WEBVTT
Kind: captions
Language: en
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so welcome to my lecture series on mathematical
methods and its applications so we were in
00:00:25.850 --> 00:00:29.920
the last lecture we were discussing about
convolution theorem i i mean convolution
00:00:29.920 --> 00:00:36.440
function what do you mean by convolution
of two functions so i told you that convolution
00:00:36.440 --> 00:00:52.000
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
00:00:52.000 --> 00:00:59.960
is how we can define convolution of two functions
so we have already we have already
00:00:59.960 --> 00:01:07.260
defined some properties of convolution
of two functions that is it is it is its satisfy
00:01:07.260 --> 00:01:13.110
commutative property its satisfy associative
property and distributive property also
00:01:13.110 --> 00:01:19.650
we have seen that convolution of zero and
f is nothing but zero but convolution of
00:01:19.650 --> 00:01:26.140
one with f may not be one may not be f sorry
convolution of one with any function f may
00:01:26.140 --> 00:01:31.780
not be f that we have already studied now
we will come to the main topic that is the
00:01:31.780 --> 00:01:38.140
convolution theorem for laplace transforms
now what this theorem states its states that
00:01:38.140 --> 00:01:46.490
if laplace transform of f t is say f p and
laplace transform of g t is say g p then laplace
00:01:46.490 --> 00:01:51.720
transform of convolution f and g which is
nothing but laplace transform of integral
00:01:51.720 --> 00:02:01.500
zero to t f t g t minus u d u is nothing but
f p into g p that is what convolution theorem
00:02:01.500 --> 00:02:07.299
is states now it has now what what is the
proof of this theorem let us see what is
00:02:07.299 --> 00:02:12.330
a proof of convolution theorem
so convolution theorem basically states that
00:02:12.330 --> 00:02:18.590
laplace of f star g which is a convolution
of two functions f and g is nothing but is
00:02:18.590 --> 00:02:25.030
equal to laplace of so this is the convolution
definition of convolution it is zero to
00:02:25.030 --> 00:02:39.849
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
00:02:39.849 --> 00:02:49.420
transform of f t and this g p is nothing but
laplace transform of g t so now let us
00:02:49.420 --> 00:02:58.760
see the proof of this theorem convolution
theorem so what is laplace of zero to t f
00:02:58.760 --> 00:03:06.310
u g t minus u d u
so this is nothing this expression is nothing
00:03:06.310 --> 00:03:13.969
but convolution of f and g thats why we call
it convolution theorem ok so how we define
00:03:13.969 --> 00:03:20.019
laplace theorem laplace of some function f
t so laplace is nothing but it is zero to
00:03:20.019 --> 00:03:28.889
infinity e k power minus p t and instead of
f t we have this entire expression so this
00:03:28.889 --> 00:03:43.400
is nothing but integral zero to t f u g t
minus u d u and whole multiplied by d t this
00:03:43.400 --> 00:03:50.030
integral is over t so this is what we define
laplace transform of this function say
00:03:50.030 --> 00:04:00.230
capital f t ok now this can be further written
as zero to infinity zero to t e k power minus
00:04:00.230 --> 00:04:14.609
p t f u g t minus u d u d t
so this is how we can define this expression
00:04:14.609 --> 00:04:20.829
now basically we have to show that this expression
is nothing but laplace transform of f t into
00:04:20.829 --> 00:04:29.410
laplace transform of g t this we have to prove
ok so so in order to evaluate this expression
00:04:29.410 --> 00:04:35.070
we have to simplify this expression we will
use change of order of integration now this
00:04:35.070 --> 00:04:40.889
is the limit for u first we have a d u this
these are the limit for u and this is d t
00:04:40.889 --> 00:04:49.389
these are the limit for t now let us draw
the like area first what is this area
00:04:49.389 --> 00:04:57.840
represent this is u this is t now u is varying
from zero to t u equals to zero is along this
