496
8 Direct Methods of Variational Problems
Thus, the approximate value of the first eigenvalue is quite accurate, but the error
of the approximate value of the second eigenvalue is relatively large.
Example 8.7.2 Let G be the domain x
2
+ y
2
≤ a
2 , Γ is the boundary. Find the
first and second characteristic value and corresponding characteristic function of the
definite problem u + λu = 0, u| Γ = 0.
Solution Choosing the polar coordinate system (r, θ), the relationships between it
and the rectangular coordinates are
x = r cos θ, y = sin θ, r
2
= x
2
+ y
2
, θ = arctan
y
x
(1)
From which getting
∂r
∂ x
=
x
r
= cos θ,
∂r
∂ y
=
y
r
= sin θ
(2)
∂θ
∂ x
=
∂
∂ x
arctan
y
x
= −
sin θ
r
,
∂θ
∂ y
=
∂
∂ y
arctan
y
x
=
cos θ
r
(3)
∂u
∂ x
=
∂u
∂r
∂r
∂ x
+
∂u
∂θ
∂θ
∂ x
= cos θ
∂u
∂r
−
sin θ
r
∂u
∂θ
(4)
∂u
∂ y
=
∂u
∂r
∂r
∂ y
+
∂u
∂θ
∂θ
∂ y
= sin θ
∂u
∂r
+
cos θ
r
∂u
∂θ
(5)
∂
2 u
∂ x 2 =
cos θ
∂
∂r
−
sin θ
r
∂
∂θ
cos θ
∂u
∂r
−
sin θ
r
∂u
∂θ
= cos
2
θ
∂
2 u
∂r 2 + sin
2
θ
1
r
∂u
∂r
+
1
r 2
∂
2 u
∂θ 2
+ 2 sin θ cos θ
1
r 2
∂u
∂θ
−
1
r
∂
2 u
∂r ∂θ
(6)
∂
2 u
∂ y 2 =
sin θ
∂
∂r
+
cos θ
r
∂
∂θ
sin θ
∂u
∂r
+
cos θ
r
∂u
∂θ
= sin
2
θ
∂
2 u
∂r 2 + cos
2
θ
1
r
∂u
∂r
+
1
r 2
∂
2 u
∂θ 2
− 2 sin θ cos θ
1
r 2
∂u
∂θ
−
1
r
∂
2 u
∂r ∂θ
(7)
Adding the above two equations, we obtain
u =
∂
2 u
∂ x 2 +
∂
2 u
∂ x 2 =
∂
2 u
∂r 2 +
1
r
∂u
∂r
+
1
r 2
∂
2 u
∂θ 2
(8)
For the axisymmetric problem, the last term of Eq. (8) is zero. The original
equation can be transformed to