so welcome to the lecture series on mathematical methods and the applications so in the last
lecture we have deal applications of fourier series that how fourier series we can apply
to boundary value problems now in this lecture we will see applications of fourier
integrals and fourier transforms however the main heading is applications to fourier
transforms but in this we will see applications of fourier series fourier transforms and fourier
integrals ok now take this equation heat equation this
is a use of fourier integrals consider heat equation this now the heat equation is given
by del u by del t is equal to c square del square u by del x square ok suppose this is
the heat equation now this is a bar which extend both to infinite on both the sides
that is bar is not of finite length it is it is tends to infinite from both the sides
ok now in this case we do not have any boundary conditions like we have in finite
bar when we have x from zero to l that in that case we have a boundary condition but
when the bar is tending to infinite from both the sides we dont have we dont have any
boundary conditions however we have the initial condition which is given by u x zero is equal
to f x and of course x varying from minus infinity to plus infinity ok
so here f x is a given initial temperature of the bar now if you want to solve this problems
if you want to solve this problems here its an infinite bar so how can we proceed again
we will use separation of variables so we will take u which is function of x and
t as f x into g t so what will be del u by del t it will be nothing but f x into g
dash t and del square u by del x square will be nothing but f double dash x into g t
so when we substitute these two in this equation we will get f double dash x ok it is left
hand side we have f x into g dash x g dash t will be c square f double dash x into g
t so this implies f double dash upon f will be one by c square g dash upon g
now again this is a function of x only f is a function of x only and this is a function
of t only and both are equal so both are equal only when its a constant quantity so it is
equal to k again as as we did in the previous lectures when k is zero or k is p square so
we discard both the cases because if suppose here here we dont have any boundary condition
still if k is a positive quantity so f will be nothing but c one e k power p s plus
c two e k power minus p x which will tends to infinite as x tends to as x increases
so which is which does not have any practical significance you see that if you take a equal
to p square suppose and f double dash will be nothing but f double dash will be nothing
but p square f so this will be nothing f from here we get e k power p x plus c two e k power
minus p x and s as x increases so this increase exponentially because of the presence of this
term ok so this has no any does not have any practical significance so we discard the
the two cases when k is zero or k is equal to p square so we take a equal to minus p
square so if k equal to minus p square so f double
dash will be nothing but minus p square times f which implies f double dash plus
p square f equal to zero and this implies f will be nothing but c one cos p x plus
c two sin p x ok c one cos p x and c two sin p x and and g will be nothing but from
this condition g dash will be nothing but it is minus p square c square g so this
implies g will be nothing but some k times e k power minus p square c square t because
when we integrate both the sides we will get g equal to this
so what will be u so u x t will be nothing but the product of these two this is nothing
but e k power minus p square c square t into c one cos p x plus c two sin p x ok this
is this is not ok this is u x t yeah this is u x t with a with a variables p so we
write it like this with a variables p ok ok so we we obtain u x t p like this now since
its an infinite bar x is varying from minus infinity to plus infinity so so the solution
u x can be obtained and p is positive quantity so u x t can be obtained by integrating all
the terms zero to infinite it is e k power minus p square c square t and it is
c one cos p x plus c two sin p x into d p now this c one c two are arbitrary constants
here so we can take them as a function of p also ok so this c one c two we can take
as a function of p so this is integral zero to infinite e k power minus p square
c square t and this is c one some function of p cos p x plus c two some function of p
into sin p x whole with d p this k will merge with when we take a port of these two
this k will merge with c one and c two ok so there is no need of writing k in this expression
now now u x zero is given as f x so this implies f x will be equal to so zero to infinity
c one p cos p x because substitute t equal to zero here plus c two p sin p x whole with
d p so this is for a integral and c one p and c two p can be obtained by where c one
p and c two p are nothing but so c one p is nothing but it will be c one p will be
given by minus infinity plus infinity minus minus plus infinite f x cos p x into d x ok
it is one by pi times i think ok it is one by pi times and c two p will be nothing but
again one by pi times minus infinity to plus infinity f x sin p x into d p
so let us verify it one by pi or what so we can simplify like this f x is obtained like
this and f two and g g t is nothing but this expression so when we put it here
and integrate over this we obtain this thing now u x zero is this thing so yeah it is
so a p and b p here it is c one p and c two p ok so c one p is this expression and c two
p is this expression which we obtained here also the same thing ok now we substitute back
to this this expressions in this u x t so what we will obtain
so u x t will be zero to infinity e k power minus p square c square t and c one is nothing