00:04:57.840 --> 00:05:07.099
line along t axis and u equal to t is this
line this is something equal to t
00:05:07.099 --> 00:05:13.039
now t is varying from zero to infinity that
is t is varying from zero to infinity means
00:05:13.039 --> 00:05:19.440
this entire expression t is varying from zero
to infinity now first we have d u here that
00:05:19.440 --> 00:05:26.830
means we have to take strip along u axis
so if you take a strip along u axis say over
00:05:26.830 --> 00:05:32.930
here either it is here or it is here because
both both expressions are in positive side
00:05:32.930 --> 00:05:38.490
it is zero to t that is in the positive side
of u it is zero to infinity that is on the
00:05:38.490 --> 00:05:44.199
positive side of t that means it lies in the
first quadrant area whatever area this this
00:05:44.199 --> 00:05:48.800
satisfy line in the first quadrant
now in the first quadrant either this is this
00:05:48.800 --> 00:05:55.379
part or this part ok now to see check whether
this is this part or this part first let us
00:05:55.379 --> 00:06:02.471
strip parallel to u axis so take a strip parallel
to u axis so i am assuming that this is this
00:06:02.471 --> 00:06:10.169
part if this is not satisfied i go directly
go to this part ok now if you take a strip
00:06:10.169 --> 00:06:15.900
parallel to u axis over here so what is u
here zero what is u here t u is varying from
00:06:15.900 --> 00:06:21.380
zero to t and t is varying from zero to infinity
so that means this is the entire region which
00:06:21.380 --> 00:06:26.699
is covered in this region covered in this
double integral
00:06:26.699 --> 00:06:34.319
now in order to change the order of integration
i have to make a strip parallel to t axis
00:06:34.319 --> 00:06:39.479
so take a strip parallel to t axis now so
take a strip parallel to t axis now suppose
00:06:39.479 --> 00:06:46.409
this is a strip parallel to t axis so this
is the further equal to now double integral
00:06:46.409 --> 00:06:53.419
ok this double integral now this is these
are the limit for t this are limit for u now
00:06:53.419 --> 00:07:04.250
what is t here t here is u what is t what
is upper limit of t it goes up to infinity
00:07:04.250 --> 00:07:12.180
and u varying from zero to infinity u varying
from u varying from zero to infinity entire
00:07:12.180 --> 00:07:20.240
reach ok
so it will be e k power minus p t f u g t
00:07:20.240 --> 00:07:32.120
minus u now it is d t into d u so this is
how we can change the order of integration
00:07:32.120 --> 00:07:39.360
first we will see that a it is d u to mark
that which region this double integral
00:07:39.360 --> 00:07:46.169
have you take a strip parallel to u axis first
ok now you take a strip parallel to u axis
00:07:46.169 --> 00:07:53.370
you fix whether this limits are covered or
not if if not then take the other region if
00:07:53.370 --> 00:07:58.930
yes then mark this region and in order to
change the order of integration take a strip
00:07:58.930 --> 00:08:06.919
parallel to t axis now ok so this is equal
to this now this expression is further equal
00:08:06.919 --> 00:08:13.949
to now in this we make a substitution take
t minus u equal to some other variable z so
00:08:13.949 --> 00:08:26.860
now let t minus u as some variable z so what
will be d t it is d z ok
00:08:26.860 --> 00:08:32.370
so this expression this expression i am talking
about this expression this expression which
00:08:32.370 --> 00:08:42.279
is u equal to zero to infinity t from u
to infinity and it is e k power minus p t
00:08:42.279 --> 00:08:52.780
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
00:08:52.780 --> 00:09:03.470
is now t when t is u z is zero and when t
is infinity z also tends to infinity so it
00:09:03.470 --> 00:09:15.960
is infinity it is e k power minus p t is nothing
but t from here is u plus z u plus z into
00:09:15.960 --> 00:09:24.700
f of u remain as it is because now we are
not changing u we are changing t ok so g of
00:09:24.700 --> 00:09:36.620
