Weak Convergence Methods for Nonlinear Partial Differential ...

Figure 3.5: Characteristics running into a shock.
Proof. By shifting and adding constants we may without loss of generality assume
that u 2 = 0, u 1 = u ≥ 0 and F(0) = Φ(0) = Ψ(0) = 0. We need to show that
uΨ(u) ≥ F(u)Φ(u) for u ≥ 0.
As both sides are equal to zero for u = 0, this is implied by
Ψ(u) + uΨ ′ (u) ≥ F ′ (u)Φ(u) + F(u)Φ ′ (u), u ≥ 0.
Again both sides are equal to zero for u = 0, so this equation holds true if
2Ψ ′ (u) + uΨ ′′ (u) ≥ F ′′ (u)Φ(u) + 2F ′ (u)Φ ′ (u) + F(u)Φ ′′ (u), u ≥ 0.
Now using that Ψ ′ = F ′ Φ ′ and hence also Ψ ′′ = F ′′ Φ ′ + F ′ Φ ′′ , we arrive at the
equivalent inequalities
u(F ′′ (u)Φ ′ (u) + F ′ (u)Φ ′′ (u)) ≥ F ′′ (u)Φ(u) + F(u)Φ ′′ (u)
⇔ F ′′ (u)(uΦ ′ (u) − Φ(u)) + Φ ′′ (uF ′ (u) − F(u)) ≥ 0
As F ′′ (u), Φ ′′ (u) ≥ 0, this inequality is implied by the inequalities
uΦ ′ (u) − Φ(u) = 0 and uF ′ (u) − F(u) ≥ 0,
which in turn follow from the convexity of Φ and F: They are true for u = 0 and
with f ∈ {Φ, F }:
(uf ′ − f) ′ = f ′ + uf ′′ − f ′ = uf ′′ ≥ 0.
□
Also observe that in general equality does not hold true: For F(y) = 1 2 y2 , Φ(y) =
1
2 y2 we have Ψ(y) = 1 3 y3 and (with u 2 = 0, u = u 1 > 0)
uΨ(u) = 1 3 u4 > 1 4 u4 = 1 2 u2 · 1
2 u2 = F(u)Φ(u)
61