Hi there.

In the previous section detail above

a curve of the vector field ie for the projection of t

taking the sum of accumulation, We saw integration.

t, u and v components and , t at time t d

s d x and d y component also with your When we were coming summer following structure.

The importance of these two structures are different.

On the left, what is the event tells more beautiful.

We're taking the projection of u t.

But it is not so easy to account.

However, the calculations easy right To make a favorable construction.

Here you can see the most important issues one given on a curve

With a pair of functions u and v or from A to B with vector

when the orbit is connected integrally whether it was one of the most important issues.

This condition is quite simple We have seen that in anything.

v is the partial derivative with respect to x,

the second component i.e. the first According to the variant of partial derivatives.

According to the second variable of the first component, There is a contingency that is here,

equally independent of the orbit was happening.

In this case the vector function f from an

gradient could be produced f of this potential were saying.

u were saying u vector potential.

V x is equal to that condition and Y of f as a very simple

x to y after before After the first derivative of y with respect to x

derivative to be equal We saw that arise.

So it's actually quite behind half a page A full summary of the issues that we see as well.

All main ideas are here.

The mixed derivative of the derivative of this equation x

and derivatives of y to y the order is not important.

That and first derivatives of f be continuous with such fine condition

soft, provided simply was being achieved by running.

We called it the Clairaut's theorem.

So the following question comes to mind:What we integrals projecting length of this t

We also take care of the interest on, projecting why go integrals, we take care of?

It had a very simple reason.

Physically, they are very different things.

One of them accumulation along the curve,

One one side of the curve shows the transition to the other side.

If you think of an example of heat transfer, the other by a wall of heat

It is important to go to the side, but a electric current, because of a gas stream

If you think that accumulation along a line, integration is important.

Therefore, this is not trivial.

As important as the other.

t'l as important.

But mathematically n d y d x is negative and saw.

Place them by the the structure of the integral comes.

As you can see the changes in a row integrals in the same structure.

So here is the end of v'yl If we change it here

V replaces negative, becomes the integral over.

Now the important question of water.

How in the orbit integral with the t'l

independence was important, the same question can be asked here.

This integral orbit Is it independent?

What is this condition can be asked.

Progress following the course I told you before.

These mathematical structures are the same for given that this integral is always the

well defined physical problem You can examine this structure or

A problem of the mathematical structure You can examine this structure.

So, here, making comments on the v's, here u minus v

Reviewed by making everything we find here We can do calculations in the same way.

Of course, the interpretation of the subject is different.

But a difference in calculation impasse.

Now the question may come to mind.

I wonder if the integral of vertical flux Is it independent from orbit?

Even beyond this, two more Is it more independent from orbit

the question naturally comes to mind.

The answer to this problem is easy.

Of this integral, which is on the This integral is now mathematically

were also more favorable than the right side, be independent from orbit it

the complete right side depends on differential.

Thus a complete right side diferansiyels have a fun.

f is d.

According to this f x of f V is derived based on data,

minus derivative of f to y gives u.

Yet if we look at here f of the second row

therefore different derivatives of the equality of Clairaut's theorem so that

v is the derivative of y with respect to x derivative equal to revenue minus sign.

Therefore, we orbit possible to ensure independence.

But as you can see here slightly different conditions.

First derivative of the first component, Getting derivative with respect to the first variable,

second second component variable derivative is taken.

Now you can call the answer to the following question.

This is the first integrals, ie for t curve

by integrals along the other by a curve

integrally related to the transition to the side of the two independent of the way one be?

What is this condition?

The answer to that is open.

Because first of t'l entegra from orbit was said to be independent.

Secondly, the integration of the orbit n'l need that was providing independent,

was sufficient and necessary condition.

E provides both are This also means that if the two conditions

We need to make one.

See, this is a very easy process.

Now here first in order Let us take the derivative with respect to y,

Let us take the second derivative with respect to x and v x v y x is equal to y, again

Due to the change in the order of derivative the famous theorem of Clairaut equation again

We just found using the equation, to ensure that both conditions.

See this derivative to y We got here u y has occurred.

We got the second derivative with respect to x, X X would be negative here.

This is the second from the first subtracted

modeled either as you see them If we return to the left of sync and u x x, we

How you do, u x x plus We see that u y y is zero.

Let them second left side will be zero.

plus u x x is equal to y will be zero.

E it from laplacian anything else is not.

So if u laplacian new zero There is a second condition, of course.

X on the same operations before this time derivative here, right Y X occurs.

In the latter condition to y Get derivative of x minus y occurs.

And when we gather them together again, there

v y y x x plus zero see that.

This v is of laplacian requires to be zero.

So what does this mean?

U will be a potential for and v components thus

The vector f of the potential and be calculated as the gradient.

This will include one, but two conditions In order to do it well f

Random Function is not enough.

f the need to provide the two conditions.

One of them is very easy provided.

This mixed derivatives are equal.

This continuity in the functions always provided but

brings more than a second condition.

f y f x minus y is equal to x.

You see that the first condition for f v f y was going to come from,

derivative thereof br by further wherein x is y again for a further differentiation with respect to y.

Here u f x has to say.

Y any more Bunund Take the derivative f

x x occurs and that brings an additional requirement.

We're here because you want a lot more.

Both of the two integral multiple to be independent from orbit.

That of f, i.e. the potential of providing also requires that an additional condition.

This new zero laplacian to be required.

Such events are a bit You'll see that the complex

We will emphasize this more in value.

Already in these conditions, we see here the real part of these two conditions,

have an imaginary part of complex that it provides the essential functions

To conditions other than Cosilinum We see that there is nothing.

But there is a sign difference in it so important.

He pointed out the difference between the new We can go on defining.

Therefore, this more complicated for us functions are connected.

And complex functions exceptionally favorable

function turns out to be kind.

Because of this dependence on orbit, independence of paramount importance.

Now the complex functions 'll pass then.

But here I want to take a break.

The work we do is just t'l i.e., the cycle of the integral

The integral along the orbit not be independent.

Also integral with the N's, perpendicular to the orbit from the orbit of the integral

become independent of the conditions Cosilinum saw me come to terms.

So it complicated for us functions relating to the conversion of

now that reminds me.

In the next lesson we will see these issues.

Bye for now.