so welcome to lecture series on mathematical methods and its applications so we have already
discussed fourier series fourier integrals and fourier transforms some problem faced
on that we are already did with now let us come to applications of fourier transforms
to bvp so in this section we also include applications of fourier series and fourier
integrals also ok now equations involving one or more independent
variables are called partial differential equation this we already know and such equation
frequently arise in varies engineering and science problems some important linear partial
differential equation of second order are as follows that we already know that one dimensional
wave equations given by this expression one dimensional heat equation is given by this
del u by del t equal to c square del square u upon del x square two dimensional laplace
equation is given by this del square u by del x square plus del square u upon del y
square equal to zero two dimensional poisson equation is given by del square u upon del
x square plus del square u upon del y square equal to f x y
similarly two two dimensional wave equations you know this expression and three dimensional
laplace equations given by this expression ok these are the important partial differential
equations now can we you solve a given partial differential equation using fourier
series first of all we will see using fourier series ok so we use separation of variables
now suppose suppose a partial differential equation is given to you any partial differential
equation it may be a one dimensional wave equation a one dimensional heat equation or
two dimensional wave equation anything so how you solve it
you assume that u x t which is a function of two independent variable is simply f
x into g x f x into g t this we assume so we basically adopt this method which we
call as separation of variables ok now we find del u by del x what it is it will be
nothing but f dash x into g t because t his function is only a function of t ok suppose
you find del u by del t so that will be nothing but f x into g dash t so del square u by
del x square that will be nothing but f double dash x into g g t and similarly del square
u by del t square will be nothing but f x into g double dash t its ok ok
now we substitute this in a given partial differential equation whatever partial differential
equation is given to us we substitute these values in that equation now when you substitute
this in that given partial differential equation we get back two ordinary differential equations
which we can easily solve now we apply initial and boundary conditions whatever given to
us and hence we can find u x t using fourier series ok then using fourier series we can
find u x t so let us illustrate this i mean this method
by some examples ok so this is all we have discussed now first of all consider one dimensional
wave equation using fourier series so consider an elastic string of length l which is fastened
at its ends on the x axis at x equal to zero and at x equal to l the string is displaced
and then released to vibrate in the x t plane where u x t denotes the vertical displacement
of the vibrating string u x t is nothing but the vertical displacement of the of the
string so the differential equation governing this is given by what is a differential
equation governing this at is nothing but del square u by del t square is equal to c
square del square u by del x square so it is x is varying from zero to l and t
is greater than zero ok so these are the boundary condition and initial conditions given to
us ok now we want to solve this this partial differential equation which is one dimensional
wave equation how can you solve it using fourier series so we assume that u x t is nothing
but a function of two variables a function of product of two functions f x into g t so
what will be del square u upon del t square it is nothing but f x into g double dash t
and what is del square u upon del x square it is nothing but f double dash x into g t
ok you substitute this two these things over
here so what we obtain it is f into g double dash f x into g double dash t which is equals
to c square f double dash x into g t so this implies f double dash x upon f x is equal
to one upon c square times g double dash t upon g t ok now this is a function of x alone
and this is a function of t alone and both are equal so both will be equal only when
it is equal to some constant its ok c k where k is a constant ok because one side we have
function of x and other side we have function of t and both are equal so both will be equal
only when its a constant quantity c k ok now this k have three possibilities either
k is zero or k is positive or k is negative now let us take all the three cases case one
suppose k is zero ok if k is zero so f double dash will be zero ok f double dash x will
be zero and this implies f x will be nothing but a x plus b where a and b are arbitrary
constants now using the boundary condition when you use the boundary conditions what
are boundary conditions ok see the boundary conditions it is u zero a t is zero
so this implies when u x t is f x into g t so it is nothing but f zero into g t is
zero and zero now u l t is also zero so this implies f l into g t equal to zero so from
these two conditions we get a either this equal to zero or this zero or this zero this
zero ok suppose g t is zero if g t is zero so u will be zero which is not possible which
is the no interest because u must be non zero ok g t cant be zero because if g t is
zero then u will be zero ok now this implies f zero will be zero and
f l will be zero now when f zero is zero when you apply this over here f zero will equal
to zero so this implies b equal to zero and f l equal to zero this implies a equal to
zero so that means f equal to zero and when f equal to zero means y equal to zero again
which is which which is of no interest ok because we want u as a non zero function
ok so now case two so this case is not possible
because from if we take a equal to zero you are taking u equal to zero which is of no
interest ok now case two suppose k is equals to a positive quantity suppose p square suppose
k is some positive quantity p square if k equal to p square so this implies f double
dash will be equal to a k into f and this implies d square minus k into f k is p
square ok k is p square so this is p square so this equal to zero and this implies f will
be nothing but c one e k power p x plus c two e k power minus p x
now again f zero equal to zero so this implies c one plus c two equal to zero and f l equal
to zero so this implies c one e k power p plus c two e k power minus p equal to zero
so when you simplify these two equations on c one and c two so this implies c one equal
to c two equal to zero so hence f equal to zero again this implies u equal to zero so
