VP LectureNotes 2010
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Transcript of VP LectureNotes 2010
Variational Principles Mark Jackson, Jan 10
0. Preliminaries Problem 1
Find the curve of shortest length going from to .
Problem 2
Find curve such that , of length which gives the maximum area.
We will develop a systematic way of obtaining solutions to such problems by showing that any solution must satisfy a specific differential equation.
Analogous to the relationship in calculus between stationary points of , given by , and the minimum and maximum values of .
In problem 2 the area is
and the length is
is a functional, i.e. a function on a space of functions.
This class of problems is called the calculus of variations.
0.2 Direct method Problem 3
Show that there exists such that , for any .
Find such that . By the intermediate value theorem, there exists a solution .
Alternative method;
Define
Claim; such that . Given this claim, we know that
.
Proof of claim.
Step 1; . On the
other hand, let ; then
But
So if . This shows the minimum value is attained inside if anywhere.
Step 2; By Analysis I, with .
This is called the direct method for variational problems.
1. Functions on
1.1 Functions Let
Then is linear if for , . Also
where .
Definition 1.1.1. A function is differentiable at if it can be well approximated by a linear function close to in the sense that
where .
i.e. , such that
Relation with difference quotients (
,
In Definition 1.1.1, put , then identical argument to the case shows that if is
differentiable at then
exists and is finite, and is equal to . This is because if such that
when , this implies
because linear.
Proposition 1.1.2. (i) If is differentiable at , then the partial derivatives
exist and the linear map (in definition 1.1.1) is given by
(ii) If all partial derivatives exist exist and are continuous on , then is differentiable
at each and is given by
We will define
to be the set of real valued continuous functions on all of whose partial derivatives are continuous on
is the set of real valued continuous functions on all of whose partial derivatives up to order are continuous on .
The definition of differentiability can now be written as
In Analysis II, you’ll see a proof that implies that is differentiable at all
.
Lemma 1.1.3. If , then . Also, if , then . This is a 1st order necessary condition for max or min.
Remark. This is also true if it holds only for in some ball around , i.e. .
Terminology. If then is a stationary or critical point. If is a maximum or minimum then it is an extremum (or extreme value).
1.2 Second-order conditions for extreme values Definition. A symmetric matrix (positive) if for all non-
zero. Similarly, (non-negative) if for all non-zero.
Exercise. Show that if a symmetric matrix is , then all its eigenvalues are , and the same with signs.
Theorem 1.2.1. If and , then
(i) is a local minimum , and a local maximum
(ii) is a strict local minimum, and is a strict local
maximum. ‘Strict’ means that, around the point which is a minimum, the function is strictly greater than at (or smaller, if is a maximum).
Imagine which has a strict local minimum at . Then and . Assume is the only stationary point, i.e. . Is a global minimum?
Yes - by Rolle’s theorem, because otherwise, with , and then Rolle’s theorem would imply with , contradicting the assumption that there is only one stationary point.
In , , there are functions with only one stationary point which is a strict local minimum but not a global minimum.
Exercise. Find such a function for .
1.3 Convexity Definition. A set is convex if, given , , we have
.
A function is convex if
i.e. the line between two points on the curve lies above the curve.
Remark. Define the epigraph
Then convex convex.
Proof ( ). Let . Let and
be in , so and . For
Hence, as was to be proved.
Proof ( ). Exercise.
Remark. Since the definition of convexity involves only line segment joining two points, to check a function is convex you can check the functions are convex .
Proposition 1.3.1. If then the following are equivalent;
(i) is convex
(ii)
(iii)
Proof that (i) (ii). Let . Then
and
By the chain rule, , which is (ii).
Proof that (ii) (i). Let and . Then
Choosing gives the definition of convexity.
Proof that (i) (iii). We have and
. Adding, as
required.
Proof that (iii) (i). Exercise.
To visualise the second condition, work in .
is graph of
Normal vector
Tangent plane at is
To have a picture for (iii) consider ;
This says that is monotone non-decreasing. Thus
Proposition 1.3.1 (in words). is convex lies above all its tangent planes is monotone.
1.3.4 Strict convexity
Definition. is strictly convex if
whenever and .
Proposition 1.3.2. If , the following are equivalent;
(i) strictly convex
(ii) for
(iii) for
Corollary to 1.3.1. If is convex with stationary point , then is a global minimiser for , i.e. we have .
Proof. and use (ii).
Corollary to 1.3.2. If is strictly convex and is a stationary point, then it is a strict global minimiser for , i.e. .
Recall that to solve you can hope to achieve this by minimising .
Corollary to 1.3.2. If is strictly convex, then the equation has at most one solution.
Proof. Assume and , so by
1.3.2(ii).
