WEBVTT
Kind: captions
Language: en
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so welcome to the lecture series on mathematical
methods and the applications so in the last
00:00:23.750 --> 00:00:30.270
lecture we have deal applications of fourier
series that how fourier series we can apply
00:00:30.270 --> 00:00:36.930
to boundary value problems now in this
lecture we will see applications of fourier
00:00:36.930 --> 00:00:42.830
integrals and fourier transforms however
the main heading is applications to fourier
00:00:42.830 --> 00:00:47.870
transforms but in this we will see applications
of fourier series fourier transforms and fourier
00:00:47.870 --> 00:00:53.300
integrals ok
now take this equation heat equation this
00:00:53.300 --> 00:00:59.150
is a use of fourier integrals consider heat
equation this now the heat equation is given
00:00:59.150 --> 00:01:11.490
by del u by del t is equal to c square del
square u by del x square ok suppose this is
00:01:11.490 --> 00:01:19.240
the heat equation now this is a bar which
extend both to infinite on both the sides
00:01:19.240 --> 00:01:26.680
that is bar is not of finite length it is
it is tends to infinite from both the sides
00:01:26.680 --> 00:01:35.770
ok now in this case we do not have any boundary
conditions like we have in finite
00:01:35.770 --> 00:01:40.600
bar when we have x from zero to l that in
that case we have a boundary condition but
00:01:40.600 --> 00:01:45.900
when the bar is tending to infinite from
both the sides we dont have we dont have any
00:01:45.900 --> 00:01:54.799
boundary conditions however we have the initial
condition which is given by u x zero is equal
00:01:54.799 --> 00:02:03.820
to f x and of course x varying from minus
infinity to plus infinity ok
00:02:03.820 --> 00:02:09.890
so here f x is a given initial temperature
of the bar now if you want to solve this problems
00:02:09.890 --> 00:02:17.360
if you want to solve this problems here its
an infinite bar so how can we proceed again
00:02:17.360 --> 00:02:23.240
we will use separation of variables so we
will take u which is function of x and
00:02:23.240 --> 00:02:34.370
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
00:02:34.370 --> 00:02:45.620
dash t and del square u by del x square will
be nothing but f double dash x into g t
00:02:45.620 --> 00:02:53.640
so when we substitute these two in this equation
we will get f double dash x ok it is left
00:02:53.640 --> 00:03:03.050
hand side we have f x into g dash x g dash
t will be c square f double dash x into g
00:03:03.050 --> 00:03:13.489
t so this implies f double dash upon f
will be one by c square g dash upon g
00:03:13.489 --> 00:03:18.470
now again this is a function of x only f is
a function of x only and this is a function
00:03:18.470 --> 00:03:24.549
of t only and both are equal so both are equal
only when its a constant quantity so it is
00:03:24.549 --> 00:03:35.180
equal to k again as as we did in the previous
lectures when k is zero or k is p square so
00:03:35.180 --> 00:03:43.699
we discard both the cases because if suppose
here here we dont have any boundary condition
00:03:43.699 --> 00:03:50.739
still if k is a positive quantity so f will
be nothing but c one e k power p s plus
00:03:50.739 --> 00:03:57.459
c two e k power minus p x which will tends
to infinite as x tends to as x increases
00:03:57.459 --> 00:04:03.840
so which is which does not have any practical
significance you see that if you take a equal
00:04:03.840 --> 00:04:09.560
to p square suppose and f double dash will
be nothing but f double dash will be nothing
00:04:09.560 --> 00:04:17.570
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
00:04:17.570 --> 00:04:25.940
minus p x and s as x increases so this increase
exponentially because of the presence of this
00:04:25.940 --> 00:04:34.180
term ok so this has no any does not have
any practical significance so we discard the
