Two-Point Boundary Value Problems - Numerical Methods in Engineering with MATLAB - page 313

Second-Order Differential Equation

Consider the second-orderdifferentialequation
y = y ) f ( x , y ,

with the boundary conditions

or y ( a )

y ( a ) = α = α

or y ( b ) y ( b ) = β = β

Approximating the derivatives at the mesh points by finite differences, the prob-
lembecomes

f x i ,

,

y i − 1 − 2 y i + y i + 1 − y i + 1 y i − 1

= y i , i = 1 , 2 ,..., n (8.9)

h 2 2 h

y 2 − y 0

y 1 = α or = α (8.10a)

2 h

y n + 1 − y n − 1

y n = β or = β (8.10b)

2 h

Note the presence of y 0 and y n + 1 , which are associatedwith points outside the solution
domain ( a

b ). This “spillover” can beeliminatedbyusing the boundary conditions.
But before we dothat, let us rewrite Eqs. (8.9) as

,

h 2 f x 1 ,

y 2 − y 0

y 0 − 2 y 1 + y 2 − y 1 , = 0 (a)

2 h

h 2 f x i ,

y i + 1 − y i − 1

y i − 1 − 2 y i + y i + 1 − y i , = 0, i = 2 , 3 ,..., n − 1 (b)

2 h

h 2 f x n

y n + 1 − y n − 1

y n − 1 − 2 y n + y n + 1 − , y n , = 0 (c)

2 h

The boundary conditionson y areeasilydealt with:Eq. (a) issimplyreplaced
by y 1 0. If y are prescribed, weobtain − α = 0 and Eq. (c) is replacedby y n − β =

fromEqs. (8.10) y 0 = , which are then substitutedinto

Eqs. (a) and (c), respectively. Hence we finish up with n equations in the unknowns
y i , i

y 2 − 2 h α and y n + 1 = y n − 1 + 2 h β

= 1 , 2 ..., n :

y 1 − α = 0 if y ( a ) = α

(8.11a)

− 2 y 1 + 2 y 2 − h 2 f ( x 1 , y 1 ,α − 2 h α = 0 if y ( a ) = α

)

h 2 f x i ,

y i + 1 − y i − 1

y i − 1 − 2 y i + y i + 1 − y i , = 0 i = 2 , 3 ,..., n − 1 (8.11b)

2 h

y n − β = 0 if y ( b ) = β

(8.11c)

h 2 f ( x n , 0 if y ( b ) 2 y n − 1 − 2 y n − y n ,β + 2 h β = = β

)