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Modeling of pipe–to–pipe intersections

pipes. We state the sets in terms of u= Iρˆ.

A+0 :={(ρ1, ρ2, I)∈R+×R+×R:u+a2 ≥0 and u≤0}, A0 :={(ρ1, ρ2, I)∈R+×R+×R: u≤0},

A#0 :={(ρ1, ρ2, I)∈R+×R+×R:u− a2 ≤0 and u≥0}.

(3.10)

The following elementary result characterizes the speed of forward 1-shock and backward 3-shock for subsonic initial data and will be used later for solutions at a pipe–to–pipe intersection.

Lemma 3.1. Let U0 be in the interior of A#0 or in the interior of A0 (respectively, A+0) and assume thatU0 is not a vacuum state. Then, the velocity of a 1–shock wave (respectively, 3–shock wave) connecting U0 and U has non–positive (respectively, non–negative) speed provided that kU−U0k is sufficiently small.

Proof. Consider a (right) stateU =L+1(ξ;U0) connected to the left stateU0 by a 1–Lax–shock, hence ξ ≥1. The shock speed is

s1(ξ;U0) =u0− a

√2

pξ≤u0− a

√2 ≤0.

Similarly, we obtain s3(ξ;U0) ≥ 0, for ξ ≥ 1, for U0 ∈ A+0 and a (left) state U =L3(ξ;U0)

Note that any state connected toU0 in the interior ofA+0 by a 3–wave has non–

negative speed due to (3.6). Similarly, any state U0 in the interior of A0 connected to a left stateU by a 1– or 2– wave has non–positive speed. This will be a key point for verifying well–posedness for pipe–to–pipe conditions.

cross–section of the pipe. Each j = 1, . . . , n belongs either to the set δ ⊂ N of incoming arcs or to the set δ+ ⊂N of outgoing arcs, see Figure 3.2 for an example.

We assume that there is at least one incoming and one outgoing pipe, which means that |δ±| ≥ 1, where |A| is the number of elements of the set A. The jth pipe is parameterized by x ∈ R, if j ∈ δ and by x ∈ R+, if j ∈δ+, see Figure 3.2. The parametrization is such that the vertex is located atx= 0 for all pipes. Along each

ν1

νn1

νn1+1

νn (0,0)

Figure 3.2: Junction with n pipes demonstrating the parametrization: the cross section νj and sets δ± with δ ={1, . . . , n1} and δ+={n1+ 1, . . . , n}

pipe j ∈ {1,2, . . . , n}, we assume the flow is modeled by a no-slip drift-flux model, that is, forx∈R(j ∈δ) or x∈R+(j ∈δ+) andt > 0,

t



 ρj1 ρj2 Ij



+∂x



ρj1uj ρj2uj ˆ

ρj

(uj)2+ a22



=



 0 0 0



 (3.11)

along with initial conditions

Uj(x,0) = (ρj1,0, ρj2,0, I0j)(x), ∀x∈R±(j). (3.12) We further prescribe algebraic conditions at the junction x = 0 coupling the dynamics on adjacent edges. Several possibilities for prescribing such conditions exist. Our conditions in the context of the two–component model are motivated as follows: It is assumed that neither mass of component one nor mass of component

two is lost when passing through the junction. This yields the two conditions (M1) and (M2) below. Furthermore, we assume that the junction does not introduce an acceleration to the combined velocity. It follows that for the two gas components in the adjacent pipes near the junctions, velocity differences vanish. This can be modeled by assuming that the flux of the momentum remains constant along all pipes near the intersection leading to the condition (Q). These assumptions could also be obtained in a rigorous way by a similar procedure as in [77]:

(M1) Conservation of mass of phase 1: for t >0 a.e., X

jδ

jk(ρj1uj)(0, t) = X

jδ+

jk(ρj1uj)(0, t).

(M2) Conservation of mass of phase 2: for t >0 a.e., X

jδ

jk(ρj2uj)(0, t) = X

jδ+

jk(ρj2uj)(0, t).

