9.6 Some differential properties 271
Then
Similarly,
Therefore
0 Exercise 49
EXAMPLE 9.19Show that the functionf 1 = 1 x
21 − 1 y
2satisfies the Laplace equation
(i) In cartesian coordinates,
and
Therefore,
(ii) In polar coordinates,f 1 = 1 r
2(cos
2θ 1 − 1 sin
2θ) 1 = 1 r
21 cos 12 θ,
∂
∂
=− ,
∂
∂
=− =−
f
r
f
rf
θ
θ
θ
22 4 24θ
2222sin cos
∂
∂
==,
∂
∂
==
f
r
r
f
r
f
r
f
r
22
2
22
2
222cos θθcos
∂
∂
∂
∂
=
22220
f
x
f
y
∂
∂
=− ,
∂
∂
=−
f
y
y
f
y
22
22∂
∂
=,
∂
∂
=
f
x
x
f
x
22
22∂
∂
∂
∂
=
∂
∂
∂
∂
∂
∂
222222222f 11
x
f
y
f
r
r
f
r
rf
θ
∂
∂
=
∂
∂
∂
∂
∂
∂
2222222222f
y
f
r
r
f
r
rf
sin
cos cos
θ
θθ
θ
−−
∂
∂
−
∂
∂∂
21
2sin cosθθ
rrθθ
ff
r
=
∂
∂
∂
∂
∂
∂
cos +
sin sin sin
222222222
θ
θθ
θ
f θ
r
r
f
r
r
f ccosθ
rrθθ
ff
r
1
2∂
∂
−
∂
∂∂
−−
∂
∂
∂
∂∂
−
∂
∂
−
sin
sin cos
θ cos sin
θθ
θ
θ
θ
θ
r
f
r
f
rr
f
r
2∂∂
∂
22f
θ
=
∂
∂
∂
∂
−
∂
∂∂
cos cos
sin sin
θθ
θ
θ
θ
θ
2222f
rr
f
r
f
r
=
∂
∂
∂
∂
−
∂
∂
−
∂
∂
cos cos
sin sin
θθ co
θ
θ
θ
r θ
f
rr
f
r
ss
sin
θ
θ
θ
∂
∂
−
∂
∂
f
rr
f
∂
∂
=
∂
∂
−
∂
∂
∂
∂
−
22f
x
rr
f
rr
cos
sin
cos
sin
θ
θ
θ
θ
θ ∂∂
∂
f
θ