70.5. A FRICTION CONTACT PROBLEM 2423
Therefore, for a.e.t, ψ (t,x,ω) is in the subgradient of the function φ η (r) = |ηr| for a.e.x∈ΓC at the point r =
∣∣vkT − U̇T∣∣. In particular, ψ ∈ [−η ,η ] so that ν
(∣∣vkT − U̇T∣∣)−ψ is
between µs (0) and µ0 if∣∣vkT − U̇T
∣∣ = 0. If this quantity is positive, then ψ = η andν(∣∣vkT − U̇T
∣∣)−ψ reduces to µs(∣∣vkT − U̇T
∣∣) . Thus(∣∣vkT − U̇T∣∣ ,ν (∣∣vkT − U̇T
∣∣)−ψ)
is in the graph of µ a.e. Similar reasoning based on strong convergence and 70.5.59 impliesthat for a.e.t, ξ ∈ ∂γ where γ (y) = |y| at the point vkT − U̇T for a.e.x ∈ ΓC.
Consider the friction terms in 70.5.58. Letting w ∈V and recalling that µ(1/k) (r) =ν (r)−h′(1/k) (r) , ∫ T
0
∫ΓC
F((ukn−g)+
)µ(1/k)
(∣∣vkT − U̇T∣∣)ξ k ·wT dαdt
=∫ T
0
∫ΓC
F((ukn−g)+
)(ν(∣∣vkT − U̇T
∣∣)−h′(1/k)
(∣∣vkT − U̇T∣∣))ξ k ·wT dαdt
=∫ T
0
∫ΓC
F((ukn−g)+
)(ν(∣∣vkT − U̇T
∣∣)−ψ)
ξ k ·wT dαdt (70.5.60)
+∫ T
0
∫ΓC
F((ukn−g)+
)(ψ−h′(1/k)
(∣∣vkT − U̇T∣∣))ξ k ·wT dαdt (70.5.61)
Now consider the first integral. The strong convergence yields that this integral in70.5.60 converges to∫ T
0
∫ΓC
F((un−g)+
)(ν(∣∣vT − U̇T
∣∣)−ψ)
ξ ·wT dαdt
where ν(∣∣vT − U̇T
∣∣)−ψ is in the graph of µ a.e.Consider the second integral in 70.5.61.∫ T
0
∫ΓC
F((ukn−g)+
)(ψ−h′(1/k)
(∣∣vkT − U̇T∣∣))ξ k ·wT dαdt
≤∫ T
0
∫ΓC
F((ukn−g)+
)(ψ−h′(1/k)
(∣∣vkT − U̇T∣∣)) ·(∣∣vkT − U̇T +wT
∣∣− ∣∣vkT − U̇T∣∣)dαdt
Similarly,
−∫ T
0
∫ΓC
F((ukn−g)+
)(ψ−h′(1/k)
(∣∣vkT − U̇T∣∣))ξ k ·wT dαdt
≤∫ T
0
∫ΓC
F((ukn−g)+
)(ψ−h′(1/k)
(∣∣vkT − U̇T∣∣)) ·(∣∣vkT − U̇T −wT
∣∣− ∣∣vkT − U̇T∣∣)dαdt