Solution of Homogeneous&Non Homogeneous Equations Video Lecture - Computer Science Engineering (CSE)

welcome to the lecture series on
differential equations today top today's
topic is solution of homogeneous and non
homogeneous second-order partial
differential equations we have learned
in partial differential equations the
second-order partial differential
equations the method of characteristics
changing the normal form and we have
done one-dimensional wave equation all
these things we had learned for
homogeneous equation
today we'll move to non-homogeneous
equation as well in today's lecture we
will cover the three things
one-dimensional non homogeneous wave
equations two dimensional homogeneous
laplace equation and two dimensional non
homogeneous Poisson equation person
equation is actually the non homogenous
form of Laplace equation so let us start
with the one-dimensional wave equation
with non homogeneous out of the right
hand side is not sealed so
one-dimensional wave equation we had
learned that it is del 2 u over del T 2
minus C square del 2 u over del X 2 is
equal to 0 that was homogeneous 1 now we
are having on the right hand side its XT
a function of X and T in the whole
reason defined on the whole real line
that is minus 20 to plus infinity and T
positive the initial conditions we do
her that you it's at 0 is 0 and del u
over del T at T is equal to 0 is 0 that
is initially at the time T is equal to 0
the function unknown function U is 0 on
the derivative of with respect to T at T
is equal to 0 is also 0 what we if you
do remember we have done the or in this
homogeneous wave equation with initial
conditions where we had that initially
the deflection was
effects and initially the velocity was
GX now because we are talking about here
this non-homogeneous equation we are
using first this homogeneous initial
conditions now the way how do we solve
this equation if you remember in the
ordinary differential equations when we
are talking about the non homogeneous
equations we first find out the general
solution from the homogeneous equation
and for the non homogenous equation we
used one particular solution and then
using the superposition principle we add
up though both the solutions we get the
solution of non-homogeneous equation
that is the method here also but let us
try one more method which is based on
the dermals principle what is that it
actually does is that it reduces this
non-homogeneous problem to a special
homogeneous problem with non-homogeneous
initial conditions see the non
homogenous problem is del 2 u over del T
2 minus C square del 2 u over del X 2 is
equal to H X T in the reason it's
belonging to R and T positive and
initial conditions we are homogeneous
conditions that is u X comma 0 is 0 and
del u over del T at T is equal to 0 is 0
now we change reduce it to a special
homogeneous problem with non-homogeneous
initial condition what is that a special
homogeneous problem we take del tuyú
that is now I am taking another function
capital u del 2 u over del T 2 minus C
square del 2 u over del X 2 is equal to
0 that is the equation we have we said
is that a equation in unknown function
capital u this is homogeneous very
equation in the whole reason it's in the
minus 1 D 2 plus infinity and T positive
and the initial conditions we are having
this the first UX comma 0 comma s is
equal to 0 and derivative of U X comma T
comma s with respect to T at T is equal
to 0 is its X comma s for all X
belonging to R
and we say that if this a special
homogeneous problem initial value
problem here
we do have del 2 u over del T 2 minus C
square del 2 u over del X 2 is equal to
0 this is homogeneous equation now what
we are taking this function capital u is
in the three variables now XT and yes
and if we can solve this and here the
second initial condition about the
velocity is its its Commerce where you
do see is that is we do have initially
that problem in the non-homogeneous
problem this was the right hand side
function of my non-homogeneous very
collision so which we are having this X
comas
now if this problem can be solved so now
this is actually homogeneous one with
non-homogeneous initial conditions of
course we do know so if this a special
homogeneous problem does as a solution
for each parameter yes now s is the
parameter here we had made if it is can
be solved for each parameter is then
dermals principle says we can find out
the solution of this which can be
related with the solution of this now
let us see what is the Dumas principle
sis Dumas principle says if the solution
of this special homogeneous problem u X
comma T gamma s is twice differentiable
and continuous in X and T that is the
function UX t Commerce which is in the
three variable is continuous with
respect to X and T and it is twice
differentiable that is it has the second
derivative with respect to X as well as
with respect to T and continuous in is
then the solution of homogeneous a non
homogeneous problem that is U XT can be
given as integral 0 to T capital u X
comma t minus s comma s DS this is the
