74 Applied calculus of variations for engineers

The coeﬃcient of the solution is also a straightforward generalization as

c

mnr

=

8

abc

c

0

b

0

a

0

f(x, y, z)sin

mπx

a

sin

nπy

b

sin

rπz

c

dxdydz.

These last two solutions were the analytic solutions to the two- and three-

dimensional heat conduction problems. The computational solution of the

two-dimensional problem will be further addressed in Chapter 12.

Let us now solve a problem of two spatial variables again, but with a tem-

poral variable whose second derivative is present:

h

2

(

∂

2

u

∂x

2

+

∂

2

u

∂y

2

)=

∂

2

u

∂t

2

.

We assume constant boundary conditions:

u(0,y,t)=u(a, y, t)=0,

and

u(x, 0,t)=u(x, b, t)=0.

The a, b are the dimensions of the domain. We seek the solution in the sepa-

rated form of

u(x, y, t)=e

iλt

v(x, y).

Note the presence of the imaginary unit i in the exponent for later conve-

nience. We introduce the constant

k

2

=

λ

2

h

2

.

Substitution yields the new diﬀerential equation

∂

2

v

∂x

2

+

∂

2

v

∂y

2

+ k

2

v =0.

The new boundary conditions are

v(0,y)=v(a, y)=0,

and

v(x, 0) = v(x, b)=0.

Furthermore, we separate the variables of this equation as

v(x, y)=u

1

(x)u

2

(y).

This leads to the system of equations

1

u

1

d

2

u

1

dx

2

= −

1

u

2

d

2

u

2

dy

2

− k

2

= −m

2

.

Analytic solutions of variational problems 75

Here m is another yet unknown constant. The now familiar system of ordi-

nary diﬀerential equations arises again

d

2

u

1

dx

2

+ m

2

u

1

=0,

and

d

2

u

2

dy

2

+ q

2

u

2

=0,

with q

2

= k

2

− m

2

. The new boundary conditions are

u

1

(0) = u

1

(a)=0,

and

u

2

(0) = u

2

(b)=0.

Following the road paved earlier in this section, the ﬁrst equation yields

u

1

(x)=A

1

sin(mx),

with ma = nπ, n =1, 2, 3, ... and the second equation

u

2

(y)=A

2

sin(qy),

with qb = rπ, r =1, 2, 3, .... Exploiting the relation

k

2

= m

2

+ q

2

= π

2

(

n

2

a

2

+

r

2

b

2

)

we obtain the original parameter of the transformation

λ

nr

= hπ

n

2

a

2

+

r

2

b

2

.

Finally by substituting and using Euler’s formula we obtain

u(x, y, t)=Σ

∞

n=1

Σ

∞

r=1

c

nr

cos(λ

nr

t)sin

nπx

a

sin

rπy

b

.

The initial displacement represented by f(x, y) aids in ﬁnding the ﬁnal coef-

ﬁcient as

c

nr

=

4

ab

b

0

a

0

f(x, y)sin

nπx

a

sin

rπy

b

dxdy.

This is the analytic solution of the problem of the vibrating membrane. A

more general solution of this problem with variable boundary conditions is

presented in Chapter 11.

Finally, let us consider an Euler-Lagrange equation of the ﬁrst order with

many independent variables in the implicit form of

F (x

1

,x

2

, ...x

n

,u,

∂u

∂x

1

,

∂u

∂x

2

, ...

∂u

∂x

n

)=0,

76 Applied calculus of variations for engineers

whose generic solution is

u(x

1

,x

2

, ...x

n

; a

1

,a

2

, ..., a

n

)=0.

The solution to such problems may be found by a repeated use of the separa-

tion of variables and the constant k. Let us ﬁrst separate one variable as

u(x

1

,x

2

, ...x

n

)=u

1

(x

1

)+u

2

(x

2

,x

3

, ...x

n

).

This corresponds to the following diﬀerential equation

F

1

(x

1

,u

1

,

∂u

1

∂x

1

)=F

2

(x

2

,x

3

...x

n

,u

2

,

∂u

2

∂x

2

,

∂u

2

∂x

3

, ...

∂u

2

∂x

n

)=k

2

.

The equation may be satisﬁed by solving a pair of equations with an unknown

constant:

F

1

(x

1

,u

1

,

∂u

1

∂x

1

)=k

2

,

and

F

2

(x

2

,x

3

...x

n

,u

2

,

∂u

2

∂x

2

,

∂u

2

∂x

3

, ...

∂u

2

∂x

n

)=k

2

.

The ﬁrst equation becomes an ordinary diﬀerential equation whose solution

is easily obtained. The second equation may again be further separated and

the same process continued.

6.3 Complete integral solutions

For certain types of problems a complete integral solution is available [16].

The complete integral form presents a parametric family of general solutions.

The particular solution of a speciﬁc problem can then be obtained from the

general complete integral solution by selection of the parameters.

We will ﬁrst demonstrate generating a complete integral solution by ex-

ploiting the concept of separation of variables introduced in the last section.

For the simplicity of the discussion, and without loss of generality, we will do

this with an example of only two independent variables.

F (x, y, u,

∂u

∂x

,

∂u

∂y

)=0.

We seek the complete integral solution as

u(x, y, u, a, b)=0.

Analytic solutions of variational problems 77

where the a, b are yet unknown coeﬃcients. Let us generate the complete

integral solution for the equation of

(

∂u

∂x

)

2

+(

∂u

∂y

)

2

=1.

The separated solution is of the form

u(x, y)=u

1

(x)+u

2

(y).

The ﬁrst diﬀerential equation with a constant k is then

F

1

(x, u

1

,

du

1

dx

)=(

du

1

dx

)

2

= k

2

.

The solution comes by

du

1

= kdx,

from which

u

1

= kx + k

1

emerges. Similarly the second, in this case also ordinary equation is

F

2

(y, u

2

,

du

2

dy

)=1−(

du

2

dy

)

2

= k

2

.

The solution of

du

2

=

1 − k

2

dy

yields

u

2

=

1 − k

2

y + k

2

.

Finally the complete integral solution for this problem is

u(x, y, a, b)=ax +

1 − a

2

y + b,

with a = k and b = k

1

+ k

2

. The complete integral solution satisﬁes the

original problem

(

∂u

∂x

)

2

+(

∂u

∂y

)

2

=1,

since

∂u

∂x

= a,

∂u

∂y

=

1 − a

2

,

and

a

2

+1− a

2

=1.

This is a two-parameter family of solutions from which any particular solu-

tion may be obtained. Surely any selection of the parameter b will satisfy

the original equation. As far as the parameter a is concerned, selecting for

Get *Applied Calculus of Variations for Engineers, 2nd Edition* now with the O’Reilly learning platform.

O’Reilly members experience live online training, plus books, videos, and digital content from nearly 200 publishers.