4  Formulaire

4.1 Coordonnées cartésiennes orthonormées

\[\overrightarrow{OM} = x\overrightarrow {e}_x + y\overrightarrow {e}_y + z\overrightarrow {e}_z\]

* Soit f une fonction scalaire, alors

\[\overrightarrow {grad}(f) = \nabla f = \frac{\partial f}{\partial x_i}\;\overrightarrow {e}_i = f_{,i}\;\overrightarrow {e}_i=\left\{ {{\begin{array}{c} {\displaystyle{{\partial f} \over {\partial x}}} \hfill \\ {\displaystyle{{\partial f} \over {\partial y}}} \hfill \\ {\displaystyle{{\partial f} \over {\partial z}}} \hfill \\ \end{array} }} \right\}\]
et
\[\Delta f = div\left( {\nabla f} \right) = \frac{\partial ^2f}{\partial x_j \partial x_j } = f_{,jj} = \frac{\partial ^2f}{\partial x^2} + \frac{\partial ^2f}{\partial y^2} + \frac{\partial ^2f}{\partial z^2}\]

* Soit \(\overrightarrow {v} = v_x \,\overrightarrow {e}_x + v_y \,\overrightarrow {e}_y + v_z \,\overrightarrow {e}_z\) un vecteur, alors

\[\nabla \overrightarrow {v} = \frac{\partial v_i }{\partial x_j }\;\overrightarrow {e}_i \otimes \overrightarrow {e}_j = v_{i,j}\;\overrightarrow {e}_i \otimes \overrightarrow {e}_j =\left[ {{\begin{array}{*{20}c} {\displaystyle{{\partial v_x } \over {\partial x}}} \hfill & {\displaystyle{{\partial v_x } \over {\partial y}}} \hfill & {\displaystyle{{\partial v_x } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial v_y } \over {\partial x}}} \hfill & {\displaystyle{{\partial v_y } \over {\partial y}}} \hfill & {\displaystyle{{\partial v_y } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial v_z } \over {\partial x}}} \hfill & {\displaystyle{{\partial v_z } \over {\partial y}}} \hfill & {\displaystyle{{\partial v_z } \over {\partial z}}} \hfill \\ \end{array} }} \right]\]

et

\[div \;\overrightarrow {v}= v_{i,i} = Tr\left( {\nabla \overrightarrow {v}} \right) = \nabla\overrightarrow{v}:\overline{\overline {I}}=\frac{\partial v_x }{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z }{\partial z}\]

\[\Delta \, \overrightarrow {v} = div\left( {\nabla \overrightarrow {v} } \right) = \frac{\partial ^2 v_i }{\partial x_j \partial x_j }\; \overrightarrow {e}_i = v_{i,jj}\; \overrightarrow {e}_i = \Delta v_x \; \overrightarrow {e}_x + \Delta v_y \; \overrightarrow {e}_y + \Delta v_z \; \overrightarrow {e}_z\]

\[rot\overrightarrow {v} = \nabla \wedge \overrightarrow {v} = \left\{ {{\begin{array}{*{20}c} {\displaystyle{{\partial v_z } \over {\partial y}} - \displaystyle{{\partial v_y } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial v_x } \over {\partial z}} - \displaystyle{{\partial v_z } \over {\partial x}}} \hfill \\ {\displaystyle{{\partial v_y } \over {\partial x}} - \displaystyle{{\partial v_x } \over {\partial y}}} \hfill \\ \end{array} }} \right\} = \varepsilon_{ijk} v_{k,j} \overrightarrow {e}_i\]

* Soit \(\overline{\overline T} = T_{ij} \;\overrightarrow {e}_i \otimes \overrightarrow {e}_j = \left[ {{\begin{array}{*{20}c} {T_{xx} } \hfill & {T_{xy} } \hfill & {T_{xz} } \hfill \\ {T_{yx} } \hfill & {T_{yy} } \hfill & {T_{yz} } \hfill \\ {T_{zx} } \hfill & {T_{zy} } \hfill & {T_{zz} } \hfill \\ \end{array} }} \right]\) un tenseur du deuxième ordre, alors:

