MATH20132 Calculus of Several Variable.

Solutions 6: Implicit & Inverse functions

v2 2017

1. Let f : R2 → R2 be the function

f(x) =

(exp(x) cos(y)

exp(x) sin(y)

),

where x = (x, y)T ∈ R2.

i. Prove that dfa : R2 → R2 has an inverse for all a ∈ R2 but that f doesnot have an inverse.

ii. Let f0 : U = { (x, y)T ∈ R2 | −π < y < π} → R2 be the restriction of f .Prove that f0 is an injection.

iii. If g : f0(U) → U is inverse of f0 find the derivative dgb : R2 → R2 atb = f0 (a) .

Hint: Use the Chain Rule.

Solution i. The Jacobian matrix of f is

Jf(x) =

(ex cos y −ex sin y

ex sin y ex cos y

),

which has determinant e2x(cos2 y + sin2 y

)= e2x 6= 0 for all x ∈ R2. Thus,

for any a ∈ R2, Jf (a) is invertible, in which case dfa has an inverse, namelythe linear function associated with the matrix Jf (a)−1.

Becausef(

(x, y + 2π)T)

= f(

(x, y)T)

the function is not an injection and thus does not have an inverse.

ii. To see that f0 is an injection assume f0 (x1) = f0 (x2) for some x1 =(x1, y1)

T ,x2 = (x2, y2)T ∈ U . Then

ex1 = |f0 (x1)| = |f0 (x2)| = ex2 ,

so x1 = x2 = x, say. Next(ex cos y1ex sin y1

)= f0 (x1) = f0 (x2) =

(ex cos y2ex sin y2

).

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so cos y1 = cos y2 and sin y1 = sin y2. The only solution of these in therestricted range −π < y1, y2 < π is y1 = y2. Hence

x1 = (x1, y1)T = (x2, y2)

T = x2,

and so f0 is injective.

iii. Since g ◦ f0 = 1U the Chain Rule gives Jg (f0(a)) Jf0(a) = I2, so, asnoted above,

Jg (b) = Jf0(a)−1 =

(ea cos b −ea sin b

ea sin b ea cos b

)−1

= e−2a

(ea cos b ea sin b

−ea sin b ea cos b

)=

(e−a cos b e−a sin b

−e−a sin b e−a cos b

)

This is the matrix associated with the derivative dgb.

2. In the notes we showed that every graph is an image set and every graphis a level set. Here you are asked this in the case for a particular graph.

Define φ : R2 → R2, (x, y)T → (xy2, x2 + y)T

.

i. The graphGφ is the image of some function F. Find F and it’s Jacobianmatrix.

ii. The graph Gφ can be expressed as a level set of a system of equa-tions. Find the system of equations and find the Jacobian matrix ofthe system.

Solution i. F : R2 → R4 is given by

F(x) =

(x

φ (x)

), (1)

for x ∈ R2. Equivalently,

F

((x

y

))=

x

y

xy2

x2 + y

2

From this second form we see that

JF

((x

y

))=

1 0

0 1

y2 2xy

2x 1

=

(I2

Jφ(x)

),

where I2 is the 2×2 identity matrix. Hopefully you could have derived thislast form for JF(x) directly from (1).

ii. Following an Example in the notes, given φ : Rn−m → Rm define the levelset on Rn as the vectors(

x

y

)∈ Rn, x ∈ Rn−m, y ∈ Rm satisfying y − φ(x) = 0.

Alternatively, the level set is the set of vectors t, where

t =

t1

t2

...tn

∈ Rn satisfying

tn−m+1

tn−m+2

...tn

− f

t1

t2

...tn−m

= 0.

In the present case the system is the set of t = (s, t, u, v)T ∈ R4 satisfying(uv

)− φ

((st

))= 0, that is,

(u− st2

v − s2 − t

)= 0.

Thus the level set is

u− st2 = 0

v − s2 − t = 0.

The Jacobian matrix of this system is(−t2 −2st 1 0

−2s −1 0 1

). (2)

Note that the system was given as y−φ (x) = 0 and not φ (x)− y = 0 forthe latter would have given (2) but with the identity matrix, seen in the lasttwo columns, multiplied by −1.

