79.3. AN EXAMPLE 2703
Letting t = T, it follows that ψ has a unique progressively measurable fixed point inL2 (Ω;C ([0,T ] ;V )). The fixed point is the limit of the sequence {ψnw} and each func-tion in the sequence is progressively measurable. This is the unique solution to the integralequation.
This is interesting because it is an example of a stochastic equation which is secondorder in time. If one were to include a point mass on the beam at x0, then this would leadto an evolution equation of the form
Bv(t)− v0 +α
∫ t
0Lv(s)ds+
∫ t
0Lwds+
∫ t
0Nwds =
∫ t
0f ds+B
∫ t
0σ (v)dW
where B = I +P with ⟨Pu,v⟩ = u(x0)v(x0). If this were done, you would need to adjustW so that B is the Riesz map from W to W ′. You would use the same kind of fixed pointargument that was just given. If you wished to consider quasistatic motion of the nonlinearbeam in which the acceleration term wtt were neglected, this would involve letting W =V and your operator B in Theorem 79.2.1 would be given by ⟨Bw,u⟩ =
∫ 10 wxxuxx. Thus
Theorem 79.2.1 gives a way to study second order in time equations, and implicit equationsin which the leading coefficient involves a self adjoint operator B. Without the assumptionthat B is one to one, we can give an even more general theorem in case σ does not depend onthe solution, Corollary 79.1.8. This one allows the time differentiated term to even vanishso the differential inclusion could be degenerate. Many other examples are available. Onehas only to consider set valued problems, for example.
Next we consider the elimination of the truncation function. As mentioned, this is easyfor a deterministic equation but not so much for the situation here. Begin with 79.3.36 -79.3.37 and use the Ito formula. Then
|v(t)|2H +2α
∫ t
0∥v∥2
V ds+2∫ t
0⟨Lw,v⟩ds
+2∫ t
0⟨Nw,v⟩ds−
∫ t
0∥σ (v)∥2 ds = 2
∫ t
0⟨ f ,v⟩ds+M (t)
where M (t) is a martingale whose total variation satisfies
[M] (t) =∫ t
0∥σ (v)∥2 |v|2H ds
Then, using estimates and simple manipulations, for M∗ (t)≡ sups≤t |M (s)| ,
12|v(t)|2H +2α
∫ t
0∥v∥2
V ds+∥w(t)∥2V +2
∫ t
0
∫ 1
0(Q(wx)−wx)vxdxds
≤∫ t
0
(C+C |v|2H
)ds+C ( f ,w0,ω)+M∗ (t)
Now let Ψ′ = Q so that Ψ(r)≥ 0 and is quadratic for large |r|. Then the above implies
12|v(t)|2H +2α
∫ t
0∥v∥2
V ds+∥w(t)∥2V +
∫ 1
0Ψ(wx)−|wx (t)|2 dx