Example 2
Approximate the solution of -y¢¢+ 2y = x, 0 £ x £ 1, and y(0) = y(1) = 0.
Using x0 = 0, x1 = 0.2, x2 = 0.4, x3 = 0.6, x4 = 0.8, and x5 = 1

Analysis: The equation is the same as in Example 1. But this time we divide the interval into 5 subintervals.
So we have p(x) = 1, q(x) = 2, and f(x) = x. Since n is one less than the number of subintervals n = 4.

We have five subintervals [0, 0.2], [0.2, 0.4], [0.4, 0.6], [0.6, 0.8], [0.8, 1] and the length of subintervals are the same, namely
h0 = h1 = h2 = h3 = h4 = 0.2.
With n = 4, we have four base functions

f1(x) =  ì
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î
 x 0.2 ,     0 < x £ 0.2
 0.4-x 0.2 ,     0.2 < x £ 0.4
 0,           0.4 £ x £ 1
f2(x) =  ì
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 0,           0 £ x £ 0.2
 x-0.2 0.2 ,     0.2 < x £ 0.4
 0.6-x 0.2 ,     0.4 < x £ 0.6
 0     0.6 < x £ 1
,
f3(x) =  ì
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î
 0,     0 < x £ 0.4
 x-.04 0.2 ,     0.4 < x £ 0.6
 0.8-x 0.2 ,     0.6 < x £ 0.8
 0,           0.8 £ x £ 1
and
f4(x) =  ì
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 0,           0 £ x £ 0.6
 x-0.6 0.2 ,     0.6 < x £ 0.8
 1-x 0.2 ,     0.8 < x £ 1
.

Our approximation to the solution will be u(x) = c1f1(x) + c2f2(x)+ c3f3(x)+ c4f4(x),
where c1, c2, c3, and c4 are the unknown parameters that are obtained by minimizing the functional

 I[f] = I[c1f1(x) + c2f2(x)+ c3f3(x)+ c4f4(x)]
with
 I[f]= ó õ 1 0 [c1f1¢(x) +c2f2¢(x) +c3f3¢(x)+c4f4¢(x)]2 + 2[c1f1(x)  c2f2(x)+c3f3(x)+c4f4(x)]2 - 2x(c1f1(x) +c2f2(x) +c3f3(x)+c4f4(x))}dx
.

From theoretical analysis, we know c = (c1, c2, c3, c4) is the solution of the linear system

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 a11     a12     a13     a14
 a21     a22     a23     a24
 a31     a32     a33     a34
 a41     a42     a43     a44
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 c1
 c2
 c3
 c4
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 b1
 b2
 b3
 b4
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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 x2 x1 (x - x1)2q(x) dx + ( 1 h2 )2 ó õ x3 x2 (x3 - x)2q(x) dx
 a33
 =
 ó õ 1 0 {p(x)[f3¢(x)]2 + q(x)[f3(x)]2} dx
 =
 ( 1 h2 )2 ó õ x3 x2 p(x) dx + ( 1 h3 )2 ó õ x4 x3 p(x) dx
 + ( 1 h2 )2 ó õ x3 x2 (x - x2)2q(x) dx + ( 1 h3 )2 ó õ x4 x3 (x4 - x)2q(x) dx
 a44
 =
 ó õ 1 0 {p(x)[f4¢(x)]2 + q(x)[f4(x)]2} dx
 =
 ( 1 h3 )2 ó õ x4 x3 p(x) dx + ( 1 h4 )2 ó õ x5 x4 p(x) dx
 + ( 1 h3 )2 ó õ x4 x3 (x - x3)2q(x) dx + ( 1 h4 )2 ó õ x5 x4 (x5 - 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
 a23
 =
 a32
 =
 ó õ 1 0 {p(x)f2¢(x)f3¢(x) + q(x)f2(x)f3(x)} dx
 =
 -( 1 h2 )2 ó õ x3 x2 p(x) dx + ( 1 h2 )2 ó õ x3 x2 (x3 - x)(x - x2)q(x) dx
 a34
 =
 a43
 =
 ó õ 1 0 {p(x)f3¢(x)f4¢(x) + q(x)f3(x)f4(x)} dx
 =
 -( 1 h3 )2 ó õ x4 x3 p(x) dx + ( 1 h3 )2 ó õ x4 x3 (x4 - x)(x - x3)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
 b3
 =
 ó õ 1 0 f(x)f3(x) dx
 =
 1 h2 ó õ x3 x2 (x - x2) f(x) dx+ 1 h3 ó õ x4 x3 (x4 - x) f(x) dx
and
 b4
 =
 ó õ 1 0 f(x)f4(x) dx
 =
 1 h3 ó õ x4 x3 (x - x3) f(x) dx+ 1 h4 ó õ x5 x4 (x5 - 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.)

We can get the linear system for c as

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 6.3
 -2.36667
 -2.36667
 6.3
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 -4.93333
 0
 0
 0
 0
 -4.93333
 0
 10.2667
 -4.93333
 -4.93333
 10.2667
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 c1
 c2
 c3
 c4
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 0.116667
 0.233333
 0.12
 0.16
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And the solution is c = (0.0260791, 0.0461646, 0.053777, 0.0414253).
Therefore we obtain the approximate solution.
Hence the approximate solution is:
 u(x) = 0.0260791f1(x) + 0.0461646f2(x) +0.053777f3(x) + 0.0414253f4(x)