z Z b g f t b φ fk ( x − b g f t, y )( b ϕ kt + b g f b ϕ kx ) dxdydt,A := N X k =1 Z τ Z Z b g f t a g f ( u f ( t − x b g f )) τ gf φ fk ( t − x b g f , , y )( b ϕ kt + b g f b ϕ kx ) dxdydt,A := N X k =1 Z τ Z Z e ω f,t e φ fk ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ ( e ϕ kt + e g f e ϕ kx + e h f e ϕ ky − e λ e ϕ k ) dxdydt,A := N X k =1 Z τ Z Z e ω f,t φ fk ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ ( e ϕ kt + e g f e ϕ kx + e h f e ϕ ky − e λ e ϕ k ) dxdydt,A := N X k =2 Z τ Z Z e ω f,t e φ fk − ( τ , , β ) e − R tτ [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ ( e ϕ kt + e g f e ϕ kx + e h f e ϕ ky − e λ e ϕ k ) dxdydt. Let us consider the first term A as an instance. By the change of variable ( x, y ) → ( x , y )and noting (6.21) of Lemma 6.1 in Appendix 6.3, we have A = N X k =1 Z τ Z β Z α φ fk ( x , y ) e − R t λ ( y ,U )( σ ) dσ (cid:16) ϕ kt ( t,x ( t ) ,y ( t ))+ g f ϕ kx ( t,x ( t ) ,y ( t ))+ h f ϕ ky ( t, x ( t ) , y ( t )) − λϕ k ( t, x ( t ) , y ( t )) (cid:17) dx dy dt, where (see Fig 4 (a) and (b)) α := 1 − Z t g f ( u f ( σ )) dσ := f ( t ) , β := 1 − Z t h f ( y , u f )( σ ) dσ := g ( t ) . (3.27)14learly, α is a function of β , suppose that α = h ( β ). After changing the order of integration, A can be rewritten as A = N X k =1 nZ g ( τ )0 Z f ( τ )0 Z τ + Z g ( τ )0 Z f ( τ ) Z f − ( α )0 + Z g ( τ ) Z f ( τ )0 Z g − ( β )0 + Z g ( τ ) Z h ( β ) Z f − ( α )0 + Z f ( τ ) Z h − ( α ) Z g − ( β )0 o φ fk ( x , y ) d (cid:16) e − R t λ ( y ,U )( σ ) dσ ϕ k ( t, x ( t ) , y ( t )) (cid:17) dt dtdx y = N X k =1 n − Z g ( τ )0 Z f ( τ )0 φ fk ( x , y ) ϕ k (0 , x , y ) dx dy − Z g ( τ )0 Z f ( τ ) φ fk ( x , y ) ϕ k (0 , x , y ) dx dy − Z g ( τ ) Z f ( τ )0 φ fk ( x , y ) ϕ k (0 , x , y ) dx dy − Z g ( τ ) Z h ( β ) φ fk ( x , y ) ϕ k (0 , x , y ) dx dy − Z f ( τ ) Z h − ( α ) φ fk ( x , y ) ϕ k (0 , x , y ) dy dx + Z g ( τ )0 Z f ( τ ) φ fk ( x , y ) e − R f −
11 ( α )0 λ ( y ,U )( σ ) dσ ϕ k ( f − ( α ) , x ( f − ( α )) , y ( f − ( α ))) dx dy + Z g ( τ ) Z f ( τ )0 φ fk ( x , y ) e − R g −
11 ( β )0 λ ( y ,U )( σ ) dσ ϕ k ( g − ( β ) , x ( g − ( β )) , y ( g − ( β ))) dx dy + Z g ( τ ) Z h ( β ) φ fk ( x , y ) e − R f −
11 ( α )0 λ ( y ,U )( σ ) dσ ϕ k ( f − ( α ) , x ( f − ( α )) , y ( f − ( α ))) dx dy + Z f ( τ ) Z h − ( α ) φ fk ( x , y ) e − R g −
11 ( β )0 λ ( y ,U )( σ ) dσ ϕ k ( g − ( β ) , x ( g − ( β )) , y ( g − ( β ))) dx dy o . Changing the order of integration again (see Fig 4 (c)), we obtain − Z f ( τ ) Z h − ( α ) φ fk ( x , y ) ϕ k (0 , x , y ) dy dx = − Z g ( τ ) Z h ( β ) f ( τ ) φ fk ( x , y ) ϕ k (0 , x , y ) dx dy . (3.28)By (3.28) and noting that x ( f − ( α )) = y ( g − ( β )) = 1, we get A = − N X k =1 Z Z φ fk ( x , y ) ϕ k (0 , x , y ) dx dy . (3.29)15imilar to A , we can prove that A = − N X k =2 Z τ Z τ gf a b φ fk − ( τ , , β ) ϕ k ( τ , , β ) dβ dτ , (3.30) A = − N X k =1 Z Z b φ fk ( x , y ) b ϕ k (0 , x , y ) dx dy , (3.31) A = − N X k =1 Z τ Z a b g f g f ( u f ( τ )) τ gf φ fk ( τ , , β ) b φ k ( τ , , β ) dβ dτ , (3.32) A = − N X k =1 Z Z e φ fk ( x , y ) e ϕ k (0 , x , y ) dx dy , (3.33) A = − N X k =1 Z τ Z a a e h f (0 , u f ( t )) φ fk ( t , a a α , e ϕ k ( t , α , dα dt , (3.34) A = − N X k =2 Z τ Z e g f e φ fk − ( τ , , β ) e ϕ k ( τ , , β ) dβ dτ . (3.35) (a) (b) (c) Figure 4: (a) For any fixed t ∈ [0 , τ ], the characteristics ξ , ξ and the definition of α, β ; (b)The illustration of functions α = f ( t ) and β = g ( t ), t ∈ [0 , τ ]; (c) For changing the order ofintegration.By (3.29)-(3.35), we have proven that the vector function ~φ f defined by (3.21)-(3.25)satisfies (2.5). Next we prove that ~φ f ∈ C ([0 , δ ]; L ((0 , )). Moreover, we can prove that ~φ f even belongs to C ([0 , δ ]; L p ((0 , )), for all p ∈ [1 , ∞ ). Lemma 3.2.