this case is also not possible so take the third case case three when k equals
to minus p square so these are the only three cases either k is zero k is positive or k
is negative if k is zero we are getting u equal to zero which is not possible if k equal
to p square [FL] positive quantity again we are getting u equal to zero and if k equal
to now let us try for the k equal to minus p square ok if k equal to minus p square f
double dash will be nothing but minus p square k so this implies d square plus p square times
ok f ok f will be equal to zero so this implies f will be nothing but c one
so it is cos p x plus c two sin p x ok now when f zero is zero f zero is zero implies
c one is equal to zero and f l equal to zero implies zero equal to c two sin p
l now if c two equal to zero again f will be zero which in turn give u equal to zero
ok so that case is omitting so this implies sin p l will be zero because it we dont
want c two to be zero because if c two is zero f is zero and which in turn gives u equal
to zero ok so sin p l equal to zero so this implies p
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
possible and from this case what will be f what is the f f will be nothing but c one
is zero and we have c two c two sin p is nothing but n pi by l into x so since n is
involved since n is involved so we take it f n or we take it c n here also we take it
c n ok ok because n is involved for different n we get different f so we are we are calling
it f n ok now we now for this p for this k here k equal
to minus p square find g what will be g so it is g double dash will be equals to it is
minus p square c square g because k is nothing but minus p square so from here again
we will get we will get g as suppose something d d cos p c t plus e sin p c t ok where
d and e are arbitrary constant now this p is nothing but pi n pi by l
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
t by l because n is involved so we are calling it d n and e n and here we are calling as
g n so what will be u now so what will be u now this u is nothing but product of f and
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
n into c n into this quantity so this constant will merge to d n and e n so finally we will
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
e n dash sin n pi c t by l ok now this is n may be one may be two may
be three and all other solution of this differential equation when you substitute n equal to one
you get one solution that is a solution of this differential equation when you substitute
n equal to two u one u two u three up to so on and so on are all the solution of this
given differential equation so we know that but the super position principle if
u one u two u three and so on are the solution of a given differential equation then the
linear combination of those will also be the solution of the given differential equation
given partial differential equation so what will be the so we are suppose u x
t as summation n one fund to infinity u n x t e which is nothing but summation n from
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
n pi c t by l ok so this will be the solution of this differential equation by the super
position principle now let us apply initial conditions now u x zero is f x so u x zero
is f x which is equal to summation n from one to infinity when you substitute t equal
to zero so when you substitute t equal to zero we get back to this expression
now this is nothing but fourier sine series ok this is nothing but is equal to fourier
sine series half range half range fourier sine series so from here what will be d n
dash so d n dash will be nothing but two upon l integral zero to l f x sin n pi x upon l
into d x so this is by the fourier series ok so so basically d n dash and e n dash is
to find out finally ok so to find d n dash we substitute we take this boundary condition
from this boundary condition we take we get this equal to this this is nothing but
sine series fourier sine series ok fourier sine series half range fourier sine series
so from here we get d n dash as two upon l zero two l f x sin by x to d x
now to get e n dash we apply another initial condition which is del u by del t at x comma
zero is is g x so this implies now what is del u by del t del u by del t is nothing
but when you differentiate respect to t so this would be sine which is zero when t equal
to zero and this is cos when which is one when t equal to zero so we finally obtain
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
upon l ok because because when you take when you differentiate partial with respect to
t and put t equal to zero so this is sins which is zero and this is cos which is one
so only the coefficient will come here ok now this is again fourier sine series half
range fourier sine series so the constant term which is this term the entire term this
implies e n dash into n pi c upon l will be nothing but two upon l integral zero to l
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
but two upon n pi c integral zero to l f x sin n pi x by l into d x
so these are the values of these are the values of d n dash and e n dash which when substitute
here you will get back to u x t which as solution of this different partial differential equation
so hence using fourier series we can solve such type of problems ok now now hence find
the solution of one dimensional wave equation which is this problem corresponding to a triangular
initial deflection suppose the u x zero which is f x is given by this expression and
the initial velocity is zero initial velocity zero means this this we are calling as del
u by del t which is nothing but the initial velocity at t equal to zero means initial
velocity is zero means u x is zero and f x is given by f x is given by this expression
so we can easily find out if g x is ok here it is g x sorry because because it is equal
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
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
expression we simply substitute f x over here find out d n dash and we substitute d n dash
over here we will get back to the value of u x t so this integral we can easily find
out ok now come back to next problem come to next problem its is a heat equation
again solution by a fourier series consider thin homogeneous bar of wire wire of length
l let the bar coincide with x axis we have a bar basically which coincide with x axis
so this is some bar of length l ok now u x t with a temperature distribution within the
bar the boundary value modelling of this temperature distribution u x t is given by this so this
is nothing but u t equals to c square u x x where this is a finite bar or finite length
and t greater than zero so these are the initial and boundary conditions given to us find u
first find u x t ok first find u x t means find the tempe[rapture]- distribution of the
