Uploaded by commutant on 21.08.2011

Transcript:

Hello. I'd like to spend the next few years talking about the 'Wave equation'.

So we saw the Wave equation already in the second video.

In that video I made a small simplication

I said the Wave equation was: 'u_tt' = 'u_xx'

so I left out this 'c^2' here

that you would usually see in the with equation when you see it..

..in a text book.

And so we put that back in there. (we'll put 'c^2' in front of 'u_xx')

So the 'c^2' is a constant and we have a whole 'family' of Wave questions

..one for every choice of 'c'.

Now it may look a little funny at first, the fact this is a 'c^2',

(rather than just a 'c')

and there's a couple of reasons for that:

First reason is: that you can think of it as a reminder

..that this constant, in front of 'u_xx', needs to be 'positive'.

If I were to take something like: 'u_tt' = -'u_xx'

..that would be a much different kind of equation: it would have

different 'kinds' of solutions

(..and we would not be able to call it the 'Wave equation')

The other reason that I take 'c^2' is that 'c' itself is a quantity of interest:

it's called the 'wave speed' and we'll see there is a very 'easy' explanation

..why I would call it the 'wave speed'. (we'll see that in a little while)

So let's start with some intuition:

Let's suppose that you and I are playing with a string.

Here you are holding onto one end of the string and..

here I am holding on to the other end.

We stretch the string out tightly between the two of us.

(so there it is: stretched out tight)

Now, we play a game..

'wiggle the string'.. so

..you give the end of the string a 'wiggle'

..and we get this wiggle

..going down the string.

(so this 'wave', going down the string)

(Okay)

Now let's connect this game we're playing with the variables we have here:

Let's call the displacement of the string from its equilibrium..

..the 'vertical' displacement of the string

(let's call that the variable 'u')

So this is: 'u(x,t)'

Now 't' is a measurement of time of course.. (this is all in motion)

and 'x' we'll just think of as a measurement of the position..

..along the string horizontally: so you could start off at 'x' = '0' (here)

and measure it as 'positive' to the right (for instance)

So now let's 'zoom in' on this wiggle thing that we've got going on here..

Let's 'zoom in' and see if we can reason a little bit about it?

(and have the Wave equation 'come out' of that reasoning)

Okay so here we are, zoomed-in on the

wiggly part of the string.. (I'll do my best to draw a 'wiggle' here)

[not bad]

(I'll draw the 'equilibrium line' in here)

So let's look first at what are, essentially..

..the easiest parts of this picture to understand.

Let's look at the 'peak' (here). and the 'valley' (there)

So what's happening to this little piece of string at the peak of this 'mountain'?

the string around

the 'top' of that mountain.. that's curved 'downward' and so

the tension will be pulling this point at the very 'peak' downward..

..so there's a downward 'force'.

(I'll 'show' that by drawing an arrow) there's a downwards force acting on that..

..little piece of string right there.

Similarly over here at the at the bottom of this 'valley',

the string around that little piece.. that string is curved 'upward'

and so it's exerting an upward force.

(the 'tension' of the string is pulling it up)

Now near these points

we will have

similar forces but they'll be smaller in magnitude

and that's simply because the string is not quite as curved

as we move away from the peak of the 'mountain'..

(or the very bottom the 'valley')

..we have a little less 'curvature' of the string.

At some point ( though )..

the 'downward forces' and the 'upward forces' have to meet

..and where they meet, in the middle, there will be 'zero' force.

(there's no up-or-down 'force' acting at that point)

Well, what are we saying here really?

We're making an assertion..

about the relationship between

the 'concavtiy' of the string,..

thinking of this string as a 'graph' of a 'function' (of 'x'):

the 'concavity' of that graph

and the forces acting on the string.

Let's make a 'leap of faith' here, and let's say:

the force is proportional to the concavity, (it's clearly related:.

..the greater the concavity, the greater the force)

and let's make that 'leap' by saying that it's proportional.

so: " force is proportional "..

.." to concavity "

of the string.

So this is (essentially) a 'reasonable' conclusion..

..from this picture that we've drawn.

Now let's take it one step further

and let's look at what we mean by 'concavity'.

So,what's 'concavity'?

it is the 'second' derivative of this graph, as a function of x:

so it's simply: 'u_xx',

(and why is it the 'partial' derivative?)

we're looking at a fixed point in 'time'

so 't' is fixed..

.. and we're taking the derivative with respect to 'x' twice

..to get the concavity of this graph.

Now what's force?

Well force is: "mass times acceleration",

(by Newton's 2nd law): 'F' = 'm*a'

and Newton's 2nd law, you would think of as being something that you use on (say)

'a particle'.

So what we can do is we can think of each little point on this string

as being some 'particle' with some fixed 'mass',

..(say) the mass has nice 'uniform density.'

(let's not worry about what the 'mass' is)

So we're talking about the acceleration of a little fixed point on this string..

..thinking of it like 'a particle'.

Well what would that be (in terms of 'the displacement')?

The acceleration in terms of the displacement is simply the

'second derivative' with respect time:

(so this is)

'm' times 'u_tt'

I'm fixing 'x' and looking at the way the little piece of string

is moving up and down..