00:04:34.180 --> 00:04:40.020
the two cases when k is zero or k is equal
to p square so we take a equal to minus p
00:04:40.020 --> 00:04:45.449
square
so if k equal to minus p square so f double
00:04:45.449 --> 00:04:53.340
dash will be nothing but minus p square
times f which implies f double dash plus
00:04:53.340 --> 00:05:02.860
p square f equal to zero and this implies
f will be nothing but c one cos p x plus
00:05:02.860 --> 00:05:18.699
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
00:05:18.699 --> 00:05:28.199
this condition g dash will be nothing but
it is minus p square c square g so this
00:05:28.199 --> 00:05:35.020
implies g will be nothing but some k times
e k power minus p square c square t because
00:05:35.020 --> 00:05:40.229
when we integrate both the sides we will get
g equal to this
00:05:40.229 --> 00:05:46.830
so what will be u so u x t will be nothing
but the product of these two this is nothing
00:05:46.830 --> 00:06:01.830
but e k power minus p square c square t
into c one cos p x plus c two sin p x ok this
00:06:01.830 --> 00:06:10.120
is this is not ok this is u x t yeah this
is u x t with a with a variables p so we
00:06:10.120 --> 00:06:21.099
write it like this with a variables p ok ok
so we we obtain u x t p like this now since
00:06:21.099 --> 00:06:29.300
its an infinite bar x is varying from minus
infinity to plus infinity so so the solution
00:06:29.300 --> 00:06:37.599
u x can be obtained and p is positive quantity
so u x t can be obtained by integrating all
00:06:37.599 --> 00:06:46.230
the terms zero to infinite it is e k
power minus p square c square t and it is
00:06:46.230 --> 00:06:59.259
c one cos p x plus c two sin p x into d
p now this c one c two are arbitrary constants
00:06:59.259 --> 00:07:05.349
here so we can take them as a function of
p also ok so this c one c two we can take
00:07:05.349 --> 00:07:12.569
as a function of p so this is integral
zero to infinite e k power minus p square
00:07:12.569 --> 00:07:22.560
c square t and this is c one some function
of p cos p x plus c two some function of p
00:07:22.560 --> 00:07:34.560
into sin p x whole with d p this k will
merge with when we take a port of these two
00:07:34.560 --> 00:07:39.729
this k will merge with c one and c two ok
so there is no need of writing k in this expression
00:07:39.729 --> 00:07:52.569
now now u x zero is given as f x so this implies
f x will be equal to so zero to infinity
00:07:52.569 --> 00:08:04.470
c one p cos p x because substitute t equal
to zero here plus c two p sin p x whole with
00:08:04.470 --> 00:08:13.789
d p so this is for a integral and c one p
and c two p can be obtained by where c one
00:08:13.789 --> 00:08:23.570
p and c two p are nothing but so c one p is
nothing but it will be c one p will be
00:08:23.570 --> 00:08:37.800
given by minus infinity plus infinity minus
minus plus infinite f x cos p x into d x ok
00:08:37.800 --> 00:08:45.690
it is one by pi times i think ok it is one
by pi times and c two p will be nothing but
00:08:45.690 --> 00:08:54.529
again one by pi times minus infinity to plus
infinity f x sin p x into d p
00:08:54.529 --> 00:09:02.279
so let us verify it one by pi or what so we
can simplify like this f x is obtained like
00:09:02.279 --> 00:09:08.110
this and f two and g g t is nothing but
this expression so when we put it here
00:09:08.110 --> 00:09:16.610
and integrate over this we obtain this thing
now u x zero is this thing so yeah it is
00:09:16.610 --> 00:09:22.139
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
00:09:22.139 --> 00:09:27.709
p is this expression which we obtained here
also the same thing ok now we substitute back
00:09:27.709 --> 00:09:36.339
to this this expressions in this u x
t so what we will obtain
00:09:36.339 --> 00:09:47.399
so u x t will be zero to infinity e k power
minus p square c square t and c one is nothing
