Calculus 65

Calculus

Lesson 65

Calculus of Variations

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Thanks to Dr. Helliwell of Harvey Mudd College - this lesson essentially comes from the class notes from his Theoretical Mechanics class at Mudd in 1992-1993.

Let's say that we know that there is some function that depends

on the path we take. It doesn't matter what the function

is: it could be friction, or work, or earned wages - whatever.

, where is the path taken. So if the function

were wages, perhaps the wages depend on which path is taken, and

how fast each section of the path is tread.

If we wanted to add up all of the wages, we would have to do

an integral:

is called a "Functional." It is NOT necessarily a function!

Now, let us suppose that we want to evaluate . In theory, we can,

at the very least, numerically integrate it if we know what is.

The question is: if we know what outcome we want, can we find a ?

More specifically, what if we want to minimize or maximize ? Can

we find what must be? Does such a function even exist?

Well, let us, for the moment, assume that it does exist, and we'll call it . Let's

also say that any other function , where is

the difference between and at any given point. It will be

easier, for our purposes, actually substitute into instead

of , since we have a way to vary as we shall see. Note that if

, that so we should be able to make the substitution

without any ill effects, as long as at the end.

We want some way to plug into our functional and to manipulate it to reduce

it to when we are minimizing the functional. Thus we will want to

see how . So, let's say

which is a power series. is called the "smallness parameter." As we make

, we certainly make ; however, note that we also make

the higher order functions unimportant, since they are multiplied by

higher powers of . Doing the above is similar to the statement we always

make about a function looking a like a line if we zoom in really close to it. In

the same way, as we zoom in on as , it looks like a

multiple of the same function . Therefore, it doesn't matter which

we use, it is just some multiple of the function as long as

we zoom in close enough. So we have reduced the problem from one with

an infinite number of s to one where we always have the same

small difference function, , and we only vary the size of a constant, .

Now, let's concentrate on . We know some things about it.

since at the endpoints. Plugging

everything into our functional, then,

where we know

SO. What is the method we will use to find the extremum of our functional?

The same thing we always do! Take the derivative with respect to something

that must go to zero:

This is especially convenient since is not in either integration limit so we can

bring the derivative inside the integral.

So what we have so far is:

We can integrate the second integral by parts.

This term=0 since at the endpoints

So, the derivative of our functional is:

We are looking for an extremum, so combining the integrals and

setting the integral = 0, we get:

The above equation must be true for ANY arbitrary (provided that

it goes to zero at the endpoints), so the only way that the integral

will ALWAYS be zero, is if what is in the parentheses is always zero.

Note that if we take the limit now as , we arrive at

our extremum path instead of one that is merely close to our

extremum path.

This is called Euler's Equation (EI). For the integral to be an extremum,

the function inside the integral MUST obey this equation.

Think back to what we were doing to recap: We are trying to find

the function that extremizes the integral. By saying that we were

using a function close to, but not quite, the minimizing function,

we were able to derive an equation that the minimizing function

must obey. We haven't done any examples yet, but at least we know

now what one of the conditions placed on the function must be.

Also remember that all we've done is find a condition for the integral to

be an extremum Ð we haven't specified whether that extremum

is a min, max, or stationary. However, usually we can use common

sense.

Example: if , write out Euler's Equation.

Of course we don't know how to solve this equation, but it's a start.

Let's do one we can solve: Find the shortest distance between two points.

Using EI:

where

So doing a bit of algebra:

The shortest distance between two points is a line!!!!!!

But we won't always be able to solve EI!

Let's find another formulation of Euler's Equation that might make it

easier for us to solve.

Let's consider and . Why? Because I'm telling you

that we can get an easier form of Euler's Equation if we do! So there!

But we know from EI that , so

The other part of what we're looking at:

Subtracting the two results (** - *), we get

This is (EII)

The reason EII is useful is that often we will not have a function that

explicitly a function of x, which means . When this is the case,

So, recap.

Use EII when is not explicate a function of .

Then we can move straight to

Use EI when is not explicate a function of .

