Conformally Invariant Variational Problems. - SAM

well posed problem in W 1,2 ∩L ∞ - due to Wente’s theorem -
⎧
⎨ div(∇εP) = div((id+ε) ∇ ⊥ ξP) in D 2
(IX.48)
⎩
ε = 0 on ∂D 2
Posing∇ ⊥ B := ∇εP−(id+ε)∇ ⊥ ξP equation(IX.45)becomes
equivalent to
−div(A∇u) = ∇ ⊥ B ·∇u . (IX.49)
Trying now to reproduce this procedure in one space dimension
leads to the following. We aim to solve
−u ′′ = Ωu ′
(IX.50)
where u : [0,1] → R m and Ω : [0,1] → so(m). We
then construct a solution P to (IX.47) which in one dimension
becomes
(P ′ P −1 ) ′ = (P ΩP −1 ) ′ .
A special solution is given by P solving
P −1 P ′ = Ω
(IX.51)
Computing now (Pu ′ ) ′ gives the 1-D analogue of (IX.47) which
is
(Pu ′ ) ′ = 0 , (IX.52)
indeed the curl operator in one dimension is trivial, and jacobians
too, since curl-like vector fields corresponds to functions
f satisfying divf = f ′ = 0 and then can be taken equal to zero.
At this stage it is not necessary to go to the ultimate step and
perturb P into (id+ε)P since the r.h.s of (IX.52) is already a
pure 1D jacobian (that is zero !) and since we have succeeded
in writing equation (IX.50) in conservative form.
The reader has then noticed that the 1-D analogue of our
approach to write equation (IX.45) in conservative form is the
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