Variational Problems ∗ Yan-Bin Jia Nov 13, 2014

Variational Problems∗
(Com S 477/577 Notes)
Yan-Bin Jia
Nov 13, 2014
d
Fy′ (x, y, y ′ )
The variational derivative of a functional J[y] can be defined as δJ/δy = Fy (x, y, y ′ )− dx
[1, pp. 27–29]. Euler’s equation essentially states that the variational derivative of the functional
must vanish at an extremum. This is analogous to the well-known result from calculus that the
derivative of a function must vanish at an extremum.
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Variable End Point Problem
In this section, we consider a simple case of the variable end point problem, which is stated as
follows: Among all curves whose end points lie on two vertical lines x = a and x = b, find the curve
for which the functional
Z
b
F (x, y, y ′ ) dx
J[y] =
(1)
a
has an extremum.
We determine the variation of the functional (1), which is the linear component of the increment
Z b
F (x, y + h, y ′ + h′ ) − F (x, y, y ′ ) dx,
∆J = J[y + h] − J[y] =
a
due to an increment of h in y. The Taylor expansion immediately leads to
Z b
(Fy h + Fy′ h′ ) dx.
δJ =
a
Unlike the fixed end point problem, the function h(x) no longer vanishes at the points x = a and
x = b. Integration by parts now yields
Z b
d
δJ =
Fy −
Fy′ dx + Fy′ h(x)|x=b
x=a
dx
a
Z b
d
Fy′ dx + Fy′ |x=b h(b) − Fy′ |x=a h(a).
(2)
=
Fy −
dx
a
Consider all functions h(x) with h(a) = h(b) = 0 first. The condition δJ = 0 implies that
Fy −
d
Fy′ = 0.
dx
∗
The material is adapted from the book Calculus of Variations by I. M. Gelfand and S. V. Fomin, Prentice Hall
Inc., 1963; Dover, 2000.
1
This means that the solution y of the end point problem must be a solution of Euler’s equation.
Suppose y is a solution. The integral in (2) must therefore vanish, reducing δJ = 0 to
Fy′ |x=b h(b) − Fy′ |x=a h(a) = 0.
Since h is arbitrary, it follows from the above equation that
Fy′ |x=a = 0
and
Fy′ |x=b = 0.
(3)
In summary, to solve the variable end point problem, we first find a general solution to Euler’s
equation (2), and then use the conditions (3) to determine the constants in the general solution.
Example 1. Starting at the origin, a particle slides down a curve in the vertical plane. Find the curve such
that the particle reaches the vertical line
√ x = b, where b 6= 0, in the shortest time.
The velocity v of motion equals 2gy from the conservation of the total energy, where g < 0 is the
gravitational acceleration. Meanwhile, it is also determined along the curve as
v=
from which we immediately have
dx
ds p
= 1 + y ′2
,
dt
dt
p
√
1 + y ′2
1 + y′2
dt =
dx = √
dx.
v
2gy
The above gives us the transit time as a functional
J=
Z
b
0
p
1 + y ′2
√
dx.
2gy
p
√
The integrand F = 1 + y ′2 / 2gy does not depend on x. So this is Case 2 where Euler’s equation can be
reduced to
(4)
F − y ′ Fy′ = C.
Since
1
y′
,
Fy′ = √
·p
2gy
1 + y ′2
several steps from Euler’s equation (4) lead to
y ′2 =
a−y
,
y
for some a < 0.
Given the downward sliding motion, we have that y ′ ≤ 0, thus
r
a−y
′
y =−
,
y
which leads to
dx = −
Next, make the substitution y = a sin2
θ
2
r
y
dy.
a−y
over [0, π] with
θ
θ 1
θ
θ
dy
= 2a sin cos · = a sin cos .
dθ
2
2 2
2
2
2
Under the above and the substitution,
dx
= −
s
θ
2
a cos2 θ2
a sin2
· a sin
θ
θ
cos dθ
2
2
θ
= −a sin2 dθ
2
1 − cos θ
= −a
dθ
2
= r(1 − cos θ) dθ,
where a constant r = −a/2 is introduced in the last step. The solution to Euler’s equation (4) is a family of
cycloids
x = r(θ − sin θ) + c,
y = −r(1 − cos θ).
Since the curve passes through the origin, the constant c = 0. The first boundary condition Fy′ |x=0 = 0
in (3), which is essentially y ′ (0) = 0, is thus satisfied. To determine the value of r, we apply the second
boundary condition Fy′ |x=b = 0, which reduces to y ′ = 0 at x = b, i.e.,
0=
dy/dθ
r sin θ
sin θ
dy
=
=
=
.
dx
dx/dθ
r − r cos θ
1 − cos θ
The above condition states that the tangent to the curve at its right end point must be horizontal. Hence
θ = π. Substituting this value into r(θ − sin θ) = b, we obtain r = πb . Hence the curve trajectory is
(x, y) =
b
(θ − sin θ, cos θ − 1).
