Implicit Function Theorem in several variables
2014-01-29, MAT237 tutorial, Fabian Parsch
Last week, we talked about the first version of the implicit function theorem: We had one equation
F (x1 , x2 , . . . , xn , y) = 0 and we wanted to express y locally as the graph of a function y = f (x1 , . . . , xk ).
Now, we are dealing with the multidimensional case, i.e. something like this:


0
 0 


F (x1 , . . . , xn , y1 , . . . , yk ) = 0 =  . 
 .. 
0
Keep in mind: Although we are lazy and might just write F (x, y) = 0, the zero on the right side now is a
k-dimensional vector with zeroes. Hence, F is a vector valued function with k coordinates.
Another way to think of this: You have as many equations as you have variables that you want to solve for
(= that you want to describe locally as the graph of a function of the other variables).
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Can we solve for y = f (x)?
In the multivariable case, at the moment, we mostly care about the question if we can locally solve the
equation for y. We don’t need to find the actual y = f (x). To check if this is possible, we need to derive
F with respect to the variables we want to solve for and check the determinant. Let’s have a look at the
following exercise:
1.1 Exercise. Investigate the possibility of solving the equations
u3 + xv − y = 0
v 3 + yu − x = 0
for any two of the variables as functions of the other two near the point (x, y, u, v) = (0, 1, 1, −1)
The function we are considering is
F (x, y, u, v) =
u3 + xv − y
v 3 + yu − x
=0
For the Implicit Function Theorem to be applicable we need F (0, 1, 1, −1) = 0 (which is easy to check and
true) and – this is the interesting part – the determinant of the derivative with respect to the variables we
want to solve for is nonzero (Please make sure that you understand this sentence before you proceed).
There are 42 = 6 options to pick two of the four variables. We will only check two of them here, that should
give you an idea how it works:
∂F (x, y, u, v)
=
∂(x, y)
v
−1
−1
u
=
−1
−1
1
−1
1
⇒ det
∂F (x, y, u, v)
= −2 6= 0
∂(x, y)
∂F (x, y, u, v)
=
∂(u, v)
3u2
y
x
3v 2
3
1
=
0
3
⇒ det
∂F (x, y, u, v)
= 9 6= 0
∂(u, v)
So we can locally describe (x, y) = f (u, v) and also (u, v) = f (x, y). Checking the other determinants, you
will get that all other options are also possible, e.g. we can describe (x, u) = f (y, v) and so on.
The following exercise from the textbook is a more abstract question:
1.2 Exercise. Suppose F (x, y) is a C 1 function such that F (0, 0) = 0. What conditions on F will guarantee
that the equation F (F (x, y), y) = 0 can be solved for y as a C 1 function of x near (0, 0)?
Let G(x, y) := F (F (x, y), y). First we need to check that G(0, 0) = 0. This is indeed true since we know
that F (0, 0) = 0:
G(0, 0) = F (F (0, 0), 0) = F (0, 0) = 0
Additionally, we need ∂y G(0, 0) 6= 0. To make the chain rule applicable in a nice way, let’s define
H(x, y) = (H1 (x, y), H2 (x, y)) = (F (x, y), y)
Then:
G(x, y) = F (H(x, y))
Using the chain rule, we get:
∂x H1 (x, y) ∂y H1 (x, y)
DG(x, y) = ∂x F (H(x, y)) ∂y F (H(x, y)) ·
∂x H2 (x, y) ∂y H2 (x, y)
∂x F (x, y) ∂y F (x, y)
= ∂x F (F (x, y), y) ∂y F (F (x, y), y) ·
0
1
= something we don’t care about ∂x F (F (x, y), y) · ∂y F (x, y) + ∂y F (F (x, y), y) · 1
Note that there are many ways to get the derivative ∂y G(x, y). Using the chain rule might be a bit of a
“sledgehammer-method” here, but it is a nice exercise to practice the chain rule for multivariable functions.
So we can use F (0, 0) = 0 to rewrite our condition:
0 6= ∂y G(0, 0) = ∂x F (F (0, 0), 0) · ∂y F (0, 0) + ∂y F (F (0, 0), 0) = ∂x F (0, 0) · ∂y F (0, 0) + ∂y F (0, 0)
= ∂y F (0, 0) · (∂x F (0, 0) + 1)
This is true if and only if ∂y F (0, 0) 6= 0 and ∂x F (0, 0) 6= −1. Hence, this is the condition on F (x, y) such
that the Implicit Function Theorem applies.
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