so welcome to my lecture series on mathematical methods and its applications so we were in
the last lecture we were discussing about convolution theorem i i mean convolution
function what do you mean by convolution of two functions so i told you that convolution
of f and g is nothing but is nothing but it is zero to t f u g t minus u d u so this
is how we can define convolution of two functions so we have already we have already
defined some properties of convolution of two functions that is it is it is its satisfy
commutative property its satisfy associative property and distributive property also
we have seen that convolution of zero and f is nothing but zero but convolution of
one with f may not be one may not be f sorry convolution of one with any function f may
not be f that we have already studied now we will come to the main topic that is the
convolution theorem for laplace transforms now what this theorem states its states that
if laplace transform of f t is say f p and laplace transform of g t is say g p then laplace
transform of convolution f and g which is nothing but laplace transform of integral
zero to t f t g t minus u d u is nothing but f p into g p that is what convolution theorem
is states now it has now what what is the proof of this theorem let us see what is
a proof of convolution theorem so convolution theorem basically states that
laplace of f star g which is a convolution of two functions f and g is nothing but is
equal to laplace of so this is the convolution definition of convolution it is zero to
t this is f u g t minus u d u is equal to f p into g p so this f p is nothing but laplace
transform of f t and this g p is nothing but laplace transform of g t so now let us
see the proof of this theorem convolution theorem so what is laplace of zero to t f
u g t minus u d u so this is nothing this expression is nothing
but convolution of f and g thats why we call it convolution theorem ok so how we define
laplace theorem laplace of some function f t so laplace is nothing but it is zero to
infinity e k power minus p t and instead of f t we have this entire expression so this
is nothing but integral zero to t f u g t minus u d u and whole multiplied by d t this
integral is over t so this is what we define laplace transform of this function say
capital f t ok now this can be further written as zero to infinity zero to t e k power minus
p t f u g t minus u d u d t so this is how we can define this expression
now basically we have to show that this expression is nothing but laplace transform of f t into
laplace transform of g t this we have to prove ok so so in order to evaluate this expression
we have to simplify this expression we will use change of order of integration now this
is the limit for u first we have a d u this these are the limit for u and this is d t
these are the limit for t now let us draw the like area first what is this area
represent this is u this is t now u is varying from zero to t u equals to zero is along this
line along t axis and u equal to t is this line this is something equal to t
now t is varying from zero to infinity that is t is varying from zero to infinity means
this entire expression t is varying from zero to infinity now first we have d u here that
means we have to take strip along u axis so if you take a strip along u axis say over
here either it is here or it is here because both both expressions are in positive side
it is zero to t that is in the positive side of u it is zero to infinity that is on the
positive side of t that means it lies in the first quadrant area whatever area this this
satisfy line in the first quadrant now in the first quadrant either this is this
part or this part ok now to see check whether this is this part or this part first let us
strip parallel to u axis so take a strip parallel to u axis so i am assuming that this is this
part if this is not satisfied i go directly go to this part ok now if you take a strip
parallel to u axis over here so what is u here zero what is u here t u is varying from
zero to t and t is varying from zero to infinity so that means this is the entire region which
is covered in this region covered in this double integral
now in order to change the order of integration i have to make a strip parallel to t axis
so take a strip parallel to t axis now so take a strip parallel to t axis now suppose
this is a strip parallel to t axis so this is the further equal to now double integral
ok this double integral now this is these are the limit for t this are limit for u now
what is t here t here is u what is t what is upper limit of t it goes up to infinity
and u varying from zero to infinity u varying from u varying from zero to infinity entire
reach ok so it will be e k power minus p t f u g t
minus u now it is d t into d u so this is how we can change the order of integration
first we will see that a it is d u to mark that which region this double integral
have you take a strip parallel to u axis first ok now you take a strip parallel to u axis
you fix whether this limits are covered or not if if not then take the other region if
yes then mark this region and in order to change the order of integration take a strip
parallel to t axis now ok so this is equal to this now this expression is further equal
to now in this we make a substitution take t minus u equal to some other variable z so
now let t minus u as some variable z so what will be d t it is d z ok
so this expression this expression i am talking about this expression this expression which
is u equal to zero to infinity t from u to infinity and it is e k power minus p t
then it is f u then g t minus u then it is d d t d u will be equal to u remain as it
is now t when t is u z is zero and when t is infinity z also tends to infinity so it
is infinity it is e k power minus p t is nothing but t from here is u plus z u plus z into
f of u remain as it is because now we are not changing u we are changing t ok so g of
t minus u is d z i mean g z and d t is d z and it is d u ok and this expression is
