2.5 Several Special Cases of the Euler Equation and Their Integrals
125
contain one or two among x, y and y
, the problems are likely to get simplified. This
section will discuss some special forms of the integrand F(x, y, y
) in the functional
(2.4.1).
(1) F does not depend on y
or only relies on y, namely F = F(x, y) or F = F(y)
At this time, F y ≡ 0, so the Euler equation F y (x, y) = 0 or F y (y) = 0, this is a
function equation, the solution does not contain arbitrary constant. The solution of the
function equation doesn’t satisfy the boundary conditions: y(x 0 ) = y 0 , y(x 1 ) = y 1 ,
the variational problem has no solution. In rare cases, for instance, only when the
solution of F y (x, y) = 0 or F y (y) = 0 passes through points (x 0 , y 0 ) and (x 1 , y 1 ),
it can become an extremal curve. If the problem has a solution, there will be no
additional boundary conditions.
Example 2.5.1 Known the functional J [y] = π
x 1
x 0
y
2 dx, y(x 0 ) = y 0 , y(x 1 ) = y 1 ,
find the extremum of the functional J [y].
Solution The Euler equation of the functional is 2y = 0 or y = 0, if and only if
y 0 = y 1 = 0, y = 0 the value of the functional J [y] is minimum. Otherwise the
minimum of the functional J [y(x)] can not be reached in the continuous function
class.
Example 2.5.2 Known the functional J [y] =
π
0 y(2x − y)dx, y(0) = 0, y(π) =
1, find the extremal curve of the functional.
Solution The Euler equation of the functional is 2x − 2y = 0, namely y = x.
Because the boundary conditions are satisfied, so on the straight line y = x, the
extremal value of the functional can be obtained. For the other boundary conditions,
such as y(0) = 0, y(π) = 1, the straight line y = x does not pass through the given
boundary points (0, 0) and (π, 1), the variational problem has no solution.
(2) F linearly depends on y
, namely F(x, y, y
) = M(x, y) + N (x, y)y
At this moment, the Euler equation is
∂ M
∂ y
+
∂ N
∂ y
y
−
dN
dx
= 0
Expand the Euler equation, we obtain
∂ M
∂ y
+
∂ N
∂ y
y
−
∂ N
∂ x
−
∂ N
∂ y
y
= 0
Arrange it, we obtain
∂ M
∂ y
−
∂ N
∂ x
= 0
125
contain one or two among x, y and y
, the problems are likely to get simplified. This
section will discuss some special forms of the integrand F(x, y, y
) in the functional
(2.4.1).
(1) F does not depend on y
or only relies on y, namely F = F(x, y) or F = F(y)
At this time, F y ≡ 0, so the Euler equation F y (x, y) = 0 or F y (y) = 0, this is a
function equation, the solution does not contain arbitrary constant. The solution of the
function equation doesn’t satisfy the boundary conditions: y(x 0 ) = y 0 , y(x 1 ) = y 1 ,
the variational problem has no solution. In rare cases, for instance, only when the
solution of F y (x, y) = 0 or F y (y) = 0 passes through points (x 0 , y 0 ) and (x 1 , y 1 ),
it can become an extremal curve. If the problem has a solution, there will be no
additional boundary conditions.
Example 2.5.1 Known the functional J [y] = π
x 1
x 0
y
2 dx, y(x 0 ) = y 0 , y(x 1 ) = y 1 ,
find the extremum of the functional J [y].
Solution The Euler equation of the functional is 2y = 0 or y = 0, if and only if
y 0 = y 1 = 0, y = 0 the value of the functional J [y] is minimum. Otherwise the
minimum of the functional J [y(x)] can not be reached in the continuous function
class.
Example 2.5.2 Known the functional J [y] =
π
0 y(2x − y)dx, y(0) = 0, y(π) =
1, find the extremal curve of the functional.
Solution The Euler equation of the functional is 2x − 2y = 0, namely y = x.
Because the boundary conditions are satisfied, so on the straight line y = x, the
extremal value of the functional can be obtained. For the other boundary conditions,
such as y(0) = 0, y(π) = 1, the straight line y = x does not pass through the given
boundary points (0, 0) and (π, 1), the variational problem has no solution.
(2) F linearly depends on y
, namely F(x, y, y
) = M(x, y) + N (x, y)y
At this moment, the Euler equation is
∂ M
∂ y
+
∂ N
∂ y
y
−
dN
dx
= 0
Expand the Euler equation, we obtain
∂ M
∂ y
+
∂ N
∂ y
y
−
∂ N
∂ x
−
∂ N
∂ y
y
= 0
Arrange it, we obtain
∂ M
∂ y
−
∂ N
∂ x
= 0