6

Analytic solutions of variational problems

This chapter presents a handful of analytic methods for solving variational

problems. They include the methods of Laplace transformation, separation

of variables, complete integrals, and Poisson’s integral formula. The method

of gradients, with high relevance to engineering optimization, concludes the

chapter.

6.1 Laplace transform solution

The ﬁrst method we discuss in this chapter transforms the original variational

problem by applying the Laplace transform and producing an auxiliary dif-

ferential equation.

Let us consider the variational problem of

I(t, x)=

f(t, x)dt = extremum,

and apply the Laplace transform to the function as

∞

0

e

−st

f(t, x)dt.

During this transform we regard time as the independent variable and x as a

parameter. Note that the transformation of the boundary conditions is also

required to obtain the complete auxiliary problem.

Let us illustrate this by the Euler-Lagrange diﬀerential equation of one spa-

tial and one temporal independent variable of the form

a

2

∂

2

u

∂x

2

−

∂u

∂t

=0,

with initial condition

u(t =0,x)=0,

69

70 Applied calculus of variations for engineers

and boundary conditions

u(t, x =0)=0

and

u(t, x =1)=B.

Executing the Laplace transform on the boundary conditions results in

u(s, 0) =

∞

0

e

−st

u(t, 0)dt =

∞

0

e

−st

0dt =

0

s

=0

and

u(s, 1) =

∞

0

e

−st

u(t, 1)dt =

∞

0

e

−st

Bdt =

B

s

.

Transforming the yet unknown solution as

u(s, x)=

∞

0

e

−st

u(t, x)dt,

we produce the auxiliary equation in the form of

a

2

d

2

u

dx

2

− su =0.

This equation is now ordinary. By integrating twice and applying Euler’s

formula the solution of the auxiliary equation becomes

u(s, x)=

B

s

sinh(x

s

a

)

sinh(

s

a

)

.

Finally, inverse Laplace transformation yields the solution of the original prob-

lem in the form of

u(t, x)=B(x +

2

π

Σ

∞

k=1

(−1)

k

k

e

−(kπa)

2

t

sin(kπx)).

This is the analytic solution of the one-dimensional heat conduction prob-

lem with constant temperature (B) at the boundary. The two- and three-

dimensional heat conduction problems will be the subject of further discus-

sion in the next section.

Let us now consider another Euler-Lagrange equation whose temporal deriva-

tive is also of second order

a

2

∂

2

u

∂x

2

−

∂

2

u

∂t

2

=0.

The initial conditions are

u(t =0,x)=0,

∂u

∂t

(t =0,x)=0.

Analytic solutions of variational problems 71

The boundary conditions are

u(t, x =0)=0,

∂u

∂x

(t, x =1)=B.

The boundary conditions are transformed again as

u(x =0)=0

and

d

u

dx

(x =1)=

B

s

,

where s is the Laplace variable. The auxiliary equation becomes an ordinary

diﬀerential equation of

d

2

u

dx

2

=

s

2

a

2

u,

Integrating this equation we obtain the result in the form

u(s, x)=

aB

s

2

sinh(

sx

a

)

cosh(

s

a

)

.

Finally the inverse transformation yields the solution of the original problem

at any point in the domain at any time as

u(t, x)=B(x −

8

π

2

Σ

∞

k=0

(−1)

k

2k +1

sin(

πx

2

(2k +1))cos(

πat

2

(2k + 1))).

This is the analytic solution to the problem of the compression of a unit length

beam along its longitudinal axis. The coeﬃcient a and boundary condition

B represent the physical characteristics of the beam and the problem, respec-

tively. They will be introduced in connection with the solution of a beam

bending under its weight, presented in Section 11.3.

6.2 Separation of variables

The method discussed here has a resemblance to the Laplace transform so-

lution since it also uses a transformation of the solution. We address the

problem of

∂u

∂t

= h

2

(

∂

2

u

∂x

2

+

∂

2

u

∂y

2

).

We impose uniformly zero boundary conditions as

u(x, 0,t)=u(x, b, t)=0;0≤ x ≤ a,

72 Applied calculus of variations for engineers

and

u(0,y,t)=u(a, y, t)=0;0≤ y ≤ b.

The initial solution is given as a non-zero function of the spatial coordinates

u(x, y, 0) = f(x, y).

The separation of variables method seeks a solution in the form of

u(x, y, t)=e

−λt

u

1

(x)u

2

(y),

where λ is a yet unknown constant. Substitution and diﬀerentiation yields

−λu

1

u

2

= h

2

(u

1

u

2

+ u

1

u

2

).

Conveniently reordering produces

u

1

u

1

+

λ

h

2

= −

u

2

u

2

= k

2

,

where k is a constant since the left-hand side is independent of y and the

right-hand side is independent of x. Introducing

q

2

=

λ

h

2

− k

2

,

we obtain a system of ordinary diﬀerential equations:

u

1

+ q

2

u

1

=0,

and

u

2

+ k

2

u

2

=0.

Their solutions are easily obtained as

u

1

(x)=a

1

sin(qx)+b

1

cos(qx),

and

u

2

(y)=a

2

sin(ky)+b

2

cos(ky).

The boundary conditions imply that b

1

= b

2

= 0 as well as

sin(qa)=0,

and

sin(kb)=0.

Here a, b are the original spatial boundaries. Due to the periodic nature of

the trigonometric functions

q =

mπ

a

,m=1, 2,..

Analytic solutions of variational problems 73

and

k =

nπ

b

,n=1, 2,..

Substituting produces the unknown variable as

λ

mn

= h

2

((

mπ

a

)

2

+(

nπ

b

)

2

),

and the solution function of

u(x, y, t)=Σ

∞

m=1

Σ

∞

n=1

c

mn

e

−λ

mn

t

sin

mπx

a

sin

nπy

b

.

The ﬁnal unknown coeﬃcient c

mn

is obtained by the satisfaction of the initial

condition:

f(x, y)=Σ

∞

m=1

Σ

∞

n=1

c

mn

sin

mπx

a

sin

nπy

b

,

from which the value of

c

mn

=

4

ab

b

0

a

0

f(x, y)sin

mπx

a

sin

nπy

b

dxdy

emerges. It is easy to generalize this solution to the three-dimensional prob-

lem of

∂u

∂t

= h

2

(

∂

2

u

∂x

2

+

∂

2

u

∂y

2

+

∂

2

u

∂z

2

).

We impose uniformly zero boundary conditions on three spatial dimensions as

u(x, y, 0,t)=u(x, y, c, t)=0;0≤ x ≤ a, 0 ≤ y ≤ b.

u(x, 0,z,t)=u(x, b, z, t)=0;0≤ x ≤ a, 0 ≤ z ≤ c.

and

u(0,y,z,t)=u(a, y, z, t)=0;0≤ y ≤ b, 0 ≤ z ≤ c.

The initial solution is given as a non-zero function of the three spatial coor-

dinates,

u(x, y, z, 0) = f(x, y, z).

The solution with

λ

mnr

= h

2

((

mπ

a

)

2

+(

nπ

b

)

2

)+(

rπ

c

)

2

),

becomes

u(x, y, z, t)=Σ

∞

m=1

Σ

∞

n=1

Σ

∞

r=1

c

mnr

e

−λ

mnr

t

sin

mπx

a

sin

nπy

b

sin

rπz

c

.

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