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
00:09:36.620 --> 00:09:44.560
nothing but laplace of convolution of f and
g ok so this is equal to this expression and
00:09:44.560 --> 00:09:49.600
this is equal to this expression
now it is further equal to now if this expression
00:09:49.600 --> 00:09:58.650
is further equal to so this is further equal
to zero to infinity zero to infinity again
00:09:58.650 --> 00:10:07.440
it is these are the limit for z these are
the limit for u e k power minus p u f u ok
00:10:07.440 --> 00:10:18.510
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
00:10:18.510 --> 00:10:23.960
u variable and z variables are separated and
limits are constant so we can always write
00:10:23.960 --> 00:10:36.060
it as zero to infinity e k power minus p u
f u d u into zero to infinity e k power minus
00:10:36.060 --> 00:10:48.570
p z g z d z and this is nothing but laplace
transform of f t
00:10:48.570 --> 00:10:55.860
ok you say f t or you say f u both are same
because its its a change of only variable
00:10:55.860 --> 00:11:03.900
ok and this is nothing but laplace transform
of g t you call g t or g t or g z all are
00:11:03.900 --> 00:11:13.380
same ok so this is nothing but f p into g
p so we have shown that convolution of f and
00:11:13.380 --> 00:11:21.080
g which is equal to this expression is equal
to laplace of f t into laplace of g t hence
00:11:21.080 --> 00:11:27.690
we have proved this convolution theorem
now similarly we can also state on the
00:11:27.690 --> 00:11:34.910
we can also state inverse laplace transforms
for this theorem now if laplace inverse
00:11:34.910 --> 00:11:43.070
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
00:11:43.070 --> 00:11:49.850
is a product of two p function functions of
two ps i mean is nothing but convolution of
00:11:49.850 --> 00:11:57.770
f and g that is zero to t f t t g t minus
u d u that that is directly from the convolution
00:11:57.770 --> 00:12:03.250
theorem itself ok
so first let us again rewrite the convolution
00:12:03.250 --> 00:12:13.940
theorem here that is laplace transform of
zero to t f u g t minus u d u is nothing but
00:12:13.940 --> 00:12:21.551
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
00:12:21.551 --> 00:12:26.510
of g t this is by convolution theorem now
let us use convolution theorem to find out
00:12:26.510 --> 00:12:34.970
laplace inverse of some functions so the first
problem so this is nothing but p upon p square
00:12:34.970 --> 00:12:43.510
plus four whole square so this can be solved
this is equals to laplace inverse of you can
00:12:43.510 --> 00:12:51.740
easily find p square by four into one upon
p square by four you can easily split this
00:12:51.740 --> 00:12:58.590
function into two sub functions f p and g
p you can call anyone as f p you can call
00:12:58.590 --> 00:13:04.300
the other function as g p suppose this is
f p suppose i am calling this as f p and suppose
00:13:04.300 --> 00:13:11.880
i am calling this as g p
so what is laplace inverse of f p this is
00:13:11.880 --> 00:13:18.450
nothing but laplace inverse of p upon p square
plus four which is directly we can say it
00:13:18.450 --> 00:13:30.040
is cos two t so suppose you are calling
it f t ok laplace inverse of f p we call it
00:13:30.040 --> 00:13:39.350
f t now second is laplace inverse of g p and
g p is nothing but one upon p square plus
00:13:39.350 --> 00:13:52.570
four so you multiply and divide by two and
this is nothing but sin two t so this is
00:13:52.570 --> 00:14:01.690
one by two sin two t so say it is g t now
again to find out laplace inverse part of
00:14:01.690 --> 00:14:07.800
these two we use if you use convolution
theorem so this is nothing but the laplace
00:14:07.800 --> 00:14:15.810
inverse of this part is nothing but zero to
t f t g t minus u d u so laplace inverse of
00:14:15.810 --> 00:14:22.620
this expression p upon p square plus four
into one upon p square plus four is nothing
00:14:22.620 --> 00:14:30.570
but zero to t now what is f t f t is laplace