nothing but laplace of convolution of f and g ok so this is equal to this expression and
this is equal to this expression now it is further equal to now if this expression
is further equal to so this is further equal to zero to infinity zero to infinity again
it is these are the limit for z these are the limit for u e k power minus p u f u ok
and it is into e k power minus p z d z g z d z d u so since now the variables are separated
u variable and z variables are separated and limits are constant so we can always write
it as zero to infinity e k power minus p u f u d u into zero to infinity e k power minus
p z g z d z and this is nothing but laplace transform of f t
ok you say f t or you say f u both are same because its its a change of only variable
ok and this is nothing but laplace transform of g t you call g t or g t or g z all are
same ok so this is nothing but f p into g p so we have shown that convolution of f and
g which is equal to this expression is equal to laplace of f t into laplace of g t hence
we have proved this convolution theorem now similarly we can also state on the
we can also state inverse laplace transforms for this theorem now if laplace inverse
of f p is f t and laplace inverse of g p is g t then laplace inverse of f p into g p which
is a product of two p function functions of two ps i mean is nothing but convolution of
f and g that is zero to t f t t g t minus u d u that that is directly from the convolution
theorem itself ok so first let us again rewrite the convolution
theorem here that is laplace transform of zero to t f u g t minus u d u is nothing but
f t into g p and what is f p here f p is laplace transform of f t and g p is laplace transform
of g t this is by convolution theorem now let us use convolution theorem to find out
laplace inverse of some functions so the first problem so this is nothing but p upon p square
plus four whole square so this can be solved this is equals to laplace inverse of you can
easily find p square by four into one upon p square by four you can easily split this
function into two sub functions f p and g p you can call anyone as f p you can call
the other function as g p suppose this is f p suppose i am calling this as f p and suppose
i am calling this as g p so what is laplace inverse of f p this is
nothing but laplace inverse of p upon p square plus four which is directly we can say it
is cos two t so suppose you are calling it f t ok laplace inverse of f p we call it
f t now second is laplace inverse of g p and g p is nothing but one upon p square plus
four so you multiply and divide by two and this is nothing but sin two t so this is
one by two sin two t so say it is g t now again to find out laplace inverse part of
these two we use if you use convolution theorem so this is nothing but the laplace
inverse of this part is nothing but zero to t f t g t minus u d u so laplace inverse of
this expression p upon p square plus four into one upon p square plus four is nothing
but zero to t now what is f t f t is laplace inverse of this function
this is nothing but cos two t cos two u into so it is cos two u into one by two sin
two t minus u d u ok now in order to simplify this expression this is nothing but one by
four integral zero to t two times cos two u into sin two t minus u d u so this is two
sin a cos v two sin a cos b one by four integral zero to t so it is sin a plus p so sin a plus
b is two t so you can calling it as a f b and sin a plus p plus sin a minus p into d
u ok this is nothing but one by four so it is sin two t will come out so integral
of d u is u from zero to t ok and sin two t minus four u is nothing but integral is
minus cos two t minus four u with respect of u this is minus four come in denominator
and it is zero to t so to if we simplify we will get the final
answer so final answer will be nothing but one by four sin two t into t and minus minus
plus one by four multiplication with when you take u as t it is cos two t when you take
u as zero it is it is minus cos two t so this term will cancel out so this will be
the final answer of this problem now similarly the last problem we can solved using convolution
theorem so what is it is laplace inverse of one upon p minus two into p plus two
whole square so this is laplace inverse of f p into g p so laplace inverse of f p will
be nothing but laplace inverse of one upon p minus two which is nothing but e k power
two t and laplace inverse of one upon p plus two whole square will be nothing but e k power
minus two t you will apply shifting property e k minus two t in to laplace inverse of
one by p square which is nothing but t so it is t e k power minus two t
so this is you can say anyone as f t and anyone as g t ok because convolution of two functions
satisfy commutative property ok so laplace inverse of f p into g p is same as laplace
inverse of g p into f p ok so this is nothing but zero to t f u g t minus u g t minus
u d u so this is by convolution theorem you can use this also no problem ok the only
thing is i have simplify it this because otherwise t minus u comes in both the in both the
expressions ok so this will be nothing but e k power two t will come out so it is zero
to t u into e k power minus four u d u and now you will apply integration by parts so
it is nothing but e k power two t so when you integrate by a part it is u e
k power minus four u upon minus four minus this derivative integral of second is e k
power minus four u upon sixteen and the whole expression from zero to t so when you integrate
from zero to t so the final answer of this expression will be nothing but e k power two
t will come out when you take when you substitute p u as t it is minus e k power minus four
t and when it is zero so it is minus minus plus ok when when u is zero it is zero now
when it is t it is minus of e k power minus four t upon sixteen and when it is zero so
minus minus plus it is one by sixteen so i simplify this the this will be the final
answer of this problem this is how we can find out laplace inverse of this expression
