MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

∂u
∂θ = ∂u
∂x .∂x ∂θ + ∂u
∂y .∂y ∂θ = u x(−r sin θ) + u y (r cos θ) = −rv y sin θ − rv x cos θ
12-21 Harmonic functions:
12. u = xy
∂v
∂r = ∂v
∂x .∂x ∂r + ∂v
∂y .∂y ∂r = v x cos θ + v y sin θ
∂v
∂θ = ∂v
∂x .∂x ∂θ + ∂v
∂y .∂y ∂θ = v x(−r sin θ) + v y (r cos θ)
=⇒ u r = 1 r v θ and v r = − 1 r u θ.
u xx = 0 and u yy = 0 =⇒ u xx + u yy = 0 =⇒ u is harmonic.
u x = v y =⇒ y = v y =⇒ v = y2
2 + h(x) =⇒ v x = dh
dx
u y = −v x =⇒ x = − dh
x2
y2
dx
=⇒ h(x) = −
2
+ c =⇒ f(z) = xy + (
2 − x2
2 + c)i.
13. v = xy
v xx = 0, v yy = 0 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
v x = −u y =⇒ u y = −y =⇒ u = − y2
2 + h(x) =⇒ u x = dh
dx .
Since u x = v y = x, dh
x2
dx
= x =⇒ h(x) =
2 + c.
=⇒ f(z) = − y2
2 + x2
2 + c + xyi.
14. v = − y
x 2 +y 2
−u y = v x =
2xy , u
(x 2 +y 2 ) 2 x = v y = y2 −x 2
(x 2 +y 2 ) 2
=⇒ v xx = −6x2 y+2y 3
(x 2 +y 2 ) 3 and v yy = 6x2 y−2y 3
(x 2 +y 2 ) 3 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
u =
x
x 2 +y 2 + c =⇒ f(z) =
15. u = ln |z| = ln √ x 2 + y 2
v y = u x =
x
2(x 2 +y 2 ) , −v x = u y =
x
x 2 +y 2 + c + i
y
2(x 2 +y 2 )
−y
.
x 2 +y 2
=⇒ u xx = y2 −x 2
2(x 2 +y 2 ) 2 and u yy = x2 −y 2
2(x 2 +y 2 ) 2 =⇒ u xx + u yy = 0 =⇒ u is harmonic.
v = 1 2 arctan( y x ) + h(x) =⇒ v x = −
=⇒ f(z) = ln |z| + 1 2 arctan( y x ) + c.
16. v = ln |z| = ln √ x 2 + y 2
−u y = v x =
x
2(x 2 +y 2 ) , u x = v y =
y
2(x 2 +y 2 ) + dh
dx
y
2(x 2 +y 2 )
dh
=⇒
dx
= 0 =⇒ h(x) = c.
=⇒ v xx = y2 −x 2
2(x 2 +y 2 ) 2 and v yy = x2 −y 2
2(x 2 +y 2 ) 2 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
u = −1
2 arctan( y x ) + h(x) =⇒ u x =
=⇒ f(z) = −1
2 arctan( y x
) + c + i ln |z|.
17. u = x 3 − 3xy 2
v y = u x = 3x 2 − 3y 2 , −v x = u y = −6xy
y
2(x 2 +y 2 ) + dh
dx
=⇒ h(x) = c.
=⇒ u xx = 6x and u yy = −6x =⇒ u xx + u yy = 0 =⇒ u is harmonic.
v = 3x 2 y − y 3 + h(x) =⇒ v x = 6xy + dh
dx
=⇒ h(x) = c.
15