4.1 Variational Problems of the Simplest Functional
249
Example 4.1.2 Find the extremal curve of the functional J [y] =
x 1
x 0
1 + y 2 dx,
the constraint condition is that the left endpoint is fixed, y(x 0 ) = y 0 , the right endpoint
is undetermined on the straight line y = kx + b, and k = 0. And find the extremal
curve and extremum of the functional J [y] =
x 1
0
1 + y 2 dx. Let the left endpoint
y(0) = 1, the right end point is undetermined on the curve y = ψ(x) = 2 − x.
Solution The problem is boiled down to finding the minimum of the functional
J [y] =
x 1
x 0
1 + y 2 dx. The constraint condition is that the left endpoint is fixed,
y(x 0 ) = y 0 , the right endpoint is on the straight line y = ψ(x) = kx + b, and
ψ
(x 1 ) = k. The general solution of the Euler equation of the functional is the
straight line y = c 1 x + c 2 . The system of equations about three unknowns x 1 , c 1 and
c 2 is
⎧
⎨
⎩
c 1 x 0 + c 2 = y 0
c 1 x 1 + c 2 = kx 1 + b
1 + kc 1 = 0
Solving this system of equations, we get c 1 = −
1
k
, c 2 = y 0 −
x 0
k
, x 1 =
k(y 0 −b)−x 0
k 2 +1
.
The ordinate corresponding to point x 1 is y 1 =
k
2 y 0 +b−kx 0
k 2 +1
. Thus the extremal curve
is y = −
x+x 0
k
+ y 0 .
For the functional J [y] =
x 1
0
1 + y 2 dx, the left endpoint is y(0) = 1, the right
endpoint is the undetermined extremal curve on the curve y = ψ(x) = 2 − x, which
is equivalent to the case of x 0 = 0, y 0 = 1, k = −1 and b = 2, substituting them
into the equation of the extremal curve, we get
y = −
x
k
+ y 0 = x + 1
The coordinates of the intersection are
x 1 =
k(y 0 − b)
k 2 + 1
=
−1 × (1 − 2)
(−1) × (−1) + 1
=
1
2
y 1 =
k
2 y 0 + b
k 2 + 1
=
(−1) × (−1) × 1 + 2
(−1) × (−1) + 1
=
3
2
Substituting the extremal curve y = x + 1 and x 1 =
1
2
into the original functional,
the value of the functional is
J [y] =
x 1
0
1 + y 2 dx =
1
2
0
1 + 1 2 dx =
√
2
2
249
Example 4.1.2 Find the extremal curve of the functional J [y] =
x 1
x 0
1 + y 2 dx,
the constraint condition is that the left endpoint is fixed, y(x 0 ) = y 0 , the right endpoint
is undetermined on the straight line y = kx + b, and k = 0. And find the extremal
curve and extremum of the functional J [y] =
x 1
0
1 + y 2 dx. Let the left endpoint
y(0) = 1, the right end point is undetermined on the curve y = ψ(x) = 2 − x.
Solution The problem is boiled down to finding the minimum of the functional
J [y] =
x 1
x 0
1 + y 2 dx. The constraint condition is that the left endpoint is fixed,
y(x 0 ) = y 0 , the right endpoint is on the straight line y = ψ(x) = kx + b, and
ψ
(x 1 ) = k. The general solution of the Euler equation of the functional is the
straight line y = c 1 x + c 2 . The system of equations about three unknowns x 1 , c 1 and
c 2 is
⎧
⎨
⎩
c 1 x 0 + c 2 = y 0
c 1 x 1 + c 2 = kx 1 + b
1 + kc 1 = 0
Solving this system of equations, we get c 1 = −
1
k
, c 2 = y 0 −
x 0
k
, x 1 =
k(y 0 −b)−x 0
k 2 +1
.
The ordinate corresponding to point x 1 is y 1 =
k
2 y 0 +b−kx 0
k 2 +1
. Thus the extremal curve
is y = −
x+x 0
k
+ y 0 .
For the functional J [y] =
x 1
0
1 + y 2 dx, the left endpoint is y(0) = 1, the right
endpoint is the undetermined extremal curve on the curve y = ψ(x) = 2 − x, which
is equivalent to the case of x 0 = 0, y 0 = 1, k = −1 and b = 2, substituting them
into the equation of the extremal curve, we get
y = −
x
k
+ y 0 = x + 1
The coordinates of the intersection are
x 1 =
k(y 0 − b)
k 2 + 1
=
−1 × (1 − 2)
(−1) × (−1) + 1
=
1
2
y 1 =
k
2 y 0 + b
k 2 + 1
=
(−1) × (−1) × 1 + 2
(−1) × (−1) + 1
=
3
2
Substituting the extremal curve y = x + 1 and x 1 =
1
2
into the original functional,
the value of the functional is
J [y] =
x 1
0
1 + y 2 dx =
1
2
0
1 + 1 2 dx =
√
2
2