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80 2. Boundary Integral Equations

is satisfied for any choice of a smooth stream function u. One can verify
directly that u satisfies (2.4.1) by taking the curl of the balance of momentum
equation in the Stokes system. We note that in terms of the stream function,
the vorticity is equal to ωk = ∇× u = ∆uk, where k is the unit vector
perpendicular to the (x 1 ,x 2 )− plane of the flow. For the homogeneous Stokes
system, the vorticity is a harmonic function, and as a consequence, u satisfies
the biharmonic equation (2.4.1).
To discuss boundary value problems for the biharmonic equation (2.4.1),
it is best to begin with Green’s formula for the equation in Ω. As is well
known, for fourth-order differential equations, the Green formula generally
varies and depends on the choice of boundary operators, i.e., how to apply
the integration by parts formulae. In order to include boundary conditions
2
arising for the thin plate, we rewrite ∆ u in terms of the Poisson ratio ν in
the form
2
2
2
∂ 2 ∂ u ∂ u ∂ 2 ∂ u ∂ 2 ∂ u ∂ u
2
2
2
∆ u = 2 2 +ν 2 +2(1−ν) + 2 2 +ν 2 .
∂x 1 ∂x 1 ∂x 2 ∂x 1 ∂x 2 ∂x 1 ∂x 2 ∂x 2 ∂x 2 ∂x 1
Now integration by parts leads to the first Green formula in the form

∂v
2
(∆ u) vdx = a(u, v) − Mu + vNu ds , (2.4.2)
∂n
Ω Γ
where the bilinear form a(u, v) is defined by
2 2
2 ∂ u ∂ v

a(u, v):= ν∆u ∆v +(1 − ν) dx ; (2.4.3)
∂x i ∂x j ∂x i ∂x j
i,j=1
Ω
and M and N are differential operators defined by
2
Mu := ν∆u +(1 − ν) (n(z) ·∇ x ) u)| z=x , (2.4.4)
∂ d

Nu := − ∆u +(1 − ν) (n(z) ·∇ x )(t(z) ·∇ x )u(x) (2.4.5)
∂n ds | z=x
where t = n ⊥ is the unit tangent vector, i.e. t 1 = −n 2 ,t 2 = n 1 . Then
∂ u 2 ∂ u ∂ u 2
# 2 2 2 $
Mu = ν∆u +(1 − ν) 2 n +2 n 1 n 2 + 2 n 2 ,
1
∂x 1 ∂x 1 ∂x 2 ∂x 2
2
2
∂ d ∂ u ∂ u ∂ u 2 2
2
Nu = − ∆u +(1 − ν) 2 − 2 n 1 n 2 − (n − n ) .
1
2
∂n ds ∂x ∂x
1 2 ∂x 1 ∂x 2
For the interior boundary value problems for (2.4.1), the starting point is the
representation formula

∂E

u(x)= {E(x, y)Nu(y)+ (x, y) Mu(y)}ds y (2.4.6)
∂n y
Γ

∂u

− { M y E(x, y) + N y E(x, y) u(y)}ds y for x ∈ Ω,
∂n y
Γ