2592 CHAPTER 76. IMPLICIT STOCHASTIC EQUATIONS

one end. Suppose for sake of illustration that the left end is clamped, u(0, t) = ux (0, t) = 0,while the right end which has the attached mass is free to move, ux (1, t) = 0, and the beamoccupies the interval [0,1] in material coordinates. Then the stress is σ =−uxxx and balanceof momentum is

utt = σ x + f

where f is a body force. Thus, letting

w ∈V ≡{

w ∈ H2 (0,1) : w(0) = wx (0) = 0,wx (1) = 0}

be a test function,∫ 1

0uttwdx = σw|10 +

∫ 1

0(−σ)wxdx+

∫ 1

0f wdx

= −mutt (1, t)w(1, t)+∫ 1

0uxxxwxdx+

∫ 1

0f wdx

Doing another integration by parts and using the boundary conditions, it follows that anappropriate variational formulation for this problem is∫ 1

0uttwdx+mγ1uttγ1w+

∫ 1

0uxxwxxdx =

∫ 1

0f wdx

where here γ1 is the trace map on the right end.Letting

u(t) = u0 +∫ t

0v(s)ds,

where u(0, t) = u0, we can write the above variational equation in the form

(Bv)′+Au = f , Bv(0) = Bv0

where we assume that v0 ∈W where W is the closure of V in H1 (0,1) and the operatorsare given by

B : W →W ′, ⟨Bu,w⟩ ≡∫ 1

0uwdx+mγ1uγ1w

A : V →V ′, ⟨Au,w⟩ ≡∫ 1

0uxxwxxdx

Thus in terms of an integral equation, this would be of the form

Bv(t)−Bv0 +∫ t

0A(u)ds =

∫ t

0f ds

This suggests a stochastic version of the form

Bv(t)−Bv0 +∫ t

0A(u)ds =

∫ t

0f ds+

∫ t

0ΦdW