Different speed along the vertices - Transport on network structures.

Until now, we have considered flow problems withc_j = 1for allj. In this section, we show that the flow problem can still be solved using the results we have obtained already by converting it to a problem with c_j ≡1. In this section, we adopt the assumption from [33] that the speeds are linearly dependent over the field of rational numbersQ. That is, for all j= 1, . . . , m,

c_j ∈Nfor some N ∈N. (4.31)

If c_j can be written in the form c_j = _l¹

j, where l_j ∈ N for all j, we make the transformation

x= (1/c_j)x. Then, the flow problem becomes:











∂tuj(˜x, t) =∂x˜uj(˜x, t), x˜∈[0,_c¹

j], t≥0, u_j(x,0) =f_j(x),

φ⁻_ijcjuj(_c¹

j, t) =wi,jPm

k=1φ⁺_i,k(c_ku_k(0, t)), ∀i.

(4.32)

If the speed c_j can be written in the formc_j =N/l_j for some N, l_j ∈ Nand j ∈ {1, . . . , m}, then we first rescale time using the transformationτ =N t. This will put the flow problem into the form where0<˜c_j <1, considered above,











∂_τu_j(x, τ) = _l¹

j∂_xu_j(x, τ), x∈(0,1), τ ≥0, u_j(x,0) =f_j(x),

φ⁻_ijc_ju_j(1, τ) =w_i,jPm

k=1φ⁺_i,k(c_ku_k(0, τ)), ∀i.

(4.33)

Notice that in this, case,˜c_j = 1/l_j, which is the first case we considered. Then we can apply the transformation x˜=xl_j = (1/˜c_j)x and this will put the problem into the form given in (4.32).

From now on, we assume that the flow problem is in the form (4.32).

We then divide the interval[0,_c¹

j]into unit intervals by creating artificial vertices along the edge e_j. The number of artificial vertices created along e_j will be (_c¹

j −1). We label the edges

joining these vertices as ej1, ej2, . . . , ej₁_/cj. In this way, we have created a larger network with n^′ vertices and m^′ edges and each edge is of unit length; that iss∈[0,1]with

m^′ = Xm

j=1

1 cj

; n^′ =n+ Xm

j=1

1 cj −1

=n+m^′−m.

The density of particles on edge e_j_i will be denoted by u_j_i for all j = 1, . . . , m and i = 1,2, . . . ,_c¹

For eachi= 1, . . . ,_c¹

j,i−1≤x˜≤i, we introduce a new variable s, which is defined on [0,1]

and depends on x˜ through the following equation:

s= ˜x−i+ 1, (4.34)

for eachiandx˜ defined above.

Remark 4.6.1. On the larger network, we shall again represent the outgoing incidence matrix as Φ⁻ and the incoming incidence matrix as Φ⁺. These matrices are n^′ ×m^′ matrices. Every artificial vertex has exactly one incoming and outgoing edge, and the weight on the outgoing edge of an artificial vertex is 1. If v_i is not an artificial vertex and e_j is an outgoing edge ofv_i, then the weight on ej,wij, is the same as in the original network.

The flow problem on this larger network is simply the flow problem in (4.10) with a few modi- fications:











∂_tu_j_i(s, t) =∂_su_j_i(s, t), s∈[0,1], t≥0, u_j_i(s,0) =ϕ_j_i(s),

φ⁻_lj

ic_j_iu_j_i(1, t) =w_l,j_iPm k=1φ⁺_l,k

cj1

c_k₁

u_k₁

(0, t), ∀l= 1, . . . , n^′.

(4.35)

where

ϕji(s) =











f_j(˜x) if0≤x˜≤1; i= 1, f_j(˜x) if1≤x˜≤2; i= 2, ... ...

f_j(˜x) if _c¹

j −1≤x˜≤ _c¹_j; i= _c¹

Since each artificial vertex has exactly one incoming and one outgoing edge and the speed on all the artificial edges is the same, the boundary condition on each of these artificial vertices is

φ⁻_lj

ic_j_iu_j_i(1, t) =c_ju_j_i(1, t) =φ⁺_lj

i−1c_j_i−1u_j_i−1(0, t) =c_ju_j_i−1(0, t) (4.36)

for all i= 2, . . . ,_c¹

j and j= 1,2, . . . , m. If we define the operator (A, D(A))where D(A) ={v∈W₁¹([0,1])^m^′ :v(1) =C⁻¹BCv(0)}

and Av = v^′, then the flow problem on the new graph is equivalent to the abstract Cauchy problem







v^′(t) =Av(t), v(0) =ϕ.

