 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
x
a
sin
y
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 = , 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
=
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
x
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
x
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 .
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

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