En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : DifférentiabilitéCalcul différentiel/Exercices/Différentiabilité », n'a pu être restituée correctement ci-dessus.
On se propose d'étendre aux espaces vectoriels normés le théorème « limite de la dérivée » , et de le renforcer en allégeant ses hypothèses. Soient
E
{\displaystyle E}
F
{\displaystyle F}
U
{\displaystyle U}
E
{\displaystyle E}
a
{\displaystyle a}
U
{\displaystyle U}
Soit
f
:
U
→
F
{\displaystyle f:U\to F}
a
{\displaystyle a}
U
∖
{
a
}
{\displaystyle U\setminus \{a\}}
a
{\displaystyle a}
lim
x
→
a
d
f
x
=
L
∈
L
(
E
,
F
)
{\displaystyle \lim _{x\to a}\mathrm {d} f_{x}=L\in {\mathcal {L}}\left(E,F\right)}
f
{\displaystyle f}
a
{\displaystyle a}
d
f
a
=
L
{\displaystyle \mathrm {d} f_{a}=L}
Application : montrer que la fonction
f
:
R
2
→
R
(
x
,
y
)
↦
{
x
y
ln
(
x
2
+
y
2
)
si
(
x
,
y
)
≠
(
0
,
0
)
0
sinon
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} \;(x,y)\mapsto {\begin{cases}xy\ln \left(x^{2}+y^{2}\right)&{\text{si }}(x,y)\neq (0,0)\\0&{\text{sinon}}\end{cases}}}
1 .
Soit
g
:
U
∖
{
a
}
→
F
{\displaystyle g:U\setminus \{a\}\to F}
a
{\displaystyle a}
critère de Cauchy pour une fonction , démontrer que si
F
{\displaystyle F}
dim
E
>
1
{\displaystyle \dim E>1}
g
{\displaystyle g}
a
{\displaystyle a}
1 en ce point).
Quelle variante de la question 3 peut-on énoncer si
E
=
R
{\displaystyle E=\mathbb {R} }
Solution
1°) Posons, pour tout
h
∈
E
{\displaystyle h\in E}
a
+
h
∈
U
{\displaystyle a+h\in U}
g
(
h
)
:=
f
(
a
+
h
)
−
L
(
h
)
{\displaystyle g\left(h\right):=f\left(a+h\right)-L\left(h\right)}
Alors,
g
{\displaystyle g}
0
{\displaystyle 0}
h
≠
0
{\displaystyle h\neq 0}
d
g
h
=
d
f
a
+
h
−
L
{\displaystyle \mathrm {d} g_{h}=\mathrm {d} f_{a+h}-L}
ε
:
h
↦
sup
x
∈
]
a
,
a
+
h
[
|
|
|
d
g
x
|
|
|
{\displaystyle \varepsilon :h\mapsto \sup _{x\in \left]a,a+h\right[}|\!|\!|\mathrm {d} g_{x}|\!|\!|}
(définie pour
h
≠
0
{\displaystyle h\neq 0}
lim
h
→
0
ε
(
h
)
=
0
{\displaystyle \lim _{h\to 0}\varepsilon \left(h\right)=0}
Enfin, d'après l'inégalité des accroissements finis,
‖
f
(
a
+
h
)
−
f
(
a
)
−
L
(
h
)
‖
=
‖
g
(
h
)
−
g
(
0
)
‖
≤
ε
(
h
)
‖
h
‖
{\displaystyle \left\|f\left(a+h\right)-f\left(a\right)-L\left(h\right)\right\|=\left\|g\left(h\right)-g\left(0\right)\right\|\leq \varepsilon \left(h\right)\left\|h\right\|}
ce qui prouve que
d
f
a
=
L
{\displaystyle \mathrm {d} f_{a}=L}
2°)
f
{\displaystyle f}
∞ sur
R
2
∖
{
(
0
,
0
)
}
{\displaystyle \mathbb {R} ^{2}\setminus \{(0,0)\}}
0
{\displaystyle 0}
En tout point
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
∂
f
∂
x
=
y
ln
(
x
2
+
y
2
)
+
2
x
2
y
x
2
+
y
2
{\displaystyle {\frac {\partial f}{\partial x}}=y\ln \left(x^{2}+y^{2}\right)+{\frac {2x^{2}y}{x^{2}+y^{2}}}}
lim
(
x
,
y
)
→
(
0
,
0
)
(
x
,
y
)
≠
(
0
,
0
)
∂
f
∂
x
=
lim
r
→
0
+
2
sin
θ
(
r
ln
r
+
r
cos
2
θ
)
=
0
{\displaystyle \lim _{(x,y)\to (0,0) \atop (x,y)\neq (0,0)}{\frac {\partial f}{\partial x}}=\lim _{r\to 0^{+}}2\sin \theta \left(r\ln r+r\cos ^{2}\theta \right)=0}
lim
(
x
,
y
)
→
(
0
,
0
)
(
x
,
y
)
≠
(
0
,
0
)
∂
f
∂
y
=
0
{\displaystyle \lim _{(x,y)\to (0,0) \atop (x,y)\neq (0,0)}{\frac {\partial f}{\partial y}}=0}
D'après la question 1) ceci prouve que
d
f
(
0
,
0
)
=
0
{\displaystyle \mathrm {d} f_{(0,0)}=0}
f
{\displaystyle f}
1 en
(
0
,
0
)
{\displaystyle (0,0)}
Mais pas deux fois différentiable, car
lim
y
→
0
y
≠
0
1
y
∂
f
∂
x
(
0
,
y
)
=
lim
y
→
0
y
≠
0
ln
(
y
2
)
=
−
∞
{\displaystyle \lim _{y\to 0 \atop y\neq 0}{\frac {1}{y}}{\frac {\partial f}{\partial x}}(0,y)=\lim _{y\to 0 \atop y\neq 0}\ln \left(y^{2}\right)=-\infty }
∂
2
f
∂
y
∂
x
(
0
,
0
)
{\displaystyle {\frac {\partial ^{2}f}{\partial y\partial x}}(0,0)}
∂
2
f
∂
x
∂
y
(
0
,
0
)
{\displaystyle {\frac {\partial ^{2}f}{\partial x\partial y}}(0,0)}
3°) Il s'agit de démontrer que
lim
h
,
k
→
0
‖
g
(
a
+
h
)
−
g
(
a
+
k
)
‖
=
0
{\displaystyle \lim _{h,k\to 0}\left\|g\left(a+h\right)-g\left(a+k\right)\right\|=0}
Puisque
d
g
{\displaystyle \mathrm {d} g}
a
{\displaystyle a}
M
,
r
>
0
{\displaystyle M,r>0}
0
<
‖
h
‖
≤
r
⇒
a
+
h
∈
U
et
|
|
|
d
g
a
+
h
|
|
|
≤
M
{\displaystyle 0<\|h\|\leq r\Rightarrow a+h\in U{\text{ et }}|\!|\!|\mathrm {d} g_{a+h}|\!|\!|\leq M}
Pour
0
<
‖
h
‖
,
‖
k
‖
≤
ε
≤
r
{\displaystyle 0<\|h\|,\|k\|\leq \varepsilon \leq r}
0
∉
]
h
,
k
[
{\displaystyle 0\notin \left]h,k\right[}
‖
g
(
a
+
h
)
−
g
(
a
+
k
)
‖
≤
M
‖
h
−
k
‖
≤
2
M
ε
{\displaystyle \left\|g\left(a+h\right)-g\left(a+k\right)\right\|\leq M\left\|h-k\right\|\leq 2M\varepsilon }
Si
0
∈
]
h
,
k
[
{\displaystyle 0\in \left]h,k\right[}
ℓ
≠
0
{\displaystyle \ell \neq 0}
h
{\displaystyle h}
k
{\displaystyle k}
E
{\displaystyle E}
>
1
{\displaystyle >1}
≤
ε
{\displaystyle \leq \varepsilon }
‖
g
(
a
+
h
)
−
g
(
a
+
k
)
‖
≤
‖
g
(
a
+
h
)
−
g
(
a
+
ℓ
)
‖
+
‖
g
(
a
+
ℓ
)
−
g
(
a
+
k
)
‖
≤
4
M
ε
{\displaystyle \left\|g\left(a+h\right)-g\left(a+k\right)\right\|\leq \left\|g\left(a+h\right)-g\left(a+\ell \right)\right\|+\left\|g\left(a+\ell \right)-g\left(a+k\right)\right\|\leq 4M\varepsilon }
4°) Si
g
:
]
a
,
b
[
→
F
{\displaystyle g:\left]a,b\right[\to F}
a
{\displaystyle a}
g
{\displaystyle g}
a
{\displaystyle a}
]
a
,
b
[
{\displaystyle \left]a,b\right[}
]
b
,
a
[
{\displaystyle \left]b,a\right[}
Soit
A
{\displaystyle A}
algèbre de Banach unifère, c'est-à-dire une
R
{\displaystyle \mathbb {R} }
sous-multiplicative (c'est-à-dire telle que
∀
x
,
y
∈
A
‖
x
y
‖
≤
‖
x
‖
‖
y
‖
{\displaystyle \forall x,y\in A\quad \left\|xy\right\|\leq \left\|x\right\|\left\|y\right\|}
pour laquelle
A
{\displaystyle A}
(par exemple : l'algèbre
L
(
E
)
{\displaystyle {\mathcal {L}}\left(E\right)}
norme
|
|
|
⋅
|
|
|
{\displaystyle |\!|\!|\cdot |\!|\!|}
, où
E
{\displaystyle E}
Démontrer que :
∀
u
∈
A
‖
u
‖
<
1
⇒
(
1
−
u
)
−
1
=
∑
n
=
0
+
∞
u
n
{\displaystyle \forall u\in A\quad \left\|u\right\|<1\Rightarrow \left(1-u\right)^{-1}=\sum _{n=0}^{+\infty }u^{n}}
1
{\displaystyle 1}
A
{\displaystyle A}
u
0
=
1
{\displaystyle u^{0}=1}
dans
A
{\displaystyle A}
groupe
A
×
{\displaystyle A^{\times }}
l'application
I
n
v
:
A
×
→
A
,
a
↦
a
−
1
{\displaystyle \mathrm {Inv} :A^{\times }\to A,\ a\mapsto a^{-1}}
d
I
n
v
a
(
h
)
=
−
a
−
1
h
a
−
1
{\displaystyle \mathrm {d} \,\mathrm {Inv} _{a}\left(h\right)=-a^{-1}ha^{-1}}
l'application
I
n
v
{\displaystyle \mathrm {Inv} }
∞ .