\[div(\overline{\overline T} ) = \frac{\partial T_{ij} }{\partial x_j }\;\overrightarrow {e}_i = T_{ij,j}\;\overrightarrow {e}_i = \left\{ {{\begin{array}{*{20}c} {\displaystyle{{\partial T_{xx} } \over {\partial x}} + \displaystyle{{\partial T_{xy} } \over {\partial y}} + \displaystyle{{\partial T_{xz} } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial T_{yx} } \over {\partial x}} + \displaystyle{{\partial T_{yy} } \over {\partial y}} + \displaystyle{{\partial T_{yz} } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial T_{zx} } \over {\partial x}} + \displaystyle{{\partial T_{zy} } \over {\partial y}} + \displaystyle{{\partial T_{zz} } \over {\partial z}}} \hfill \\ \end{array} }} \right\}\]
et \[\Delta \overline {\overline {T}} = \frac{\partial^2 T_{ij}}{\partial x_k \partial x_k } \; \overrightarrow {e}_i \otimes \overrightarrow {e}_j = T_{ij,kk} \;\overrightarrow {e}_i \otimes \overrightarrow {e}_j = \left[ {{\begin{array}{ccc} \Delta T_{xx} & \Delta T_{xy} & \Delta T_{xz} \\ \Delta T_{yx} & \Delta T_{yy} & \Delta T_{yz} \\ \Delta T_{zx} & \Delta T_{zy} & \Delta T_{zz} \\ \end{array} }} \right]\]

4.2 Coordonnées cylindriques

\[\overrightarrow{OM} = r\overrightarrow {e}_r + z\overrightarrow {e}_z \quad \mbox{et}\quad \frac{\partial \overrightarrow{OM} }{\partial r} = \overrightarrow {e}_r \;,\quad \frac{1}{r}\frac{\partial \overrightarrow{OM} }{\partial \theta } = \overrightarrow {e}_\theta \;,\quad \frac{\partial \overrightarrow{OM} }{\partial z} = \overrightarrow {e}_z\]

\[d\overrightarrow{OM} = \overrightarrow {e}_r \, dr + r \, d\theta \; \overrightarrow {e}_\theta + \overrightarrow {e}_z \,dz\]

\[{\begin{array}{*{20}c} {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial r} = 0} & , & {\displaystyle\frac{\partial \overrightarrow{e}_\theta }{\partial r} = 0} & , & {\displaystyle\frac{\partial\overrightarrow {e}_z }{\partial r}= 0} \\ {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial \theta } = \overrightarrow {e}_\theta } & , & {\displaystyle\frac{\partial \overrightarrow {e}_\theta }{\partial \theta } = -\overrightarrow {e}_r } & , & {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial \theta } = 0} \\ {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial z} = 0} & , & {\displaystyle\frac{\partial \overrightarrow{e}_\theta }{\partial z} = 0} & , & {\displaystyle\frac{\partial \overrightarrow {e}_z }{\partial z} = 0} \\ \end{array} }\]

* Soit f une fonction scalaire, alors \[\overrightarrow {grad}(f) = \nabla f = \frac{\partial f}{\partial r}\; \overrightarrow {e}_r+ \frac{1}{r}\frac{\partial f} {\partial \theta}\;\overrightarrow {e}_\theta+ \frac{\partial f} {\partial z}\;\overrightarrow {e}_z\]

et

\[\Delta f = div\left( \nabla f \right) = \frac{\partial ^2f}{\partial r^2} + \frac{1}{r}\frac{\partial f}{\partial r} +\frac{1}{r^2}\frac{\partial ^2f}{\partial \theta^2} + \frac{\partial ^2f}{\partial z^2}\]

* Soit \(\overrightarrow {v} = v_r \, \overrightarrow {e}_r + v_\theta \, \overrightarrow {e}_\theta + v_z \, \overrightarrow {e}_z\) un vecteur, alors

\[\nabla(\overrightarrow {v}) = \left[ {{\begin{array}{*{20}c} {\displaystyle{{\partial v_r } \over {\partial r}}} \hfill & {\displaystyle{1\over r}\left( {\displaystyle{{\partial v_r } \over {\partial \theta }} - v_\theta } \right)} \hfill & {\displaystyle{{\partial v_r } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial v_\theta } \over {\partial r}}} \hfill & {\displaystyle{1 \over r}\left( {\displaystyle{{\partial v_\theta } \over {\partial \theta }} + v_r } \right)} \hfill & {\displaystyle{{\partial v_\theta } \over {\partial z}}} \hfill \\ {\displaystyle{{\partial v_z } \over {\partial r}}} \hfill & {\displaystyle{1 \over r}\displaystyle{{\partial v_z } \over {\partial \theta }}} \hfill & {\displaystyle{{\partial v_z } \over {\partial z}}} \hfill \\ \end{array} }} \right]\]

