You have f : X × Λ → Y where here X,Λ are Banach spaces. Suppose
(0,0)
∈ X × Λ
and f
(0,0)
= 0. Then if D_{1}f
(0,0)
^{−1} is in ℒ
(Y,X)
, the implicit function theorem says
that there exists x
(λ)
a C^{p} function such that locally f
(x(λ),λ)
= 0. So what if
D_{1}f
(0,0)
fails to be one to one? Sometimes this case is also considered. It may be that
D_{1}f
(0,0)
is one to one on some subspace and other nice things happen. In particular,
suppose the following.
Letting X_{2}≡ kerD_{1}f
(0,0)
assume
X = X ⊕ X , dim (X ) < ∞
1 2 2
where X_{1} is a closed subspace. Thus D_{1}f
(0,0)
is one to one on X_{1}. We let
Y1 = D1f (0,0) (X1 )
and suppose that Y = Y_{1}⊕ Y_{2} where dim
(Y )
2
< ∞,Y_{1} also a closed subspace.
X1 D1f→(0,0)Y1 = D1f (0,0)(X1 ), Y1 closed
Y = Y1 ⊕Y2, dim (Y2) < ∞ < ∞
By the open mapping theorem, D_{1}f
(0,0)
^{−1} is also continuous.
Let Q be a continuous projection onto Y_{1} which is assumed to
exist^{2}
so that
(I − Q )
is a projection onto Y_{2}. Then the equation f
(x(λ),λ)
= 0 can be
written as the pair
Qf (x,λ) = 0
(I − Q)f (x,λ) = 0
Consider the top. For x = x_{1} + x_{2} where x_{i}∈ X_{i}, this is
Qf (x1 + x2,λ) = 0
Then if g
(x ,x ,λ)
1 2
= Qf
(x + x ,λ)
1 2
, one has g : X_{1}× X_{2}× Λ → Y_{1}
D1g (x1,x2,λ)h = D1Qf (x1 + x2,λ)h, h ∈ X1.
Thus D_{1}g
(0,0,0)
^{−1} is continuous by the open mapping theorem (D_{1}f
(0,0)
is
one to one on X_{1}), and by the implicit function theorem, there is a solution
to
Qf (x1 + x2,λ) = 0
for x_{1} = x_{1}
(x ,λ )
2
. (Note how it is important that X_{1} and Y_{1} be Banach spaces.) Then
the other equation yields
(I − Q)f (x1(x2,λ) + x2,λ) = 0
and so for fixed λ, this is a finite set of equations of a variable in a finite dimensional
space.
This depends on being able to write X = X_{1}⊕ X_{2} where X_{1} is closed,
X_{2} = kerD_{1}f
(0,0)
, a similar situation for Y = Y_{1}⊕Y_{2}. So when does this happen? Are
there conditions on D_{1}f
(0,0)
which will cause it to occur?
There are such conditions. For example, D_{1}f
(0,0)
could be a Fredholm operator
defined in Definition 15.6.7. The following are some easy examples in which all that
nonsense about things being finite dimensional and part of a direct sum does not need to
be considered.
Example 20.12.1Say X = ℝ^{2}and Λ = ℝ. Let f
(x,y,λ)
= x + xy + y^{2} + λ.Then
D1f (0,0,0) = (1,0)
this 1 × 2 matrix mapping ℝ^{2}to ℝ. Thus X_{2} =
(0,α)
^{T} : α ∈ ℝ and X_{1} =
(α,0)
^{T} : α ∈ ℝ.In this case, Y_{1} = ℝ and so Q = I. Thus the above reduces to the single equation
f ((α,0)+ (0,β),λ) = 0
and so since D_{1}f
(0,0,0)
is one to one, x_{1} =
(α,0)
= x_{1}
((0,β),λ)
. Of course this iscompletely obvious because if you consider f in the natural way as a function of threevariables, then the implicit function theorem immediately gives x = x
(y,λ)
which isessentially the same result. We just write
(α,0)
in place of α. The first independentvariable is a function of the other two.
Example 20.12.2Here is another easy example. f : ℝ^{2}× ℝ → ℝ^{2}
from the first equation. Both of these are easily
seen to be 0. Now consider x_{yy}. After some computations, this is seen to be
x_{yy} = −2. Similarly, x_{yλ}
(0,0)
= 0,x_{λλ}
(0,0)
= 0 also. Thus up to terms of degree
3,
1
x (y,λ) = − y2 = 2 (− 2)y2
Place this in the bottom equation.
1y2λ − 1y2sin λ− 1 y3 + 1y4 = 0
2 2 2 2
Now the idea is to find y = y
(λ)
, hopefully nonzero. Divide by y^{2} and multiply by
2.
y2 − y + λ− sin λ = 0
Then for small λ this is approximately equal to
3
y2 − y+ λ- = 0
6
Then a solution for y for small λ is
∘ -------
1 + 1 − 2λ3
y = ---------3--
2
Of course there is another solution as well, when you replace the + with a minus sign.
This is the one we want because when λ = 0 it reduces to y = 0. This shows that
there exist solutions to the equations f
where higher order terms are neglected. Thus there exist other solutions than
the zero solution even though λ may be nonzero. Note that in this example,
f