Dans le cas particulier
A
=
M
n
(
R
)
{\displaystyle A=\mathrm {M} _{n}\left(\mathbb {R} \right)}
Solution
1°) La série
∑
n
∈
N
u
n
{\displaystyle \sum _{n\in \mathbb {N} }u^{n}}
absolument convergente car
‖
u
n
‖
{\displaystyle \left\|u^{n}\right\|}
‖
u
‖
n
{\displaystyle \left\|u\right\|^{n}}
série géométrique convergente . La multiplication
A
×
A
→
A
{\displaystyle A\times A\to A}
norme d'application bilinéaire vaut
1
{\displaystyle 1}
(
1
−
u
)
∑
n
∈
N
u
n
=
∑
n
∈
N
u
n
−
∑
n
∈
N
u
n
+
1
=
u
0
=
1
{\displaystyle \left(1-u\right)\sum _{n\in \mathbb {N} }u^{n}=\sum _{n\in \mathbb {N} }u^{n}-\sum _{n\in \mathbb {N} }u^{n+1}=u^{0}=1}
(
∑
n
∈
N
u
n
)
(
1
−
u
)
=
1
{\displaystyle \left(\sum _{n\in \mathbb {N} }u^{n}\right)\left(1-u\right)=1}
2°) Soient
a
∈
A
×
{\displaystyle a\in A^{\times }}
r
=
‖
a
−
1
‖
−
1
{\displaystyle r=\left\|a^{-1}\right\|^{-1}}
k
∈
A
{\displaystyle k\in A}
<
r
{\displaystyle <r}
‖
−
a
−
1
k
‖
<
1
{\displaystyle \left\|-a^{-1}k\right\|<1}
1
+
a
−
1
k
{\displaystyle 1+a^{-1}k}
a
{\displaystyle a}
a
(
1
+
a
−
1
k
)
=
a
+
k
{\displaystyle a\left(1+a^{-1}k\right)=a+k}
B
(
a
,
r
)
⊂
A
×
{\displaystyle B\left(a,r\right)\subset A^{\times }}
A
×
{\displaystyle A^{\times }}
3°) Montrons d'abord que
I
n
v
{\displaystyle \mathrm {Inv} }
1
{\displaystyle 1}
d
I
n
v
1
=
−
i
d
A
{\displaystyle \mathrm {d} \,\mathrm {Inv} _{1}=-\mathrm {id} _{A}}
h
∈
A
{\displaystyle h\in A}
<
1
{\displaystyle <1}
I
n
v
(
1
+
h
)
=
1
−
h
+
∑
n
≥
2
(
−
h
)
n
{\displaystyle \mathrm {Inv} \left(1+h\right)=1-h+\sum _{n\geq 2}\left(-h\right)^{n}}
‖
∑
n
≥
2
(
−
h
)
n
‖
≤
∑
n
≥
2
‖
h
‖
n
=
‖
h
‖
2
1
−
‖
h
‖
=
o
(
‖
h
‖
)
{\displaystyle \left\|\sum _{n\geq 2}\left(-h\right)^{n}\right\|\leq \sum _{n\geq 2}\left\|h\right\|^{n}={\frac {\left\|h\right\|^{2}}{1-\left\|h\right\|}}=o\left(\left\|h\right\|\right)}
Pour
a
∈
A
×
{\displaystyle a\in A^{\times }}
r
{\displaystyle r}
k
∈
A
{\displaystyle k\in A}
<
r
{\displaystyle <r}
I
n
v
(
a
+
k
)
=
I
n
v
(
a
(
1
+
a
−
1
k
)
)
=
(
I
n
v
(
1
+
a
−
1
k
)
)
a
−
1
{\displaystyle \mathrm {Inv} \left(a+k\right)=\mathrm {Inv} \left(a\left(1+a^{-1}k\right)\right)=\left(\mathrm {Inv} \left(1+a^{-1}k\right)\right)a^{-1}}
I
n
v
{\displaystyle \mathrm {Inv} }
a
{\displaystyle a}
h
∈
A
{\displaystyle h\in A}
d
I
n
v
a
(
h
)
=
(
d
I
n
v
1
(
a
−
1
h
)
)
a
−
1
=
−
a
−
1
h
a
−
1
{\displaystyle \mathrm {d} \,\mathrm {Inv} _{a}\left(h\right)=\left(\mathrm {d} \,\mathrm {Inv} _{1}\left(a^{-1}h\right)\right)a^{-1}=-a^{-1}ha^{-1}}
4°)
d
I
n
v
{\displaystyle \mathrm {d} \,\mathrm {Inv} }
I
n
v
{\displaystyle \mathrm {Inv} }
∞ : l'application
Δ
:
A
→
A
×
A
,
x
↦
(
x
,
x
)
{\displaystyle \Delta :A\to A\times A,\ x\mapsto \left(x,x\right)}
A
×
A
→
L
(
A
)
,
(
x
1
,
x
2
)
↦
−
x
1
i
d
A
x
2
{\displaystyle A\times A\to {\mathcal {L}}\left(A\right),\ \left(x_{1},x_{2}\right)\mapsto -x_{1}\,\mathrm {id} _{A}\,x_{2}}
I
n
v
{\displaystyle \mathrm {Inv} }
n n : c'est acquis pour n = 0 et si
I
n
v
{\displaystyle \mathrm {Inv} }
n
d
I
n
v
{\displaystyle \mathrm {d} \,\mathrm {Inv} }
I
n
v
{\displaystyle \mathrm {Inv} }
n +1
5°) Si
A
=
M
n
(
R
)
{\displaystyle A=\mathrm {M} _{n}\left(\mathbb {R} \right)}
A
×
{\displaystyle A^{\times }}
G
L
n
(
R
)
{\displaystyle \mathrm {GL} _{n}\left(\mathbb {R} \right)}
I
n
v
{\displaystyle \mathrm {Inv} }
∞ sur cet ouvert puisque l'expression d'une matrice inverse
M
−
1
{\displaystyle M^{-1}}
est une fonction rationnelle des coefficients de
M
{\displaystyle M}
I
n
v
(
M
)
⋅
M
=
I
n
{\displaystyle \mathrm {Inv} \left(M\right)\cdot M=\mathrm {I} _{n}}
d
I
n
v
M
(
H
)
⋅
M
+
I
n
v
(
M
)
⋅
H
=
0
{\displaystyle \mathrm {d} \,\mathrm {Inv} _{M}\left(H\right)\cdot M+\mathrm {Inv} \left(M\right)\cdot H=0}
d'où
d
I
n
v
M
(
H
)
=
−
I
n
v
(
M
)
⋅
H
⋅
M
−
1
{\displaystyle \mathrm {d} \,\mathrm {Inv} _{M}\left(H\right)=-\mathrm {Inv} \left(M\right)\cdot H\cdot M^{-1}}
Références
François Laudenbach, Calcul différentiel et intégral , éd. École Polytechnique, 2000 (ISBN 978-2-73020724-9 [lire en ligne ] , p. 58 Jean Dieudonné, Éléments d'analyse , t. I
Fondements de l'analyse moderne Ralph Abraham, Jerrold E. Mardsen et Tudor Ratiu, Manifolds, Tensor Analysis, and Applications [lire en ligne ] , p. 104-105
Soient
E
{\displaystyle E}
Ω
{\displaystyle \Omega }
R
×
E
{\displaystyle \mathbb {R} \times E}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
f
:
Ω
→
E
{\displaystyle f:\Omega \to E}
d
2
f
{\displaystyle \operatorname {d} _{2}f}
Ω
{\displaystyle \Omega }
r
>
0
{\displaystyle r>0}
F
{\displaystyle F}
u
:
[
−
r
,
r
]
→
E
{\displaystyle u:\left[-r,r\right]\to E}
V
{\displaystyle V}
∀
t
∈
[
−
r
,
r
]
(
t
,
u
(
t
)
)
∈
Ω
{\displaystyle \forall t\in \left[-r,r\right]\quad \left(t,u\left(t\right)\right)\in \Omega }
Démontrer que l'application
Φ
:
V
→
F
{\displaystyle \Phi :V\to F}
Φ
u
(
t
)
=
f
(
t
,
u
(
t
)
)
{\displaystyle \Phi _{u}\left(t\right)=f\left(t,u\left(t\right)\right)}
d
Φ
u
h
(
t
)
=
d
2
f
(
t
,
u
(
t
)
)
(
h
(
t
)
)
{\displaystyle \operatorname {d} \Phi _{u}h\left(t\right)=\operatorname {d} _{2}f_{\left(t,u\left(t\right)\right)}\left(h\left(t\right)\right)}
Solution
Fixons
u
∈
V
{\displaystyle u\in V}
et montrons d'abord que l'application
R
:
F
→
F
{\displaystyle R:F\to F}
définie (pour
u
+
h
∈
V
{\displaystyle u+h\in V}
) par
R
h
(
t
)
=
Φ
u
+
h
(
t
)
−
Φ
u
(
t
)
−
d
2
f
(
t
,
u
(
t
)
)
(
h
(
t
)
)
{\displaystyle Rh\left(t\right)=\Phi _{u+h}\left(t\right)-\Phi _{u}\left(t\right)-\operatorname {d} _{2}f_{\left(t,u\left(t\right)\right)}\left(h\left(t\right)\right)}
vérifie :
R
h
=
o
(
‖
h
‖
)
{\displaystyle Rh=o\left(\left\|h\right\|\right)}
.
En notant
x
=
u
(
t
)
{\displaystyle x=u\left(t\right)}
k
=
h
(
t
)
{\displaystyle k=h\left(t\right)}
‖
f
(
t
,
x
+
k
)
−
f
(
t
,
x
)
−
d
2
f
(
t
,
x
)
(
k
)
‖
≤
‖
k
‖
sup
z
∈
[
x
,
x
+
k
]
|
|
|
d
2
f
(
t
,
z
)
−
d
2
f
(
t
,
x
)
|
|
|
{\displaystyle \left\|f\left(t,x+k\right)-f\left(t,x\right)-\operatorname {d} _{2}f_{\left(t,x\right)}\left(k\right)\right\|\leq \left\|k\right\|\sup _{z\in \left[x,x+k\right]}|\!|\!|\operatorname {d} _{2}f_{\left(t,z\right)}-\operatorname {d} _{2}f_{\left(t,x\right)}|\!|\!|}
donc
‖
R
h
‖
≤
‖
h
‖
ε
(
‖
h
‖
)
{\displaystyle \left\|Rh\right\|\leq \left\|h\right\|\varepsilon \left(\left\|h\right\|\right)}
où la fonction
ε
{\displaystyle \varepsilon }
est définie par :
ε
(
η
)
=
sup
(
t
,
x
)
∈
(
id
,
u
)
(
[
−
r
,
r
]
)
,
‖
z
−
x
‖
≤
η
|
|
|
d
2
f
(
t
,
z
)
−
d
2
f
(
t
,
x
)
|
|
|
{\displaystyle \varepsilon \left(\eta \right)=\sup _{\left(t,x\right)\in \left(\operatorname {id} ,u\right)\left(\left[-r,r\right]\right),\ \left\|z-x\right\|\leq \eta }|\!|\!|\operatorname {d} _{2}f_{\left(t,z\right)}-\operatorname {d} _{2}f_{\left(t,x\right)}|\!|\!|}
.
