calculus of variation and its application

Calculus of Variation and its Application

Tien-Tsan Shieh

Institute of MathematicsAcademic Sinica

July 14, 2011

Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 1 / 1

Outline

1 Basic ideas in the Calculus of VariationsI Some review on calculusI Classical variational problemsI The indirect method:

F The first variation and the Euler-Lagrange equationF The second variation

I The direct method

2 Applications of the Calculus of Variations in Physics and ChemistryI The Fermat’s Principles in OpticsI The principle of least actionI The law of maximal entropy


CalculusCalculus was developed in the late 17th century by Newton andLeibniz individually.It is a mathematical discipline focus on limit, derivative, integral andinfinite series.Derivative of a function at the point x0:

f ′(x0) = limx→x0

f (x)− f (x0)

x − x0.

We also denote by dfdx (x0) as well.

Integral of a function:∫ b

af (x) dx = lim

n→∞,∆xi→0,

n−1∑i=0

f (xi )∆xi

where ∆xi = xi+1 − xi .Fundamental theorem of calculus∫ b

af ′(x) dx = f (b)− f (a).


Finding the Maximum and Minimum

Finding maximums and minimums for a giving a smooth function f (x):Considerer the Taylor expansion of the smooth function f (x):

f (x) = f (x0) + f ′(x0)(x − x0) +f ′′(x0)

2(x − x0)2 + higher oreder terms

f ′(x0) > 0 implies the function f (x) is increasing around x0.

f ′(x0) < 0 implies the function f (x) is decreasing around x0.

f ′(x0) = 0 and f ′′(x0) > 0 implies x0 is the local minimum.

f ′(x0) = 0 and f ′′(x0) < 0 implies x0 is the local maximum.


Review on Multi-dimensional CalculusSuppose f is a C 2-function on R2.

The directional derivative of the function f at −→x along thedirection −→v is defined as

D−→v f (−→x ) ≡ limt→0

f (−→x + t−→v )− f (−→x )

t= ∇f (−→x ) · −→v

Direction derivative can also be denoted by

∇vf (x), ∂f (x)

∂v, f ′v (x), Dvf (x).

The derivative of f at x ∈ R2 along the direction v:

D2v f (x) ≡ lim

t→0

f (x + tv)− f (x)− tv · ∇f (x)

t2

= fx1x1(x)v21 + 2fx1x2(x)v1v2 + fx2x2(x)v2

2

Here x = (x1, x2) and v = (v1, v2).


Taylor expansion of a multi-dimensional function

f (x + tv) = f (x) + t Dvf (x) +t2

2D2

v f (x) + higher order terms

= f (x) + t (fx1 (x)v1 + fx2(x)v2)

+t2

2

(fx1x1(x)v2

1 + 2fx1x2(x)v1v2 + fx2x2(x)v22

)+ higher order terms


Johann Bernoulli’s Challenge

The calculus of variations as a recognizable as a part of mathematics hadits origins in Johann Bernoulli’s challenge in 1696 to the mathematiciansof Europe to find the curve of quickest descent, or brachistochrone. Thebrachistochrone means the ”shortest time” in Greek.


Brachistochrone Curve

Newton was challenged to solve the problem, and did so the very nextday.

In fact, the solution, which is a segment of a cycloid, was found byLeibniz, L’Hospital, Newton, and the Johann Bernoullis and JakobBernoullis.


Formulation of the brachistochrone problemThe time of travel from a point P1 to a point P2 is given by

Time =

∫ P2

P1

ds

v

where s is the arclength and v is the speed.

1

2mv2 = mgy ⇒ v =

√2gy

ds =√

dx2 + dy2 =

√1 + (

dy

dx)2 dx

Therefore we obtain the following functional

F (y) :=

∫ P2

P1

√1 + y ′2√

2gydx

You may think functional F as a function defined on a class of paths(x , y(x)) which connect two points P1 and P2.


Geodesic Problems

Problem: Find a curve, joining points A and B, with a shortest distance.

Geodesic problem on a plane.Answer: a straight line joining the points A and B.

Geodesic problem on a sphere.Answer: a great circle joining the points A and B.


