WEBVTT
Kind: captions
Language: en
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so welcome to a series of lectures on mathematical
methods and its applications we were discussing
00:00:25.100 --> 00:00:31.640
laplace transform and their properties so
in this lecture we will see that what are
00:00:31.640 --> 00:00:37.750
the initial and final value theorems on laplace
transforms and how they are important to find
00:00:37.750 --> 00:00:46.470
some laplace transforms of some functions
ok now what are what are what is initial value
00:00:46.470 --> 00:00:52.690
theorem let us see let laplace transform
of f t is suppose f p and f dash t be piecewise
00:00:52.690 --> 00:00:59.460
continuous and is of exponential order for
t non negative t greater than equal to zero
00:00:59.460 --> 00:01:02.610
ok
if limit p tends to infinity p f p exists
00:01:02.610 --> 00:01:14.140
then this theorem is states that limit p tend
to infinity p f p is nothing but limit
00:01:14.140 --> 00:01:25.510
t tend into zero f t so this is initial value
theorem now what is the proof of this theorem
00:01:25.510 --> 00:01:38.500
let us see so what is what is laplace of
f dash t it is nothing but we already know
00:01:38.500 --> 00:01:49.770
that it is nothing but p f p minus f zero
laplace f dash t is nothing but p f p minus
00:01:49.770 --> 00:02:00.620
f zero where where this f p is nothing but
laplace of f t ok so this f p is nothing but
00:02:00.620 --> 00:02:10.590
laplace of f t now let us compute laplace
of f dash t by the definition and take
00:02:10.590 --> 00:02:15.591
limit p tend in to infinity because in initial
value theorem we have to take limit p tend
00:02:15.591 --> 00:02:22.810
into infinity ok so what is laplace of f dash
t by the main definition of laplace it is
00:02:22.810 --> 00:02:36.230
nothing but zero to infinity e k power minus
p t f dash t into d t ok and this is which
00:02:36.230 --> 00:02:46.560
is equal to by this it is equals to p f p
minus f zero
00:02:46.560 --> 00:02:55.290
so take limit p tend to infinity both the
sides so it is limit p tend to infinity of
00:02:55.290 --> 00:03:06.780
zero to infinity e k power minus p t f dash
t d t will be equal to limit p tend into infinity
00:03:06.780 --> 00:03:22.430
p f p minus f zero now let us compute this
part let us compute this part now f dash t
00:03:22.430 --> 00:03:30.810
is of exponential order and is piecewise
continuous so mode of f dash t will be
00:03:30.810 --> 00:03:38.160
nothing but less than equals to some m
into e k power m alpha t there will exists
00:03:38.160 --> 00:03:43.880
some m and alpha such that mode of f dash
t will be less than equal to m into e k power
00:03:43.880 --> 00:03:48.040
alpha t
since it is exponential order alpha so
00:03:48.040 --> 00:03:57.459
we can easily write that this value the left
hand side ok the left hand side zero to infinity
00:03:57.459 --> 00:04:09.430
e k power minus p t f dash t d t and its modulus
will be less than or equals to zero to infinity
00:04:09.430 --> 00:04:17.470
e k power minus p t mode f dash t into d t
and this mode f dash t less than equal to
00:04:17.470 --> 00:04:23.880
this because it is of exponential order of
alpha so it is again less than equals to zero
00:04:23.880 --> 00:04:33.570
to infinity m into e k power minus p t into
e k power alpha t d t and this quantity is
00:04:33.570 --> 00:04:45.400
nothing but m into integral zero to infinity
e k power minus p minus alpha t into d t so
00:04:45.400 --> 00:04:58.280
this this integral is nothing but this integral
is nothing but it is m into e k power minus
00:04:58.280 --> 00:05:06.440
p minus alpha times t upon minus p minus alpha
times t zero to infinity
00:05:06.440 --> 00:05:13.900
so when p is greater than alpha so this value
is nothing but one m upon p minus alpha because
00:05:13.900 --> 00:05:21.330
when p greater than alpha and t tend to infinity
it will tends to zero and at zero this will
00:05:21.330 --> 00:05:27.810
tends to one so this will nothing but m upon
p minus alpha so basically this expression
00:05:27.810 --> 00:05:37.370
this expression is less than equal to this
expression ok and when p tending to infinity