bar if these are the initial boundary conditions so again we will apply separation of variables
we will take u as f x into g t we will substitute it here so it is g dash t into
f x is equals to c square f double dash x into g t which implies f double dash upon
f is equals to one by c square g dash upon t again its a function of x its a function
of t and both are equal so both will be equal only when its a constant quantity say k ok
now again we will take three cases so the boundary condition are similar to
the previous problem ok it is zero at both the ends when x equal to zero and x equal
to n it is zero so it is similar to the previous case one dimensional wave equation so again
when k is zero u comes out to be zero and when k equal to p square u again come out
come out to be zero this is boundary condition of this problem and the previous problem and
this problem are same it is also zero when x equal to zero and x equal to l and it is
also zero when you x equal to zero and e equal to l so boundary condition are same
so when k equal to zero or k equal to p square u comes out to comes out to be zero
so we dont take these cases so take a equal to minus p square solution will exist
only when k equal to minus p square ok so when k equal to minus p square f double dash
upon f will be equals to minus p square which is implies f will be equal to c one cos
p x plus c two sin p x again f zero is zero this implies c one is zero and f l
equal to zero so this will imply c two sin p l equal to zero and since c two cant be
zero because the c two equal to zero again f will be zero which in turn gives u equal
to zero ok because if f equal to zero then u is u will be zero and u cant be zero
so this implies p l will be n pi and this implies p will be n pi by l so from here we
will get f n equal to sin n pi l by x ok now we get back to here this expression
so what will be g dash so g dash will be nothing but minus p square c square g when you
integrate its a first derivative only so when we integrate both side it is log g which
is minus p square c square t plus log some k which is which implies g is equals to k
e k power minus p square c square d now when you substitute p equal to pi n by
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
pi by l whole square into c square t so g n will be given as k n e k power suppose it
is minus lambda n square into t where lambda n is pi this expression ok so what will be
what will be u n now u n will be f n into g n so the multiplication of these two will
be nothing but this gives k n into sin n pi x upon l into e k power minus lambda
n square into t now again by the super position principle u x t which is solution of this
differential equation will be nothing but this partial differential equation so u x
t will be nothing but summation n from one to infinity u n x t
so which is nothing but summation n from one to infinity u x t is k n sin n pi x
pi l into e k power minus lambda n square into t where lambda n is n pi c pi l now apply
the apply the initial condition that u zero u x zero is equals to f x so this implies
this implies f x is equal to summation n from one to infinity k n sin n pi x by l
now again this is a fourier sin series ok so hence k n will be given by so this implies
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
if you substitute this k n over here this k n over here so we will get a the solution
u x t of this partial differential equation ok now also find the solution of the boundary
condition or replace by the condition that both ends of the bar are insulated
so the problem is same the same in the same problem the boundary condition which is a
these conditions are replaced by the condition that both ends are insulated that means no
heat is passing through both the ends ok so what does it mean it means that now the
boundary conditions are u x at zero t equals to u x at l t are zero ok these are the
these are the new boundary condition for this problems all all other conditions are same
so how what will be the solution of this of this problem under these boundary
conditions and initial conditions so again we will substitute u as the same
thing f into g f x into g t so we will get back to f double dash upon f which is equals
to one upon c square g dash upon g which is k k cant be k is on the only possibilities
is k equal to minus p square that we have already discussed so k equal to minus p square
so when k equal to minus p square f double dash will be equals to minus p square f which
implies f is equals to c one cos p x plus c two sin p x now
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
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
zero so g t cant be zero g t cant be zero because if g t equal to zero then u equal
to zero which is not possible so g t cant be zero so this implies f dash zero is equal
to zero and f dash l equal to zero so when we apply this thing over here
so what will be get so we will get now f dash what is a f dash it is minus p c one sin
p x plus p c two cos p x and f dash zero equal to zero implies c two equal to zero because
p cant be zero and f dash l equal to zero implies minus p c one sin p l equal to zero
so this cant be zero because p if p is zero k equal to zero that k is already omitted
and if c one equal to zero then c two is already zero then f will be zero this implies u equal
to zero this is also not possible so only the only possibility if sin p l equal
to zero if sin p l equal to zero the p l will be equal to n pi so p will be nothing but
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
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
same what what will be g n g what g be obtained the last last problems k n e k power minus
lambda n square into t where lambda n is nothing but n pi c by l ok
so what will be u now u n will be nothing but f n into g n so that will be nothing but
k n cos n pi l into x into e k power minus lambda n square into t ok so again by
the super position principle what will be the solution so the solution will be nothing
but u x t will be nothing but summation n from one to infinity u n x t which is nothing
but summation n from one to infinity it is k n cos n pi x by l into e k power minus lambda
n square into t where lambda n is n pi c by l ok
now apply the same initial condition that u x zero equal to f x so this implies f x
will be equal to summation n from one to infinity k n cos n pi x by l so now it is a fourier
sin series i mean cosine series ok here is a cosine series so though then k n will
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
substitute this k n over here so we will get back to the solution u x t which is a solution
of this partial differential equations so hence the partial differential equation
can be solved using using fourier series ok so these are the some applications of fourier
series which we have discussed here so thats all for this lecture so
thank you