WEBVTT
Kind: captions
Language: en
00:00:18.820 --> 00:00:23.830
so welcome to lecture series on mathematical
methods and its applications so we have already
00:00:23.830 --> 00:00:30.070
discussed fourier series fourier integrals
and fourier transforms some problem faced
00:00:30.070 --> 00:00:35.870
on that we are already did with now let us
come to applications of fourier transforms
00:00:35.870 --> 00:00:44.350
to bvp so in this section we also include
applications of fourier series and fourier
00:00:44.350 --> 00:00:51.920
integrals also ok
now equations involving one or more independent
00:00:51.920 --> 00:00:56.660
variables are called partial differential
equation this we already know and such equation
00:00:56.660 --> 00:01:02.190
frequently arise in varies engineering and
science problems some important linear partial
00:01:02.190 --> 00:01:07.710
differential equation of second order are
as follows that we already know that one dimensional
00:01:07.710 --> 00:01:13.140
wave equations given by this expression one
dimensional heat equation is given by this
00:01:13.140 --> 00:01:19.940
del u by del t equal to c square del square
u upon del x square two dimensional laplace
00:01:19.940 --> 00:01:25.700
equation is given by this del square u by
del x square plus del square u upon del y
00:01:25.700 --> 00:01:32.240
square equal to zero two dimensional poisson
equation is given by del square u upon del
00:01:32.240 --> 00:01:36.490
x square plus del square u upon del y square
equal to f x y
00:01:36.490 --> 00:01:40.610
similarly two two dimensional wave equations
you know this expression and three dimensional
00:01:40.610 --> 00:01:46.259
laplace equations given by this expression
ok these are the important partial differential
00:01:46.259 --> 00:01:53.850
equations now can we you solve a given
partial differential equation using fourier
00:01:53.850 --> 00:01:59.030
series first of all we will see using fourier
series ok so we use separation of variables
00:01:59.030 --> 00:02:04.119
now suppose suppose a partial differential
equation is given to you any partial differential
00:02:04.119 --> 00:02:07.950
equation it may be a one dimensional wave
equation a one dimensional heat equation or
00:02:07.950 --> 00:02:12.430
two dimensional wave equation anything so
how you solve it
00:02:12.430 --> 00:02:19.700
you assume that u x t which is a function
of two independent variable is simply f
00:02:19.700 --> 00:02:26.610
x into g x f x into g t this we assume
so we basically adopt this method which we
00:02:26.610 --> 00:02:33.830
call as separation of variables ok now we
find del u by del x what it is it will be
00:02:33.830 --> 00:02:41.730
nothing but f dash x into g t because t
his function is only a function of t ok suppose
00:02:41.730 --> 00:02:50.949
you find del u by del t so that will be nothing
but f x into g dash t so del square u by
00:02:50.949 --> 00:02:58.420
del x square that will be nothing but f double
dash x into g g t and similarly del square
00:02:58.420 --> 00:03:06.720
u by del t square will be nothing but f x
into g double dash t its ok ok
00:03:06.720 --> 00:03:11.519
now we substitute this in a given partial
differential equation whatever partial differential
00:03:11.519 --> 00:03:17.150
equation is given to us we substitute these
values in that equation now when you substitute
00:03:17.150 --> 00:03:24.880
this in that given partial differential equation
we get back two ordinary differential equations
00:03:24.880 --> 00:03:30.920
which we can easily solve now we apply initial
and boundary conditions whatever given to
00:03:30.920 --> 00:03:38.640
us and hence we can find u x t using fourier
series ok then using fourier series we can
00:03:38.640 --> 00:03:44.229
find u x t
so let us illustrate this i mean this method
00:03:44.229 --> 00:03:52.700
by some examples ok so this is all we have
discussed now first of all consider one dimensional
00:03:52.700 --> 00:04:00.850
wave equation using fourier series so consider
an elastic string of length l which is fastened
00:04:00.850 --> 00:04:07.560
