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Conformally Invariant Variational Problems. - SAM

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∂<br />

∂x k<br />

(e 2λ δ ij ) = ∂<br />

∂x k<br />

< ∂u<br />

∂x i<br />

, ∂u<br />

∂x j<br />

><br />

= X j ki eλ +X i kj eλ<br />

From this identity we deduce<br />

∀i,j,k ∈ {1...n}<br />

X j ik<br />

= δ ij<br />

∂λ<br />

∂x k<br />

e λ +δ jk<br />

∂λ<br />

∂x i<br />

e λ −δ ik<br />

∂λ<br />

∂x j<br />

e λ<br />

(II.12)<br />

Combining (II.11) and (II.12) for a pair of indices<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

∂ 2 u<br />

∂x k ∂x i<br />

= ∂λ<br />

∂x k<br />

∂ 2 u<br />

∂x i<br />

2 = 2∂λ ∂x i<br />

∂u<br />

+ ∂λ<br />

∂x i ∂x i<br />

n∑<br />

∂u<br />

∂x i<br />

−<br />

j=1<br />

∂u<br />

∂x k<br />

∂λ<br />

∂x j<br />

∂u<br />

∂x j<br />

∀k ≠ i<br />

∀i<br />

(II.13)<br />

Multiplying the first line of (II.13) by e −λ and taking the ∂<br />

∂x j<br />

derivative gives<br />

∂e −λ<br />

∂x j<br />

+ ∂e−λ<br />

∂x k<br />

∂ 2 u<br />

∂x k ∂x i<br />

+e −λ<br />

∂ 2 u<br />

∂x i ∂x j<br />

+ ∂e−λ<br />

∂x i<br />

∂ 3 u<br />

+ ∂2 e −λ<br />

∂x k ∂x i ∂x j ∂x j ∂x k<br />

∂ 2 u<br />

∂x k ∂x j<br />

= 0<br />

∂u<br />

∂x i<br />

(II.14)<br />

9

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