..thinking about it as a little 'particle')

What can we conclude from all this?

Well we know that force is proportional to: 'u_tt',

it's 'm' times 'u_tt'

'Concavity' on the other hand it's simply: 'u_xx'

so the 'big' conclusion here is:

'u_tt' is proportional to 'u_xx'

and I'll write the proportionality as simply 'k': 'u_tt' = 'k'*'u_xx'.

What's the sign of 'k'?

If I'm concave 'down' my acceleration is downward..

If I'm concave 'up' my acceleration is upward..

So 'k' is positive and I could, if I wanted to..

write it as 'a square'.

I could say something like: 'k' = 'c'^2

Okay so we finished our slick little argument

and here we are..

..staring at: the 'Wave equation'.

Now there's a lot of things sort of wrong with this argument..

..it's what I call a: 'fast and loose'

So see if you can just reflect on it..

..and think about all the assumptions that are lurking in here?

(I 'll give you a couple)

well, one thing is we assumed very clearly that:

force is 'proportional to' concavity

there's no reason, a priori, that

it should be a ( simple ) proportionality.

maybe there's some other sort of functional relationship that looks like this?

Another assumption that we made is:

the motion is straight up-and-down ('vertical').

(maybe there's some side-to-side motion going on there..

..things might get much more complicated?)

Alright so there's 'lots of problems' but you can get some intuition from this.

Now let me conclude with a little 'nugget of information' about 'c'.

So what I'm going to do is just a quick

'dimensional analysis'

and this is something you usually learn in Physics classes.

(but it's also useful in math classes)

What are the 'dimensions' of 'u_tt'?

Well the dimensions of 'u' are 'length'.

(it's a displacement, so it's 'length')

The second derivative with respect to 'time'..

..the dimensions of 'u_tt' are:

'length' / 'time-squared'

So 'u_tt'..

has dimensions..

'L' / 'T'^2

On the other hand 'u_xx'..

..that's 'length' / 'length-squared'

so that's simply: '1 / L'

So what does that mean the dimensions of 'c'^2 will be?

Well that implies that in order for the 'dimensionality' of 'u_tt' to equal

the dimensionality of 'c^2'*'u_xx' we must have the

dimensions of 'c'^2 be:

'length-squared' / 'time-squared': ( 'c*c' <=> 'L*L' / 'T*T' )

Well that means that 'c' has dimensions

'length' / 'time': ( 'c' <=> 'L' / 'T' )

That's 'a velocity'.

So it's reasonable to expect that 'c' (in a 'physical' problem) should be

some kind of 'velocity'.

So we saw the Wave equation already in the second video.

In that video I made a small simplication

I said the Wave equation was: 'u_tt' = 'u_xx'

so I left out this 'c^2' here

that you would usually see in the with equation when you see it..

..in a text book.

And so we put that back in there. (we'll put 'c^2' in front of 'u_xx')

So the 'c^2' is a constant and we have a whole 'family' of Wave questions

..one for every choice of 'c'.

Now it may look a little funny at first, the fact this is a 'c^2',

(rather than just a 'c')

and there's a couple of reasons for that:

First reason is: that you can think of it as a reminder

..that this constant, in front of 'u_xx', needs to be 'positive'.

If I were to take something like: 'u_tt' = -'u_xx'

..that would be a much different kind of equation: it would have

different 'kinds' of solutions

(..and we would not be able to call it the 'Wave equation')

The other reason that I take 'c^2' is that 'c' itself is a quantity of interest:

it's called the 'wave speed' and we'll see there is a very 'easy' explanation

..why I would call it the 'wave speed'. (we'll see that in a little while)

So let's start with some intuition:

Let's suppose that you and I are playing with a string.

Here you are holding onto one end of the string and..

here I am holding on to the other end.

We stretch the string out tightly between the two of us.

(so there it is: stretched out tight)

Now, we play a game..

'wiggle the string'.. so

..you give the end of the string a 'wiggle'

..and we get this wiggle

..going down the string.

(so this 'wave', going down the string)

(Okay)

Now let's connect this game we're playing with the variables we have here:

Let's call the displacement of the string from its equilibrium..

..the 'vertical' displacement of the string

(let's call that the variable 'u')

So this is: 'u(x,t)'

Now 't' is a measurement of time of course.. (this is all in motion)

and 'x' we'll just think of as a measurement of the position..

..along the string horizontally: so you could start off at 'x' = '0' (here)

and measure it as 'positive' to the right (for instance)

So now let's 'zoom in' on this wiggle thing that we've got going on here..

Let's 'zoom in' and see if we can reason a little bit about it?

(and have the Wave equation 'come out' of that reasoning)

Okay so here we are, zoomed-in on the

wiggly part of the string.. (I'll do my best to draw a 'wiggle' here)

[not bad]

(I'll draw the 'equilibrium line' in here)

So let's look first at what are, essentially..

..the easiest parts of this picture to understand.

Let's look at the 'peak' (here). and the 'valley' (there)

So what's happening to this little piece of string at the peak of this 'mountain'?

the string around

the 'top' of that mountain.. that's curved 'downward' and so

the tension will be pulling this point at the very 'peak' downward..