00:05:37.370 --> 00:05:45.850
so this will definitely tend to zero so hence
this tends to zero ok because because this
00:05:45.850 --> 00:05:51.139
expression is less than this expression less
than equal to this expression m upon p
00:05:51.139 --> 00:05:58.450
minus alpha and as p tending to infinity this
expression will be less than equals to limit
00:05:58.450 --> 00:06:03.610
t tend to infinity this expression and as
p tend to infinity this is tend in to zero
00:06:03.610 --> 00:06:13.960
which is tend in to zero as p tend in to infinity
so hence we can say that limit p tending
00:06:13.960 --> 00:06:23.460
to infinity zero to infinity e k power minus
p t f dash t d t is equal to zero
00:06:23.460 --> 00:06:31.060
now using this fact in this expression what
you will obtain so if you use this fact over
00:06:31.060 --> 00:06:39.450
here so this is nothing but zero ok we have
just proved that this expression is nothing
00:06:39.450 --> 00:06:51.130
but zero so so from here we will obtain that
limit t tend into infinity p f p is nothing
00:06:51.130 --> 00:07:00.289
but f zero now when we proved this property
of laplace transform laplace transform of
00:07:00.289 --> 00:07:05.380
derivatives that laplace transform of f dash
t is nothing but this expression so there
00:07:05.380 --> 00:07:12.460
we are assume that function is continuous
for all t greater than equal to zero ok so
00:07:12.460 --> 00:07:18.720
its function is not continuous thats why we
are obtaining f zero over here because we
00:07:18.720 --> 00:07:28.319
are assuming f is continuous throughout
non negative side of real axis ok
00:07:28.319 --> 00:07:35.840
so thats why we obtain f zero over here
if it if function is not continuous t equal
00:07:35.840 --> 00:07:44.819
to zero then instead of obtaining f zero we
will obtain limit t tending to zero f zero
00:07:44.819 --> 00:07:52.530
over here f t over here instead of this expression
over here we will obtain this expression if
00:07:52.530 --> 00:08:01.400
f is not continuous at t equal to zero ok
so so basically so basically things are
00:08:01.400 --> 00:08:06.669
same if [finc/function] function is continuous
at t equal to zero so this is nothing but
00:08:06.669 --> 00:08:14.639
f zero however a function is not continuous
at t equal to zero so this is nothing but
00:08:14.639 --> 00:08:22.449
this expression nothing but f zero plus on
the right hand side of t equal to zero
00:08:22.449 --> 00:08:29.130
ok it is a right hand limit of t equal to
zero that is f zero plus
00:08:29.130 --> 00:08:35.919
so therefore to generalize this theorem
we are writing on the right hand side we are
00:08:35.919 --> 00:08:41.889
writing t tending to zero f t because if function
is continuous so right hand side will definitely
00:08:41.889 --> 00:08:47.199
be f zero if function is not continuous at
t equal to zero so right hand side is nothing
00:08:47.199 --> 00:08:53.790
but f zero plus so in both the case it will
work ok so thats why we are writing limit
00:08:53.790 --> 00:09:01.570
t tend into zero f t in the right hand side
instead of writing f zero ok so this is initial
00:09:01.570 --> 00:09:07.470
value theorem now now we will come to final
value theorem now what final value theorem
00:09:07.470 --> 00:09:14.949
state let us see if laplace transform of f
t is f p suppose and f dash t is piecewise
00:09:14.949 --> 00:09:20.980
continuous for t greater than equal to zero
and if limit tend in to infinity f t exists
00:09:20.980 --> 00:09:28.670
then this result hold or now what this
result is so this was initial value theorem
00:09:28.670 --> 00:09:38.110
initial value theorem and what what is final
value theorem it is limit p tend into to zero
00:09:38.110 --> 00:09:50.779
f p f p is nothing but limit t tend in to
infinity f t so this is final value theorem
00:09:50.779 --> 00:09:59.899
so let us now prove the final value theorem
how we will obtain final value theorem let
00:09:59.899 --> 00:10:05.559
us see
now again again we will go to the same expression
00:10:05.559 --> 00:10:16.300
what is that limit laplace transform
of f dash t is nothing but p f p minus f zero