at its ends on the x axis at x equal to zero
and at x equal to l the string is displaced
00:04:07.560 --> 00:04:12.820
and then released to vibrate in the x t plane
where u x t denotes the vertical displacement
00:04:12.820 --> 00:04:18.370
of the vibrating string u x t is nothing but
the vertical displacement of the of the
00:04:18.370 --> 00:04:27.620
string so the differential equation governing
this is given by what is a differential
00:04:27.620 --> 00:04:37.720
equation governing this at is nothing but
del square u by del t square is equal to c
00:04:37.720 --> 00:04:47.410
square del square u by del x square
so it is x is varying from zero to l and t
00:04:47.410 --> 00:04:53.960
is greater than zero ok so these are the boundary
condition and initial conditions given to
00:04:53.960 --> 00:05:01.300
us ok now we want to solve this this partial
differential equation which is one dimensional
00:05:01.300 --> 00:05:07.580
wave equation how can you solve it using fourier
series so we assume that u x t is nothing
00:05:07.580 --> 00:05:15.720
but a function of two variables a function
of product of two functions f x into g t so
00:05:15.720 --> 00:05:23.600
what will be del square u upon del t square
it is nothing but f x into g double dash t
00:05:23.600 --> 00:05:31.500
and what is del square u upon del x square
it is nothing but f double dash x into g t
00:05:31.500 --> 00:05:35.070
ok
you substitute this two these things over
00:05:35.070 --> 00:05:43.910
here so what we obtain it is f into g double
dash f x into g double dash t which is equals
00:05:43.910 --> 00:05:56.431
to c square f double dash x into g t so this
implies f double dash x upon f x is equal
00:05:56.431 --> 00:06:08.870
to one upon c square times g double dash t
upon g t ok now this is a function of x alone
00:06:08.870 --> 00:06:15.410
and this is a function of t alone and both
are equal so both will be equal only when
00:06:15.410 --> 00:06:23.650
it is equal to some constant its ok c k where
k is a constant ok because one side we have
00:06:23.650 --> 00:06:29.591
function of x and other side we have function
of t and both are equal so both will be equal
00:06:29.591 --> 00:06:36.510
only when its a constant quantity c k ok
now this k have three possibilities either
00:06:36.510 --> 00:06:43.720
k is zero or k is positive or k is negative
now let us take all the three cases case one
00:06:43.720 --> 00:06:56.870
suppose k is zero ok if k is zero so f double
dash will be zero ok f double dash x will
00:06:56.870 --> 00:07:08.470
be zero and this implies f x will be nothing
but a x plus b where a and b are arbitrary
00:07:08.470 --> 00:07:13.840
constants now using the boundary condition
when you use the boundary conditions what
00:07:13.840 --> 00:07:22.919
are boundary conditions ok see the boundary
conditions it is u zero a t is zero
00:07:22.919 --> 00:07:32.180
so this implies when u x t is f x into
g t so it is nothing but f zero into g t is
00:07:32.180 --> 00:07:45.190
zero and zero now u l t is also zero so this
implies f l into g t equal to zero so from
00:07:45.190 --> 00:07:50.410
these two conditions we get a either this
equal to zero or this zero or this zero this
00:07:50.410 --> 00:08:00.050
zero ok suppose g t is zero if g t is zero
so u will be zero which is not possible which
00:08:00.050 --> 00:08:10.550
is the no interest because u must be non
zero ok g t cant be zero because if g t is
00:08:10.550 --> 00:08:18.699
zero then u will be zero ok
now this implies f zero will be zero and
00:08:18.699 --> 00:08:26.580
f l will be zero now when f zero is zero when
you apply this over here f zero will equal
00:08:26.580 --> 00:08:34.839
to zero so this implies b equal to zero and
f l equal to zero this implies a equal to
00:08:34.839 --> 00:08:40.990
zero so that means f equal to zero and when
f equal to zero means y equal to zero again
00:08:40.990 --> 00:08:49.829
which is which which is of no interest
ok because we want u as a non zero function
00:08:49.829 --> 00:08:54.690
ok
so now case two so this case is not possible
00:08:54.690 --> 00:08:59.080
