welcome to the lecture series on mathematical methods and its applications so let us discuss
some more application of fourier transforms so we have already discussed some application
in last lectures fourier series fourier integral fourier transforms now now this
this we have already discussed that how we can find out fourier derivative of
sin or cosine transform of derivatives this we have already discussed that if we want
to find out fourier cosine of f dash x so that will be nothing but omega fourier
sin of f x minus f zero and similarly if you want to find out fourier sin of f dash
x it will nothing but minus omega fourier cosine of f dash f x
so when you replace f by f f dash in the first expression and f by f dash in the second
expression you will get back to these two expression so what are what are these two
expression so let us this we have already derived in the last lectures so what is fourier
cosine of f double dash x fourier cosine of f double dash will be nothing but x dash we
have already derived it is minus omega square fourier cosine of f x minus f dash zero and
fourier sin of f double f double dash x is nothing but omega square fourier fourier
fourier sin of f x ok it is minus omega square fourier sin of this plus omega f zero
so these two results these we have already discussed in the last lectures now what how
they are important and why i am discussing this these things over here let us see now
in this problem for the equation u t equal to del u by del t is equal to two del
square u by del x square this equation we have to solve now x is greater than zero and
t is greater than zero now x is varying from zero to infinity so that means we have
either to apply a fourier cosine or fourier sin transform because if x is this x is not
varying from minus infinity to plus infinity if x is varying if any one of variable is
varying from minus infinity to plus infinity then we have to apply fourier transform
b ut this x is varying only in half range from zero to infinity so that means we have
to apply either fourier cosine or or fourier sin ok now now which we have to apply
fourier sin or cosine how we will decide we will decide by seeing the conditions if we
apply a fourier cosine in f double dash here f double dash if this this term we are
we are applying fourier transform fourier sin or cosine transform respect to x ok so
we have a double derivative so when we have a double derivative we apply suppose fourier
cosine so in this we need f dash zero so here here we need here we need del u
by del x del u by del x at t equal to zero at x equal to zero sorry ok we need the derivative
condition so which is not here it is only u equal to zero when x equal to zero the first
condition so if we now observe fourier sin transform they are we need only f zero ok
when when x is zero ok so so in this way we conclude that we have to apply fourier
sin transform in this expression ok so basically whenever we have a half range we have to
apply either fourier sin or fourier cosine when the initial conditions when in the
initial condition we have the partial derivative at x equal to zero so then we have to apply
fourier cosine otherwise we have to apply fourier sin transform
so in this particular problem we have to apply fourier sin transform so take fourier sin
transform both the sides because we have the condition when x equal to zero u equal
to zero if we have derivative condition at x equal to zero then we apply fourier sin
fourier cosine transform now take fourier sin transform both the side what will be the
fourier sin transform of del u by del t it will be nothing but zero to infinity del
u by del t of sin omega x into d omega ok so that will be nothing but d by d t because
this term is free from t so zero to infinity u into sin omega x into d omega which is nothing
but d by d t of fourier sin transform of u x t ok so now when you take fourier sin transform
both the side it will be nothing but d by d t of fourier sin transform of u x t which
is equal to two times minus omega square apply this result so fourier transform of sin transform
of u x t plus omega into zero because u is zero when x is zero
now let fourier sin transform of u x t as suppose you take it for f of s it is omega
t ok because we are applying fourier transform with respect to x so that is it is d upon
f x upon d t which is minus minus plus two omega square f s which is equals to zero so
that implies d f s upon f s is equals to minus two omega square d t when w e integrate both
sides f x is nothing but you obtain as k times e k power minus two omega square d
so again this k will be nothing but function of omega so now we have and the next condition
where when t equal to zero u equal to e k power minus x where this condition so now
apply this condition over here so what we obtain so u u x zero is e k power minus
x so apply fourier sin transform both the sides so fourier sin transform of this expression
