110514 calculus of variation 1

7/31/2019 110514 Calculus of Variation 1

1/56

Calculus of Variation and its Application

Tien-Tsan Shieh

Institute of MathematicsAcademic Sinica

July 14, 2011

ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)

Calculus of Variation and its Application July 14, 2011 1 / 1
http://find/http://goback/


2/56

Outline

1 Basic ideas in the Calculus of Variations Some review on calculus Classical variational problems The indirect method:

The first variation and the Euler-Lagrange equation The second variation

The direct method

2 Applications of the Calculus of Variations in Physics and Chemistry The Fermats Principles in Optics

The principle of least action The law of maximal entropy


http://find/


3/56

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

a

f(x) dx = lim

n, xi0,

n1

i=0

f(xi)xi

where xi = xi+1 xi.Fundamental theorem of calculus

b

a

f(x) dx = f(b)

f(a).


http://find/


4/56

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.

ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 4 / 1
http://find/


5/56

Review on Multi-dimensional CalculusSuppose f is a C2-function on R2.

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

Dv f(x ) lim

t0f(x + tv ) f(x )

t= f(x ) v

Direction derivative can also be denoted by

vf(x), f(x)v

, fv (x), Dvf(x).

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

D2v f(x) limt0

f(x + tv) f(x) tv f(x)t2

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

22

Here x = (x1, x2) and v = (v1, v2).ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 5 / 1
http://find/


6/56

Taylor expansion of a multi-dimensional function

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

2D2v f(x) + higher order terms

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

+t2

2

fx1x1 (x)v

21 + 2fx1x2 (x)v1v2 + fx2x2 (x)v

22

+ higher order terms



7/56

Johann Bernoullis Challenge

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



8/56

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, LHospital, Newton, and the Johann Bernoullis and Jakob

Bernoullis.



9/56

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

Time =P2P1

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) :=

P2P1

1 + y2

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.ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 9 / 1


10/56

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.



11/56

Geodesic problem on a sphere

Using spherical coordinate,

Y = (Rcos sin , Rsin sin , Rcos )

A given the curve joining A to B could berepresented by a pair of function ((t), (t))

for t [0, 1] with (0) = (1) = 0; (1) = 1The resulting curve defined by

Y(t) = (Rcos (t)sin (t), Rsin (t)sin (t), Rcos (t))

and its derivative is

Y(t) = R sin (sin ) + cos (cos ), cos (sin ) + sin (cos ),(sin ) (t).



12/56

Geodesic problem on a sphereThe length of the curve is

L(Y) := 10 |

Y(t)|

dt = R10sin2 (t)(t)2 + (t)2 dt

The Problem

Find a minimum of the functional L among a curve Y(t), represented by

((t), (t)), satisfying (0) = (1) = 0; (1) = 1

min L(Y) = min R

10

sin2 (t)(t)2 + (t)2 dt

L(Y) R

1

0(t) dt = R(t)

10

= R1

Let Y1(t) is a curve represented by (t) = 0, (t) = t1. Then

L(Y1) = R1

0

sin

2

(t1) 02

+ 21 dt = R1.



13/56

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 = for some givenfunction defined on the boundary

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

minuA S(u) = minuAD

1 + u2x + u

2y dxdy

where A is a collection of continuous differentiable functions

A=

{u

C1(D) : u = g on D

}.

Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 13 / 1


14/56

Plateaus Problem

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

This problem is called Plateaus 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, Plateaus problem is to show the existence of aminimal surface with a given boundary.



15/56

The simplest One-dimensional Example:

The variational problem is

minyA

E(y) = minyA

10|dy

dx|2 dx

where the collection A of functions is given by

A = {y C1(0, 1) : y(0) = 0, y(1) = 2}.

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

E(y + t) =

10|dy

dx+ t

d

dx|2 dx



16/56

The first variation

E(y, ) dE(y + t)dt t

=0Since y is a minimizer of the functional E, the following derivative is zero.

E(y, ) = 2

10

dy

dx+ t

d

dx

d

dx

t=0

dx = 2

10

dy

dx

d

dxdx = 0

For any

C1

[0,

1] with

(0) =

(1) = 0, the following it true10

dy

dx

d

dxdx = 0.

By using the integration by part,

0 =1

0

dydx

ddx

dx = dydx

1

0

10

d2

ydx2

dx = 10

d2

ydx2

dx.

