Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 356

u ∗ . The PDS (7.50) is thus

Consider a function transformation u ( x , y )= v ( x , y )+

transformed into

⎧
⎨ x 2 y 2 Δ v = 0 , + < 1 ,

⎩

15
4

v | x 2 = .

y 2

+ = 1

By the extremum principle, we have

15
4 ,

x 2 y 2 v ( x , y )= + ≤ 1 ,

so that

x 2 y 2

15
4 +

+

u = .

4

Example 2. Find the solution of
Δ

x 2

x 2 y 2 u = + y 2 , + < 1 ,

(7.51)

u | x 2 = 2 .

y 2

+ = 1

Solution. It is not straightforward to find u ∗ , a particular solution of the Pois-
son equation in PDS (7.51). In a polar coordinate system, the Poisson equation in
PDS (7.51) reads

2 u 2 u

∂θ

∂ 1

r ∂

u

r 2 ∂

1

r 2 + r +

2 =

r ,

∂ ∂

whose nonhomogeneous term is independent of θ . We can thus consider a particular

r 3
9

. A simple test shows that u ∗ = solution independent of θ is a particular solution.

r 3

By the function transformation u ( r , θ )= v ( r , θ )+

9 , we obtain

⎨

Δ v ( r , θ )= 0 , 0 < r < 1 ,

17
9 .
Its solution is, by the extremum principle,

⎩

v ( 1 , θ )=

17
9 ,

v = r ≤ 1 ,

so the solution of PDS (7.51) is

(

17
9 +

x 2 + y 2 )

3

u = .

9

Example 3 . Find the solution of
u xx

x 2 y 2 z 2 + u yy + u zz = 1 , + + < 1 ,

(7.52)

u | x 2 + y 2 + z 2 = 1 = 9 .