22.06.2013 Views

Polyharmonic boundary value problems

Polyharmonic boundary value problems

Polyharmonic boundary value problems

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

2.4 Hilbert space theory 37<br />

Introduction of a suitable Hilbert triple. Divide the <strong>boundary</strong> operators in (2.33)<br />

into two classes. If m j < m we say that the <strong>boundary</strong> operator B j(x;D) is stable<br />

while if m j ≥ m we say that it is natural. Assume that there are p stable <strong>boundary</strong><br />

operators, with p being an integer between 0 and m. If p = 0 all the <strong>boundary</strong> operators<br />

are natural, whereas if p = m all <strong>boundary</strong> operators are stable. In view of<br />

(2.30) the stable operators correspond to indices j ≤ p. Then we define the space<br />

V := {v ∈ H m (Ω); B j(x,D)v = 0 on ∂Ω for j = 1,..., p}. (2.34)<br />

Clearly, if p = 0 we have V = Hm (Ω) while if p = m we have V = Hm 0 (Ω) (provided<br />

the assumption (2.36) below holds). In particular, in the case of Dirichlet <strong>boundary</strong><br />

conditions (2.20) we have<br />

V = H m 0 (Ω),<br />

in the case of Navier <strong>boundary</strong> conditions (2.21) we have<br />

V = H m <br />

ϑ (Ω) := v ∈ H m (Ω); ∆ j v = 0 on ∂Ω for j < m<br />

<br />

, (2.35)<br />

2<br />

in the case of Steklov <strong>boundary</strong> conditions (2.22) we have<br />

V = H 2 ∩ H 1 0 (Ω) = H 2 ϑ (Ω).<br />

In any case, the space V is well-defined since each B j contains trace operators of<br />

maximal order m j < m. Moreover, V is a closed subspace of H m (Ω) which satisfies<br />

H m 0 (Ω) ⊂ V ⊂ Hm (Ω) with continuous embedding. Therefore, V inherits the scalar<br />

product and the Hilbert space structure from H m (Ω). If we put H = L 2 (Ω), then<br />

V ⊂ H ⊂ V ′ forms a Hilbert triple with compact embeddings.<br />

Assumptions on the <strong>boundary</strong> operators. Assume that<br />

and that the orders of the B j’s satisfy<br />

{B j(x;D)} m j=1 forms a normal system (2.36)<br />

mi + m j = 2m − 1 for all i, j = 1,...,m. (2.37)<br />

This assumption is needed since we are not free to choose the orders of the B j’s. For<br />

every k = 0,...,m − 1 there must be exactly one m j in the set {k,2m − k − 1}.<br />

Let p denote the number of stable <strong>boundary</strong> operators. In view of (2.30) we know<br />

that these operators are precisely {B j} p<br />

j=1<br />

and, of course, they also form a normal<br />

system of <strong>boundary</strong> operators. By Proposition 2.12, there exists a family of normal<br />

operators {S j} m j=p+1 such that {B1,...,Bp,Sp+1,...,Sm} forms a Dirichlet system of<br />

order m. We relabel this system and define<br />

{Fj} m j=1 ≡ {B1,...,Bp,Sp+1,...,Sm} (2.38)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!