et

\[div\overrightarrow {v} = Tr\left({\nabla\overrightarrow {v}} \right) = \nabla\overrightarrow{v}:\overline{\overline{I}} =\frac{v_r}{r}+\frac{\partial v_r }{\partial r} + \frac{1}{r}\frac{\partial v_\theta }{\partial \theta} + \frac{\partial v_z }{\partial z}\]

\[\Delta \overrightarrow {v} = div\left( {\nabla\overrightarrow {v}} \right) = \left({\Delta v_r - \frac{2}{r^2}\frac{\partial v_\theta }{\partial \theta } - \frac{v_r }{r^2}} \right)\overrightarrow {e}_r +\left({\Delta v_\theta + \frac{2}{r^2}\frac{\partial v_r }{\partial \theta } -\frac{v_\theta }{r^2}} \right)\overrightarrow {e}_\theta + \Delta v_z \overrightarrow {e}_z\]

\[rot\overrightarrow {v} = \nabla \wedge \overrightarrow {v} = \left\{ \begin{array}{c} \displaystyle\frac{1}{r} \displaystyle\frac{\partial v_z }{\partial \theta} - \displaystyle\frac{\partial v_{\theta}}{\partial z} \\ \displaystyle\frac{\partial v_r }{\partial z} - \displaystyle\frac{\partial v_z }{\partial r} \\ \displaystyle\frac{\partial v_{\theta}}{\partial r} + \displaystyle\frac{v_{\theta}}{r} - \displaystyle\frac{1}{r} \displaystyle\frac{\partial v_r }{\partial \theta} \end{array} \right\}\]

* Soit \(\overline{\overline T} = \left[ {{\begin{array}{*{20}c} {T_{rr} } \hfill & {T_{r\theta} } \hfill & {T_{rz} } \hfill \\ {T_{\theta r} } \hfill & {T_{\theta \theta} } \hfill & {T_{\theta z} } \hfill \\ {T_{zr} } \hfill & {T_{z\theta} } \hfill & {T_{zz} } \hfill \\ \end{array} }} \right]\) un tenseur du deuxième ordre, alors: \[div(\overline{\overline T} ) = \left\{ {{\begin{array}{c} {\displaystyle\frac{\partial T_{rr}}{\partial r} +\frac {1}{r} \displaystyle\frac{\partial T_{r\theta}}{\partial \theta} + \displaystyle\frac {\partial T_{rz}}{\partial z}+\displaystyle\frac{T_{rr}-T_{\theta \theta}}{r}} \\ {\displaystyle\frac{\partial T_{\theta r}} {\partial r} + \displaystyle\frac {1}{r}\frac{\partial T_{\theta \theta}}{\partial \theta} + \displaystyle\frac{\partial T_{\theta z}}{\partial z} +\displaystyle\frac{T_{r\theta}}{r}+\displaystyle\frac{T_{\theta r}}{r}}\\ {\displaystyle\frac{\partial T_{zr}}{\partial r} + \displaystyle\frac {1}{r}\displaystyle\frac{\partial T_{z\theta}}{\partial \theta} + \displaystyle\frac{\partial T_{zz} }{\partial z} + \displaystyle\frac{T_{zr}}{r}} \\ \end{array} }} \right\}\]