00:09:47.399 --> 00:09:54.620
but c one p is nothing but it is integral
minus infinity to plus infinity one by pi
00:09:54.620 --> 00:10:06.860
will can come out minus infinity to plus infinity
it is f x and it is it is a you can
00:10:06.860 --> 00:10:13.140
you can put it f psi or something f psi because
x is also here and x is also here so it will
00:10:13.140 --> 00:10:23.930
be a confusion so you can take f psi here
f psi cos p psi and it is cos p x d
00:10:23.930 --> 00:10:34.620
psi this term come here plus plus e k power
minus p square c square t and with this minus
00:10:34.620 --> 00:10:47.269
infinity to plus infinity again it is f psi
cos or sin it is sin p psi and sin p x
00:10:47.269 --> 00:10:58.720
d psi and whole multiplied by d p
so in this way we will get by this expression
00:10:58.720 --> 00:11:04.029
go on when we simplify this nothing but one
by pi integral zero to infinity e k power
00:11:04.029 --> 00:11:11.360
minus p square square t so it is again one
more integral minus infinity to plus infinity
00:11:11.360 --> 00:11:16.790
and it is f psi is common from both the terms
and it is cos a cos b plus sin a sin b which
00:11:16.790 --> 00:11:27.759
is nothing but cos a minus b cos a minus b
or d psi into d p so this will be the required
00:11:27.759 --> 00:11:34.190
solution of this differential equation when
we have an infinite bar ok
00:11:34.190 --> 00:11:40.990
so in this way we can solve we can solve
such type of problems we can solve such type
00:11:40.990 --> 00:11:46.170
of problems when the bar is tending to
infinity from both the sides using fourier
00:11:46.170 --> 00:11:57.220
integrals ok now come to a use of fourier
transforms so let us discuss this problems
00:11:57.220 --> 00:12:03.120
that temperature distribution u x t in a thin
homogeneous infinite bar can be modelled by
00:12:03.120 --> 00:12:12.750
the initial boundary value problem this so
now how can we use fourier transforms so
00:12:12.750 --> 00:12:18.850
we have already discuss the applications of
fourier series and fourier integrals that
00:12:18.850 --> 00:12:28.430
how can we use fourier series or fourier integrals
now let us discuss fourier transforms now
00:12:28.430 --> 00:12:39.390
in this problem in this problem it is
u t it is equal to c square u x x ok u t equal
00:12:39.390 --> 00:12:47.140
to u x x this and minus infinite less than
x less than infinite and t greater than zero
00:12:47.140 --> 00:12:51.800
so it is again an infinite bar x is varying
from minus infinity to plus infinity and u
00:12:51.800 --> 00:12:58.550
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
00:12:58.550 --> 00:13:04.690
x tending to plus or minus infinity so what
will be u x t u x t is a temperature distribution
00:13:04.690 --> 00:13:11.110
function so how can you find u x t again now
since x is varying from both the sides it
00:13:11.110 --> 00:13:16.940
is minus infinity to plus infinity so take
fourier transform both the sides now here
00:13:16.940 --> 00:13:23.769
x is varying from minus infinity to plus infinity
take take fourier transform respect to x so
00:13:23.769 --> 00:13:29.959
first where we have to identify that with
because it contain two variable because u
00:13:29.959 --> 00:13:36.709
is nothing but function of two variable
so first we have to decide that respect
00:13:36.709 --> 00:13:42.230
to which variable fourier transform is to
be taken so now it is x is varying from
00:13:42.230 --> 00:13:48.269
minus infinity to plus infinity so take fourier
transform respect to x ok so when you take
00:13:48.269 --> 00:13:55.610
fourier transforms here fourier transform
of fourier transform of u t which is
00:13:55.610 --> 00:14:04.630
nothing but fourier transform of del u by
del t so it is nothing but minus into plus
00:14:04.630 --> 00:14:10.709
infinity del u by del t of because we are
taking fourier transform with respect to x
00:14:10.709 --> 00:14:24.459