00:10:16.300 --> 00:10:22.889
so again this f zero will be limit t tending
to zero f t if function is not continuous
00:10:22.889 --> 00:10:34.029
at t equal to zero ok and this this is this
laplace of f dash t is nothing but from the
00:10:34.029 --> 00:10:44.860
definition of laplace transforms zero to infinity
e k power minus p t f dash t into d t now
00:10:44.860 --> 00:10:54.439
we want now we want limit p tend into zero
ok this side so same here same on this side
00:10:54.439 --> 00:11:04.559
also so this is equal to p f p minus f
zero here we are assuming f is continuous
00:11:04.559 --> 00:11:14.069
at t equal to zero take limit p tend in to
zero both the sides
00:11:14.069 --> 00:11:22.820
so what will obtain limit p tend in to zero
zero to infinity e k power minus p t f dash
00:11:22.820 --> 00:11:32.220
t d t and the right hand side will remain
the same limit p tend in to zero p f p minus
00:11:32.220 --> 00:11:41.610
f zero ok so now if p tend into zero this
will tend to one so this side is nothing but
00:11:41.610 --> 00:11:49.149
this implies zero to infinity f dash t d t
which is equals to limit p tend in to zero
00:11:49.149 --> 00:11:59.160
p f p minus f zero now let us see the left
hand side what is left hand side left hand
00:11:59.160 --> 00:12:09.649
side is zero to infinity f dash t d t this
is right left hand side so what this expression
00:12:09.649 --> 00:12:18.079
is basically this is something limit k
tend into infinity zero to k something f dash
00:12:18.079 --> 00:12:26.160
t d t now since derivative integral is anti
derivative we already know this so this is
00:12:26.160 --> 00:12:35.420
nothing but limit k tending to infinity f
t from zero to k
00:12:35.420 --> 00:12:48.499
so this is nothing but limit k tend into infinity
f k minus f zero ok and which is nothing but
00:12:48.499 --> 00:12:56.820
we can simply write limit lim t tend in
to infinity f t minus f zero this we can simply
00:12:56.820 --> 00:13:02.459
write like this so this is left hand side
so left hand side is left hand side is equal
00:13:02.459 --> 00:13:09.579
to this expression so when we equate both
the sides so this will be this will be
00:13:09.579 --> 00:13:19.209
equal to from this side is nothing but limit
p tend into zero p f p minus f zero so f zero
00:13:19.209 --> 00:13:25.970
f zero cut cancels out both the sides so this
we obtain this result which is final value
00:13:25.970 --> 00:13:32.139
theorem ok if function is continuous at
t equal to zero we are having f zero over
00:13:32.139 --> 00:13:37.759
here if it is not continuous at t equal to
zero we will be having f zero plus here also
00:13:37.759 --> 00:13:41.420
here also
so in both the cases whether function is continuous
00:13:41.420 --> 00:13:48.720
or not equal to zero f zero or f zero plus
will cancel from both the sides ok so finally
00:13:48.720 --> 00:13:54.579
we will be having this expression which you
which we are calling as final value theorem
00:13:54.579 --> 00:14:01.929
ok now let us see some of the problem based
on initial final value theorem first of
00:14:01.929 --> 00:14:10.109
all let us see the simple example of f
t equal to two e k power minus six t
00:14:10.109 --> 00:14:16.239
let us illustrate this using these two theorems
so one can easily illustrate this very
00:14:16.239 --> 00:14:22.919
easily you see that first is suppose initial
value theorem suppose first is initial value
00:14:22.919 --> 00:14:30.769
theorem initial value theorem first suppose
you want to illustrate this theorem using
00:14:30.769 --> 00:14:39.949
this example for this example sorry so what
is f t f t is two e k power minus six t so
00:14:39.949 --> 00:14:49.860
what is laplace of f t very simple it is two
upon p plus six so this we are calling as
00:14:49.860 --> 00:14:58.439
f p now see initial value theorem the left
hand side of initial value theorem is limit
00:14:58.439 --> 00:15:08.600
p tending to infinity p f p
so what is limit tend into infinity p f p
00:15:08.600 --> 00:15:19.239
p into f p that is two p upon p plus six and
as p tends to infinity this will tends to
00:15:19.239 --> 00:15:32.609
it divide numerator denominator by p this