4.3 Coordonnées sphériques

\[\overrightarrow{OM} = r\overrightarrow {e}_r \quad \mbox{et}\quad \frac{\partial \overrightarrow{OM} }{\partial r} = \overrightarrow {e}_r \;,\quad \frac{1}{r}\frac{\partial \overrightarrow{OM} }{\partial \theta } = \overrightarrow {e}_\theta \;,\quad \frac{1}{r\sin\theta}\;\frac{\partial \overrightarrow{OM} }{\partial \varphi} = \overrightarrow {e}_\varphi\]

\[d\overrightarrow{OM} = dr \, \overrightarrow {e}_r + \, r \,d\theta \, \overrightarrow {e}_\theta + \,r\, \sin\theta\;d\varphi\;\overrightarrow {e}_\varphi\]

\[{\begin{array}{*{20}c} {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial r} = 0} & , & {\displaystyle\frac{\partial \overrightarrow {e}_\theta }{\partial r} = 0} & , & {\displaystyle\frac{\partial \overrightarrow {e}_\varphi }{\partial r}= 0} \\ {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial \theta } = \overrightarrow {e}_\theta } & , & {\displaystyle\frac{\partial \overrightarrow {e}_\theta }{\partial \theta } = -\overrightarrow {e}_r } & , & {\displaystyle\frac{\partial \overrightarrow {e}_\varphi }{\partial \theta } = 0} \\ {\displaystyle\frac{\partial \overrightarrow {e}_r }{\partial \varphi} = \sin\theta \; \overrightarrow{e}_\varphi} & , & {\displaystyle\frac{\partial \overrightarrow{e}_\theta }{\partial \varphi} = \cos\theta \; \overrightarrow{e}_\varphi} & , & {\displaystyle\frac{\partial \overrightarrow{e}_\varphi }{\partial \varphi} = \sin\theta \; \overrightarrow{e}_r} - \cos\theta \; \overrightarrow{e}_\theta \\ \end{array} }\]

* Soit f une fonction scalaire, alors \[\overrightarrow {grad}(f) = \nabla f = \left\{ {{\begin{array}{*{20}c} {\displaystyle\frac{\partial f}{\partial r}} \\ {\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta }} \\ {\displaystyle\frac{1} {r\sin \theta }\displaystyle\frac{\partial f}{\partial \varphi }} \\ \end{array} }} \right\}\]
et

\[\Delta f = div\left( {\nabla f} \right) = \frac{\partial^2f}{\partial r^2} + \frac{2}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial ^2f}{\partial\theta ^2} + \frac{1}{r^2} \cot\theta \;\frac{\partial f}{\partial\theta } + \frac{1}{r^2\sin ^2\theta }\frac{\partial ^2f}{\partial\varphi ^2}\]

* Soit \(\overrightarrow {v} = v_r \,\overrightarrow {e}_r + v_\theta \,\overrightarrow {e}_\theta + v_\varphi \,\overrightarrow {e}_\varphi\) un vecteur, alors

\[\nabla\overrightarrow {v} = \left[ {{\begin{array}{*{20}c} {\displaystyle\frac {\partial v_r }{\partial r}} \hfill & {\displaystyle\frac {1}{r} \left( { \displaystyle\frac{\partial v_r }{\partial \theta} - v_\theta } \right)} \hfill & {\displaystyle\frac {1}{r}\left( {\displaystyle\frac {1}{\sin\theta} \frac {\partial v_r}{\partial \varphi}-v_\varphi} \right )} \hfill \\ {\displaystyle\frac {\partial v_\theta } {\partial r}} \hfill & {\displaystyle\frac{1}{r} \left( {\displaystyle\frac{\partial v_\theta } {\partial \theta } + v_r } \right)} \hfill & {\displaystyle\frac {1}{r}\left( {\displaystyle\frac {1}{\sin\theta} \displaystyle\frac {\partial v_\theta} {\partial \varphi}- \cot\theta \; v_\varphi } \right )} \hfill \\ {\displaystyle\frac {\partial v_\varphi } {\partial r}} \hfill & {\displaystyle\frac {1}{r} \frac {\partial v_\varphi } {\partial \theta }}\hfill & {\displaystyle\frac {1}{r} \left( {\displaystyle\frac {1}{\sin\theta} \displaystyle\frac {\partial v_\varphi } {\partial \varphi}+ \cot\theta \; v_\theta +v_r} \right )} \hfill \\ \end{array} }} \right]\]

et

\[div\; \overrightarrow {v} = \nabla \overrightarrow {v}:\overline{\overline I} = \frac{\partial v_r }{\partial r} + 2\frac{v_r }{r} + \frac{1}{r}\frac{\partial v_\theta }{\partial \theta } + \frac{1}{r \sin \theta }\frac{\partial v_\varphi }{\partial \varphi } + \cot \theta \frac{v_\theta }{r}\]