Geodesic problem on a sphere

Using spherical coordinate,

Y = (R cos θ sinϕ,R sin θ sinϕ,R cosϕ)

A given the curve joining A to B could berepresented by a pair of function (θ(t), ϕ(t))for t ∈ [0, 1] with ϕ(0) = θ(1) = 0; ϕ(1) = ϕ1

The resulting curve defined by

Y (t) = (R cos θ(t) sinϕ(t),R sin θ(t) sinϕ(t),R cosϕ(t))

and its derivative is

Y ′(t) = R(− sin θ(sinϕ)θ′ + cos θ(cosϕ)ϕ′, cos θ(sinϕ)θ′ + sin θ(cosϕ)ϕ′,−(sinϕ)ϕ′

)(t).


Geodesic problem on a sphereThe length of the curve is

L(Y ) :=

∫ 1

0|Y ′(t)| dt = R

∫ 1

0

√sin2 ϕ(t)θ′(t)2 + ϕ′(t)2 dt

The Problem

Find a minimum of the functional L among a curve Y (t), represented by(θ(t), ϕ(t)), satisfying ϕ(0) = θ(1) = 0; ϕ(1) = ϕ1

min L(Y ) = minR

∫ 1

0

√sin2 ϕ(t)θ′(t)2 + ϕ′(t)2 dt

L(Y ) ≥ R

∫ 1

0ϕ′(t) dt = Rϕ(t)

∣∣∣∣10

= Rϕ1

Let Y1(t) is a curve represented by θ(t) = 0, ϕ(t) = tϕ1. Then

L(Y1) = R

∫ 1

0

√sin2(tϕ1) 02 + ϕ2

1 dt = Rϕ1.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 12 / 1

Minimal Area ProblemFind a least-area surface among allsmooth surface which can berepresented as a graph of a smoothfunction u = u(x , y) on a planardomain D and satisfies the boundarycondition u|∂D = γ for some givenfunction defined on the boundary∂D.

Let S(u) be the associated surface area. The problem is formulated in thefollowing form:

minu∈A

S(u) = minu∈A

∫D

√1 + u2

x + u2y dxdy

where A is a collection of continuous differentiable functions

A = u ∈ C1(D) : u = g on ∂D.


Plateau’s Problem

Suppose C is a closed curve in space. What is the surface S of smallestarea having boundary C?

This problem is called Plateau’s problem in honor of the physicist,J. Plateau (1801-1883), who experimentally determined a number ofthe geometric properties of soap films and soap bubbles throughinteresting experiments with soap films.

In mathematics, Plateau’s problem is to show the existence of aminimal surface with a given boundary.


The simplest One-dimensional Example:

The variational problem is

miny∈A

E (y) = miny∈A

∫ 1

0|dydx|2 dx

where the collection A of functions is given by

A = y ∈ C 1(0, 1) : y(0) = 0, y(1) = 2.

Suppose y ∈ A is a minimizer of the functional E (y). Let ϕ(x) be anycontinuous differentiable function with ϕ(0) = ϕ(1) = 0.

E (y + tϕ) =

∫ 1

0|dydx

+ tdϕ

dx|2 dx


The first variation

δE (y , ϕ) ≡ dE (y + tϕ)

dt

∣∣∣∣t=0

Since y is a minimizer of the functional E , the following derivative is zero.

δE (y , ϕ) = 2

∫ 1

0

(dy

dx+ t

dϕ

dx

)dϕ

dx

∣∣∣∣t=0

dx = 2

∫ 1

0

dy

dx

dϕ

dxdx = 0

For any ϕ ∈ C 1[0, 1] with ϕ(0) = ϕ(1) = 0, the following it true∫ 1

0

dy

dx

dϕ

dxdx = 0.

By using the integration by part,

0 =

∫ 1

0

dy

dx

dϕ

dxdx =

dy

dxϕ

∣∣∣∣10

−∫ 1

0

d2y

dx2ϕ dx = −

∫ 1

0

d2y

dx2ϕ dx .

The Euler-Lagrange equation:d2ydx2 = 0 in (0, 1)y(0) = 0, y(1) = 2

=⇒ y(x) = 2x .


The second variation

The above discussion only guarantee the solution y = 2x is a criticalpoint for the functional E among all possible variation ϕ.

