welcome to the session now we will be solving
the problem that it generalized three dimensional
problem that that we are solving we we have
taken up in the last class will be completing
it we we have already completed the the sub
problem theta one in the sub problem theta
two we have completely solve the ah study
state part which is nothing but an elliptical
partial differential equation
now let us look into the time varying part
if you look into the solution of the time
varying part it will be having a non zero
initial condition and the rest boundary conditions
are all homogenous and dirichlet so therefore
ah we already saw this type of problem in
the first sub problem and i will be directly
writing the solution of that so theta two
tau will be nothing but summation of m summation
over n c m n exponential minus lambda m n
square tau sin m pie x sin m pie y and ah
at tau is equal to zero we have the initial
condition theta tau is nothing but minus theta
two s which will be the solution of the steady
state part
so we we ah put it put it there so these becomes
minus theta two s is equal to summation over
m summation over n c m n sin m pie x sin m
pie y and we will be evaluating the the constants
c m n is by using auto[mal]- property of the
sin function and the ah eigen functions we
have done already before that ah what is the
solution of theta two s theta two s is minus
two theta one naught n is equal to one infinity
one minus cosine n pie divided by n pie sin
hyperbolic k n pie x divided by sin hyperbolic
k n pie sin n pie y
so that was the solution of the steady state
part that should be equal to double summation
one over m another over n c m n sin m pie
x and sin n pie y so we multiplied by both
side by sin m pie x and sin m pie y dx dy
integrate after opening the summation series
only one term will survive in the right hand
side and ah that will be ah and other will
be ah lost will be gone so it will be minus
two theta one naught some function of x and
y so i write this as some function of x and
y is equal to summation m n c m n sin m pie
x sin m pie y
so this will be nothing but minus two theta
one naught double integral over x over y f
of x y ah sin m pie x sin n pie y dx d y and
the right hand side we will be getting c m
n there will be one integral sin square pie
x will be half integrals you know integral
sign m sin square pie y d y will be half so
there will be half half there so if we do
that then we will be getting an expression
of c m n and c m n will be minus eight theta
one naught from zero to one integral from
y is equal to from zero to one f of x y sin
m pie x sin n pie y dx dy and where the index
m was associated with y and index n was associated
with x in this particular problem so we convert
them back to n and m our normal variable
so these becomes eight theta one naught x
y ah f of x y sin m pie x sin n pie y d x
d y this will be minus eight theta one naught
zero to one zero to one now let us put ah
f of x y f of x y is nothing but ah one minus
cosine n pie divided by n pie sin hyperbolic
k n pie x sin hyperbolic k n pie sin n pie
y sin n pie x sin m pie y d x d y so we will
be will be getting so this will be giving
a you know as some some constant some value
so you will be we can obtain the c m n and
accurately so once we will be getting the
estimating the c m n similarly the other sub
problems which will be having a zero initial
condition and non zero boundary a boundary
conditions they can be converted into a steady
state part and it will be divided into steady
state part and and a transient part and that
will be we will be you will be able to solve
them
so once we do that then let us so that that
completes the problem that we are dealing
with now let us go to the other problem now
will be taking a typical problem that we have
not talked about earlier so this is a two
dimensional problem so we take up another
example let say a two dimensional problem
a parabolic one two dimensional parabolic
partial differential equation like del u del
t it will be is equal to del square u del
x square and at t is equal to zero we have
let say u is equal to u naught and the boundary
at x is equal to zero it is insulated so it
will be del u by del x will be equal to zero
and boundary at x is equal to one we have
a mixed boundary condition del u del x plus
beta u is equal to zero
so if we remember that in the previous we
we have looked into the problem where the
both x is equal to zero and x is equal to
one we have a dirichlet boundary condition
that is the problem of first kind problem
of the second kind was at x is equal to zero
del u del is equal to zero and x is equal
to one u is equal to zero and problem of third
kind was at del was at x is equal to zero
u was equal to zero and del u del x plus beta
u will be zero x is equal to one so it's a
mixed boundary condition that is prevailing
at x is equal to zero
now this particular problem we have a neumann
boundary condition at x is equal to zero and
they have mixed boundary condition at x is
equal to one so that we we have not attempted
this problem earlier so now so let us look
into the solution of this particular problem
so again we will be going ahead with a separation
of variable type of solution u is a function
of function of space and it's a sole function
of time so if you put it there so these becomes
x dt dt and x and t d square x d x square
now if you derived by x t and separate out
the variable this becomes one over t d t d
t is equal to one over x d square x d x square
