Practice English Speaking&Listening with: PDE 1 | Introduction

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Hello and welcome to the first video in my video series

on partial differential equations.

So I'll start with a definition, and that is:

a PDE is "an equation relating the partial derivatives of some unknown function".

So hopefully you've studied some ordinary differential equations already,

and you know that an 'ordinary differential equation' relates the 'ordinary' derivatives.

(when there's only 'one' independent variable)

Now we have (perhaps) 'many' independent variables,

so we have 'partial' derivatives

and the equation will relate those partial derivatives.

Now oftentimes the goal when we study PDEs

is to learn something about this 'unknown function'.

This unknown function we'll call: 'a solution'.

The solution is often something that we want to find an explicit formula for..

..or sometimes it's something we just want to know something about.

(we want to know some of its properties)

Now we will have three basic goals in our study of partial differential equations.

Let me describe those now:

The first goal is one that's kind of easily overlooked

(and that is..) to learn 'how to formulate' partial differential equations.

So what do I mean by this?

Well if you're studying some field of science or engineering

you often encounter some kind of dynamical process:

you may be observing data from laboratory experiments,

you may be deriving something from physical principles,

but you're going to encounter a dynamical process

..and you're going to want to convert that into a mathematical description of the process.

and that is the 'art' of formulating a PDE,

..coming from a dynamical process based on perhaps laboratory data

..and writing down a mathematical equation to describe that process.

(So this is a very useful skill)

The second goal will be pretty much

the most obvious one:

to 'solve' the partial differential equations..

That is: to find the unknown functions that satisfy the relation the parital differential equation describes.

(So we're going to spend most of our time doing this)

We won't be able to solve every PDE that we encounter,

but we will be able to develop techniques..

that will solve large classes of important equations,

and so that will be most of our focus.

Now the third goal is:

to study the solutions to partial differential equations.

Now I don't necessarily mean.. study the 'formulas' that we get from part two:

oftentimes we will have trouble obtaining formulas or..

we may get formulas which don't give us much information

So we would like to have techniques for learning things

'about' the solutions to partial differential equations

without doing a lot of hard work

to find formulas for the solutions.

(So things we can learn about a PDE

..just from the equation itself)

I'll introduce now a little bit notation that I'll use in the rest of the videos

(to be consistent)

So for independent variables I'll use things like 'x' and 'y' and 'z' and 't'.

(So these are the independent variables)

For dependent variables I'll use letters like 'u' and perhaps 'v' (if I have more than one)

..and I'll use subscripts for partial derivatives so when I write something like:

'u' (subscript) 'x',

I mean the partial derivative of 'u' with respect to 'x'

or if I write:

'u' (subscript) 'xt',

I mean the 2nd partial derivative..

with respect to the 'x' - and then with respect to 't'.

(Okay so these will be)

.." subscripts for partial derivatives "

And one thing I should point out here is

I will use letters like: 'x', 'y' and 'z' for 'spatial' coordinates..

..and something like 't' ..will be 'time'.

We will often have partial differential equations

which describe some relationship between

derivatives with respect to 'space' and with respect to 'time'

So right now we should just try to get our feet wet a little bit

by looking at a particular partial differential equation

and examples of solutions to that PDE.

So the equation we're going to look at is:

'u_t' = 'u_xx'

This equation is called the 'Heat equation'

and I'll explain in the next video why we would want to call it the 'Heat equation'

but for now let's just take it as it is:

'u_t' = 'u_xx'.

Now I claim first of all that the function:

'u' = one half 'x'^2 plus 't'

is indeed a solution

to the Heat equation.

Let's check it:

(ok: so, "check")

How do we check?

Well we take this particular function and we compute 'u_t'... and 'u_xx'

..and compare them, and see that they are equal and we know

that it will be a solution to that point.

So first let's compute 'u_t'

'u_t' is the partial derivative of 'u' with respect to 't'

Now that is ..'1'

because the partial derivative (with respect to t) of 't'..

is '1' and the partial derivative of

one half 'x'^2 with respect to 't' is '0'

Now we'll compute 'u_xx' ..but first let's compute 'u_x'.

(and then we'll take one more partial derivative)

So 'u_x' is: partial derivative of 'u' with respect x

and that is.. well, 't' goes away because the derivative of t with respect to x is '0'

and I get a: 'two times one-half x'. (so that's just 'x')

So taking one more derivative, I get 'u_xx'

which is the second partial with respect to x: (that's just '1')

So checking that the Heat equation is true

amounts to checking that: 'u_t' = 'u_xx'

(and this simply the identity '1' = '1')

so, we get a solution.

So we've checked this is a solution to the Heat equation.

Now (just for fun) let me point out

that 'this guy' right here.. this particular solution to the Heat equation.

this is called ..a 'heat polynomial'

and the reason it's called a 'heat polynomial'

is because it is a 'polynomial' in x and t and,,

.. it is a solution to the Heat equation!

There are many heat polynomials and I challenge you to find another one find a different heat polynomial.

(they're not that hard to find)

Okay now let's do something just a little bit different: let's see if we can

find a solution to the Heat equation

simply by making 'a guess'.

