Analyse vectorielle/Exercices/Opérateurs vectoriels

Gradient cartésien et gradient cylindrique

Le gradient d'un champ scalaire $f$ est défini de telle sorte que pour toute variation de coordonnées $\mathrm {d} {\overrightarrow {r}}$ , on ait :

\mathrm {d} f={\overrightarrow {\mathrm {grad} }}f\cdot \mathrm {d} {\overrightarrow {r}}

.

Exprimer $\mathrm {d} {\overrightarrow {r}}$ et $\mathrm {d} f$ dans un système de coordonnées cartésien. En déduire que le gradient s'écrit :
${\overrightarrow {\mathrm {grad} }}f={\frac {\partial f}{\partial x}}{\overrightarrow {e_{x}}}+{\frac {\partial f}{\partial y}}{\overrightarrow {e_{y}}}+{\frac {\partial f}{\partial z}}{\overrightarrow {e_{z}}}$ .
Exprimer $\mathrm {d} {\overrightarrow {r}}$ dans un système de coordonnées cylindriques. En déduire l’expression du gradient dans ce système.

Solution question 1

Dans un repère cartésien :

\mathrm {d} {\overrightarrow {r}}=\mathrm {d} x{\overrightarrow {e_{x}}}+\mathrm {d} y{\overrightarrow {e_{y}}}+\mathrm {d} z{\overrightarrow {e_{z}}}

;

\mathrm {d} f={\frac {\partial f}{\partial x}}\mathrm {d} x+{\frac {\partial f}{\partial y}}\mathrm {d} y+{\frac {\partial f}{\partial z}}\mathrm {d} z

.

On pose :

{\overrightarrow {\mathrm {grad} }}f=X{\overrightarrow {e_{x}}}+Y{\overrightarrow {e_{y}}}+Z{\overrightarrow {e_{z}}}

.

Il vient :

\mathrm {d} f={\overrightarrow {\mathrm {grad} }}f\cdot \mathrm {d} {\overrightarrow {r}}=X\mathrm {d} x+Y\mathrm {d} y+Z\mathrm {d} z

,

d'où :

{\overrightarrow {\mathrm {grad} }}f={\frac {\partial f}{\partial x}}{\overrightarrow {e_{x}}}+{\frac {\partial f}{\partial y}}{\overrightarrow {e_{y}}}+{\frac {\partial f}{\partial z}}{\overrightarrow {e_{z}}}

.

Solution question 2

Dans un repère cylindrique :

\mathrm {d} {\overrightarrow {r}}=\mathrm {d} \rho {\overrightarrow {e_{\rho }}}+\rho \mathrm {d} \theta {\overrightarrow {e_{\theta }}}+\mathrm {d} z{\overrightarrow {e_{z}}}

;

\mathrm {d} f={\frac {\partial f}{\partial \rho }}\mathrm {d} \rho +{\frac {\partial f}{\partial \theta }}\mathrm {d} \theta +{\frac {\partial f}{\partial z}}\mathrm {d} z

.

Ainsi, pour que

\mathrm {d} f={\overrightarrow {\mathrm {grad} }}f\cdot \mathrm {d} {\overrightarrow {r}}

,

il faut que

{\overrightarrow {\mathrm {grad} }}f={\frac {\partial f}{\partial \rho }}{\overrightarrow {e_{\rho }}}+{\frac {1}{\rho }}{\frac {\partial f}{\partial \theta }}{\overrightarrow {e_{\theta }}}+{\frac {\partial f}{\partial z}}{\overrightarrow {e_{z}}}

.

Composition d'opérateurs

Démontrer les quatre égalités suivantes :
1. ${\mbox{div}}({\overrightarrow {\nabla }}f)=\Delta f$ (en prenant comme définition : $\Delta ={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}$ ) ;
2. ${\overrightarrow {\mbox{rot}}}({\overrightarrow {\nabla }}f)={\overrightarrow {0}}$ ;
3. ${\mbox{div}}({\overrightarrow {\mbox{rot}}}({\overrightarrow {F}}))=0$ ;
4. ${\overrightarrow {\mbox{rot}}}({\overrightarrow {\mbox{rot}}}({\overrightarrow {F}}))=\nabla ({\mbox{div}}({\overrightarrow {F}}))-\Delta {\overrightarrow {F}}$ .
Montrer, en développant les produits vectoriels sur la base $({\overrightarrow {e_{x}}},{\overrightarrow {e_{y}}},{\overrightarrow {e_{z}}})$ , que
$({\overrightarrow {\nabla }}\wedge {\overrightarrow {v}})\wedge {\overrightarrow {v}}=({\overrightarrow {v}}\cdot {\overrightarrow {\nabla }}){\overrightarrow {v}}-{\frac {1}{2}}{\overrightarrow {\nabla }}v^{2}$ .

