Page 143 - Schaum's Outline of Theory and Problems of Advanced Calculus

134 PARTIAL DERIVATIVES [CHAP. 6

@z @z @u @z @z @u 2
¼ f ðuÞ 2xy; ¼ f ðuÞ x
0
0
@x ¼ @u @x @y ¼ @u @y
@z 2 @z 2 @z @z
Then x ¼ f ðuÞ 2x y; 2y ¼ f ðuÞ 2x y and so x ¼ 2y :
0
0
@x @y @x @y
Another method:
2
2
2
2
We have dz ¼ f ðx yÞ dðx yÞ¼ f ðx yÞð2xy dx þ x dyÞ:
0
0
@z @z
Also, dz ¼ dx þ dy:
@x @y
@z 2 @z 3 2
Then ¼ 2xy f ðx yÞ; ¼ x f ðx yÞ.
0
0
@x @y
2
Elimination of f ðx yÞ yields x @z ¼ 2y @z .
0
@x @y
p
6.25. If for all values of the parameter and for some constant p, Fð x; yÞ¼ Fðx; yÞ identically,
where F is assumed differentiable, prove that xð@F=@xÞþ yð@F=@yÞ¼ pF.
Let x ¼ u, y ¼ v. Then
p
Fðu; vÞ¼ Fðx; yÞ ð1Þ
The derivative with respect to of the left side of (1)is
@F @F @u @F dv @F @F
y
@ ¼ @u @ þ @v @ ¼ @u x þ @v
The derivative with respect to of the right side of (1)is p p 1 F. Then
@F @F p 1
x þ y ¼ p F ð2Þ
@u @v
Letting ¼ 1in(2), so that u ¼ x; v ¼ y,we have xð@F=@xÞþ yð@F=@yÞ¼ pF.
4 2
6.26. If Fðx; yÞ¼ x y sin 1 y=x, show that xð@F=@xÞþ yð@F=@yÞ¼ 6F.
2
6 4 2
6
4
Since Fð x; yÞ¼ ð xÞ ð yÞ sin 1 y= x ¼ x y sin 1 y=x ¼ Fðx; yÞ,the result follows from Pro-
blem 6.25 with p ¼ 6. It can of course also be shown by direct differentiation.
2
2
2
2
2
6.27. Prove that Y ¼ f ðx þ atÞþ gðx atÞ satisfies @ Y=@t ¼ a ð@ Y=@x Þ, where f and g are assumed
to be at least twice differentiable and a is any constant.
Let u ¼ x þ at; v ¼ x at so that Y ¼ f ðuÞþ gðvÞ. Then if f ðuÞ df =du, g ðvÞ dg=dv,
0
0
@Y @Y @u @Y @v @Y @Y @u @Y @v
¼ af ðuÞ ag ðvÞ; ¼ þ ¼ f ðuÞþ g ðvÞ
0
0
0
0
þ
¼
@t @u @t @v @t @x @x @x @v @x
2
2
2
2
By further differentiation, using the notation f ðuÞ d f =du , g ðvÞ d g=dv ,we have
00
00
2
@ Y @Y t @Y t @u @Y t @v @ @
0 0 0 0
@t @t @u @t @v @t @u @v
ð1Þ 2 ¼ ¼ þ ¼ faf ðuÞ ag ðvÞgðaÞþ faf ðuÞ ag ðvÞg ð aÞ
2 00 2 00
¼ a f ðuÞþ a g ðvÞ
2
@ Y @Y x @Y x @u @Y x @v @ @
0 0 0 0
ð2Þ 2 ¼ ¼ þ ¼ f f ðuÞþ g ðvÞg þ f f ðuÞþ g ÞðvÞg
@x @x @u @x @v @x @u @v
00 00
¼ f ðuÞþ g ðvÞ
2
2
2
2
2
Then from (1) and (2), @ Y=@t ¼ a ð@ Y=@x Þ.