Page 147 - Schaum's Outline of Theory and Problems of Advanced Calculus

138 PARTIAL DERIVATIVES [CHAP. 6

Differentiating (2)with respect to x and using (1), we have
2
2
2
@ z 1 @z 1 6z½z=ð3z xފ 3z þ x
6z
@x @y ¼ 2 2 @x 1 ¼ 2 2 ¼ 2 3
ð3z xÞ ð3z xÞ ð3z xÞ
The result can also be obtained by differentiating (1)with respect to y and using (2).
6.35. Let u ¼ f ðx; yÞ and v ¼ gðx; yÞ, where f and g are continuously differentiable in some region r.
Prove that a necessary and sufficient condition that there exists a functional relation between u
and v of the form ðu; vÞ¼ 0is the vanishing of the Jacobian, i.e., @ðu; vÞ ¼ 0 identically.
@ðx; yÞ
Necessity. We have to prove that if the functional relation ðu; vÞ¼ 0exists, then the Jacobian @ðu; vÞ ¼ 0
identically. To do this, we note that @ðx; yÞ

d ¼ u du þ v dv ¼ u ðu x dx þ u y dyÞþ v ðv x dx þ v y dyÞ
¼ð u u x þ v v x Þ dx þð u u y þ v v y Þ dy ¼ 0

Then ð1Þ u u x þ v v x ¼ 0 ð2Þ u u y þ v v y ¼ 0
Now u and v cannot be identically zero since if they were, there would be no functional relation,

u x @ðu; vÞ
contrary to hypothesis. Hence it follows from (1) and (2)that v x ¼ 0identically.
¼
u y v y
@ðx; yÞ
Sufficiency. We have to prove that if the Jacobian @ðu; vÞ ¼ 0identically, then there exists a functional
@ðx; yÞ
relation between u and v, i.e., ðu; vÞ¼ 0.
Let us first suppose that both u x ¼ 0 and u y ¼ 0. In this case the Jacobian is identically zero and u is a
constant c 1 ,sothat the trival functional relation u ¼ c 1 is obtained.
Let us now assume that we do not have both u x ¼ 0 and u y ¼ 0; for definiteness, assume u x 6¼ 0. We
may then, according to Theorem 1, Page 124, solve for x in the equation u ¼ f ðx; yÞ to obtain x ¼ Fðu; yÞ,
from which it follows that
ð1Þ u ¼ f fFðu; yÞ; yg ð2Þ v ¼ gfFðu; yÞ; yg
From these we have respectively,
du ¼ u x dx þ u y dy ¼ u x ðF u du þ F y dyÞþ u y dy ¼ u x F u du þðu x F y þ u y Þ dy
ð3Þ
dv ¼ v x dx þ v y dy ¼ v x ðF u du þ F y dyÞþ v y dy ¼ v x F u du þðv x F y þ v y Þ dy
ð4Þ
From (3), u x F u ¼ 1 and u x F y þ u y ¼ 0or (5) F y ¼ u y =u x . Using this, (4)becomes

u x v y u y v x
dy:
ð6Þ dv ¼ v x F u du þfv x ð u y =u x Þþ v y g dy ¼ v x F u du þ
u x

u x
@ðu; vÞ
But by hypothesis u y ¼ u x v y u y v x ¼ 0identically, so that (6)becomes dv ¼ v x F u du.
¼
v x v y

@ðx; yÞ
This means essentially that referring to (2), @v=@y ¼ 0 which means that v is not dependent on y but depends
only on u, i.e., v is a function of u, which is the same as saying that the functional relation ðu; vÞ¼ 0exists.
x þ y 1 1 @ðu; vÞ
and v ¼ tan x þ tan y, find .
1 xy @ðx; yÞ
6.36. (a)If u ¼
(b) Are u and v functionally related? If so, find the relationship.