L'image par
(
id
,
u
)
{\displaystyle \left(\operatorname {id} ,u\right)}
[
−
r
,
r
]
{\displaystyle \left[-r,r\right]}
K
⊂
V
{\displaystyle K\subset V}
d
2
f
{\displaystyle \operatorname {d} _{2}f}
V
{\displaystyle V}
théorème de Heine ) :
∀
α
>
0
∃
β
>
0
∀
a
∈
K
∀
b
∈
V
(
‖
b
−
a
‖
≤
β
⇒
|
|
|
d
2
f
b
−
d
2
f
a
|
|
|
≤
α
)
{\displaystyle \forall \alpha >0\quad \exists \beta >0\quad \forall a\in K\quad \forall b\in V\quad \left(\left\|b-a\right\|\leq \beta \Rightarrow |\!|\!|\operatorname {d} _{2}f_{b}-\operatorname {d} _{2}f_{a}|\!|\!|\leq \alpha \right)}
Ceci prouve que
lim
η
→
0
ε
(
η
)
=
0
{\displaystyle \lim _{\eta \to 0}\varepsilon \left(\eta \right)=0}
R
h
=
o
(
‖
h
‖
)
{\displaystyle Rh=o\left(\left\|h\right\|\right)}
d
Φ
{\displaystyle \operatorname {d} \Phi }
|
|
|
d
Φ
u
+
h
−
d
Φ
u
|
|
|
≤
ε
(
‖
h
‖
)
{\displaystyle |\!|\!|\operatorname {d} \Phi _{u+h}-\operatorname {d} \Phi _{u}|\!|\!|\leq \varepsilon \left(\left\|h\right\|\right)}
Soient
I
:=
[
−
1
,
1
]
{\displaystyle I:=\left[-1,1\right]}
E
{\displaystyle E}
F
{\displaystyle F}
I
{\displaystyle I}
E
{\displaystyle E}
1 et nulles en
0
{\displaystyle 0}
‖
x
‖
1
:=
‖
x
′
‖
0
{\displaystyle \left\|x\right\|_{1}:=\left\|x'\right\|_{0}}
‖
‖
0
{\displaystyle \left\|~\right\|_{0}}
V
:
F
×
I
→
E
,
(
x
,
t
)
↦
x
(
t
)
{\displaystyle V:F\times I\to E,\ \left(x,t\right)\mapsto x\left(t\right)}
1 .
Solution
Montrons que les deux dérivées partielles de
V
{\displaystyle V}
D
1
V
{\displaystyle D_{1}V}
x
{\displaystyle x}
D
2
V
{\displaystyle D_{2}V}
t
{\displaystyle t}
D
1
V
{\displaystyle D_{1}V}
t
∈
I
{\displaystyle t\in I}
V
(
⋅
,
t
)
{\displaystyle V\left(\cdot ,t\right)}
‖
V
(
x
,
t
)
‖
≤
‖
x
‖
0
≤
‖
x
‖
1
{\displaystyle \left\|V\left(x,t\right)\right\|\leq \left\|x\right\|_{0}\leq \left\|x\right\|_{1}}
x
∈
F
{\displaystyle x\in F}
D
1
V
(
x
,
t
)
=
V
(
⋅
,
t
)
{\displaystyle D_{1}V\left(x,t\right)=V\left(\cdot ,t\right)}
x
{\displaystyle x}
D
1
V
{\displaystyle D_{1}V}
t
↦
V
(
⋅
,
t
)
{\displaystyle t\mapsto V\left(\cdot ,t\right)}
I
{\displaystyle I}
L
(
F
,
E
)
{\displaystyle {\mathcal {L}}\left(F,E\right)}
1
{\displaystyle 1}
x
∈
F
{\displaystyle x\in F}
‖
V
(
x
,
t
2
)
−
V
(
x
,
t
1
)
‖
=
‖
∫
t
1
t
2
x
′
(
t
)
d
t
‖
≤
|
∫
t
1
t
2
‖
x
′
(
t
)
‖
d
t
|
≤
|
t
2
−
t
1
|
‖
x
‖
1
{\displaystyle \left\|V\left(x,t_{2}\right)-V\left(x,t_{1}\right)\right\|=\left\|\int _{t_{1}}^{t_{2}}x'\left(t\right)\ \mathrm {d} t\right\|\leq \left|\int _{t_{1}}^{t_{2}}\left\|x'\left(t\right)\right\|\ \mathrm {d} t\right|\leq \left|t_{2}-t_{1}\right|\left\|x\right\|_{1}}
|
|
|
V
(
⋅
,
t
2
)
−
V
(
⋅
,
t
1
)
|
|
|
≤
|
t
2
−
t
1
|
{\displaystyle |\!|\!|V\left(\cdot ,t_{2}\right)-V\left(\cdot ,t_{1}\right)|\!|\!|\leq \left|t_{2}-t_{1}\right|}
D
2
V
{\displaystyle D_{2}V}
D
2
V
(
x
,
t
)
=
x
′
(
t
)
{\displaystyle D_{2}V\left(x,t\right)=x'\left(t\right)}
D
2
V
{\displaystyle D_{2}V}
x
,
x
0
∈
F
{\displaystyle x,x_{0}\in F}
t
,
t
0
∈
I
{\displaystyle t,t_{0}\in I}
‖
x
′
(
t
)
−
x
0
′
(
t
0
)
‖
≤
‖
x
′
(
t
)
−
x
0
′
(
t
)
‖
+
‖
x
0
′
(
t
)
−
x
0
′
(
t
0
)
‖
≤
‖
x
−
x
0
‖
1
+
‖
x
0
′
(
t
)
−
x
0
′
(
t
0
)
‖
→
0
{\displaystyle \left\|x'\left(t\right)-x_{0}'\left(t_{0}\right)\right\|\leq \left\|x'\left(t\right)-x_{0}'\left(t\right)\right\|+\left\|x'_{0}\left(t\right)-x_{0}'\left(t_{0}\right)\right\|\leq \left\|x-x_{0}\right\|_{1}+\left\|x'_{0}\left(t\right)-x_{0}'\left(t_{0}\right)\right\|\to 0}
x
→
x
0
{\displaystyle x\to x_{0}}
t
→
t
0
{\displaystyle t\to t_{0}}
Soient
E
{\displaystyle E}
F
{\displaystyle F}
f
:
E
→
F
{\displaystyle f:E\to F}
homogène de degré
m
∈
N
{\displaystyle m\in \mathbb {N} }
∀
x
∈
E
∀
t
∈
R
f
(
t
x
)
=
t
m
f
(
x
)
{\displaystyle \forall x\in E\quad \forall t\in \mathbb {R} \qquad f(tx)=t^{m}f(x)}
Parmi les fonctions homogènes de degré
0
{\displaystyle 0}
0
{\displaystyle 0}
Montrer que si
f
{\displaystyle f}
m
>
0
{\displaystyle m>0}
f
{\displaystyle f}
0
{\displaystyle 0}
Montrer que si
f
{\displaystyle f}
m
≥
1
{\displaystyle m\geq 1}
0
{\displaystyle 0}
f
{\displaystyle f}
m
=
1
{\displaystyle m=1}
m
>
1
{\displaystyle m>1}
d
f
0
=
0
{\displaystyle \mathrm {d} f_{0}=0}
Montrer que si
f
{\displaystyle f}
m
>
1
{\displaystyle m>1}
f
{\displaystyle f}
0
{\displaystyle 0}
d
f
0
=
0
{\displaystyle \mathrm {d} f_{0}=0}
Application : étudier la continuité et la différentiabilité en
(
0
,
0
)
{\displaystyle (0,0)}
f
,
g
,
h
:
R
2
→
R
{\displaystyle f,g,h:\mathbb {R} ^{2}\to \mathbb {R} }
f
(
x
,
y
)
=
0
{\displaystyle f(x,y)=0}
y
≠
0
{\displaystyle y\neq 0}
f
(
x
,
0
)
=
x
{\displaystyle f(x,0)=x}
g
(
x
,
y
)
=
x
3
x
2
+
y
2
{\displaystyle g(x,y)={\frac {x^{3}}{x^{2}+y^{2}}}}
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
g
(
0
,
0
)
=
0
{\displaystyle g(0,0)=0}
h
(
x
,
y
)
=
x
p
y
q
x
2
−
x
y
+
y
2
{\displaystyle h(x,y)={\frac {x^{p}y^{q}}{x^{2}-xy+y^{2}}}}
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
h
(
0
,
0
)
=
0
{\displaystyle h(0,0)=0}
p
,
q
∈
N
∗
{\displaystyle p,q\in \mathbb {N} ^{*}}
Solution
Les fonctions constantes (pour que
∀
x
∈
E
f
(
0
)
=
lim
t
→
0
t
≠
0
f
(
t
x
)
=
f
(
x
)
{\displaystyle \forall x\in E\quad f(0)=\lim _{t\to 0 \atop t\neq 0}f(tx)=f(x)}
Sous ces hypothèses, on a bien
lim
x
→
0
f
(
x
)
=
lim
r
→
0
‖
u
‖
=
1
f
(
r
u
)
=
lim
r
→
0
‖
u
‖
=
1
r
m
f
(
u
)
=
0
=
f
(
0
)
.