but c one p is nothing but it is integral minus infinity to plus infinity one by pi
will can come out minus infinity to plus infinity it is f x and it is it is a you can
you can put it f psi or something f psi because x is also here and x is also here so it will
be a confusion so you can take f psi here f psi cos p psi and it is cos p x d
psi this term come here plus plus e k power minus p square c square t and with this minus
infinity to plus infinity again it is f psi cos or sin it is sin p psi and sin p x
d psi and whole multiplied by d p so in this way we will get by this expression
go on when we simplify this nothing but one by pi integral zero to infinity e k power
minus p square square t so it is again one more integral minus infinity to plus infinity
and it is f psi is common from both the terms and it is cos a cos b plus sin a sin b which
is nothing but cos a minus b cos a minus b or d psi into d p so this will be the required
solution of this differential equation when we have an infinite bar ok
so in this way we can solve we can solve such type of problems we can solve such type
of problems when the bar is tending to infinity from both the sides using fourier
integrals ok now come to a use of fourier transforms so let us discuss this problems
that temperature distribution u x t in a thin homogeneous infinite bar can be modelled by
the initial boundary value problem this so now how can we use fourier transforms so
we have already discuss the applications of fourier series and fourier integrals that
how can we use fourier series or fourier integrals now let us discuss fourier transforms now
in this problem in this problem it is u t it is equal to c square u x x ok u t equal
to u x x this and minus infinite less than x less than infinite and t greater than zero
so it is again an infinite bar x is varying from minus infinity to plus infinity and u
x zero is equal to f x it is given to us it is also given to us that u x t is finite as
x tending to plus or minus infinity so what will be u x t u x t is a temperature distribution
function so how can you find u x t again now since x is varying from both the sides it
is minus infinity to plus infinity so take fourier transform both the sides now here
x is varying from minus infinity to plus infinity take take fourier transform respect to x so
first where we have to identify that with because it contain two variable because u
is nothing but function of two variable so first we have to decide that respect
to which variable fourier transform is to be taken so now it is x is varying from
minus infinity to plus infinity so take fourier transform respect to x ok so when you take
fourier transforms here fourier transform of fourier transform of u t which is
nothing but fourier transform of del u by del t so it is nothing but minus into plus
infinity del u by del t of because we are taking fourier transform with respect to x
so it will be e k power minus eta omega x d omega eta omega x d x ok so since it is
free from t so we can take del by del t outside so it will be d by d t of minus infinity to
plus infinity u into e k minus eta omega x d x which is nothing but d by d t of fourier
transform of u fourier transform of u x t ok so thats why if we take fourier transform
of this side because we are taking fourier transform with respect to x so when we take
fourier transform this side this is nothing but d by d t of suppose fourier transform
of u x t is f omega t ok suppose we are fourier transform of u x t we are taking
as f omega t so it is f omega t fourier transform of u x t is f omega t which is equal to
c square times now fourier transform u x x now fourier transform we already know that
fourier transform of nth derivative of t is nothing but nth derivative of t is
nothing but eta omega e k power n and fourier transform of f t
so we are taking fourier transforms respect to x so here it will be it is double derivative
so it is eta omega square fourier transform of u x t which i taking as f omega t ok so
when we simplify this so this is d f upon d t where f is a function of omega and t is
equal to minus c square omega square into f so it is d f upon f is equals to minus c
square omega square d t so when we integrate this it will be x is equals to some k times
e k power minus c square omega square t so when we integrate both the sides we get
this expression so this f is a function of this f is a function of omega and t
so this k will be nothing but here k will be nothing but function of omega ok so in
this way we obtain we obtain f omega t which is nothing but k omega e k power minus
c square omega square t now it is given to us that u zero x u x zero is equals to f x
so u x zero is equal to f x now take fourier transform both the sides so fourier transform
of u x zero will be fourier transform of f x and will be equal to minus infinity to
plus infinity f x e k power e k power minus eta omega x into d x ok and that will be
that will be nothing but f omega zero that you can simply that because you are putting
t equal to zero because fourier transforms of u x t is u x t is f omega t and when
we put t equal to zero both the sides so f omega zero will be nothing but fourier transform
of u x zero so fourier transform of u x zero will be nothing
but f omega zero so when you take f omega zero over here u get k omega so this k omega
will be nothing but integral minus infinity to plus infinity f x e k power minus eta omega
x into d x ok so now take inverse fourier transforms both the sides in this expression
so so what i what we are having so what we obtained basically so now if you take
so now f omega t is nothing but k omega e k power minus c square omega square t here
k omega is given by this expression now taking inverse fourier transform both the side u