is this quantity and this will tends to two
00:15:32.609 --> 00:15:40.989
now the left hand side limit t tending
to zero f t so limit t tend into zero what
00:15:40.989 --> 00:15:48.370
is f t to e k power minus six t so it is nothing
but two so hence initial value theorem has
00:15:48.370 --> 00:15:54.870
been verified because left hand side is equal
to right hand side ok now come to final value
00:15:54.870 --> 00:16:07.029
theorem for the same problem so now the
final value theorem so what is left hand side
00:16:07.029 --> 00:16:11.959
for final value theorem limit p tend into
zero p f p
00:16:11.959 --> 00:16:20.619
so it is limit p tend into zero p f p is two
p upon p plus six and it will is equal to
00:16:20.619 --> 00:16:30.720
zero equal to zero now what is limit t tend
in to infinity f t it is limit t tend in to
00:16:30.720 --> 00:16:39.119
infinity f t is two e k power minus six t
which is obviously zero because at t tends
00:16:39.119 --> 00:16:45.809
to infinity e k power minus t tends to zero
so that will tends to zero so hence since
00:16:45.809 --> 00:16:49.310
left hand side equal to right hand side hence
the final value theorem has been verified
00:16:49.310 --> 00:16:58.699
so the first problem has been solved now come
to second problem given at f p is this ok
00:16:58.699 --> 00:17:06.569
find f zero so we already know f p which is
laplace transform of some f t ok now without
00:17:06.569 --> 00:17:14.459
finding laplace inverse we can easily find
what is f zero how using initial value theorem
00:17:14.459 --> 00:17:19.160
so here here we are assuming that function
is continuous at t equal to zero
00:17:19.160 --> 00:17:30.340
ok thats then then only we can obtain f zero
ok so what is what is f p for this problem
00:17:30.340 --> 00:17:41.039
what is f p given to us it is p plus three
upon p plus two whole square plus three square
00:17:41.039 --> 00:17:48.830
this is f p now what will be what is f zero
f zero is nothing but by the initial value
00:17:48.830 --> 00:17:57.179
theorem is f is continuous at t equal to zero
f zero is nothing but limit p tending to infinity
00:17:57.179 --> 00:18:13.820
p f p so it is nothing but limit p tending
to infinity p into f p p into f p
00:18:13.820 --> 00:18:20.050
so this you can easily see that this limit
will tends to one so it is equal to one so
00:18:20.050 --> 00:18:26.740
what is a value of f zero this one so without
finding laplace inverse of this function we
00:18:26.740 --> 00:18:34.679
can easily find out what is the value of f
zero using initial value theorem ok similarly
00:18:34.679 --> 00:18:40.580
if you want to find out limit t tend to infinity
f t so that also we can find out using final
00:18:40.580 --> 00:18:47.139
value theorem
so limit t tend in to infinity f t will be
00:18:47.139 --> 00:18:52.639
nothing but using final value theorem it is
equal to this expression limit p tend in to
00:18:52.639 --> 00:19:04.809
zero p f p and it is nothing but limit p tend
into zero p in to f p is p plus three upon
00:19:04.809 --> 00:19:15.571
p plus two whole square plus nine and it is
nothing but is equal to zero so limit t tend
00:19:15.571 --> 00:19:22.289
to infinity f t is nothing but zero so
that also we can find out using initial and
00:19:22.289 --> 00:19:30.519
final value theorems now let us solve some
problems which we are already discussed
00:19:30.519 --> 00:19:36.220
in the previous lectures that how to find
laplace transform of these functions but let
00:19:36.220 --> 00:19:40.110
us find these function using initial value
theorem let us [sim/solve] solve first problem
00:19:40.110 --> 00:19:47.190
ok so i am telling these problem so are here
because these theorems will play an
00:19:47.190 --> 00:19:56.670
important role while we solve some deferential
equations so thats why we must have information
00:19:56.670 --> 00:20:01.730
about what initial final value theorem are
and how they are important and how we can
00:20:01.730 --> 00:20:07.570
solve problem based on this ok
so what how to find laplace transform of this
00:20:07.570 --> 00:20:23.259
function zero to t it is sin u upon u d u
