474
8 Direct Methods of Variational Problems
The first approximate solution is
z 1 = a 1 ϕ 1 (x) = a 1 (x − 1)(2 − x) = a 1 (−2 + 3x − x
2
)
(4)
Derive and obtain
z
1 = (3 − 2x)a 1 , z
1 = −2a 1
(5)
Substituting the two derivatives of Eqs. (5) into (1), the residual is obtained
x
2 z
+ xz
+ (x
2
− 1)z + x
3
= (2 − 5x
2
+ 3x
3
− x
4
)a 1 + x
3
(6)
The Galerkin equation is
2
1
[(2 − 5x
2
+ 3x
3
− x
4
)a 1 + x
3
](x − 1)(2 − x)dx = 0
( 7 )
The integral result is
−
311
420
a 1 +
3
5
= 0
( 8 )
Solve for a 1
a 1 =
252
311
(9)
Thus, we get
z 1 =
252
311
(x − 1)(2 − x)
(10)
Substituting it back into the original variable, the first approximate solution of the
given boundary value problem is
y =
252
311
(x − 1)(2 − x) + x
(11)
Example 8.5.3 Use the Galerkin method to solve the rocket flight problem of
Example 8.3.3.
Solution Choosing the base function such that it has the same form as the Ritz
method
v(m) = βm
(1)
8 Direct Methods of Variational Problems
The first approximate solution is
z 1 = a 1 ϕ 1 (x) = a 1 (x − 1)(2 − x) = a 1 (−2 + 3x − x
2
)
(4)
Derive and obtain
z
1 = (3 − 2x)a 1 , z
1 = −2a 1
(5)
Substituting the two derivatives of Eqs. (5) into (1), the residual is obtained
x
2 z
+ xz
+ (x
2
− 1)z + x
3
= (2 − 5x
2
+ 3x
3
− x
4
)a 1 + x
3
(6)
The Galerkin equation is
2
1
[(2 − 5x
2
+ 3x
3
− x
4
)a 1 + x
3
](x − 1)(2 − x)dx = 0
( 7 )
The integral result is
−
311
420
a 1 +
3
5
= 0
( 8 )
Solve for a 1
a 1 =
252
311
(9)
Thus, we get
z 1 =
252
311
(x − 1)(2 − x)
(10)
Substituting it back into the original variable, the first approximate solution of the
given boundary value problem is
y =
252
311
(x − 1)(2 − x) + x
(11)
Example 8.5.3 Use the Galerkin method to solve the rocket flight problem of
Example 8.3.3.
Solution Choosing the base function such that it has the same form as the Ritz
method
v(m) = βm
(1)