π
Figure 1 shows the optimal trajectory of the sliding particle to reach the line x = 2π, represented by the
cycloid 2(θ − sin θ, cos θ − 1) over [0, π].
2π
Figure 1: The cycloid 2(θ − sin θ, cos θ − 1) with 0 ≤ θ ≤ π is the minimum time trajectory along which a
particle starting at the origin slides pass the vertical line x = 2π under gravity.
2
The Case of Several Variables
Many problems involve functionals that depend on functions of several independent variables, for
example, surfaces in 3D depending on two parameters. Here we only look at how the solution to the
3
case of single-variable variational problems would carry over to the case of functionals depending
on surfaces. We focus on the case of two independent variables but refer to [1] for the case of more
than two variables.
Let F (x, y, z, p, q) be twice continuously differentiable with respect to all five variables, and
consider
ZZ
F (x, y, z, zx , zy ) dx dy,
(5)
J[z] =
R
where R is some closed region, and zx and zy are the partial derivatives of z(x, y) with respect to
x and y. We are looking for a function z(x, y) that
1. twice continuously differentiable with respect to x and y in R;
2. assumes given values on the boundary Γ of R;
3. yields an extremum of the functional (5).
Theorem 1 in the notes Calculus of Variations does not depend on the form of the functional J.
Therefore, a necessary condition for the functional (5) to have an extremum is that its variation
vanishes. We must first calculate the variation δJ. Let h(x, y) be an arbitrary function with
continuous first and second derivatives in the region R and vanishes on it boundary Γ. Then, if
the surface z(x, y) satisfies conditions 1–3, so does z(x, y) + h(x, y). We apply the Taylor series to
the change in the functional as follows:
∆J
= J[z + h] − J[z]
ZZ F (x, y, z + h, zx + hx , zy + hy ) − F (x, y, z, zx , zy ) dx dy
=
Z ZR
(Fz h + Fzx hx + Fzy hy ) dx dy + · · · .
=
(6)
R
Meanwhile, we have that
ZZ
(Fzx hx + Fzy hy ) dx dy
R
ZZ ZZ ∂
∂
∂
∂
(Fzx h) +
(Fzy h) dx dy −
Fzx +
Fz
h dx dy
=
∂y
∂y y
R ∂x
R ∂x
ZZ
Z
∂
∂
(Fzx h dy − Fzy h dx) −
=
Fzx +
Fz
h dx dy.
∂y y
R ∂x
Γ
(7)
where the last step used Green’s theorem:
Z
ZZ ∂Q ∂P
−
dx dy = (P dx + Q dy).
∂x
∂y
Γ
In (7), the integral along Γ is zero since h vanishes on the boundary. Substituting (7) into (6), we
obtain
ZZ ∂
∂
δJ =
Fz −
Fzx −
Fzy h(x, y) dx dy.
(8)
∂x
∂y
R
4
Thus, the condition δJ = 0 implies that the double integral above vanishes for any h(x, y) with
continuous derivatives up to the second order and vanishing on the boundary Γ. This leads to the
second-order partial differential equation below 1 , also known as Euler’s equation:
Fz −
∂
∂
Fzx −
Fz = 0.
∂x
∂y y
Next, we generalize the functional (5) to allow second order partial derivatives of z to appear in
the integrand function F . More specifically, let F (x, y, z, p, q, r, s, t) be a function with continuous
first and second partial derivatives with respective to all arguments. The function z = z(x, y) has
continuous derivatives up to the fourth order, and has given values on the boundary Γ of R. The
functional
ZZ
F (x, y, z, zx , zy , zxx , zxy , zyy ) dx dy
J˜[z] =
R
has an extremum at z = z(x, y) only if the following version of Euler’s equation holds:
Fz −
∂
∂2
∂2
∂2
∂
Fzx −
Fzy + 2 Fzxx +
Fzxy + 2 Fzyy = 0.
∂x
∂y
∂x
∂x∂y
∂y
(9)
Example 2. Surface z = z(x, y) of least area spanned by a given contour. The problem reduces to finding
the minimum of the functional
ZZ q
J[z] =
1 + zx2 + zy2 dx dy,
R
for which Euler’s equation has the form
zxx (1 + zy2 ) − 2zxy zx zy + zyy (1 + zx2 ) = 0.