so so we already see that how to find laplace
00:20:23.259 --> 00:20:29.990
of this we first find laplace of sin u sin
t which is one upon p square plus one [FL]
00:20:29.990 --> 00:20:40.799
sin t upon t is giver by integral p to
infinity f f p d p ok and then for zero to
00:20:40.799 --> 00:20:47.690
t it is nothing but f p by p which is f p
is laplace of this entire function so that
00:20:47.690 --> 00:20:52.200
we have already discussed in the last lectures
now how to find laplace of these functions
00:20:52.200 --> 00:20:58.100
using initial value theorem ok so let us see
now let us let us call this whole function
00:20:58.100 --> 00:21:11.790
as f t so f t is equals to zero to t sin u
upon u d u so what is f dash u derivative
00:21:11.790 --> 00:21:18.059
of f respect to t that we can find out using
leibniz theorem so what will be the derivative
00:21:18.059 --> 00:21:26.850
of this it will be sin t by t into one minus
zero ok so that is quite obvious that this
00:21:26.850 --> 00:21:34.379
is will be this and of course one can easily
see that f zero is nothing but zero
00:21:34.379 --> 00:21:40.139
because when you take t tend into zero
or t equal to zero so this lower and upper
00:21:40.139 --> 00:21:52.059
limit both are same value will be zero ok
now here t f dash t is nothing but sin t t
00:21:52.059 --> 00:22:00.389
f dash t is nothing but sin t ok take laplace
both the sides so laplace of t f dash t is
00:22:00.389 --> 00:22:11.520
nothing but laplace of sin t ok where interested
to find out laplace of f t ok we have to find
00:22:11.520 --> 00:22:19.090
out laplace of f t using initial value theorem
ok now laplace of t in to some function is
00:22:19.090 --> 00:22:24.730
nothing but we already know that laplace of
t k power n in to some function say f t
00:22:24.730 --> 00:22:32.990
it is nothing but minus one k power n d by
d p of f p where f p is nothing but laplace
00:22:32.990 --> 00:22:38.779
transform n th derivative sorry n th derivative
of f p where f p is nothing but laplace of
00:22:38.779 --> 00:22:43.269
this f t
now here in place of f t we have f dash t
00:22:43.269 --> 00:22:50.289
so what will be this expression will be minus
one k power one n is one so it is d by d p
00:22:50.289 --> 00:23:02.330
of laplace of f dash t and what is laplace
of sin t it is one upon p square plus one
00:23:02.330 --> 00:23:13.139
now this is further equal to minus of d by
d p of laplace of f dash t is laplace of this
00:23:13.139 --> 00:23:27.850
t is p f p minus f zero is equals to one upon
p square plus one ok so this expression further
00:23:27.850 --> 00:23:40.549
can be written as minus d by d p of p f p
is equal to one upon p square plus one derivative
00:23:40.549 --> 00:23:47.690
of this this respect to the p will be zero
its a constant quantity to derivative will
00:23:47.690 --> 00:23:55.759
be zero now it is nothing but minus p f p
is equals to take integration both the sides
00:23:55.759 --> 00:24:02.080
respect to p so it is tends to inverse p plus
some arbitrary constant c because it is an
00:24:02.080 --> 00:24:10.720
indefinite integral ok now how to find c so
to find c take limit p tending to infinity
00:24:10.720 --> 00:24:16.169
both the sides ok limit p tend in to infinity
n both the sides
00:24:16.169 --> 00:24:25.830
so it is limit p tend to infinity minus p
f p it is limit p tend in to infinity tan
00:24:25.830 --> 00:24:35.019
inverse p plus c now this quantity is by the
initial value theorem is nothing but limit
00:24:35.019 --> 00:24:47.590
t tend to zero f t and limit t tend to zero
f t which is f zero here is nothing but zero
00:24:47.590 --> 00:24:56.740
ok which is nothing but zero so we can say
that this is zero because this quan[tity]
00:24:56.740 --> 00:25:02.690
this expression p f p limit t tend into infinity
is nothing but t tend to zero f t and limit
00:25:02.690 --> 00:25:12.499
t is tend to zero f t is f zero which is zero
ok so this is zero and this is pi by two plus
00:25:12.499 --> 00:25:23.030
c so c is nothing but minus pi by two so what
will be f p from here so f p will be nothing
00:25:23.030 --> 00:25:39.880
but minus one by p times tan inverse p minus
