Hello, this is the first lecture of the set
of 20 lectures on calculus of variations.
This
lecture, I will start with the introduction
of calculus of variation. Here in this, I
will be following the reference book or the
books. This is calculus of variations with
applications. On calculus of variations, we
will be covering many aspects of calculus
of variations, how the study of calculus of
variation started, and what are the applications
of the theory developed for this, that is
what we will be discussing in this.
So, in this first lecture, we will give some
introductions of problems arising in calculus
of variations. The study of calculus of variation
started in 1696 by John Bernoulli. He proposed
a problem, which is known as Brachisto chrone
problem. In this problem, what he proposed
to find path along which the particle slides
under gravity then it takes the least time.
So, mathematically this problem can be stated
like this, that in a vertical plane, you have
2 points A and B, not at the same level at
different levels. And then if you join this
by a smooth curve like this and if particle
slides from A to B under the influence of
gravity then suddenly the time taken by this
particle from coming from A to B is function
of this y, where y is this curve as a function
of x. So, this is the x axis, and this is
y axis and this y is a function of x smooth
function of x.
So that at each point, a tangent is defined
and so here this time taken by the particle
under the influence of gravity sliding from
A to the point B is denoted by t y and this
will be given by here, let us say this is
the point x 1 and this is the point x 2. Here,
so this point is (x 1, y 1) and this point
is (x 2, y 2). Then here this is actually
can be seen. This is like integral from x
1 to x 2 d s upon the velocity, that is what
will d s is the element arc length along this
curve, the integral is from x 1 to x 2 square
root of 1 plus y prime square upon this the
d s distance and upon the velocity which is
a function of y. And this functional is what
called the time taken by the particle sliding
from A to B under the influence of gravity.
Here, this time, here will be actually given
by this integral x 1 to x 2 square root of
1 plus y prime means d y by d x upon the velocity.
Here this actually can be seen. This is like
a integral from x 1 to x 2 d s upon the velocity
that is what will d s is the element arc length
along this curve. So that d s, the time taken
from this point to this point, so in this
element area d s upon the velocity that will
be the time taken by this particle coming
from this point to this point. Now here, so
if this total time taken by this particle
from A to B will be given by the integral
x 1 to x 2 integral of d s upon v. So, d s
is actually a square root of 1 plus y prime
square and the velocity v will be taken as
a function of y. So, that is the time taken
by the particle from side of a sliding from
point A to B.
Now, this is what is called the problem of
Brachisto chrone and we are asked to find
a curve along which the states the least time.
So, that is what is called the problem of
quickest decent quickest decent. So, we have
to find y such that time taken in sliding
time taken by the particle the particle 
in sliding under gravity from A to B is least.
So, this is the problem of Brachisto chrone.
Now, this problem was proposed by the famous
mathematician John Bernoulli and it was in
the same year 1696 was solved by several mathematicians
like Newton, Lebanese, Weierstrass, and John
Bernoulli himself and some other mathematicians
of that time. So, that the problem of calculus
of variation is started with that. Now, there
are many other problems, which can be posed
as the problem of calculus of variations
like you are in the plane. Suppose you have
a 2 points A and B and then the length of
the curve is the function y of x. We know
that the length of this curve is given by
the integral x 1 to x 2 square root of 1 plus
y prime square d x which is again actually
given like x 1 to x 2 d s. So, again here
d s is the element area element length in
terms of arc length s. So, integration of
this d s element length from x 1 to x 2, here
again this is point x 1 and this point is
x 2. So, this is (x 1, y 1) and (x 2, y 2).
So, this is what this integral x 1 to x 2
square root of 1 plus y prime square d x which
is nothing but the integral of d s element
length x 1 to x 2. So, this gives the length
of the curve y. So, this is a function of
y. If we change y then its length will change.
Now the problem of calculus of variation in
relation to this is to find the curve such
that the length is least. So, we know the
answer is that the state line joining these
2 points will be having. So, straight line
joining A and B has least length, but how
we prove it mathematically? Again it is a
problem of calculus of variations.
