Order Form - FENOMEC

38 CHAPTER II. THE RIEMANN PROBLEM
To show the first two statements in (3.2), note that the differentials of the functions
E(u − ,u + )anda(u − ,u + ) are closely related. Indeed, a calculation based on
differentiating (1.8) and (3.1) with respect to u + yields
∂ u+ E(u − ,u + )=b(u − ,u + ) ∂ u+ a(u − ,u + ),
b(u − ,u + ):=U(u − ) − U(u + ) − U ′ (u + )(u − − u + ) > 0
for u − ≠ u + and
∂ u+ a(u − ,u + )= f ′ (u + ) − a(u − ,u + )
. (3.8)
u + − u −
In view of (3.8) and (2.5) it is clear that
∂ u+ a(u − ,u + ) < 0, u + 0, u + >ϕ ♮ (u − ).
This leads us to the conclusions listed in (3.2). Then, in view of (3.2) there exists
ϕ ♭ 0(u − ) satisfying E(u − ,ϕ ♭ 0(u − )) = 0.
By definition, for any u ≠0
E ( u, ϕ ♭ 0(u) ) =0,
u ≠ ϕ ♭ 0(u)
and, since a similar result as (3.2) holds for negative left-hand side,
E ( ϕ ♭ 0(u),ϕ ♭ 0(ϕ ♭ 0(u)) ) =0,
ϕ ♭ 0(u) ≠ ϕ ♭ 0(ϕ ♭ 0(u)).
Thus, (3.4) follows from the fact that the entropy dissipation has a single “non-trivial”
zero and from the symmetry property
E ( ϕ ♭ 0(u),u ) = −E ( u, ϕ ♭ 0(u) ) =0.
Finally, by the implicit function theorem it is clear that the function ϕ ♭ 0 is smooth,
at least away from u = 0. (The regularity at u = 0 is discussed in Remark 4.4 below.)
Using again the symmetry property E(u − ,u + )=−E(u + ,u − )wehave
(
∂u− E )( u, ϕ ♭ 0(u) ) = − ( ∂ u+ E )( ϕ ♭ 0(u),u ) . (3.9)
Thus, differentiating the identity E ( u, ϕ ♭ 0(u) ) = 0 we obtain
(
dϕ ♭ 0
du (u) =− ∂u− E )( u, ϕ ♭ 0(u) ) (
(
∂u+ E )( u, ϕ ♭ 0 (u)) = ∂u+ E )( ϕ ♭ 0(u),u )
(
∂u+ E )( u, ϕ ♭ 0 (u)),
where we used (3.9). For u>0 we have already established (see (3.2)) that
(
∂u+ E )( u, ϕ ♭ 0(u) ) < 0,
and since a similar result as (3.2) holds for negative left-hand side and that we have
(ϕ ♭ 0 ◦ ϕ ♭ 0)(u) =u, we conclude that
(
∂u+ E )( ϕ ♭ 0(u),u ) > 0
and therefore dϕ ♭ 0/du < 0.
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