Page 145 - Schaum's Outline of Theory and Problems of Advanced Calculus

136 PARTIAL DERIVATIVES [CHAP. 6

dF ¼ F x dx þ F y dy þ F u du þ F v dv ¼ 0
ð1Þ
dG ¼ G x dx þ G y dy þ G u du þ G v dv ¼ 0
ð2Þ
Also, since u and v are functions of x and y,
ð3Þ du ¼ u x dx þ u y dy ð4Þ dv ¼ v x dx þ v y dy:

Substituting (3) and (4)in (1) and (2)yields
dF ¼ðF x þ F u u x þ F v v x Þ dx þðF y þ F u u y þ F v v y Þ dy ¼ 0
ð5Þ
dG ¼ðG x þ G u u x þ G v v x Þ dx þðG y þ G u u y þ G v v y Þ dy ¼ 0
ð6Þ
Since x and y are independent, the coefficients of dx and dy in (5) and (6)are zero. Hence we obtain

F u u x þ F v v x ¼ F x F u u y þ F v v y ¼ F y
ð7Þ ð8Þ
G u u x þ G v v x ¼ G x G u u y þ G v v y ¼ G y
Solving (7) and (8)gives

F x F v F u F x
@ðF; GÞ @ðF; GÞ

@u G x G v @v G u G x
@ðx; vÞ @ðu; xÞ
@x F u F v @ðF; GÞ @x F u F v @ðF; GÞ
ðaÞ u x ¼ ¼ ¼ ðbÞ v x ¼ ¼ ¼

G u G v @ðu; vÞ G u G v @ðu; vÞ

F y F u
F v F y
@ðF; GÞ @ðF; GÞ

@u G y G v @ðy; vÞ @v G u G y @ðu; yÞ
ðcÞ u y ¼ ¼ ¼ ðdÞ v y ¼ ¼ ¼
@y F u F v @ðF; GÞ @y F u F v @ðF; GÞ

G u G v @ðu; vÞ G u G v @ðu; vÞ

F; G
F u @ðF; GÞ
The functional determinant F v ,denoted by or J ,isthe Jacobian of F and G with
u; v
G u G v
@ðu; vÞ
respect to u and v and is supposed 6¼ 0.
Note that it is possible to devise mnemonic rules for writing at once the required partial derivatives in
terms of Jacobians (see also Problem 6.33).
2
2
6.32. If u v ¼ 3x þ y and u 2v ¼ x 2y, find (a) @u=@x; ðbÞ @v=@x; ðcÞ @u=@y; ðdÞ @v=@y.
Method 1: Differentiate the given equations with respect to x,considering u and v as functions of x and y.
Then
@u @v @u @v
2u ¼ 3 4v ¼ 1
@x @x @x @x
ð1Þ ð2Þ
@u 1 12v @v 2u 3
Solving, ¼ ; ¼ :
@x 1 8uv @x 1 8uv
Differentiating with respect to y,we have
@u @v @u @v
2u ¼ 1 4v ¼ 2
@y @y @y @y
ð3Þ ð4Þ
@u 2 4v @v 4u 1
Solving, ¼ ; ¼ :
@y 1 8uv @y 1 8uv
We have, of course, assumed that 1 8uv 6¼ 0.
2
2
Method 2: The given equations are F ¼ u v 3x y ¼ 0, G ¼ u 2v x þ 2y ¼ 0. Then by Problem
6.31,