Polyharmonic boundary value problems

1.4 Paradoxes for the hinged plate 17
An example where two different solutions appear naturally from two straightforward
settings goes as follows. Both fourth order boundary value problems, hinged or
Steklov (1.10) as well as Navier (1.12) boundary conditions, allow a reformulation
as a coupled system, see (1.14) and (1.13), respectively. In the latter case, one tends
to solve by an iteration of the Green operator for the second order Poisson problem.
This approach works fine for bounded smooth domains, but whenever the domain
has a nonconvex corner, one does not necessarily get the solution one is looking for.
Indeed, for the fourth order problem the natural setting for a weak solution to the
Navier boundary value problem would be H2 ∩ H1 0 (Ω). The second Navier boundary
condition ∆u = 0 would follow naturally on smooth boundary parts from the
weak formulation where u satisfies
(∆u∆ϕ − f ϕ) dx = 0 for all ϕ ∈ H 2 ∩ H 1 0 (Ω). (1.27)
Ω
However, for the system in (1.13) the natural setting is that one looks for function
pairs (u,v) ∈ H1 0 (Ω)×H1 0 (Ω). In [320] it is shown that for domains with a reentrant
corner both problems have a unique solution but the solutions u1 to (1.13) and u2
to (1.27) are different. Indeed, there exist a constant c f and a nontrivial biharmonic
function b that satisfies (1.13) with zero Navier boundary condition except in the
corner such that u1 = u2 + c f b. The related problem for domains with edges is considered
in [319]. We refer to Section 2.7 for more details and an explicit example.
1.4.2 The Babuˇska paradox
In the original Babuˇska or polygon-circle paradox one considers problem (1.10) for
f = 1 and when Ω = Pm ⊂ B (m ≥ 3) is the interior of the regular polygon with
corners e 2kπi/m for k ∈ N, namely
∆ 2 u = 1 in Pm,
u = ∆u = 0 on ∂Pm.
If um denotes the solution of this problem extended by 0 in B\Pm, it can be shown
that the sequence (um) converges uniformly to
u∞(x) := 3 1
−
64 16 |x|2 + 1
64 |x|4
which is not the solution to the “limit problem” (where κ = 1), namely
∆ 2u = 1 in B,
u = ∆u − (1 − σ)κ ∂u
∂ν = 0 on ∂B
unless σ = 1, see Figure 1.3.
For more details on this Babuˇska paradox see Section 2.7.