0:21

As a third example, we now claim that both functions

first the y is equal to the constant function 0.

And y is equal to x to the 4 over 16

are solutions of the initial value problem,

y prime is equal to xy to the one half, and

y of 0 is equal to 0, right.

0:56

A couple of minutes before, right.

Okay, okay, do you remember?

Okay, we have a differential equation y is equal to xy to the one half.

And I said, for this one, phi of x is equal to x squared

over 4 plus arbitrary constant of the n squared,

this is a general solution of it, right.

Can you remind it?

1:29

When c is equals to 0 we get this solution, x to the 4 over 16, right.

That's the solution to this differentiate equation,

that's a particular solution, okay.

1:53

Y is equal to identical 0 satisfy the given differential equation trivially.

Moreover, both the functions y is equal to 0 and

y is equal to x to the 4 over 16, satisfy this initial condition, right.

y over 0 is equal to 0, from the trivia,

force trivia from the second choice, right.

So this example shows that, okay, think about this

initial value problem y prime is equal to xy to the one half.

The initial condition y over 0 is equal to 0, right.

2:34

For this single initial value problem,

we have at least two distinct solutions say,

y is equal to 0 and y is equal to x to the 4 over 16, right.

Okay, so this example shows that some initial value

problem may have more than one solution, okay.

Therefore, there arise two very natural questions.

Say, does every initial value problem have a solution?

3:20

First question is a question on existence.

Second question is a question on uniqueness, okay,

uniqueness of the solution, okay.

3:33

The partial answer to these questions is given

by the following celebrated theorem, so

called Picard's theorem on unique existence

of a solution to initial value problem, okay.

3:55

Here we consider the first order initial value problem.

Say y prime is equal to f of x,y, satisfying the initial

condition y of x-naught equal to y-naught, okay.

x-naught and y-naught are arbitrary two numbers, okay.

4:18

If there is a rectangle capital R,

it's a collection of points x,y in the plane,

where x is between a and b, y is between c and d, okay.

Containing the initial point x-naught,y-naught, okay.

Such that, both functions f(x, y) and

df over dy(x,y) are continuous

in this rectangle R, okay.

Then the given initial value problem has a unique solution phi of x,

in some interval around x is 0, okay.

Given by x0- h from x0- h to x0 + h for

some positive constant to h, okay.

That's to celebrate the Picard's theorem on the unique

existence of a solution to initial value problems, okay.

5:30

Pay attention to these conditions, all right?

You need the open rectangle R containing

the initial point x-naught, y-naught,

in which both neither f of x and y, and df over dy x and

y are continuous in the rectangle R, okay?

Then we have a unique solution of the given initial value problem,

valid on some interval containing x-naught,

okay, That's the conclusion, okay.

Okay, let me explain the Picard's theorem through the picture, okay.

Here we have xy plane, here's x axis and

here's y axis, right, okay.

And remind our problem, our initial value

problem as y prime is equal to f of xy and

y of x-naught equal to y-naught, okay.

And the condition we require is, we need a rectangle

R containing the initial point x-naught, y-naught, right.

Containing this initial point the inside, right.

Okay, so here, right.

The rectangle is a kind of the open rectangle x is moving from a to b, right.

Okay, y is moving from c to the d, right, okay.

For example, then you get

a rectangle like this and then.

7:27

This is a rectangle R, right, open rectangle R, okay.

Containing this initial point x-naught, y-naught,

so x-naught is somewhere between a and b, right.

And y-naught is somewhere between c and d, okay.

So that the initial point is right there, right.

Okay.

8:01

What's the conditions?

Okay, we require, okay, require both f of x and

y and df over dy, x and y, okay.

Are continuous, continuous on the rectangle R, right.

Okay, I'm assuming the one, okay.

Then what was the conclusion then?

Okay, then, okay, there is a unique way,

this is the mathematical shortcut for

the theories or their existence, okay.

There is only one solution y is equal to phi of x, right, satisfying this, right.

Satisfying the above initial value problem, okay.

8:54

And valid, okay, the solution is valid on some interval including x-naught, okay.

That I express it by an open interval from x0- h 2x0 + h, right.

Okay, so here, right.

What I'm saying is the conclusion is, right.

If both conditions are satisfied then,

there is some open interval containing x0,

from x0- h to the x0 + h, right.

9:31

Over this open interval, some solution exists, right.

Unique solution exist,

which must to be a certain curve passing through this initial point, right.

Okay, passing through this initial point, okay.

I'm excluding this to end the point,

because the function field axis defined only on this opening curve, right.

So this is right here, okay and that is right there, okay.

So what is this curve, right?

What is this curve?

10:06

This curve is the graph of the y is equal to phi of x,

that is the unique solution of this given initial value problem, okay.

That's the conclusion of this Picard's theorem, okay.

Do you get some idea what the theorem means?