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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