will be nothing but fourier sin transform of e k power minus x which is nothing but
zero to infinity e k power minus x sin omega x into d omega into d x sorry it will
be d x so it will be equal to one upon one plus omega square and it is minus e k power
minus x to integrate by parts and minus e k power minus x omega cos omega x from zero
to infinity so this is nothing but one upon one plus omega
square so when x is tend to infinity both will ten to zero when x is tends to zero it
will zero it will be nothing but minus minus plus so it is omega so that will be nothing
but omega upon one plus omega square ok now when you take when you take t equal to zero
here so fourier transform fourier sin transform of u x zero will be nothing but f s omega
zero so this is nothing but basically f x omega zero
so when you take f x omega zero here f x when you substitute t equal to zero here basically
so it will be k omega and k omega is nothing but from here will be nothing but omega upon
omega upon one plus omega square ok so therefore therefore what will be f s f s omega t it
will be omega upon one plus omega square e k power minus two omega square t now take
inverse fourier sin transform both the side to find out the solution so this implies
u x t will be nothing but two upon pi integral zero to infinity omega upon one plus omega
square e k power minus two omega t and it is sin omega x d omega
so this will be the final solution of this problem ok so here is a here is a solution
the same solution i have discussed here so this is a solution two upon pi f zero to infinity
omega upon one plus omega square e k power minus two omega square d sin omega x into
d omega ok now in the next problem it is again in half range now here the half range
is in y y is varying from zero to infinity so that means we have to apply fourier sin
or fourier cosine transform with respect to y first we have to decide that for which
variable we have to apply fourier sin or cosine transform and that depends on the range here
x is in zero to l and y is varying from zero to infinity so we have to apply fourier sin
or cosine transform with respect to y number one number two we have we have to apply
fourier sin or fourier cosine that will decide by the conditions
now here we will apply fourier sin or cosine transform with respect to y and the
partial derivative with respect to y when y is zero is zero it is given to us that means
we will apply a fourier cosine transform here because when you see because when you see
this expression in fourier cosine transform the second derivative here we have the
derivative at t equal to at x equal to zero ok so that means we have to apply fourier
cosine transform when when we have the derivative term in the initial conditions
in the conditions ok so here del u by del y at y equal to
zero is given that means we have to apply fourier cosine transform so now in this problem
apply fourier cosine transform both the sides so what is the problems basically now the
problems here is del square u upon del x square plus del square u upon del y square equal
to zero so you will apply fourier cosine transform that is clear and this respect to variable
y that you have to apply apply fourier cosine transform
with respect to y so what we will obtain so x will be three
so it is d square f upon d x square now let us suppose fourier cosine transform of u x
t u x y will be nothing but suppose it is f c x omega because we are apply with respect
to y ok so x we will treat as it is so it will be f c x omega plus now to find out
now when we apply fourier cosine transform here so we will apply that result that
result that will be nothing that is nothing but minus omega square fourier sin fourier
sin cosine transform of x omega plus and minus it is del u by del y when y equal to at x
comma zero equal to zero ok this condition we have applied for fourier cosine transform
so what we obtain finally oh sorry so basically now this this term is zero so what we obtained
d square f c upon d x square minus omega square f c will be zero because this is zero
ok so this gives d square minus omega square f c is equal to zero where d is nothing but
d by d x so f c will be nothing but c one e k power or you can break in terms of hyperbolic
functions so it is sin hyperbolic omega x ok plus c two sin hyperbolic omega y and
again c one and two are the arbitrary functions and f c is a functions of x and omega so c
one and c two are the nothing but functions of omega
ok so it is c one sin hyperbolic omega ok
it is sorry sorry it is cos ok so it is cos hyperbolic omega y now now use a initial
use the conditions conditions are u zero y equal to e k power minus two y is given
to you ok you apply a fourier cosine transform both the sides so it is fourier cosine transform
of e k power minus two y so which is given as zero to infinity e k power minus two y
and it is cos omega y into d y so when you integrate it what we will obtain it is
one upon four plus omega square and again you integrate so it is minus two minus two