HereB is the adjacency matrix for the line graph of the new expanded network.

Define a functionv(s) = (v₁(s), . . . , v_m(s)), wherev_j = (v_j₁, . . . , v_j₁

) is defined as

v_j =v_j_i(s), s= ˜x−i+ 1,1≤i≤c⁻¹_j , 1≤j≤m. (4.37) Then we define u(x) by

uj(x) =vj(cjx)˜ (4.38)

Theorem 4.6.2. Let S:L₁([0,1])^m^′ → L₁([0,1])^m be the transformation u =Sv defined by (4.38). ThenS is an isomorphism.

Proof. We note first thatS is invertible, since

vj(s) =











u_j(˜x) if0≤x˜≤1; i= 1 u_j(˜x) if1≤x˜≤2; i= 2 ... ...

u_j(˜x) if _c¹

j −1≤x˜≤ _c¹_j; i= _c¹

(4.39)

wheresis as defined in (4.34) for all j= 1, . . . , m, defines the inverse of S. Then kSvk=

j=1

Z 1

0 |u_j(x)|dx

= Xm

j=1

c_j Z ¹

0 |v_j(˜x)|d˜x

= Xm

j=1

c_j

Xcj

i=1

Z i

i−1|v_j_i(˜x)|d˜x

= Xm

j=1

c_j

Xcj

i=1

Z 1

0 |v_j_i(s)|ds

≤max{c_j}kvk.

On the other hand,kS⁻¹uk=kuk kS⁻¹uk=

j=1

Z 1

0 |v_j₁(s)|ds+ Z 1

0 |v_j₂(s)|ds+· · ·+ Z 1

0 |v_j₁

(s)|ds

= Xm

j=1

Xcj

i=1

Z 1

0 |v_j_i(s)|ds

= Xm

j=1

Xcj

i=1

Z i

i−1|v_j(˜x)|d˜x

= Xm

j=1

Z ¹

0 |v_j(˜x)|d˜x= Xm

j=1

1 c_j

Z 1

0 |u_j(x)|dx

≤max

1 c_j

kuk. We conclude thatS is an isomorphism.

Lemma 4.6.3. The function v∈D(A) if and only if u=Sv∈D(A).

Proof. From the preceding results (in particular Theorem 4.4.1), we know that (4.35) admits semigroup solutions. Suppose that

v(s) = v1,1(s), . . . , v_1,1/c₁(s), . . . , vm,1(s), vm,2(s), . . . , v_m,1/c_m

∈D(A).

Then Equation (4.36) gives us continuity of the flow at the artificial vertices. That is,v_j_i(1) = v_j_i−1(0) for all i= 2, . . . ,_c¹

j. So the function v_j(s), defined in equation (4.37), is continuous on[0,_c¹

j]for all j= 1, . . . , mand Xm

j=1

Z ¹

0 |v_j(˜x)|d˜x= Xm

j=1



 Z 1

0 |v_j(˜x)|d˜x+ Z 2

1 |v_j(˜x)|d˜x+· · ·+ Z ¹

cj 1

cj−1|v_j(˜x)|d˜x





= Xm

j=1

1 c_j

Z 1

0 |vj1(x)|dx+ Z 1

0 |vj2(x)|dx+· · ·+ Z 1

0 |vj1 cj

(x)|dx

<∞ since each vji(x) ∈ L1[0,1]. So vj ∈ L1([0,_c¹

j]) for any j = 1, . . . , m. Next, we show that v_j ∈W₁¹([0,_c¹

j]), j = 1, . . . , m. Since eachv_j_i ∈W₁¹([0,1]), there exists g_j_i ∈L₁([0,1]) such