pi by two so and what is f p is laplace transform
00:25:39.880 --> 00:25:45.610
of f t so hence laplace transform of this
f t is nothing but this
00:25:45.610 --> 00:25:50.350
so basically initial final value theorem have
important to find out this arbitrary constant
00:25:50.350 --> 00:25:58.090
c otherwise it is difficult to find ok hence
this laplace transform of this f t is nothing
00:25:58.090 --> 00:26:09.620
but this expression now let us compute the
second problem this problem can be solved
00:26:09.620 --> 00:26:21.399
also using initial value theorem e k power
it is u sin square u upon cos upon u sorry
00:26:21.399 --> 00:26:32.590
upon u into d u so again you will take this
function as say f t and try to find out
00:26:32.590 --> 00:26:38.740
it laplace transform using initial value theorem
on the same lines we did earlier so take this
00:26:38.740 --> 00:26:51.159
is f t what is f t f t is zero to t e k
power u into sin square u upon u into d u
00:26:51.159 --> 00:27:04.679
ok so what is f dash t is e k power t sin
square t by t by leibniz theorem and of course
00:27:04.679 --> 00:27:17.169
limit t tend to zero f t or we are calling
as f zero is nothing but zero ok so take
00:27:17.169 --> 00:27:25.591
t in the right hand left hand side
so it is this expression take laplace transform
00:27:25.591 --> 00:27:36.230
both the sides
laplace transform of left hand side is minus
00:27:36.230 --> 00:27:45.150
one k power one d by d p of laplace transform
of f dash t now to find laplace transform
00:27:45.150 --> 00:27:49.549
of this expression first in laplace transform
of sin square t and then apply first shifting
00:27:49.549 --> 00:27:57.200
property so what is laplace transform of sin
square t laplace transform of sin square t
00:27:57.200 --> 00:28:04.460
this we can finding out laplace transform
of one minus cos two t by two and it is nothing
00:28:04.460 --> 00:28:13.559
but one by two p minus one by two in to p
upon p square plus four so it is nothing but
00:28:13.559 --> 00:28:22.429
and again apply shifting property so you
replace p by p minus one so it is one upon
00:28:22.429 --> 00:28:35.249
two times one upon p minus one minus p minus
one upon p minus one whole square plus four
00:28:35.249 --> 00:28:44.640
ok you replace p by p minus one
now minus of d by d p of it is p f p minus
00:28:44.640 --> 00:28:54.149
f zero which is one by two of one upon p minus
one minus p minus one upon p minus one whole
00:28:54.149 --> 00:29:07.990
square plus four derivative of this respect
to p zero so it is minus p f p you integrate
00:29:07.990 --> 00:29:15.309
both sides respect to p and derivative of
this is zero ok so this is one by two comes
00:29:15.309 --> 00:29:23.700
out this is log of p minus one and multiply
and divide by two so it is minus one by two
00:29:23.700 --> 00:29:35.169
log of p minus one whole square plus four
so from here f p is nothing but and plus c
00:29:35.169 --> 00:29:45.080
also because it is an indefinite integral
ok so f p is nothing but minus one by two
00:29:45.080 --> 00:29:55.940
p ok take p over here only and one by two
it is log p minus one upon under root p minus
00:29:55.940 --> 00:30:04.440
one whole square plus four and plus c
now you can use initial value theorem to find
00:30:04.440 --> 00:30:12.049
out the value of c take limit p tending to
infinity both the sides as p tend to infinity
00:30:12.049 --> 00:30:18.800
this side will tends to this side will be
equal to limit t tending to zero f t which
00:30:18.800 --> 00:30:25.070
is nothing but zero so this side will be zero
as p tend to infinity this will tends to one
00:30:25.070 --> 00:30:37.210
and log one is zero so c is nothing but zero
so c will be zero ok and f p can be find out
00:30:37.210 --> 00:30:43.659
the f p will be nothing but you divide
this by minus p so f and f p is nothing but
00:30:43.659 --> 00:30:49.260
laplace transform of this f t so hence we
can find out laplace transform of this function
00:30:49.260 --> 00:30:55.580
so similarly the third problem can also be
solved using the same concept and initial
00:30:55.580 --> 00:30:57.580
value theorem so
thank you very much