1550251515-Classical_Complex_Analysis__Gonzalez_

316 Chapter6

= [u(x + b.x, y + b.y) - u(x, y + b.y)] + [u(x, y + b.y) - u(x, y)]

= ux(x + .Ab.x, y + b.y)b.x + [u(x, y + b.y) - u(x, y)] (6.4-3)

where the mean-value theorem for functions of one variable has been

applied to the first bracket, .A being such that 0 < .A < 1.

By the continuity of Ux at ( x, y) we have

Ux(x +A b.x, y + b.y) = ux(x, y) + c1 (6.4-4)

where c 1 --t 0 as .6.x and b.y --t 0. On the other hand, by the existence

of Uy at (x, y) we have

u(x,y+b.y)-u(x,y) · ( )

f}.y = Uy x, y + c2 (6.4-5)

where c 2 --t 0 as b.y --t 0. By making use of (6.4-4) and (6.4-5) in (6.4-3),

we get

b.u = Ux ~x + Uy b.y + c1 b.x + c2 b.y

which establishes the differentiability of u at ( x, y ).

The conditions stated in Theorems 6.3 and 6.4, although sufficient, are

by no means necessary. The following example illustrates this by showing

that a function can be differentiable at a point even when both Ux and Uy

are discontinuous at that point.

Example Let

u(x,y) = (x^2 +y^2 )sin 2

1

x +y^2 for (x,y) =/= (0,0)

.and u(O, 0) = 0. Here

Ux (0 0) ' -- Ll.;~o r D.x2 sin(l/ b.x b.x2) =^0

and similarly, uy(O,O) = 0. For (x,y) =/= (0,0),

. 1 2x 1
ux(x,y) = 2xs1n 2 2 - 2 2 cos 2 2

x +y x +y x +y

which has no finite limit, as (x,y) --t (0,0). Hence, Ux is not continuous

at the origin, and similarly for Uy.

Yet u is differentiable at (0, 0) since

and

b.u=O·b.x+O·b.y+(b.x^2 +b.y^2 )sin b. 2

1

x +^6 y^2

I ~z I = ../ D.x2 + D.y2 ,sin D.x2: D.y2 ' :::; lb.zl