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8 Direct Methods of Variational Problems
V
(T u n − f )W vi dV = 0 (i = 1, 2, . . . , n)
(8.5.9)
Solving the above algebraic equations, parameters a i (i = 1, 2, . . . , n) can be
obtained, Substituting a i back into the approximate solution (8.5.3), then the
approximate solution of Eq. (8.5.1) is obtained.
It can be seen from the above description of the fundamental principle of weighted
residual method that, the application of the weighted residual method can be divided
into the three steps: (1) Choosing the trial functions; (2) Substituting them into the
operator equation and solving the residual expression; (3) Choosing the weighted
functions, making of the inner product of the residual expressions and weighted
functions, and making they be orthogonal to eliminate residuals.
In the weighted residual method, the method of choice for the weighted functions
is very important, it has close relation with the precision of the approximate solution found, the different weighted function will derive the different approximation
methods, also represent the different error distribution. If the weighted functions W vi
are equal to the trial functions ϕ i , namely putting W vi = ϕ i , then there is
V
(T u n − f )ϕ i dV = 0 (i = 1, 2, . . . , n)
(8.5.10)
The expression (8.5.10) is called the Galerkin(’s) equations, solve for
a i (i = 1, 2, . . . , n) from the equations, at this point a i (i = 1, 2, . . . , n) are called
the Galerkin coefficient, substituting them into the approximate solution (8.5.3),
the approximate solutions of the boundary value problems (8.5.1) and (8.5.2) are
obtained. Because the Galerkin method has high precision, the labor of calculation
is not very large, it is widely used.
Note that the Galerkin method is only applicable to the homogeneous boundary
conditions. If the boundary conditions are non homogeneous, through the appropriate
variable substitution the nonhomogeneous boundary conditions are transformed into
the homogeneous boundary conditions.
Example 8.5.1 Use Galerkin method to solve the boundary value problem y
+ y =
2x, y(0) = 0, y(1) = 0.
Solution Choosing the coordinate function sequence
ϕ k (x) = x
k
(1 − x) (k = 1, 2, . . .)
Taking the former two terms, then the approximate solution is
y 2 = a 1 x(1 − x) + a 2 x
2
(1 − x)
Thus, there are
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