y it is cos omega y minus minus plus e k power minus two y omega sin omega y from zero to
infinity at infinity both will tend to zero and at
zero this is zero and this will be one and minus minus plus so it is two upon four plus
omega square so this we will obtain here as a fourier cosine transform of this term and
again when you substitute x equal to zero both the sides this will be nothing but is
equal to f c of zero omega so when you take f c of zero omega here f c of zero omega
here so sin hyperbolic zero is zero and cos hyperbolic zero is one that is nothing
but c two ok it is nothing but it is it is also actually x it is also x ok we are
putting x equal to zero so when you put x equal to zero it is zero
and it is one total c two and c two will be nothing but two upon four plus omega square
ok so in this way c two can be obtained now for c one we apply another condition another
condition is u at l y is zero now again take fourier cosine transform both the sides
so fourier cosine transform of u l y will be nothing but fourier cosine transform
of zero which is nothing but zero so and this is also equal to when you substitute x equal
to l both the sides this is nothing but f c l omega so when you find f c l omega from
here f c l omega so that will be nothing but c one sin hyperbolic omega l plus c two sin
c two cos hyperbolic omega l which is equal to zero from here
now now we know we know c two c one can be find out so what will be c one c one
will be nothing but minus c two cos hyperbolic w l upon sin hyperbolic w l and c two is nothing
but this term so it is minus two upon four plus omega square into cos hyperbolic omega
l upon sin hyperbolic omega l so in this way once we obtain c one and c two which is here
then the only thing is you take inverse fourier cosine transform both the sides to find out
the final solution ok that is the only thing remain now
so what is f c omega t f c omega t here is nothing but ok f c x omega sorry because
f c is a functions of x in omega so x and omega is nothing but c one sin hyperbolic
omega x plus c two cos hyperbolic omega x here c one and c two are given by these two
expressions so the case are inverse cosine transform both the sides so u x y will be
nothing but two upon pi integral zero to infinity integral zero to infinity and that will be
f c x omega into e k power eta eta omega y into d omega so that will be two upon pi
integral zero to infinity and f c is nothing but c one times c one is this term so this
is minus two upon four plus omega square cos hyperbolic omega l upon sin hyperbolic omega
l c one at this term into sin hyperbolic omega x hyperbolic omega x ok and plus c two c two
is two upon two upon four plus omega square into cos cos hyperbolic omega x and whole
multiplied by e k power eta omega y in o d y d omega
so we can simplify this expression and that will give the final solution of final solution
u x you which is a solution of this partial differential equation ok which is a solution
of this partial differential equation ok so hence we can solve such type of problems ok
so whenever the half range is given to us you first you first make sure that respect
to which variable you will have to apply fourier transform x or y and that will be decided
by the conditions given to us ok now you apply either fourier sin or fourier cosine transforms
and for for finding the arbitrary constants c one and c two you have the conditions given
the problem and final take initial sin or cosine transform to find out the final solution
of the given partial differential equation now so this is the this is what i have done
here also this is when you simplify this when you simplify this expression so we will get
this form of answer this one can easily solve ok so this is the final answer which
we can obtained after solving this ok ok it is ok ok sorry it is a one one thing
i have missed here it is not e k power eta y it is basically because we are
taking inverse co inverse cosine transform it will be nothing but cos omega y and
it is d y ok d omega so similarly it it is not it will be cos omega y into d y
into d omega because we are taking inverse fourier cosine transform it will be cosine
term will be here ok so that we have missed so in this way we will get the final answer
ok now the last problem the equation for the
vibration of the string is given as this and initially the string is at rest and the initial
displacement is f x now here x is ranging from minus infinity to plus infinity so that
means we have to apply fourier transform ok so initially string is at rest these are
the conditions given to us so simply apply fourier transform both the sides to solve
this problem so what what is a condition what is the problem del square u upon del t square
is equals to c square del square u upon del x square ok and initially the finite infinity
string is at rest so that means del by del t of u at x comma zero is zero and