solution of non-homogeneous problem so
what we are doing this we
changing our non-homogeneous problem
that is non-homogeneous wave equation
with homogeneous initial conditions we
are changing it to a special homogeneous
equation - with non-homogeneous initial
conditions that we do know how to solve
that we are doing is with one parameter
s as well and then with that parameter s
we are integrating it and we are saying
is that this is the solution of this let
us say that is whether it is solution of
our non-homogeneous problem let us prove
this principle we are saying is that
capital u is the solution of this
problem ux t we are saying is that this
should be this one now let us try to see
that can be satisfied this is satisfying
the our non-homogeneous equation or not
so first find out its derivative with
respect to G now when we are finding out
the derivative of this is small u X
comma T with respect to T we see that u
XT is actually integral of from 0 to T
capital u X T - is s DS that is we are
having a function which is depending on
T as well as the limit is also depending
on T we want the derivative of this
integral we have already done in fast
course this kind of integrate this kind
of differentiation under the integral
sign where the limit is itself a
function of the variable so using that
one we do have it it is u X comma 0
comma T plus integral 0 to T the partial
derivative of this function with respect
to T that is U T we are writing that is
a partial derivative with respect to T
of this function XT minus s is
integrated with respect to Y s now my
initial condition says u X comma 0 comma
s is 0 for all is so
when I'm keeping easy here SST this uux
comma zero comma T this would be
actually zero
so what I would get is that first
partial derivative of U XT is integral 0
to T the partial derivative of u capital
u xt minus s Khmers with integration
with respect to yes similarly i go again
to find out the second derivative that
is again differentiate this function
with respect to t again we'll use the
same formula so what we would get UT x
comma 0 comma t plus integral 0 to t u
TT because now i want the partial
derivative of this function with respect
to t again that is it is second partial
derivative with respect to t that is it
is Delta u over del T to with of course
what arguments if the function is of XT
minus s is integrated with respect to s
now this second condition is about the
second initial condition is del u over
del T of this function at T is equal to
0 is its X comma so that will substitute
over here so what we get the second
derivative derivative with respect to T
of UX T's its X comma T plus 0 to T
second derivative UT T of X comma T
minus s comma is integrated with respect
to Y s similarly we can find out for
this derivative with respect to X now
when we will find out the derivative
respectives then this integral does have
the fix limit that is then the limits
are not variable again using that
formula for the differentiation under
the integral sign doubles two times we
would be using that is twice
differentiable we do get the second
derivative of U X comma T is integral 0
to t second derivative partial
derivative with respect to X of capital
u the function is of X comma T minus s
comma s integrate it with respect to s
in the whole range 0 to T now substitute
this u GT on u xx in our non-homogeneous
equation our equation is Del 2 u over
del T 2 minus C square del 2 u over del
X 2 so if I substitute it here what I do
get the first second derivative with
respect to T we have got HX comma T plus
integral 0 to 2 u TT X comma t minus s
SD is minus C square times second
derivative with respect to X that is
from here integral 0 to t u X X X comma
T minus s comma s DS now rearrange these
terms what I do get h xt are such plus
these two integrals we could write it
out as the UT T XT minus s comma 's
minus UT t X comma t minus s comma s yes
now if we see capital u is the solution
of our homogeneous problem or
homogeneous equation homogeneous
equation was so here it has to be C
square or also the C square is also
going to come over here so this C square
u x6
that would be now a special homogeneous
problem which says is del 2 u over del T
2 minus C square del 2 u over del X 2 is
0 since capital u is satisfying that
equation that means this equation must
be 0 since what we do get is this is
nothing but H X T that is it is
satisfying our non-homogeneous wave
equation that says is u X comma T
integral 0 to T capital u X comma t
minus s comma s integrate with respect
to s is a solution of non-homogeneous
condition non-homogeneous very equation
moreover it is going to satisfy our
initial conditions us with now how to
find out the complete solution to find
out the solution of that special
homogeneous problem we will try to use
the D alembert's solution because that
is only initial value problem and only
initial value problem when we are
talking about we do find out the daily
umber
solution is more nice and more elegant
way after finding out the solution so we
do have this problem homogeneous wave
