4Multivariate calculus of variations

IB Variational Principles



4 Multivariate calculus of variations
So far, the function
x
(
t
) we are varying is just a function of a single variable
t
.
What if we have a more complicated function to consider?
We will consider the most general case
y
(
x
1
, ··· , x
m
)
R
n
that maps
R
m
R
n
(we can also view this as
n
different functions that map
R
m
R
).
The functional will be a multiple integral of the form
F [y] =
Z
···
Z
f(y, y, x
1
, ··· , x
m
) dx
1
···dx
m
,
where y is the second-rank tensor defined as
y =
y
x
1
, ··· ,
y
x
m
.
In this case, instead of attempting to come up with some complicated generalized
Euler-Lagrange equation, it is often a better idea to directly consider variations
δy of y. This is best illustrated by example.
Example
(Minimal surfaces in
E
3
)
.
This is a natural generalization of geodesics.
A minimal surface is a surface of least area subject to some boundary conditions.
Suppose that (
x, y
) are good coordinates for a surface
S
, where (
x, y
) takes
values in the domain
D R
2
. Then the surface is defined by
z
=
h
(
x, y
), where
h is the height function.
When possible, we will denote partial differentiation by suffices, i.e.
h
x
=
h
x
.
Then the area is given by
A[h] =
Z
D
q
1 + h
2
x
+ h
2
y
dA.
Consider a variation of h(x, y): h 7→ h + δh(x, y). Then
A[h + δh] =
Z
D
q
1 + (h
x
+ (δh)
x
)
2
+ (h
y
+ (δh)
y
)
2
dA
= A[h] +
Z
D
h
x
(δh)
x
+ h
y
(δh)
y
q
1 + h
2
x
+ h
2
y
+ O(δh
2
)
dA
We integrate by parts to obtain
δA =
Z
D
δh
x
h
x
q
1 + h
2
x
+ h
2
y
+
y
h
y
q
1 + h
2
x
+ h
2
y
dA + O(δh
2
)
plus some boundary terms. So our minimal surface will satisfy
x
h
x
q
1 + h
2
x
+ h
2
y
+
y
h
y
q
1 + h
2
x
+ h
2
y
= 0
Simplifying, we have
(1 + h
2
y
)h
xx
+ (1 + h
2
x
)h
yy
2h
x
h
y
h
xy
= 0.
This is a non-linear 2nd-order PDE, the minimal-surface equation. While it is
difficult to come up with a fully general solution to this PDE, we can consider
some special cases.
There is an obvious solution
h(x, y) = Ax + By + C,
since the equation involves second-derivatives and this function is linear.
This represents a plane.
If |∇h|
2
1, then h
2
x
and h
2
y
are small. So we have
h
yy
+ h
yy
= 0,
or
2
h = 0.
So we end up with the Laplace equation. Hence harmonic functions are
(approximately) minimal-area.
We might want a cylindrically-symmetric solution, i.e.
h
(
x, y
) =
z
(
r
), where
r =
p
x
2
+ y
2
. Then we are left with an ordinary differential equation
rz
00
+ z
0
+ z
03
= 0.
The general solution is
z = A
1
cosh(Ar) + B,
a catenoid.
Alternatively, to obtain this,this we can substitute
h
(
x, y
) =
z
(
r
) into
A
[
h
]
to get
A[z] = 2π
Z
r
p
1 + (h
0
(r))
2
dr,
and we can apply the Euler-Lagrange equation.
Example
(Small amplitude oscillations of uniform string)
.
Suppose we have a
string with uniform constant mass density ρ with uniform tension T .
y
Suppose we pull the line between
x
= 0 and
x
=
a
with some tension
T
. Then we
set it into motion such that the amplitude is given by y(x; t). Then the kinetic
energy is
T =
1
2
Z
a
0
ρv
2
dx =
ρ
2
Z
a
0
˙y
2
dx.
The potential energy is the tension times the length. So
V = T
Z
d` = T
Z
a
0
p
1 + (y
0
)
2
dx = (Ta) +
Z
a
0
1
2
T (y
02
) dx.
Note that y
0
is the derivative wrt x while ˙y is the derivative wrt time.
The
T a
term can be seen as the ground-state energy. It is the energy initially
stored if there is no oscillation. Since this constant term doesn’t affect where
the stationary points lie, we will ignore it. Then the action is given by
S[y] =
ZZ
a
0
1
2
ρ ˙y
2
1
2
T (y
0
)
2
dx dt
We apply Hamilton’s principle which says that we need
δS[y] = 0.
We have
δS[y] =
ZZ
a
0
ρ ˙y
t
δy T y
0
x
δy
dx dt.
Integrate by parts to obtain
δS[y] =
ZZ
a
0
δy(ρ¨y T y
00
) dx dt + boundary term.
Assuming that the boundary term vanishes, we will need
¨y v
2
y
00
= 0,
where
v
2
=
T
. This is the wave equation in two dimensions. Note that this is a
linear PDE, which is a simplification resulting from our assuming the oscillation
is small.
The general solution to the wave equation is
y(x, t) = f
+
(x vt) + f
(x + vt),
which is a superposition of a wave travelling rightwards and a wave travelling
leftwards.
Example
(Maxwell’s equations)
.
It is possible to obtain Maxwell’s equations
from an action principle, where we define a Lagrangian for the electromagnetic
field. Note that this is the Lagrangian for the field itself, and there is a separate
Lagrangian for particles moving in a field.
We have to first define several quantities. First we have the charges:
ρ
represents the electric charge density and
J
represents the electric current
density.
Then we have the potentials:
φ
is the electric scalar potential and
A
is the
magnetic vector potential.
Finally the fields:
E
=
−∇φ
˙
A
is the electric field, and
B
=
× A
is the
magnetic field.
We pick convenient units where
c
=
ε
0
=
µ
0
= 1. With these concepts in
mind, the Lagrangian is given by
S[A, φ] =
Z
1
2
(|E|
2
|B|
2
) + A · J φρ
dV dt
Varying A and φ by δA and δφ respectively, we have
δS =
Z
E ·
δφ +
t
δA
B · × δA + δA · J ρδφ
dV dt.
Integrate by parts to obtain
δS =
Z
δA · (
˙
E × B + J) + δφ( · E ρ)
dV dt.
Since the coefficients have to be 0, we must have
× B = J +
˙
E, · E = ρ.
Also, the definitions of E and B immediately give
· B = 0, ×E =
˙
B.
These four equations are Maxwell’s equations.