because from if we take a equal to zero you
are taking u equal to zero which is of no
00:08:59.080 --> 00:09:07.210
interest ok now case two suppose k is equals
to a positive quantity suppose p square suppose
00:09:07.210 --> 00:09:13.740
k is some positive quantity p square if k
equal to p square so this implies f double
00:09:13.740 --> 00:09:24.589
dash will be equal to a k into f and this
implies d square minus k into f k is p
00:09:24.589 --> 00:09:31.279
square ok k is p square so this is p square
so this equal to zero and this implies f will
00:09:31.279 --> 00:09:38.390
be nothing but c one e k power p x plus c
two e k power minus p x
00:09:38.390 --> 00:09:45.580
now again f zero equal to zero so this implies
c one plus c two equal to zero and f l equal
00:09:45.580 --> 00:09:53.029
to zero so this implies c one e k power p
plus c two e k power minus p equal to zero
00:09:53.029 --> 00:09:58.350
so when you simplify these two equations on
c one and c two so this implies c one equal
00:09:58.350 --> 00:10:08.550
to c two equal to zero so hence f equal to
zero again this implies u equal to zero so
00:10:08.550 --> 00:10:18.180
this case is also not possible
so take the third case case three when k equals
00:10:18.180 --> 00:10:24.350
to minus p square so these are the only three
cases either k is zero k is positive or k
00:10:24.350 --> 00:10:31.080
is negative if k is zero we are getting u
equal to zero which is not possible if k equal
00:10:31.080 --> 00:10:36.940
to p square [FL] positive quantity again
we are getting u equal to zero and if k equal
00:10:36.940 --> 00:10:42.890
to now let us try for the k equal to minus
p square ok if k equal to minus p square f
00:10:42.890 --> 00:10:50.960
double dash will be nothing but minus p square
k so this implies d square plus p square times
00:10:50.960 --> 00:10:58.550
ok f ok f will be equal to zero
so this implies f will be nothing but c one
00:10:58.550 --> 00:11:12.220
so it is cos p x plus c two sin p x ok now
when f zero is zero f zero is zero implies
00:11:12.220 --> 00:11:25.850
c one is equal to zero and f l equal to
zero implies zero equal to c two sin p
00:11:25.850 --> 00:11:34.270
l now if c two equal to zero again f will
be zero which in turn give u equal to zero
00:11:34.270 --> 00:11:45.110
ok so that case is omitting so this implies
sin p l will be zero because it we dont
00:11:45.110 --> 00:11:50.210
want c two to be zero because if c two is
zero f is zero and which in turn gives u equal
00:11:50.210 --> 00:11:55.830
to zero ok
so sin p l equal to zero so this implies p
00:11:55.830 --> 00:12:06.550
l equal to n pi or p equals to n pi by l p
equal to n pi by l so this is the only case
00:12:06.550 --> 00:12:15.709
possible and from this case what will be f
what is the f f will be nothing but c one
00:12:15.709 --> 00:12:27.800
is zero and we have c two c two sin p is
nothing but n pi by l into x so since n is
00:12:27.800 --> 00:12:35.010
involved since n is involved so we take it
f n or we take it c n here also we take it
00:12:35.010 --> 00:12:45.610
c n ok ok because n is involved for different
n we get different f so we are we are calling
00:12:45.610 --> 00:12:52.450
it f n ok
now we now for this p for this k here k equal
00:12:52.450 --> 00:13:02.330
to minus p square find g what will be g so
it is g double dash will be equals to it is
00:13:02.330 --> 00:13:10.920
minus p square c square g because k is
nothing but minus p square so from here again
00:13:10.920 --> 00:13:29.290
we will get we will get g as suppose something
d d cos p c t plus e sin p c t ok where
00:13:29.290 --> 00:13:35.740
d and e are arbitrary constant now this p
is nothing but pi n pi by l
00:13:35.740 --> 00:13:48.650
so when you replace you call it d n g n and
d n cos n pi c t by l plus e n sin n pi c
00:13:48.650 --> 00:13:55.270
t by l because n is involved so we are calling
it d n and e n and here we are calling as
00:13:55.270 --> 00:14:03.709
g n so what will be u now so what will be
u now this u is nothing but product of f and
00:14:03.709 --> 00:14:15.700
g ok so we are calling it u n so it is nothing