\[\Delta \overrightarrow {v} = div\left( {\nabla\overrightarrow {v}} \right) = \left\{ {{\begin{array}{*{20}c} {\Delta v_r - \displaystyle\frac{2}{r^2}\left( {v_r + \displaystyle\frac{1}{\sin \theta }\displaystyle\frac{\partial (\sin \theta v_\theta )}{\partial \theta } + \displaystyle\frac{1}{\sin \theta }\displaystyle\frac{\partial v_\varphi }{\partial \varphi }} \right)} \hfill \\ {\Delta v_\theta + \displaystyle\frac{2}{r^2}\left( {\displaystyle\frac{\partial v_r }{\partial \theta } - \displaystyle\frac{v_\theta }{2\sin ^2\theta } - \displaystyle\frac{\cos \theta }{\sin ^2\theta }\displaystyle\frac{\partial v_\varphi }{\partial \varphi }} \right)} \hfill \\ {\Delta v_\varphi + \displaystyle\frac{2}{r^2\sin \theta }\left( {\displaystyle\frac{\partial v_r }{\partial \varphi } + \cot\theta \,\displaystyle\frac{\partial v_\theta }{\partial \varphi } - \displaystyle\frac{v_\varphi }{2\sin \theta }} \right)} \hfill \\ \end{array} }} \right\}\]

\[rot\overrightarrow {v} = \nabla \wedge \overrightarrow {v} = \left\{ \begin{array}{c} \displaystyle\frac{1}{r\,\sin\theta} \left( \displaystyle\frac{\partial (\sin\theta \, v_{\varphi}) }{\partial \theta} - \displaystyle\frac{\partial v_{\theta}}{\partial \varphi} \right) \\ \displaystyle\frac{1}{r\,\sin\theta} \left( \displaystyle\frac{\partial v_r }{\partial \varphi} - \displaystyle\frac{\partial (r \, \sin\theta \, v_{\varphi}) }{\partial r} \right) \\ \displaystyle\frac{1}{r} \left( \displaystyle\frac{\partial (r\, v_{\theta})}{\partial r} - \displaystyle\frac{\partial v_r }{\partial \theta} \right) \end{array} \right\}\]

* Soit \(\overline{\overline T} = \left[ {{\begin{array}{*{20}c} {T_{rr} } \hfill & {T_{r\theta} } \hfill & {T_{r\varphi} } \hfill \\ {T_{\theta r} } \hfill & {T_{\theta \theta} } \hfill & {T_{\theta \varphi} } \hfill \\ {T_{\varphi r} } \hfill & {T_{\varphi \theta} } \hfill & {T_{\varphi \varphi} } \hfill \\ \end{array} }} \right]\) un tenseur symétrique du deuxième ordre, alors:

\[div(\overline{\overline T} ) = \left\{ {{\begin{array}{*{20}c} {\displaystyle{{\partial T_{rr} } \over {\partial r}} + \frac{1}{r}\displaystyle{{\partial T_{r\theta } } \over {\partial \theta }} + \frac{1}{r\sin \theta }\displaystyle{{\partial T_{r\varphi } } \over {\partial \varphi }} + \frac{1}{r}\left( {2T_{rr} - T_{\theta \theta } - T_{\varphi \varphi } + T_{r\theta } \, \cot\theta } \right)} \hfill \\ {\displaystyle{{\partial T_{\theta r} } \over {\partial r}} + \frac{1}{r}\displaystyle{{\partial T_{\theta \theta } } \over {\partial \theta }} + \frac{1}{r\sin \theta }\displaystyle{{\partial T_{\theta \varphi } } \over {\partial \varphi }} + \frac{1}{r}\left( {(T_{\theta \theta} - T_{\varphi \varphi } )\cot\theta + 3T_{r \theta } } \right)} \hfill \\ {\displaystyle{{\partial T_{\varphi r} } \over {\partial r}} + \frac{1}{r}\displaystyle{{\partial T_{\varphi \theta } } \over {\partial \theta }} + \frac{1}{r\sin \theta }\displaystyle{{\partial T_{\varphi \varphi } } \over {\partial \varphi }} + \frac{1}{r}\left( {2T_{\theta \varphi } \, \cot\theta + 3T_{r\varphi } } \right)} \hfill \\ \end{array} }} \right\}\]

