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2 Variational Problems with Fixed Boundaries
y = y(x) solved from the above expression is still a function equation, it does
not necessarily satisfy the boundary conditions. So the solution of the variational
problem discussed here usually does not belong to the continuous function class.
If
∂ M
∂ y
−
∂ N
∂ x
≡ 0, then the expression Mdx + N dy is the total differential. The
integral
x 1
x 0
(Mdx + N dy) has nothing to do with the integral path at this time, thus
the value of the functional
J [y] =
x 1
x 0
M + N
dy
dx
dx
does not depend on the selection of the curve y = y(x), namely the functional on
the allowable curve is a fixed value, the variational problem loses its meaning.
Example 2.5.3 Find the extremal curve of the functional J [y]
=
x 1
x 0
(x y
2
+ x
2 yy
)dx, the boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution Since
∂ M
∂ y
= 2x y =
∂ N
∂ x
, namely
∂ M
∂ y
≡
∂ N
∂ x
, so, the integrand is total
differential, and the integral
J [y] =
1
2
x 1
x 0
d(x
2 y
2
) =
1
2
(x
2
1 y
2
1 − x
2
0 y
2
0 )
has nothing to do with the integral path, only depends on the boundary conditions.
The variational problem does not make sense.
Example 2.5.4 Discuss whether the extremal curve of functional J [y] =
1
0 (y
2
+ x
2 y
)dx exists, the boundary conditions are y(0) = 0, y(1) = a.
Solution Since
∂ M
∂ y
= y,
∂ N
∂ x
= x, so the Euler equation is
∂ M
∂ y
−
∂ N
∂ x
= 0, namely
y − x = 0. The extremal curve satisfies the first boundary condition y(0) = 0, and
the second boundary condition is satisfied only when a = 1, if a = 1, then the
extremal curve which satisfies the boundary conditions does not exist.
Example 2.5.5 Discuss whether the functional J [y] =
x 1
x 0
(ax + by + cx y
)dx
makes sense.
Solution Since
∂ M
∂ y
= b,
∂ N
∂ x
= c, so the Euler equation is
∂ M
∂ y
−
∂ N
∂ x
= 0, namely b −
c = 0, unless b = c, otherwise the functional or the Euler equation is unreasonable.
Even if b = c, the functional relation of y and x could not be obtained, Such functional
is unable to solve.
Under normal circumstances, if the integrand of a functional is linear combination
of derivatives of unknown functions, then the corresponding Euler equation is not a
differential equation, and its solution is meaningless.
2 Variational Problems with Fixed Boundaries
y = y(x) solved from the above expression is still a function equation, it does
not necessarily satisfy the boundary conditions. So the solution of the variational
problem discussed here usually does not belong to the continuous function class.
If
∂ M
∂ y
−
∂ N
∂ x
≡ 0, then the expression Mdx + N dy is the total differential. The
integral
x 1
x 0
(Mdx + N dy) has nothing to do with the integral path at this time, thus
the value of the functional
J [y] =
x 1
x 0
M + N
dy
dx
dx
does not depend on the selection of the curve y = y(x), namely the functional on
the allowable curve is a fixed value, the variational problem loses its meaning.
Example 2.5.3 Find the extremal curve of the functional J [y]
=
x 1
x 0
(x y
2
+ x
2 yy
)dx, the boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution Since
∂ M
∂ y
= 2x y =
∂ N
∂ x
, namely
∂ M
∂ y
≡
∂ N
∂ x
, so, the integrand is total
differential, and the integral
J [y] =
1
2
x 1
x 0
d(x
2 y
2
) =
1
2
(x
2
1 y
2
1 − x
2
0 y
2
0 )
has nothing to do with the integral path, only depends on the boundary conditions.
The variational problem does not make sense.
Example 2.5.4 Discuss whether the extremal curve of functional J [y] =
1
0 (y
2
+ x
2 y
)dx exists, the boundary conditions are y(0) = 0, y(1) = a.
Solution Since
∂ M
∂ y
= y,
∂ N
∂ x
= x, so the Euler equation is
∂ M
∂ y
−
∂ N
∂ x
= 0, namely
y − x = 0. The extremal curve satisfies the first boundary condition y(0) = 0, and
the second boundary condition is satisfied only when a = 1, if a = 1, then the
extremal curve which satisfies the boundary conditions does not exist.
Example 2.5.5 Discuss whether the functional J [y] =
x 1
x 0
(ax + by + cx y
)dx
makes sense.
Solution Since
∂ M
∂ y
= b,
∂ N
∂ x
= c, so the Euler equation is
∂ M
∂ y
−
∂ N
∂ x
= 0, namely b −
c = 0, unless b = c, otherwise the functional or the Euler equation is unreasonable.
Even if b = c, the functional relation of y and x could not be obtained, Such functional
is unable to solve.
Under normal circumstances, if the integrand of a functional is linear combination
of derivatives of unknown functions, then the corresponding Euler equation is not a
differential equation, and its solution is meaningless.