now let us see some more problems based on convolution theorem which which otherwise
is very difficult to actually find out ok so let us see laplace inverse of these functions
also using convolution theorem so what is a first problem let us see laplace inverse
of p upon p square plus nine whole cube now how to find laplace inverse of this expression
ok so so it can be break into two functions suppose one function is p upon p square upon
nine and suppose other function is one upon p square plus nine whole square now if you
split the in to this form if you split into this form so finding laplace inverse of this
is difficult we take p over here and one over here ok so
that because otherwise one upon p square plus nine whole square laplace inverse is difficult
ok so ok now now let us suppose we already know that laplace inverse of this is suppose
this f p and suppose this is g p so laplace inverse of f p we already know it is nothing
but one by three sin three t and that we can easily find out ok now you to find laplace
inverse of this expression we have two methods either we again apply convolution theorem
in this expression separately find laplace inverse of g p which again can be written
as function of a product of two functions are you getting my point i i want to say
that let us write this as p upon p square by nine in two one upon p square plus nine
ok and suppose this is some function this is some function and again apply convolution
theorem to find out laplace inverse of this expression or the other way out is whatever
matters we have studied whatever properties we have studied for laplace transforms find
out laplace inverse of this function using those properties and directly substitute it
over here so choice is choice is ours ok so let us find
out laplace inverse of this function using the earlier properties which we have already
discussed so suppose though this is suppose this is suppose some some k p function
of k p lets suppose because f p we have already used we are using some other function of p
is one upon p square plus nine ok so k dash p is nothing but minus two p upon p square
plus nine whole square so now laplace inverse of k dash p is nothing
but minus two times laplace inverse of p upon p square plus nine whole square you want to
find out laplace inverse of this expression so what is what is f dash ok ok a laplace
of t f t is nothing but minus one k power one f dash p so actually we have to use this
property ok actually we have to use this property so laplace inverse of f dash p is nothing
but minus t into f t and f t is laplace inverse of k p if minus two laplace inverse of p upon
p square plus nine whole square so minus minus cancels out ok so this is nothing but t
into now what is laplace inverse of this expression this is one by three sin three t so it is
two laplace inverse of this expression so laplace inverse of this expression is nothing
but t by six sin three t so this is laplace inverse of this expression
so basically laplace inverse of g p is nothing but t by six sin three t so this you can also
find out using convolution theorem i told you this we can easily find out in convolution
theorem also ok now now to find laplace inverse of the product of these two that is f p into
g p laplace inverse of g p into f p here f p into g p both are same so f p into g p is
nothing but zero to t now the first function is this this is f u u by six sin
three u into into g t minus u d u into g t minus u that is one by three sin three t minus
u into d u so this is nothing but u by ok one by eighteen
you again multiply and divided by two integrate this so it is one by thirty six i can write
here one by thirty six it is zero to t u now when you multiply and divided by two it is
two sin a sin b which is nothing but cos a minus b minus cos a plus v so it is cos a
minus b cos a minus b is nothing but six u minus three t and minus cos a plus b so
a plus b is nothing but three t into d u so now this you can easily integrate using method
of using by parts and you can easily find out laplace inverse of the first problem
ok now now second problem let us say second problem
ok laplace inverse of second problem laplace inverse of p square upon p square plus one
whole square into p square plus four again we have to split into two in such a way that
laplace inverse of both the functions you can easily find out then only we can apply
convolution theorem ok so it is p square upon p square plus one whole square into
p square plus four so it is laplace inverse of p upon p square plus so either you can
split in to in to three and apply convolution theorem two times it is again it is our choice
all you split into two and [lapl/laplace] laplace theorem only once i mean convolution
theorem only once so this you can do like this this into this and this into this
now laplace inverse of this we already know this is nothing but cos two t which is same
which is f t or g t anything ok now laplace inverse of this ok laplace inverse of this
again you can find out using the previous problems as in the previous problems we did
finding f dash ok we can easily find out because you can take f p as one upon
p square plus one and then find the derivative as f dash p is equals to minus two p upon
p square plus one whole square then taking the laplace inverse both the side it is minus
t f p is equals to two minus two times laplace inverse of p upon p square plus one whole
square and minus t f t is nothing but laplace inverse of this is sin t it equal to minus
two laplace inverse of p upon p square plus one whole square
so laplace inverse of this is nothing but t sin t by two so this is where this is f
t so to find out laplace inverse of these two is nothing but zero to t f t and g t minus
f u sorry f u the first always involved u f u g t minus u that is cos of two t minus
u into d u so one can easily integrate it again by integration by parts and find
out the laplace inverse of this expression so
thank you very much