JUST THE MATHS SLIDES NUMBER 14.8 PARTIAL DIFFERENTIATION 8 - - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.8 PARTIAL DIFFERENTIATION 8 (Dependent and independent functions) by A.J.Hobson

14.8.1 The Jacobian

UNIT 14.8 PARTIAL DIFFERENTIATION 8 DEPENDENT AND INDEPENDENT FUNCTIONS 14.8.1 THE JACOBIAN Suppose that u ≡ u(x, y) and v ≡ v(x, y) are two functions of two independent variables, x and y. Then, it is not normally possible to express u solely in terms of v, nor v solely in terms of u. However, it may sometimes be possible ILLUSTRATIONS

• 1. If

u ≡ x + y x and v ≡ x − y y , then, u ≡ 1 + x y and v ≡ x y − 1.

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This gives (u − 1)(v + 1) ≡ x y.y x ≡ 1. Hence, u ≡ 1 + 1 v + 1 and v ≡ 1 u − 1 − 1.

• 2. If

u ≡ x + y and v ≡ x2 + 2xy + y2, then, v ≡ u2 and u ≡ ±√v. If u and v are not connected by an identical relationship, they are said to be “independent functions” THEOREM Two functions, u(x, y) and v(x, y) are independent if and

• nly if

J ≡

• ∂u

∂x ∂u ∂y ∂v ∂x ∂v ∂y

• ≡ 0.

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Proof: We prove that u(x, y) and v(x, y) are dependent if and

• nly if

J ≡

• ∂u

∂x ∂u ∂y ∂v ∂x ∂v ∂y

• ≡ 0.

(a) Suppose that v ≡ v(u). Then, ∂v ∂x ≡ dv du.∂u ∂x and ∂v ∂y ≡ dv du.∂u ∂y. Thus, ∂v ∂x ÷ ∂u ∂x ≡ ∂v ∂y ÷ ∂u ∂y ≡ dv du

• r

∂u ∂x.∂v ∂y − ∂v ∂x.∂u ∂y ≡ 0.

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That is, J ≡ 0. (b) Secondly, suppose that J ≡

• ∂u

∂x ∂u ∂y ∂v ∂x ∂v ∂y

• ≡ 0.

In theory, we could express v in terms of u and x by eliminating y between u(x, y) and v(x, y). We assume that v ≡ A(u, x) and show that A(u, x) does not contain x. We have

  ∂v

∂x

  

y ≡

  ∂A

∂u

  

x .

  ∂u

∂x

  

y +

  ∂A

∂x

  

u

and

   ∂v

∂y

   

x

≡

  ∂A

∂u

  

x .

   ∂u

∂y

   

x

. Hence, if J ≡ 0,

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• ∂u

∂x

• y
• ∂u

∂y

• x

∂A

∂u

• x .
• ∂u

∂y

• y +

∂A

∂x

• u

∂A

∂u

• x .
• ∂u

∂y

• x
• ≡ 0.

On expansion, this gives

   ∂u

∂y

   

x

.

  ∂A

∂x

  

u ≡ 0.

If the first of these two is equal to zero, then u contains

• nly x.

Hence, x could be expressed in terms of u giving v as a function of u only. If the second is equal to zero, then A contains no x’s and, again, v is a function of u only. Notes: (i) The determinant J ≡

• ∂u

∂x ∂u ∂y ∂v ∂x ∂v ∂y

• may also be denoted by

∂(u, v) ∂(x, y).

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It is called the “Jacobian determinant” or simply the “Jacobian” of u and v with respect to x and y. (ii) Similar Jacobian determinants may be used to test for the dependence or independence of three functions of three variables, four functions of four variables, and so

• n.

For example, the three functions u ≡ u(x, y, z), v ≡ v(x, y, z) and w ≡ w(x, y, z) are independent if and only if J ≡ ∂(u, v, w) ∂(x, y, z) ≡

• ∂u

∂x ∂u ∂y ∂u ∂z ∂v ∂x ∂v ∂y ∂v ∂z ∂w ∂x ∂w ∂y ∂w ∂z

• ≡ 0.

ILLUSTRATIONS 1. u ≡ x + y x and v ≡ x − y y are not independent, since J ≡

• − y

x2 1 x 1 y

− x

y2

• ≡ 1

xy − 1 xy ≡ 0.

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2. u ≡ x + y and v ≡ x2 + 2xy + y2 are not independent, since J ≡

• 1

1 2x + 2y 2x + 2y

• ≡ 0.

3. u ≡ x2 + 2y and v ≡ xy are independent, since J ≡

• 2x

2 y x

• ≡ 2x2 − 2y ≡ 0.

4. u ≡ x2−2y+z, v ≡ x+3y2−2z, and w ≡ 5x+y+z2 are not independent, since J ≡

• 2x

−2 1 1 6y −2 5 1 2z

• ≡ 24xyz+4x−30y+4z+25 ≡ 0.

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