Solution question 1

${\mbox{div}}({\overrightarrow {\nabla }}f)={\mbox{div}}{\begin{pmatrix}{\frac {\partial f}{\partial x}}\\{\frac {\partial f}{\partial y}}\\{\frac {\partial f}{\partial z}}\end{pmatrix}}={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}=\Delta f$ .
${\overrightarrow {\mbox{rot}}}({\overrightarrow {\nabla }}f)={\overrightarrow {\mbox{rot}}}{\begin{pmatrix}{\frac {\partial f}{\partial x}}\\{\frac {\partial f}{\partial y}}\\{\frac {\partial f}{\partial z}}\end{pmatrix}}={\begin{pmatrix}{\frac {\partial ^{2}f}{\partial y\partial z}}-{\frac {\partial ^{2}f}{\partial z\partial y}}\\{\frac {\partial ^{2}f}{\partial z\partial x}}-{\frac {\partial ^{2}f}{\partial x\partial z}}\\{\frac {\partial ^{2}f}{\partial x\partial y}}-{\frac {\partial ^{2}f}{\partial y\partial x}}\end{pmatrix}}={\overrightarrow {0}}$ d'après le théorème de Schwarz (qui s'applique à $f$ deux fois différentiable).
${\mbox{div}}({\overrightarrow {\mbox{rot}}}({\overrightarrow {F}}))={\frac {\partial }{\partial x}}\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)+{\frac {\partial }{\partial y}}\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)+{\frac {\partial }{\partial z}}\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)=$
${\frac {\partial ^{2}F_{z}}{\partial x\partial y}}-{\frac {\partial ^{2}F_{z}}{\partial y\partial x}}+{\frac {\partial ^{2}F_{y}}{\partial z\partial x}}-{\frac {\partial ^{2}F_{y}}{\partial x\partial z}}+{\frac {\partial ^{2}F_{x}}{\partial y\partial z}}-{\frac {\partial ^{2}F_{x}}{\partial z\partial y}}=0$ d'après le théorème de Schwarz.
${\overrightarrow {\mbox{rot}}}({\overrightarrow {\mbox{rot}}}({\overrightarrow {F}}))={\overrightarrow {\mbox{rot}}}\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}},\;\;\;{\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}},\;\;\;{\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)={\begin{pmatrix}{\frac {\partial }{\partial y}}\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)-{\frac {\partial }{\partial z}}\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\\{\frac {\partial }{\partial z}}\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)-{\frac {\partial }{\partial x}}\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\\{\frac {\partial }{\partial x}}\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)-{\frac {\partial }{\partial y}}\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\end{pmatrix}}$ $={\begin{pmatrix}{\frac {\partial }{\partial x}}\left({\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\right)-{\frac {\partial ^{2}F_{x}}{\partial x^{2}}}-{\frac {\partial ^{2}F_{x}}{\partial y^{2}}}-{\frac {\partial ^{2}F_{x}}{\partial z^{2}}}\\{\frac {\partial }{\partial y}}\left({\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\right)-{\frac {\partial ^{2}F_{y}}{\partial x^{2}}}-{\frac {\partial ^{2}F_{y}}{\partial y^{2}}}-{\frac {\partial ^{2}f_{2}}{\partial z^{2}}}\\{\frac {\partial }{\partial z}}\left({\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\right)-{\frac {\partial ^{2}F_{z}}{\partial x^{2}}}-{\frac {\partial ^{2}F_{z}}{\partial y^{2}}}-{\frac {\partial ^{2}F_{z}}{\partial z^{2}}}\end{pmatrix}}={\begin{pmatrix}{\frac {\partial }{\partial x}}({\mbox{div}}({\overrightarrow {F}}))-\Delta F_{x}\\{\frac {\partial }{\partial y}}({\mbox{div}}({\overrightarrow {F}}))-\Delta F_{y}\\{\frac {\partial }{\partial z}}({\mbox{div}}({\overrightarrow {F}}))-\Delta F_{z}\end{pmatrix}}=\nabla ({\mbox{div}}({\overrightarrow {F}}))-\Delta {\overrightarrow {F}}$ .

Solution question 2

On pose :

{\overrightarrow {v}}=v_{x}{\overrightarrow {e_{x}}}+v_{y}{\overrightarrow {e_{y}}}+v_{z}{\overrightarrow {e_{z}}}

.