And next problem can be thought as suppose
you have this surface area double integral
over D then you have surface area d s, now
which is bounded by so this surface area over
D. So, this is what is the smallest surface
area what I mean the curve is given to you
and then here let us say this is s surface
bounded by this curve and now this length
is length of the curve is fixed. Now we want
to find what is that surface which will have
the least surface area? We know that it will
be that flat area, if it is a plane curve.
So, this is also a problem of calculus of
variation.
Now so with this, here the other problem of
surface of area for a given surface z x y,
it can be written as double integral D square
root of 1 plus z x square plus z y square
d x d y. So, here we have to choose that surface
S surface z x y such that this has least surface
area, so this also a problem of calculus of
variations. Now, here this is certainly a
generalization of a minimization or maximization
of a curve, where you want minimization, maximization
of points at which the function has minimum
or maximum; for example, here if you consider
this curve.
So, let us say this is a point a, and this
is the point b here. So, this f is defined
from a to b into x into a to b into R and
this function has at least continuous and
we may like to have more smoothness property
on this function like differentiability at
all points in the interval [a, b]. Here we
can see that this, at this point f a is maximum
of f x between a to b. So, f a is the maximum
here and f b, similarly f b is a minimum of
f x between. So, this is a global maximum
f a is global maximum and f b is a global
minimum then there are other points like this,
let us say this is point x 1, and this is
point x 2, this is point x 3. So, here in
this, f of x 1; similarly f of x 2 and f of
x 3, now these are the points of local minimum
or maximum.
For example, f, this is local minimum, because
in the neighborhood of this, here these are
the least value. Similarly, in the neighborhood
of x 2, here this has the maximum value. So,
this is local maximum. Similarly, f 3 is also
local minimum, but we have this, f a as global
maximum and f b is global minimum. So, at
these points, here we can see that the if
the function is a smooth, if it does not have
corners like this like this, if it is a smooth
function like this, where the derivative exists
then we can see that here tangent becomes
horizontal. Similarly, the tangent becomes
here horizontal; that means, derivative is
zero, a prime at this point x 1 is zero, a
prime at x 2 is 0. So, a prime at x 1 equal
to zero, similarly a prime at x 2 is zero
and a prime at x 3 is also 0. So, these are
actually if the function has continuous derivative
at all points in the interval a to b then
we can see that at these local minimum or
local maximum, we have the derivative of these
functions becoming 0. So, here given a functional
like this.
So, we have a functional; for example, here
l of y where l is a function from this is
function space. So, we call it the admissible
function class from here interval a to b.
So, this into R, where A is the admissible
class a is called the admissible class of
functions. So, l y is actually called functional.
So, l y, l is for each y here, so each y in
a b A a to b, we have associated l y which
is a function in, which is the number R. So,
given a function in this admissible class,
this l which is called functional is a function
of functioning loosely speaking. So, here
given any so here the argument of l is a function
actually whereas, in the previous case, here
this f was from number to number. Here in
this case, l is from a given function to a
number. So, it associates a function to a
number. So, such a thing is called functional.
So, l is called functional.
For example, this length of y given by this
x 1 to x 2 square root 1 plus y prime square
d x. this gives you the length of y. So, l
for a given function continuously differentiable
from 
the set or the from the function space we
call it C 1 x 1 to x 2 from the space C 1
to C 2. C 1 means continuously differentiable
the set of the space of all continuously differentiable
function from the interval x 1 to x 2 into
R. So, this is the space of all continuously
differentiable function, let us say usually
we will consider C k, k being integer nonnegative
integer C k a to b which will denote the space
of all continuous functions whose k eth derivative
will be continuous. So, for k equal to 1,
we will call it C 1 x 1 to x 2 and the values
are real numbers. So, we are considering only
the real valued functions here, and so an
element here will be called an admissible
function here. So, this is the admissible
class like a comma b. So, here this a is x
1, b is x 2 and this A here is actually C
1. So, that is the class of that is the admissible
class of functions for this functional to
l to be defined here. So, here in this case,
for any given function y from this is space
l assigns a number l y. So, that is what is
the actually functional which assigns to each
function y in this is space admissible space
of functions here for this functional to be
defined.