Furthermore, we require as an additional condition which correspond to an equal momentum along P

jνj :

(Q) The flux of the momentum density remains constant at the intersection: for t >0 a.e.,

ˆ ρj

(uj)2 +a2 2

(0, t) =

ˆ ρi

(ui)2+a2 2

(0, t) =P(t), ∀i6=j.

Conditions (M1) and (M2) are compulsory and resemble Kirchoff’s law at the intersection. They can be obtained by considering the weak formulation of (3.11).

Condition (Q) is also obtained from the weak formulation of (3.11) using a special class of test functions as in [77]. However, other conditions can be proposed to replace(Q), see also [41, 61, 7, 95]. Clearly, in the case n= 2, ν1 =−ν2 =−(1,0)T, (Q) is equivalent to assuming:

(Q’) There is a single pressure p at the intersection a2

2 ρˆj

(0, t) = p, ∀j.

If we consider the sum of(M1)and(M2)we obtain X2

j=1

jkρˆjuj = 0 and therefore 2p

a2 kνk X2

j=1

uj = 0 and X2

j=1

uj = 0. Combining this last equality and the equality of momentum immediately yields (Q).

Remark 3.4. We present some discussion on condition (Q) similar to [32]. Con- dition (Q) implies that the momentum over time [t1, t2]:

Z t2

t1

Pjdt:=

Z t2

t1

ˆ

ρj((uj)2+ a2

2)(0, t)dt= Z t2

t1

Pdt≡κ.

Therefore, we obtain

Z t2

t1

X

jδ±

Pjνjdt=κX

jδ±

νj. This is equivalent to the following:

∀η ∈

X

jδ±

νj

,

X

jδ±

Pjνj

·η= 0.

Hence, the linear momentum orthogonal to X

jδ±

νj is conserved and the constraint acts parallel to X

jδ±

νj.

In recent years there has been an intense discussion on existence of solutions to coupled systems of hyperbolic conservation laws. Without giving a complete list of references we mention the publications by Colombo et. al. [41, 33, 32]. Therein, for gas and traffic flow networks, existence of solutions to a Riemann problem and, depending on the application, to the Cauchy problem has been proven. We apply the technique derived in [41] to prove existence to a Riemann problem under conditions (M1)–(Q)for the no–slip multiphase model. As expected, the result is essentially a perturbation result for constant data. The assertions are restrictive since the data has to belong to certain sub–critical sets. Nevertheless, the importance of the result

is the construction of a solution at the junction. This construction is used and implemented later in the numerical scheme to compute the boundary values at the junction and prevent boundary layer effects.

Proposition 3.1. Given n distinct vectors νj ∈ R2\{0}, νj,1 ≥ 0, and coupling conditions given by (M1), (M2) and (Q). Let j = 1 ∈ δ and assume constant initial dataj,0 with the following properties:1,0 ∈A+0,j,0

A#0 for j ∈δ\{1},j,0 ∈ A+0 for j ∈ δ+ and let the constant statesj,0 satisfy the conditions (M1), (M2) and (Q). Moreover, assume that the initial data satisfies the technical condi- tion detM 6= 0 for M given by (3.14) below.

Then, there exists δ, C > 0 such that, for all states Vj with kVj −U¯j,0k ≤ δ, there exists self–similar functions Uj(x, t) satisfying the weak formulation of (3.4a, 3.4b, 3.4c), the initial conditionUj(x,0) = ¯Uj and such that the trace of Uj atx= 0 satisfies (M1), (M2) and (Q); furthermore, U satisfies the stability condition

kUj −Uj,0kL

(R×[0,),

R+×

R+×R)≤CkVj −Uj,0k, for all j ∈δ∪δ+. (3.13) The matrix M in Proposition 3.1 is given by

M :=











A0 A1 A2 A3 . . . An

B0 B1 B2 B3 . . . Bn

b0 b1 0 0 . . . −bn

0 0 b2 0 . . . −bn

... ... ... ... ... ...

0 . . . bn1 −bn











(3.14)

where

A0 =kν12(U1,01,01 , B0 =kν12(U1,01,02 , b02(U1,0)I1,0, i∈δ, i≥1 : Ai =kνi1(Ui,0i,01 , Bi=kνi1(Ui,0i,02 , bi21(Ui,0)ˆρi,0, i∈δ+, i≥1 : Ai =−kνi3(Ui,0i,01 , Bi =−kνi3(Ui,0i,02 , bi23(Ui,0)ˆρi,0.