that Z 1

v_j_iφ^′ds=− Z 1

g_j_iφds, ∀φ∈C₀^∞([0,1]), for eachj= 1, . . . , mandi= 1, . . . ,_c¹

j. On the other hand, let ξ∈C₀^∞([0,1/c_j]). Then Z ¹

v_jξ^′d˜x= Z 1

v_jξ^′d˜x+ Z 2

v_jξ^′d˜x+· · ·+ Z ¹

cj 1 cj−1

v_jξ^′d˜x

From this equation, using Corollary 8.10 of [8], we get Z ¹

v_jξ^′d˜x=− Z ₁

g_j₁ξd˜x+ (v_j₁ξ)|¹₀− Z ₂

g_j₂ξd˜x+ (v_j₂ξ)|¹₀− · · · − Z ¹

cj 1 cj−1

ˆ g_j₁

ξd˜x

+ (vj1 cj

ξ)

1 cj 1 cj−1

− Z ₁

g_j₁ξd˜x+v_j₁(1)ξ(1)

− Z ₂

g_j₂ξd˜x+v_j₂(2)ξ(2)−v_j₂(1)ξ(1)

+· · · +



− Z ¹

cj 1 cj−1

ˆ g_j₁

ξdx˜−v_j₁

1 c_j −1



,

wheregˆ_j_i(˜x) =g_j_i(s)with s= ˜x−i+ 1for j= 1, . . . , mandi= 1, . . . ,1/c_j. By the Kirchoff law at the artificial vertices and continuity of the test functions ξ on [0,_c¹

j], all the boundary terms in the integral cancel out. That is,

v_j₁(1) =v_j₂(1), v_j₂(2) =v_j₃(2),· · · , v_j₁

1 cj −1

=v_j₁

cj−1

1 cj −1

Hence

Z ¹

v_jξ^′d˜x=− Z 1

g_j₁ξdx˜− Z 2

g_j₂ξd˜x− · · · − Z ¹

cj 1 cj−1

ˆ g_j₁

ξd˜x

=− Z ¹

g_jv_jd˜x;

where

g_j(˜x) =









 ˆ

gj1(˜x), if 0≤x˜≤1, ˆ

g_j₂(˜x), if 1≤x˜≤2, ... ...

ˆ g_j₁

(˜x), if _c¹

j −1≤x˜≤ _c¹_j. Sinceg_j(˜x)∈L₁([0,_c¹

j]),v_j(˜x, t)∈W₁¹([0,_c¹

j])for each j= 1, . . . , m, hencev_j ∈W₁¹([0,_c¹

j]), j= 1, . . . , m. We have u∈W₁¹([0,1])^m, whereu is defined by (4.38). Since the Kirchoff laws at the original vertices have not been changed, we must have u∈D(A₀).

Thus, the solution to (4.35) is given by

(T(t)f)(s) =C⁻¹BⁿCψ(t+s−n), n∈N₀, 0≤t+s−n <1s∈[0,1],

where

ψ(s) =







ϕ_1,1(s) ... ϕ_1,1

(s) ...

ϕ_m,1(s) ... ϕ_m, 1

cm(s)







and from this, we obtain the solution to the original problem through equation (4.38) above.

Using the transformation in Theorem 4.6.2, we have

T(t)u=S⁻¹T(N t)Su. (4.40)

That is, in order to get u₁(x, t), we take the submatrix of B which contains only the first _c¹

rows of B and multiply it withϕ. To get u₂(x, t), we take the next _c¹

2 rows by starting from the(_c¹

1 + 1)^th row up to the(_c¹

1 +_c¹

2)^th row and so on.

A Graph Theoretic Point of View

5.1 Introduction

In the preceding chapter, we proved the existence of semigroup solutions to the flow problem on the graph with no sinks. In this chapter, we extend the results of Chapter 4 to graphs with non trivial acyclic part. We will show that asymptotically, the flow will remain in certain parts of the graph with cycles and that these subgraphs where the flow remains asymptotically are those cycles with no outgoing flow. We also show that the flow on the edges in the acyclic part of the graph will be depleted in finite time while the flow in the cyclic parts of the graph with both incoming and outgoing flow will be depleted asymptotically. We start with certain useful graph descriptions which will enable asymptotic description of the flow to be easier.

Dalam dokumen Transport on network structures. (Halaman 78-84)