When the two players keep the same common support

We look first at the situation when players i and j have the same support, and this support is the same before and after the tranfer. In this case, we can describe explicitely the flows.

Proposition 6.20. Let ~x = ~xN E(d) and ~y = ~xN E(dδ) and suppose that

supp(xi) = supp(xj) = supp(yi) = supp(yj). Then Q(x) = Q(y).

Proof. Denote by S0 be the common support. Let δmin = min(δ0, δ1), where

δ0 = min a∈S0 xi_ac0_a(xa)
X a∈S0 1 c0 a(xa) , δ1 =  min b /∈S0 cb(xb) − πj(~x)  X a∈S0 1 c0 a(xa) . Note that δmin _{is well defined since c}0

a(x) > 0 for every x > 0, and is nonneg-

ative, according to Corollary6.12. We show that when δ ≤ δmin_{, the players}

keep the same support after the transfer.

Let then δ ≤ δmin_{. We consider the flow ~z defined by}

z_ak= xk_a+ d k δ − d k βc0 a(xa)

1{a∈S0} for every player k, where β = X a∈S0 1 c0 a(xa) .

Then we show that ~z = ~y. Since za = xa for every arc a, it will give the

result.

We first check that ~z is an admissible flow. We have more precisely z_ai = xi_a− δ βc0 a(xa) 1{a∈S0} z_aj = xj_a+ δ βc0 a(xa) 1{a∈S0} z_ak = xk_a for k 6= i, j. For each player k and arc a, we have zk

a ≥ 0 since δ ≤ δ0. Furthermore, we

can check easily that for every player k,P

a∈Az k

a = dkδ. Hence, the flow ~z is

admissible.

We check now that ~z is an equilibrium flow, by checking that it satisfies the conditions of Proposition6.10with πk_{(~z) = π}k_(~x)+dkδ−d

k

β for every player

k. Consider first a player k ∈ {i, j}. For every arc a ∈ S0.

ca(za) + zakc 0 a(za) = ca(xa) + xkac 0 a(xa) + dk_δ − dk β = πk(~x) + d k δ − dk β with (6.7).

Consider now an arc a /∈ S0. In particular, zak = xka = 0. We have

ca(za) = ca(xa) ≥ πj(~x) + δ β ≥ π k (~x) + d k δ − dk β ,

where the first equality holds since za= xa, the first inequality since δ ≤ δ1,

and the last inequality for player i since πj(~x) ≥ πi(~x), according to Lemma 6.18. The conditions of Proposition6.10are satisfied for any player k ∈ {i, j}.

Consider then a player k 6= {i, j}, we have dk_δ = dk and we retrieve ca(za) + zkac

0

a(za) = πk(~x) for a ∈ supp(xk)

ca(za) ≥ πk(~x) for a /∈ supp(xk).

Then, using Proposition6.10, ~z is an equilibrium, with πk_{(~z) = π}k_(~x)+dkδ−d k

β

for every player k. By uniqueness of the equilibrium, ~z = ~y.

Remark 6.22. We can extend Proposition 6.20 in a more general case. It remains valid for any demand ˜d in the neighbourhood of d defined by

        

there exists S0 s.t. supp(xk) = S0 for all k s.t. ˜dk 6= dk,

dk− ˜dk ≤ mina∈S0 x k ac0a(xa)
β, ˜ dk− dk _{≤ min} a /∈S0ca(xa) − π k_(x a)
β, ˜ d =P k∈[K]d˜ k₌P k∈[K]d k_{= d.}

We can also extend this result in the case where players have specific cost functions. It remains valid as soon as for all players k such that ˜dk _{6= d}k_,

P a∈S0 1 ck a 0

(xa) does not depend on k. For example when the cost functions are the same up to an additive constant.

6.6.3 When the two players keep the same different support

We suppose now that the players keep the same support, but there is at least an arc in the support of one player which is not in the support of the other one.

Since the players have the same cost functions, we can precise the Corol- lary6.16 coming from Proposition 6.15.

Corollary 6.23. Suppose that there are two arcs or two players. Let δ0 and

δ > 0 and note ~x = ~xN E(dδ0), ~y = ~x

N E

(dδ0+δ). Suppose that supp(x

i_{) =}

supp(yi_{) 6= supp(x}j_{) = supp(y}j_{). Suppose that x 6= y, then}

{a ∈ A, ya< xa} = supp(xi),

{a ∈ A, ya> xa} = supp(xj) \ supp(xi).

Proof. When there are two arcs, Corollary 6.16 gives that {a ∈ A, ya <

xa} ⊆ supp(xi), and {a ∈ A, ya > xa} ⊆ supp(xj) \ supp(xi). A cardinality

argument gives the equalities.

When there are two players, let a ∈ supp(xi_{). Lemma 6.18 gives that}

a ∈ supp(xj_{), and we have then}

2ca(xa) + xac0a(xa) = πi(~x) + πj(~x). (6.10)

Since supp(~x) = supp(~y), we have also

Since x 6= y, there is an arc a such that ya < xa. According to Corol-

lary6.16, we have a ∈ supp(xi_{). Then, since the function u 7→ 2c}

a(u)+uc0a(u)

is increasing, we get with Equations (6.10) and (6.11) πi(~y) + πj(~y) < πi(~x) + πj(~x).

In particular, if there is an arc a ∈ supp(xi_{) with y}

a ≥ xa, Equa-

tions (6.10) and (6.11) give πi(~y) + πj(~y) ≥ πi(~x) + πj(~x), which is not possible. Then ya < xa for all arcs of supp(xi). Since Corollary 6.16 gives

that {a ∈ A, ya < xa} ⊆ supp(xi), we can conclude that {a ∈ A, ya < xa} =

supp(xi).

