Conformally Invariant Variational Problems. - SAM
Conformally Invariant Variational Problems. - SAM
Conformally Invariant Variational Problems. - SAM
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order to hope to construct energy controlled liftings. However,<br />
if one removes the requirement to control the energy, every map<br />
⃗n ∈ W 1,2 (D 2 ,Gr m−2 (R m )) admits a W 1,2 lifting. The following<br />
theorem is proved in [He] chapter 5.2.<br />
Theorem X.9. Let⃗n in W 1,2 (D 2 ,Gr m−2 (R m )), then thereexists<br />
⃗e 1 ,⃗e 2 ∈ W 1,2 (D 2 ,S m−1 ) such that<br />
⃗e 1 ∧⃗e 2 = ⋆⃗n .<br />
Wecanthenmakethefollowingobservation. Let⃗n ∈ W 1,2 (D 2 ,S m−1 )<br />
and ⃗e 1 ,⃗e 2 ∈ W 1,2 (D 2 ,S m−1 ) given by the previous theorem.<br />
Consider θ ∈ W 1,2 to be the unique solution of<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
∆θ = −div in D 2<br />
〈<br />
∂θ<br />
∂ν = − ⃗e 1 , ∂⃗e 〉<br />
2<br />
∂ν<br />
then the lifting ⃗ f = ⃗ f 1 +i ⃗ f 2 given by<br />
⃗f 1 +i ⃗ f 2 = e iθ (⃗e 1 +i⃗e 2 ) ,<br />
∫<br />
is Coulomb. Let λ such that −∇ ⊥ λ =< f ⃗ 1 ,∇f ⃗ 2 > and such that<br />
∂D<br />
λ = 0, one easily sees that λ satisfies<br />
2<br />
⎧<br />
⎨ −∆λ =< ∇ ⊥ f1 ⃗,∇f ⃗ 2 > in D 2<br />
(X.153)<br />
⎩<br />
λ = 0 on ∂D 2<br />
Observe that since ∇⃗e 1 is perpendicular to ⃗e 1 and since ∇⃗e 2 is<br />
perpendicular to ⃗e 2 one has<br />
< ∇ ⊥ ⃗ f1 ,∇ ⃗ f 2 >=< π ⃗n (∇ ⊥ ⃗ f1 ),π ⃗n (∇ ⃗ f 2 ) > (X.154)<br />
✷<br />
168