4
Higher order derivatives
The fundamental problem of the calculus of variations involved the first deriva-
tive of the unknown function. In this chapter we will allow the presence of
higher order derivatives.
4.1 The Euler-Poisson equation
First let us consider the variational problem of a functional with a single func-
tion, but containing its higher derivatives:
I(y)=
x
1
x
0
f(x, y, y
,...,y
(m)
)dx.
Accordingly, boundary conditions for all derivatives will also be given as
y(x
0
)=y
0
,y(x
1
)=y
1
,
y
(x
0
)=y
0
,y
(x
1
)=y
1
,
y
(x
0
)=y
0
,y
(x
1
)=y
1
,
and so on until
y
(m−1)
(x
0
)=y
(m−1)
0
,y
(m−1)
(x
1
)=y
(m−1)
1
.
As in the past chapters, we introduce an alternative solution of
Y (x)=y(x)+η(x),
where the arbitrary auxiliary function η(x) is continuously differentiable on
the interval x
0
≤ x ≤ x
1
and satisfies
η(x
0
)=0,η(x
1
)=0.
The variational problem in terms of the alternative solution is
I()=
x
1
x
0
f(x, Y, Y
,...,Y
(m)
)dx.
49
50 Applied calculus of variations for engineers
The differentiation with respect to follows
dI
d
=
x
1
x
0
d
d
f(x, Y, Y
,...,Y
(m)
dx,
and by using the chain rule the integrand is reshaped as
∂f
∂Y
dY
d
+
∂f
∂Y
dY
d
+
∂f
∂Y
dY
d
+ ...+
∂f
∂Y
(m)
dY
(m)
d
.
Substituting the alternative solution and its derivatives with respect to the
integrand yields
∂f
∂Y
η +
∂f
∂Y
η
+
∂f
∂Y
η
+ ...+
∂f
∂Y
(m)
η
(m)
.
Hence the functional becomes
dI
d
=
x
1
x
0
(
∂f
∂Y
η +
∂f
∂Y
η
+
∂f
∂Y
η
+ ...+
∂f
∂Y
(m)
η
(m)
)dx.
Integrating by terms results in
dI
d
=
x
1
x
0
∂f
∂Y
ηdx+
x
1
x
0
∂f
∂Y
η
dx+
x
1
x
0
∂f
∂Y
η
dx+ ...+
x
1
x
0
∂f
∂Y
(m)
η
(m)
dx,
and integrating by parts produces
dI
d
=
x
1
x
0
η
∂f
∂Y
dx −
x
1
x
0
η
d
dx
∂f
∂Y
dx +
x
1
x
0
η
d
2
dx
2
∂f
∂Y
dx−
...(−1)
m
x
1
x
0
η
d
(m)
dx
(m)
∂f
∂Y
(m)
dx.
Factoring the auxiliary function and combining the terms again simplifies to
dI
d
=
x
1
x
0
η(
∂f
∂Y
−
d
dx
∂f
∂Y
+
d
2
dx
2
∂f
∂Y
− ...(−1)
m
d
(m)
dx
(m)
∂f
∂Y
(m)
)dx.
Finally the extremum at = 0 and the fundamental lemma produces the
Euler-Poisson equation
∂f
∂y
−
d
dx
∂f
∂y
+
d
2
dx
2
∂f
∂y
− ...(−1)
m
d
(m)
dx
(m)
∂f
∂y
(m)
=0.
The Euler-Poisson equation is an ordinary differential equation of order 2m
and requires the aforementioned 2m boundary conditions, where m is the
highest order derivative contained in the functional.
For example, the simple m = 2 functional
I(y)=
x
1
x
0
(y
2
− (y
)
2
)dx
Higher order derivatives 51
results in the derivatives
∂f
∂y
= −2y
,
and
∂f
∂y
=2y.
The corresponding Euler-Poisson equation derivative term is
d
2
dx
2
∂f
∂y
=
d
2
dx
2
(−2y
)=−2
d
4
dx
4
y,
and the equation, after cancellation by −2, becomes
d
4
dx
4
y − y =0.