If we have odd derivatives appearing here
then we will take R order function spaces.
So, this for given y, the length of y will
be given by this integral here. So, if we
change for example, here you have this x 1,
x 2 here. So, this the point A x 1 y 1, and
this the point B x 2 y 2, and this the curve
y x here, which joins these 2 points a and
b. So, this functional will give the length
of this y, if you change this let say this
is y 1, if we consider the another function
y 2 here. So, then we will l of y 1 will be
given by integral x 1 to x 2 square root 1
plus y 1 prime square x d x.
Similarly, l of y 2 will be integral x 1 to
x 2 square root 1 plus y 2 prime x square
d x like that. So, we change this y, this
l will be changed. Now we want to find what
is that y for which we get the least number,
which will be the least length? The length
of the curve for joining these 2 points a
to b such that its length is the least one.
We know the answer that it has to be a straight
line joining these 2 points, but how do we
prove that. So, that will be a problem of
calculus of variations. So, now before going
into the calculus of variations, we need to
develop certain tools for that. Similarly,
we will be having this higher order.
So, here the functional will be of this type
like the integral I going from x 1 to x 2
certain function x, certain function f, which
is function of these variables x is an independent
variable y as a function of x and then d y
by d x that is y prime x d x. Now, again here
this f is function of 3 variables x, y, and
y prime. So, here and these functions like
these arguments of course, x is an independent
variable which is varying, so x is lying between
x 1 to x 2, and y is a function from x 1 to
x 2 into R such that y prime and y being C
1 x 1 to x 2 like this. So, that y prime will
also be a function from x 1 to x 2 and it
will be continuous. So, that y prime is continuous
here and this f is continuous of all its arguments.
So, there are 3 arguments for this function
x, y and y prime and here this f is a continuous
function of this variable, this variable and
this and y in turn is continuous function
between x 1 to x 2 and y prime is also a continuous
function. So, over all this integral becomes
a continuous function from x 1 to x 2, and
then by this remain hypothesis, we know that
this integral I exist as a remain integral.
So, here the problem of calculus of variation
is then to find y function y such that it
is derivative is also continuous and this
value is minimum or maximum. So, we call such
a thing as optimum value optimal value and
why we call as optimal curve for or extremely
is called extremal and this integral I will
be optimal. So, these are the terms we use
that what is that extremal for which I is
actually having the value extremal value.
So, such a functions y and which are from
the admissible class. Here admissible class
is C 1 class of functions from x 1 to x 2
into R.
So, here the generalization of this could
be like I x 1 to x 2 and f has more variables
now like x, y 1 x, y 2 x and similarly this
y n x and then there derivatives y 1 prime
x, y 2 prime x and so on up to y n prime x
d x. So this, the generalization of this,
so this can be treated as a particular case
of the more general problem where again you
have these y 1, y 2, y , they are from the
admissible class like, you, we are considering
here derivatives. So, they can, derivative
should be continuous here in order this integral
to exist as a remain integral. So, here f
will be actually a continuous function of
all these arguments and these depend this
x is an independent variable here and y 1,
y 2, y n they are dependent variables and
these derivatives are also appearing here.
So, again the problem of calculus of variation
is to find these functions y 1, y 2, y n and
such that this integral has either minimum
value or maximum value depending upon the
problem posed here and in addition to this
like, we have in the this case, we have the
curve joining these 2 points a and b. So,
here the condition is that it should pass
through (x 1, y 1) and (x 2, y 2). So that
means, here y 1 at x 1 should be this y 1
here or y at, so y at, so if y joins this.
So y, so this y at x 1 should be y 1. Similarly,
y at x 2 should be y 2.
So, that this curve passes through those given
points so and these points are fixed here.