Consider the second variation

δ2E (y , ϕ) ≡ d2E (y + tϕ)

dt2

∣∣∣∣t=0

= 2

∫ 1

0

(dϕ

dx

)2

dx > 0

Suppose we set Φ(t) = E (y + tϕ). According to the Taylor expansionof the function Φ(t) at t = 0, we find

Φ(t) = Φ(0) + Φ′(0) t +1

2Φ′′(0) t2 + high order terms,

that is

E (y + tϕ) = E (y) + δE (y , ϕ) t +1

2δ2E (y , ϕ) t2 + high order terms

We could conclude thaty = 2x

is a minimizer.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 17 / 1

Typical One-dimensional Variational Problems

Let L(x , y , z) a function with continuous first and second (partial)derivatives in x , y , z . Let E (y) be a functional defined by

E (y) =

∫ 1

0L(x , y(x), y ′(x)) dx

on the class A of functions

A = y ∈ C 1[0, 1] : y(0) = a , y(1) = b.

Then the extreme y(x) of the functional E (y) satisfies the Euler-Lagrangeequation:

∂L

∂y− d

dx

∂L

∂y ′= 0 in on (0, 1).

with the boundary condition y(0) = a, y(1) = b.


The simplest 2-dimensional problem

Let D be some smooth bounded domain and g is a function defined onthe boundary ∂D. Set

A = u ∈ C 1(D) : u(x , y) = g(x , y) for (x , y) ∈ ∂D.

The following is one of the simplest 2-d minimization problem

minu∈A

E (u) = minu∈A

∫Du2x + u2

y dxdy

The Euler-Lagrange equation:

∂2u

∂x2+∂2u

∂x2= 0 in D

with the boundary condition

u(x , y) = g(x , y) on (x , y) ∈ ∂D.


The typical 2-dimensional variational problem

Let D be some smooth bounded domain and g is a function defined onthe boundary ∂D. Set

A = u ∈ C 1(D) : u(x , y) = g(x , y) for (x , y) ∈ ∂D.

Let L(x , y , z , p1, p2) be a C 2 function in all arguments. Let E (y) be afunctional defined by

E (u) =

∫DL(x , y , u, ux , uy ) dxdy

Then the extreme u(x , y) ∈ A of the functional E (u) satisfies theEuler-Lagrange equation:

∂L

∂y− d

dx

(∂L

∂ux

)− d

dy

(∂L

∂uy

)= 0 in D

with the boundary condition u(x , y) = g(x , y) on (x , y) ∈ ∂D.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 20 / 1

The Area Minimizing ProblemLet S(u) be the associated surfacearea.

minu∈A

S(u) = minu∈A

∫D

√1 + u2

x + u2y dxdy

where A is a collection of continuousdifferentiable functions

A = u ∈ C1(D) : u = g on ∂D.

The Euler-Lagrange equation

∇ · (ux , uy )√1 + u2

x + u2y

= 0

with the boundary condition

u(x , y) = g(x , y) for (x , y) ∈ ∂D.


Ideas from Calculus

In calculus, suppose f (x) is a C 2-continuous function.

f ′(x0) = 0 ⇒ x0 is a critical point.

In order to check whether x0 is a maximal point or a minimal pointwe need to check the sign of the second derivative.

f ′′(x0) > 0 ⇒ f (x0) is a local minimal.

f ′′(x0) < 0 ⇒ f (x0) is a local maximum.

Suppose f (x , y) is a C 2-continuous function. The Taylor expansion ofthe function f (x , y) at (x0, y0) is

f (x0 + tu, y0 + tv) = f (x0, y0) + [fx(x0, y0)u + fy (x0, y0)v ]t

+1

2[fxx(x0, y0)u2 + 2fxy (x0, y0)uv + fyy (x0, y0)v2]t2

+higher order terms


Analogy to Calculus of Variation

Suppose E (y) ≡∫ 1

0 L(x , y(x), y ′(x)) dx andA = y ∈ C 1[0, 1] : y(0) = a , y(1) = b.

The same issue happen in the calculus of variation.

δE (u, ϕ) = 0 for all possible variation ϕ

i.e.∂L

∂y− d

dx

∂L

∂y ′= 0 in on (0, 1).

only impliesy is a local extreme point in A.

In order to tell whether the extreme y of the functional E (y) is alocal minimum or a local maximum, we need check the secondvariation of E (y).