but f n into g n and f n is nothing but c
00:14:15.700 --> 00:14:23.200
n into c n into this quantity so this constant
will merge to d n and e n so finally we will
00:14:23.200 --> 00:14:39.769
get sin n pi by l into x into d n the new
d n suppose d n dash cos n pi c t by l plus
00:14:39.769 --> 00:14:53.530
e n dash sin n pi c t by l ok
now this is n may be one may be two may
00:14:53.530 --> 00:14:59.649
be three and all other solution of this differential
equation when you substitute n equal to one
00:14:59.649 --> 00:15:04.279
you get one solution that is a solution of
this differential equation when you substitute
00:15:04.279 --> 00:15:10.950
n equal to two u one u two u three up to so
on and so on are all the solution of this
00:15:10.950 --> 00:15:16.870
given differential equation so we know
that but the super position principle if
00:15:16.870 --> 00:15:22.690
u one u two u three and so on are the solution
of a given differential equation then the
00:15:22.690 --> 00:15:26.529
linear combination of those will also be the
solution of the given differential equation
00:15:26.529 --> 00:15:31.760
given partial differential equation
so what will be the so we are suppose u x
00:15:31.760 --> 00:15:40.820
t as summation n one fund to infinity u n
x t e which is nothing but summation n from
00:15:40.820 --> 00:15:55.130
one to infinity sin n pi x by l into d
n dash cos n pi c t by l plus e n dash sin
00:15:55.130 --> 00:16:06.209
n pi c t by l ok so this will be the solution
of this differential equation by the super
00:16:06.209 --> 00:16:15.040
position principle now let us apply initial
conditions now u x zero is f x so u x zero
00:16:15.040 --> 00:16:21.620
is f x which is equal to summation n from
one to infinity when you substitute t equal
00:16:21.620 --> 00:16:30.029
to zero so when you substitute t equal to
zero we get back to this expression
00:16:30.029 --> 00:16:37.279
now this is nothing but fourier sine series
ok this is nothing but is equal to fourier
00:16:37.279 --> 00:16:43.100
sine series half range half range fourier
sine series so from here what will be d n
00:16:43.100 --> 00:16:55.570
dash so d n dash will be nothing but two upon
l integral zero to l f x sin n pi x upon l
00:16:55.570 --> 00:17:03.720
into d x so this is by the fourier series
ok so so basically d n dash and e n dash is
00:17:03.720 --> 00:17:10.179
to find out finally ok so to find d n dash
we substitute we take this boundary condition
00:17:10.179 --> 00:17:15.919
from this boundary condition we take we get
this equal to this this is nothing but
00:17:15.919 --> 00:17:21.880
sine series fourier sine series ok fourier
sine series half range fourier sine series
00:17:21.880 --> 00:17:27.309
so from here we get d n dash as two upon l
zero two l f x sin by x to d x
00:17:27.309 --> 00:17:36.749
now to get e n dash we apply another initial
condition which is del u by del t at x comma
00:17:36.749 --> 00:17:47.169
zero is is g x so this implies now what
is del u by del t del u by del t is nothing
00:17:47.169 --> 00:17:54.960
but when you differentiate respect to t so
this would be sine which is zero when t equal
00:17:54.960 --> 00:18:02.600
to zero and this is cos when which is one
when t equal to zero so we finally obtain
00:18:02.600 --> 00:18:17.919
g x is equal to summation n from one to infinity
e n dash into n pi c by l into sin n pi x
00:18:17.919 --> 00:18:27.509
upon l ok because because when you take when
you differentiate partial with respect to
00:18:27.509 --> 00:18:33.730
t and put t equal to zero so this is sins
which is zero and this is cos which is one
00:18:33.730 --> 00:18:39.350
so only the coefficient will come here ok
now this is again fourier sine series half
00:18:39.350 --> 00:18:44.840
range fourier sine series so the constant
term which is this term the entire term this
00:18:44.840 --> 00:18:53.679
implies e n dash into n pi c upon l will be
nothing but two upon l integral zero to l
00:18:53.679 --> 00:19:04.830
f x sin n pi x upon l into d x so l l cancels
out so from here e n dash will be nothing