Remark. There is a corresponding notion of concavity, with the inequalities reversed, i.e.
etc. etc. E.g. on .
Lemma 1.3.3. If then
(i)
(ii)
Exercise. Think of an example to show that the converse of (ii) is false.
Proof of (i) ( ).
Note this is using the chain rule in the form
Warning – when we say a symmetric matrix is , we mean
Example. Show that
is a concave function on the set of all with and (a
probability distribution on ).
Note that if are both probability distributions on this set of points, then so is , since
.
Calculate the matrix of second partial derivatives;
since
This is a negative definite matrix. QED.
1.4 Constraints and Lagrange multipliers Example. Maximise subject to the constraint set
Answer; , attained at .
But notice is perpendicular to (unit circle) at these points. To see why, solve the problem
by parametrising as . Then
At maximum values we must have . Apply the chain rule;
Since is the tangent vector to , this means when is normal to .
General situation; if we have a constraint set , then assume admits
local parametrisation . If we find a point
where a function has a maximum/minimum on , then
Then has an unconstrained maximum at
, so
is perpendicular to the tangent vectors
Alternatively, , i.e. such that
the Lagrange multiplier. Use function .
Example. Find the rectangle in the plane inscribed in the unit circle which has the largest possible area.
Area .
Augmented function where is the Lagrange multiplier.
Stationary points of ;
Assume neither nor are , then
i.e. stationary point is a square.
If say , we just have a line inside the circle, so and this is a minimum value of amongst inscribed rectangles.
For , recall that . We will assume , and that we can write in parametric form, so
Example.
To search for stationary points of a function constrained to , we look at
written as .
First-order necessay condition for stationary point;
where the are tangent vectors to .
If is perpendicular to all tangent vectors, it is to , i.e. such that at a stationary point.
For ,
Assume is a stationary point at which . Then
But
using the summation convention. Thus
Theorem 1.4.1. Let be functioms and . Let ; then if has maximum at , then for some , and
If has minimum, then the second inequality has a . Here we use the notion of a symmetric matrix being defined above.
Moral – do usual calculus on .
To solve problems with two or more constraints, , do the same thing with
Examples. (i) Find the probability distribution on which maximises
Constraint is , so , so
(ii) For , , find on the unit circle and on the line to minimise/maximise . satisfies
, so
Then .
Using
, we get
As always, give the constraints .
In vector form we have
Notice cannot be zero, as this would imply , contradicting constraints since and , .
From the first pair
But . Since , we get . Clearly gives the minimum value.
Example of 2nd derivative test. Let , and . As we proved, we need to look at
We want to check this is a negative definite matrix at the solution
so the matrix of 2nd partial derivatives is
Find eigenvalues of , i.e.
which satisfies the necessary 2nd order condition for being a maximum.
1.5 The Legendre transform For , given a function, we define its Legendre transform by
This is only defined for such that this is finite.
Examples. (i) Let , , then is defined for all and attained at where
At that point
Now define
If is convex, then .
(ii) , , then where this is finite. Here the domain of is because the is nowhere defined.
(iii) . Then , so the domain of is with .
(iv) . Then when defined, so the domain of is with .
Proposition 1.5.1. is convex (on its domain).
Proof.
So
So if lie in the domain of , then so does , and
Theorem 1.5.2. If is convex, then .
Proof. Prove for with . Then
is attained for all at unique
such that . (Recall the corollary to 1.3.2.)
Given , there is a unique point where the tangent line to has slope .
Consider, for given , the tangent line to
with slope . It has equation
or
Recall that a convex function lies above its tangent lines (or planes). Call . For given , the line lies below the graph of , and the supremum is attained when , and supremum is , i.e.
i.e. . QED.
Definition. An affine function is one of the form
Corollary. If convex, then it is a supremum of a family of affine functions.
Example. Consider . Then
Domain of is ; supremum is attained when
Then
(Young’s inequality). In fact we always have .
For we define the Legendre transform by
Exercise. Calculate the Legendre transform of
where is a positive symmetric matrix.
Applications
(i) Classical mechanics. The Lagrangian
Hamiltonian
To calculate; supremum attained when
Exercise. Newton equations are equivalent to
(ii) Economics. A company buys imports and produces a product which is sold with revenue . Let be the price of a unit of the th good. Then the total cost is and the profit
is . To maximise profit, they try to choose to get
(iii) Thermodynamics. An engine has internal energy where is the volume and is the entropy. In idealised ‘reversible change’, the heat flow . Also where is the temperature and the pressure. Then
(Maxwell relations).
Suppose the system is immersed in a constant temperature reservoir. In this case the system is described by a function . This is called the Helmholtz free energy, and is defined by
The supremum is attained at such that
This defines , and we can substitute in to get
In Helmholtz description
(Maxwell relation).