The Euler-Lagrange equation:

d2y

dx2= 0 in (0, 1)

y(0) = 0, y(1) = 2=

y(x) = 2x.

http://find/


17/56

The second variation

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

Consider the second variation

2E(y, ) d2E(y + t)

dt2

t=0

= 2

10

d

dx

2dx > 0

Suppose we set (t) = E(y + t). According to the Taylor expansion

of the function (t) at t = 0, we find

(t) = (0) + (0) t +1

2(0) t2 + high order terms,

that is

E(y + t) = E(y) + E(y, ) t + 12

2E(y, ) t2 + high order terms

We could conclude thaty = 2x

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

O d


18/56

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) =

10

L(x, y(x), y(x)) dx

on the class A of functionsA = {y C1[0, 1] : y(0) = a , y(1) = b}.

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

equation: L

y d

dx

L

y= 0 in on (0, 1).

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


Th i l 2 di i l bl


19/56

The simplest 2-dimensional problem

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

A = {u C1(D) : u(x, y) = g(x, y) for (x, y) D}.

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

minuAE(u) = minuA

D

u2x + u2y 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.ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 19 / 1

Th i l 2 di i l i i l bl


20/56

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 C1(D) : u(x, y) = g(x, y) for (x, y) D}.

Let L(x, y, z, p1, p2) be a C2 function in all arguments. Let E(y) be a

functional defined by

E(u) =

D

L(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.


Th A Mi i i i P bl
http://find/


21/56

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

minuA S(u) = minuA

D

1 + u2x + 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 + u2x + u2y

= 0

with the boundary condition

u(x, y) = g(x, y) for (x, y) D.Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 21 / 1

Id f C l l
http://find/


22/56

Ideas from Calculus

In calculus, suppose f(x) is a C2-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 C2-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)u

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

+higher order terms


A alog to Calc l s of Va iatio
http://find/


23/56

Analogy to Calculus of Variation

Suppose E(y)

10 L(x, y(x), y

(x)) dx and

A = {y

C1

[0,

1] :y

(0) =a , y

(1) =b}

.

The same issue happen in the calculus of variation.

E(u, ) = 0 for all possible variation

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)


24/56

The Second Variation ofE(y)

2E(y, ) = d2

E(y + t)dt2t=0

=

10

d2

dt2L(x, y + t, y + t)

t=0

dx

=1

0(Lyy2 + 2Lyy + Lyy2) dx

Suppose y(x) is a local extreme, i.e. E(y, ) = 0 for all C10 [0, 1]. Wehave the following conclusion:

If Lyy(x, y(x), y(x)) > 0 for all C10 [0, 1] and 2E(y, ) > 0 forall C10 [0, 1], then y(x) is a weak minimum.If Lyy(x, y(x), y

(x)) < 0 for all C10 [0, 1] and 2E(y, ) < 0 forall C10 [0, 1], then y(x) is a weak maximum.


Indirect Methods
http://find/


25/56

Indirect Methods

The classical indirect method of variational problems is based on the

optimistic 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 the

Euler-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
http://find/


26/56

The Direct Method for area minimizing problem

A surface u1 with area + 1

A surface u2 with area +14

A surface u3 with area +1

16

Let F be the a functional defined on anunit disk.

F(u) =

D

1 + u2x + u

2y dxdy

We can find an area minimizing sequence

{un} such thatlimnF(un) = minu F(u) = .

If the area minimizing sequence is nice

enough, we could see the sequence {un}is getting closer and closer to the unit disGuess: The minimal surface is the unitdisk.


The Main Idea of Direct Method


27/56

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

minu

AE(u)

Suppose there exist a minimizing sequence {un} in A. i.e.limnE(un) = minuA

E(u) < +.

Suppose we could find an element v0 in A such thatun v0 as n .

Suppose the functional E has some kind of continuity. Therefore

limnE(un) = E(v0).

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

E(v0) = limnE(un) = minuA

E(u).


The Difficulty of Direct Method
http://find/


28/56

The Difficulty of Direct Method

A surface with area + 1

A surface with area + 14

A surface with area + 116A surface with area + 164

http://find/


29/56

Existence of a Minimizer by the Direct method


30/56

Existence of a Minimizer by the Direct method

Suppose {vn} is an minimizing sequence.

limnE(vn) = infvA

E(v) < +

If we are possible to show {vn} is bounded in A, there there exist asubsequence

{vnk

}and v0

Asuch that

vnk v0 as k.

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

E(v0) lim infk E(Enk) = limnE(vn) = infvAE(v)This implies v0 is a minimizer.


A minimum problem without a minimizer
http://find/


31/56

A minimum problem without a minimizer

minu

A 1

0(u2x 1)2 + u2 dx

whereA = {u C1[0, 1] : u(0) = u(1) = 0}

Choose a minimizing sequence {un} which are zigzag functions with slope1 and un 0 as n .We have

limnE(un) = 0

butE(0) = 1.

Therefore the minimum is not attained.ien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 31 / 1

Variational principles in physics
http://find/


32/56

Variational principles in physics

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

The Fermats principle in optics

The principle of least action This principle is equivalent to the

Newton second law of motion in a mechanical system.