..so there's a downward 'force'.

(I'll 'show' that by drawing an arrow) there's a downwards force acting on that..

..little piece of string right there.

Similarly over here at the at the bottom of this 'valley',

the string around that little piece.. that string is curved 'upward'

and so it's exerting an upward force.

(the 'tension' of the string is pulling it up)

Now near these points

we will have

similar forces but they'll be smaller in magnitude

and that's simply because the string is not quite as curved

as we move away from the peak of the 'mountain'..

(or the very bottom the 'valley')

..we have a little less 'curvature' of the string.

At some point ( though )..

the 'downward forces' and the 'upward forces' have to meet

..and where they meet, in the middle, there will be 'zero' force.

(there's no up-or-down 'force' acting at that point)

Well, what are we saying here really?

We're making an assertion..

about the relationship between

the 'concavtiy' of the string,..

thinking of this string as a 'graph' of a 'function' (of 'x'):

the 'concavity' of that graph

and the forces acting on the string.

Let's make a 'leap of faith' here, and let's say:

the force is proportional to the concavity, (it's clearly related:.

..the greater the concavity, the greater the force)

and let's make that 'leap' by saying that it's proportional.

so: " force is proportional "..

.." to concavity "

of the string.

So this is (essentially) a 'reasonable' conclusion..

..from this picture that we've drawn.

Now let's take it one step further

and let's look at what we mean by 'concavity'.

So,what's 'concavity'?

it is the 'second' derivative of this graph, as a function of x:

so it's simply: 'u_xx',

(and why is it the 'partial' derivative?)

we're looking at a fixed point in 'time'

so 't' is fixed..

.. and we're taking the derivative with respect to 'x' twice

..to get the concavity of this graph.

Now what's force?

Well force is: "mass times acceleration",

(by Newton's 2nd law): 'F' = 'm*a'

and Newton's 2nd law, you would think of as being something that you use on (say)

'a particle'.

So what we can do is we can think of each little point on this string

as being some 'particle' with some fixed 'mass',

..(say) the mass has nice 'uniform density.'

(let's not worry about what the 'mass' is)

So we're talking about the acceleration of a little fixed point on this string..

..thinking of it like 'a particle'.

Well what would that be (in terms of 'the displacement')?

The acceleration in terms of the displacement is simply the

'second derivative' with respect time:

(so this is)

'm' times 'u_tt'

I'm fixing 'x' and looking at the way the little piece of string

is moving up and down..

..thinking about it as a little 'particle')

What can we conclude from all this?

Well we know that force is proportional to: 'u_tt',

it's 'm' times 'u_tt'

'Concavity' on the other hand it's simply: 'u_xx'

so the 'big' conclusion here is:

'u_tt' is proportional to 'u_xx'

and I'll write the proportionality as simply 'k': 'u_tt' = 'k'*'u_xx'.

What's the sign of 'k'?

If I'm concave 'down' my acceleration is downward..

If I'm concave 'up' my acceleration is upward..

So 'k' is positive and I could, if I wanted to..

write it as 'a square'.

I could say something like: 'k' = 'c'^2

Okay so we finished our slick little argument

and here we are..

..staring at: the 'Wave equation'.

Now there's a lot of things sort of wrong with this argument..

..it's what I call a: 'fast and loose'

So see if you can just reflect on it..

..and think about all the assumptions that are lurking in here?

(I 'll give you a couple)

well, one thing is we assumed very clearly that:

force is 'proportional to' concavity

there's no reason, a priori, that

it should be a ( simple ) proportionality.

maybe there's some other sort of functional relationship that looks like this?

Another assumption that we made is:

the motion is straight up-and-down ('vertical').

(maybe there's some side-to-side motion going on there..

..things might get much more complicated?)

Alright so there's 'lots of problems' but you can get some intuition from this.

Now let me conclude with a little 'nugget of information' about 'c'.

So what I'm going to do is just a quick

'dimensional analysis'

and this is something you usually learn in Physics classes.

(but it's also useful in math classes)

What are the 'dimensions' of 'u_tt'?

Well the dimensions of 'u' are 'length'.

(it's a displacement, so it's 'length')

The second derivative with respect to 'time'..

..the dimensions of 'u_tt' are:

'length' / 'time-squared'

So 'u_tt'..

has dimensions..

'L' / 'T'^2

On the other hand 'u_xx'..

..that's 'length' / 'length-squared'

so that's simply: '1 / L'

So what does that mean the dimensions of 'c'^2 will be?

Well that implies that in order for the 'dimensionality' of 'u_tt' to equal

the dimensionality of 'c^2'*'u_xx' we must have the

dimensions of 'c'^2 be:

'length-squared' / 'time-squared': ( 'c*c' <=> 'L*L' / 'T*T' )

Well that means that 'c' has dimensions

'length' / 'time': ( 'c' <=> 'L' / 'T' )

That's 'a velocity'.

So it's reasonable to expect that 'c' (in a 'physical' problem) should be

some kind of 'velocity'.