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3. Level sets are locally graphs; the Implicit Function Theorem: Suppose thatf : U → Rm is a C1-function on an open set U ⊆ Rn where 1 ≤ m < n, andthere exists p ∈ U such that f(p) = 0, Jf(p) has full-rank m, and the finalm columns of Jf(p) are linearly independent. Write

p =

(qr

),

where q ∈ Rn−m and r ∈ Rm. Then there exists

• an open set V : q ∈ V ⊆ Rn−m,

• a C1-function φ : V → Rm and

• an open set W : p ∈ W ⊆ U

such that for(xT ,yT

)T ∈ W,f

((xy

))= 0 if, and only if x ∈ V and y = φ(x) .

i. a. Prove, using the Implicit Function Theorem, that for the solutionsof

x2 + y2 + 2uv = 4

x3 + y3 + u3 − v3 = 0,

there exists an open subset of R4 containing the solution p =(−1, 1, 1, 1) in which the u and v can be given as functions of xand y, with (x, y)T in some open subset of R2 containing the pointq = (−1, 1)T .

b. Find the partial derivatives of u and v with respect to x and y atq.

c. Is there any open subset of R4 containing p in which y and u canbe given as functions of x and v? What happens if you attemptto find the partial derivatives of y and u as functions of x and vat this point?

ii. Do the same calculation of partial derivatives for the point (−1, 1,−1,−1)T .

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Solution i.a. Define f : R4 → R2 by

f(x) =

(x2 + y2 + 2uv − 4

x3 + y3 + u3 − v3

).

with x = (x, y, u, v)T . Then

Jf(p) =

(2x 2y 2v 2u

3x2 3y2 3u2 −3v2

)x=p

=

(−2 2 2 2

3 3 3 −3

), (3)

when p = (−1, 1, 1, 1)T . The last two columns of this matrix (2, 3)T and(2,−3)T are linearly independent and so we need no rearrangement of columnsto apply the Implicit Function Theorem. In the notation of the Theorem,q = (−1, 1)T . Thus there exists an open set V : q ∈ V ⊆ R2, a C1-functionφ : V → Rm and an open set W : p ∈ W ⊆ R4 such for x ∈ W, f(x) = 0 if,and only if

x =

xyuv

with

(xy

)∈ V and

(uv

)= φ

((xy

)).

Our required functions are u = φ1 and v = φ2, the component functionsof vector-valued function φ.

b. Restrict to (x, y)T ∈ V . Differentiating the equations given in the questionwith respect to x gives

2x+ 2∂u

∂xv + 2

∂v

∂xu = 0

3x2 + 3u2∂u

∂x− 3v2

∂v

∂x= 0,

At the point (x, y)T = q = (−1, 1)T this gives the system

−2 + 2∂u

∂x(q) + 2

∂v

∂x(q) = 0

3 + 3∂u

∂x(q)− 3

∂v

∂x(q) = 0,

remembering that u and v, and thus their derivatives, depend only on x andy. Solve this to give

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∂u

∂x(q) = 0 and

∂v

∂x(q) = 1.

Similarly

∂u

∂y(q) = −1 and

∂v

∂y(q) = 0.

Lesson: The Implicit Function Theorem says that u = u (x, y) and v =v (x, y) exist though it doesn’t say what these functions are. Nonethelesswe can find the partial derivatives of these functions.

c. In the Jacobian matrix in (3) the second and third column, correspondingto y and u are not linearly independent. The Implicit Function Theoremdoes not allow us to conclude that y and u can be given locally as functionsof x and v.

If you assume that y and u can be given locally as functions of x and v,you can attempt to follow the above. Restrict to (x, v)T in an open subsetof R2 containing q′ = (−1, 1) (Note that q′ looks identical to q above, but qcontains the first and second coordinates of p, while q′ the first and fourth.)

Differentiating the equations given in the question with respect to x

2x+ 2y∂y

∂x+ 2v

∂u

∂x= 0

3x2 + 3y2∂y

∂x+ 3u2

∂u

∂x= 0.

At q′ = (−1, 1)

−2 + 2∂y

∂x(q′) + 2

∂u

∂x(q′) = 0

3 + 3∂y

∂x(q′) + 3

∂u

∂x(q′) = 0.

This system is quickly seen to be inconsistent; no solutions exist for ∂y/∂xand ∂u/∂x.

ii. Labelling p2 = (−1, 1,−1,−1)T , we have

Jf(p2) =

(−2 2 −2 −2

3 3 3 −3

).

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The last two columns are still linearly independent so we can express uand v as functions of x and y in an open set containing p2. The same methodas above gives

∂u

∂x(q2) = −1 and

∂v

∂x(q2) = 0,

and∂u

∂y(q2) = 0 and

∂v

∂y(q2) = 1,

where q2 = (−1, 1)T .

Note that the answers are different to those found in Part i.a. becauseu = u (x, y) and v = v (x, y) near p are different functions to those near p2.We could have labelled the solutions in Part i.a. as u1, v1 and those frompart i.c. as u2, v2. Then for (x, y)T close to (−1, 1)T the points (x, y, u1, v1)

T

lie close to p while (x, y, u2, v2)T lie close to p2.