Similarly here, this y 1 x 1 must be some
given value like, so again if you have several
curves y 1, y 2, y n and they should pass
through the these given points A and B. So,
this is y 1 x 1. There will be conditions
on this like, you have y 1 y 1. Similarly
y 1 x 2 will be y 2 and so this should be
for all those y 2 x 1 will be rather y 1 1,
y 1 0. You can put here y 1 1, y 2 at x 2
is y 2 1 and so on. You have y n x 1 as y
n 1 and y n x 2 is y n 2. So, these are the
additional conditions to be satisfied, so
that the curve pass through those given points
here.
For that generalization could be, here you
consider higher order derivatives. So, I again
it will be. So, in the previous case, this
I is a function of I is a function of y 1
y 2, and y n, and here in this case, let us
say I is a 
or a special case like you have I as y only.
But then you have higher order derivatives
appearing x 1 to x 2 f of x y y prime y double
prime, and up to y n of the derivative is
appearing there. So, like this. So, again
here the curve y is expected to it is passing
through those given points A and B here. So,
this is y x and it has a more smoothness now.
So, admissible class has to be c n x 1 to
x 2. So, y has to be actually in this space
of functions c n. So, up to nth order derivative
should exist and it should be a continuous
function from x 1 to x 2. So, again here you
will have certain additional conditions like
y at x 1 is y 1, and y at x 2 equal to y 2
like 0 0. And then y prime at x 1 is something
y 1 1, y prime at x 2 is y 1 2 and so on,
there will be up to y n minus 1 at x 1 y n
minus 1 1 ,and y n minus 1 at x 2 y n minus
1 2.
So, these are additional conditions given.
So, that this y has to satisfy all these conditions
and it has to optimize this integral depending
upon whether to minimize or to maximize. Then
further generalization could be like I y.
Here you have more independent variables,
for example, you have let say 2. So, then
you have double integral and f is function
of let say x 1 x 2, and then some function
of x 1 x 2, and then dou z by dou x 1 x 1
x 2, and dou z by dou x 2 x 1 x 2 and dx.
Here you have two-dimensional situation here,
this is x 1, x 2 and there is some domain
d here and on this, z is the surface given
like this. So, this is the surface z equal
to z x y. So far a given any point, here this
is the value z x y x y. Here this is z x y
and z x y, this point is x y and z x y. Let
us say this is z here like this.
So, here this function f is a smooth function
here of all these arguments and this z x y
is denoting this surface here. So, here again
the problem of sorry this should be z here.
So, this integral I is a function of z, if
you change this surface this value is going
to change and we need to find that z such
that it actually the boundaries fixed here.
So, long this, z will be satisfying certain
boundary condition that z has the fixed boundary
here and that z is actually optimizing this
functional.
So, that is the condition of that is the problem
of optimization, which we will be considering
here. So, we will be considering these many
such cases where we will have one of these
situations here. Now, we will consider certain
preliminaries of the concepts, which will
be required subsequently.
So, we start with preliminaries. So, here
we had been considering continuity of a function.
So, what do you mean by that? A function f
from this x 1 to x 2 to R is called continuous,
if continuous at x equal to x 0, if this limit
extending to x 0 of f x exist and is equal
to f of x 0. For example, here you consider
this function. So, here this is x 1, this
x 2 and then there is some point here x 0
here.
So, here you can see that let us say, if we
consider this is a straight line except at
that this point x 0, function is not defined
here. So, it cannot be continuous, because
it has to be defined at that point. So, let
us say at this, the function is here. So,
at x, at all these points, function is like
this and here all these points on the right,
the function is like this up to x 2. But at
this point x 0, this value is here. So, this
is, although the limit extending to x 0 exist
and here that will not be equal to the value
of the function which is actually this. So,
this is a discontinuous function.
Again the other example is, suppose the function
is 1 here and this point onwards 2 here. So,
here up to 0 to 1 the value is so f x equal
to 1 0 less than x less than 1 and it is 2
for 1 less than equal to x and let us say
up to 2 1 to 2. So, this function is discontinuous
at this point. Here the left limit exist,
right limit exist, but the that is not equal
to the value of the function here. So, the
left limit here is 1 at this point. So, f
x limit extending to x 0 minus that is what.