inverse of this function
00:14:30.570 --> 00:14:41.710
this is nothing but cos two t cos two u into
so it is cos two u into one by two sin
00:14:41.710 --> 00:14:52.970
two t minus u d u ok now in order to simplify
this expression this is nothing but one by
00:14:52.970 --> 00:15:11.070
four integral zero to t two times cos two
u into sin two t minus u d u so this is two
00:15:11.070 --> 00:15:22.200
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
00:15:22.200 --> 00:15:37.610
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
00:15:37.610 --> 00:15:53.980
u ok this is nothing but one by four so it
is sin two t will come out so integral
00:15:53.980 --> 00:16:05.110
of d u is u from zero to t ok and sin two
t minus four u is nothing but integral is
00:16:05.110 --> 00:16:12.590
minus cos two t minus four u with respect
of u this is minus four come in denominator
00:16:12.590 --> 00:16:19.800
and it is zero to t
so to if we simplify we will get the final
00:16:19.800 --> 00:16:26.870
answer so final answer will be nothing but
one by four sin two t into t and minus minus
00:16:26.870 --> 00:16:35.940
plus one by four multiplication with when
you take u as t it is cos two t when you take
00:16:35.940 --> 00:16:45.350
u as zero it is it is minus cos two t so
this term will cancel out so this will be
00:16:45.350 --> 00:16:50.240
the final answer of this problem now similarly
the last problem we can solved using convolution
00:16:50.240 --> 00:17:02.240
theorem so what is it is laplace inverse
of one upon p minus two into p plus two
00:17:02.240 --> 00:17:14.510
whole square so this is laplace inverse of
f p into g p so laplace inverse of f p will
00:17:14.510 --> 00:17:23.520
be nothing but laplace inverse of one upon
p minus two which is nothing but e k power
00:17:23.520 --> 00:17:33.380
two t and laplace inverse of one upon p plus
two whole square will be nothing but e k power
00:17:33.380 --> 00:17:40.310
minus two t you will apply shifting property
e k minus two t in to laplace inverse of
00:17:40.310 --> 00:17:47.340
one by p square which is nothing but t so
it is t e k power minus two t
00:17:47.340 --> 00:17:57.270
so this is you can say anyone as f t and anyone
as g t ok because convolution of two functions
00:17:57.270 --> 00:18:09.650
satisfy commutative property ok so laplace
inverse of f p into g p is same as laplace
00:18:09.650 --> 00:18:27.680
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
00:18:27.680 --> 00:18:41.560
u d u so this is by convolution theorem
you can use this also no problem ok the only
00:18:41.560 --> 00:18:49.780
thing is i have simplify it this because otherwise
t minus u comes in both the in both the
00:18:49.780 --> 00:18:56.160
expressions ok so this will be nothing but
e k power two t will come out so it is zero
00:18:56.160 --> 00:19:07.190
to t u into e k power minus four u d u and
now you will apply integration by parts so
00:19:07.190 --> 00:19:13.350
it is nothing but e k power two t
so when you integrate by a part it is u e
00:19:13.350 --> 00:19:21.450
k power minus four u upon minus four minus
this derivative integral of second is e k
00:19:21.450 --> 00:19:31.280
power minus four u upon sixteen and the whole
expression from zero to t so when you integrate
00:19:31.280 --> 00:19:37.480
from zero to t so the final answer of this
expression will be nothing but e k power two
00:19:37.480 --> 00:19:49.240
t will come out when you take when you substitute
p u as t it is minus e k power minus four
00:19:49.240 --> 00:20:00.850
t and when it is zero so it is minus minus
plus ok when when u is zero it is zero now
00:20:00.850 --> 00:20:10.250
when it is t it is minus of e k power minus
four t upon sixteen and when it is zero so
00:20:10.250 --> 00:20:20.090
minus minus plus it is one by sixteen
so i simplify this the this will be the final
00:20:20.090 --> 00:20:29.260
answer of this problem this is how we can
find out laplace inverse of this expression
00:20:29.260 --> 00:20:35.480
now let us see some more problems based