4. i. Does the equation x = sin(xyz) determine x as a function of y and zin any open subset of R3 containing the point p = (1, 1, π/2)T ?

ii. Does the equation x = sin(xyz) determine z as a function of x and yin a open subset of R3 containing the point p = (1, 1, π/2)T ?

Solution In both parts of this question we are concerned with the level setf−1 (0) with f(x) = sin (xyz)− x where x = (x, y, z)T . The Jacobian matrixfor f is

Jf(x) = (yz cos (xyz)− 1, xz cos (xyz) , xy cos (xyz)) .

ThenJf(p) = (−1, 0, 0).

i. The first column in Jf(p) (corresponding to the x variable) is linearlyindependent (i.e. non-zero). To apply the Implicit Function Theorem asstated in Question 3 we should rearrange the variables so the last column islinearly independent.

Or we just note that, in the notation in Question 3, the vector r consistsof the coordinates of p corresponding to the independent columns. Herethere is only one such column so r is a scalar, in fact r = 1. The vector qcontains all other coordinates of p, so q = (1, π/2)T .

The conclusion is that there exists an open set V : q ∈ V ⊆ R2, a C1-function φ : V → R and an open set W : p ∈ W such for (x, y, z)T ∈ W,

sin (xyz)− x = 0 if, and only if (y, z)T ∈ V andx = φ(

(y, z)T)

.

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Hence x can be given as a function of y and z in some open subset of R3

containing the point p = (1, 1, π/2)T .

ii. Since the last column in Jf(p) is linearly dependent (because it is zero),the Implicit Function Theorem tells us nothing; z may be a function of xandy or it may not.

We have approach the problem differently. For example, we can attemptto solve sin (xyz) − x = 0 giving z =

(sin−1 x

)/xy. Yet in any open set

containing (1, 1, π/2)T there will be points (x, y, z)T with x > 1. But sin−1 xis not defined for such x hence z cannot be given as a function of x and y inany open set containing (1, 1, π/2)T .

5. Recall the Inverse Function Theorem: Suppose that f : U ⊆ Rn → Rn

is a C1-function on an open subset U ⊆ Rn such that for some a ∈ U theJacobian matrix Jf(a) is of full rank then f is locally invertible.

Define the function f : R3 → R3 by x 7→ (yz, xz, xy)T .

i. Prove that f has a local inverse at a = (a, b, c)T (i.e. has an differentiableinverse when restricted to an open set containing a) if and only ifabc 6= 0.

ii. Find the Jacobian matrix Jg(f(a)

)of the local inverse g = f−1 at f(a),

when it exists.

Solution i. The Jacobian matrix of f is

Jf(a) =

0 c bc 0 ab a 0

and det Jf(a) = abc.

Thus Jf(a) is invertible if, and only if, abc 6= 0.

In particular abc 6= 0 implies Jf(a) is invertible which in turn implies fis locally invertible by the Inverse Function Theorem.

Conversely if f is locally invertible there exists g such that g◦f is the iden-tity when the Chain Rule implies Jg

(f(a)

)Jf (a) = I, so Jf (a) is invertible

which implies abc 6= 0.

Hence f is locally invertible if, and only if, abc 6= 0.

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ii. For the inverse g, from Jg(f(a)

)Jf (a) = I we get

Jg(f(a)

)= Jf (a)−1 =

1

2abc

−a2 ab acab −b2 bcac bc −c2

.

Find the inverse by using row operations or, perhaps, the adjoint matrix(the transpose of the cofactor matrix).

6. Define the function f : R2 → R2 by f((x, y)T

)= (x3 − 2xy2, x+ y)T . Is f

locally invertible at a = (1,−1)T ?

If f is locally invertible at a what is the best affine approximation to theinverse function near f (a)?

Solution To answer the first question, calculate the Jacobian matrix

Jf(a) =

(3x2 − 2y2 −4xy

1 1

)(1,−1)T

=

(1 41 1

).

The matrix is of full-rank and so, by the Inverse Function Theorem, f isinvertible in some open set containing a (i.e. it is locally invertible). As wehave seen in the notes and the previous question, the Jacobian of the inverseis the inverse of the Jacobian. That is, with b = f (a), the Jacobian of theinverse is

Jf−1(b) = Jf(a)−1 =1

3

(−1 4

1 −1

).

Then the best affine approximation to f−1near b = f(a) = (−1, 0)T is, asa function of u = (u, v)T ∈ R2,

f−1(b) + Jf−1(b) (u− b) = a+Jf(a)−1 (u− f(a))

=

(1−1

)+

1

3

(−1 4

1 −1

)(u+ 1v

)

=1

3

(2− u+ 4v−2 + u− v

).