(10)
To understand the geometric meaning of equation (10), we note that the mean curvature of the surface
is given as
EN − 2F M + GL
,
H=
2(EG − F 2 )
where E, F, G are the coefficients of the first fundamental form of the surface, and L, M, N are the coefficients
of its second fundamental form. We obtain that
E = 1 + zx2 ,
zxx
,
2 +z 2
1+zx
y
L= √
F = zx zy ,
zxy
,
2 +z 2
1+zx
y
M= √
G = 1 + zy2 ,
zyy
,
2 +z 2
1+zx
y
N= √
and hence,
H=
zxx (1 + zy2 ) − 2zxy zx zy + zyy (1 + zx2 )
.
2(1 + zx2 + zy2 )3/2
Thus, Euler’s equation (10) implies that the mean curvature of the solution surface is zero everywhere.
A surface with zero mean curvature everywhere is thus called a minimal surface. Figure 2 plots
a minimal surface — a catenoid (2 cos u cosh(v/2), 2 sin u cosh(v/2), v) for 0 ≤ u ≤ 2π, −4 ≤ v ≤ 4.
1
To be rigorous, a simple proof can be derived from (8) and that h is arbitrary under stipulated conditions.
5
Figure 2: Catenoid (2 cos u cosh(v/2), 2 sin u cosh(v/2), v) over [0, 2π] × [−4, 4].
3
Several Unknown Functions
Let F (x, y1 , . . . , yn , z1 , . . . , zn ) be a function with continuous first and second derivatives with respect to all its arguments. We now derive necessary conditions for an extremum of the functional
J[y1 , . . . , yn ] =
Z
b
F (x, y1 , . . . , yn , y1′ , . . . , yn′ ) dx,
a
where y1 (x), . . . , yn (x) are continuously differentiable and satisfy the boundary conditions
yi (a) = ai ,
yi (b) = bi ,
for i = 1, . . . , n.
Namely, we are looking for an extremum of J defined on the set of smooth curves joining two points
(a, a1 , . . . , an ) and (b, b1 , . . . , bn ) in the (n + 1)-dimensional space Rn+1 .
Similar to the derivation of Euler’s equations for the variational problems studied before, we
add a continuously differentiable function hi (x) to each yi (x) such that the resulting yi (x) + hi (x)
still satisfy the boundary conditions. Therefore, hi (a) = hi (b) = 0, for i = 1, . . . , n. The variation
of J[y1 , . . . , yn ] is found to be
Z bX
n
(Fyi hi + Fyi′ h′i ) dx.
δJ =
a i=1
At each step, setting all but one of the hi to zero, we obtain the following system of Euler’s equations
that must be satisfied at an extremum:
Fyi −
d
F ′ = 0,
dx yi
6
i = 1, . . . , n.
(11)
Example 3. Geodesics. Given a parametric surface
σ = σ(u, v),
the variational problem here is to find the minimum distance between p and q on the surface.
Parametrize a curve from p and q as α(t) = σ(u(t), v(t)) so that p = α(t0 ) and q = α(t1 ). We denote
by ‘.’ differentiation with respect to t. The arc length between the two points is
Z t1 p
J[u, v] =
E u˙ 2 + 2F u˙ v˙ + Gv˙ 2 dt,
(12)
t0
where E, F, G are the coefficients of the first fundamental form of the surface given as
E = σu · σu ,
F = σu · σv ,
G = σv · σv .
Euler’s equations for the functional (12) become
d
E u˙ + F v˙
Eu u˙ 2 + 2Fu u˙ v˙ + Gu v˙ 2
√
√
=
−
dt
2 E u˙ 2 + 2F u˙ v˙ + Gv˙ 2
E u˙ 2 + 2F u˙ v˙ + Gv˙ 2
d
Ev u˙ 2 + 2Fv u˙ v˙ + Gv v˙ 2
F u˙ + Gv˙
√
√
=
−
dt
2 E u˙ 2 + 2F u˙ v˙ + Gv˙ 2
E u˙ 2 + 2F u˙ v˙ + Gv˙ 2
0,
(13)
0.
(14)
Since the parametrization does not change the total length of the curve, we may use the arc length
parametrization. From surface geometry, we know that this means
E u˙ 2 + 2F u˙ v˙ + Gv˙ 2 = 1.
Then equations (13) and (14) reduce to
d
(E u˙ + F v)
˙ =
dt
d
(F u˙ + Gv)
˙ =
dt
1
(Eu u˙ 2 + 2Fu u˙ v˙ + Gv v˙ 2 ),
2
1
(Ev u˙ 2 + 2Fv u˙ v˙ + Gv v˙ 2 ).