Then we can move straight to

The reason this is called variational calculus is that small variations

of the path do not make a large change in our functional. Think of it like

at the top of a parabola, small variations in x do not greatly change the

slope. However, when we are NOT near the vertex, a small change has

a large effect on the slope.

Note that when we look at our previous line example,

is an explicit function of neither nor , so we could use either

EI or EII (although we would still use EI since although we know

, we don't know so we wouldn't be able

to solve EII.

Now, let's see the problem that caused Newton to essentially invent the

calculus of variations: the brachistochrone.

When only gravity is considered, what is the fastest path between two points?

It's not a line. Note that the sooner we speed up, the faster we'll be.

The answer isn't straight down first since then we waste time by taking

a path that's too long. So. We know what it should look like,

at least approximately: A

Let's minimize time:

We know is the differential path length.

We can use the Work-Energy Equation (transferring potential to

kinetic energy, which we have done innumerable times) to find that

at any given point.

Plugging this into EII (since has no explicit -

dependence,

We will call to simplify things. Also, we only choose the +ve answer since

y is +ve down!

I'll tell you the substitution to make:

We can at least see why this might work!

We can solve for by saying that are starting point is . This gives

Since we have already said what is, we now have a parametric solution

to the problem:

These equations describe what is called a "cycloid." It is the shape that

is traced by any point on a circle as it rolls without slipping on a plane. If we

are given a final , we can solve for the constant, .

B

A

A B

How would we find the time it takes to get from A to B?

Where

After lot's o' algebra...

Note that the time it would take to get back to the bottom is when

Where is the bottom of the bowl (from our parametric equations).

Note that this is obviously a minimum time path. A maximum time path

could go out to infinity and back!

Example: A curve connects A and B, and this curve is rotated about an axis.

Find the curve that minimizes the surface area. This is the

lamp-shade problem. The lamp-shade manufacturers want

to use the least amount of material possible while maintaining

the connection between the small diameter top to the large

diameter bottom.

Are any of these right? Which one looks like the least?

There is no explicit dependence on y, so use EI:

This shape is called a "catenary" and also happens to be the shape of a hanging chain.

Example: As we saw in the reading, the speed of light travels in such a way

as to make the time an extremum. Between two points,

this means it takes the shortest distance possible, a line. Between

two points such that it must hit a parabolic mirror first means that

it travels to hit the mirror and then on to the next point in

a straight line, but that path is a "stationary" path: it is shortest

in one sense that it essentially travels in a line, but it is longest

in another sense since there are other straight-line paths

that would take a shorter amount of time.

longest straight-line path

So, we know that we are looking to find a place where small

variations in the path do not change the path time significantly.

Let's derive Snell's Law.

We already know that light travels in a straight line, so the time it takes

to travel in that line is:

The total time, then is: .

We will want to extremize the total time.

The endpoints are fixed, so and are fixed, but and will

vary depending on where the light hits the glass. To reduce our equation

to a single variable, we will use the notation for and instead of

we will use since is also fixed.

The total travel time becomes:

The extreme time path can be found by taking the derivative with respect

to the only value that is changing!

Setting this equal to zero, we find that

Which of course we know as Snell's Law!

Finally (last one!), what happens as light goes through the atmosphere? As

the air gets more and more dense (due to gravity), the index of refraction

of air changes continuously. We could model it like this:

Of course, in the limit that the thickness of the layers goes to zero, it

will look like this:

Our goal is to derive a formula to find q_fgiven that we know q₀ and the

function for the index of refraction, n(y).

It's actually pretty much the same procedure!

Use EII since there is no explicit x-dependence!

But at any given point, we know that at any given point since the

light travels in a straight line in a differential sheet of atmosphere and

is measured compared to the vertical.

Which is just Snell's Law!! So all we need to know is the index of refraction

at the top and bottom, and the angle at the top.

The index of refraction at the top of the atmosphere is , so all we really need to

know is the index of refraction of the level at which we want to look, and the

incident angle.

If we want to solve for or , then the only additional thing we must do

is go back and integrate the derivative we found:

We can solve for when we integrate and put in the initial condition of

whatever is.

If, for example, we know that , where is constant.

Since for any point, then

So we have found a function for the x-position given the y-position and

the initial angle.

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