so it will be e k power minus eta omega x
d omega eta omega x d x ok so since it is
00:14:24.459 --> 00:14:31.690
free from t so we can take del by del t outside
so it will be d by d t of minus infinity to
00:14:31.690 --> 00:14:41.540
plus infinity u into e k minus eta omega x
d x which is nothing but d by d t of fourier
00:14:41.540 --> 00:14:53.220
transform of u fourier transform of u x t
ok so thats why if we take fourier transform
00:14:53.220 --> 00:14:58.370
of this side because we are taking fourier
transform with respect to x so when we take
00:14:58.370 --> 00:15:05.041
fourier transform this side this is nothing
but d by d t of suppose fourier transform
00:15:05.041 --> 00:15:18.880
of u x t is f omega t ok suppose we
are fourier transform of u x t we are taking
00:15:18.880 --> 00:15:30.209
as f omega t so it is f omega t fourier transform
of u x t is f omega t which is equal to
00:15:30.209 --> 00:15:39.360
c square times now fourier transform u x x
now fourier transform we already know that
00:15:39.360 --> 00:15:47.579
fourier transform of nth derivative
of t is nothing but nth derivative of t is
00:15:47.579 --> 00:15:55.089
nothing but eta omega e k power n and fourier
transform of f t
00:15:55.089 --> 00:16:00.330
so we are taking fourier transforms respect
to x so here it will be it is double derivative
00:16:00.330 --> 00:16:13.959
so it is eta omega square fourier transform
of u x t which i taking as f omega t ok so
00:16:13.959 --> 00:16:20.490
when we simplify this so this is d f upon
d t where f is a function of omega and t is
00:16:20.490 --> 00:16:29.310
equal to minus c square omega square into
f so it is d f upon f is equals to minus c
00:16:29.310 --> 00:16:36.709
square omega square d t so when we integrate
this it will be x is equals to some k times
00:16:36.709 --> 00:16:43.560
e k power minus c square omega square t
so when we integrate both the sides we get
00:16:43.560 --> 00:16:53.350
this expression so this f is a function
of this f is a function of omega and t
00:16:53.350 --> 00:17:02.839
so this k will be nothing but here k will
be nothing but function of omega ok so in
00:17:02.839 --> 00:17:11.189
this way we obtain we obtain f omega t
which is nothing but k omega e k power minus
00:17:11.189 --> 00:17:20.230
c square omega square t now it is given to
us that u zero x u x zero is equals to f x
00:17:20.230 --> 00:17:30.890
so u x zero is equal to f x now take fourier
transform both the sides so fourier transform
00:17:30.890 --> 00:17:39.910
of u x zero will be fourier transform of
f x and will be equal to minus infinity to
00:17:39.910 --> 00:17:56.090
plus infinity f x e k power e k power minus
eta omega x into d x ok and that will be
00:17:56.090 --> 00:18:07.460
that will be nothing but f omega zero that
you can simply that because you are putting
00:18:07.460 --> 00:18:20.120
t equal to zero because fourier transforms
of u x t is u x t is f omega t and when
00:18:20.120 --> 00:18:25.541
we put t equal to zero both the sides so f
omega zero will be nothing but fourier transform
00:18:25.541 --> 00:18:28.870
of u x zero
so fourier transform of u x zero will be nothing
00:18:28.870 --> 00:18:38.070
but f omega zero so when you take f omega
zero over here u get k omega so this k omega
00:18:38.070 --> 00:18:44.720
will be nothing but integral minus infinity
to plus infinity f x e k power minus eta omega
00:18:44.720 --> 00:18:55.580
x into d x ok so now take inverse fourier
transforms both the sides in this expression
00:18:55.580 --> 00:19:09.520
so so what i what we are having so what
we obtained basically so now if you take
00:19:09.520 --> 00:19:19.160
so now f omega t is nothing but k omega e
k power minus c square omega square t here
00:19:19.160 --> 00:19:24.780
k omega is given by this expression now taking
inverse fourier transform both the side u
00:19:24.780 --> 00:19:34.240
