Une page de Wikiversité, la communauté pédagogique libre.
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Opérateurs vectorielsAnalyse vectorielle/Exercices/Opérateurs vectoriels », n'a pu être restituée correctement ci-dessus.
Le gradient d'un champ scalaire
f
{\displaystyle f}
est défini de telle sorte que pour toute variation de coordonnées
d
r
→
{\displaystyle \mathrm {d} {\overrightarrow {r}}}
, on ait :
d
f
=
g
r
a
d
→
f
⋅
d
r
→
{\displaystyle \mathrm {d} f={\overrightarrow {\mathrm {grad} }}f\cdot \mathrm {d} {\overrightarrow {r}}}
.
Exprimer
d
r
→
{\displaystyle \mathrm {d} {\overrightarrow {r}}}
et
d
f
{\displaystyle \mathrm {d} f}
dans un système de coordonnées cartésien. En déduire que le gradient s'écrit :
g
r
a
d
→
f
=
∂
f
∂
x
e
x
→
+
∂
f
∂
y
e
y
→
+
∂
f
∂
z
e
z
→
{\displaystyle {\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
d
r
→
{\displaystyle \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 :
d
r
→
=
d
x
e
x
→
+
d
y
e
y
→
+
d
z
e
z
→
{\displaystyle \mathrm {d} {\overrightarrow {r}}=\mathrm {d} x{\overrightarrow {e_{x}}}+\mathrm {d} y{\overrightarrow {e_{y}}}+\mathrm {d} z{\overrightarrow {e_{z}}}}
;
d
f
=
∂
f
∂
x
d
x
+
∂
f
∂
y
d
y
+
∂
f
∂
z
d
z
{\displaystyle \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 :
g
r
a
d
→
f
=
X
e
x
→
+
Y
e
y
→
+
Z
e
z
→
{\displaystyle {\overrightarrow {\mathrm {grad} }}f=X{\overrightarrow {e_{x}}}+Y{\overrightarrow {e_{y}}}+Z{\overrightarrow {e_{z}}}}
.
Il vient :
d
f
=
g
r
a
d
→
f
⋅
d
r
→
=
X
d
x
+
Y
d
y
+
Z
d
z
{\displaystyle \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 :
g
r
a
d
→
f
=
∂
f
∂
x
e
x
→
+
∂
f
∂
y
e
y
→
+
∂
f
∂
z
e
z
→
{\displaystyle {\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 :
d
r
→
=
d
ρ
e
ρ
→
+
ρ
d
θ
e
θ
→
+
d
z
e
z
→
{\displaystyle \mathrm {d} {\overrightarrow {r}}=\mathrm {d} \rho {\overrightarrow {e_{\rho }}}+\rho \mathrm {d} \theta {\overrightarrow {e_{\theta }}}+\mathrm {d} z{\overrightarrow {e_{z}}}}
;
d
f
=
∂
f
∂
ρ
d
ρ
+
∂
f
∂
θ
d
θ
+
∂
f
∂
z
d
z
{\displaystyle \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
d
f
=
g
r
a
d
→
f
⋅
d
r
→
{\displaystyle \mathrm {d} f={\overrightarrow {\mathrm {grad} }}f\cdot \mathrm {d} {\overrightarrow {r}}}
,
il faut que
g
r
a
d
→
f
=
∂
f
∂
ρ
e
ρ
→
+
1
ρ
∂
f
∂
θ
e
θ
→
+
∂
f
∂
z
e
z
→
{\displaystyle {\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}}}}
.