equation del 2 u over del T 2 minus C
square del 2 u over del X 2 is equal to
0 our initial condition u X comma 0
comma s is 0 and UT X comma 0 comma s is
its axis for all X belonging to R on
this equation is on the whole real line
with T positive so if I compare it by
usable that is what we have already done
there we had this our wave equation did
del 2 u over del T 2 minus C square del
2 u over del X 2 is equal to 0 and the
initial conditions were u X comma 0 is
equal to F X and del u over del T T
comma 0 is equal to G X now if I compare
my wave equation is alright
the same similar kind of things the only
thing is that here we are having capital
u here we are having a small u if I
compare the initial condition my FX is
coming as 0 and this second initial
condition G exist it's X comma s so what
we will do get the solution for this the
solution was if you do remember we have
done that your X comma T the solution of
this initial problem was initial value
problem 1/2 times f of X minus CT plus
half of plus f of X plus CT plus 1 by 2
C integral minus X minus CT 2 X plus CT
Z's I desire GX was the second initial
condition initial velocity now if I
compare it my F is 0 here and G X here
is H X s so what we do get the solution
capital u xt comma s as 1 upon 2 C
integral X minus CT 2 X plus CT h GI
comma s DS I so this is the solution of
my special homogenous problem
so what will be the non-homogeneous
problem solution initial value problem
solution that we can find it out using
our domeless principle which says is
that it should be integrated around so
we do get us integrate this whole thing
from zero to T with respect to s so this
one upon two C is constant that I have
taken out integral 0 to T integral off
now the function has to be you its t
minus if you do remember
so here T minus s if I am writing here
that is that the place of key I have to
replace it by t minus s so what I am
getting is the limit says X minus C
times t minus 2x plus a times t minus s
and the integrand is s Z is DZ and this
whole thing whatever this integral we
are getting that we have to integrate
with respect to Y s so this is what is
the solution of non-homogeneous wave
equation with homogeneous initial
conditions now if I do have a problem in
which I do have non-homogeneous initial
conditions what we do use is since we
had find out the solution with respect
to the homogeneous initial conditions we
do know that in the similar manner we
can find out the solution for the
non-homogeneous initial conditions how
we could do is before that let us see
this is that is this solution would be
existing if my non this right-hand side
function is sixties continuous and
differentiable in X and continuous in T
then this uxt solution of
non-homogeneous initial value problem
that is what the condition we do require
is only that these integrals must exist
for integrals to be existing I do
require that this integral that is this
function must be continuous and since we
are using double integral and partial
derivative with respect to T we do
require that is
it should be absentee that is it it just
requires that with respect to its it
should be differentiable and with
respect to J it should be only
continuous now if I do have this
non-homogeneous wave equation del 2 u
over del T 2 minus C square del 2 u over
del X 2 is equal to H X comma T and
non-homogeneous initial conditions as
well that u X comma 0 is equal to a
fixed and u T X comma 0 is equal to G X
then we can find out the solution first
what we do is we take the homogeneous
initial conditions and then take this
then we can change it to the especial
homogenous problem with non-homogeneous
initial conditions in that
non-homogeneous initial conditions we
can fix up these non-homogeneous initial
conditions of years from here itself and
then we can find out the solution of
homogeneous wave equation with initial
conditions as non homogeneous we can get
it using the D alembert's formula and
then we can use the principle this
doramas principle for finding out the
solution of this non-homogeneous
equation on that we can add it up that's
what it simply says it whenever we do
have non-homogeneous problem the basic
method says is we convert it to the
homogenous that is we just simply put
right hand side 0 and solve it that
gives me solution of homogeneous one and
for non homogeneous one we find out a
particular solution and that V art and
that we can do using the superposition
principle
it's say the superposition principle is
ourself for the homogeneous ones but
that acclimates is that if it is non
homogeneous and we are adding the
homogeneous solution solution of
homogeneous equation and non particular
solution of non-homogeneous equation we
are getting the solution of
non-homogeneous equation that principles
can be extended with the initial
conditions as well that is if the
initial conditions are homogeneous or
non homogeneous we can still change it
that is we can use it we can change this
initial condition so homogeneous that is
we write the right-hand side is 0
