79.3. AN EXAMPLE 2705

Thus, letting t = T,

E(∥v∥2

C([0,T ];H)

)+E

(∥w∥2

C([0,T ];V )

)+E

(α

∫ T

0∥v∥2

V ds)≤C ( f ,w0,T ) (79.3.40)

Now let wn,vn correspond in the above to qn where qn (r) = r2 for |r| ≤ 2n and qn (r) =4n for |r| ≥ 2n. Let Nn be the operator resulting from qn and let N be the operator definedby

⟨Nw,u⟩ ≡∫ 1

0

(w3

x

3−wx

)uxdx

By embedding theorems, ∥wnx∥C([0,1]) ≤C∥wxx∥V . We define stopping times.

τn (ω)≡ inf{

t ∈ [0,T ] : C∥wnxx (·,ω)∥C([0,t]) > 2n}

inf( /0) defined as ∞. Then

vτnn (t)− v0 +α

∫ t

0X[0,τn] (s)Lvn (s)ds+

∫ t

0X[0,τn] (s)Lwnds

+∫ t

0X[0,τn] (s)Nnwnds =

∫ t

0X[0,τn] (s) f ds+

∫ t

0X[0,τn] (s)σ (v)dW

Does limn→∞ τn = ∞? If not so for some ω, then there is a subsequence still denoted withn such that for all n, τn (ω)≤ T . This implies

C2 ∥wnxx (·,ω)∥2C([0,T ]) > 4n

but the set of ω for which the above holds has measure no more than C2C ( f ,w0,T )/4n

thanks to 79.3.40 and so there is a further subsequence and a set of measure zero such thatfor ω not in this set, eventually, for all n large enough,

C∥wnxx (·,ω)∥C([0,T ]) ≤ 2n,

which requires τn =∞, contrary to the construction of this subsequence. Thus τn convergesto ∞ off a set of measure zero. Using uniqueness, define v = vn whenever τn = ∞ and w =wn whenever τn = ∞. Then from the embedding theorem mentioned above, wx (t,ω)2 =qn (wx (t,ω)) and so

v(t)− v0 +α

∫ t

0Lv(s)ds+

∫ t

0Lwds

+∫ t

0Nwds =

∫ t

0f ds+

∫ t

0σ (v)dW (79.3.41)

where

⟨Nw,u⟩=∫ 1

0

((w3

x

3

)−wx

)uxdx

Of course it would be very interesting in this example to see if you can pass to a limit asα→ 0. We do this in the non probabilistic version of this problem quite easily, but whetherit can be done in the stochastic equations considered here is not clear.