the left and side of function time and the
right hand side is a function of space they
are equal and they will be equal to some constant
in order to have a non trivial solution
now let us constitute the eigen problem in
the x direction the eigen problem solution
is d square x d x square plus lambda square
x is equal to zero and at z is equal to zero
d x d x is equal to zero and at x is equal
to one we have d x d x plus beta x is equal
to zero so these formulates the standard eigen
value problem in the x direction and let us
look into the solution of this now as you
all of us know the solution of x varying part
will be constituted by combination of sin
and cosine function so x n is equal to c one
sin lambda n x plus c two cosine lambda x
lambda n x
so eight correspond to nth eigen value we
will be getting this solution so now d x n
d x will be nothing but c one lambda cosine
lambda n x minus c two lambda sin lambda n
x and d x n d x evaluated at x is equal to
zero sin zero is zero so the term that will
survive this c one lambda cosine lambda ok
so that will be equal to zero at this boundary
so therefore lambda n c one lambda n cosine
lambda n will be equal to zero
so in order to have a ah no cosine at at x
is equal to zero so cos zero is one so this
will be multiplied by zero so you will be
will be having cos zero so c one lambda n
will be equal to zero so for a non trivial
solution ah lambda n cannot be equal to zero
if it is equal to zero then we will be landing
up with a trivial solution so the solution
is c one is equal to zero if c one is equal
to zero let say what is the solution of this
solution is c two cosine lambda n x
now these are the eigen function of the this
particular problem now let us look into the
eigen values how the eigen values are obtained
the eigen values are obtained by the boundary
condition we will be putting at x is equal
to one so at x is equal to one it should satisfy
the original boundary condition boundary condition
of the original problem so it will be d d
x d x plus beta x is equal to zero if you
look into the n eth eigen value so there will
be d x n d x plus beta x n value equal to
zero or x n equal to this now let us say what
d x n d x d x n d x will be nothing but minus
c two lambda n sin lambda n x and at x is
equal to one this becomes d x n d x is equal
to minus c two lambda n sin lambda n and x
n at x is equal to one becomes c two cosine
lambda n
so then we will be substituting this into
the boundary condition at x is equal to one
and let us see what we get what we will be
getting is that minus c two lambda n sin lambda
n plus beta c two cosine lambda n is equal
to zero so therefore c two is in common so
ah beta cosine lambda n minus lambda n sin
lambda n will be equal to zero so c two is
equal to beta minus lambda n tan lambda n
will be equal to zero so in order to have
a no trivial solution c two must not be equal
to zero so the only option is lambda n tan
lambda n is equal to beta so lambda n is nothing
but the eigen values of this transcendental
equation roots of this transcendental equation
equations are the eigen values
now this lambda n tan lambda n equal to beta
it will be it will be intersecting the lambda
n axis at n n number of locations so if you
really plot it so this will be there will
be it will be cutting the x axis at infinite
point and each of the intersection point is
a root of this particular equation this can
be also solved numerically by using a newton
and method so typically these type of equation
the the roots are appearing in an arithmetic
progression of common difference three so
one can take take request to newton and and
having an initial guess and then once that
will be that will be newton for a particular
guess is conversed then one can put a outer
loop where the initial guess will given an
increment ah plus three the outer loop will
be completed
so if you calculate this four times one will
be automatically landing up with four roots
of this governing equation so by by bring
by taking a request to a numerical method
one can really ah obtain the you know first
five roots of first ten roots of this particular
transcendental equation so once we get that
then we will be formulating the time varying
part t n is equal to c three exponential minus
lambda n square t and u will be as a function
of x and t will be nothing but n is equal
to one to infinity c n exponential minus lambda
n square t cosine lambda n x ok
now the constant c n has to be evaluated from
the initial non zero initial condition that
at t is equal to zero u u is equal to u naught
so let us ah go head with that if you really
do it then you will be getting at t is equal
to zero u is equal to u naught and therefore
u naught will be is equal to summation cn
cosine lambda n x where n is equal to from
one to infinity so you utilize the orthogonal
property of the of the eigen functions so
u zero zero to one cosine ah lambda n x d
x is equal to ah c n cos square lambda n x
d x so left hand side will be getting u zero
sin lambda n divided by lambda n and the right
hand side we are having c n by two zero to
one two cos square lambda n x d x
so ah we can further simplified it sin lambda
n by lambda n c n by two this can be written
as zero one one plus cosine two lambda n x
d x if that is the case then we can integrate
it out on the right hand side and what will
be getting is c n by two is equal to one plus
sin two lambda n x divided by two lambda n
zero to one so we will be getting c n by two