and then if we can find out the solution
of this non-homogeneous and we can add
it up and that will give me the solution
with this initial condition nan who uses
the initial condition so that principle
would be using here what it says is for
this we can use the solution is because
for non-homogeneous conditions the D
alembert's solution for the homogeneous
wave equation is half of FX minus C T
plus f of X plus C T plus 1 upon 2 C
integral X minus CT 2 X plus CT d ZZ
Jesus where my F is this initial
condition and J is the function
difference in giving the second initial
condition then we just add up the
solution of this non-homogeneous problem
which we just have find out using the
diamond spin spell which says is it
should be 1 upon 2 C integral 0 to T up
integral X minus C minus C times t minus
s 2 X plus C times t minus s it is I s
diesel and coal integrate with respect
to yes now let us do again the solution
would exist if I do have that my the
function f G and H are continuous and
differentiable that is my function UX T
should be continuous and twice
differentiable so that it is it can
satisfy these equations let us do one
example solve the initial-value problem
del 2 u over del T 2 minus 4 times del 2
u over del X 2 is equal to 2 T with
non-homogeneous initial conditions UX
comma 0 is equal to X square and UT X
comma 0 is 1 now what we do is first we
will solve the homogeneous equation with
this non-homogeneous initial condition
using the D alembert's principle and
then using the de this Dumas principle
would write the solution of this
non-homogeneous equation with
homogeneous initial conditions and that
would
the solution of the whole problem by
adding up so we are being given here
right hand side that we have been noted
HX comma T that is 2t ant initial
condition we had used that u X comma 0
should be fixed so FX here is X square
on UT X comma 0 is 1 so GX is 1
moreover in this initial condition if in
the wave equation here we used to have C
square so see we put says 2 because in
solutions we are having is X plus C T
and X minus CT kind of thing so C is 2
here now let us first get the D
alembert's solution for the homogeneous
problem with the so corresponding a
special homogenous problem
del 2 u over del T 2 minus 4 del 2 u
over del X 2 is equal to 0 with u X
comma 0 comma s is equal to X square +
UT X comma 0 comma s is equal to 2 years
so now what we have done is we have
changed our non-homogeneous problem to
the homogenous problem with non
homogenous condition after the initial
conditions itself for non homogenous
still have done it the only thing what
we have changed is that is rather than
using that there GX I am using here is
true is so using it
Dalembert solution if we do go ahead
what we will do is we will first find
out the characteristics of this equation
and then change the transform it and do
it so characteristics for this here B
Square is 0 B is 0 is minus 4 and C is 1
so we would be getting his elliptic 1 DT
by DX would be getting is b square minus
4ac is plus minus 4 by 8 that is 1 by 2
so characteristic cop you would get 2 T
is equal to XA 1 & 2 t is equal to minus
X plus C 2 him here we can transform it
using the transformation V as X minus 2
T and J does X plus 2 T we are changing
it to the canonical form so for this we
do know that del V over del X would be 1
del V over del T is equal to minus 2
Delta over del X would be 1 Delta by
Delta Y would be 2 using this
change will tell you our del X we could
write del u over del V times del V over
del X plus del u over del J times del u
over del X just putting del V over del X
has 1 and del u over del X is equal to 1
we care it as del u over del V Plus del
u over del J again the second derivative
simply says is that differentiate this
with respect to X again again we would
use the chain rule that is del upon del
V of this whole function Del E upon del
u plus del u upon del J times del V over
del X plus derivative with respect to
charge of this whole function del u upon
del y plus del u upon delicious times
bleed del Jade over del X now again del
V over del X and del Jade over del X we
are nothing but the one so we would get
it as del 2 u over del V 2 plus 2 times
del 2 u over del V del J plus del 2 u
over del J 2 since we had assumed that
our X a function U is continuous in X
and T that guardant is that we can get
that del 2 u over del v over del J our
del 2 u over del G del V that would be
same similarly we can go ahead with the
derivative with respect to T again using
the chain rule in the similar manner
first differentiating with respect to B
then derivative up with respect to T and
then differentiating with respect to
check on the derivative of Z with
respect to T so DV over DT here is minus
2 and del V over Delta Delta door del T
is plus 2 we do get is minus 2 times del
u over del V minus del u over del J
again use the second derivative that is
again differentiated with respect to T
we do get it should be four times del 2
u over del V 2 minus 2 times del 2 u
over del V del J plus del 2 u over del J