Proof. (of Proposition 3.1) Assume δ :={1, . . . , n1}andδ+ :={n1+1, . . . , n}. Consider a perturbation of ¯Uj,0, Vj, such that V1 ∈ A+0, Vj

A#0 for j ∈ δ\{1},

Vj ∈ A+0 for j ∈ δ+. Let Uj = (ρj1, ρj2, Ij) and consider the map ˜Ψ : R3n → Rn+1 given by

Ψ((U˜ j)nj=1) :=







 Pn1

j=1jj1uj−Pn

j=n1+1jj1uj Pn1

j=1jj2uj−Pn

j=n1+1jj2uj ˆ

ρ1((u1)2+a22)−ρˆn((un)2a22) ...

ˆ

ρn1((un1)2 +a2)−ρˆn((un)2a22)









Thanks to Lemma 3.1 we obtain for any fixed stateV, parameters (σ, ξ1, . . . , ξn) = (σ,(ξ1, . . . , ξn1),(ξn1+1, . . . , ξn))∈(R+)n+1 such that

Ψ((σ, ξ1, . . . , ξn),(Vj)) := ˜Ψ(L2(σ;L+11, V1)), L+12, V2), . . .

. . . , L+1n1, Vn1), L3n1+1, Vn1+1), . . . , L3n, Vn)) = 0. (3.15) Note that the function Ψ depends on the parameterization σ, ξ1, . . . , ξn and the perturbed state Vi. For σ = ξ1 = · · · = ξn = 1 and Vi = ¯Uj,0 we have that Ψ vanishes due to the assumption that ¯Uj,0 satisfies the coupling conditions. We want to apply the implicit function theorem and obtain a parameterization in terms of the perturbed state Vi. We compute the determinant of D(σ,ξ1,...,ξn)Ψ at σ = ξ1 =

· · ·=ξn = 1 and obtain

det Ψ(·) = det















A0 A1 . . . An

B0 B1 . . . Bn

b0 b1 0 . . . −bn

0 0 b2 0 . . . −bn

0 0 0 . .. −bn

...

0 0 0 0 . . . bn1 −bn















where

A0 =kν12(U1,01,01 ≤0, B0 =kν12(U1,01,02 ≤0,

b02(U1,0)I1,0 ≥0,

i∈δ, i≥1 Ai =kνi1(Ui,0i,01 ≤0, Bi =kνi1(Ui,0i,02 ≤0, bi21(Ui,0)ˆρi,0 ≥0, i≥1

i∈δ+, i≥1 : Ai =−kνi3(Ui,0i,01 ≤0, Bi =−kνi3(Ui,0i,02 ≤0, i≥1, bi23(Ui,0)ˆρi,0 ≥0, i≥1

Since the initial data is in the interior of the subsonic sets (3.10), the inequalities are strict, i.e., Ai, Bi are negative andbi are positive. Due to the assumption, this determinant is non–zero. Hence, for any perturbation Vi of Uj,0 with kVi −Uj,0k sufficiently small, we obtain values σ = ξ1 = · · · = ξn such that the coupling conditions are fulfilled. The solution to the Riemann problem at the junction is now constructed by using the states:

Ve1 =L2(σ;L+11;V1)), Vej =L+1j;Vj) for j ∈δ\ {1}, Vej =L3j;Vj) for j ∈δ+ (3.16) for σ, ξj given by (3.15). Then for j ∈ {1, . . . , n}, the solution Uj is given by the restriction to the real half–line of the solution to the Riemann problem

Uj = (ρj1, ρj2, uj)(x,0) =

( Vj, x≤0

Vej, x >0 j ∈δ and (3.17) Uj = (ρj1, ρj2, uj)(x,0) =

( Vej, x <0

Vj, x≥0 j ∈δ+. (3.18) By construction, the trace of the solution at the junction satisfies the coupling conditions. The stability estimate (3.13) is derived from the C1–regularity of the map Ψ.