456
8 Direct Methods of Variational Problems
Let the boundary of D be Γ , and the surface u(x, y) passes through Γ , namely on
Γ u(x, y) = 0. All the admissible surfaces are continuous in D, and are piecewise
smooth.
Solution Obviously, the extremal function of this functional is u(x, y) = 0. Because
no matter under what circumstances, there is u
2
x + u
2
y ≥ 0, as long as u(x, y) in D
has the value at a point which is not zero, then there will be a neighborhood near the
point, in this neighborhood, there exist u x and u y which are not all zero, such that
u
2
x + u
2
y ≥ 0, and the functional J [u] is greater than zero. So the extremal function
of the problem u(x, y) in D is zero everywhere, and the functional J [u] is also zero.
For this problem, a minimizing sequence can be found, the limits of various order
approximate extrema of the functional are zero, but the function of various order
approximate extrema is not a true extremal function, that is u(x, y) = 0.
Let there be a circle of radius R in the boundary Γ , the center is at the origin of
coordinates. In the domain between the circle R and the boundary Γ , u(x, y) is zero
everywhere. Let there be another circle of radius R
2 , where R < 1, then the circle
r = R
2 must be in the circle r = R. Let approximate extremal function be
u(x, y) =
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
0
(in Γ and out r = R)
A
ln R
ln
x 2 + y 2
R
=
A
ln R
ln
r
R
(R
2
≤ r ≤ R, R < 1)
A
(0 ≤ r ≤ R
2
)
(2)
Equation (2) has adopted the polar coordinates (r, θ), the pole is the origin of
coordinates. The graph of the function is as shown in Fig. 8.1, it is continuous, and
is piecewise smooth.
Taking the partial derivatives of u with respect to x and y respectively, we get
u x =
A
ln R
x
x 2 + y 2 =
A
ln R
x
r 2
(3)
u y =
A
ln R
y
x 2 + y 2 =
A
ln R
y
r 2
(4)
Substituting Eqs. (3) and (4) into the functional (1), out r = R and in r = R
2 ,
u
2
x + u
2
y ≡ 0, therefore Eq. (2) can be expressed by the polar coordinates as
J [u(R)] =
A
2
(ln R) 2
R
R 2
1
r 2 2πr dr = −
2πA
2
ln R
(5)
It can thus be seen that when the value of R reduces gradually, a minimizing
sequence can be obtained, that is Eq. (2), the limit is
lim
R→0
J [u(R)] = 0
( 6 )
8 Direct Methods of Variational Problems
Let the boundary of D be Γ , and the surface u(x, y) passes through Γ , namely on
Γ u(x, y) = 0. All the admissible surfaces are continuous in D, and are piecewise
smooth.
Solution Obviously, the extremal function of this functional is u(x, y) = 0. Because
no matter under what circumstances, there is u
2
x + u
2
y ≥ 0, as long as u(x, y) in D
has the value at a point which is not zero, then there will be a neighborhood near the
point, in this neighborhood, there exist u x and u y which are not all zero, such that
u
2
x + u
2
y ≥ 0, and the functional J [u] is greater than zero. So the extremal function
of the problem u(x, y) in D is zero everywhere, and the functional J [u] is also zero.
For this problem, a minimizing sequence can be found, the limits of various order
approximate extrema of the functional are zero, but the function of various order
approximate extrema is not a true extremal function, that is u(x, y) = 0.
Let there be a circle of radius R in the boundary Γ , the center is at the origin of
coordinates. In the domain between the circle R and the boundary Γ , u(x, y) is zero
everywhere. Let there be another circle of radius R
2 , where R < 1, then the circle
r = R
2 must be in the circle r = R. Let approximate extremal function be
u(x, y) =
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
0
(in Γ and out r = R)
A
ln R
ln
x 2 + y 2
R
=
A
ln R
ln
r
R
(R
2
≤ r ≤ R, R < 1)
A
(0 ≤ r ≤ R
2
)
(2)
Equation (2) has adopted the polar coordinates (r, θ), the pole is the origin of
coordinates. The graph of the function is as shown in Fig. 8.1, it is continuous, and
is piecewise smooth.
Taking the partial derivatives of u with respect to x and y respectively, we get
u x =
A
ln R
x
x 2 + y 2 =
A
ln R
x
r 2
(3)
u y =
A
ln R
y
x 2 + y 2 =
A
ln R
y
r 2
(4)
Substituting Eqs. (3) and (4) into the functional (1), out r = R and in r = R
2 ,
u
2
x + u
2
y ≡ 0, therefore Eq. (2) can be expressed by the polar coordinates as
J [u(R)] =
A
2
(ln R) 2
R
R 2
1
r 2 2πr dr = −
2πA
2
ln R
(5)
It can thus be seen that when the value of R reduces gradually, a minimizing
sequence can be obtained, that is Eq. (2), the limit is
lim
R→0
J [u(R)] = 0
( 6 )