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


Fermats Principle


33/56

p

In optics, the Fermats principle, also called the principle of theleast time, states that a path taken by a ray of light between two

points is the least-time path among all possible paths.

The law of reflection and the law of refraction could be derived fromthe Fermats principle.

Reflection Refraction

The Fermats principle motivate the principle of least action inmechanic systems.


The Principle of Least Action


34/56

p

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

action 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) =

T0

L(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


35/56

Lets consider the one-particle system in a conservative field. The relationbetween the potential energy of the particle and the force acting on the

particle is given by F = (F1, F2, F3) = V = (Vx, Vy, Vz) or

Work = V(A) V(B) =BA

F 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 = 12

m(x2 + y2 + z2) V(x, y, z).


Consider the variational problem
http://find/


36/56

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

T0

1

2m(x2 + y2 + z2) V(x, y, z) dt

S =T

0m(xx + yy + zz) (Vxx + Vyy + Vzz) dt

S =

T0m(xx + yy + zz) + (F1x + F2y + F3z) dt

S =T

0(F1 mx)x + (F2 my)y + (F3 mz)z dt

According to the principle of least action S = 0, we obtain

m x = F1

m y = F2m z = F3.

This is the Newton second law of motion,

F = ma.


Lagrangian Mechanics


37/56

The 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

Lq1

ddt

Lq1

= 0... =

...

Lqn ddt Lqn = 0.

We usually called F ( Lq1 , . . . , Lqn ) a generalized force andP ( Lq1 , . . . , Lqn ) a generalized momentum. Therefore, we derive aNewton second law, P = F, in the general coordinate.

http://find/


38/56

The principle of least action

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

T0

L(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


39/56

A Legendre transformationof a convex function f is

defined 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


40/56

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 have f(x) + f(p) = xp.

And alsof = f


Hamiltonian Mechanics


41/56

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

H(q, p) = maxq (pq L(q, q)).This is equivalent to

H(q, p) = pq L(q, q) and p = Lq

(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 = Hp

(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) = pq

L(q, q) =

p2

m (

p2

2mV(q)) =

p2

2m

+V(q) = Total energy


Hamiltons equations
http://goback/http://find/http://find/http://goback/


42/56

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

t2t1

L dt = t2t1

[pq H(q, p)] dt.

We will find

t2t1qp+ pq

H

pp

H

qq dt = 0,

t2t1

q H

p

p+

p H

q

q dt = 0.

This leads us to Hamiltons equationsq = Hp ,

p = Hq .


Thermodynamics
http://find/


43/56

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

Thermodynamics concerns phenomena that are in thermal

equilibrium. 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, materials

science, and economics.


Laws of Thermodynamics


44/56

Zeroth 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 = QWQ = QHQC

Second law of thermodynamics: is also called the law of increaseof entropy. It states that if a closed system is in a configuration that

is 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
http://goback/http://find/http://goback/


45/56

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 ways

which a system could be arranged. The entropy S is defined as

S = kB ln

where 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


46/56

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 between

a isolated thermal system andits surrounding.


Callens Postulates of Thermodynamics


47/56

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 X

1, X

2, . . . ,

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 the

absence 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

USX1,X2,... = 0.


The Internal Energy of a system


48/56

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

Entropy Maximum Principle and Energy Minimum Principle


49/56

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


50/56

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=0PkXk

.

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)

sk=0

PkXk

= min

Xs+1,...,XtU

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


Some Classical Thermodynamic Free EnergiesSuppose the internal energy take the form


51/56

Suppose 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.

http://find/


52/56

Equilibrium state


53/56

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

minc

G(c) = minc

|c|2 + f(c) dx.

Here we setf(c) = (1 c2)2.

The Euler-Lagrange equation is

2c + fc

= 0 in

and the boundary condition

n c = 0 on .


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


54/56

Suppose interface is one dimensional. We have to consider the problem

minc

|dc

dx|2 + f(c) dx

and c(x) 1 as x and c(x) 1 as x.

2

d2c

dx2

+df

dc

= 0

2 d

2c

dx2+

df

dc

dc

dx= 0

d

dxdcdx

2

+ f(c)

= 0

dc

dx

2+ f(c) = constant = 0


1/2dc

d=

f(c) = 1 c2


55/56

dx

( )

1/2

1 c2 dcdx = 1

c(x) =exp( x

) exp( x

)

exp(

x

) + exp(x

)

= tanhx

This tells us the thickness of the interface is

.

The energy concentrate near the interface.

We may guess

G(c) Area functional of the interface

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

http://find/


56/56

Thank you for your attention!

http://find/

110514 calculus of variation 1

Documents