The weak solution ~φ f to Cauchy problem (1.1) - (1.5) belongs to C ([0 , δ ]; L p ((0 , )) for all p ∈ [1 , ∞ ) . Proof:
From the definition (3.21)-(3.25) of ~φ f , we get easily that ~φ f ∈ L ∞ ((0 , δ ) × (0 , ).Next we prove that the vector function ~φ f belongs to C ([0 , δ ]; L p ((0 , )) for all p ∈ [1 , ∞ ),16.e., for every ˜ t, t ∈ [0 , δ ] with ˜ t ≥ t (the case that ˜ t ≤ t can be treated similarly), we need toprove k ~φ f (˜ t, · ) − ~φ f ( t, · ) k L p ((0 , ) → , as | ˜ t − t | → , ∀ p ∈ [1 , ∞ ) . In order to do that, we estimate k φ fk (˜ t, · ) − φ fk ( t, · ) k L p ((0 , ) , k b φ fk (˜ t, · ) − b φ fk ( t, · ) k L p ((0 , ) and k e φ fk (˜ t, · ) − e φ fk ( t, · ) k L p ((0 , ) ( k = 1 , · · · , N ) separately.Suppose that the characteristic passing through ( t, ,
0) intersects time ˜ t plane at (˜ t, q , q )(see Fig 5 (a)), we have q = Z ˜ tt g f ( u f ( σ )) dσ, q = Z ˜ tt h f ( y, u f )( σ ) dσ, where y ( s ) , t ≤ s ≤ ˜ t satisfies dyds = h f ( y, u f )( s ) , y ( t ) = 0 . We get easily that q ≤ C | ˜ t − t | , q ≤ C | ˜ t − t | . (3.36)Here and hereafter in this section, we denote by C various constants which do not dependon ˜ t , t , x or y . Let S := [ q , × [ q ,
1] (see Fig 5 (a)), following our method to construct thesolution φ fk , when ( x, y ) ∈ S , we have | φ fk (˜ t, x, y ) − φ fk ( t, x, y ) | = (cid:12)(cid:12)(cid:12) φ fk ( t, ˜ x, ˜ y ) e − R ˜ tt [ λ (˜ y ,U )+ ∂hf∂y (˜ y ,u f )]( σ ) dσ − φ fk ( t, x, y ) (cid:12)(cid:12)(cid:12) ≤ | φ fk ( t, ˜ x, ˜ y ) − φ fk ( t, x, y ) | e − R ˜ tt [ λ ( e y ,U )+ ∂hf∂y ( e y ,u f )]( σ ) dσ + C | φ fk ( t, x, y ) || ˜ t − t | . Here e ξ = ( e x , e y ) denotes the characteristic passing through (˜ t, x, y ) that intersects time t plane at ( t, ˜ x, ˜ y ). Hence (˜ x, ˜ y ) is defined by˜ x = x − Z ˜ tt g f ( u f ( σ )) dσ, ˜ y = y − Z ˜ tt h f ( e y , u f )( σ ) dσ. We have | ˜ x − x | ≤ C | ˜ t − t | , | ˜ y − y | ≤ C | ˜ t − t | . (3.37)Since φ fk ∈ L ∞ ((0 , δ ) × (0 , ), for every l >
0, there exists C l such that for every t ∈ [0 , δ ],there exists φ flk ( t, · ) ∈ C ([0 , ) satisfying k φ flk ( t, · ) − φ fk ( t, · ) k L p ((0 , ) ≤ l , k φ flk ( t, · ) k C ([0 , ) ≤ C l . (3.38)Here and hereafter in this section, we denote by C l various constants that may depend on l (the index of the corresponding approximating sequences φ flk ( t, · ), b φ flk ( t, · ) and e φ flk ( t, · )) butare independent of 0 ≤ t ≤ ˜ t ≤ δ . 17hus, we have | φ fk (˜ t, x, y ) − φ fk ( t, x, y ) |≤ C | φ flk ( t, ˜ x, ˜ y ) − φ fk ( t, ˜ x, ˜ y ) | + C | φ flk ( t, x, y ) − φ fk ( t, x, y ) | + C | φ fk ( t, x, y ) || ˜ t − t | + C l | ˜ t − t | . (3.39)By (3.38) and (3.39), we have Z Z S | φ fk (˜ t, x, y ) − φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p . (3.40)Noting (3.36), we have Z Z [0 , \ S | φ fk (˜ t, x, y ) − φ fk ( t, x, y ) | p dxdy ≤ C ( q + q ) ≤ C | ˜ t − t | . (3.41)Combining (3.40) with (3.41), we obtain Z Z | φ fk (˜ t, x, y ) − φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p + C | ˜ t − t | . (3.42) (a) (b) (c) Figure 5: (a) For Phase 1, the definition of S and characteristic e ξ which connects (˜ t, x, y )with ( t, ˜ x, ˜ y ), characteristic passing through ( t, ,
0) intersects time ˜ t plane at (˜ t, q , q ); (b)For Phase 2, the definition of b S ; (c) For Phase 3, the definition of e S and characteristic e ξ which connects (˜ t, x, y ) with ( t, ˜ x, ˜ y ), characteristic passing through ( t, ,
0) intersects time ˜ t plane at (˜ t, v, v ) and characteristic passing through ( t, ,
1) intersects time ˜ t plane at (˜ t, v, v ).Next we estimate k b φ fk (˜ t, · ) − b φ fk ( t, · ) k L p ((0 , ) . For Phase 2, the characteristic passingthrough ( t, ,
0) intersects time ˜ t plane at (˜ t, b g f (˜ t − t ) , b S := [ b g f (˜ t − t ) , × [0 ,
1] (seeFig 5 (b)), when ( x, y ) ∈ b S , we have | b φ fk (˜ t, x, y ) − b φ fk ( t, x, y ) | = | b φ fk ( t, ˜ x, ˜ y ) − b φ fk ( t, x, y ) | = | b φ fk ( t, x − b g f (˜ t − t ) , y ) − b φ fk ( t, x, y ) | . b φ fk ∈ L ∞ ((0 , δ ) × (0 , ), for every l >