{\displaystyle \lim _{x\to 0}f(x)=\lim _{r\to 0 \atop \|u\|=1}f(ru)=\lim _{r\to 0 \atop \|u\|=1}r^{m}f(u)=0=f(0).}
Sous ces hypothèses (et en utilisant l'exercice précédent),
d
f
0
(
x
)
=
lim
t
→
0
t
≠
0
f
(
t
x
)
−
f
(
0
)
t
=
lim
t
→
0
t
≠
0
t
m
−
1
f
(
x
)
{\displaystyle \mathrm {d} f_{0}(x)=\lim _{t\to 0 \atop t\neq 0}{\frac {f(tx)-f(0)}{t}}=\lim _{t\to 0 \atop t\neq 0}t^{m-1}f(x)}
f
(
x
)
{\displaystyle f(x)}
m
=
1
{\displaystyle m=1}
0
{\displaystyle 0}
m
>
1
{\displaystyle m>1}
Sous ces hypothèses,
ε
(
r
u
)
:=
f
(
r
u
)
r
=
r
m
−
1
f
(
u
)
{\displaystyle \varepsilon (ru):={\frac {f(ru)}{r}}=r^{m-1}f(u)}
0
{\displaystyle 0}
r
→
0
{\displaystyle r\to 0}
u
{\displaystyle u}
f
{\displaystyle f}
g
{\displaystyle g}
1
{\displaystyle 1}
(
0
,
0
)
{\displaystyle (0,0)}
h
{\displaystyle h}
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
x
2
−
x
y
+
y
2
=
(
x
−
y
/
2
)
2
+
3
y
2
/
4
>
0
{\displaystyle x^{2}-xy+y^{2}=(x-y/2)^{2}+3y^{2}/4>0}
h
{\displaystyle h}
p
+
q
−
2
{\displaystyle p+q-2}
f
{\displaystyle f}
g
{\displaystyle g}
p
{\displaystyle p}
q
{\displaystyle q}
(
0
,
0
)
{\displaystyle (0,0)}
p
+
q
>
2
{\displaystyle p+q>2}
p
+
q
>
3
{\displaystyle p+q>3}
Soient
u
,
v
:
R
→
R
{\displaystyle u,v:\mathbb {R} \to \mathbb {R} }
1 et
f
:
R
2
→
R
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} }
1 . On considère
F
(
x
)
:=
∫
u
(
x
)
v
(
x
)
f
(
x
,
t
)
d
t
{\displaystyle F(x):=\int _{u(x)}^{v(x)}f\left(x,t\right)\,\mathrm {d} t}
Montrer que
F
{\displaystyle F}
R
{\displaystyle \mathbb {R} }
1 et calculer
F
′
(
x
)
{\displaystyle F'(x)}
Application :
F
(
x
)
=
∫
1
1
+
x
2
ln
(
x
2
+
t
2
)
d
t
{\displaystyle F(x)=\int _{1}^{1+x^{2}}\ln \left(x^{2}+t^{2}\right)\,\mathrm {d} t}
Solution
F
(
x
)
=
G
(
x
,
u
(
x
)
,
v
(
x
)
)
{\displaystyle F(x)=G\left(x,u(x),v(x)\right)}
G
:
R
3
→
R
{\displaystyle G:\mathbb {R} ^{3}\to \mathbb {R} }
G
(
x
,
u
,
v
)
:=
∫
u
v
f
(
x
,
t
)
d
t
{\displaystyle G\left(x,u,v\right):=\int _{u}^{v}f\left(x,t\right)\,\mathrm {d} t}
La fonction
G
{\displaystyle G}
1 (car d'après ce qui suit, ses trois dérivées partielles sont définies et continues grâce aux hypothèses) donc
F
{\displaystyle F}
∂
G
∂
v
(
x
,
u
,
v
)
=
f
(
x
,
v
)
{\displaystyle {\frac {\partial G}{\partial v}}\left(x,u,v\right)=f\left(x,v\right)}
∂
G
∂
u
(
x
,
u
,
v
)
=
−
f
(
x
,
u
)
{\displaystyle {\frac {\partial G}{\partial u}}\left(x,u,v\right)=-f\left(x,u\right)}
∂
G
∂
x
(
x
,
u
,
v
)
=
∫
u
v
∂
f
∂
x
(
x
,
t
)
d
t
{\displaystyle {\frac {\partial G}{\partial x}}\left(x,u,v\right)=\int _{u}^{v}{\frac {\partial f}{\partial x}}\left(x,t\right)\,\mathrm {d} t}
donc (dérivée d'une fonction composée) :
F
′
(
x
)
=
f
(
x
,
v
(
x
)
)
v
′
(
x
)
−
f
(
x
,
u
(
x
)
)
u
′
(
x
)
+
∫
u
(
x
)
v
(
x
)
∂
f
∂
x
(
x
,
t
)
d
t
{\displaystyle F'(x)=f\left(x,v(x)\right)v'(x)-f\left(x,u(x)\right)u'(x)+\int _{u(x)}^{v(x)}{\frac {\partial f}{\partial x}}\left(x,t\right)\,\mathrm {d} t}
u
(
x
)
=
1
{\displaystyle u(x)=1}
v
(
x
)
=
1
+
x
2
{\displaystyle v(x)=1+x^{2}}
f
(
x
,
t
)
=
ln
(
x
2
+
t
2
)
{\displaystyle f\left(x,t\right)=\ln \left(x^{2}+t^{2}\right)}
u
′
(
x
)
=
0
{\displaystyle u'(x)=0}
v
′
(
x
)
=
2
x
{\displaystyle v'(x)=2x}
∂
f
∂
x
(
x
,
t
)
=
2
x
x
2
+
t
2
{\displaystyle {\frac {\partial f}{\partial x}}\left(x,t\right)={\frac {2x}{x^{2}+t^{2}}}}
F
′
(
x
)
=
2
x
ln
(
x
2
+
(
1
+
x
2
)
2
)
+
2
x
∫
1
1
+
x
2
d
t
x
2
+
t
2
=
2
x
ln
(
1
+
3
x
2
+
x
4
)
+
2
(
arctan
1
+
x
2
x
−
arctan
1
x
)
=
2
x
ln
(
1
+
3
x
2
+
x
4
)
+
2
arctan
x
3
1
+
2
x
2
.
{\displaystyle {\begin{aligned}F'(x)&=2x\ln \left(x^{2}+\left(1+x^{2}\right)^{2}\right)+2x\int _{1}^{1+x^{2}}{\frac {\mathrm {d} t}{x^{2}+t^{2}}}\\&=2x\ln \left(1+3x^{2}+x^{4}\right)+2\left(\arctan {\frac {1+x^{2}}{x}}-\arctan {\frac {1}{x}}\right)\\&=2x\ln \left(1+3x^{2}+x^{4}\right)+2\arctan {\frac {x^{3}}{1+2x^{2}}}.\end{aligned}}}
somme de deux arctan .)
Soient
ϕ
=
(
ϕ
1
,
ϕ
2
)
:
R
2
→
R
2
{\displaystyle \phi =\left(\phi _{1},\phi _{2}\right):\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
f
:
R
2
→
R
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} }
f
∘
ϕ
{\displaystyle f\circ \phi }
f
{\displaystyle f}
ϕ
{\displaystyle \phi }
Application à
ϕ
:
]
0
,
+
∞
[
×
]
−
π
,
π
[
→
R
2
∖
{
(
0
,
0
)
}
,
(
r
,
θ
)
↦
(
r
cos
θ
,
r
sin
θ
)
{\displaystyle \phi :\left]0,+\infty \right[\times \left]-\pi ,\pi \right[\to \mathbb {R} ^{2}\setminus \{\left(0,0\right)\},\;\left(r,\theta \right)\mapsto \left(r\cos \theta ,r\sin \theta \right)}
f
{\displaystyle f}
R
2
∖
{
(
0
,
0
)
}
{\displaystyle \mathbb {R} ^{2}\setminus \{\left(0,0\right)\}}
On pose
Θ
f
(
x
,
y
)
=
x
∂
f
∂
x
(
x
,
y
)
+
y
∂
f
∂
y
(
x
,
y
)
{\displaystyle \Theta _{f}(x,y)=x{\frac {\partial f}{\partial x}}(x,y)+y{\frac {\partial f}{\partial y}}(x,y)}
Ψ
f
(
x
,
y
)
=
−
y
∂
f
∂
x
(
x
,
y
)
+
x
∂
f
∂
y
(
x
,
y
)
{\displaystyle \Psi _{f}(x,y)=-y{\frac {\partial f}{\partial x}}(x,y)+x{\frac {\partial f}{\partial y}}(x,y)}
Θ
f
∘
ϕ
{\displaystyle \Theta _{f}\circ \phi }
Ψ
f
∘
ϕ
{\displaystyle \Psi _{f}\circ \phi }
f
∘
ϕ
{\displaystyle f\circ \phi }
opérateurs différentiels
Θ
{\displaystyle \Theta }
Ψ
{\displaystyle \Psi }
Plus généralement, exprimer
∂
f
∂
x
∘
ϕ
{\displaystyle {\frac {\partial f}{\partial x}}\circ \phi }
∂
f
∂
y
∘
ϕ
{\displaystyle {\frac {\partial f}{\partial y}}\circ \phi }
f
∘
ϕ
{\displaystyle f\circ \phi }
Exprimer de même le « laplacien en coordonnées polaires », c'est-à-dire
(
Δ
f
)
∘
ϕ
{\displaystyle (\Delta f)\circ \phi }
Δ
f
=
∂
2
f
∂
x
2
+
∂
2
f
∂
y
2
{\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}}
f
∘
ϕ
{\displaystyle f\circ \phi }
Trouver toutes les fonctions différentiables
f
:
R
2
∖
{
(
0
,
0
)
}
→
R
{\displaystyle f:\mathbb {R} ^{2}\setminus \{(0,0)\}\to \mathbb {R} }
Θ
f
(
x
,
y
)
=
x
2
+
y
2
{\displaystyle \Theta _{f}(x,y)={\sqrt {x^{2}+y^{2}}}}
(
0
,
0
)
{\displaystyle (0,0)}
Solution
En développant
J
f
∘
ϕ
(
s
,
t
)
=
(
J
f
)
ϕ
(
s
,
t
)
×
(
J
ϕ
)
(
s
,
t
)
{\displaystyle \operatorname {J} _{f\circ \phi }(s,t)=(J_{f})_{\phi (s,t)}\times (J_{\phi })_{(s,t)}}
∂
(
f
∘
ϕ
)
∂
s
(
s
,
t
)
=
∂
f
∂
x
(
ϕ
(
s
,
t
)
)
∂
ϕ
1
∂
s
(
s
,
t
)
+
∂
f
∂
y
(
ϕ
(
s
,
t
)
)
∂
ϕ
2
∂
s
(
s
,
t
)
{\displaystyle {\frac {\partial (f\circ \phi )}{\partial s}}(s,t)={\frac {\partial f}{\partial x}}(\phi (s,t)){\frac {\partial \phi _{1}}{\partial s}}(s,t)+{\frac {\partial f}{\partial y}}(\phi (s,t)){\frac {\partial \phi _{2}}{\partial s}}(s,t)}
∂
(
f
∘
ϕ
)
∂
t
(
s
,
t
)
=
∂
f
∂
x
(
ϕ
(
s
,
t
)
)
∂
ϕ
1
∂
t
(
s
,
t
)
+
∂
f
∂
y
(
ϕ
(
s
,
t
)
)
∂
ϕ
2
∂
t
(
s
,
t
)
{\displaystyle {\frac {\partial (f\circ \phi )}{\partial t}}(s,t)={\frac {\partial f}{\partial x}}(\phi (s,t)){\frac {\partial \phi _{1}}{\partial t}}(s,t)+{\frac {\partial f}{\partial y}}(\phi (s,t)){\frac {\partial \phi _{2}}{\partial t}}(s,t)}
J
ϕ
(
r
,
θ
)
=
(
cos
θ
−
r
sin
θ
sin
θ
r
cos
θ
)
{\displaystyle \operatorname {J} _{\phi }(r,\theta )={\begin{pmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{pmatrix}}}
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
=
∂
f
∂
x
(
r
cos
θ
,
r
sin
θ
)
cos
θ
+
∂
f
∂
y
(
r
cos
θ
,
r
sin
θ
)
(
−
r
sin
θ
)
{\displaystyle {\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )={\frac {\partial f}{\partial x}}(r\cos \theta ,r\sin \theta )\cos \theta +{\frac {\partial f}{\partial y}}(r\cos \theta ,r\sin \theta )(-r\sin \theta )}
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
=
∂
f
∂
x
(
r
cos
θ
,
r
sin
θ
)
sin
θ
+
∂
f
∂
y
(
r
cos
θ
,
r
sin
θ
)
r
cos
θ
{\displaystyle {\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta )={\frac {\partial f}{\partial x}}(r\cos \theta ,r\sin \theta )\sin \theta +{\frac {\partial f}{\partial y}}(r\cos \theta ,r\sin \theta )r\cos \theta }
(
Θ
f
∘
ϕ
)
(
r
,
θ
)
=
r
cos
θ
∂
f
∂
x
(
r
cos
θ
,
r
sin
θ
)
+
r
sin
θ
∂
f
∂
y
(
r
cos
θ
,
r
sin
θ
)
=
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
{\displaystyle \left(\Theta _{f}\circ \phi \right)\left(r,\theta \right)=r\cos \theta {\frac {\partial f}{\partial x}}(r\cos \theta ,r\sin \theta )+r\sin \theta {\frac {\partial f}{\partial y}}(r\cos \theta ,r\sin \theta )=r{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )}
(
Ψ
f
∘
ϕ
)
(
r
,
θ
)
=
−
r
sin
θ
∂
f
∂
x
(
r
cos
θ
,
r
sin
θ
)
+
r
cos
θ
∂
f
∂
y
(
r
cos
θ
,
r
sin
θ
)
=
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
{\displaystyle \left(\Psi _{f}\circ \phi \right)\left(r,\theta \right)=-r\sin \theta {\frac {\partial f}{\partial x}}(r\cos \theta ,r\sin \theta )+r\cos \theta {\frac {\partial f}{\partial y}}(r\cos \theta ,r\sin \theta )={\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta )}
(
∂
(
f
∘
ϕ
)
∂
r
∂
(
f
∘
ϕ
)
∂
θ
)
=
(
∂
f
∂
x
∘
ϕ
∂
f
∂
y
∘
ϕ
)
×
J
ϕ
{\displaystyle {\begin{pmatrix}{\frac {\partial (f\circ \phi )}{\partial r}}&{\frac {\partial (f\circ \phi )}{\partial \theta }}\end{pmatrix}}={\begin{pmatrix}{\frac {\partial f}{\partial x}}\circ \phi &{\frac {\partial f}{\partial y}}\circ \phi \end{pmatrix}}\times \operatorname {J} _{\phi }}
(
∂
f
∂
x
∘
ϕ
∂
f
∂
y
∘
ϕ
)
(
r
,
θ
)
=
(
∂
(
f
∘
ϕ
)
∂
r
∂
(
f
∘
ϕ
)
∂
θ
)
(
r
,
θ
)
×
(
J
ϕ
)
−
1
(
r
,
θ
)
=
(
∂
(
f
∘
ϕ
)
∂
r
∂
(
f
∘
ϕ
)
∂
θ
)
(
r
,
θ
)
×
(
cos
θ
sin
θ
−
sin
θ
r
cos
θ
r
)
=
(
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
cos
θ
−
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
sin
θ
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
sin
θ
+
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
cos
θ
r
)
.