The Second Variation of E (y)

δ2E (y , ϕ) =d2E (y + tϕ)

dt2

∣∣∣∣t=0

=

∫ 1

0

d2

dt2L(x , y + tϕ, y ′ + tϕ′)

∣∣∣∣t=0

dx

=

∫ 1

0(Lyyϕ

2 + 2Lyy ′ϕϕ′ + Ly ′y ′ϕ

′2) dx

Suppose y(x) is a local extreme, i.e. δE (y , ϕ) = 0 for all ϕ ∈ C 10 [0, 1]. We

have the following conclusion:

If Ly ′y ′(x , y(x), y ′(x)) > 0 for all ϕ ∈ C 10 [0, 1] and δ2E (y , ϕ) > 0 for

all ϕ ∈ C 10 [0, 1], then y(x) is a weak minimum.

If Ly ′y ′(x , y(x), y ′(x)) < 0 for all ϕ ∈ C 10 [0, 1] and δ2E (y , ϕ) < 0 for

all ϕ ∈ C 10 [0, 1], then y(x) is a weak maximum.


Indirect Methods

The classical indirect method of variational problems is based on theoptimistic idea that every minimum problem has a solution. In orderto determine this solution, one looks for conditions which have tosatisfies by a minimizer. An analysis of the necessary conditions oftenpermits one to eliminate many candidates and eventually identifies aunique solution. For example, we only have one solution of theEuler-Lagrange equation satisfying all prescribed conditions. Wetempted to infer that this solution is also be a solution to the originalminimum problem.

The above approach is false. We have to prove the existence ofminimizer before we could conclude the uniqueness of solutions to theEuler-Lagrange equation.

Usually, in higher dimension solving the Euler-Lagrange equation ismore difficult than finding a minimizer of a functional .


The Direct Method for area minimizing problem

A surface u1 with area π + 1



Let F be the a functional defined on anunit disk.

F (u) =

∫D

√1 + u2

x + u2y dxdy

We can find an area minimizing sequenceun such that

limn→∞

F (un) = minu

F (u) = π.

If the area minimizing sequence is ”niceenough”, we could see the sequence unis getting closer and closer to the unit disk.Guess: The minimal surface is the unitdisk.


The Main Idea of Direct MethodConsider a minimization problem on some class A of functions:

minu∈A

E (u)

Suppose there exist a minimizing sequence un in A. i.e.

limn→∞

E (un) = minu∈A

E (u) < +∞.

Suppose we could find an element v0 in A such that

”un → v0” as n→∞.

Suppose the functional E has some kind of continuity. Therefore

limn→∞

E (un) = E (v0).

We conclude that v0 is a minimizer of the functional E because

E (v0) = limn→∞

E (un) = minu∈A

E (u).


The Difficulty of Direct Method

A surface with area π + 1





The Setting of the Direct Method

Consider a minimization problem on some class A of functions:

minu∈A

E (u)

Suppose we have following properties on E and A.

1 (Lower-semi-continuity) For every sequence ”un → u0”, we have

E (u0) ≤ lim infn→∞

E (un).

2 (Compactness) For a bounded sequence un in A, there exist aconvergent subsequence unk to some u0 ∈ A.


Existence of a Minimizer by the Direct method

Suppose vn is an minimizing sequence.

limn→∞

E (vn) = infv∈A

E (v) < +∞

If we are possible to show vn is bounded in A, there there exist asubsequence vnk and v0 ∈ A such that

vnk → v0 as k →∞.

By the lower-semi-continuity of the functional E , we have

E (v0) ≤ lim infk→∞

E (Enk ) = limn→∞

E (vn) = infv∈A

E (v)

This implies v0 is a minimizer.


A minimum problem without a minimizer

minu∈A

∫ 1

0(u2

x − 1)2 + u2 dx

whereA = u ∈ C 1[0, 1] : u(0) = u(1) = 0

Choose a minimizing sequence un which are zigzag functions with slope±1 and un → 0 as n→∞.We have

limn→∞

E (un) = 0

butE (0) = 1.

Therefore the minimum is not attained.The direct method does not always work.


Variational principles in physics

There are many laws of physics which are written as variational principles.

The Fermat’s principle in optics

The principle of least action This principle is equivalent to theNewton second law of motion in a mechanical system.

The law of maximum entropy: Thermal equilibrium will reach themaximum entropy of the system.


Fermat’s Principle

In optics, the Fermat’s principle, also called the principle of theleast time, states that a path taken by a ray of light between twopoints is the least-time path among all ”possible” paths.