4.4 Comment retrouver les formules

Nous nous plaçons par exemple en coordonnées cylindriques. On note

\(\overrightarrow {v} = v_r \overrightarrow {e}_r + v_\theta \overrightarrow {e}_\theta + v_z \overrightarrow {e}_z =v_i \overrightarrow{e}_i\) avec i = r, \(\theta\), z et \(,i=\displaystyle\frac{\partial}{\partial r}\;,\;\;\displaystyle\frac{1}{r}\frac{\partial}{\partial \theta}\;,\;\;\displaystyle\frac{\partial}{\partial z}\)

Donc, avec cette convention

\[\overrightarrow {e}_{r,\theta } = \frac{\overrightarrow {e}_\theta }{r} \;\;\text{et}\;\; \overrightarrow {e}_{\theta ,\theta } = - \frac{\overrightarrow {e}_r }{r}\]

Chercher le gradient d’un tenseur consiste à augmenter l’ordre de ce tenseur, donc en appliquant Equation 3.2, on obtient: \[\nabla \overrightarrow v = (v_i \; \overrightarrow {e}_i )_{,j} \otimes \overrightarrow {e}_j \;.\]

On peut alors développer, mais sans oublier de dériver les vecteurs de base, car nous sommes dans un système de coordonnées cylindrique,

\[\begin{aligned} \nabla\overrightarrow {v} &= v_{i,j} \;\overrightarrow {e}_{i} \otimes \overrightarrow {e}_j + v_i \; \overrightarrow {e}_{i,j} \otimes \overrightarrow {e}_j = v_{i,j} \;\overrightarrow {e}_{i} \otimes \overrightarrow {e}_j + v_i \; \overrightarrow {e}_{i,\theta} \otimes \overrightarrow {e}_\theta \\ &= v_{i,j} \;\overrightarrow {e}_{i} \otimes \overrightarrow {e}_j + v_r \; \overrightarrow {e}_{r,\theta} \otimes \overrightarrow {e}_\theta + v_\theta \; \overrightarrow {e}_{\theta,\theta} \otimes \overrightarrow {e}_\theta \\ &= v_{i,j} \;\overrightarrow{e}_{i} \otimes \overrightarrow {e}_j + \frac {v_r}{r} \; \overrightarrow {e}_{\theta} \otimes \overrightarrow {e}_\theta - \frac {v_\theta}{r} \; \overrightarrow {e}_{r} \otimes \overrightarrow {e}_\theta \end{aligned}\]

Pour obtenir l’opérateur divergence, il suffit de contracter doublement avec le tenseur unité d’ordre 2,comme décrit par Equation 3.1 soit dans le cas d’un vecteur: \[div\overrightarrow {v} = \nabla (\overrightarrow {v} ):\overline{\overline I} = v_{i,i}+\frac {v_r}{r} =\frac{v_r}{r}+\frac{\partial v_r }{\partial r} + \frac{1}{r}\frac{\partial v_\theta }{\partial \theta} + \frac{\partial v_z }{\partial z}\]

et donc l’opérateur Laplacien pour un scalaire \[\Delta \phi= div \left( \nabla \phi \right) = \phi_{,ii}+\frac {\phi_{,r}}{r}=\frac {\partial ^2\phi}{\partial r^2}+\frac {1}{r}\frac {\partial \phi}{\partial r}+\frac {1}{r^2}\frac {\partial ^2 \phi}{\partial \theta^2}+\frac {\partial ^2\phi}{\partial z^2}\] Appliquons maintenant cette méthodologie à un tenseur d’ordre 2.

