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 =
a
,m=1, 2,..
Analytic solutions of variational problems 73
and
k =
b
,n=1, 2,..
Substituting produces the unknown variable as
λ
mn
= h
2
((
a
)
2
+(
b
)
2
),
and the solution function of
u(x, y, t)=Σ
m=1
Σ
n=1
c
mn
e
λ
mn
t
sin
x
a
sin
y
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
x
a
sin
y
b
,
from which the value of
c
mn
=
4
ab
b
0
a
0
f(x, y)sin
x
a
sin
y
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
((
a
)
2
+(
b
)
2
)+(
c
)
2
),
becomes
u(x, y, z, t)=Σ
m=1
Σ
n=1
Σ
r=1
c
mnr
e
λ
mnr
t
sin
x
a
sin
y
b
sin
rπz
c
.

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