0, there exists C l such that for every t ∈ [0 , δ ],there exists b φ flk ( t, · ) ∈ C ([0 , ) satisfying k b φ flk ( t, · ) − b φ fk ( t, · ) k L p ((0 , ) ≤ l , k b φ flk ( t, · ) k C ([0 , ) ≤ C l . (3.43)Similar to (3.40), we can prove that Z Z b S | b φ fk (˜ t, x, y ) − b φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p . (3.44)It is easy to check that Z Z [0 , \ b S | b φ fk (˜ t, x, y ) − b φ fk ( t, x, y ) | p dxdy ≤ C | ˜ t − t | . (3.45)Combining (3.44) with (3.45), we get Z Z | b φ fk (˜ t, x, y ) − b φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p + C | ˜ t − t | . (3.46)Finally, we estimate k e φ fk (˜ t, · ) − e φ fk ( t, · ) k L p ((0 , ) . Suppose that the characteristic passingthrough ( t, ,
0) intersects time ˜ t plane at (˜ t, v, v ), and the characteristic passing through( t, ,
1) intersects time ˜ t plane at (˜ t, v, v ) (see Fig 5 (c)). Similar to (3.36), we can prove that v ≤ C | ˜ t − t | , v ≤ C | ˜ t − t | , − v ≤ C | ˜ t − t | . (3.47)Let e S := [ v, × [ v , v ] (see Fig 5 (c)), when ( x, y ) ∈ e S , we have | e φ fk (˜ t, x, y ) − e φ fk ( t, x, y ) | = | e φ fk ( t, ˜ x, ˜ y ) e − R ˜ tt [ e λ ( e y ,U )+ ∂ e hf∂y ( e y ,u f )]( σ ) dσ − e φ fk ( t, x, y ) | . Here e y = ( e x , e y ) denotes the characteristic passing through (˜ t, x, y ) that intersects time t plane at ( t, ˜ x, ˜ y ). Similar to (3.37), we can prove that | ˜ x − x | ≤ C | ˜ t − t | , | ˜ y − y | ≤ C | ˜ t − t | . Since e φ fk ∈ L ∞ ((0 , δ ) × (0 , ), for every l >
0, there exists C l such that for every t ∈ [0 , δ ],there exists e φ flk ( t, · ) ∈ C ([0 , ) satisfying k e φ flk ( t, · ) − e φ fk ( t, · ) k L p ((0 , ) ≤ l , k e φ flk ( t, · ) k C ([0 , ) ≤ C l . (3.48)Similar to (3.40), we can prove that Z Z e S | e φ fk (˜ t, x, y ) − e φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p . (3.49)Noting (3.47), we have Z Z [0 , \ e S | e φ fk (˜ t, x, y ) − e φ fk ( t, x, y ) | p dxdy ≤ C ( v + v + 1 − v ) ≤ C | ˜ t − t | . (3.50)19ombining (3.49) with (3.50), we get Z Z | e φ fk (˜ t, x, y ) − e φ fk ( t, x, y ) | p dxdy ≤ Cl p + C l | ˜ t − t | p + C | ˜ t − t | . (3.51)Therefore letting l be large enough and then | ˜ t − t | be small enough, the right hand side of(3.42), (3.46) and (3.51) can be arbitrarily small. Above all, we prove that the vector function ~φ f belongs to C ([0 , δ ]; L p ((0 , )) for all p ∈ [1 , ∞ ).Let us recall Definition 2.1 of a weak solution, we prove that the vector function ~φ f definedby (3.21)-(3.25) is indeed a weak solution to Cauchy problem (1.1)-(1.5). In this section, we prove the uniqueness of the solution.
Lemma 4.1.
The weak solution to Cauchy problem (1.1) - (1.5) is unique. Proof:
Assume that ~φ f = ( φ f , · · · , φ fN , b φ f , · · · , b φ fN , e φ f , · · · , e φ fN ) T ∈ C ([0 , δ ]; L ((0 , )) is aweak solution to Cauchy problem (1.1)-(1.5). Similar to Lemma 2.2 in [3, Section 2.2], wecan prove that for any fixed t ∈ [0 , δ ] and any ~ Φ( τ, x, y ) := (Φ , · · · , Φ N , b Φ , · · · , b Φ N , e Φ , · · · , e Φ N ) T ∈ C ([0 , t ] × [0 , ) with ~ Φ( τ, , y ) = 0 , ∀ ( τ, y ) ∈ [0 , t ] × [0 , , Φ ( τ, , y ) = e Φ ( τ, , y ) = 0 , ∀ ( τ, y ) ∈ [0 , t ] × [0 , , Φ k ( τ, x,
0) = b Φ k ( τ, x,
0) = Φ k ( τ, x,
1) = b Φ k ( τ, x,
1) = 0 , ∀ ( τ, x ) ∈ [0 , t ] × [0 , , one has Z t Z Z ~φ f ( τ, x, y ) · ( ~ Φ τ ( τ, x, y )+ A f ~ Φ x ( τ, x, y )+ B f ~ Φ y ( τ, x, y )+ C ~ Φ( τ, x, y )) dxdydτ (4.1)+ Z Z ~φ f ( x, y ) · ~ Φ(0 , x, y ) dxdy + N X k =1 Z t Z a a e h f (0 , u f ( τ )) φ fk ( τ, a a x, e Φ k ( t, x, dxdτ + N X k =2 Z t Z τ gf a b φ fk − ( τ, , y )Φ k ( τ, , y ) dydτ + N X k =2 Z t Z e g f e φ fk − ( τ, , y ) e Φ k ( τ, , y ) dydτ + N X k =1 Z t Z a b g f g f ( u f ( τ )) τ gf φ fk ( τ, , y ) b Φ k ( τ, , y ) dydτ = Z Z ~φ f ( t, x, y ) · ~ Φ( t, x, y ) . In (4.1), the velocity matrices A f , B f and C are defined the same way as A f , B f and C butwith M f and M instead of M f and M , where M f ( t ) = N X k =1 Z Z a γ s yφ fk ( t, x, y ) dx dy + N X k =1 Z Z ( a − a ) γ s y b φ fk ( t, x, y ) dx dy + N X k =1 Z Z a γ ( γ y + γ s ) e φ fk ( t, x, y ) dx dy, M ( t )= P nf =1 M f ( t ). Let ~ϕ ( x, y )=( ϕ , · · · , ϕ N , b ϕ , · · · , b ϕ N , e ϕ , · · · , e ϕ N ) T ∈ C ((0 , ),and choose the test function as the solution to the following backward linear Cauchy problem ~ Φ τ + A f ~ Φ x + B f ~ Φ y = − C ~ Φ , ≤ τ ≤ t, ( x, y ) ∈ [0 , ,~ Φ( t, x, y ) = ~ϕ ( x, y ) , ( x, y ) ∈ [0 , ,~ Φ( τ, , y ) = 0 , ∀ ( τ, y ) ∈ [0 , t ] × [0 , , Φ ( τ, , y ) = e Φ ( τ, , y ) = 0 , ∀ ( τ, y ) ∈ [0 , t ] × [0 , , Φ k ( τ, x, b Φ k ( τ, x, k ( τ, x, b Φ k ( τ, x, , ∀ ( τ, x ) ∈ [0 , t ] × [0 , . (4.2)For