2 now substitute this in our given
equation so we do get the canonical form
up del 2 u over del T 2 minus 4 times
del 2 u over del X 2 is equal to 0 in
this we are going to substitute now the
derivative with respect to V and we said
so we are getting is four times del 2 u
over del V 2 plus 2 times del 2 u over
del v del J plus del 2 u over del Del 2
that is roughly the derivative of del 2
u
- so this I'm keeping movius this should
be equal to four times del 2 u over del
V 2 minus 2 times del 2 u over del V del
X plus del 2 u over don't you - now if
we just simplify it we to get is that
del 2 u over del V del V is equal to 0
that is it has been changes has changed
to the canonical form we do get very
easily that is the terms involving the
second derivative with respect to V and
second derivative with respect to Jed
they are equal on both sides so they
would be cancelling it out now this
equation we can solve by first
differentiating with respect to G and
then differentiating with respect to V
so integrating with respect to Chet
first we do get del u over del V as Phi
of V why because this is zero so we'll
get the constant with respect to Z but
that may contain a function of me
because it is a partial derivative with
respect to J again integrating it with
respect to V we do get us integral of
Phi V DV plus a constant constant main
well the function involve the Vario
budget so let's write that the function
of sysop this we have already done this
is again just for you could say is your
practice that is how we are using this
error D alembert's and this canonical
form to solve the equations since this
integral would be only a function of V
let me write it out another Phi so Phi
of V plus sorry now change pocket to in
X and T so what we have got u X comma T
comma s as Phi of X minus 2 T plus Phi
of X plus 2 T now what would be this far
inside for that we require to determine
this Phi n Sarita unknown functions
general functions we use the initial
conditions our initial conditions are
being given as us X comma 0 comma s as X
square + UT X comma 0 comma s 2 s so u X
comma 0 comma s as Phi X + IX is equal
to XC square + the derivative of this
you XTS that would be minus
- five - X minus two T plus two x side -
X plus two T because - means the
derivative with respect to this
evaluated at T is equal to zero this is
giving us one yeah DX was giving us once
of here it should be one when we are
keeping it over here we do get it as Phi
X minus sorry X is equal to K X not plus
half integral X naught to X 1 with
respect to Z that is DS IDs IDs a so now
we have got one equation over here in
volume Phi inside and another involving
Phi inside from these two ones we do get
the solution and TX naught is of course
a constant when we are integrating it
with respect to x we are getting is 2
Phi X is equal to FX minus K X not minus
1/2 X naught to X DS - say X is equal to
FX plus K X not plus 1/2 X naught to X D
Z so this you can this sa we are just
taking out rather than we are using is I
because this s may should not confuse
with our parameter is so what we get the
solution is U X comma T commas as Phi of
X - 2 T plus Phi of X plus 2 T now
replace Phi inside in the terms of U and
F and G we do get half of X - 2 T Square
minus half of K X not minus 1 by 4 X
naught to X minus 2 T DZ and Phi X plus
2 T is half of X plus 2 T square plus
half of K X not plus 1 by 4 times X
naught to X minus 2 T DZ now add it up
what we do get it X square plus 4 T
Square it should be 2 d square X is quad
plus 2 T Square yeah so it says X minus
2 T whole Square and X plus 2 T whole
square so it should be X is cot plus 4 T
Square on this gate this constant would
it be cancelling out because the same
constant on both this one on this
integral that is
it's not X minus 2 T and minus sign I
could write it as X minus 2 T 2 X naught
so we do get finally the integral 1 by 4
times integral X minus 2 T 2 X plus 2 T
DS I now evaluate this integral this
integral DS I that we would get Zi Zi is
evaluated from x minus 2 T 2 X plus 2 T
what we would you get 1 by 4 X plus 2 t
minus X minus 2 T we add it up we just
get it only T so we have got the
solution you X comma T comma is as X is
car plus 4 T square plus T now we just
go with the solution of non-homogeneous
problem using the dermals principle
which says is that it should be 1 upon 2
C 0 2 T integral X minus C mine times t
minus s 2 X plus C times T minuses its I
s DS I DS no it's is this nothing but 2
s so we get it has to is integral from X
minus C times to my t minus s with
respect to Z so that is we are not
having any function of Z over there so I
would get simply to us and then X plus C
I say t minus C is minus X minus C T
plus CS and that would give me only say
times t minus s twice three times t
minus yes
twice see that would be getting canceled
it out
what we would be left is 2 times 2 s is
there so 2 s multiplied with t minus s
so integral to I have taken outside
integral 0 to t s times t minus s that
is TS minus s this car integrated with
respect to s now integrated here we have