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
00:19:34.240 --> 00:19:47.260
pi integral minus infinity to plus infinity
it is f omega t e k power eta omega x into
00:19:47.260 --> 00:19:58.140
d omega ok so that will be equal to one upon
twp pi integral minus infinity to plus infinity
00:19:58.140 --> 00:20:08.180
it is k omega e k power minus c square omega
square t into e k power eta omega x d omega
00:20:08.180 --> 00:20:14.660
now k omega is given by this expression to
substitute it over here you substitute it
00:20:14.660 --> 00:20:21.600
over here so what we will obtain it is nothing
but one upon two pi integral minus infinity
00:20:21.600 --> 00:20:29.080
to plus infinity it is again minus infinity
to plus infinity suppose it is suppose
00:20:29.080 --> 00:20:42.290
it is f f zeta ok f zeta e k power minus
eta omega zeta into e k power minus c square
00:20:42.290 --> 00:20:52.130
omega square t into e k power eta omega x
d zeta into d omega ok
00:20:52.130 --> 00:20:59.520
so when we simplify this so now we further
simplify this so one upon two pi integral
00:20:59.520 --> 00:21:09.150
minus infinity to plus infinity minus infinity
to plus infinity it is f zeta and e k power
00:21:09.150 --> 00:21:21.220
minus eta omega it is i minus x ok e k power
minus c square omega square t d psi d omega
00:21:21.220 --> 00:21:30.250
ok now e k power eta zeta is e k power minus
eta zeta is cos theta minus eta sin theta
00:21:30.250 --> 00:21:40.400
and it will be nothing but cos cos this
expression is nothing but cos omega psi minus
00:21:40.400 --> 00:21:51.690
x minus eta sin psi minus x omega times ok
now this is even function omega square
00:21:51.690 --> 00:21:59.740
given here so this is even now sin is odd
in omega ok it is it is even even into odd
00:21:59.740 --> 00:22:04.370
is odd so this will be zero from minus infinity
to plus infinity so we will left with only
00:22:04.370 --> 00:22:09.500
one term which is one upon two pi integral
minus infinity to plus infinity minus infinity
00:22:09.500 --> 00:22:20.160
to plus infinity it is f zeta into cos omega
psi minus x e k power minus c square omega
00:22:20.160 --> 00:22:26.910
square t d psi p omega so this will be the
final solution of this will be the final
00:22:26.910 --> 00:22:34.190
solution of this so hence we will obtain the
solution of the given differential equation
00:22:34.190 --> 00:22:42.080
is it ok
so here we have solve the same problem
00:22:42.080 --> 00:22:50.260
the technique which i have solved so when
we simplify we obtained the same solution
00:22:50.260 --> 00:22:56.430
which we obtained over here so thats all for
this problems now see one more problem
00:22:56.430 --> 00:23:03.720
using fourier transforms ok this is laplace
equation del square u upon del x square plus
00:23:03.720 --> 00:23:08.950
del x u upon del y square equal to zero and
x is tending from minus infinity to plus infinity
00:23:08.950 --> 00:23:14.070
u is finite u is varying from zero to pi so
we will apply fourier transforms respect to
00:23:14.070 --> 00:23:19.290
x ok so let us see how so what ia a equation
given to us
00:23:19.290 --> 00:23:31.330
now see it is u x x plus u y y equal to zero
basically u is a function of x and y here
00:23:31.330 --> 00:23:37.260
now x is varying from minus infinity to
plus infinity as given and u y is varying
00:23:37.260 --> 00:23:44.130
from zero to pi we have to apply fourier transforms
respect to x so apply fourier transform both
00:23:44.130 --> 00:23:52.370
the sides so here you apply fourier transform
it is eta omega whole square fourier transform
00:23:52.370 --> 00:24:02.070
of fourier transform of u x y which
i am taking as omega y ok so fourier transform
00:24:02.070 --> 00:24:08.630
of u x y because i am applying fourier transform
respect to x
00:24:08.630 --> 00:24:19.770
so it will be nothing but i am assuming as
f omega y ok plus now why we are not taking
