54 Applied calculus of variations for engineers
and
y
− y + z =0.
The elimination of the Lagrange multiplier results in the system of
y − z + z
=0,
and
y
− y + z =0,
whose solution follows from classical calculus.
4.4 Linearization of second order problems
It is very common in engineering practice that the highest derivative of interest
is of second order. Accelerations in engineering analysis of motion, curvature
in description of space curves, and other important application concepts are
tied to the second derivative.
This specific case of quadratic problems may be reverted to a linear prob-
lem involving two functions. Consider
I(y)=
x
1
x
0
f(x, y, y
,y
)dx = extremum.
with the following boundary conditions given
y(x
0
),y(x
1
),y
(x
0
),y
(x
1
).
By introducing a new function
z(x)=y
(x),
we can reformulate the unconstrained second order variational problem as a
variational problem of the first order with multiple functions in the integrand
I(y,z)=
x
1
x
0
f(x, y, z, z
)dx = extremum,
but subject to a constraint involving the derivative
g(x, y, z)=z − y
=0.
Using a Lagrange multiplier function in the form of
h(x, y, z, z
,λ)=f(x, y, z, z
)+λ(x)(z − y
),
Higher order derivatives 55
and following the process laid out in the last section we can produce a system
of three Euler-Lagrange differential equations.
∂h
∂y
−
d
dx
∂h
∂y
=
∂f
∂y
−
dλ
dx
=0,
∂h
∂z
−
d
dx
∂h
∂z
=
∂f
∂z
+ λ −
d
dx
∂f
∂z
=0,
and
∂h
∂λ
−
d
dx
∂h
∂λ
= z − y
.
This may, of course, be turned into the Euler-Poisson equation by expressing
λ =
d
dx
∂f
∂z
−
∂f
∂z
from the middle equation and differentiating as
dλ
dx
=
d
2
dx
2
∂f
∂z
−
d
dx
∂f
∂z
.
Substituting this and the third equation into the first yields the Euler-Poisson
equation we could have achieved, had we approached the original quadratic
problem directly:
∂f
∂y
−
d
dx
∂f
∂y
+
d
2
dx
2
∂f
∂y
=0.
Depending on the particular application circumstance, the linear system of
Euler-Lagrange equations may be more conveniently solved than the quadratic
Euler-Poisson equation.
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