2
They are essentially the geodesic equations. This proves what we have learned before that the shortest curve
on σ connecting two surface points p and q is a geodesic between the two points.
Sometimes, it is more convenient to use one of the surface parameters to parametrize the geodesic. In
such a case, we still need to fall back on equations (13) and (14).
To illustrate, we now find the geodesics of the circular cylinder
σ = (a cos φ, a sin φ, z).
First, we obtain the coefficients of the first fundamental form:
E = a2 ,
F = 0,
Substitute these coefficients into (13) and (14):


d 
a2 φ˙

q
dt
2
2
2
˙
a φ + z˙


z˙
d 

q
dt
2
2
2
˙
a φ + z˙
7
G = 1.
= 0,
= 0,
Integrate the two equations above:
a2 φ˙
q
a2 φ˙ 2 + z˙ 2
z˙
q
a2 φ˙ 2 + z˙ 2
= C1 ,
= C2 .
Dividing the last equation by the one before, we obtain
dz
= D1 ,
dφ
which has the solution
z = D1 φ + D2 .
The geodesic between two points on a cylinder thus is a helix lying on the cylinder. Given two points
σ(φ0 , z0 ) and σ(φ1 , z1 ), the helix is described as
z0 − z1
φ0 z1 − φ1 z0
α(φ) = a cos φ, a sin φ,
φ+
.
φ0 − φ1
φ0 − φ1
Figure 3 shows a geodesic (cos φ, sin φ, 2φ) connecting two points with φ0 = 0 and φ1 =
(cos φ, sin φ, z) which is plotted over the subdomain [0, 2π] × [−1, 4].
3
2
on a cylinder
Figure 3: A geodesic on a cylinder.
4
Variational Problems in Parametric Form
We have considered functionals of curves given by equations of the form y = y(x). Often, it is more
convenient to consider functionals of curves in parametric form, say, (x(t), y(t)). Given a functional
8
of such a curve
Z
x1
F (x, y, y ′ ) dx,
(15)
x0
over [t0 , t1 ] such that x(t0 ) = x0 and x(t1 ) = x1 , we can rewrite it as
Z t1
Z t1 y(t)
˙
Φ(x, y, x,
˙ y)
˙ dt.
F x(t), y(t),
x˙ dt =
x(t)
˙
t0
t0
Here the overdot denotes differentiation with respect to t. This becomes a functional depending on
two unknown functions x(t) and y(t). The function Φ does not have t as an explicit argument. It
is positive-homogeneous of degree 1 in x˙ and y:
˙
Φ(x, y, cx,
˙ cy)
˙ ≡ cΦ(x, y, x,
˙ y),
˙
for any c > 0.
Conversely, let
Z
t1
Φ(x, y, x,
˙ y)
˙ dt
t0
be a functional whose integrand Φ is positive-homogeneous of degree 1 in x˙ and y.
˙ We now show that
the value of the functional is independent of the curve parameterization. Suppose we reparametrize
the curve with τ over [τ0 , τ1 ] such that t = t(τ ), and dt/dτ > 0 over [τ0 , τ1 ]. Then
Z τ1 Z τ1 dt dt
dx dy
Φ x, y, x˙ , y˙
dτ =
dτ
Φ x, y, ,
dτ dτ
dτ dτ
τ0
τ0
Z t1
dt
Φ(x, y, x,
˙ y)
˙
=
dτ
dτ
t0
Z t1
Φ(x, y, x,
˙ y)
˙ dt.
=
t0
Now, suppose the curve y = y(x) has a parameterization in t that reduces the functional (15)
to the form
Z t1
Φ(x, y, x,
˙ y)
˙ dt.
t0
Applying Euler’s equation (11), we end up with two equations:
d
Φx˙ = 0,
dt
d
Φy − Φy˙ = 0.
dt
Φx −
(16)
We can solve them for x and y.
The two equations in (16) must be equivalent to the single Euler equation:
d
Fy′ = 0,
dx
which results from the variational problem for the original functional (15). Hence the two equations
are not independent.
We refer to [1] for other variational problems such as involving higher derivatives, with subsidiary
conditions, involving multiple integrals, etc. Also, we refer to [2] for application of the calculus of
variations in theory of elasticity, quantum mechanics, and electrostatics.
Fy −
9
References
[1] I. M. Gelfand and S. V. Fomin. Calculus of Variations Prentice-Hall, Inc., 1963; Dover Publications, Inc., 2000.
[2] R. Weinstock. Calculus of Variations: With Applications to Physics and Engineering. Dover
Publications, Inc., 1974.
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