x t will be given by u x t will be given by one by pi one by two pi i think one by two
pi integral minus infinity to plus infinity it is f omega t e k power eta omega x into
d omega ok so that will be equal to one upon twp pi integral minus infinity to plus infinity
it is k omega e k power minus c square omega square t into e k power eta omega x d omega
now k omega is given by this expression to substitute it over here you substitute it
over here so what we will obtain it is nothing but one upon two pi integral minus infinity
to plus infinity it is again minus infinity to plus infinity suppose it is suppose
it is f f zeta ok f zeta e k power minus eta omega zeta into e k power minus c square
omega square t into e k power eta omega x d zeta into d omega ok
so when we simplify this so now we further simplify this so one upon two pi integral
minus infinity to plus infinity minus infinity to plus infinity it is f zeta and e k power
minus eta omega it is i minus x ok e k power minus c square omega square t d psi d omega
ok now e k power eta zeta is e k power minus eta zeta is cos theta minus eta sin theta
and it will be nothing but cos cos this expression is nothing but cos omega psi minus
x minus eta sin psi minus x omega times ok now this is even function omega square
given here so this is even now sin is odd in omega ok it is it is even even into odd
is odd so this will be zero from minus infinity to plus infinity so we will left with only
one term which is one upon two pi integral minus infinity to plus infinity minus infinity
to plus infinity it is f zeta into cos omega psi minus x e k power minus c square omega
square t d psi p omega so this will be the final solution of this will be the final
solution of this so hence we will obtain the solution of the given differential equation
is it ok so here we have solve the same problem
the technique which i have solved so when we simplify we obtained the same solution
which we obtained over here so thats all for this problems now see one more problem
using fourier transforms ok this is laplace equation del square u upon del x square plus
del x u upon del y square equal to zero and x is tending from minus infinity to plus infinity
u is finite u is varying from zero to pi so we will apply fourier transforms respect to
x ok so let us see how so what ia a equation given to us
now see it is u x x plus u y y equal to zero basically u is a function of x and y here
now x is varying from minus infinity to plus infinity as given and u y is varying
from zero to pi we have to apply fourier transforms respect to x so apply fourier transform both
the sides so here you apply fourier transform it is eta omega whole square fourier transform
of fourier transform of u x y which i am taking as omega y ok so fourier transform
of u x y because i am applying fourier transform respect to x
so it will be nothing but i am assuming as f omega y ok plus now why we are not taking
respect to y so it remains as it is d square upon d square y of fourier transform of omega
y is equals to zero fourier transform of u u x y ok so that will be zero ok so it will
be nothing but d square f upon d y square minus omega square f equal to zero so this
is d square minus omega square f equal to zero where d is nothing but d upon d y when
you simply so f omega y comes out to be c one e k power omega y plus c two e k power
minus omega y again c one and c two are arbitrary constant and its a function of two variable
omega and y so we can take c one as a function of omega and c two as a function of omega
ok now we will apply the conditions given to
us now u x zero is u x zero is given to us as e k power minus two x u naught t take
fourier transform of both the sides so fourier transform of u x zero will be nothing but
one upon two plus eta omega and that is nothing but when you substitute y equal to zero here
so that will be nothing but f omega zero so when you take f omega zero here so when you
take f omega zero here so f omega zero is nothing but c one plus c two which will be
equals one upon two plus eta omega ok now second condition is u x pi equal to zero
now u x pi equal to zero so when you take fourier transform both the sides u x pi it
will be zero which is equals to f of omega pi because when you substitute y equal
to pi both the side we get back to this expression ok so from here when you take f of omega
pi which is equal to zero so this implies c one e k power omega pi plus c two e k power
minus omega pi will be equal to zero now solving these two equations we can get
the values of c one and c two so c one and c two we can obtain from here ok we can easily
solve these two equations and find the values of c one and c two which we can substitute
over here ok so what are c one and c two that we can obtain ok i am not solving for c
one and c two now f x f omega y is nothing but c one omega e k power omega y plus c two
omega e k power minus omega y so here c one and c two are governed by these two equations
we can simply we can solve these two equations find the values of c one and c two ok
so the inverse take the inverse fourier transforms both the sides so it will be u x y will be
equal to one upon two pi times integral minus infinity to plus infinity f omega y e k
power eta omega x into d omega so it will be one upon two pi integral minus into plus
infinite f omega y i c one omega e k power omega y plus c two omega e k power minus omega
y into e k power eta omega x d omega so whatever c one c two we obtain from these two equations
we simply substitute it over here and we get the final solution we get the final answer
u x y of this laplace equation ok so in this way whenever we have an x or y ranging
from minus infinity to plus infinity we can solve those equations using fourier transforms
ok so thats all for this lecture so thank you