Finally, let a ∈ supp(xj_{) \ supp(x}i_{). Since there are two players and}

supp(~x) = supp(~y), we have xa = xja and ya = yja, and Proposition 6.15

gives that ya > xa. Conversely every arc a such that ya > xa belongs to

supp(xj_{) according to Corollary 6.16, but not to supp(x}i_{) as we just proved.}

Then {a ∈ A, ya > xa} = supp(xj) \ supp(xi).

In order to prove Theorem 6.3 we use the following lemmas. Lemma 6.24. Let δ0 and δ > 0 and note ~x = ~xN E(dδ0), ~y = ~x

N E

(dδ0+δ). Suppose that supp(xi_{) = supp(y}i_{) 6= supp(x}j_{) = supp(y}j_{). For every a ∈}

supp(xi_{) and b ∈ supp(x}j_{) \ supp(x}i_),

∂Qa

∂zb

(x−a)(yb− xb) ≤ 0.

Proof. Since b /∈ supp(xi_{), Corollary 6.16 gives that y}

b ≥ xb and since a ∈

supp(xi_{), Corollary 6.12 gives that c}

b(xb) > ca(xa).

Consider the set Ib of players k such that b ∈ supp(xk). Then Ib is

nonempty and for every player k ∈ Ib we have, according to Proposition6.10,

cb(xb) + xkbc 0

b(xb) ≤ ca(xa) + xkac 0 a(xa).

By summing over these players, we get |Ib|cb(xb) + xbc0b(xb) ≤ |Ib|ca(xa) +  xa− X k /∈Ib xk_a  c0_a(xa) < |Ib|ca(xa) + xac0a(xa).

Since |Ib| ≥ 1, and cb(xb) > ca(xa) we must have

cb(xb) + xbc0b(xb) < ca(xa) + xac0a(xa),

i.e. ∂Qa

∂zb(x

Lemma 6.25. Suppose that there are only two players. Let δ0 and δ > 0 and

note ~x = ~xN E(dδ0), ~y = ~x

N E

(dδ0+δ). Suppose that supp(x

i_{) = supp(y}i_{) 6=}

supp(xj) = supp(yj). For δ small enough, there exists a ∈ supp(xi) such that for every b ∈ supp(xi_{) \ {a},}

∂Qa

∂zb

(x−a)(yb− xb) ≤ 0.

Proof. Consider an arc a ∈ arg max{ca0(x_a0), a0 ∈ supp(xi)}. Let then b ∈ supp(xi) \ {a}.

Since supp(xi_{) ⊆ supp(x}j_{) according to Lemma6.18, a and b ∈ supp(x}j_).

Summing Equations (6.7) for both players, we get

2ca(xa) + xac0a(xa) = 2cb(xb) + xbc0b(xb). (6.12)

According to the definition of a, we have ca(xa) ≥ cb(xb) and then

ca(xa) + xac0a(xa) ≤ cb(xb) + xbc0b(xb), i.e.

∂Qa

∂xb

(x−a) ≥ 0. (6.13) Furthermore, Corollary 6.23 gives that yb < xb.

We can now prove the result.

Proposition 6.26. Suppose that there are two arcs or two players. The cost at equilibrium ¯Q is nonincreasing on the set of transfers such that the support of each player remains constant after the transfer.

Proof. Let then δ0 such that the players keep the same support after a trans-

fer of δ0. We prove that ¯Q0(δ) ≤ 0 for any δ ∈ [0, δ0].

When there are two arcs, the result is an immediate corollary of Propo- sition 6.20 or Lemmas 6.19 and 6.24.

When there are two players, if they have the same support, the result is a corollary of Proposition 6.20. If they have a different support, let a ∈ supp(xi_{) as in Lemma6.25. Then, we prove that Equation (6.9) is satisfied.}

Indeed, we use Lemma 6.24 for the sum over arcs in supp(xj) \ supp(xi), Lemma6.25 for the sum over arcs in supp(xi_{) \ {a}. The sum over arcs not}

in supp(xj_{) is zero since there is no flow on these arcs before and after the}

transfer. We can conlude with Lemma 6.19.

Note that when there are two arcs, Qa is a function of only one variable.

We can directly take its derivative, without using Lemma6.19. Then, we do not need to use the differentiability of ~xN E. The result holds then even if we do not assume that the cost functions are twice continuously differentiable.

6.6.4 Proof of Theorem 6.3

According to Theorem 6.2, and more precisely Corollary 6.16, the support of player i is nonincreasing with respect to the inclusion, and the support of player j is nondecreasing. Hence, when δ goes from 0 to di_{, the two players}

change supports only a finite number of times. According to Proposition6.26, the equilibrium cost ¯Q is nonincreasing on a finite number of sub-intervals of [0, di_{] whose closure is [0, d}i_{] itself.}

Since ¯Q is continuous, using Theorem 6.1, we can conclude that ¯Q is nonincreasing on [0, di_{]. It proves Theorem 6.3.}

6.7

Discussion

A natural question is whether Theorems 6.2 and 6.3 are valid when there are three arcs or three players. Another question is whether Theorem 6.3re- mains valid when we allow player-specific cost functions. The answer is “no” for Theorem 6.3 and partially for Theorem 6.2, as shown by the following examples.

6.7.1 When there are three arcs and three players

In document Congestion games with player-specific cost functions (Page 136-141)