any fixed t ∈ [0 , δ ], we introduce three new subsets ω f,t , ω f,t and ω f,t of [0 , ω f,t := n ( x, y ) | Z t g f ( u f ( σ )) dσ ≤ x ≤ , η ( t, Z t g f ( u f ( σ )) dσ ) ≤ y ≤ o ,ω f,t := n ( x, y ) | ≤ x ≤ Z t g f ( u f ( σ )) dσ, η ( t, x ) ≤ y ≤ o ,ω f,t := [0 , \ ( ω f,t ∪ ω f,t ) . Here y = η ( t, x ) satisfies dηds = h f ( η, u f )( s ) , η ( θ ) = 0 , θ ≤ s ≤ t, with θ defined by x = Z tθ g f ( u f ( σ )) dσ .For any fixed t ∈ [0 , δ ] and ( x, y ) ∈ [0 , . If ( x, y ) ∈ ω f,t , we define ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . Let us then define ( x , y ) := ( x (0) , y (0)) . (4.3)If ( x, y ) ∈ ω f,t , we define ξ = ( x , y ) by dx ds = g f ( u f ( s )) , dy ds = h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique τ such that x ( τ ) = 0, so that we can define β := y ( τ ) . (4.4)By (4.1) and (4.2), we obtain that Z Z ~φ f ( t, x, y ) · ~ϕ f ( x, y ) dxdy (4.5)= Z Z ~φ f ( x, y ) · ~ Φ(0 , x, y ) dxdy + N X k =1 Z t Z a a e h f (0 , u f ( τ )) φ fk ( τ, a a x, e Φ k ( t, x, dxdτ + N X k =2 Z t Z τ gf a b φ fk − ( τ, , y )Φ k ( τ, , y ) dydτ + N X k =2 Z t Z e g f e φ fk − ( τ, , y ) e Φ k ( τ, , y ) dydτ + N X k =1 Z t Z a b g f g f ( u f ( τ )) τ gf φ fk ( τ, , y ) b Φ k ( τ, , y ) dydτ. ξ and also the definition (4.3) of ( x , y ).By solving the backward linear Cauchy problem (4.2) and noting (6.21) of Lemma 6.1 inAppendix 6.3, we have Z Z ~φ f ( x, y ) · ~ Φ(0 , x, y ) dxdy = N X k =1 Z Z ω f,t φ fk ( x , y ) ϕ k ( x, y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ dxdy + Z Z b g f t b φ fk ( x − b g f t, y ) b ϕ k ( x, y ) dxdy + N X k =1 Z Z e φ fk ( x , y ) e Φ k (0 , x , y ) dx dy . (4.6)As for the last term of (4.5), by solving the backward linear Cauchy problem (4.2), we have N X k =1 Z t Z a b g f g f ( u f ( τ )) τ gf φ fk ( τ, , y ) b Φ k ( τ, , y ) dydτ = N X k =1 Z Z b g f t a g f ( u f ( t − x b g f )) τ gf φ fk ( t − x b g f , , y ) b ϕ k ( x, y ) dxdy. Since ~ϕ ∈ C ((0 , ) and t ∈ [0 , δ ] are both arbitrary, we obtain in C ([0 , δ ]; L ((0 , ))that for k = 1 , · · · , N b φ fk ( t, x, y ) = a g f ( u f ( t − x b g f )) τ gf φ fk ( t − x b g f , , y ) , if ( x, y ) ∈ [0 , b g f t ] × [0 , , b φ fk ( t, x, y ) = b φ fk ( x − b g f t, y ) = b φ fk ( t, x, y ) , if ( x, y ) ∈ [ b g f t, × [0 , . (4.7)Let us recall the definition of the characteristic ξ and also the definition (4.4) of ( τ , β ).By solving the backward linear Cauchy problem (4.2), noting (4.7) and (6.22) of Lemma 6.1in Appendix 6.3, we have N X k =2 Z t Z τ gf a b φ fk − ( τ, , y )Φ k ( τ, , y ) dydτ = N X k =2 Z t Z τ gf a b φ fk − ( τ , , β )Φ k ( τ , , β ) dβ dτ = N X k =2 Z Z ω f,t τ gf b φ fk − ( τ , , β ) a g f ( u f ( τ )) ϕ k ( x, y ) e − R tτ [ λ ( y ,U )+ ∂hf∂y ( y ,u f )] dxdy. Since ~ϕ ∈ C ((0 , ) and t ∈ [0 , δ ] are both arbitrary, we obtain in C ([0 , δ ]; L ((0 , ))that for k = 1 φ f ( t, x, y ) = φ f ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , , else . (4.8)22or k = 2 , · · · , N , we have φ fk ( t, x, y ) = φ fk ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , τ gf b φ k − ( τ , , β ) a g f ( u f ( τ )) e − R tτ [ λ ( y ,U )+ ∂hf∂y ( y ,u f )] , if ( x, y ) ∈ ω f,t , , else . (4.9)For any fixed t ∈ [0 , δ ], we introduce four new subsets e ω f,t , e ω f,t , e ω f,t and e ω f,t of [0 , e ω f,t := n ( x, y ) | e g f t ≤ x ≤ , e η ( t, e g f t ) ≤ y ≤ e η ( t, e g f t ) o , e ω f,t := n ( x, y ) | ≤ x ≤ e g f t, ≤ y ≤ e η ( t, x ) o ∪ n ( x, y ) | e g f t ≤ x ≤ , ≤ y ≤ e η ( t, e g f t ) o , e ω f,t := n ( x, y ) | ≤ x ≤ e g f t, e η ( t, x ) ≤ y ≤ e η ( t, x ) o , e ω f,t :=[0 , \ ( e ω f,t ∪ e ω f,t ∪ e ω f,t ) . Here y = e η ( t, x ) and y = e η ( t, x ) satisfy d e η ds = e h f ( e η , u f )( s ) , e η ( t − x e g f ) = 0 , t − x e g f ≤ s ≤ t,d e η ds = e h f ( e η , u f )( s ) , e η ( t − x e g f ) = 1 , t − x e g f ≤ s ≤ t. For any fixed t ∈ [0 , δ ] and ( x, y ) ∈ [0 , . If ( x, y ) ∈ e ω f,t , we define ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . Let us then define ( x , y ) := ( x (0) , y (0)). By solving the backward linear Cauchy problem(4.2) and noting (6.24) of Lemma 6.1 in Appendix 6.3, the last term of (4.6) can be rewrittenas N X k =1 Z Z e φ fk ( x , y ) e Φ k (0 , x , y ) dx dy = N X k =1 Z Z e ω f,t e φ fk ( x , y ) e ϕ k ( x, y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ dxdy. If ( x, y ) ∈ e ω f,t , we define ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique t such that y ( t ) = 0, so that we can define α := x ( t ). By solvingthe backward linear Cauchy problem (4.2), and noting (6.25) of Lemma 6.1 in Appendix 6.3,23e have N X k =1 Z t Z a a e h f (0 , u f ( τ )) φ fk ( τ, a a x, e Φ k ( τ, x, dxdτ = N X k =1 Z t Z a a e h f (0 , u f ( t )) φ fk ( t , a a α , e Φ k ( t , α , dα dt = N X k =1 Z Z e ω f,t φ fk ( t , a a α , e ϕ k ( x, y ) e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ dxdy. If ( x, y ) ∈ e ω f,t , we define ξ = ( x , y ) by dx ds = e g f , dy ds = e h f ( y , u f )( s ) , ξ ( t ) = ( x, y ) . There exists a unique τ such that x ( τ ) = 0, so that we can define β := y ( τ ). By solvingthe backward linear Cauchy problem (4.2) and noting (6.28) of Lemma 6.1 in Appendix 6.3,we have N X k =2 Z t Z e g f e φ fk − ( τ, , y ) e Φ k ( τ, , y ) dydτ = N X k =2 Z t Z e g f e φ fk − ( τ , , β ) e Φ k ( τ , , β ) dβ dτ = N X k =2 Z Z e ω f,t e φ fk − ( τ , , β ) e ϕ k ( x, y ) e − R tτ [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ dxdy. Since ~ϕ ∈ C ((0 , ) and t ∈ [0 , δ ] are both arbitrary, we obtain in C ([0 , δ ]; L ((0 , ))that for k = 1 e φ f ( t, x, y ) = e φ f ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ f ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (4.10)For k = 2 , · · · , N , we get e φ fk ( t, x, y ) = e φ fk ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ f ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , e φ fk − ( τ , , β ) e − R tτ [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (4.11)By (3.11), we claim that ~M ( t ) ∈ Ω δ,K and ~M ( t ) satisfies the same contraction mappingfunction ~G . Since ~M is the unique fixed point of ~G in Ω δ,K , therefore we have ~M ≡ ~M .24onsequently, we have ξ i ≡ ξ i ( i = 1 , · · · , e η i ≡ e η i ( i = 1 ,
2) and η ≡ η , so that ( x , y ) ≡ ( x , y ), ( t , α ) ≡ ( t , α ), ( τ , β ) ≡ ( τ , β ), ω f,ti ≡ ω f,ti ( i = 1 , ,
3) and e ω f,ti ≡ e ω f,ti ( i =1 , , , ~φ f with the definition (3.21)-(3.25) of ~φ f , we obtain ~φ f ≡ ~φ f . This gives us the uniqueness of the weak solution for smalltime. Let us now prove the existence of global solution to Cauchy problem (1.1)-(1.5). Notingthe definition (3.10) of G f and the definition (3.21)-(3.25) of the local solution ~φ f , it is easyto check that the following two estimates hold for all t ∈ [0 , δ ]0 ≤ M ( t ) ≤ K, (5.1) k ~φ f ( t, · ) k L ∞ ((0 , ) ≤ e N +1) T K max k n ( 2 K K + a K τ gf ) N k φ fk k , ( 2 K K + 2 τ gf a K ) N k b φ fk k , k e φ fk k o , (5.2)where K is defined by (3.1), K is defined by (3.2) and K is defined by (3.3). In order toobtain a global solution, we suppose that we have solved Cauchy problem (1.1)-(1.5) up tothe moment τ ∈ [0 , T ] with the weak solution ~φ f ∈ C ([0 , τ ]; L p ((0 , )). It follows from ourway to construct the weak solution, we know that for any 0 ≤ t ≤ τ , the weak solution isgiven by for k = 1 φ f ( t, x, y ) := φ f ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , , else . (5.3)For k = 2 , · · · , Nφ fk ( t, x, y ) := φ fk ( x , y ) e − R t [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , τ gf b φ fk − ( τ , , β ) a g f ( u f ( τ )) e − R tτ [ λ ( y ,U )+ ∂hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ ω f,t , , else . (5.4)For k = 1 , · · · , N b φ fk ( t, x, y ) := b φ fk ( x − b g f t, y ) , if 0 ≤ t ≤ x b g f , y ∈ [0 , ,a g f ( u f ( t − x b g f )) τ gf φ fk ( t − x b g f , , y ) , if x b g f ≤ t ≤ T, y ∈ [0 , . (5.5) e φ f ( t, x, y ) := e φ f ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ f ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (5.6)25or k = 2 , · · · , N e φ fk ( t, x, y ) := e φ fk ( x , y ) e − R t [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t ,φ fk ( t , a a α , e − R tt [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , e φ fk − ( τ , , β ) e − R tτ [ e λ ( y ,U )+ ∂ e hf∂y ( y ,u f )]( σ ) dσ , if ( x, y ) ∈ e ω f,t , , else . (5.7)As time increases, for each cell cycle k ( k = 1 , · · · , N ) in Phase 1, the characteristicpassing through the origin may have two possible cases. It may either intersect with thefront face (see Fig 6 (a)) or intersect with the right face (see Fig 6 (b)). For each cell cycle k ( k = 1 , · · · , N ) in Phase 3, the characteristic passing through the origin will definitelyintersect with the front face due to the fact that e h f (1 , u f ) < M f ( t ) := N X k =1 Z Z ω f,t a γ s yφ fk ( t, x, y ) dxdy + N X k =1 Z Z ( a − a ) γ s y b φ fk ( t, x, y ) dxdy + N X k =1 Z Z e ω f,t a γ ( γ y + γ s ) e φ fk ( t, x, y ) dxdy + N X k =1 Z Z e ω f,t a γ ( γ y + γ s ) e φ fk ( t, x, y ) dxdy (a) (b) (c) Figure 6: For t large enough. Case (a): the characteristic passing through the origin intersectsthe front face at ( τ , , β ) in Phase 1; Case (b): the characteristic passing through the originintersects the right face at ( t , α ,