{\displaystyle {\begin{aligned}{\begin{pmatrix}{\frac {\partial f}{\partial x}}\circ \phi &{\frac {\partial f}{\partial y}}\circ \phi \end{pmatrix}}(r,\theta )&={\begin{pmatrix}{\frac {\partial (f\circ \phi )}{\partial r}}&{\frac {\partial (f\circ \phi )}{\partial \theta }}\end{pmatrix}}(r,\theta )\times (\operatorname {J} _{\phi })^{-1}(r,\theta )\\&={\begin{pmatrix}{\frac {\partial (f\circ \phi )}{\partial r}}&{\frac {\partial (f\circ \phi )}{\partial \theta }}\end{pmatrix}}(r,\theta )\times {\begin{pmatrix}\cos \theta &\sin \theta \\-{\frac {\sin \theta }{r}}&{\frac {\cos \theta }{r}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )\cos \theta -{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\sin \theta }{r}}&{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )\sin \theta +{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\cos \theta }{r}}\end{pmatrix}}.\end{aligned}}}
∂
f
∂
x
(
x
,
y
)
=
1
x
2
+
y
2
(
x
Θ
f
(
x
,
y
)
−
y
Ψ
f
(
x
,
y
)
)
{\displaystyle {\frac {\partial f}{\partial x}}(x,y)={\frac {1}{x^{2}+y^{2}}}\left(x\Theta _{f}(x,y)-y\Psi _{f}(x,y)\right)}
∂
f
∂
x
∘
ϕ
(
r
,
θ
)
=
1
r
2
(
r
cos
θ
×
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
−
r
sin
θ
×
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
)
=
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
cos
θ
−
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
sin
θ
r
{\displaystyle {\frac {\partial f}{\partial x}}\circ \phi (r,\theta )={\frac {1}{r^{2}}}\left(r\cos \theta \times r{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )-r\sin \theta \times {\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta )\right)={\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )\cos \theta -{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\sin \theta }{r}}}
∂
f
∂
y
(
x
,
y
)
=
1
x
2
+
y
2
(
y
Θ
f
(
x
,
y
)
+
x
Ψ
f
(
x
,
y
)
)
{\displaystyle {\frac {\partial f}{\partial y}}(x,y)={\frac {1}{x^{2}+y^{2}}}\left(y\Theta _{f}(x,y)+x\Psi _{f}(x,y)\right)}
∂
f
∂
y
∘
ϕ
(
r
,
θ
)
=
1
r
2
(
r
sin
θ
×
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
+
r
cos
θ
×
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
)
=
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
sin
θ
+
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
cos
θ
r
{\displaystyle {\frac {\partial f}{\partial y}}\circ \phi (r,\theta )={\frac {1}{r^{2}}}\left(r\sin \theta \times r{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )+r\cos \theta \times {\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta )\right)={\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )\sin \theta +{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\cos \theta }{r}}}
∂
2
f
∂
x
2
∘
ϕ
(
r
,
θ
)
=
∂
(
∂
f
∂
x
∘
ϕ
)
∂
r
(
r
,
θ
)
cos
θ
−
∂
(
∂
f
∂
x
∘
ϕ
)
∂
θ
(
r
,
θ
)
sin
θ
r
=
∂
∂
r
(
∂
(
f
∘
ϕ
)
∂
r
cos
θ
−
∂
(
f
∘
ϕ
)
∂
θ
sin
θ
r
)
(
r
,
θ
)
cos
θ
−
∂
∂
θ
(
∂
(
f
∘
ϕ
)
∂
r
cos
θ
−
∂
(
f
∘
ϕ
)
∂
θ
sin
θ
r
)
(
r
,
θ
)
sin
θ
r
=
∂
2
(
f
∘
ϕ
)
∂
r
2
(
r
,
θ
)
cos
2
θ
−
∂
2
(
f
∘
ϕ
)
∂
r
∂
θ
(
r
,
θ
)
sin
θ
cos
θ
r
+
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
sin
θ
cos
θ
r
2
−
∂
2
(
f
∘
ϕ
)
∂
θ
∂
r
(
r
,
θ
)
cos
θ
sin
θ
r
+
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
sin
2
θ
r
+
∂
2
(
f
∘
ϕ
)
∂
θ
2
(
r
,
θ
)
sin
2
θ
r
2
+
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
cos
θ
sin
θ
r
2
=
∂
2
(
f
∘
ϕ
)
∂
r
2
(
r
,
θ
)
cos
2
θ
−
2
∂
2
(
f
∘
ϕ
)
∂
r
∂
θ
(
r
,
θ
)
sin
θ
cos
θ
r
+
∂
2
(
f
∘
ϕ
)
∂
θ
2
(
r
,
θ
)
sin
2
θ
r
2
+
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
sin
2
θ
r
+
2
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
sin
θ
cos
θ
r
2
{\displaystyle {\begin{aligned}{\frac {\partial ^{2}f}{\partial x^{2}}}\circ \phi (r,\theta )&={\frac {\partial ({\frac {\partial f}{\partial x}}\circ \phi )}{\partial r}}(r,\theta )\cos \theta -{\frac {\partial ({\frac {\partial f}{\partial x}}\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\sin \theta }{r}}\\&={\frac {\partial }{\partial r}}\left({\frac {\partial (f\circ \phi )}{\partial r}}\cos \theta -{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\sin \theta }{r}}\right)(r,\theta )\cos \theta -{\frac {\partial }{\partial \theta }}\left({\frac {\partial (f\circ \phi )}{\partial r}}\cos \theta -{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\sin \theta }{r}}\right)(r,\theta ){\frac {\sin \theta }{r}}\\&={\frac {\partial ^{2}(f\circ \phi )}{\partial r^{2}}}(r,\theta )\cos ^{2}\theta -{\frac {\partial ^{2}(f\circ \phi )}{\partial r\partial \theta }}(r,\theta ){\frac {\sin \theta \cos \theta }{r}}+{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\sin \theta \cos \theta }{r^{2}}}\\&-{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta \partial r}}(r,\theta ){\frac {\cos \theta \sin \theta }{r}}+{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta ){\frac {\sin ^{2}\theta }{r}}+{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta ^{2}}}(r,\theta ){\frac {\sin ^{2}\theta }{r^{2}}}+{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\cos \theta \sin \theta }{r^{2}}}\\&={\frac {\partial ^{2}(f\circ \phi )}{\partial r^{2}}}(r,\theta )\cos ^{2}\theta -2{\frac {\partial ^{2}(f\circ \phi )}{\partial r\partial \theta }}(r,\theta ){\frac {\sin \theta \cos \theta }{r}}+{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta ^{2}}}(r,\theta ){\frac {\sin ^{2}\theta }{r^{2}}}\\&+{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta ){\frac {\sin ^{2}\theta }{r}}+2{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\sin \theta \cos \theta }{r^{2}}}\end{aligned}}}
∂
2
f
∂
y
2
∘
ϕ
(
r
,
θ
)
=
∂
(
∂
f
∂
y
∘
ϕ
)
∂
r
(
r
,
θ
)
sin
θ
+
∂
(
∂
f
∂
y
∘
ϕ
)
∂
θ
(
r
,
θ
)
cos
θ
r
=
∂
∂
r
(
∂
(
f
∘
ϕ
)
∂
r
sin
θ
+
∂
(
f
∘
ϕ
)
∂
θ
cos
θ
r
)
(
r
,
θ
)
sin
θ
+
∂
∂
θ
(
∂
(
f
∘
ϕ
)
∂
r
sin
θ
+
∂
(
f
∘
ϕ
)
∂
θ
cos
θ
r
)
(
r
,
θ
)
cos
θ
r
=
∂
2
(
f
∘
ϕ
)
∂
r
2
(
r
,
θ
)
sin
2
θ
+
∂
2
(
f
∘
ϕ
)
∂
r
∂
θ
(
r
,
θ
)
cos
θ
sin
θ
r
−
∂
(
f
∘
ϕ
)
∂
θ
cos
θ
sin
θ
r
2
+
∂
2
(
f
∘
ϕ
)
∂
θ
∂
r
(
r
,
θ
)
sin
θ
cos
θ
r
+
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
cos
2
θ
r
+
∂
2
(
f
∘
ϕ
)
∂
θ
2
(
r
,
θ
)
cos
2
θ
r
2
−
∂
(
f
∘
ϕ
)
∂
θ
sin
θ
cos
θ
r
2
=
∂
2
(
f
∘
ϕ
)
∂
r
2
(
r
,
θ
)
sin
2
θ
+
2
∂
2
(
f
∘
ϕ
)
∂
r
∂
θ
(
r
,
θ
)
cos
θ
sin
θ
r
+
∂
2
(
f
∘
ϕ
)
∂
θ
2
(
r
,
θ
)
cos
2
θ
r
2
+
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
cos
2
θ
r
−
2
∂
(
f
∘
ϕ
)
∂
θ
(
r
,
θ
)
cos
θ
sin
θ
r
2