Remark. The entropy is determined implicitly by
This determines uniquely as long as
Constant volume heat capacity heat input needed to increase temperature at fixed volume
So is a convex function of you need heat to raise the temperature.
2. Calculus of variations
2.1 Differentiating functionals A functional is a function any space of functions or .
Examples. (i) Let and consider , the ‘Dirac functional’.
(ii) Let functions with . Then
Recall. If , then the directional derivative at in the direction of is
We will apply this to the functionals;
To define something analogous to , we use
or for complex functions we will often use
Now . In this case, we call the functional derivative. More
generally, for any functional we have
when such a function exists. Let’s try to apply this to .
In Methods we will write
Example.
because
Therefore
Example 3. Let . Then
Now . When we calculate a derivative, we consider smooth with , and look at
since vanishes at the end points of the integral; . So
using the notation e.g.
Problem. Let . Minimise
amongst all smooth -periodic functions .
Solution. Assume such a minimising function does exist. Obtain a condition which determines it. We’ll then show that there exists a function satisfying this condition, then finally we’ll show that this function minimises .
Assume is a solution, then for any , where is the set of smooth -periodic functions. In particular, if for any , then .
, i.e. .
Exercise. Show that
So if does minimise , then
Lemma 2.1.1. If (i)
for all smooth functions , with if is outside some sub-interval ;
(ii)
then in .
Proof. Assume, to the contrary, that with . Without loss of generality, take ; since is continuous, such that on .
Consider the function
Now , because .
Consider
Then
which is a contradiction.
Exercise. Find (for the previous example) such that – .
Jargon.
We define closure of set where .
We say is properly supported in or if .
Problem 1. Minimise
over set of smooth -periodic functions, and .
Solution. 1st stage (last time); assume there is such a minimising function , then for all and
.
Therefore
for all and
By general formula ( ),
Therefore
By Lemma 2.1.1, , i.e. if exists it solves this equation. This is called the
Euler-Lagrange equation.
When we get, with periodic,
Remark. To prove there is only one -periodic solution of – , imagine is
another, where – with . Subtract .
Multiply by and integrate;
by parts. Middle term is by periodicity and then we get , i.e. .
Final stage of minimisation problem; consider
Integrating the middle term by parts we get
for any , . So really minimises over smooth -periodic
functions.
Problem 2. Find the curve of shortest length, joining the two points in the plane, and .
Mathematically
We need to minimise amongst such curves.
Assume is a minimising curve, then exactly as before, at . Now
The Euler-Lagrange equation is therefore
constantm so . Fix so that and . This is our
candidate minimising curve.
Claim that is strictly convex
By lemma 1.3.2,
. Now
which if and . So if is another -curve joining to it has greater length than the straight line.
2.3 Multi-dimensional integrals Moral. Integration by parts in one dimension converts to using Green identities in more than
one dimension.
Consider
where some region in . Then
To achieve this we recall
Assume vanishes on , then
i.e.
but so we get Poisson’s equation; . This gives the minimum value of ‘energy’, . This gives gravitational potential of mass distribution , or electrostatic potential of charge.
Problem. Tourist El Cheapo wants to drive 3000 miles from New York to San Francisco in car, taking 100 days. Car uses gallons per mile, with . Fuel costs cents per gallon. How should he choose his speed to spend least money?
For we have with and .
Assume he pays for his fuel as he uses it (continuous ), and that .
Cost to go a distance at time is . So total cost is
A minimising will be a stationary point, so .
Note that the integrand so the functional derivative is
already, and then choose such that .
Notice that if is independent of then is constant, by the Euler-Lagrange equation.
Concepts in theoretical physics; an action in would be of the form
with and . Writing so that , the Euler-
Lagrange equation becomes
using the summation convention.
Derivation. Let be a smooth ‘variation’ function, with near the boundary of . Then
since near the boundary of . Alternative form;
Example. Let an action be
so that and . Then
Euler-Lagrange equation;
the wave equation.
2.4 Lagrange multipliers DIDO problem; find a curve such that the area under the curve and above the -axis is
maximised, for a fixed length. I.e. maximise
subject to
Consider ‘augmented function’ where is the Lagrange multiplier. Then
.
Substitute so that , i.e. which is a
circle. Therefore the solution is is arc of circle passing through of length .
Remark. The formulation of the problem disallows curves that go back on themselves in the -direction, because this is not a graph ! To get over this, we can instead consider
parametrised curves . Then integral functional is
even with .
Euler-Lagrange equations;
for .
Exercise. Find closed curve in the plane such that to maximise
the area inside the curve, with
Finite number of constraints with . Consider as for finite-dimensional problems.