1) in Phase 1; Case (c): due to the fact that e h f (1 , u f ) < e g f , , β ).We can prove that estimate (5.1) holds for every t ∈ [0 , T ] by tracing back φ fk ( t, x, y ), b φ fk ( t, x, y ) and e φ fk ( t, x, y ) along the characteristics to the initial data at most N times. More-26ver, noting the definition (5.3)-(5.7) of the global solution, it is easy to check that the uniforma priori estimate (5.2) holds for every t ∈ [0 , T ]. Hence we can choose δ ∈ [0 , T ] independent of τ . Applying Lemma 3.2 and Lemma 4.1 again, the weak solution ~φ f ∈ C ([0 , τ ]; L p ((0 , ))is extended to the time interval [ τ, τ + δ ] ∩ [ τ, T ]. Step by step, we finally obtain a uniqueglobal weak solution ~φ f ∈ C ([0 , T ]; L p ((0 , )). This finishes the proof of the existence of aglobal solution to Cauchy problem (1.1)-(1.5). In this section, we first recall the systems of equations describing the dynamics of the celldensity of a follicle in the model of F. Cl´ement [1, 5, 6]. The cell population in a follicle f isrepresented by cell density functions φ fj,k ( t, a, γ ) defined on each cellular phase Q fj,k with age a and maturity γ , which satisfy the following conservation laws ∂φ fj,k ∂t + ∂ ( g f ( u f ) φ fj,k ) ∂a + ∂ ( h f ( γ, u f ) φ fj,k ) ∂γ = − λ ( γ, U ) φ fj,k , in Q fj,k (6.1) Q fj,k := Ω j,k × [0 , T ] , for j = 1 , , , f = 1 , · · · , n, whereΩ ,k := [( k − a , ( k − a + a ] × [0 , γ s ] , Ω ,k := [( k − a + a , ka ] × [0 , γ s ] , Ω ,k := [( k − a , ka ] . Here k = 1 , · · · , N , and N is the number of consecutive cell cycles (see Fig 7).Define the maturity operator M as M ( ϕ )( t ) := Z γ max Z a max γϕ ( t, a, γ ) dadγ. (6.2)Then M f := X j =1 N X k =1 Z γ max Z a max γφ fj,k ( t, a, γ ) dadγ (6.3)is the global follicular maturity on the follicular scale, while M := n X f =1 3 X j =1 N X k =1 Z γ max Z a max γφ fj,k ( t, a, γ ) dadγ (6.4)is the global maturity on the ovarian scale.The velocity and source terms are all smooth functions of u f and U , where u f is the localcontrol and U is the global control. In this paper, we consider close/open loop problem, thatis to say u f and U are functions of M f , M and t . As an instance of close loop problem, thevelocity and source terms are given by the following expressions in [1] (all parameters are27ositive constants) g f ( u f ) = τ gf (1 − g (1 − u f )) , in Ω ,k ,g f ( u f ) = τ gf , in Ω j,k , j = 2 , ,h f ( γ, u f ) = τ hf ( − γ + ( c γ + c )(1 − e − ufu )) , in Ω j,k , j = 1 , ,h f ( γ, u f ) = λ ( γ, U ) = 0 , in Ω ,k ,λ ( γ, U ) = Ke − ( γ − γsγ ) (1 − U ) , in Ω j,k , j = 1 , .U = S ( M ) + U = U + U s + 1 − U s e c ( M − m ) ,u f = b ( M f ) U = min { b + e b M f b , } · [ U + U s + 1 − U s e c ( M − m ) ] . The initial conditions are given as follows φ fj,k (0 , a, γ ) = φ fk ( a, γ ) | Ω j,k , j = 1 , , . (6.5)The boundary conditions are given as follows g f ( u f ) φ f ,k ( t, ( k − a , γ ) = τ gf φ f ,k − ( t, ( k − a , γ ) , for k ≥ , , for k = 1 , for γ ∈ [0 , γ s ] .φ f ,k ( t, a,
0) = 0 , for a ∈ [( k − a , ( k − a + a ] ,τ gf φ f ,k ( t, ( k − a + a , γ ) = g f ( u f ) φ f ,k ( t, ( k − a + a , γ ) , for γ ∈ [0 , γ s ] ,φ f ,k ( t, a,
