Mathematical Physics - Volume II - Numerical Methods

1.1 Finite difference method

27

In an analogous manner, we obtain

β u ( R ) − ( 1 + β ) u ( P )+ u ( r ) h 2 β ( β + 1 ) / 2

u yy ( P ) =

+ O ( h ) .

(1.80)

By summing we obtain an approximation of O ( h ) order of Poisson’s equations in P :

U ( R ) β + 1 −

1 β

h 2 2

U ( Q ) α + 1

1 α +

U ( q ) α ( α + 1 )

U ( r ) β ( β + 1 )

U ( P )+

f ( P ) .

(1.81)

+

+

=

Based on the expression derived above, the problem can be solved as follows: u xx + u yy = 0 na x 2 + y 2 < 1 y > 0 ,

(1.82) (1.83)

u ( x , y ) = 100 x 2 + y 2 = 1 y > 0 , u ( x , y ) = 0 y = 0 − 1 < x < 1 ,

(1.84) By using a square mesh, with h = 0 . 5. Thanks to its y -axis symmetry, the number of unknowns in the problem is reduced from three to two, since the semi-circular domain can be reduced to a quarter-circle, Fig. 1.3. Based on boundary conditions, we obtain: U 00 = 0 , U 10 = 0 , U 02 = 100 , U ( q ) = 100 , U ( r ) = 100 , U 11 = U − 1 , 1 .

y

1 = y

2

r

u = 0 x

u = 100

P

Q

y 1

q

S

R

x

x -1

x 0

x 1

x =1 2

u = 0

Figure 1.3: Half circle domain, reduced to circle quarter.

The only nodes in the mesh where the solution needs to be approximated are P ( x 1 , y 1 ) and Q ( x 0 , y 1 ) , which is achieved by using a finite difference method in Q : U 11 + 100 − 4 U 01 + U − 1 , 1 + 0 = 0 . Taking into account the boundary conditions, we obtain 2 U 01 − U 11 = 50 . Coordinates q and r are ( √ 3 h , h ) and ( h , h √ 3 ) ⇒ α = β = √ 3 − 1, thus the following is obtained: U 01 √ 3 + U 10 √ 3 − 2 U 11 √ 3 − 1 + U ( q ) 3 − √ 3 + U ( r ) 3 − √ 3 = 0 ,