on convolution theorem which which otherwise
00:20:35.480 --> 00:20:41.500
is very difficult to actually find out ok
so let us see laplace inverse of these functions
00:20:41.500 --> 00:20:47.110
also using convolution theorem so what is
a first problem let us see laplace inverse
00:20:47.110 --> 00:20:59.940
of p upon p square plus nine whole cube now
how to find laplace inverse of this expression
00:20:59.940 --> 00:21:11.180
ok so so it can be break into two functions
suppose one function is p upon p square upon
00:21:11.180 --> 00:21:25.170
nine and suppose other function is one upon
p square plus nine whole square now if you
00:21:25.170 --> 00:21:32.600
split the in to this form if you split into
this form so finding laplace inverse of this
00:21:32.600 --> 00:21:38.530
is difficult
we take p over here and one over here ok so
00:21:38.530 --> 00:21:46.910
that because otherwise one upon p square plus
nine whole square laplace inverse is difficult
00:21:46.910 --> 00:21:58.130
ok so ok now now let us suppose we already
know that laplace inverse of this is suppose
00:21:58.130 --> 00:22:07.480
this f p and suppose this is g p so laplace
inverse of f p we already know it is nothing
00:22:07.480 --> 00:22:16.870
but one by three sin three t and that we can
easily find out ok now you to find laplace
00:22:16.870 --> 00:22:26.280
inverse of this expression we have two methods
either we again apply convolution theorem
00:22:26.280 --> 00:22:34.840
in this expression separately find laplace
inverse of g p which again can be written
00:22:34.840 --> 00:22:43.520
as function of a product of two functions
are you getting my point i i want to say
00:22:43.520 --> 00:22:53.780
that let us write this as p upon p square
by nine in two one upon p square plus nine
00:22:53.780 --> 00:22:59.200
ok and suppose this is some function this
is some function and again apply convolution
00:22:59.200 --> 00:23:05.050
theorem to find out laplace inverse of this
expression or the other way out is whatever
00:23:05.050 --> 00:23:09.691
matters we have studied whatever properties
we have studied for laplace transforms find
00:23:09.691 --> 00:23:15.770
out laplace inverse of this function using
those properties and directly substitute it
00:23:15.770 --> 00:23:21.330
over here
so choice is choice is ours ok so let us find
00:23:21.330 --> 00:23:28.510
out laplace inverse of this function using
the earlier properties which we have already
00:23:28.510 --> 00:23:40.550
discussed so suppose though this is suppose
this is suppose some some k p function
00:23:40.550 --> 00:23:46.390
of k p lets suppose because f p we have already
used we are using some other function of p
00:23:46.390 --> 00:23:57.060
is one upon p square plus nine ok so k
dash p is nothing but minus two p upon p square
00:23:57.060 --> 00:24:09.860
plus nine whole square
so now laplace inverse of k dash p is nothing
00:24:09.860 --> 00:24:20.620
but minus two times laplace inverse of p upon
p square plus nine whole square you want to
00:24:20.620 --> 00:24:34.700
find out laplace inverse of this expression
so what is what is f dash ok ok a laplace
00:24:34.700 --> 00:24:43.150
of t f t is nothing but minus one k power
one f dash p so actually we have to use this
00:24:43.150 --> 00:24:49.040
property ok actually we have to use this property
so laplace inverse of f dash p is nothing
00:24:49.040 --> 00:25:01.260
but minus t into f t and f t is laplace inverse
of k p if minus two laplace inverse of p upon
00:25:01.260 --> 00:25:12.429
p square plus nine whole square so minus minus
cancels out ok so this is nothing but t
00:25:12.429 --> 00:25:19.850
into now what is laplace inverse of this expression
this is one by three sin three t so it is
00:25:19.850 --> 00:25:32.920
two laplace inverse of this expression so
laplace inverse of this expression is nothing
00:25:32.920 --> 00:25:42.320
but t by six sin three t so this is laplace
inverse of this expression
00:25:42.320 --> 00:25:53.500