0) = 0 , for a ∈ [( k − a + a , ka ] ,φ f ,k ( t, ( k − a , γ ) = φ f ,k − ( t, ( k − a , γ ) , for k ≥ , , for k = 1 , for γ ∈ [ γ s , γ m ] ,φ f ,k ( t, a, γ s ) = φ f ,k ( t, a, γ s ) , for a ∈ [( k − a , ( k − a + a ] , , for a ∈ [( k − a + a , ka ] . For sake of simplicity, we denote by γ := γ m − γ s > γ s of the maturity in Phase 1 and the maximum γ m of the maturityin Phase 3. In this section, we perform a mathematical reformulation of the original model introducedin Appendix 6.1. Notice that in system (6.1), all the unknowns are defined on different28 ✻ · · ·· · · γ s γ m γ a a a a + a a ( k − a ( k − a + a ka ( N − a ( N − a + a Na φ f , φ f , φ f , φ f , φ f , φ f , φ f ,k φ f ,k φ f ,k φ f ,N φ f ,N φ f ,N · · ·· · · ✒❘ ✒❘ ✒❘ ✒❘ ✒❘✒ ✒ ✲ ✒ ✒ ✲ ✒ ✒ ✲ ✒ ✒ Figure 7: The illustration of N cell cycles for follicle f ; a represents the age of the cell and γ represents the maturity of the cell. The top of the domain corresponds to the differentiationphase and the bottom to the the proliferation phase.domains, so that one has to solve the equations successively. Here we transform system (6.1)into a regular one where the unknowns are defined on the same domain [0 , T ] × [0 , . Wedenote by φ fk the density functions and by g f and h f the age and maturity velocities for Phase1; b φ fk the density functions and b g f the age velocities for Phase 2; e φ fk the density functionsand e g f , e h f the age and maturity velocities for Phase 3.Let φ fk ( t, x, y ) := φ f ,k ( t, a, γ ) , ( a, γ ) ∈ Ω ,k , (6.6)where x := a − ( k − a a , y := γγ s , (6.7)so that we get ( a, γ ) ∈ Ω ,k ⇐⇒ ( x, y ) ∈ [0 , , and( φ fk ) t + ( g f φ fk ) x + ( h f φ fk ) y = − λ φ fk , (6.8)with g f ( u f ) := g f ( u f ) a , h f ( y, u f ) := h f ( γ s y, u f ) γ s , λ ( y, U ) := λ ( γ s y, U ) . (6.9)Let b φ fk ( t, x, y ) := φ f ,k ( t, a, γ ) , ( a, γ ) ∈ Ω ,k , (6.10)where x := a − ( k − a − a a − a , y := γγ s . (6.11)So that we get ( a, γ ) ∈ Ω ,k ⇐⇒ ( x, y ) ∈ [0 , , and( b φ fk ) t + b g f ( b φ fk ) x = 0 , b g f := τ gf a − a . (6.12)Let e φ fk ( t, x, y ) := φ f ,k ( t, a, γ ) , ( a, γ ) ∈ Ω ,k , (6.13)where x := a − ( k − a a , y := γ − γ s γ , (6.14)so that we get ( a, γ ) ∈ Ω ,k ⇐⇒ ( x, y ) ∈ [0 , , and( e φ fk ) t + ( e g f e φ fk ) x + ( e h f e φ fk ) y = − e λ e φ fk , (6.15)29ith e g f := τ gf a , e h f ( y, u f ) := h f ( γ y + γ s , u f ) γ , e λ ( y, U ) := λ ( γ y + γ s , U ) . (6.16)Accordingly, let us denote the initial conditions (6.5) with new notations by ~φ f = ( φ f ( x, y ) , · · · , φ fN ( x, y ) , b φ f ( x, y ) , · · · b φ fN , e φ f ( x, y ) , · · · , e φ fN ( x, y )) , (6.17)where φ fk ( x, y ) := φ f ,k (0 , a, γ ) , b φ fk ( x, y ) := φ f ,k (0 , a, γ ) , e φ fk ( x, y ) := φ f ,k (0 , a, γ ) . (6.18)Let ~φ f = ( φ f , · · · , φ fN , b φ f , · · · , b φ fN , e φ f , · · · , e φ fN ) T , we have ~φ f ( t, x, y ) t + ( A f ~φ f ( t, x, y )) x + ( B f ~φ f ( t, x, y )) y = C ~φ f ( t, x, y ) , (6.19) t ∈ [0 , T ] , ( x, y ) ∈ [0 , , where A f : = diag { N z }| { g f , · · · , g f , N z }| {b g f , · · · , b g f N z }| {e g f , · · · , e g f } ,B f : = diag { N z }| { h f , · · · , h f , N z }| { , · · · , N z }| {e h f , · · · , e h f } ,C : = − diag { N z }| { λ, · · · , λ, N z }| { , · · · , N z }| {e λ, · · · , e λ } . From the original expression of h f ( γ, u f ) (see Appendix 6.1), let us define γ ± ( u f ) := c (1 − e − ufu ) ± q c (1 − e − ufu ) + 4 c (1 − e − ufu )2 . It is easy to see that γ + ( u f ) is an increasing function of u f , and h f ( γ, u f ) = τ hf ( γ + ( u f ) − γ )( γ − γ − ( u f )) . Hence, when γ = γ + ( u f ), we have h f ( γ, u f ) = 0, moreover h f ( γ, u f ) > , if 0 ≤ γ < γ + ( u f ) ,h f ( γ, u f ) < , if γ > γ + ( u f ) . Furthermore, under the assumption that the local control u f satisfies γ + ( u f ) > γ s , and0 ≤ γ + ( u f ) ≤ γ + (1) ≃ (1 − e − u ) c + p c + 4 c < γ m . From the fact that 0 < γ ≤ γ s in Phase 1, and γ s ≤ γ ≤ γ m in Phase 3, we have h f ( γ, u f ) > , < γ ≤ γ s ,h f ( γ s , u f ) > , h f ( γ m , u f ) < . h f ( y, u f ) > , ∀ y ∈ [0 , , e h f (0 , u f ) > , e h f (1 , u f ) < . (6.20)Since M ( φ f ,k ) = Z Z Ω ,k γφ f ,k ( t, a, γ ) da dγ = Z Z a γ s yφ fk ( t, x, y ) dx dy,M ( φ f ,k ) = Z Z Ω ,k γφ f ,k ( t, a, γ ) da dγ = Z Z ( a − a ) γ s y b φ fk ( t, x, y ) dx dy,M ( φ f ,k ) = Z Z Ω ,k γφ f ,k ( t, a, γ ) da dγ = Z Z a γ ( γ y + γ s ) e φ fk ( t, x, y ) dx dy. We have M f ( t ) = N X k =1 M ( φ f ,k ) + N X k =1 M ( φ f ,k ) + N X k =1 M ( φ f ,k )= N X k =1 Z Z a γ s yφ fk ( t, x, y ) dxdy + N X k =1 Z Z ( a − a ) γ s y b φ fk ( t, x, y ) dxdy + N X k =1 Z Z a γ ( γ y + γ s ) e φ fk ( t, x, y ) dx dy. The following lemma is used to prove the existence of the weak solution to Cauchy problem(1.1)-(1.5), when we derive the contraction mapping function ~G , change variables in certainintegrals (see Section 3) and prove the uniqueness of the solution (see Section 4). Lemma 6.1.