00:19:04.830 --> 00:19:15.919
but two upon n pi c integral zero to l f x
sin n pi x by l into d x
00:19:15.919 --> 00:19:21.309
so these are the values of these are the values
of d n dash and e n dash which when substitute
00:19:21.309 --> 00:19:28.610
here you will get back to u x t which as solution
of this different partial differential equation
00:19:28.610 --> 00:19:39.390
so hence using fourier series we can solve
such type of problems ok now now hence find
00:19:39.390 --> 00:19:45.059
the solution of one dimensional wave equation
which is this problem corresponding to a triangular
00:19:45.059 --> 00:19:52.200
initial deflection suppose the u x zero
which is f x is given by this expression and
00:19:52.200 --> 00:19:57.489
the initial velocity is zero initial velocity
zero means this this we are calling as del
00:19:57.489 --> 00:20:01.919
u by del t which is nothing but the initial
velocity at t equal to zero means initial
00:20:01.919 --> 00:20:08.999
velocity is zero means u x is zero and f x
is given by f x is given by this expression
00:20:08.999 --> 00:20:15.499
so we can easily find out if g x is ok here
it is g x sorry because because it is equal
00:20:15.499 --> 00:20:23.159
to g x so it is g x sorry it is g x ok
now if g x equal to zero so e n dash will
00:20:23.159 --> 00:20:29.840
be zero so e n dash is zero ok and to calculate
d n dash so we know that f x is given by that
00:20:29.840 --> 00:20:34.859
expression we simply substitute f x over here
find out d n dash and we substitute d n dash
00:20:34.859 --> 00:20:41.179
over here we will get back to the value of
u x t so this integral we can easily find
00:20:41.179 --> 00:20:53.239
out ok now come back to next problem come
to next problem its is a heat equation
00:20:53.239 --> 00:21:00.269
again solution by a fourier series consider
thin homogeneous bar of wire wire of length
00:21:00.269 --> 00:21:08.909
l let the bar coincide with x axis we have
a bar basically which coincide with x axis
00:21:08.909 --> 00:21:18.889
so this is some bar of length l ok now u x
t with a temperature distribution within the
00:21:18.889 --> 00:21:24.039
bar the boundary value modelling of this temperature
distribution u x t is given by this so this
00:21:24.039 --> 00:21:34.049
is nothing but u t equals to c square u x
x where this is a finite bar or finite length
00:21:34.049 --> 00:21:40.059
and t greater than zero so these are the initial
and boundary conditions given to us find u
00:21:40.059 --> 00:21:46.109
first find u x t ok first find u x t means
find the tempe[rapture]- distribution of the
00:21:46.109 --> 00:21:52.350
bar if these are the initial boundary conditions
so again we will apply separation of variables
00:21:52.350 --> 00:22:03.830
we will take u as f x into g t we will
substitute it here so it is g dash t into
00:22:03.830 --> 00:22:13.419
f x is equals to c square f double dash x
into g t which implies f double dash upon
00:22:13.419 --> 00:22:20.129
f is equals to one by c square g dash upon
t again its a function of x its a function
00:22:20.129 --> 00:22:26.850
of t and both are equal so both will be equal
only when its a constant quantity say k ok
00:22:26.850 --> 00:22:33.619
now again we will take three cases
so the boundary condition are similar to
00:22:33.619 --> 00:22:41.049
the previous problem ok it is zero at both
the ends when x equal to zero and x equal
00:22:41.049 --> 00:22:47.789
to n it is zero so it is similar to the previous
case one dimensional wave equation so again
00:22:47.789 --> 00:22:56.750
when k is zero u comes out to be zero and
when k equal to p square u again come out
00:22:56.750 --> 00:23:02.580
come out to be zero this is boundary condition
of this problem and the previous problem and
00:23:02.580 --> 00:23:08.309
this problem are same it is also zero when
x equal to zero and x equal to l and it is
00:23:08.309 --> 00:23:12.799
also zero when you x equal to zero and e equal
to l so boundary condition are same
00:23:12.799 --> 00:23:19.499
so when k equal to zero or k equal to p
square u comes out to comes out to be zero
