3
Multivariate functionals
3.1 Functionals with several functions
The variational problem of multiple dependent variables is posed as
I(y
1
,y
2
,...,y
n
)=
x
1
x
0
f(x, y
1
,y
2
,...,y
n
,y
1
,y
2
,...,y
n
)dx
with a pair of boundary conditions given for all functions:
y
i
(x
0
)=y
i,0
and
y
i
(x
1
)=y
i,1
for each i =1, 2,...,n. The alternative solutions are:
Y
i
(x)=y
i
(x)+
i
η
i
(x); i =1,...,n
with all the arbitrary auxiliary functions obeying the conditions:
η
i
(x
0
)=η
i
(x
1
)=0.
The variational problem becomes
I(
1
,...,
n
)=
x
1
x
0
f(x,...,y
i
+
i
η
i
,...,y
i
+
i
η
i
,...)dx,
whose derivative with respect to the auxiliary variables is
∂I
∂
i
=
x
1
x
0
∂f
∂
i
dx =0.
Applying the chain rule we get
∂f
∂
i
=
∂f
∂Y
i
∂Y
i
∂
i
+
∂f
∂Y
i
∂Y
i
∂
i
=
∂f
∂Y
i
η
i
+
∂f
∂Y
i
η
i
.
Substituting into the variational equation yields, for i =1, 2,...,n:
I(
i
)=
x
1
x
0
(
∂f
∂Y
i
η
i
+
∂f
∂Y
i
η
i
)dx.
37
38 Applied calculus of variations for engineers
Integrating by parts and exploiting the alternative function form results in
I(
i
)=
x
1
x
0
η
i
(
∂f
∂y
i
−
d
dx
∂f
∂y
i
)dx.
To reach the extremum, based on the fundamental lemma, we need the solu-
tion of a set of n Euler-Lagrange equations of the form
∂f
∂y
i
−
d
dx
∂f
∂y
i
=0;i =1,...,n.
3.2 Variational problems in parametric form
Most of the discussion heretofore was focused on functions in explicit form.
The concepts also apply to problems posed in parametric form. The explicit
form variational problem of
I(y)=
x
1
x
0
f(x, y, y
)dx
may be reformulated with the substitutions
x = u(t),y= v(t).
The parametric variational problem becomes of the form
I(x, y)=
t
1
t
0
f(x, y,
˙y
˙x
)˙xdt,
or
I(x, y)=
t
1
t
0
F (t, x, y, ˙x, ˙y)dt.
The Euler-Lagrange differential equation system for this case becomes
∂F
∂x
−
d
dt
∂F
∂ ˙x
=0,
and
∂F
∂y
−
d
dt
∂F
∂ ˙y
=0.
It is proven in [9] that an explicit variational problem is invariant under pa-
rameterization. In other words, independently of the algebraic form of the
parameterization, the same explicit solution will be obtained.
Parametrically given problems may be considered as functionals with sev-
eral functions. As an example, we consider the following twice differentiable
Multivariate functionals 39
functions
x = x(t),y = y(t),z = z(t).
The variational problem in this case is presented as
I(x, y, z)=
t
1
t
0
f(t, x, y, z, ˙x, ˙y, ˙z)dx.
Here the independent variable t is the parameter, and there are three depen-
dent variables : x, y, z. Applying the steps just explained for this specific case
results in the system of Euler-Lagrange equations
∂f
∂x
−
d
dt
∂f
∂ ˙x
=0,
∂f
∂y
−
d
dt
∂f
∂ ˙y
=0,
and
∂f
∂z
−
d
dt
∂f
∂ ˙z
=0.
The most practical applications of this case are variational problems in
three-dimensional space, presented in parametric form. This is usual in many
geometry problems and will be exploited in Chapters 8 and 9.
3.3 Functionals with two independent variables
All our discussions so far were confined to a single integral of the functional.
The next step of generalization is to allow a functional with multiple indepen-
dent variables. The simplest case is that of two independent variables, and
this will be the vehicle to introduce the process. The problem is of the form
I(z)=
y
1
y
0
x
1
x
0
f(x, y, z, z
x
,z
y
)dxdy = extremum.
Here the derivatives are
z
x
=
∂z
∂x
and
z
y
=
∂z
∂y
.
The alternative solution is also a function of two variables
Z(x, y)=z(x, y)+η(x, y).
40 Applied calculus of variations for engineers
The now familiar process emerges as
I()=
y
1
y
0
x
1
x
0
f(x, y, Z, Z
x
,Z
y
)dxdy = extremum.
The extremum is obtained via the derivative
∂I
∂
=
y
1
y
0
x
1
x
0
∂f
∂
dxdy.
Differentiating and substituting yields
∂I
∂
=
y
1
y
0
x
1
x
0
(
∂f
∂Z
η +
∂f
∂Z
x
η
x
+
∂f
∂Z
y
η
y
)dxdy.
The extremum is reached when =0:
∂I
∂
|
=0
=
y
1
y
0
x
1
x
0
(
∂f
∂z
η +
∂f
∂z
x
η
x
+
∂f
∂z
y
η
y
)dxdy =0.
Applying Green’s identity for the second and third terms produces
y
1
y
0
x
1
x
0
(
∂f
∂z
−
∂
∂x
∂f
∂z
x
−
∂
∂y
∂f
∂z
y
)ηdxdy +
∂D
(
∂f
∂z
x
dy
ds
−
∂f
∂z
y
dx
ds
)ηds =0.
Here ∂D is the boundary of the domain of the problem and the second integral
vanishes by the definition of the auxiliary function. Due to the fundamental
lemma of calculus of variations, the Euler-Lagrange differential equation be-
comes
∂f
∂z
−
∂
∂x
∂f
∂z
x
−
∂
∂y
∂f
∂z
y
=0.
3.4 Application: minimal surfaces
Minimal surfaces occur in intriguing applications. For example, soap films
spanned over various types of wire loops intrinsically attain such shapes, no
matter how difficult the boundary curve is. Various biological cell interactions
also manifest similar phenomena.
From a differential geometry point of view a minimal surface is a surface
for which the mean curvature of the form
κ
m
=
κ
1
+ κ
2
2
vanishes, where κ
1
and κ
2
are the principal curvatures. A subset of minimal
surfaces are the surfaces of minimal area, and surfaces of minimal area passing
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