The law of reflection and the law of refraction could be derived fromthe Fermat’s principle.

Reflection Refraction

The Fermat’s principle motivate the principle of least action inmechanic systems.


The Principle of Least Action

In Physics, the principle of least action states that the motion of amechanical system will follow the trajectory which minimize theaction of the system.

In classical mechanics, the action is defined as an integral along anactual or virtual space-time trajectory q(t) connecting two space-timeevents, initial event A ≡ (qA, tA = 0) and final eventB ≡ (qB , tB = T ). The action is defined as

S(q) =

∫ T

0L(q(t), q(t)) dt

where L(q, q) is the Lagrangian, and q = dqdt . The Lagrangian is given

as the difference of kinetic energy K and potential energy V alongthe trajectory q(t):

L(q, q) = K − V .


Classical Mechanics

Let’s consider the one-particle system in a conservative field. The relationbetween the potential energy of the particle and the force acting on theparticle is given by F = (F1,F2,F3) = −∇V = −(Vx ,Vy ,Vz) or

Work = V (A)− V (B) =

∫ B

AF · ds.

The kinetic energy of the particle is

K =1

2m(x2 + y2 + z2).

The associated Lagrangian is

L(x , y , z , x , y , z) = K − V =1

2m(x2 + y2 + z2)− V (x , y , z).


Consider the variational problem

minS(x(t), y(t), z(t)) = min

∫ T

0

1

2m(x2 + y2 + z2)− V (x , y , z) dt

δS =

∫ T

0m(xδx + yδy + zδz)− (Vxδx + Vyδy + Vzδz) dt

δS =

∫ T

0−m(xδx + yδy + zδz) + (F1δx + F2δy + F3δz) dt

δS =

∫ T

0(F1 −mx)δx + (F2 −my)δy + (F3 −mz)δz dt

According to the principle of least action δS = 0, we obtainm x = F1

m y = F2

m z = F3.

This is the Newton second law of motion,

F = ma.


Lagrangian MechanicsThe Lagrangian is given by

L(q1, . . . , qn, q1, . . . , qn) = K (q1, . . . , qn)− V (q1, . . . , qn, q1, . . . , qn)

where K is the kinetic energy and V is the potential energy of the systemand (q1, . . . , qn) is a generalized coordinate. The action S is defined by

S(q1, . . . , qn) =

∫ T

0L(q1, . . . , qn, q1, . . . , qn) dt.

According to the principle of least action, the Euler-Lagrange equations are∂L∂q1− d

dt∂L∂q1

= 0... =

...∂L∂qn− d

dt∂L∂qn

= 0

.

We usually called F ≡ ( ∂L∂q1, . . . , ∂L∂qn ) a generalized force and

P ≡ ( ∂L∂q1, . . . , ∂L∂qn ) a generalized momentum. Therefore, we derive a

Newton second law, P = F , in the general coordinate.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 37 / 1

The principle of least action

minS(q1, . . . , qn) = min

∫ T

0L(q1, . . . , qn, q1, . . . , qn) dt

provide us a way to find a ”suitable” coordinate system for themechanical system.

The principle of least action could also be applied in the theory ofrelativity, quantum mechanics and quantum field theory.


Legendre Transformation

A Legendre transformationof a convex function f isdefined by

f ∗(p) = maxx

(px − f (x)).

If f is differentiable, then f ∗(p) can be interpreted as the negative ofthe y -intercept of the tangent line to the graph of f that has slope p.The value of x attains the maximum has the property

f ′(x) = p.

f ∗(f ′(x)) = x f ′(x)− f (x).

Another definition: Set p = f ′(x). The inverse function theoremimplies x = x(p). Thus we define

f ∗(p) = p x(p)− f (x(p)).


Properties of Legendre Transformation

Set p = f ′(x). The inverse function theorem implies x = x(p). Thus wedefine

f ∗(p) = p x(p)− f (x(p)).

We have following relations:

p =df

dx(x),

x =df ∗

dp(p).

Also we havef (x) + f ∗(p) = xp.

And alsof = f ∗∗


Hamiltonian MechanicsThe Lagrangian of a system is given by L(q, q). We introduce theHamiltonian function

H(q, p) = maxq

(p q − L(q, q)).

