One has ∂ ∂ ∂ ∂ u 1 2 R 2 u 1 R 2 − − | u | 2 d u 1 = 2 ( R 2 − − | u | 2 ) + 4 R 2 ( u 1 ) 2 ( R 2 − − | u | 2 ) 2 d u 1 , ∂ ∂ ∂ ∂ u 2 2 R 2 u 1 R 2 − − | u | 2 d u 2 = 4 R 2 u 1 u 2 ( R 2 − − | u | 2 ) 2 d u 2 , and ∂ ∂ ∂ ∂ τ τ 2 R 2 u 1 R 2 − − | u | 2 d τ τ 2 = 0.