{\displaystyle {\begin{aligned}{\frac {\partial ^{2}f}{\partial y^{2}}}\circ \phi (r,\theta )&={\frac {\partial ({\frac {\partial f}{\partial y}}\circ \phi )}{\partial r}}(r,\theta )\sin \theta +{\frac {\partial ({\frac {\partial f}{\partial y}}\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\cos \theta }{r}}\\&={\frac {\partial }{\partial r}}\left({\frac {\partial (f\circ \phi )}{\partial r}}\sin \theta +{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\cos \theta }{r}}\right)(r,\theta )\sin \theta +{\frac {\partial }{\partial \theta }}\left({\frac {\partial (f\circ \phi )}{\partial r}}\sin \theta +{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\cos \theta }{r}}\right)(r,\theta ){\frac {\cos \theta }{r}}\\&={\frac {\partial ^{2}(f\circ \phi )}{\partial r^{2}}}(r,\theta )\sin ^{2}\theta +{\frac {\partial ^{2}(f\circ \phi )}{\partial r\partial \theta }}(r,\theta ){\frac {\cos \theta \sin \theta }{r}}-{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\cos \theta \sin \theta }{r^{2}}}\\&+{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta \partial r}}(r,\theta ){\frac {\sin \theta \cos \theta }{r}}+{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta ){\frac {\cos ^{2}\theta }{r}}+{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta ^{2}}}(r,\theta ){\frac {\cos ^{2}\theta }{r^{2}}}-{\frac {\partial (f\circ \phi )}{\partial \theta }}{\frac {\sin \theta \cos \theta }{r^{2}}}\\&={\frac {\partial ^{2}(f\circ \phi )}{\partial r^{2}}}(r,\theta )\sin ^{2}\theta +2{\frac {\partial ^{2}(f\circ \phi )}{\partial r\partial \theta }}(r,\theta ){\frac {\cos \theta \sin \theta }{r}}+{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta ^{2}}}(r,\theta ){\frac {\cos ^{2}\theta }{r^{2}}}\\&+{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta ){\frac {\cos ^{2}\theta }{r}}-2{\frac {\partial (f\circ \phi )}{\partial \theta }}(r,\theta ){\frac {\cos \theta \sin \theta }{r^{2}}}\end{aligned}}}
(
Δ
f
)
∘
ϕ
(
r
,
θ
)
=
∂
2
(
f
∘
ϕ
)
∂
r
2
(
r
,
θ
)
+
1
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
+
1
r
2
∂
2
(
f
∘
ϕ
)
∂
θ
2
(
r
,
θ
)
{\displaystyle (\Delta f)\circ \phi (r,\theta )={\frac {\partial ^{2}(f\circ \phi )}{\partial r^{2}}}(r,\theta )+{\frac {1}{r}}{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )+{\frac {1}{r^{2}}}{\frac {\partial ^{2}(f\circ \phi )}{\partial \theta ^{2}}}(r,\theta )}
r
∂
(
f
∘
ϕ
)
∂
r
(
r
,
θ
)
=
r
⇔
∂
(
f
∘
ϕ
)
∂
r
=
1
⇔
f
∘
ϕ
(
r
,
θ
)
=
r
+
g
(
θ
)
⇔
f
(
x
,
y
)
=
x
2
+
y
2
+
g
(
arg
(
x
+
i
y
)
)
{\displaystyle r{\frac {\partial (f\circ \phi )}{\partial r}}(r,\theta )=r\Leftrightarrow {\frac {\partial (f\circ \phi )}{\partial r}}=1\Leftrightarrow f\circ \phi (r,\theta )=r+g(\theta )\Leftrightarrow f(x,y)={\sqrt {x^{2}+y^{2}}}+g(\arg(x+\mathrm {i} y))}
g
{\displaystyle g}
2
π
{\displaystyle 2\pi }
(
0
,
0
)
{\displaystyle (0,0)}
g
{\displaystyle g}
Wikipedia-logo-v2.svg Soient
E
{\displaystyle E}
F
{\displaystyle F}
K
{\displaystyle K}
K
=
R
{\displaystyle K=\mathbb {R} }
C
{\displaystyle \mathbb {C} }
C
{\displaystyle C}
cône de
E
{\displaystyle E}
f
:
C
→
F
{\displaystyle f:C\to F}
α
{\displaystyle \alpha }
K
{\displaystyle K}
Montrer que
f
{\displaystyle f}
α
{\displaystyle \alpha }
∀
x
∈
C
∀
t
>
0
f
(
t
x
)
=
t
α
f
(
x
)
{\displaystyle \forall x\in C\quad \forall t>0\quad f(tx)=t^{\alpha }f(x)}
Euler :
∀
x
∈
C
d
f
x
(
x
)
=
α
f
(
x
)
{\displaystyle \forall x\in C\quad \mathrm {d} f_{x}(x)=\alpha f(x)}
Soient
f
:
R
→
R
{\displaystyle f:\mathbb {R} \to \mathbb {R} }
g
:
R
2
→
R
{\displaystyle g:\mathbb {R} ^{2}\to \mathbb {R} }
ϕ
:
R
→
R
,
x
↦
g
(
x
,
−
x
)
{\displaystyle \phi :\mathbb {R} \to \mathbb {R} ,\;x\mapsto g(x,-x)}
ψ
:
R
2
→
R
,
(
x
,
y
)
↦
g
(
y
,
x
)
{\displaystyle \psi :\mathbb {R} ^{2}\to \mathbb {R} ,\;(x,y)\mapsto g(y,x)}
h
:
R
2
→
R
,
(
x
,
y
)
↦
f
(
x
+
g
(
x
,
y
)
)
{\displaystyle h:\mathbb {R} ^{2}\to \mathbb {R} ,\;(x,y)\mapsto f(x+g(x,y))}
k
:
R
2
→
R
,
(
x
,
y
)
↦
f
(
x
y
2
g
(
x
,
y
)
)
{\displaystyle k:\mathbb {R} ^{2}\to \mathbb {R} ,\;(x,y)\mapsto f(xy^{2}g(x,y))}
ℓ
:
R
3
→
R
,
(
x
,
y
,
z
)
↦
f
(
x
2
+
y
2
+
z
2
)
{\displaystyle \ell :\mathbb {R} ^{3}\to \mathbb {R} ,\;(x,y,z)\mapsto f(x^{2}+y^{2}+z^{2})}
Solution
ϕ
=
g
∘
L
{\displaystyle \phi =g\circ L}
L
:
R
→
R
2
,
x
↦
(
x
,
−
x
)
{\displaystyle L:\mathbb {R} \to \mathbb {R} ^{2},\;x\mapsto (x,-x)}
ϕ
{\displaystyle \phi }
d
ϕ
x
=
d
g
L
(
x
)
∘
d
L
x
=
d
g
(
x
,
−
x
)
∘
L
{\displaystyle \mathrm {d} \phi _{x}=\mathrm {d} g_{L(x)}\circ \mathrm {d} L_{x}=\mathrm {d} g_{(x,-x)}\circ L}
d
ϕ
x
(
u
)
=
d
g
(
x
,
−
x
)
(
u
,
−
u
)
=
∂
g
∂
x
(
x
,
−
x
)
u
+
∂
g
∂
y
(
x
,
−
x
)
(
−
u
)
{\displaystyle \mathrm {d} \phi _{x}(u)=\mathrm {d} g_{(x,-x)}(u,-u)={\frac {\partial g}{\partial x}}(x,-x)u+{\frac {\partial g}{\partial y}}(x,-x)(-u)}
ϕ
{\displaystyle \phi }
ϕ
′
(
x
)
=
∂
g
∂
x
(
x
,
−
x
)
−
∂
g
∂
y
(
x
,
−
x
)
{\displaystyle \phi '(x)={\frac {\partial g}{\partial x}}(x,-x)-{\frac {\partial g}{\partial y}}(x,-x)}
ψ
=
g
∘
T
{\displaystyle \psi =g\circ T}
T
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
y
,
x
)
{\displaystyle T:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\;(x,y)\mapsto (y,x)}
ψ
{\displaystyle \psi }
d
ψ
(
x
,
y
)
=
d
g
T
(
x
,
y
)
∘
d
T
(
x
,
y
)
=
d
g
(
y
,
x
)
∘
T
{\displaystyle \mathrm {d} \psi _{(x,y)}=\mathrm {d} g_{\,T(x,y)}\circ \mathrm {d} T_{(x,y)}=\mathrm {d} g_{(y,x)}\circ T}
d
ψ
(
x
,
y
)
(
u
,
v
)
=
d
g
(
y
,
x
)
(
v
,
u
)
=
∂
g
∂
x
(
y
,
x
)
v
+
∂
g
∂
y
(
y
,
x
)
u
{\displaystyle \mathrm {d} \psi _{(x,y)}(u,v)=\mathrm {d} g_{(y,x)}(v,u)={\frac {\partial g}{\partial x}}(y,x)v+{\frac {\partial g}{\partial y}}(y,x)u}
h
=
f
∘
(
p
+
g
)
{\displaystyle h=f\circ (p+g)}
p
:
R
2
→
R
,
(
x
,
y
)
↦
x
{\displaystyle p:\mathbb {R} ^{2}\to \mathbb {R} ,\;(x,y)\mapsto x}
h
{\displaystyle h}
d
h
(
x
,
y
)
=
f
′
(
x
+
g
(
x
,
y
)
)
×
(
d
p
(
x
,
y
)
+
d
g
(
x
,
y
)
)
=
f
′
(
x
+
g
(
x
,
y
)
)
×
(
p
+
d
g
(
x
,
y
)
)
{\displaystyle \mathrm {d} h_{(x,y)}=f'(x+g(x,y))\times (\mathrm {d} p_{(x,y)}+\mathrm {d} g_{(x,y)})=f'(x+g(x,y))\times (p+\mathrm {d} g_{(x,y)})}
d
h
(
x
,
y
)
(
u
,
v
)
=
f
′
(
x
+
g
(
x
,
y
)
)
[
(
1
+
∂
g
∂
x
(
x
,
y
)
)
u
+
∂
g
∂
y
(
x
,
y
)
v
]
{\displaystyle \mathrm {d} h_{(x,y)}(u,v)=f'(x+g(x,y))\left[\left(1+{\frac {\partial g}{\partial x}}(x,y)\right)u+{\frac {\partial g}{\partial y}}(x,y)v\right]}
k
=
f
∘
(
q
×
g
)
{\displaystyle k=f\circ (q\times g)}
q
:
R
2
→
R
,
(
x
,
y
)
↦
x
y
2
{\displaystyle q:\mathbb {R} ^{2}\to \mathbb {R} ,\;(x,y)\mapsto xy^{2}}
d
q
(
x
,
y
)
(
u
,
v
)
=
y
2
u
+
2
x
y
v
{\displaystyle \mathrm {d} q_{(x,y)}(u,v)=y^{2}u+2xyv}
k
{\displaystyle k}
d
k
(
x
,
y
)
=
f
′
(
x
y
2
g
(
x
,
y
)
)
×
(
g
(
x
,
y
)
d
q
(
x
,
y
)
+
q
(
x
,
y
)
d
g
(
x
,
y
)
)
{\displaystyle \mathrm {d} k_{(x,y)}=f'(xy^{2}g(x,y))\times (g(x,y)\mathrm {d} q_{(x,y)}+q(x,y)\mathrm {d} g_{(x,y)})}
d
k
(
x
,
y
)
(
u
,
v
)
=
f
′
(
x
+
g
(
x
,
y
)
)
[
(
g
(
x
,
y
)
y
2
+
x
y
2
∂
g
∂
x
(
x
,
y
)
)
u
+
(
g
(
x
,
y
)
2
x
y
+
x
y
2
∂
g
∂
y
(
x
,
y
)
)
v
]
{\displaystyle {\begin{aligned}\mathrm {d} k_{(x,y)}(u,v)&=f'(x+g(x,y))\left[\left(g(x,y)y^{2}+xy^{2}{\frac {\partial g}{\partial x}}(x,y)\right)u+\left(g(x,y)2xy+xy^{2}{\frac {\partial g}{\partial y}}(x,y)\right)v\right]\end{aligned}}}