Tout d’abord, on calcule le produit vectoriel entre parenthèses :

{\overrightarrow {\nabla }}\wedge {\overrightarrow {v}}=({\frac {\partial v_{z}}{\partial y}}-{\frac {\partial v_{y}}{\partial z}}){\overrightarrow {e_{x}}}+({\frac {\partial v_{x}}{\partial z}}-{\frac {\partial v_{z}}{\partial x}}){\overrightarrow {e_{y}}}+({\frac {\partial v_{y}}{\partial x}}-{\frac {\partial v_{x}}{\partial y}}){\overrightarrow {e_{z}}}

.

Puis le deuxième produit vectoriel :

({\overrightarrow {\nabla }}\wedge {\overrightarrow {v}})\wedge {\overrightarrow {v}}={\Big (}({\frac {\partial v_{x}}{\partial z}}-{\frac {\partial v_{z}}{\partial x}})v_{z}-({\frac {\partial v_{y}}{\partial x}}-{\frac {\partial v_{x}}{\partial y}})v_{y}{\Big )}{\overrightarrow {e_{x}}}+{\Big (}({\frac {\partial v_{y}}{\partial x}}-{\frac {\partial v_{x}}{\partial y}})v_{x}-({\frac {\partial v_{z}}{\partial y}}-{\frac {\partial v_{y}}{\partial z}})v_{z}{\Big )}{\overrightarrow {e_{y}}}+{\Big (}({\frac {\partial v_{z}}{\partial y}}-{\frac {\partial v_{y}}{\partial z}})v_{y}-({\frac {\partial v_{x}}{\partial z}}-{\frac {\partial v_{z}}{\partial x}})v_{x}{\Big )}{\overrightarrow {e_{z}}}

.

Par ailleurs,

({\overrightarrow {v}}\cdot {\overrightarrow {\nabla }})\cdot {\overrightarrow {v}}={\Big (}v_{x}{\frac {\partial v_{x}}{\partial x}}+v_{y}{\frac {\partial v_{x}}{\partial y}}+v_{z}{\frac {\partial v_{x}}{\partial z}}{\Big )}{\overrightarrow {e_{x}}}+{\Big (}v_{x}{\frac {\partial v_{y}}{\partial x}}+v_{y}{\frac {\partial v_{y}}{\partial y}}+v_{z}{\frac {\partial v_{y}}{\partial z}}{\Big )}{\overrightarrow {e_{y}}}+{\Big (}v_{x}{\frac {\partial v_{z}}{\partial x}}+v_{y}{\frac {\partial v_{z}}{\partial y}}+v_{z}{\frac {\partial v_{z}}{\partial z}}{\Big )}{\overrightarrow {e_{z}}}

et

{\frac {1}{2}}{\overrightarrow {\nabla }}v^{2}={\frac {1}{2}}{\Big (}{\frac {\partial (v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{\partial x}}{\overrightarrow {e_{x}}}+{\frac {\partial (v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{\partial y}}{\overrightarrow {e_{y}}}+{\frac {\partial (v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{\partial z}}{\overrightarrow {e_{z}}}{\Big )}

.

Sans perte de généralité, nous allons nous intéresser uniquement aux composantes en ${\overrightarrow {e_{x}}}$ :

{\frac {1}{2}}{\frac {\partial (v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{\partial x}}=v_{x}{\frac {\partial v_{x}}{\partial x}}+v_{y}{\frac {\partial v_{y}}{\partial x}}+v_{z}{\frac {\partial v_{z}}{\partial x}}

.

Toujours en ne s'intéressant qu'aux composantes en ${\overrightarrow {e_{x}}}$ , on constate que

{\Big (}({\overrightarrow {v}}\cdot {\overrightarrow {\nabla }})\cdot {\overrightarrow {v}}-{\frac {1}{2}}{\overrightarrow {\nabla }}v^{2}{\Big )}\cdot {\overrightarrow {e_{x}}}={\Big (}v_{x}{\frac {\partial v_{x}}{\partial x}}+v_{y}{\frac {\partial v_{x}}{\partial y}}+v_{z}{\frac {\partial v_{x}}{\partial z}}{\Big )}-{\Big (}v_{x}{\frac {\partial v_{x}}{\partial x}}+v_{y}{\frac {\partial v_{y}}{\partial x}}+v_{z}{\frac {\partial v_{z}}{\partial x}}{\Big )}=({\frac {\partial v_{x}}{\partial z}}-{\frac {\partial v_{z}}{\partial x}})v_{z}-({\frac {\partial v_{y}}{\partial x}}-{\frac {\partial v_{x}}{\partial y}})v_{y}=(({\overrightarrow {\nabla }}\wedge {\overrightarrow {v}})\wedge {\overrightarrow {v}})\cdot {\overrightarrow {e_{x}}}

.

La démonstration est la même pour les autres composantes.

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