Compte tenu de l'expression du tenseur métrique en coordonnées cylindriques,
∇ 2 f = 1 det g ∂ i ( det g g i j ∂ j f ) {\displaystyle \nabla ^{2}f={\frac {1}{\sqrt {\det g}}}\partial _{i}\left({\sqrt {\det g}}\;g^{ij}\partial _{j}f\right)}
s'écrit
∇ 2 f = ( ∂ 2 ∂ r 2 + 1 r ∂ ∂ r + 1 r 2 ∂ 2 ∂ ϕ 2 + ∂ 2 ∂ z 2 ) f {\displaystyle \nabla ^{2}f=\left({\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial }{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)f}
...À RÉDIGER...
, le laplacien
