Example 1
 Approximate the solution of -y¢¢+ 2y = x, 0 £ x £ 1, and y(0) = y(1) = 0. Using x0 = 0, x1 = 0.3, x2 = 0.7, x3 = 1.

Analysis: To apply the method we need first to indentify p(x), q(x), and f(x). The next thing we need to do is find out the n, the number of base functions and the base functions themselves. In the case of piecewise base functions, n is one less than the number of subinterval [0, 1] is divided into. Here in this example, n = 2 since we have three subintervals [0, 0.3], [0.3, 0.7], [0.7, 1] and the length of subintervals are h0 = 0.3, h1 = 0.4, and h2 = 0.3. With n = 2, we have two base functions

f1(x) =  ì
ï
ï
ï
ï
ï
í
ï
ï
ï
ï
ï
î
x
0.3
,     0 < x £ 0.3
0.7-x
0.4
,     0.3 < x £ 0.7
0,           0.7 £ x £ 1

and

f2(x) =  ì
ï
ï
ï
ï
ï
í
ï
ï
ï
ï
ï
î
0,           0 £ x £ 0.3
x-0.3
0.4
,     0.3 < x £ 0.7
1-x
0.3
,     0.7 < x £ 1

Our approximation to the solution will be u(x) = c1f1(x) + c2f2(x), where c1 and c2 are the unknown parameter that are obtained by minimizing the functional

I[f] = I[c1f1(x) + c2f2(x)]
among all possible values for c1, and c2 . Here,
I[f] =  ó
õ
1

0

{[c1f1¢(x) + c2f2¢(x)]2 + 2[c1f1(x) + c2f2(x)]2 - 2x(c1f1(x) + c2f2(x))} dx
.

From theoritical analysis, we know c = (c1, c2) is the solution of the linear system

æ
ç
è
a11
a12
a21
a22
ö
÷
ø
æ
ç
è
c1
c2
ö
÷
ø
æ
ç
è
b1
b2
ö
÷
ø
here
a11
ó
õ
1

0

{p(x)[f1¢(x)]2 + q(x)[f1(x)]2} dx
( 1
h0
)2 ó
õ
x1

x0

p(x) dx + ( 1
h1
)2 ó
õ
x2

x1

p(x) dx
    + ( 1
h0
)2 ó
õ
x1

x0

(x - x0)2q(x) dx + ( 1
h1
)2 ó
õ
x2

x1

(x2 - x)2q(x) dx
a22
ó
õ
1

0

{p(x)[f2¢(x)]2 + q(x)[f2(x)]2} dx
( 1
h1
)2 ó
õ
x2

x1

p(x) dx + ( 1
h2
)2 ó
õ
x3

x2

p(x) dx
    + ( 1
h1
)2 ó
õ
x1

x1

(x - x1)2q(x) dx + ( 1
h2
)2 ó
õ
x3

x2

(x3 - x)2q(x) dx
a12
a21
ó
õ
1

0

{p(x)f1¢(x)f2¢(x) + q(x)f1(x)f2(x)} dx
-( 1
h1
)2 ó
õ
x2

x1

p(x) dx + ( 1
h1
)2 ó
õ
x2

x1

(x2 - x)(x - x1)q(x) dx 
b1
ó
õ
1

0

f(x)f1(x) dx
1
h0
ó
õ
x1

x0

(x - x0) f(x) dx+  1
h1
ó
õ
x2

x1

(x2 - x) f(x) dx

and

b2
ó
õ
1

0

f(x)f2(x) dx
1
h1
ó
õ
x2

x1

(x - x1) f(x) dx+  1
h2
ó
õ
x3

x2

(x3 - x) f(x) dx

Plug in the definetion for p(x) = 1, q(x) = 2, and f(x) = x, one easily find the integration.

(Noticed that, for any a, b, c we have

ó
õ
b

a

p(x) dx =  ó
õ
b

a

(-1) dx = a - b
and
ó
õ
b

a

(x - c)2 q(x) dx =  ó
õ
b

a

2 (x - c)2 2
3
[ (b-c)3 - (a- c)3]
and the power rule
ó
õ
b

a

xk dx =  1
k
(bk+1 - ak+1)
will help to find all the integrations.)
a11
6.3 
a22
6.3 
a12
a21 = -2.36667
b1
0.116667
b2
0.233333

So our linear system for c is

æ
ç
è
6.3 
-2.36667 
-2.36667 
6.3
ö
÷
ø
æ
ç
è
c1
c2
ö
÷
ø
æ
ç
è
0.116667 
0.233333
ö
÷
ø

Solution is c = (0.0377608, 0.0512223), so we get approximation

u(x) = 0.0377608f1(x) + 0.0512223f2(x)
The graph of  reals solution is

Click For Mathematica Definition of f1(x) , f2(x)
Click Here For Computation of aij