proof of implicit function theorem
We state the Theorem with a different notation:
Theorem 1.
Let Ω be an open subset of Rn×Rm and let f∈C1(Ω,Rm). Let (x0,y0)∈Ω⊂Rn×Rm. If the matrix Dyf(x0,y0) defined by
Dyf(x0,y0)=(∂fj∂yk(x0,y0))j,k j=1,…,m k=1,…,m |
is invertible, then there exists a neighbourhood U⊂Rn of x0,
a neighbourhood V⊂Rm of y0
and a function
g∈C1(U,V) such that
f(x,y)=f(x0,y0)⇔y=g(x) ∀(x,y)∈U×V. |
Moreover
Dg(x)=-(Dyf(x,g(x)))-1Dxf(x,g(x)). |
Proof.
Consider the function F∈𝒞1(Ω,ℝn×ℝm) defined by
F(x,y)=(x,f(x,y)). |
One finds that
DF(x,y)=(Im0DxfDyf). |
Being Dyf(x0,y0) invertible, DF(x0,y0) is invertible too.
Applying the inverse function Theorem to F
we find that there exist a neighbourhood U of x0 and V of y0 and
a function G∈C1(U×V,ℝn+m) such that F(G(x,y))=(x,y)
for all (x,y)∈U×V. Letting G(x,y)=(G1(x,y),G2(x,y))
(so that G1:V×W→ℝn, G2:V×W→ℝm)
we hence have
(x,y)=F(G1(x,y),G2(x,y))=(G1(x,y),f(G1(x,y),G2(x,y))) |
and hence x=G1(x,y) and y=f(G1(x,y),G2(x,y))=f(x,G2(x,y)). So we only have to set g(x)=G2(x,f(x0,y0)) to obtain
f(x,g(x))=f(x0,y0),∀x∈U. |
Differentiating with respect to x we obtain
Dxf(x,g(x))+Dyf(x,g(x))Dg(x)=0 |
which gives the desired formula for the computation of Dg.
∎
Title | proof of implicit function theorem |
---|---|
Canonical name | ProofOfImplicitFunctionTheorem |
Date of creation | 2013-03-22 13:31:23 |
Last modified on | 2013-03-22 13:31:23 |
Owner | paolini (1187) |
Last modified by | paolini (1187) |
Numerical id | 8 |
Author | paolini (1187) |
Entry type | Proof |
Classification | msc 26B10 |