Démontrer les quatre égalités suivantes :
div
(
∇
→
f
)
=
Δ
f
{\displaystyle {\mbox{div}}({\overrightarrow {\nabla }}f)=\Delta f}
(en prenant comme définition :
Δ
=
∂
2
∂
x
2
+
∂
2
∂
y
2
+
∂
2
∂
z
2
{\displaystyle \Delta ={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}}
) ;
rot
→
(
∇
→
f
)
=
0
→
{\displaystyle {\overrightarrow {\mbox{rot}}}({\overrightarrow {\nabla }}f)={\overrightarrow {0}}}
;
div
(
rot
→
(
F
→
)
)
=
0
{\displaystyle {\mbox{div}}({\overrightarrow {\mbox{rot}}}({\overrightarrow {F}}))=0}
;
rot
→
(
rot
→
(
F
→
)
)
=
∇
(
div
(
F
→
)
)
−
Δ
F
→
{\displaystyle {\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
(
e
x
→
,
e
y
→
,
e
z
→
)
{\displaystyle ({\overrightarrow {e_{x}}},{\overrightarrow {e_{y}}},{\overrightarrow {e_{z}}})}
, que
(
∇
→
∧
v
→
)
∧
v
→
=
(
v
→
⋅
∇
→
)
v
→
−
1
2
∇
→
v
2
{\displaystyle ({\overrightarrow {\nabla }}\wedge {\overrightarrow {v}})\wedge {\overrightarrow {v}}=({\overrightarrow {v}}\cdot {\overrightarrow {\nabla }}){\overrightarrow {v}}-{\frac {1}{2}}{\overrightarrow {\nabla }}v^{2}}
.
Solution question 2
On pose :
v
→
=
v
x
e
x
→
+
v
y
e
y
→
+
v
z
e
z
→
{\displaystyle {\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 :
∇
→
∧
v
→
=
(
∂
v
z
∂
y
−
∂
v
y
∂
z
)
e
x
→
+
(
∂
v
x
∂
z
−
∂
v
z
∂
x
)
e
y
→
+
(
∂
v
y
∂
x
−
∂
v
x
∂
y
)
e
z
→
{\displaystyle {\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 :
(
∇
→
∧
v
→
)
∧
v
→
=
(
(
∂
v
x
∂
z
−
∂
v
z
∂
x
)
v
z
−
(
∂
v
y
∂
x
−
∂
v
x
∂
y
)
v
y
)
e
x
→
+
(
(
∂
v
y
∂
x
−
∂
v
x
∂
y
)
v
x
−
(
∂
v
z
∂
y
−
∂
v
y
∂
z
)
v
z
)
e
y
→
+
(
(
∂
v
z
∂
y
−
∂
v
y
∂
z
)
v
y
−
(
∂
v
x
∂
z
−
∂
v
z
∂
x
)
v
x
)
e
z
→
{\displaystyle ({\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,
(
v
→
⋅
∇
→
)
⋅
v
→
=
(
v
x
∂
v
x
∂
x
+
v
y
∂
v
x
∂
y
+
v
z
∂
v
x
∂
z
)
e
x
→
+
(
v
x
∂
v
y
∂
x
+
v
y
∂
v
y
∂
y
+
v
z
∂
v
y
∂
z
)
e
y
→
+
(
v
x
∂
v
z
∂
x
+
v
y
∂
v
z
∂
y
+
v
z
∂
v
z
∂
z
)
e
z
→
{\displaystyle ({\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
1
2
∇
→
v
2
=
1
2
(
∂
(
v
x
2
+
v
y
2
+
v
z
2
)
∂
x
e
x
→
+
∂
(
v
x
2
+
v
y
2
+
v
z
2
)
∂
y
e
y
→
+
∂
(
v
x
2
+
v
y
2
+
v
z
2
)
∂
z
e
z
→
)
{\displaystyle {\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
e
x
→
{\displaystyle {\overrightarrow {e_{x}}}}
:
1
2
∂
(
v
x
2
+
v
y
2
+
v
z
2
)
∂
x
=
v
x
∂
v
x
∂
x
+
v
y
∂
v
y
∂
x
+
v
z
∂
v
z
∂
x
{\displaystyle {\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
e
x
→
{\displaystyle {\overrightarrow {e_{x}}}}
, on constate que
(
(
v
→
⋅
∇
→
)
⋅
v
→
−
1
2
∇
→
v
2
)
⋅
e
x
→
=
(
v
x
∂
v
x
∂
x
+
v
y
∂
v
x
∂
y
+
v
z
∂
v
x
∂
z
)
−
(
v
x
∂
v
x
∂
x
+
v
y
∂
v
y
∂
x
+
v
z
∂
v
z
∂
x
)
=
(
∂
v
x
∂
z
−
∂
v
z
∂
x
)
v
z
−
(
∂
v
y
∂
x
−
∂
v
x
∂
y
)
v
y
=
(
(
∇
→
∧
v
→
)
∧
v
→
)
⋅
e
x
→
{\displaystyle {\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.