00:23:19.499 --> 00:23:26.399
so we dont take these cases so take
a equal to minus p square solution will exist
00:23:26.399 --> 00:23:33.600
only when k equal to minus p square ok so
when k equal to minus p square f double dash
00:23:33.600 --> 00:23:40.600
upon f will be equals to minus p square which
is implies f will be equal to c one cos
00:23:40.600 --> 00:23:51.070
p x plus c two sin p x again f zero is
zero this implies c one is zero and f l
00:23:51.070 --> 00:23:57.899
equal to zero so this will imply c two sin
p l equal to zero and since c two cant be
00:23:57.899 --> 00:24:02.489
zero because the c two equal to zero again
f will be zero which in turn gives u equal
00:24:02.489 --> 00:24:09.989
to zero ok because if f equal to zero then
u is u will be zero and u cant be zero
00:24:09.989 --> 00:24:20.200
so this implies p l will be n pi and this
implies p will be n pi by l so from here we
00:24:20.200 --> 00:24:35.039
will get f n equal to sin n pi l by x ok
now we get back to here this expression
00:24:35.039 --> 00:24:42.310
so what will be g dash so g dash will be nothing
but minus p square c square g when you
00:24:42.310 --> 00:24:48.600
integrate its a first derivative only so when
we integrate both side it is log g which
00:24:48.600 --> 00:24:58.029
is minus p square c square t plus log some
k which is which implies g is equals to k
00:24:58.029 --> 00:25:06.210
e k power minus p square c square d
now when you substitute p equal to pi n by
00:25:06.210 --> 00:25:16.349
l so it is nothing but we got it g k n g n
and it is k n e k power minus it is a pi n
00:25:16.349 --> 00:25:25.249
pi by l whole square into c square t so g
n will be given as k n e k power suppose it
00:25:25.249 --> 00:25:37.480
is minus lambda n square into t where lambda
n is pi this expression ok so what will be
00:25:37.480 --> 00:25:44.820
what will be u n now u n will be f n into
g n so the multiplication of these two will
00:25:44.820 --> 00:25:56.559
be nothing but this gives k n into sin
n pi x upon l into e k power minus lambda
00:25:56.559 --> 00:26:04.720
n square into t now again by the super position
principle u x t which is solution of this
00:26:04.720 --> 00:26:09.379
differential equation will be nothing but
this partial differential equation so u x
00:26:09.379 --> 00:26:16.229
t will be nothing but summation n from one
to infinity u n x t
00:26:16.229 --> 00:26:26.039
so which is nothing but summation n from one
to infinity u x t is k n sin n pi x
00:26:26.039 --> 00:26:39.669
pi l into e k power minus lambda n square
into t where lambda n is n pi c pi l now apply
00:26:39.669 --> 00:26:48.619
the apply the initial condition that u
zero u x zero is equals to f x so this implies
00:26:48.619 --> 00:27:01.840
this implies f x is equal to summation
n from one to infinity k n sin n pi x by l
00:27:01.840 --> 00:27:09.610
now again this is a fourier sin series ok
so hence k n will be given by so this implies
00:27:09.610 --> 00:27:22.500
k n will be nothing but two upon l integral
zero to l f x sin n pi x by l into d x so
00:27:22.500 --> 00:27:29.269
if you substitute this k n over here this
k n over here so we will get a the solution
00:27:29.269 --> 00:27:38.870
u x t of this partial differential equation
ok now also find the solution of the boundary
00:27:38.870 --> 00:27:43.029
condition or replace by the condition that
both ends of the bar are insulated
00:27:43.029 --> 00:27:48.119
so the problem is same the same in the same
problem the boundary condition which is a
00:27:48.119 --> 00:27:54.210
these conditions are replaced by the condition
that both ends are insulated that means no
00:27:54.210 --> 00:28:00.690
heat is passing through both the ends ok
so what does it mean it means that now the
00:28:00.690 --> 00:28:16.840
boundary conditions are u x at zero t equals
to u x at l t are zero ok these are the
00:28:16.840 --> 00:28:22.039
these are the new boundary condition for this
problems all all other conditions are same
00:28:22.039 --> 00:28:31.320
so how what will be the solution of this
of this problem under these boundary