Clearly the solution of this may be achieved by classical calculus tools with
four boundary conditions. Application problems exploiting this will be ad-
dressed in Chapters 9 (the natural spline) and 11 (the bending beam).
4.2 The Euler-Poisson system of equations
In the case of a functional with multiple functions along with their higher
order derivatives, the problem gets more difficult. Assuming p functions in
the functional, the problem is posed in the form of
I(y
1
,...,y
p
)=
x
1
x
0
f(x, y
1
,y
1
,...,y
(m
1
)
1
,...,y
p
,y
p
,...,y
(m
p
)
p
)dx.
Note that the highest order of the derivative of the various functions is not
necessarily the same. This is a rather straightforward generalization of the
case of the last section, leading to a system of Euler-Poisson equations as
follows:
∂f
∂y
1
−
d
dx
∂f
∂y
1
+
d
2
dx
2
∂f
∂y
1
− ...(−1)
m
1
d
(m
1
)
dx
(m
1
)
∂f
∂y
(m
1
)
1
=0,
...,
∂f
∂y
p
−
d
dx
∂f
∂y
p
+
d
2
dx
2
∂f
∂y
p
− ...(−1)
m
p
d
(m
p
)
dx
(m
p
)
∂f
∂y
(m
p
)
p
=0.
This is a set of p ordinary differential equations that may or may not be cou-
pled, hence resulting in a varying level of ease of the solution.
52 Applied calculus of variations for engineers
4.3 Algebraic constraints on the derivative
It is also common in engineering applications to impose boundary conditions
on some of the derivatives (Neumann boundary conditions). These result in
algebraic constraints posed on the derivative, such as
I(y)=
x
1
x
0
f(x, y, y
)dx
subject to
g(x, y, y
)=0.
In order to be able to solve such problems, we need to introduce a Lagrange
multiplier as a function of the independent variable as
h(x, y, y
,λ)=f(x, y, y
)+λ(x)g(x, y, y
).
The use of this approach means that the functional now contains two unknown
functions and the variational problem becomes
I(y,λ)=
x
1
x
0
h(x, y, y
,λ)dx,
with the original boundary conditions, but without a constraint. The solu-
tion for this unconstrained, two function case is obtained by a system of two
Euler-Lagrange equations.
Derivative constraints may also be applied to the case of higher order deriva-
tives. The second order problem of
I(y)=
x
1
x
0
f(x, y, y
,y
)dx
may be subject to a constraint
g(x, y, y
,y
)=0.
In order to be able to solve such problems, we also introduce a Lagrange mul-
tiplier function as
h(x, y, y
,y
)=f(x, y, y
,y
)+λ(x)g(x, y, y
,y
).
The result is a variational problem of two functions with higher order deriva-
tives as
I(y,λ)=
x
1
x
0
h(x, y, y
,y
,λ)dx.
Higher order derivatives 53
Hence the solution may be obtained by the application of a system of two
Euler-Poisson equations.
Finally, derivative constraints may also be applied to a variational problem
originally exhibiting multiple functions, such as
I(y,z)=
x
1
x
0
f(x, y, y
,z,z
)dx
subject to
g(x, y, y
,z,z
)=0.
Here the new functional is
h(x, y, y
,z,z
,λ)=f(x, y, y
,z,z
)+λ(x)g(x, y, y
,z,z
).
Following above, this problem translates into the unconstrained form of
I(y,z, λ)=
x
1
x
0
f(x, y, y
,z,z
,λ)dx
that may be solved by a system of three Euler-Lagrange differential equations
∂h
∂y
−
d
dx
∂h
∂y
=0,
∂h
∂z
−
d
dx
∂h
∂z
=0,
and
∂h
∂λ
−
d
dx
∂h
∂λ
=0.
For example, the variational problem of
I(y,z)=
x
1
x
0
(y
2
− z
2
)dx = extremum,
under the derivative constraint of
y
− y + z =0,
results in
h(x, y, y
,z,z
,λ)=y
2
− z
2
+ λ(x)(y
− y + z).
The solution is obtained from the following three equations
2y − λ + λ
=0,
−2z + λ =0,
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