ℓ
=
f
∘
r
{\displaystyle \ell =f\circ r}
r
:
R
3
→
R
,
(
x
,
y
)
↦
x
2
+
y
2
+
z
2
{\displaystyle r:\mathbb {R} ^{3}\to \mathbb {R} ,\;(x,y)\mapsto x^{2}+y^{2}+z^{2}}
d
r
(
x
,
y
,
z
)
(
u
,
v
,
w
)
=
2
(
x
u
+
y
v
+
z
w
)
{\displaystyle \mathrm {d} r_{(x,y,z)}(u,v,w)=2(xu+yv+zw)}
ℓ
{\displaystyle \ell }
d
ℓ
(
x
,
y
,
z
)
=
f
′
(
x
2
+
y
2
+
z
2
)
d
r
(
x
,
y
,
z
)
{\displaystyle \mathrm {d} \ell _{(x,y,z)}=f'(x^{2}+y^{2}+z^{2})\mathrm {d} r_{(x,y,z)}}
d
ℓ
(
x
,
y
,
z
)
(
u
,
v
,
w
)
=
2
(
x
u
+
y
v
+
z
w
)
f
′
(
x
2
+
y
2
+
z
2
)
{\displaystyle \mathrm {d} \ell _{(x,y,z)}(u,v,w)=2(xu+yv+zw)f'(x^{2}+y^{2}+z^{2})}
Si
f
=
(
P
,
Q
)
{\displaystyle f=(P,Q)}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
∂
¯
f
=
(
∂
P
∂
x
−
∂
Q
∂
y
,
∂
P
∂
y
+
∂
Q
∂
x
)
{\displaystyle {\bar {\partial }}f=\left({\frac {\partial P}{\partial x}}-{\frac {\partial Q}{\partial y}},{\frac {\partial P}{\partial y}}+{\frac {\partial Q}{\partial x}}\right)}
Calculer
∂
¯
f
{\displaystyle {\bar {\partial }}f}
f
(
x
,
y
)
=
(
x
2
−
y
2
,
2
x
y
)
{\displaystyle f(x,y)=(x^{2}-y^{2},2xy)}
Justifier que si
f
{\displaystyle f}
g
{\displaystyle g}
∂
¯
f
=
∂
¯
g
=
0
{\displaystyle {\bar {\partial }}f={\bar {\partial }}g=0}
∂
¯
(
g
∘
f
)
=
0
{\displaystyle {\bar {\partial }}(g\circ f)=0}
Solution
∂
¯
f
(
x
,
y
)
=
(
2
x
−
2
x
,
−
2
y
+
2
y
)
=
(
0
,
0
)
{\displaystyle {\bar {\partial }}f(x,y)=\left(2x-2x,-2y+2y\right)=(0,0)}
Soient
f
=
(
P
,
Q
)
{\displaystyle f=(P,Q)}
g
=
(
R
,
S
)
{\displaystyle g=(R,S)}
∂
P
∂
x
−
∂
Q
∂
y
=
∂
P
∂
y
+
∂
Q
∂
x
=
∂
R
∂
x
−
∂
S
∂
y
=
∂
R
∂
y
+
∂
S
∂
x
=
0
{\displaystyle {\frac {\partial P}{\partial x}}-{\frac {\partial Q}{\partial y}}={\frac {\partial P}{\partial y}}+{\frac {\partial Q}{\partial x}}={\frac {\partial R}{\partial x}}-{\frac {\partial S}{\partial y}}={\frac {\partial R}{\partial y}}+{\frac {\partial S}{\partial x}}=0}
(
∂
R
∂
x
∂
P
∂
x
+
∂
R
∂
y
∂
Q
∂
x
)
−
(
∂
S
∂
x
∂
P
∂
y
+
∂
S
∂
y
∂
Q
∂
y
)
=
(
∂
R
∂
x
∂
P
∂
x
−
∂
R
∂
y
∂
P
∂
y
)
−
(
−
∂
R
∂
y
∂
P
∂
y
+
∂
R
∂
x
∂
P
∂
x
)
=
0
{\displaystyle \left({\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial x}}+{\frac {\partial R}{\partial y}}{\frac {\partial Q}{\partial x}}\right)-\left({\frac {\partial S}{\partial x}}{\frac {\partial P}{\partial y}}+{\frac {\partial S}{\partial y}}{\frac {\partial Q}{\partial y}}\right)=\left({\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial x}}-{\frac {\partial R}{\partial y}}{\frac {\partial P}{\partial y}}\right)-\left(-{\frac {\partial R}{\partial y}}{\frac {\partial P}{\partial y}}+{\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial x}}\right)=0}
(
∂
R
∂
x
∂
P
∂
y
+
∂
R
∂
y
∂
Q
∂
y
)
+
(
∂
S
∂
x
∂
P
∂
x
+
∂
S
∂
y
∂
Q
∂
x
)
=
(
∂
R
∂
x
∂
P
∂
y
+
∂
R
∂
y
∂
P
∂
x
)
+
(
−
∂
R
∂
y
∂
P
∂
x
−
∂
R
∂
x
∂
P
∂
y
)
=
0
{\displaystyle \left({\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial y}}+{\frac {\partial R}{\partial y}}{\frac {\partial Q}{\partial y}}\right)+\left({\frac {\partial S}{\partial x}}{\frac {\partial P}{\partial x}}+{\frac {\partial S}{\partial y}}{\frac {\partial Q}{\partial x}}\right)=\left({\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial y}}+{\frac {\partial R}{\partial y}}{\frac {\partial P}{\partial x}}\right)+\left(-{\frac {\partial R}{\partial y}}{\frac {\partial P}{\partial x}}-{\frac {\partial R}{\partial x}}{\frac {\partial P}{\partial y}}\right)=0}
Ou plus simplement : dire que
∂
¯
f
=
0
{\displaystyle {\bar {\partial }}f=0}
f
{\displaystyle f}
similitude directe ou la matrice nulle, or l'ensemble de ces matrices est stable par produit.
Remarque : vue comme fonction de
C
{\displaystyle \mathbb {C} }
C
{\displaystyle \mathbb {C} }
z
↦
z
2
{\displaystyle z\mapsto z^{2}}
n
{\displaystyle n}
f
{\displaystyle f}
z
↦
z
n
{\displaystyle z\mapsto z^{n}}
∂
¯
f
=
0
{\displaystyle {\bar {\partial }}f=0}
Voir aussi : Fonctions d'une variable complexe/Fonctions holomorphes .
On considère la fonction
f
:
R
→
C
,
x
↦
e
i
x
{\displaystyle f:\mathbb {R} \to \mathbb {C} ,\ x\mapsto \operatorname {e} ^{\mathrm {i} x}}
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
f
(
b
)
−
f
(
a
)
=
(
b
−
a
)
f
′
(
c
)
{\displaystyle f(b)-f(a)=(b-a)f'(c)}
Solution
Pour tout réel
c
{\displaystyle c}
|
(
b
−
a
)
f
′
(
c
)
|
=
|
(
b
−
a
)
i
e
i
c
|
=
|
b
−
a
|
{\displaystyle |(b-a)f'(c)|=\left|(b-a)\mathrm {i} \operatorname {e} ^{\mathrm {i} c}\right|=|b-a|}
alors que si
b
≠
a
{\displaystyle b\neq a}
|
f
(
b
)
−
f
(
a
)
|
=
|
e
i
(
b
+
a
)
/
2
×
2
i
sin
b
−
a
2
|
=
2
|
sin
b
−
a
2
|
<
2
|
b
−
a
2
|
=
|
b
−
a
|
{\displaystyle |f(b)-f(a)|=\left|\operatorname {e} ^{\mathrm {i} (b+a)/2}\times 2\mathrm {i} \sin {\frac {b-a}{2}}\right|=2\left|\sin {\frac {b-a}{2}}\right|<2\left|{\frac {b-a}{2}}\right|=|b-a|}
Il n'y a donc pas d'égalité des accroissements finis pour les fonctions à valeurs vectorielles. Remarquons que pour
b
−
a
∈
2
π
Z
{\displaystyle b-a\in 2\pi \mathbb {Z} }
f
(
b
)
−
f
(
a
)
=
0
{\displaystyle f(b)-f(a)=0}
Soit
F
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
cos
x
−
sin
y
,
sin
x
−
cos
y
)
{\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\;(x,y)\mapsto (\cos x-\sin y,\sin x-\cos y)}
Justifier que
F
{\displaystyle F}
Montrer que
∀
M
∈
R
2
|
|
|
d
F
M
|
|
|
≤
2
{\displaystyle \forall M\in \mathbb {R} ^{2}\quad |\!|\!|\mathrm {d} F_{M}|\!|\!|\leq {\sqrt {2}}}
|
|
|
|
|
|
{\displaystyle |\!|\!|~|\!|\!|}
R
2
{\displaystyle \mathbb {R} ^{2}}
En déduire que pour tout
M
0
∈
R
2
{\displaystyle M_{0}\in \mathbb {R} ^{2}}
M
{\displaystyle M}
∀
n
∈
N
M
n
+
1
=
1
2
F
(
M
n
)
{\displaystyle \forall n\in \mathbb {N} \quad M_{n+1}={\frac {1}{2}}F(M_{n})}
Solution
F
{\displaystyle F}
∞ car ses deux composantes
F
1
{\displaystyle F_{1}}
F
2
{\displaystyle F_{2}}
F
1
=
cos
∘
p
1
−
sin
∘
p
2
{\displaystyle F_{1}=\cos \circ p_{1}-\sin \circ p_{2}}
F
2
=
sin
∘
p
1
−
cos
∘
p
2
{\displaystyle F_{2}=\sin \circ p_{1}-\cos \circ p_{2}}
p
1
,
p
2
:
R
2
→
R
{\displaystyle p_{1},p_{2}:\mathbb {R} ^{2}\to \mathbb {R} }
La matrice jacobienne de
F
{\displaystyle F}
(
x
,
y
)
{\displaystyle (x,y)}
(
−
sin
x
−
cos
y
cos
x
sin
y
)
{\displaystyle {\begin{pmatrix}-\sin x&-\cos y\\\cos x&\sin y\end{pmatrix}}}
‖
d
F
(
x
,
y
)
(
u
,
v
)
‖
2
=
‖
(
−
u
sin
x
−
v
cos
y
,
u
cos
x
+
v
sin
y
)
‖
2
=
(
−
u
sin
x
−
v
cos
y
)
2
+
(
u
cos
x
+
v
sin
y
)
2
=
u
2
+
v
2
+
2
u
v
sin
(
x
+
y
)
≤
2
(
u
2
+
v
2
)
{\displaystyle \|\mathrm {d} F_{(x,y)}(u,v)\|^{2}=\|(-u\sin x-v\cos y,u\cos x+v\sin y)\|^{2}=(-u\sin x-v\cos y)^{2}+(u\cos x+v\sin y)^{2}=u^{2}+v^{2}+2uv\sin(x+y)\leq 2(u^{2}+v^{2})}
D'après l'inégalité des accroissements finis,
1
2
F
{\displaystyle {\frac {1}{2}}F}
2
2
{\displaystyle {\frac {\sqrt {2}}{2}}}
théorème du point fixe de Picard-Banach .