initial displacement is f x that means x comma zero is fx
so these are the conditions given to you and x is varying from minus infinity to plus
infinity you have to solve this equation so again we will apply fourier transform technique
so you will take fourier transform both the sides respect to ok because x is varying from
minus infinity to plus infinity so that means it is d square upon d t square of f which
is equals to c square eta omega square f here f is nothing but fourier transform
of u x t which is a function of basically omega and t ok
so when you simply d square f upon d t square and it is minus one so it is plus c square
omega square f equal to zero so it is d square plus c square omega square into f equal to
zero so where d is nothing but d by d t so when you simply this so f we can obtain
as c one e k power oh it is cos because so it is cos c omega t plus c two sin c omega
t again c one and c two are the functions of omega ok
now we will apply this conditions so we will take the first condition is del u upon
d t at x comma zero is zero when you take fourier transform both the sides of this expression
we will obtain fourier transform of del u by del t at x comma zero is zero so that
means d by d t of capital f at x comma zero is zero a capital if it is f is nothing
but fourier transform of u x t ok its a it is fourier transform of u x t so that means
when you take d by x by d t that is zero at t equal to zero
so when you apply this condition over here so the derivative of this will be sin minus
sin ok when t equal to zero it will be zero and it will be cos when t equal to zero that
will be one so we will left with c two c omega which is c two c omega which is
equal to zero ok c two c omega which is equal to zero so this implies of course this implies
c two equal to zero because c one c and omega are constant so c cant be zero and omega cant
be zero c two is the only thing which can be zero
so f is nothing but c one cos c omega t ok now we have one more condition to find out
the value of c one that is u x zero equal to f x ok now now u x zero is f x so take
fourier transform both the sides so fourier transform of u x zero will be nothing but
fourier transform of f x and that will be nothing but minus infinity to plus infinity
f x into e k power minus eta omega x into d omega into d x sorry so that will be nothing
but f of because you are putting t equal to zero so that will be o f omega zero
so when you take f omega zero so that will be c one and that c one will be nothing but
this expression so that will be nothing but minus [infinity] plus infinity f x e k power
minus eta omega x into d x ok so so now now we have the value of c one and c two in
our hand so we will simply take inverse fourier transform to find out the find out the final
solution so what is f f is nothing but c one cos c omega t where c is given by this expression
take inverse fourier transform both the sides so it is u x t will be nothing but one upon
two pi integral minus [infinity] to plus infinity f omega t e k power eta omega x into d omega
that will give that will give the final answer so it is nothing but one upon two pi integral
minus infinity to plus infinity and f omega f omega t is c one into cos omega cos c
omega t and c one is given by this expression so we can substitute c one also it is f psi
e k power minus eta omega psi and cos c omega t e k power eta omega x d psi d omega
ok so that will be here now when you further simplify it so it is one upon two pi integral
minus infinity to plus infinity integral minus infinity to plus infinity f psi e k power
minus eta psi minus x into omega cos c omega t into d psi into d omega
so again when you write e k power minus eta k eta as cos theta minus eta sin theta
then cos theta cos theta will be will contain omega and which is even in omega it is also
even in omega even into even remain even however when you have a sin term ok now this can be
written as cos omega psi minus x minus eta sin omega psi minus x and this into this is
a even function in omega and this into this is an odd function in omega so this will be
zero in terms of omega so we will left with only one term that is one upon pi integral
minus infinity to plus infinity minus infinity to plus infinity as psi cos omega psi minus
x cos c cos c omega t d psi into d omega so that will be the final solution of the
given differential equation so in this way we can probe we can solve several problems
several several partial differential equation using fourier series fourier integrals or
fourier transforms so i hope in this course you have learnt lot of things about mathematical
matters lot of techniques to solve ordinary differential equation partial differential
equation laplace transforms its applications fourier series fourier transform and fourier
integrals their applications that all we have seen so i hope you might have enjoyed
the course so i wish you all the best for your bright career ahead
thank you very much