The Inverse Function Theorem tells us when an inverse function existsbut not what it is. It is an example of an existence result. Nonetheless wecan find the best affine approximation to the inverse function.

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7. Prove that the Inverse Function Theorem follows from the Implicit Func-tion Theorem by applying the Implicit Function Theorem to the functionh : Rn × U ⊆ R2n → Rn defined by

h

((x

y

))= x− f (y) , where x ∈ Rn, y ∈ U.

Hint: The important observation is that by definition of h,

h

((f (a)

a

))= f (a)− f (a) = 0.

So, setting

p =

(f (a)

a

),

we have h (p) = 0 as required for the Implicit Function Theorem. To deduceanything from that Theorem it is required that Jh (p) is of full rank. Whatis Jh (p)?

Solution Assume that f : U ⊆ Rn → Rn is a C1-function on an open subsetU ⊆ Rn such that for some a ∈ U the Jacobian matrix Jf(a) is of full rank.Follow the hint and define h : Rn × U ⊆ R2n → Rn by

h

((x

y

))= x− f(y) .

Given w ∈ Rn × U write it as w =(xT ,yT

)Twith x ∈ Rn and y ∈ U .

The Jacobian matrix of h at w is then

Jh(w) = (In | − Jf(y)) .

Let

p =

(f (a)a

)∈ Rn × U,

with a the point at which Jf(a) is non-singular. Then Jh(p) = (In | − Jf(a))and, since Jf(a) is non-singular, the last n columns of Jf(p) are linearlyindependent.

Hence, by the Implicit Function Theorem, (and no permutation of coor-dinates is required), there exists

• an open set A ⊆ Rn containing f(a) ,

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• a C1-function φ : A→ Rn and

• an open set B ⊆ Rn × U containing p such for(xT ,yT

)T ∈ B, (x,y ∈Rn),

h

((x

y

))= 0 if, and only if, x ∈ A and y = φ (x) .

Note that the set B can be written as A × C for some open set C ⊆ Rn.Thus we have

x = f(y) ⇐⇒ x− f(y) = 0 ⇐⇒ h

((x

y

))= 0 ⇐⇒ y = φ (x) .

This gives the existence of a C1-inverse.

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Additional Questions

8 Further practice on the Chain Rule Use the chain rule to find the derivativeof f ◦ g at the point c for each of the following. Give your answers in theform d (f ◦ g)c (t) .

i. f((x, y)T

)= (x2y, x−y)T , g

((u, v)T

)= (3uv, u2−4v)T , c = (1,−2)T ,

ii. f((x, y, z)T

)= (4xy, 3xz)T , g

((u, v)T

)= (uv2 − 4v, u2, 4/uv)

T, c =

(−2, 3)T

iii. f((x, y)T

)= (3x+4y, 2x2y, x−y)T , g

((u, v, w)T

)= (4u−3v+w, uv2)T ,

c = (1,−2, 3)T .

Solution The Chain Rule states that d (f ◦ g)c = dfg(c) ◦ dgc, so

d (f ◦ g)c (t) = dfg(c) (dgc (t)) .

i. Let t ∈ R2, so t = (s, t)T say. Then

dgc(t) =

(3v 3u

2u −4

)(1,−2)T

(st

)=

(−6 3

2 −4

)(st

)=

(−6s+ 3t

2s− 4t

).

Next g (c) = (−6, 9)T so

dfg(c) (dgc(t)) =

(2xy x2

1 −1

)(−6,9)T

(−6s+ 3t

2s− 4t

)=

(−108 36

1 −1

)(−6s+ 3t

2s− 4t

)

=

(720s− 468t

−8s+ 7t

).

ii. Let t ∈ R2, so t = (s, t)T say. Then

dgc(t) =

v2 2uv − 4

2u 0

−4/u2v −4/uv2

(−2,3)T

t =

9 −16

−4 0

−1/3 2/9

t =

9s− 16t

−4s

−s/3 + 2t/9

.

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Next g((−2, 3)T

)= (−30, 4,−2/3)T so

fg(c) (dgc(t)) =

(4y 4x 0

3z 0 3x

)(−30,4,−2/3)T

dgc(t)

=

(16 −120 0

−2 0 −90

) 9s− 16t

−4s

−s/3− 2t/9

=

(624s− 256t

12s+ 42t

).

iii. An alternative approach is to not mention t until the end but, instead,look at the Jacobian matrices.