one plus sin two lambda n divided by two lambda
n now we can put sin two lambda n in terms
of we can express into sin lambda n into tan
lambda n are we have the equation lambda n
tan lambda n equal to beta and we can substitute
that and will be getting a simplified version
of the of the coefficient c n if you do that
we will be getting u zero sin lambda n divided
by lambda n is equal to c n by two one plus
one over two lambda n is equal to into two
tan lambda n we are substituting cos lambda
sin two lambda n by tan tan lambda
so one plus tan square lambda n so this is
will be c n by two one plus two will be canceling
out so it will be beta tan lambda n will be
beta by lambda n that is the transcendental
equation the eigen values must satisfied so
beta by lambda n divided by one plus beta
square by lambda n square so this becomes
c n over two lambda n square plus beta square
plus beta divided by lambda n square plus
beta square and we can get the expression
of c n as two u naught sin lambda n divided
by lambda n multiplied by lambda n square
plus beta square divided by lambda n square
plus beta plus beta square
so this the constant and the complete solution
is u is equal to summation n is equal to one
to infinity c n exponential minus lambda n
square t cosine lambda n x where n can transform
one to infinity so that solves this problem
completely now now let us looking into the
ah some some more problems so this completely
solves this problem it's a we can call it
a fourth kind of boundary condition where
the boundary at x is equal to zero is ah neumann
and boundary at one is a robin mixed
next while we taking up one more example in
order to demonstrate the various types of
solutions using the separation of variables
the problem that i will be taking about it
will again a practical problem where we will
be having a you know robin mixed boundary
condition ah it's a one dimensional heat conduction
transient problem this will be alpha so this
will be another example example three let
us say alpha del square t del x square at
t is equal to zero we have t is equal to t
naught at x is equal to zero t is equal to
t one and at x is equal to l we have robin
mixed boundary condition k del t del x is
equal to h t minus infinity that means at
this boundary it is opened the atmosphere
whatever the amount of heat that as been transported
here by conduction that will be taken by convection
where h is the heat transfer coefficient and
t infinity is the ah you know temperature
now we make this equation non dimensional
such that ah one of the non homogeneity will
go so x star will be x by l and theta the
non dimensional at temperature will be defined
t minus t infinity divided by t zero minus
t infinity so let us put it in the governing
equation so this becomes del theta del t is
equal to alpha del square theta divided by
l square del x star square
so this will be nothing but del theta del
tau is equal to delta square theta del x star
square where tau is equal to l square t over
ah over over alpha ah alpha t over l square
so tau is equal to nothing but so this will
be l square will be on the other side so it
will be alpha t over l square is the non dimensional
time now let us put in the boundary condition
at tau is equal to zero means at t is equal
to zero means at tau is equal to zero theta
is equal to theta naught minus t naught minus
t infinity divided by t naught by t infinity
that will be equal to one at x star is equal
to zero ah theta is equal to t minus t t is
equal to t one so t one minus t infinity divided
by t zero minus t infinity so that will be
theta one and at x star is equal to one we
have we have a mixed boundary condition
so i think ah we will discuss it in detail
so at x star is equal to one we have minus
k by l this will be del x star and this will
be del theta and that will be multiplied by
t naught minus t infinity so this will be
equal to h t minus t infinity so this will
be ah so what is the definition of the theta
theta is equal to t minus t infinity by t
naught minus t infinity so it will be t naught
minus t infinity times theta so this will
be canceling out
so what will be getting is del theta del x
star plus h l over k times theta and what
is h l by k this called the biot number it
is a non dimensional number it's a biot number
it's a ratio of heat transfer coefficient
divided by thermal conductivity it's ratio
of the thermal resistance by diffusion divided
by thermal resistance by convection so we
will be having del theta del x star plus b
i theta is equal to zero at x star is equal
to one so in this particular problem we have
two sources of ah non homogeneity so the non
homogeneity ah two sources of homogeneity
so let us look into what are the two sources
so one is at initial condition at tau is equal
to zero theta is equal to one another is at
boundary condition at x x is equal to zero
theta is equal to theta one so i divide this
ah we we call this as theta one naught in
order to make it consistent so i divide this
problem into two problem one is theta one
another is theta two considering one non homogeneity
at a time as we have done earlier so therefore
theta one sub problem one will be basically
ah del theta one del tau is equal to del square
theta one del x star square at tau is equal
to zero theta one is equal to you will put
it put the boundary ah you you keep the non
homogeneity intact and force the other to
be vanished at tau is equal to one and theta
is equal to one and at x x star is equal to
zero theta one is equal to theta one naught