so basically laplace inverse of g p is nothing
but t by six sin three t so this you can also
00:25:53.500 --> 00:25:58.120
find out using convolution theorem i told
you this we can easily find out in convolution
00:25:58.120 --> 00:26:06.170
theorem also ok now now to find laplace inverse
of the product of these two that is f p into
00:26:06.170 --> 00:26:15.160
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
00:26:15.160 --> 00:26:24.530
nothing but zero to t now the first
function is this this is f u u by six sin
00:26:24.530 --> 00:26:38.170
three u into into g t minus u d u into g t
minus u that is one by three sin three t minus
00:26:38.170 --> 00:26:49.170
u into d u
so this is nothing but u by ok one by eighteen
00:26:49.170 --> 00:26:54.059
you again multiply and divided by two integrate
this so it is one by thirty six i can write
00:26:54.059 --> 00:27:01.460
here one by thirty six it is zero to t u now
when you multiply and divided by two it is
00:27:01.460 --> 00:27:07.950
two sin a sin b which is nothing but cos a
minus b minus cos a plus v so it is cos a
00:27:07.950 --> 00:27:18.160
minus b cos a minus b is nothing but six
u minus three t and minus cos a plus b so
00:27:18.160 --> 00:27:28.280
a plus b is nothing but three t into d u so
now this you can easily integrate using method
00:27:28.280 --> 00:27:35.820
of using by parts and you can easily find
out laplace inverse of the first problem
00:27:35.820 --> 00:27:40.730
ok now
now second problem let us say second problem
00:27:40.730 --> 00:27:53.320
ok laplace inverse of second problem laplace
inverse of p square upon p square plus one
00:27:53.320 --> 00:28:04.010
whole square into p square plus four again
we have to split into two in such a way that
00:28:04.010 --> 00:28:09.570
laplace inverse of both the functions you
can easily find out then only we can apply
00:28:09.570 --> 00:28:16.090
convolution theorem ok so it is p square
upon p square plus one whole square into
00:28:16.090 --> 00:28:24.320
p square plus four so it is laplace inverse
of p upon p square plus so either you can
00:28:24.320 --> 00:28:31.640
split in to in to three and apply convolution
theorem two times it is again it is our choice
00:28:31.640 --> 00:28:36.440
all you split into two and [lapl/laplace]
laplace theorem only once i mean convolution
00:28:36.440 --> 00:28:43.350
theorem only once so this you can do like
this this into this and this into this
00:28:43.350 --> 00:28:51.340
now laplace inverse of this we already know
this is nothing but cos two t which is same
00:28:51.340 --> 00:29:01.790
which is f t or g t anything ok now laplace
inverse of this ok laplace inverse of this
00:29:01.790 --> 00:29:08.809
again you can find out using the previous
problems as in the previous problems we did
00:29:08.809 --> 00:29:17.010
finding f dash ok we can easily find out
because you can take f p as one upon
00:29:17.010 --> 00:29:22.820
p square plus one and then find the derivative
as f dash p is equals to minus two p upon
00:29:22.820 --> 00:29:28.890
p square plus one whole square then taking
the laplace inverse both the side it is minus
00:29:28.890 --> 00:29:36.150
t f p is equals to two minus two times laplace
inverse of p upon p square plus one whole
00:29:36.150 --> 00:29:42.720
square and minus t f t is nothing but laplace
inverse of this is sin t it equal to minus
00:29:42.720 --> 00:29:49.480
two laplace inverse of p upon p square plus
one whole square
00:29:49.480 --> 00:29:57.780
so laplace inverse of this is nothing but
t sin t by two so this is where this is f
00:29:57.780 --> 00:30:10.830
t so to find out laplace inverse of these
two is nothing but zero to t f t and g t minus
00:30:10.830 --> 00:30:21.110
f u sorry f u the first always involved u
f u g t minus u that is cos of two t minus
00:30:21.110 --> 00:30:27.690
u into d u so one can easily integrate it
again by integration by parts and find
00:30:27.690 --> 00:30:31.190
out the laplace inverse of this expression
so
00:30:31.190 --> 00:30:31.960
thank you very much