Jacobians

JACOBIANS

Jacobians have many important applications such as functional dependence, transformation of variable in multiple integrals, problems in partial differentiation and in the study of existence of implicit functions determined by a system of functional equations.

Definition 3.3.1 (1) If u and v are continuous functions of two independent variables x and y, having first order partial derivatives, then the determinant

(2) If u, v, w are continuous functions of three independent variables x, y, z having first order partial derivatives then the Jacobian of u, v, w with respect to x, y, z is defined as

Similarly, we can define Jacobians for functions of 4 or more variables.

1. Properties of Jacobians

For simplicity we shall prove the properties of Jacobians for two variables. However, they can be extended to any number of variables.

Property 1: If u and v are functions of x and y, then

Proof Let u = f1(x, y) and v = f2(x, y) be continuous functions of two independent variables x and y having first order partial derivatives then

The condition for these equations to be solvable for x and y is J ≠ 0.

Since u and v are independent variables from differentials (1) and (2) we get

Property 2: Jacobians of composite functions or chain rule

If u and v are functions of p and q, where p and q are functions of x and y, then

Proof If u and v are continuous functions of p and q and p and q are functions of x and y, then

Property 3: If u and v are functions of two independent variables x and y and

u and v are functionally dependent

Proof If u and v are not independent, then there is a relation between u and v. Let f(u, v) = 0 be the relation between u and v. 

Differentiating with respect to x and y we have,

Note

(1) The converse of property 3 is also true.

(2) The property can be extended to functions of more than two variables.

Property 4: If x = f(u, v), y = g(u, v) and h(x, y) = F(u, v) then

WORKED EXAMPLES

Example 1 

Solution 

and v = y2 – 2

Example 2 

Solution 

Example 3 

If x = r cos θ, y = r sin θ, then find the Jacobian of x and y with respect to r and θ.

Solution 

The Jacobian of x and y with respect to r and θ is

Note x = r cos θ, y = r sin θ transforms cartesian coordinates into polar coordinates.

Example 4 

Solution 

Since x and y are function of u, v we can find

Subtracting in (1), we get

Example 5 

If x = u (1 − v), y = uv, then compute J and J' and prove that JJ' = 1.

Solution 

We know J =

To prove JJ' = 1

Given x = u(1 − v)

We have to find u and v in terms of x and y.

x = u (1 − v) = u - uv

and y = uv

⇒ x = u - y

Example 6 

If u = 2xy, v = x2 - y2, x = r cos θ, y = r sin θ, evaluate without actual substitution.

Solution 

Given u, v are functions of x and y and x and y are functions of r and θ.

So by property 2 of composite function

Since x = r cos θ, y = r sin θ

⇒ x2 + y2 = r2 cos2 θ + r2 sin2 θ = r2

From example 3, we have

Example 7 

For the transformation x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, compute the Jacobian of x, y, z with respect to r, θ, φ.

Solution 

The Jacobian of transformation

Note This is the transformation of cartesian coordinates to spherical polar coordinates (r, θ, φ)

Example 8 

Solution 

We have

Note This is the transformation of cartesian coordinate to cylindrical coordinates (ρ, φ, z).

Example 9 

Solution 

= −1(+1−1)−1(− 1 − 1) +1(1+1)

= 0 + 2 + 2 = 4

Example 10 

If u = x + y + z, uv = y + z, uvw = z, then find

Solution 

Given u = x + y + z, uv = y + z, uvw = z

Example 11 

Solution 

Example 12 

Solution 

Example 13 

Solution 

2. Jacobian of implicit functions

If y1, y2 y3,... yn are implicitly given as functions of x1, x2, ..., xn by the functional equations f1(x1, x2, ..., xn, y1, y2 y3,... yn) = 0 for i = 1, 2, ... n, then

Example 14 

If F = xu + v - y, G = u2 + vy + w, H = zu – v + vw

compute (1)

Solution 

F = xu + v - y, G = u2 + vy + w

H = zu – v + vw

These equations implicitly define x, y, z in terms of u, v, w.

⸫ by the Jacobian of implicit functions, we have

Example 15 

If x + y + z - u = 0, y + z - uv = 0, z – uvw = 0 then find

Solution 

Given x + y + z - u = 0, y + z - uv = 0, z - uvw = 0

These equations implicitly define x, y, z interms of u, v, w

Let f1 = x + y + z - u

f2 = y + z - uv

f3 = z - uvw

By the Jacobian of implicit functions

Given f1 = x + y + z - u

f3 = z - uvw

EXERCISE 3.3

ANSWERS TO EXERCISE 3.3.