and at x star is equal to one
we have the mixed boundary condition del theta
one del x star plus b i theta one is equal
to zero now this particular sub problem these
are well posed problem well posed partial
parabolic partial differential equation and
we have already solved in great derail how
to solve this sub problem earlier now let
us look into the other sub problem the other
sub problem is for theta two so it will be
del theta two del tau is equal to del square
theta two del x star square and at tau is
equal to zero we keep this initial condition
to be homogenous and ah substitute the other
one x star equal to zero theta is equal to
theta one naught and x star is equal to one
we have the robin mixed boundary condition
del theta two del x star plus b i theta two
is equal to zero
now again since we have a zero initial condition
this problem as to be divided into two sub
problems one is the time dependent part another
is the time independent part so we can constitute
the governing equation of theta two s where
s is time dependent part d square theta two
s d x star square will be equal to zero at
x star is equal to zero we have theta two
s is equal to theta one naught we associate
the non homogenous part in the steady state
solution and x star is equal to one d theta
two s d x star plus b i theta two s is equal
to zero similarly we formulate the other part
the transient parts it will be del theta two
t del tau is equal to del square theta two
t del x star square at tau is equal to zero
theta t two will be nothing but minus theta
two s x star it's a steady state solution
solution of the steady state part and at x
star is equal to zero since we have associated
the non homogenous governed boundary condition
with the steady state solution so theta two
t will be become homogenous and at x star
one equal to one this becomes del theta two
t del x star plus b i theta two t will be
equal to zero
now again this a well posed problem of well
posed of third kind where the eigen functions
will be the sin functions and eigen value
will be coming from the transcendental equation
that we have already looked before and this
will be a straight forward part we will be
getting the straight forward ah solution of
the steady state part and the solution of
the steady state part will be substituted
here and again it's a well posed problem and
we know the solution of this so the complete
solution will be obtained as theta is equal
to theta theta one plus theta two s which
will be a function of x star only plus theta
two t which will be the solution of this part
so we will be constituting this we will be
construct the we will able to construct the
complete solution of this problem so i am
also almost come to the end of my ah end of
our course so in this course we have we have
learnt about the solution of partial differential
equations for for for for the engineering
problems and we have already looked into the
in detailed the ah how to form the what what
are the classifications of the ah partial
differential equations they are ah you know
how to define how to identify and define various
boundary conditions various partial differential
equations what is the what is the principle
of a linear super position we have developed
various theorems related to the standard eigen
value problem at joint operator and the properties
of the standard eigen value problem which
will be having a infinite eigen number of
eigen values and in and the eigen functions
of orthogonal to each other various properties
we have looked into then we have we have looked
into the solution of you know separation of
using separation of variable method for the
rectangular coordinate we have defined different
kinds of sub problems depending on the boundary
conditions
for example first kind second kind third kind
and how to handle the how to divide the problems
into sub problems in order to take care of
non homogeneities in the boundary conditions
then we have looked into the three types general
solutions of the three types of partial differential
equations that is parabolic elliptical and
hyperbolic then we have looked into the one
two dimensional problem three dimensional
problem as well as the four dimensional problem
and then we have covered the ah you know second
dimensional two dimensional three dimensional
problems in cylindrical coordinate system
as well as the spherical polar coordinate
systems which the engineers will be they will
be becoming across quite often
so these course gives a basic fundamental
the grou[nd]- of a offer the engineers of
how to tackle the you know partial differential
equation those will be appearing for different
engineering applications and how to solve
them using separation variables if the operator
is a linear operator now in in a actual case
the operator may not be linear the problem
may not be a linear so one can take request
to the numerical techniques will be the time
consuming as well as the computational intensive
but as a first case one can go ahead with
the we can one can linearize the problem assuming
the you know constant thermal physical and
the transport coefficients and one can go
ahead with the solution of the separation
using separation of variables as a as a first
ah as a first and information of the complicated
system
so i hope this course will be useful to you
for all the engineering students and particularly
for you know chemical engineering mechanical
engineering aerospace engineering students
ok
thank you very much