Déterminer les fonctions dérivables
f
:
R
→
R
{\displaystyle f:\mathbb {R} \to \mathbb {R} }
∀
(
x
,
y
)
∈
R
2
f
(
x
+
y
)
=
f
(
x
+
f
(
y
)
)
{\displaystyle \forall (x,y)\in \mathbb {R} ^{2}\quad f(x+y)=f(x+f(y))}
Pour une étude complète de cette équation fonctionnelle , voir Topologie générale/Exercices/Topologie de R ou C#Exercice 4 .
Redémontrer directement le corollaire 1 du cours (dont le corollaire 2 est une conséquence immédiate très utile) dans le cas particulier plus simple où
F
{\displaystyle F}
espace euclidien
R
n
{\displaystyle \mathbb {R} ^{n}}
Soient
a
<
b
{\displaystyle a<b}
f
:
[
a
,
b
]
→
R
n
{\displaystyle f:\left[a,b\right]\to \mathbb {R} ^{n}}
[
a
,
b
]
{\displaystyle \left[a,b\right]}
]
a
,
b
[
{\displaystyle \left]a,b\right[}
∀
t
∈
]
a
,
b
[
‖
f
′
(
t
)
‖
≤
M
{\displaystyle \forall t\in \left]a,b\right[\quad \|f'(t)\|\leq M}
‖
f
(
b
)
−
f
(
a
)
‖
≤
M
(
b
−
a
)
{\displaystyle \|f(b)-f(a)\|\leq M(b-a)}
Indication : appliquer le théorème des accroissements finis usuel à la fonction
g
:
[
a
,
b
]
→
R
,
t
↦
⟨
f
(
t
)
∣
f
(
b
)
−
f
(
a
)
⟩
{\displaystyle g:\left[a,b\right]\to \mathbb {R} ,\,t\mapsto \langle f(t)\mid f(b)-f(a)\rangle }
Solution
g
{\displaystyle g}
[
a
,
b
]
{\displaystyle \left[a,b\right]}
]
a
,
b
[
{\displaystyle \left]a,b\right[}
∀
t
∈
]
a
,
b
[
g
′
(
t
)
=
⟨
f
′
(
t
)
∣
f
(
b
)
−
f
(
a
)
⟩
≤
M
‖
f
(
b
)
−
f
(
a
)
‖
{\displaystyle \forall t\in \left]a,b\right[\quad g'(t)=\langle f'(t)\mid f(b)-f(a)\rangle \leq M\|f(b)-f(a)\|}
D'après le théorème des accroissements finis,
∃
c
∈
]
a
,
b
[
(
b
−
a
)
g
′
(
c
)
=
g
(
b
)
−
g
(
a
)
=
‖
f
(
b
)
−
f
(
a
)
‖
2
{\displaystyle \exists c\in \left]a,b\right[\quad (b-a)g'(c)=g(b)-g(a)=\|f(b)-f(a)\|^{2}}
‖
f
(
b
)
−
f
(
a
)
‖
2
≤
(
b
−
a
)
M
‖
f
(
b
)
−
f
(
a
)
‖
{\displaystyle \|f(b)-f(a)\|^{2}\leq (b-a)M\|f(b)-f(a)\|}
Wikipedia-logo-v2.svg Soient
E
{\displaystyle E}
F
{\displaystyle F}
f
:
E
→
F
{\displaystyle f:E\to F}
f
{\displaystyle f}
a
∈
E
{\displaystyle a\in E}
u
∈
E
{\displaystyle u\in E}
lim
t
→
0
t
≠
0
f
(
a
+
t
u
)
−
f
(
a
)
t
{\displaystyle \lim _{t\to 0 \atop t\neq 0}{\frac {f(a+tu)-f(a)}{t}}}
Vérifier que si cette dérivée directionnelle selon
u
{\displaystyle u}
λ
u
{\displaystyle \lambda u}
λ
{\displaystyle \lambda }
u
{\displaystyle u}
λ
{\displaystyle \lambda }
Montrer que si
f
{\displaystyle f}
a
{\displaystyle a}
a
{\displaystyle a}
u
{\displaystyle u}
d
f
a
(
u
)
{\displaystyle \mathrm {d} f_{a}(u)}
u
∈
E
{\displaystyle u\in E}
Montrer que si, pour toute courbe
γ
:
R
→
E
{\displaystyle \gamma :\mathbb {R} \to E}
γ
(
0
)
=
a
{\displaystyle \gamma (0)=a}
γ
′
(
0
)
=
u
{\displaystyle \gamma '(0)=u}
f
∘
γ
{\displaystyle f\circ \gamma }
0
{\displaystyle 0}
f
{\displaystyle f}
a
{\displaystyle a}
u
{\displaystyle u}
![{\displaystyle \varepsilon :h\mapsto \sup _{x\in \left]a,a+h\right[}|\!|\!|\mathrm {d} g_{x}|\!|\!|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6960d2954f88f6a255e7b2e4b605cefadc31e2bf)
![{\displaystyle 0\notin \left]h,k\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/654c799abdff69b755b3dfacd9b8dd46ce4e6940)
![{\displaystyle 0\in \left]h,k\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1340f870383da9788c5ce7056bc0efafb9ea31aa)
![{\displaystyle g:\left]a,b\right[\to F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9f1c9461ba74a882794845bdca95a132c1345a1)
![{\displaystyle \left]a,b\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e46a283d90c9cfc7e343f61c41081b47a4577996)
![{\displaystyle \left]b,a\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce1be5efbee244550947fdd25b357c8ce619f81f)
![{\displaystyle u:\left[-r,r\right]\to E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51a6709fdca50022dcc1f82fbc268654a52be5d0)
![{\displaystyle \forall t\in \left[-r,r\right]\quad \left(t,u\left(t\right)\right)\in \Omega }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3fe48dd728a0cd7fb56f28af8a097d93cf68603)
![{\displaystyle \left\|f\left(t,x+k\right)-f\left(t,x\right)-\operatorname {d} _{2}f_{\left(t,x\right)}\left(k\right)\right\|\leq \left\|k\right\|\sup _{z\in \left[x,x+k\right]}|\!|\!|\operatorname {d} _{2}f_{\left(t,z\right)}-\operatorname {d} _{2}f_{\left(t,x\right)}|\!|\!|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd83416a1368be433f28d2ee1af8c59125c5493b)
![{\displaystyle \varepsilon \left(\eta \right)=\sup _{\left(t,x\right)\in \left(\operatorname {id} ,u\right)\left(\left[-r,r\right]\right),\ \left\|z-x\right\|\leq \eta }|\!|\!|\operatorname {d} _{2}f_{\left(t,z\right)}-\operatorname {d} _{2}f_{\left(t,x\right)}|\!|\!|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8051af15056e4cf3ae36b2f59edfec86698d01c2)
![{\displaystyle \left[-r,r\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7a39448a1ef195b3beccc692f50b8aef0391278)
![{\displaystyle I:=\left[-1,1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90837c11799b3ad382be7de17930877bdb47195e)
![{\displaystyle \phi :\left]0,+\infty \right[\times \left]-\pi ,\pi \right[\to \mathbb {R} ^{2}\setminus \{\left(0,0\right)\},\;\left(r,\theta \right)\mapsto \left(r\cos \theta ,r\sin \theta \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/434575d52d470d716c23413e4b220f7df451dda9)
![{\displaystyle \left]0,+\infty \right[\to F,\,t\mapsto t^{-\alpha }f(tx)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1a62eb478dd176478a5554c73b2dcdf979e4b09)
![{\displaystyle \mathrm {d} h_{(x,y)}(u,v)=f'(x+g(x,y))\left[\left(1+{\frac {\partial g}{\partial x}}(x,y)\right)u+{\frac {\partial g}{\partial y}}(x,y)v\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19d6bf71cc48cd54cb17aaf36f50ded7dc3b7be8)
![{\displaystyle {\begin{aligned}\mathrm {d} k_{(x,y)}(u,v)&=f'(x+g(x,y))\left[\left(g(x,y)y^{2}+xy^{2}{\frac {\partial g}{\partial x}}(x,y)\right)u+\left(g(x,y)2xy+xy^{2}{\frac {\partial g}{\partial y}}(x,y)\right)v\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65631a644cdcba071f1d0b1df9b938eebd6e8a9c)
![{\displaystyle f:\left[a,b\right]\to \mathbb {R} ^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/236dc81ddea52e3b09be8c7e0bc02533f1fd613d)
![{\displaystyle \left[a,b\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f30926fb280a9fdf66fd931e14d4363cb824feaa)
![{\displaystyle \forall t\in \left]a,b\right[\quad \|f'(t)\|\leq M}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c22e1b73233fb70349ee396699e6f35db8619cb)
![{\displaystyle g:\left[a,b\right]\to \mathbb {R} ,\,t\mapsto \langle f(t)\mid f(b)-f(a)\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f919f4d496ced5da7bfb82073a62862965f035f)
![{\displaystyle \forall t\in \left]a,b\right[\quad g'(t)=\langle f'(t)\mid f(b)-f(a)\rangle \leq M\|f(b)-f(a)\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eb1694c8e5e1e2d1f2a9914d4c19bca0549e203)
![{\displaystyle \exists c\in \left]a,b\right[\quad (b-a)g'(c)=g(b)-g(a)=\|f(b)-f(a)\|^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e73555ad62d0152932a7cfbad2f517fc0dcae8d9)