Infinite number of constraints – need Lagrange ‘multiplier function’. E.g. in fluid mechanics. A velocity vector field on subject to an infinite number of constraints at each point; .
Minimise
subject to (incompressible). Then
Exercise. Work out .
Remark. For the notation
and
Constraint for each means there is a Lagrange multiplier function .
Green identities
by Green, assuming for large
so 2 lines above
1st term =
using the second Green identity;
Therefore
So the Euler-Lagrange equation is – . Now
Final construction; stationary for with if
Static Navier-Stokes equation, then approximate by putting all terms to zero, you get this equation where is the pressure.
3. Scientific examples Fermat’s principle – when light
reflects in a mirror, the angle of reflection is equal to the angle of incidence . This can be explained by postulating that light rays take the shortest time possible to get from A to B.
Time as path through
where speed of light.
This implies and
Typical problem; imagine the speed of light is . Then consider a light path .
Time is
This is the type of integral
we know about. A minimising path would solve
2. Mechanics A particle of mass has position , moving in a potential . The force so
This equation is the Euler-Lagrange equation for the ‘action’
where
Most ‘fundamental’ equations of classical physics come in this form, eg. Maxwell’s.
3. Geodesics A geodesic is a curve of shortest length, or more generally a stationary point for the length
.
(i) In the plane, curves from to ,
The only solution is the straight line. We can also write the curve and then
Since
by the change of variables and
Simplest to choose so that
constant, i.e. arc length.
Using these parametrisations, geodesics are curves which make stationary
For
, the Euler-Lagrange equation is (put , in the second
example.) This is exactly what you get by putting constant in .
Recall. We defined a geodesic curve to be a curve which makes stationary either the length
or the integral
The length is independent of parametrisation;
The second definition gives curves parametrised so that is constant. This is usually easier to work with.
Problem. Find geodesic curves on a cylinder
Solution 1. Use cylindrical coordinates; , , . Then
For a curve on , is fixed, and , . Therefore
A geodesic curve on is a curve which makes stationary
The Euler-Lagrange equations are
Conclusion; geodesic curves are ‘helicoidal’ curves. (See example sheet 2, question 9 – geodesics are great circles.)
Solution 2. Introduce as a constraint, and use the Lagrange multiplier function , and
The Euler-Lagrange equations are
To find and we must use constraints;
Find ; But by Euler-Lagrange,
Write , then we have and . But constant constant.
Solutions are therefore , . Again, we get helicoidal curves.
Problem. (i) Consider
(no -dependence!) and show that a curve which makes stationary satisfies
(ii) Find the curve such that a bead moving without friction on the curve takes the shortest possible time to go from to .
Solution. (i) Since makes stationary,
The last two terms come about by the chain rule, since . This is equal to
(ii) The bead moves under gravity, so the speed satisfies
Euler-Lagrange;
This is very hard to solve! Instead we use (i), to say
Choose ; then
The curve is a cycloid.
4. Conservation laws (Noether theorem)
Let be a smooth solution of
Theorem 4.1. (i) If (no dependence) then constant.
(ii) If (no dependence) then constant.
Proof. (i) constant by Euler-Lagrange equation.
(ii) See last lecture.
Special case from mechanics;
Newton’s law is Euler-Lagrange equation for
Conservation laws; (i) if does not depend on then
constant. We sometimes call a ‘cyclic coordinate’.
(ii)
since does not depend explicitly on .
5. The second variation When is a solution of the Euler-Lagrange equation
a (local) minimiser of
Possible methods; (i) use convexity. See 2.2 for example (shortest curve between two points).
(ii) look for condition analogous to the 2nd order condition in calculus.
If , then Taylor’s theorem gives
This tells us that (i)
for all sufficiently small (‘strict local minimum’).
(Positive symmetric means
(ii) a local minimum and is non-negative (i.e. ).
To extend this to a functional , let be a smooth function (or with . ( is analogous to .) Assume is smooth, then;
For all , such that then
In this case we get
Terminology.
Definition. is a weak local minimum for if for sufficiently small. Call it strict if inequality is strict for such .
Theorem 5.1. (i) If , and
for some , then is a strict weak local minimum for .
(ii) If is a weak local minimum for , then and .
Remark. A space of functions is an infinite-dimensional vector space, so we have to be careful which norm we use on a space of functions!
Example.
Euler-Lagrange equation;
There is a solution – is this a (weak) local minimum?
Exercise. The second variation is
Now
satisfies the criterion with . Therefore for , is a strict weak local minimum. Then consider
Recall . Try , then , so
since . Conclusion; is not a weak local minimum for .
Remark. In general the second variation can be written
Associated to this is the Sturm-Liouville operator
This has an infinite sequence of eigenvalues, where is an eigenfunction. So if then