So, this limit extending to 1 minus means
if you approach this point 1 from the left
side. So, this will be actually equal to 1
and if you take limit extending to 1 plus;
that means, you approach from this side. So,
of f x that we know that we are approaching
from this approaching to this point to from
this side like this, it will go to 2. So,
these two are not equal and therefore, this
limit does not exist and therefore, the function
is discontinuous which can be seen intuitively
also from this figure that there is a discontinuity
here.
Although here in this case, it is not so apparent
here, but then we can see that if we approach
to this point from the left, then it is going
to this value here and similarly if you approach
to this point like this from here. So, values
are approaching along this and it is going
to this. So, these left limit and right limit
they are same and so limit exist, but it is
not equal to the value of the function, which
is also a condition for continuity here, because
f of x 0 is here f of x 0 is here, which is
not here. So, therefore there is a discontinuous.
So, these kinds of things should not happen
for a continuous function.
So, these are the concepts of left limit and
right limit and the value of the function
they should all be equal. So, then we say
that the function is continuous. When discontinuity
can occur in many ways? But we will be considering
only discontinuous function of the first kind.
So that, here what we have, only jumped discontinuities
like this. So, in this case, we have left
limit and right limit exists and they can
have the difference here. So, that difference
is called the jump here. At this point, the
jump is this, the right you are approaching
to this point, and from the left we are approaching
to this point here. Therefore, there is a
jump discontinuity here.
So, that is what is called the discontinuity
of the first kind. The discontinuity of second
kind, where limit does not exist at all, one
of the limits left or right does not exist.
Such a thing will be called discontinuity
of the second kind. We will not be considering
such functions and we will consider only the
functions having only discontinuities of the
first kind. So, that they have jumps at a
finitely many points. So, such functions are
called piecewise continuous function.
So, we have piecewise continuous function.
So, these are the functions like, you have
left and right limits existing in the interval
like this x 1 x 2. So, here this is the point
of discontinuity, this is the point of discontinuity
here. So, let us say at a and b in the interval
x 1 to x 2. Here at this point, there is a
jump discontinuity and the here also there
is a jump discontinuity of this. So, we can
have a finitely many in any in any bounded
interval 
like x 1, x 2; there are finitely many points
of jump discontinuities. So, there are there
are only, like in this interval, there are
only 2 points here at this point and this
is at the point a and at the point b, there
are jump discontinuities here.
So, this is a piecewise, this is an example
of piecewise smooth function. So, in any interval
of finite length, there should be only finitely
many points where the function can have jump
discontinuities. So, such a function is called
piecewise continuous function. Similarly,
we will have the piecewise differentiable
function. So, piecewise differentiable function
means here you have the function will be continuous
function is continuous on x 1 to x 2 and a
prime is piecewise continuous. So, a prime
has only jump discontinuities at finitely
many points in the interval x 1 to x 2. So,
such a function is called piecewise differentiable,
if it is, if the function is continuous on
x 1 to x 2 and a prime has left derivative
and right derivative existing at all points
and there are only finitely many points where
they differ. And it will eliminate the case
of the points, where the left derivative and
the right derivative are not equal only if
they are equal then it will be actually equal
to the value of the function.
So, if there cannot be a case, in this that
the left derivative and right derivative are
equal, but they are not equal to the value
of the derivative the value of the derivative
at that point. So, such a case will be avoided
in this case, because function is continuous
and if the left derivative, right derivatives
are equal, then the derivative at that point
will also exist and it will be equal to the
value of the function.