Jg (c) =

(4 −3 1

v2 2uv 0

)c

=

(4 −3 14 −4 0

).

Next g (c) = (13, 4)T . Then

Jf(g(c)) =

3 4

4xy 2x2

1 −1

(13,4)T

=

3 4

208 338

1 −1

.

Multiplying together,

Jf(g(c)) Jg(c) =

3 4

208 338

1 −1

( 4 −3 1

4 −4 0

)=

28 −25 3

2184 −1976 208

0 1 1

.

Now introduce t ∈ R3 so t = (r, s, t)T say. Then

d (f ◦ g)c (t) =

28 −25 3

2184 −1976 208

0 1 1

rst

=

28r − 25s+ 3t

2184r − 1976s+ 208t

s+ t

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9. Return to Question 3. A purpose of the question was to highlight the factthat when conditions are satisfied the Implicit Function Theorem ensures thatfunctions exist, but gives no further information about them. Nonethelesstheir derivatives can be found. In this question we continue to find theirsecond derivatives.

Define the function f : R4 → R2 by

f(x) =

(x2 + y2 + 2uv − 4

x3 + y3 + u3 − v3

),

for x = (x, y, u, v)T . The level set f−1 (0) contains the point p = (−1, 1,−1,−1)T .The Jacobian matrix of f at p is(

−2 2 −2 −2

3 3 3 −3

).

Because the last two columns are linearly independent the Implicit FunctionTheorem applies. Thus, with q = (−1, 1)T the first two coordinates of p,there exists V : q ∈ V ⊆ R2, a C1-function φ : V → Rm and W : p ∈ W ⊆R4 such for x ∈ W, f(x) = 0 if, and only if

x =

xyuv

with

(xy

)∈ V and

(uv

)= φ

((xy

)).

Our required functions are u = φ1 and v = φ2 where φ1 and φ2 are thecomponent functions of vector-valued function φ.

Question Restrict to (x, y)T ∈ V . Find the second derivatives of u and v atq.

Solution Write the level set f−1 (0) as the system

x2 + y2 + 2uv = 4 (4)

x3 + y3 + u3 − v3 = 0.

By differentiating w.r.t.x and w.r.t y we found in Question 3 that

∂u

∂x(q) = −1

∂v

∂x(q) = 0,

∂u

∂y(q) = 0 and

∂v

∂y(q) = 1. (5)

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Differentiate the system (4) twice, w.r.t x both times to get

2 + 2∂2u

∂x2v + 2

∂u

∂x

∂v

∂x+ 2

∂2v

∂x2u+ 2

∂v

∂x

∂u

∂x= 0

6x+ 6u

(∂u

∂x

)2

+ 3u2∂2u

∂x2− 6v

(∂v

∂x

)2

− 3v2∂2v

∂x2= 0.

Then at p, using (4), we get the simpler system

∂2u

∂x2+∂2v

∂x2= 1 and

∂2u

∂x2− ∂2v

∂x2= 4.

Solve these to give

∂2u

∂x2=

5

2and

∂2v

∂x2= −3

2,

where the derivatives are evaluated at q.

Return and differentiate (4) first w.r.t x and then w.r.t. y to get

2v∂2u

∂y∂x+ 2

∂u

∂x

∂v

∂y+ 2

∂u

∂y

∂v

∂x+ 2u

∂2v

∂y∂x= 0,

6u∂u

∂y

∂u

∂x+ 3u2

∂2u

∂y∂x− 6v2

∂v

∂y

∂v

∂x− 3v2

∂2v

∂y∂x= 0.

At p, again using (5), this becomes

∂2u

∂y∂x+

∂2v

∂y∂x= 1 and

∂2u

∂y∂x− ∂2v

∂y∂x= 0.

Hence∂2u

∂y∂x=

∂2v

∂y∂x=

1

2.

I do not give the details for the mixed derivative in the reversed order,but you can check that ∂2u/∂x∂y = ∂2u/∂y∂x and the same for v. It can,in fact, be shown that the order is immaterial whenever f is C2.

Finally, differentiate the system w.r.t y twice to get

2 + 2v∂2u

∂y2+ 2

∂u

∂y

∂v

∂y+ 2

∂u

∂y

∂v

∂y+ 2u

∂2v

∂y2= 0

6y + 6u

(∂u

∂y

)2

+ 3u2∂2u

∂y2− 6v

(∂v

∂y

)2

− 3v2∂2v

∂y2= 0.

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At p this reduces to

∂2u

∂y2+∂2v

∂y2= 1 and

∂2u

∂y2− ∂2v

∂y2= −4.

Hence∂2u

∂y2= −3

2and

∂2v

∂y2=

5

2.

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