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 D1f
(0,0)
−1 is in ℒ
(Y,X)
, the implicit function theorem says
that there exists x
(λ)
a Cp function such that locally f
(x(λ),λ)
= 0. So what if
D1f
(0,0)
fails to be one to one? Sometimes this case is also considered. It may be that
D1f
(0,0)
is one to one on some subspace and other nice things happen. In particular,
suppose the following.
Letting X2≡ kerD1f
(0,0)
assume
X = X ⊕ X , dim (X ) < ∞
1 2 2
where X1 is a closed subspace. Thus D1f
(0,0)
is one to one on X1. We let
Y1 = D1f (0,0) (X1 )
and suppose that Y = Y1⊕ Y2 where dim
(Y )
2
< ∞,Y1 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, D1f
(0,0)
−1 is also continuous.
Let Q be a continuous projection onto Y1 which is assumed to
exist2
so that
(I − Q )
is a projection onto Y2. 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 = x1 + x2 where xi∈ Xi, this is
Qf (x1 + x2,λ) = 0
Then if g
(x ,x ,λ)
1 2
= Qf
(x + x ,λ)
1 2
, one has g : X1× X2× Λ → Y1
D1g (x1,x2,λ)h = D1Qf (x1 + x2,λ)h, h ∈ X1.
Thus D1g
(0,0,0)
−1 is continuous by the open mapping theorem (D1f
(0,0)
is
one to one on X1), and by the implicit function theorem, there is a solution
to
Qf (x1 + x2,λ) = 0
for x1 = x1
(x ,λ )
2
. (Note how it is important that X1 and Y1 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 = X1⊕ X2 where X1 is closed,
X2 = kerD1f
(0,0)
, a similar situation for Y = Y1⊕Y2. So when does this happen? Are
there conditions on D1f
(0,0)
which will cause it to occur?
There are such conditions. For example, D1f
(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 = ℝ2and Λ = ℝ. Let f
(x,y,λ)
= x + xy + y2 + λ.Then
D1f (0,0,0) = (1,0)
this 1 × 2 matrix mapping ℝ2to ℝ. Thus X2 =
(0,α)
T : α ∈ ℝ and X1 =
(α,0)
T : α ∈ ℝ.In this case, Y1 = ℝ and so Q = I. Thus the above reduces to the single equation
f ((α,0)+ (0,β),λ) = 0
and so since D1f
(0,0,0)
is one to one, x1 =
(α,0)
= x1
((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 : ℝ2× ℝ → ℝ2
from the first equation. Both of these are easily
seen to be 0. Now consider xyy. After some computations, this is seen to be
xyy = −2. Similarly, xyλ
(0,0)
= 0,xλλ
(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 y2 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