\[{\begin{array}{*{20}l} {\nabla\overline{\overline T} } &= \left( {T_{ij} \; \overrightarrow {e}_i \otimes \overrightarrow {e}_j } \right)_{,k} \otimes \overrightarrow {e}_k \\ &= T_{ij,k} \,\overrightarrow {e}_i \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_k + T_{ij} \,\overrightarrow {e}_{i,k} \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_k + T_{ij} \,\overrightarrow {e}_i \otimes \overrightarrow {e}_{j,k} \otimes \overrightarrow {e}_k \\ &= T_{ij,k} \,\overrightarrow {e}_i \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_k + T_{ij} \,\overrightarrow {e}_{i,\theta} \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_\theta + T_{ij} \,\overrightarrow {e}_i \otimes \overrightarrow {e}_{j,\theta } \otimes \overrightarrow {e}_\theta \\ &= T_{ij,k} \,\overrightarrow {e}_i \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_k + \displaystyle\frac{T_{rj} }{r}\,\overrightarrow {e}_\theta \otimes \overrightarrow {e}_j \otimes \overrightarrow {e}_\theta - \displaystyle\frac{T_{\theta j} }{r}\,\overrightarrow {e}_r \otimes \overrightarrow {e}_j \otimes\overrightarrow {e}_\theta \\ &+ \displaystyle\frac{T_{ir} }{r}\,\overrightarrow {e}_i \otimes \overrightarrow {e}_\theta \otimes \overrightarrow {e}_\theta - \displaystyle\frac{T_{i\theta } }{r}\,\overrightarrow {e}_i \otimes \overrightarrow {e}_r \otimes \overrightarrow {e}_\theta \\ \end{array} }\]

Pour obtenir la trace de ce tenseur d’ordre 3 on contracte les deux derniers indices:

\[{\begin{array}{*{20}l} {div \overline{\overline T} = \nabla \overline{\overline T} :\overline{\overline I} } &= T_{ij,j} \,\overrightarrow {e}_i + \displaystyle\frac{T_{r\theta } }{r}\,\overrightarrow {e}_\theta - \displaystyle\frac{T_{\theta \theta } }{r}\,\overrightarrow {e}_r + \displaystyle\frac{T_{ir} }{r}\,\overrightarrow {e}_i \\ &= \left( {\displaystyle\frac{\partial T_{rr} }{\partial r} + \displaystyle\frac{1}{r}\displaystyle\frac{\partial T_{r\theta } }{\partial \theta } + \displaystyle\frac{\partial T_{rz} }{\partial z} - \displaystyle\frac{T_{\theta \theta } }{r} + \displaystyle\frac{T_{rr} }{r}} \right)\overrightarrow {e}_r \\ &+ \left( {\displaystyle\frac{\partial T_{\theta r} }{\partial r} + \displaystyle\frac{1}{r}\displaystyle\frac{\partial T_{\theta \theta } }{\partial \theta } + \displaystyle\frac{\partial T_{\theta z} }{\partial z} + \displaystyle\frac{T_{r\theta } }{r} + \displaystyle\frac{T_{\theta r} }{r}} \right)\overrightarrow {e}_\theta \\ &+ \left( {\displaystyle\frac{\partial T_{zr} }{\partial r} + \displaystyle\frac{1}{r}\displaystyle\frac{\partial T_{z\theta } }{\partial \theta } + \displaystyle\frac{\partial T_{zz} }{\partial z} + \displaystyle\frac{T_{zr} }{r}} \right)\overrightarrow {e}_z \\ \end{array} }\] On peut donc maintenant retrouver l’opérateur Laplacien d’un vecteur : \[\begin{aligned} \Delta \overrightarrow {v}&= div \left( \nabla \overrightarrow {v} \right)\\ &= v_{i,jj}+\displaystyle\frac{v_{r,\theta}}{r}\;\overrightarrow {e}_\theta - \displaystyle\frac{v_{\theta,\theta}}{r}\;\overrightarrow {e}_r+ \displaystyle\frac{v_{r,\theta}-\displaystyle\frac{v_\theta}{r}}{r}\overrightarrow {e}_\theta -\displaystyle\frac{v_{\theta,\theta}+\displaystyle\frac{v_r}{r}}{r} \;\overrightarrow {e}_r +\displaystyle\frac{v_{i,r}}{r}\overrightarrow {e}_i\\ &= \left( \Delta v_r-\displaystyle\frac{2}{r^2}\displaystyle\frac{\partial v_\theta}{\partial \theta} - \displaystyle\frac {v_r}{r^2}\right)\overrightarrow {e}_r + \left( \Delta v_\theta + \displaystyle\frac{2}{r^2}\displaystyle\frac{\partial v_r}{\partial \theta}-\displaystyle\frac {v_\theta}{r^2}\right)\;\overrightarrow {e}_\theta +\Delta v_z\overrightarrow {e}_z \end{aligned}\]