Kenneth Kuttler

2424 CHAPTER 70. MEASURABILITY WITHOUT UNIQUENESS

Each of these integrals on the right side converge to 0 because, from the strong convergenceresults,

F((ukn−g)+

)(∣∣vkT − U̇T ±wT∣∣− ∣∣vkT − U̇T

∣∣)converges in L1

([0,T ] ,L1 (ΓC)

)and so the weak ∗ convergence to 0 of

ψ−h′(1/k)

(∣∣vkT − U̇T∣∣)

implies that these integrals converge to 0. Thus the integral in 70.5.61∫ T

∫ΓC

F((ukn−g)+

)(ψ−h′(1/k)

(∣∣vkT − U̇T∣∣))ξ k ·wT dαdt

is between two sequences each of which converges to 0 so it also converges to 0.To save space, denote by

µ̂ = ν(∣∣vT − U̇T

∣∣)−ψ

Then passing to the limit in this subsequence, we obtain for fixed ω the existence of asolution to the following integral equation.

v(t)−v0 +∫ t

0Mvds+

∫ t

0Auds+

∫ t

0Puds+

∫ t

0γ∗T F((un−g)+

)µ̂ξ ds =

∫ t

0fds

(70.5.62)where

u(t) = u0 +∫ t

0v(s)ds (70.5.63)

and(∣∣vkT − U̇T

∣∣ , µ̂) is contained in the graph of µ a.e. Also for each w ∈ V ,∫ T

∫ΓC

ξ ·wT dαds≤∫ T

∫ΓC

∣∣vkT − U̇T +wT∣∣− ∣∣vkT − U̇T

∣∣dαds (70.5.64)

The remaining issue concerns the existence of a measurable solution. However, thisfollows in the same way as before from the measurable selection theorem, Theorem 70.2.1.From the above reasoning, for fixed ω any sequence has a subsequence which leads to asolution to the integral equation 70.5.62 - 70.5.64 which is continuous into V ′. There isalso an estimate of the right sort for all of the vk. Therefore, from this theorem, there is afunction v(·,ω) in V ′ which is weakly continuous into V ′ and a sequence vk(ω) (·,ω) con-verging to v(·,ω). Then from the above argument, a subsequence converges to a solutionto the integral equation and since both are weakly continuous into V ′, it follows that thesolution to the integral equation equals this measurable function for all t, this for each ω .Thus there is a measurable solution to the stochastic friction problem. The result is statedin the following theorem.

Theorem 70.5.3 For each ω let u0 (ω) ∈V,v0 (ω) ∈H. Let f ∈ V ′. Also assume the gap gand sliding velocity U̇T are F measurable. Then there exists a solution v, to the problemsummarized in 70.5.62 - 70.5.64 for each ω . This solution (t,ω)→ v(t,ω) is measurableinto V ′,H ′ and V ′.

It only remains to check the last claim about measurability into the other spaces. Bydensity of V into H, it follows that H ′ is dense in V ′ and so a simple Pettis theorem argumentimplies right away that ω → v(t,ω) is F measurable into both V and H.

2424 CHAPTER 70. MEASURABILITY WITHOUT UNIQUENESSEach of these integrals on the right side converge to 0 because, from the strong convergenceresults,F ((Win ~8),) (|Yer —Ur wr] ~|ver —Url)converges in L! ({0,7],L'(Ic)) and so the weak * convergence to 0 ofy- Ae jp) (|Ver —Ur|)implies that these integrals converge to 0. Thus the integral in 70.5.61T[ [er (in ~ 8) (Cat (|Yer -Ur|)) €,-wrdadtis between two sequences each of which converges to 0 so it also converges to 0.To save space, denote byfi=v(\vr—Ur|) —Then passing to the limit in this subsequence, we obtain for fixed @ the existence of asolution to the following integral equation.tot t ot tv(t) —vo+ f Mvas-+ [ Auds+ | Puds+ | VF ((Un —8).) gas = | fds0 0 0 00(70.5.62)where :u(t) =u + [ v(s)ds (70.5.63)Joand (\ver —Ur ft) is contained in the graph of ys a.e. Also for each w E ¥,T T . .[lS -wraaas < [O |vir—Ur-+wr|—|ver—Ur|dads ——(70.5.64)0 Tc 0 WalThe remaining issue concerns the existence of a measurable solution. However, thisfollows in the same way as before from the measurable selection theorem, Theorem 70.2.1.From the above reasoning, for fixed @ any sequence has a subsequence which leads to asolution to the integral equation 70.5.62 - 70.5.64 which is continuous into V’. There isalso an estimate of the right sort for all of the v,. Therefore, from this theorem, there is afunction v(-,@) in Y’ which is weakly continuous into V’ and a sequence Vx(q) (-,@) con-verging to v(-,@). Then from the above argument, a subsequence converges to a solutionto the integral equation and since both are weakly continuous into V’, it follows that thesolution to the integral equation equals this measurable function for all t, this for each @.Thus there is a measurable solution to the stochastic friction problem. The result is statedin the following theorem.Theorem 70.5.3 For each @ let up (@) € V, vo (@) € H. Letf € V’. Also assume the gap gand sliding velocity Ur are ¥ measurable. Then there exists a solution v, to the problemsummarized in 70.5.62 - 70.5.64 for each @. This solution (t,@) — v(t,@) is measurableinto V',H’ and V'.It only remains to check the last claim about measurability into the other spaces. Bydensity of V into H, it follows that H’ is dense in V’ and so a simple Pettis theorem argumentimplies right away that @ + v(t, @) is ¥ measurable into both V and H.