Fourier Spectral Moving Mesh Method for Willmore Equation of ...

Proof:∂J∂ξ l = ∂ ξ l a 1 · (a 2 × a 3 ) =3∑i=1∂a i∂xi l · (a j × a k ) = J3∑i=1a i · ∂a i∂ξ l∂a i∂ξ l = − ∑ s(a i · ∂a s∂ξ l )asProof:and we also havea i · a j = δ j i ⇒ ∂∂ξ l (ai · a j ) = 0 ⇒ ( ∂ai∂ξ l ) · a j = −( ∂a j∂ξ l ) · ai3∑− (a i · ∂a ss=1∂ξ l )as · a j = −( ∂a j∂ξ ) · l ai ,Hence, as long as the Jacobian Matrix is fully ranked, a 1 , a 2 , a 3 are a group of basis in R 3and we can attain the desired equation.∂x∂t = − ∑ ia i∂ξ i∂tProof:∀, −→ x 0 = (x 1 , x 2 , x 3 ), −→ x (ξ( −→ x 0 , t), t) = −→ x 0 ⇒3∑i=1∂x∂ξ i ∂ξ i∂t + ∂x∂t = 0 ⇒ ∂x∂t = − ∑ ia i∂ξ i∂tIt is very easy to verify thatHence∇ x f =3∑i=1a i ∂f∂ξ i∆ 2 xf = ∇ x · ∇f = ∇ x · (= 1 J ∇ ξ · (JA∇ ξ f)3∑j=1a i ∂f∂ξ i ) = 1 J3∑ ∂3∑Ja j · (j=1∂ξ j j=1a i ∂f∂ξ i )16