This is equivalent to

H(q, p) = p q − L(q, q) and p =∂L

∂q(q, q),

where the second equation gives us the relation q = q(q, p).Since the double Legendre transformation is itself, we have

L(q, q) = pq − H(q, p) and q =∂H

∂p(q, p).

Example: A system of a single particle

L(q, q) =m

2q2 − V (q). p =

∂L

∂q= mq ⇒ q =

p

m

H(q, p) = p q−L(q, q) =p2

m−(

p2

2m−V (q)) =

p2

2m+V (q) = Total energy


Hamilton’s equations

Applying the principle of least action to L(q, q) = pq − H(q, p), we have

δ

∫ t2

t1

L dt = δ

∫ t2

t1

[p q − H(q, p)] dt.

We will find ∫ t2

t1

[q δp + p δq − ∂H

∂pδp − ∂H

∂qδq

]dt = 0,

∫ t2

t1

(q − ∂H

∂p

)δp +

(−p − ∂H

∂q

)δq dt = 0.

This leads us to Hamilton’s equationsq = ∂H

∂p ,

p = −∂H∂q .


Thermodynamics

Thermodynamics is a branch of physics which study physicalsystems through macroscopic quantities, such as temperature,volume, pressure, etc.

Thermodynamics concerns phenomena that are in thermalequilibrium. Studies of non-equilibrium physical processes are out ofthe scope of the classical thermodynamics.

The results of thermodynamics are essential for other fields of physicsand chemistry, chemical engineering, aerospace engineering,mechanical engineering, cell biology, biomedical engineering, materialsscience, and economics.


Laws of ThermodynamicsZeroth law of thermodynamics: states that if two systems are inthermal equilibrium with a third system, they are also in thermalequilibrium with each other.First law of thermodynamics: states that heat is a form of energy.The law is no more than a statement of the principle of conservationof energy. Energy can be transformed from one to another.

∆U = ∆Q −W

∆Q = QH − QC

Second law of thermodynamics: is also called the law of increaseof entropy. It states that if a closed system is in a configuration thatis not the equilibrium configuration, the most probable consequencewill be that the entropy of the system will increase monotonically insuccessive of time.Third law of thermodynamics: states that the entropy of a systemapproaches to a constant value as the temperature approaches zero.


Entropy

There are two definitions of entropy. One is the thermodynamicdefinition and the other is the statistic mechanics definition.

Thermodynamic entropy is a non-conserved state function ofphysical systems. Thermodynamic entropy is more generally definedfrom the statistical viewpoint, in which the molecular nature ofmatter is explicitly considered.

In statistical mechanics, statistical entropy is a measure of wayswhich a system could be arranged. The entropy S is defined as

S = kB ln Ω

where Ω is a number of ways and kB is the Boltzmann constant.

Boltzmann showed the statistical entropy is equivalent to thethermodynamic one.

The first law of thermodynamics tells us that energy is conserved in athermodynamic system. The second law tells us natural processeshave a preferred direction of progresses.


Thermal Equilibrium

A system that is in equilibrium experience no change when isisolated from its surroundings.

In thermodynamics, a system is in thermal equilibrium when it is inthermal equilibrium, mechanical equilibrium, radiative equilibrium,and chemical equilibrium and chemical equilibrium.

One of important questions inthermodynamics is what is theequilibrium state after weremove the boundary betweena isolated thermal system andits surrounding.


Callen’s Postulates of Thermodynamics

I. There exist particular states (called equilibrium states) that,macroscopically, are characterized completely by the specification ofthe internal energy U and a set of extensive parameters X1, X2, . . . ,Xt later to be specifically enumerated.

II. There exists a function (called the entropy)of the extensiveparameters, defined for all equilibrium states, and having the followingproperty. The values assumed by the extensive parameters in theabsence of a constraint are those that maximize the entropy over themanifold of constrained equilibrium states.

III. The entropy of a composite system is additive over the constituentsubsystems (whence the entropy of each constituent system is ahomogeneous first-order function of the extensive parameters). Theentropy is continuous and differentiable and is a monotonicallyincreasing function of the energy.

IV. The entropy of any system vanishes in the state for whichT ≡

(∂U∂S

)X1,X2,...

= 0.