So this guess is often called an 'ansatz' [from German]

and it's a little smarter than 'a guess'.

What it is.. it's an assumption that the solution has a 'certain form'.

So what we're going to assume is that the solution

..has the form: 'u = e^(ax+bt)'

This kind of an assumption is a lot more powerful than you might first assume.

(and we'll see that later)

So what's the question here?

The question is: what do 'a' and 'b' need to be..

in order for 'this' expression to be a solution.. the Heat equation?

So in other words:

what are 'a' and 'b'.. (or what relationship do they satisfy)?

Let's figure it out. So we plug it in to the Heat equation:

We have 'u_t' equals?

Well it's 'e^(ax+bt)' ..times (by the chain-rule)...

..the derivative of this inside function (with respect to t)

So we have: partial of '(ax+bt)' (with respect to t)

and that gives us simply: 'e^(ax+bt)' times 'b'

Now we will compute 'u_xx' (but let's do 'u_x' first.)

That gives us

'e^(ax+bt)' ..times

..partial derivative of '(ax+bt)' (now with respect to x)

So that gives us: 'e^(ax+bt)' times 'a'.

And similarly we would compute 'u_xx' to be:

'e^(ax+bt)' times 'a^2'.

So together, this information tells us that

this expression, this function, will be a solution to the Heat equation,

if the following relationship is satisfied:

So 'u_t' equals 'u_xx'

which is the same saying that:

'e^(ax+bt)' times 'b'

is equal to:

'e^(ax+bt)' times 'a^2'

(okay, well, I can cancel 'these' guys)

Let's cancel off this 'e^(ax+bt)'

Now, maybe I should say something quickly about that cancellation..

these terms cannot be '0': if I 'exponentiate'..

anything at all..

I get something which is a non-zero,

so it's perfectly legitimate to 'cancel' these terms..

(and so what i end up with is..)

'b' = 'a^2'

So if you choose any 'a' and any 'b' that satisfy this relationship

you get a solution of the Heat equation.

For instance: 'a' could be '-1' and 'b' could be '1'

(or they could both be '0')

or maybe 'a' could be 'i' ( imaginary )

and 'b' could be '-1'.

(you could think about that)

I'd like to conclude this introduction with a quick comparison

of solutions to ordinary differential equations

and solutions to partial differential equations.

So this will be just a 'comparison by example'. and a very simple one.

(but when you're starting out ..'simple is good')

So, let's start with the ODE

which will be

the 'Population equation'.

now this ODE is wriiten

'd_P (by) d_t' = 'P'.

Frequently there's a constant in front of this (solution) 'P'..

and we're going to leave it off just to keep things simple.

Now a general solution to ODE is

'P(t)' = 'C' times 'e^t'

the 'C' here is an arbitrary constant.

And this a thing that you see

very frequently when you solve

ordinary differential equations: you find arbitrary constants in the solutions.

In fact

you often will find as many arbitrary constants in the general solution as the

order of the ODE.

So it's the number of derivatives that you find in the ODE - just one derivative here.

(so just one arbitrary constant)

Now let's look at the same equation but let's just make it a PDE.

So we'll say ..a PDE that 'looks like' the Population equation.

Okay, and what do I mean by that?

Well I'll just write: 'u_t' = 'u'

and here we're going to assume that 'u'

is a function which depends on more than one independent variable.

So 'u' " depends on "

'x' and 't'.

So this is now a partial differential equation..

..and looks very similar to this ordinary differential equation.

Let's see what we can figure out about the general solution?

First of all,


that if I take

the function

(so I'll say 'question mark')

if I take the function: 'u' = 'C(e^t)'

okay so just like I took 'P'..

(Let me be explicit that this is a function 'x' and 't'.. this is 'u' ..of ..'(x,t)' )

if I take: 'u(x,t)' = 'C(e^t)'..

(then) indeed this is a solution to this equation

..because when I take the partial derivative (with respect to t)

..and I plug that in

..I will find that I get a solution to the equation

for any 'C' that I choose.

But you may ask yourself..

..why didn't I incorporate 'x' somehow? (I should be able to incorporate x)

If you think about it for a while, you realize that the partial derivative (with respect to t)

treats 'x' as a constant.

So by plugging in

'x' anywhere in this equation

it essentially acts like a constant.

So what I could do is I take this 'C' and replace it

with any function of 'x' at all.

So let's call it 'F(x)'..

,,where 'this' guy now is an arbitrary function of x.

So 'arbitrary function' (is) appearing in this general solution.

Check it and make sure that you believe that's the case:

Take the partial derivative (with respect to t)

..and compare it to the original function.

So, here in an ODE I had an 'arbitrary constant'..

here in a PDE I have an 'arbitrary function'..

appearing in the general solution.

Now this is not a 'general theorem'

but this is a phenomenon that you'll see very frequently.

You'll find arbitrary functions appearing in 'general' solutions.

So this just gives you a basic idea to the differences

between solutions to an ODE and a PDE

The Description of PDE 1 | Introduction