The characteristic ξ = ( x , y ) passing through ( t, x, y ) intersects the bottomface at (0 , x , y ) . We have ∂ ( x, y ) ∂ ( x , y ) = e R t ∂hf∂y ( y ,u f )( σ ) dσ . (6.21) The characteristic ξ = ( x , y ) passing through ( t, x, y ) intersects the back face at ( τ , , β ) .We have ∂ ( x, y ) ∂ ( τ , β ) = − g f ( u f ( τ )) · e R tτ ∂hf∂y ( y ,u f )( σ ) dσ . (6.22) The characteristic ξ = ( x , y ) passing through ( τ , , β ) intersects the bottom face at (0 , x , y ) . We have ∂ ( τ , β ) ∂ ( x , y ) = − g f ( u f ( τ )) · e R τ ∂hf∂y ( y ,u f )( σ ) dσ . (6.23)31 he characteristic ξ = ( x , y ) passing through ( t, x, y ) intersects the bottom face at (0 , x , y ) .We have ∂ ( x, y ) ∂ ( x , y ) = e R t ∂ e hf∂y ( y ,u f )( σ ) dσ . (6.24) The characteristic ξ = ( x , y ) passing through ( t, x, y ) intersects the left face at ( t , α , .We have ∂ ( x, y ) ∂ ( t , α ) = e h f ( u f ( t ) , · e R tt ∂ e hf∂y ( y ,u f )( σ ) dσ . (6.25) The characteristic ξ = ( x , y ) passing through ( t , α , intersects the bottom face at (0 , x , y ) . We have ∂ ( α , t ) ∂ ( x , y ) = − h f ( u f ( t ) , · e R t ∂hf∂y ( y ,u f )( σ ) dσ . (6.26) The characteristic ξ = ( x , y ) passing though ( t , α , intersects the back face at ( τ , , β ) .We have ∂ ( α , t ) ∂ ( τ , β ) = − g f ( u f ( τ )) h f ( u f ( t ) , · e R t τ ∂hf∂y ( y ,u f )( σ ) dσ . (6.27) The characteristic ξ = ( x , y ) passing through ( t, x, y ) intersects the back face at ( τ , , β ) .We have ∂ ( x, y ) ∂ ( τ , β ) = − e g f e R tτ ∂ e hf∂y ( y ,u f )( σ ) dσ . (6.28) The characteristic ξ = ( x , y ) passing through ( τ , , β ) intersects the bottom face at (0 , x , y ) . We have ∂ ( τ , β ) ∂ ( x , y ) = − e g f · e R τ ∂ e hf∂y ( y ,u f )( σ ) dσ . (6.29)The proof of this lemma is trivial, and can be found in [14] of 1-space dimension case, weomit here.The expressions of contraction mapping coefficients C f and C f are as following C f := a γ m (2 K − tK + 3 K + tK K + 9 K K ) K (1 − tK ) N X k =1 k φ fk k (6.30)+ 2( a − a ) γ m ( K +2 tK K +4 K K ) K (1 − tK ) N X k =2 k b φ fk k + a γ m (2 K +2 tK )1 − tK N X k =1 k e φ fk k ,C f := a γ m (2 K + 2 K + 12 K K − tK K ) K (1 − tK ) N X k =1 k φ fk k + 2( a − a ) γ m ( K + 6 K K ) K (1 − tK ) N X k =2 k b φ fk k + 4 a γ m K − tK k e φ fk k . Acknowledgements
The author would like to thank the professors Fr´ed´erique Cl´ement, Jean-Michel Coronand Zhiqiang Wang for their interesting comments and many valuable suggestions on thiswork. 32 eferences [1] F. Cl´ement. Multiscale modelling of endocrine systems: new insight on the gonadotropeaxis. ESAIM: proceedings, 27 :209-226, 2009.[2] J.-M. Coron,
Control and nonlinearity , Mathematical Surveys and Monographs ,American Mathematical Society, Providence, RI, 2007.[3] J.-M. Coron, M. Kawski, and Z. Wang. Analysis of a conservation law modeling a highlyre-entrant manufacturing system, preprint, arXiv:0907.1274v1.[4] P. Shang and Z. Wang. Analysis and control of a scalar conservation law modeling ahighly re-entrant manufacturing system, preprint, arXiv:1003.4411v1.[5] N. Echenim, D. Monniaux, M. Sorine, and F. Cl´ement. Multi-scale modeling of thefollicle selection process in the ovary. Math. Biosci., 198 :57–79, 2005.[6] N. Echenim, F. Cl´ement, and M. Sorine. Multi-scale modeling of follicular ovulation asa reachability problem. Multiscale Model. Simul., 6 :895–912, 2007.[7] S. Bianchini and A. Bressan, Vanishing viscosity solutions of nonlinear hyperbolic sys-tems, Ann. of Math., (2) 161 (2005), pp. 223–342.[8] A. Bressan,
Hyperbolic systems of conservation laws. The one dimensional Cauchy prob-lem , Oxford University Press, Oxford, 2000.[9] S. N. Kruˇzkov. First order quasilinear equations in several independent variables. Sb.Math., 10(2) :217-243, 1970.[10] P. Michel. Multiscale modeling of follicular ovulation as a mass and maturity dynamicalsystem, 2009.[11] P. Michel. Existence of measure solution to a renewal equation, 2009.[12] EA. McGee and AJ. Hsueh. Initial and cyclic recruitment of ovarian follicles. EndocrineReviews, 21(2) :200-214, 2009.[13] T.-T. Li,
Global classical solutions for quasilinear hyperbolic systems , Research in Ap-plied Mathematics , John Wiley & Sons, Chichester, 1994.[14] T.-T. Li and W. C. Yu, Boundary value problems for quasilinear hyperbolic systems , DukeUniversity Mathematics Series V, Duke University, Mathematics Department, Durham,NC, 1985.[15] T.-P. Liu and T. Yang, Well-posedness theory for hyperbolic conservation laws, Comm.Pure Appl. Math. 52 (1999), no. 12, pp. 1553–1586.3316] A. Poretta and J. Vovelle. L1