The Internal Energy of a system

U = U(X0,X1,X2,X3, . . . ,Xt) = U(S ,V ,X2, . . . ,Xt)

where X0 denote the entropy S , X1 the volume, and the remaining Xj arethe mole numbers. For non-simple systems, the Xj may representmagnetic, electric, elastic extensive parameters to the system considered.

dU = TdS +t∑

k=1

PkdXk =t∑

k=0

PkdXk

T =∂U

∂S

P =∂U

∂V

Pk =∂U

∂Xkfor k = 0, . . . t

TdS is the flux of heat and∑t

k=1 PkdXk is the work.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 48 / 1

Entropy Maximum Principle and Energy Minimum Principle

S = S(U,X1,X2, . . . ,Xt) ⇔ U = U(S ,X1,X2, . . . ,Xt)

maxX1,...,Xt

S(U0,X1, . . . ,Xt)

Entropy Maximum Principle

minX1,...,Xt

U(S0,X1, . . . ,Xt)

Energy Minimum Principle


Legendre Transformations of the Internal Energy

Experimenters frequently find that the intensive parameters are themore easily measured and controlled and therefore is likely to think ofthe intensive parameters as operationally independent variables and ofthe extensive parameters as operationally derived quantities.A partial Legendre transformation can be made by replacing thevariables X0,X1, . . . ,Xs by P0,P1, . . . ,Ps .

U∗(P0,P1, . . . ,Ps ,Xs+1, . . . ,Xt) = minX0,...,Xs

(U(X0,X1, . . . ,Xt)−

s∑k=0

PkXk

).

The equilibrium values of any unconstrained extensive parameters(Xs+1, . . . ,Xt) in a system in contact with reservoirs of constantP0,P1, . . . ,Ps minimize the thermodynamic free energy(potential) U∗.

minX0,...,Xt

(U(X0,X1, . . . ,Xt) −

s∑k=0

PkXk

)= min

Xs+1,...,Xt

U∗(P0,P1, . . . ,Ps ,Xs+1, . . . ,Xt).


Some Classical Thermodynamic Free EnergiesSuppose the internal energy take the form

U = U(S ,V ,N).

Helmholtz Free Energy

A(T ,V ,N) = minS

[U(S ,V ,N)− TS ]

minV ,N

A(T ,V ,N)

Gibbs Free Energy

G (T , p,N) = minS,V

[U(S ,V ,N) + pV − TS ]

minN

G (T , p,N)

EnthaplyH(S , p,N) = min

V[U(S ,V ,N) + pV ]

minS,N

H(S , p,N)

Remark: Here pV related to the work done to the surrounding. It could be replaced by other kinds of work in other system.


Nonuniform Binary Systems

Consider a binary system of A component and B component.The Gibbs energy of the system take the form

G (c) =

∫Ωκ|∇c|2 + ωc(1− c) + kT [c ln c + (1− c) ln(1− c)] dx

where c is the mole fraction of the A component and (1− c) is the molefraction of the B component.

T temperaturek Boltzmann constantω interaction constant between twocomponentsκ some constantWe will see that sufficient cooling may leadto phase separation. (Phase transition)


Equilibrium state

The equilibrium state c of the binary system will minimize the Gibbs energy

minc

G (c) = minc

∫Ωκ|∇c |2 + f (c) dx .

Here we setf (c) = (1− c2)2.

The Euler-Lagrange equation is

−2κ∇c +∂f

∂c= 0 in Ω

and the boundary condition

n · ∇c = 0 on ∂Ω.


One-dimensional InterfaceSuppose interface is one-dimensional. We have to consider the problem

minc

∫ ∞−∞

κ|dcdx|2 + f (c) dx

and c(x)→ −1 as x → −∞ and c(x)→ 1 as x →∞.

−2κd2c

dx2+

df

dc= 0

(−2κ

d2c

dx2+

df

dc

)dc

dx= 0

d

dx

(−κ(dc

dx

)2

+ f (c)

)= 0

−κ(dc

dx

)2

+ f (c) = constant = 0


κ1/2 dc

dx=√f (c) = 1− c2

κ1/2

1− c2

dc

dx= 1

c(x) =exp( x√

κ)− exp(− x√

κ)

exp( x√κ

) + exp(− x√κ

)= tanh

(x√κ

)

This tells us the thickness of the interface is√κ.

The energy concentrate near the interface.

We may guess

G (c)→ Area functional of the interface

as κ→ 0. It is true but the rigorous proof requires more mathematics.


Thank you for your attention!


calculus of variation and its application

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