It is now possible to differentiate the first derivatives given by equations (12.62) and
(12.64) to obtain the second derivatives
∂^2 r
∂x^2 =−sin θ∂θ
∂x
∂^2 r
∂x^2 =sin^2 θ
r,∂^2 r
∂y^2 = cosθ∂θ
∂y
∂^2 r
∂y^2=cos(^2) θ
r
(12 .65)
and
∂^2 θ
∂x^2=−cos θ
r∂θ
∂x+sin θ
r∂r
∂x
∂^2 θ
∂x^2 =2 sin θcos θ
r^2,∂^2 θ
∂y^2=−sin θ
r∂θ
∂y−cos θ
r^2∂r
∂y
∂^2 θ
∂y^2 =−2 sin θcos θ
r^2(12 .66)Substitute the first and second derivatives from equations (12.62), (12.64), (12.66)
into the equations (12.57) and (12.58) to show that after simplification the equation
(12.56) becomes
∇^2 U=∂(^2) U
∂x^2
+∂
(^2) U
∂y^2
=∂
(^2) U
∂r^2
=^1
r
∂U
∂r
+^1
r^2
∂^2 U
∂θ^2
= 0 (12 .67)
Solution 2
Write the transformation equations (12.59) as
F(x, y, r, θ ) = x−rcos θ= 0 and G(x, y, r, θ) = y−rsin θ= 0 (12 .68)
and use the notation for Jacobian determinants to write
∂r
∂x =−∂(F,G )
∂(x,θ)
∂(F,G )
∂(r,θ)=−∣∣
∣∣^1 rsin θ
0 −rcos θ∣∣
∣∣
∣∣
∣∣−cos θ r sin θ
−sin θ −rcos θ∣∣
∣∣=rcosr θ= cos θ∂θ
∂x =−∂(F,G )
∂(r,x )
∂(F,G )
∂(r,θ)=−∣∣
∣∣−sinθ −rcos θ
−sin θ 0∣∣
∣∣r =−sin θ
r∂r
∂y=−∂(F,G )
∂(y,θ)
∂(F,G )
∂(r,θ)=−∣∣
∣∣^0 rsin θ
1 −rcos θ∣∣
∣∣r= sin θ∂θ
∂y=−∂(F,G )
∂(r,y )
∂(F,G )
∂(r,θ)=−∣∣
∣∣−cos θ^0
−sin θ 1∣∣
∣∣r=cos θ
r