2
Constrained variational problems
The boundary values applied in the prior discussion may also be considered
as constraints. The subject of this chapter is to generalize the constraint con-
cept in two senses. The first is to allow more difficult, algebraic boundary
conditions, and the second is to allow constraints imposed on the interior of
the domain as well.
2.1 Algebraic boundary conditions
There is the possibility of defining the boundary condition at one end of the
integral of the variational problem with an algebraic constraint. Let the
x
1
x
0
f(x, y, y
)dx = extremum
variational problem subject to the customary boundary condition
y(x
0
)=y
0
,
on the lower end, and on the upper end an algebraic condition of the following
form is given:
g(x, y)=0.
We again consider an alternative solution of the form
Y (x)=y(x)+η(x).
The given boundary condition in this case is
η(x
0
)=0.
Then, following [9], the intersection of the alternative solution and the alge-
braic curve is
X
1
= X
1
()
25
26 Applied calculus of variations for engineers
and
Y
1
= Y
1
().
The notation is to distinguish from the fixed boundary condition values given
via x
1
,y
1
. Therefore the algebraic condition is
g(X
1
,Y
1
)=0.
This must be true for any , hence applying the chain rule yields
dg
d
=
∂g
∂X
1
dX
1
d
+
∂g
∂Y
1
dY
1
d
=0. (2.1)
Since
Y
1
= y(X
1
)+η(X
1
),
we expand the last derivative of the second term of Equation (2.1) as
dY
1
d
=
dy
dx
|
x=X
1
dX
1
d
+ η(X
1
)+
dη
dx
|
x=X
1
dX
1
d
.
Substituting into Equation (2.1) results in
dg
d
=
∂g
∂X
1
dX
1
d
+
∂g
∂Y
1
(
dy
dx
|
x=X
1
dX
1
d
+ η(X
1
)+
dη
dx
|
x=X
1
dX
1
d
)=0.
Since (X
1
,Y
1
) becomes (x
1
,y
1
)when =0,
dX
1
d
|
=0
= −
η(x
1
)
∂g
∂y
|
y=y
1
∂g
∂x
|
x=x
1
+
∂g
∂y
|
y=y
1
dy
dx
|
x=x
1
. (2.2)
We now consider the variational problem of
I()=
X
1
x
0
f(x, Y, Y
)dx.
The derivative of this is
∂I()
∂
=
dX
1
d
f|
x=X
1
+
X
1
x
0
(
∂f
∂Y
η +
∂f
∂Y
η
)dx.
Integrating by parts and taking =0yields
∂I()
∂
|
=0
=
dX
1
d
|
=0
f|
x=x
1
+
∂f
∂y
|
x=x
1
η(x
1
)+
x
1
x
0
(
∂f
∂y
−
d
dx
∂f
∂y
)ηdx.
Substituting the first expression with Equation (2.2) results in
(
∂f
∂y
|
x=x
1
−
∂g
∂y
|
y=y
1
f|
x=x
1
∂g
∂x
|
x=x
1
+
∂g
∂y
|
y=y
1
dy
dx
|
x=x
1
)η(x
1
)+
x
1
x
0
(
∂f
∂y
−
d
dx
∂f
∂y
)ηdx =0.
Constrained variational problems 27
Due to the fundamental lemma of calculus of variations, to find the con-
strained variational problem’s extremum the Euler-Lagrange differential equa-
tion of
∂f
∂y
−
d
dx
∂f
∂y
=0,
with the given boundary condition
y(x
0
)=y
0
,
and the algebraic constraint condition of the form
∂f
∂y
|
x=x
1
=
∂g
∂y
|
y=y
1
f|
x=x
1
∂g
∂x
|
x=x
1
+
∂g
∂y
|
y=y
1
dy
dx
|
x=x
1
all need to be satisfied.
2.2 Lagrange’s solution
We now further generalize the variational problem and impose both boundary
conditions as well as an algebraic condition on the whole domain as follows:
I(y)=
x
1
x
0
f(x, y, y
)dx = extremum,
with
y(x
0
)=y
0
,y(x
1
)=y
1
,
while
J(y)=
x
1
x
0
g(x, y, y
)dx = constant.
Following the earlier established pattern, we introduce an alternative solution
function, at this time, however, with two auxiliary functions as
Y (x)=y(x)+
1
η
1
(x)+
2
η
2
(x).
Here the two auxiliary functions are arbitrary and both satisfy the boundary
conditions:
η
1
(x
0
)=η
1
(x
1
)=η
2
(x
0
)=η
2
(x
1
)=0.
28 Applied calculus of variations for engineers
Substituting these into the integrals gives
I(Y )=
x
1
x
0
f(x, Y, Y
)dx,
and
J(Y )=
x
1
x
0
g(x, Y, Y
)dx.
Lagrange’s ingenious solution is to tie the two integrals together with a yet
unknown multiplier (now called the Lagrange multiplier) as follows:
I(
1
,
2
)=I(Y )+λJ(Y )=
x
1
x
0
h(x, Y, Y
)dx,
where
h(x, y, y
)=f(x, y, y
)+λg(x, y, y
).
The condition to solve this variational problem is
∂I
∂
i
=0
when
i
=0;i =1, 2.
The derivatives are of the form
∂I
∂
i
=
x
1
x
0
(
∂h
∂Y
η
i
+
∂h
∂Y
η
i
)dx.
The extremum is obtained when
∂I
∂
i
|
i
=0,i=1,2
=
x
1
x
0
(
∂h
∂Y
η
i
+
∂h
∂Y
η
i
)dx =0.
Considering the boundary conditions and integrating by parts yields
x
1
x
0
(
∂h
∂y
−
d
dx
∂h
∂y
)η
i
dx =0,
which, due to the fundamental lemma of calculus of variations, results in the
relevant Euler-Lagrange differential equation
∂h
∂y
−
d
dx
∂h
∂y
=0.
This equation contains three undefined coefficients: the two coefficients of
integration satisfying the boundary conditions and the Lagrange multiplier,
enforcing the constraint.
Constrained variational problems 29
2.3 Application: iso-perimetric problems
Iso-perimetric problems use a given perimeter of a certain object as the con-
straint of some variational problem. The perimeter may be a curve in the
two-dimensional case, as in the example of the next section. It may also be
the surface of a certain body, in the three-dimensional case.
2.3.1 Maximal area under curve with given length
This problem is conceptually very simple, but useful to illuminate the process
just established. It is also a very practical problem with more difficult geome-
tries involved. Here we focus on the simple case of finding the curve of given
length between two points in the plane. Without restricting the generality of
the discussion, we’ll position the two points on the x axis in order to simplify
the arithmetic.
The given points are (x
0
, 0) and (x
1
, 0) with x
0
<x
1
. The area under any
curve going from the start point to the endpoint in the upper half-plane is
I(y)=
x
1
x
0
ydx.
The constraint of the given length L is presented by the equation
J(y)=
x
1
x
0
1+y
2
dx = L.
The Lagrange multiplier method brings the function
h(x, y, y
)=y(x)+λ
1+y
2
.
The constrained variational problem is
I(y)=
x
1
x
0
h(x, y, y
)dx
whose Euler-Lagrange equation becomes
1 − λ
d
dx
y
1+y
2
=0.
Integration yields
λy
1+y
2
= x − c
1
.
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