00:28:31.320 --> 00:28:38.440
conditions and initial conditions
so again we will substitute u as the same
00:28:38.440 --> 00:28:45.460
thing f into g f x into g t so we will get
back to f double dash upon f which is equals
00:28:45.460 --> 00:28:53.519
to one upon c square g dash upon g which is
k k cant be k is on the only possibilities
00:28:53.519 --> 00:28:59.460
is k equal to minus p square that we have
already discussed so k equal to minus p square
00:28:59.460 --> 00:29:04.519
so when k equal to minus p square f double
dash will be equals to minus p square f which
00:29:04.519 --> 00:29:13.849
implies f is equals to c one cos p x plus
c two sin p x now
00:29:13.849 --> 00:29:25.789
now from here what will be u x u x will
be f dash x into g t and u x at zero t means
00:29:25.789 --> 00:29:38.739
f dash at zero into g t equal to zero and
u x at l t means f dash l into g t equal to
00:29:38.739 --> 00:29:44.289
zero so g t cant be zero g t cant be zero
because if g t equal to zero then u equal
00:29:44.289 --> 00:29:51.609
to zero which is not possible so g t cant
be zero so this implies f dash zero is equal
00:29:51.609 --> 00:29:59.869
to zero and f dash l equal to zero
so when we apply this thing over here
00:29:59.869 --> 00:30:08.549
so what will be get so we will get now f dash
what is a f dash it is minus p c one sin
00:30:08.549 --> 00:30:20.739
p x plus p c two cos p x and f dash zero equal
to zero implies c two equal to zero because
00:30:20.739 --> 00:30:35.779
p cant be zero and f dash l equal to zero
implies minus p c one sin p l equal to zero
00:30:35.779 --> 00:30:42.159
so this cant be zero because p if p is zero
k equal to zero that k is already omitted
00:30:42.159 --> 00:30:48.639
and if c one equal to zero then c two is already
zero then f will be zero this implies u equal
00:30:48.639 --> 00:30:53.999
to zero this is also not possible
so only the only possibility if sin p l equal
00:30:53.999 --> 00:31:00.399
to zero if sin p l equal to zero the p l will
be equal to n pi so p will be nothing but
00:31:00.399 --> 00:31:12.120
n pi by l so from here f will be nothing but
or f l will be nothing but cos p n pi by l
00:31:12.120 --> 00:31:21.589
cos n pi by l into x substitute the value
of p ok and g and g n is the same g n is the
00:31:21.589 --> 00:31:29.009
same what what will be g n g what g be obtained
the last last problems k n e k power minus
00:31:29.009 --> 00:31:38.499
lambda n square into t where lambda n is nothing
but n pi c by l ok
00:31:38.499 --> 00:31:44.820
so what will be u now u n will be nothing
but f n into g n so that will be nothing but
00:31:44.820 --> 00:31:59.269
k n cos n pi l into x into e k power
minus lambda n square into t ok so again by
00:31:59.269 --> 00:32:03.739
the super position principle what will be
the solution so the solution will be nothing
00:32:03.739 --> 00:32:12.239
but u x t will be nothing but summation n
from one to infinity u n x t which is nothing
00:32:12.239 --> 00:32:24.899
but summation n from one to infinity it is
k n cos n pi x by l into e k power minus lambda
00:32:24.899 --> 00:32:32.499
n square into t where lambda n is n pi c by
l ok
00:32:32.499 --> 00:32:41.960
now apply the same initial condition that
u x zero equal to f x so this implies f x
00:32:41.960 --> 00:32:53.679
will be equal to summation n from one to infinity
k n cos n pi x by l so now it is a fourier
00:32:53.679 --> 00:33:01.029
sin series i mean cosine series ok here
is a cosine series so though then k n will
00:33:01.029 --> 00:33:14.409
be nothing but two by l zero to l it is
f x cos n pi x by l into d x so when you
00:33:14.409 --> 00:33:20.129
substitute this k n over here so we will get
back to the solution u x t which is a solution
00:33:20.129 --> 00:33:25.670
of this partial differential equations
so hence the partial differential equation
00:33:25.670 --> 00:33:34.580
can be solved using using fourier series
ok so these are the some applications of fourier
00:33:34.580 --> 00:33:39.469
series which we have discussed here so thats
all for this lecture so
00:33:39.469 --> 00:33:39.840
thank you