8.7 Eigenvalues and Eigenfunctions of Operator Equations
497
1
r
d
dr
r
du
dr
+ λu = 0
( 9 )
Choosing the coordinate functions ϕ i = cos
(2i−1)π
2a
r , i = 1, 2, . . ., taking the first
approximate solution, there is
u 1 = a 1 cos
π
2a
r
(10)
Substituting Eq. (10) into the differential equation, the residual is obtained
R = −a 1
π
2ar
sin
π
2a
r − a 1
π
2
4a 2 cos
π
2a
r + λ cos
π
2a
r
(11)
The Galerkin equation is
2π
0
a
0
R cos
π
2a
r
r dr dθ = 0
( 1 2 )
The integral result is
π +
π
3
8
−
π
2
− πλa
2
1
2
−
2
π 2
= 0
(13)
Solving for λ from Eq. (13), we give
λ =
π
2
(π
2
+ 4)
4a 2 (π 2 − 4)
=
5.830355
a 2
(14)
Taking the second approximate solution, there is
u 2 = a 1 cos
π
2a
r + a 2 cos
3π
2a
r
(15)
According to the similar to the above steps, the residual and Galerkin equations
can be obtained, when the integral the following formulas of trigonometric function
can be used
sin 3x = 3 sin x − 4 sin
3 x, cos 3x = 4 cos
3 x − 3 cos x
(16)
The integral result of Galerkin equations can get the following homogeneous
linear equations
497
1
r
d
dr
r
du
dr
+ λu = 0
( 9 )
Choosing the coordinate functions ϕ i = cos
(2i−1)π
2a
r , i = 1, 2, . . ., taking the first
approximate solution, there is
u 1 = a 1 cos
π
2a
r
(10)
Substituting Eq. (10) into the differential equation, the residual is obtained
R = −a 1
π
2ar
sin
π
2a
r − a 1
π
2
4a 2 cos
π
2a
r + λ cos
π
2a
r
(11)
The Galerkin equation is
2π
0
a
0
R cos
π
2a
r
r dr dθ = 0
( 1 2 )
The integral result is
π +
π
3
8
−
π
2
− πλa
2
1
2
−
2
π 2
= 0
(13)
Solving for λ from Eq. (13), we give
λ =
π
2
(π
2
+ 4)
4a 2 (π 2 − 4)
=
5.830355
a 2
(14)
Taking the second approximate solution, there is
u 2 = a 1 cos
π
2a
r + a 2 cos
3π
2a
r
(15)
According to the similar to the above steps, the residual and Galerkin equations
can be obtained, when the integral the following formulas of trigonometric function
can be used
sin 3x = 3 sin x − 4 sin
3 x, cos 3x = 4 cos
3 x − 3 cos x
(16)
The integral result of Galerkin equations can get the following homogeneous
linear equations