Next, we will be considering a partial differentiation
and total differentiation. Here we consider
a function u as a function of several variables
u, let us say x 1, x 2, x n. So u, here this
u is from R n to R or in certain domain, which
is a subset of R n and the values are real
numbers here. So, and these x 1, x 2, x n
are in turn and this x 1 or x i is function
of let us say t 1, t 2, t m. So, then we will
be considering this dou u over dou t I, it
will be actually dou u over dou x j and then
dou x j over dou t i summation over j equal
to 1 to n. So, j is running from 1 to n and
this i is running from 1 to m. So, this i
equal to 1, 2, to m.
So, these are the partial derivatives. Here
u is a function of several variables x 1,
x 2, x n and these x i’s in turn are functions
of some n variables t 1, t 2, t m. Then the
partial derivative of u with respect to these
variables t i's will be given by this kind
of chain rule, summation j equal to 1 to n
dou u over dou x j and dou x j over dou t
I and summation over this j going from 1 to
n.
Now, if the in particularly if this x i’s
are, so in particular, if x i’s are just
function of one variable, x I of t only. Then
we can have, then this derivative will be
total derivative like d u by d t and you have
this summation dou u by dou x j and then d
x j by d t then j equal to 1 to n. So, this
will, here then this u itself becomes a function
of single variable t and therefore, this partial
derivative reduces to ordinary derivative
that is known as total derivative here. Although
u is function of several variables x 1, x
2, x n, but these x i’s are now in this
case function of a single variable t and therefore,
u is a function of single variable t, we get
d u by d t as a total derivative in this case.
Now the next one would be differentiation
of integral at several places, we will have
the integrals coming into our picture as functional
and we will be required to differentiate it
with respect to its arguments appearing in
the integrant. So, here like I of certain
variable t integral x 1 function of t x 2
of t and f let us say x and t d x. So, this
is x is a variable of integration here and
this as limits x 1, x 2, but these f as well
as this x i’s are functions of t. They are
continuously or piecewise a differentiable
functions of t then this d I by d t is given
by the Lebanese rule is called that is also
denoted short I prime t, which is f at the
upper limit. You have x 2 t t and then d x
2 by d t minus f at x 1, which is function
of t t d x 1 by d t and then plus integral
x 1 t x 2 t and the dou f by dou t of x t
d x. So, this is what is called the Lebanese
rule, which can be seen by the first principle,
I prime t will be actually.
See here it can be calculated by this, what
is I prime t? I prime t is by definition,
it is limit t tending to or h tending to 0
I t plus h minus I t upon h. So, we can apply
this here. So, we will have this I t plus
h where t will be replaced by t plus h; here
x 1 t plus h, x 2 t plus h. So, that is what
we will have I t plus h will be given by x
1 t plus h x 2 t plus h f of x t plus h dx,
and then subtracting here. So I of t plus
h minus I of t will be x 1 t plus h x 2 t
plus h f of x t plus h dx minus x 1 t x 2
t f of x t dx .Then here we will break it
into 2 parts, x 1 t plus h to x 1 t, and then...
So, this first integral can be written as
x 1 t plus h to x 1 t and f of x t plus h
d x plus x 1 t to x 2 t f of x t plus h. So,
this you can take with this minus f of x t
d x and plus then you have x 2 t to x 2 t
plus h f of x t plus h d x and then dividing
by h here. So, we get 1 by h here, 1 by h
here, 1 by h, then letting h tends to 0 , we
get the first term by this, that the second
term with the, because this we will write
as minus of this thing. The first term, we
will write it as minus 1 by h x 1 t to x 1
t plus h f of x t plus h d x and this the
last one is anyway, here as it is x 2 t to
x 2 t plus h f of x t plus h dx.
And the middle one will be that 1 by h of
x 1 t 2 x 2 t. This 1 by h we will take it
inside and then f of x t plus h minus f x
t d. So, 1 by h taking inside and using the
continuity or differentiability property of
f, we can see that the last term. Here this
one will be given by which will give you the
partial derivative with respect to t here.
Like in this one and this term, with the that
minus sign, here is the limit of this, and
the first term will be the limit of this.
So, we get the, this, what is called the Lebanese
rule of differentiation of the integral. The
next tool will be the integration by parts.
Thank you very much for viewing this lecture.