00:24:19.770 --> 00:24:27.600
respect to y so it remains as it is d square
upon d square y of fourier transform of omega
00:24:27.600 --> 00:24:38.970
y is equals to zero fourier transform of u
u x y ok so that will be zero ok so it will
00:24:38.970 --> 00:24:46.660
be nothing but d square f upon d y square
minus omega square f equal to zero so this
00:24:46.660 --> 00:24:56.360
is d square minus omega square f equal to
zero where d is nothing but d upon d y when
00:24:56.360 --> 00:25:08.150
you simply so f omega y comes out to be
c one e k power omega y plus c two e k power
00:25:08.150 --> 00:25:15.470
minus omega y again c one and c two are arbitrary
constant and its a function of two variable
00:25:15.470 --> 00:25:24.800
omega and y so we can take c one as a function
of omega and c two as a function of omega
00:25:24.800 --> 00:25:28.710
ok
now we will apply the conditions given to
00:25:28.710 --> 00:25:42.400
us now u x zero is u x zero is given to us
as e k power minus two x u naught t take
00:25:42.400 --> 00:25:48.000
fourier transform of both the sides so fourier
transform of u x zero will be nothing but
00:25:48.000 --> 00:25:56.350
one upon two plus eta omega and that is nothing
but when you substitute y equal to zero here
00:25:56.350 --> 00:26:03.080
so that will be nothing but f omega zero so
when you take f omega zero here so when you
00:26:03.080 --> 00:26:09.010
take f omega zero here so f omega zero is
nothing but c one plus c two which will be
00:26:09.010 --> 00:26:17.190
equals one upon two plus eta omega ok
now second condition is u x pi equal to zero
00:26:17.190 --> 00:26:26.000
now u x pi equal to zero so when you take
fourier transform both the sides u x pi it
00:26:26.000 --> 00:26:34.050
will be zero which is equals to f of omega
pi because when you substitute y equal
00:26:34.050 --> 00:26:43.330
to pi both the side we get back to this expression
ok so from here when you take f of omega
00:26:43.330 --> 00:26:51.430
pi which is equal to zero so this implies
c one e k power omega pi plus c two e k power
00:26:51.430 --> 00:27:00.410
minus omega pi will be equal to zero
now solving these two equations we can get
00:27:00.410 --> 00:27:08.380
the values of c one and c two so c one and
c two we can obtain from here ok we can easily
00:27:08.380 --> 00:27:12.490
solve these two equations and find the values
of c one and c two which we can substitute
00:27:12.490 --> 00:27:21.470
over here ok so what are c one and c two that
we can obtain ok i am not solving for c
00:27:21.470 --> 00:27:35.170
one and c two now f x f omega y is nothing
but c one omega e k power omega y plus c two
00:27:35.170 --> 00:27:41.220
omega e k power minus omega y so here c one
and c two are governed by these two equations
00:27:41.220 --> 00:27:46.310
we can simply we can solve these two equations
find the values of c one and c two ok
00:27:46.310 --> 00:27:52.620
so the inverse take the inverse fourier transforms
both the sides so it will be u x y will be
00:27:52.620 --> 00:28:04.620
equal to one upon two pi times integral minus
infinity to plus infinity f omega y e k
00:28:04.620 --> 00:28:14.740
power eta omega x into d omega so it will
be one upon two pi integral minus into plus
00:28:14.740 --> 00:28:23.790
infinite f omega y i c one omega e k power
omega y plus c two omega e k power minus omega
00:28:23.790 --> 00:28:33.460
y into e k power eta omega x d omega so whatever
c one c two we obtain from these two equations
00:28:33.460 --> 00:28:39.000
we simply substitute it over here and we get
the final solution we get the final answer
00:28:39.000 --> 00:28:48.200
u x y of this laplace equation ok so in
this way whenever we have an x or y ranging
00:28:48.200 --> 00:28:54.590
from minus infinity to plus infinity we can
solve those equations using fourier transforms
00:28:54.590 --> 00:28:59.320
ok so thats all for this lecture so
thank you