to integrate it with respect to s so it
this will give me s square by 2 K times
s square by 2 evaluated from 0 to Chile
I would get T times T Square by 2 and s
2
get in cancel it out so I would get T
cube similarly this one minus 2 s square
you could say s squared US has integral
would be s cube by 3 evaluated from 0 to
T I would get T cube by 3 so what we
have got T cube minus 2 T cube by 3 that
gives me T cube by 3 so this is what we
have got the solution of this special
homogeneous is this non homogeneous
problem with homogeneous initial
condition and previously we had find out
the solution of homogeneous problem with
non-homogeneous initial conditions so if
we just add up these two prints
solutions we do get the final solution
of our problem initial value problem as
X square plus 4 is 4 T square plus T
plus T cube by 3 now let us see that is
whether it is a solution of our given
equation or not you can just find out
its derivatives with respect to T and
with respect to X second derivatives try
to put if I am finding out the second
derivative with respect to X over here I
would get it simply 2 and if I find out
the second derivative with respect to T
you can just get it out that is here I
would be getting it it and this would be
0 and this would be a first derivative
would be 3 T Square that is a key
squirrel then we would get 2 T so we do
get is that is when we are keeping it in
our equation we can get that it is
satisfying our solution our equation we
would be getting it equal to 2t and when
does it satisfy our initial conditions
let us see is that is there a the
derivative of this with respect to t is
8 t plus 1 plus G square so from here
actually you can see that the second
derivative would be 8 plus 2 T so you
can just get it that is 8 minus 8 plus 2
T that is you
get the solution as to t so no initial
conditions are at T is equal to zero the
function should be X is first so here if
I put T is equal to zero
all these terms would be zero I would
get only X is square and UT at T is
equal to zero
it's giving us one so if I keep T is
equal to zero for T is equal to zero
over here I would get only one that is
satisfying our equation also well it is
satisfying the initial condition so this
is the solution now let us come to
another problem solve the initial
boundary value problem del 2 u over del
T 2 minus del 2 u over del X 2 is equal
to 1 with initial conditions UX comma 0
is 0 and UT X comma 0 is 0
I'm boundary conditions you commas u 0
comma T is 0 u L comma T is minus L LS
square by 2 where this equation is for X
0 to N and T is positive so you do
remember that is if we are having is
that our bound has been fixed into 0 to
L that we are getting is this boundary
condition and this initial condition so
now we are having actually
non-homogeneous equation homogeneous
initial conditions but with boundary
conditions how to solve this one we can
solve this kind of equations using just
take this non-homogeneous equation we
just take the homogenous part use these
boundary conditions unused the Fourier
series method so what we do is now we
will go with the method II which says is
that find out a particular solution
aren't find the general solution for the
homogeneous one so what we do is that is
we take one particular solution of this
non homogenous problem and the boundary
value problem we then will treat with
this homogeneous equation in the dais
boundary condition the boundary value
problem find out the solution using the
Fourier series method and then such
by the initial kundan put that our
particular problem over there and find
out the using this initial condition
find out the solution so let us do that
is how we are going to do it so first
this particular solution let's I am
taking it here in general let us say
that particular solution BUP X comma T
first then solve the boundary value
problem del 2 u over del T 2 minus del 2
u over del X 2 is equal to 0 with
boundary conditions use 0 comma T is
equal to 0 u L comma T is equal to minus
L square by 2 in last lectures we have
done that is how to solve this
one-dimensional wave equations with
boundary conditions using the foil
method using that variable separable
method the this one so we have got this
using this products method and
superposition principle if you do
remember we had got that the now I'll
call this solution of this as
homogeneous that u H X comma T that we
have got us summation n is running from
1 to infinity a n cos n PI T over L plus
n star sine n PI T over L sine n PI x
over L now if I just use the our this
solution u P X comma T as the particular
solution of our non-homogeneous equation
then the solution of non-homogeneous
boundary value problem would be u X
comma T summation n is running from 1 to
infinity n cos n PI T over L plus n star
sine n PI T over L into sine n PI x over
L u P X comma T you do remember that
there we are having C also here C is 1
so this C would be 1 now we try to
satisfy the initial conditions over here
initial condition says this at T is
equal to 0 this function should be 0 and
the derivative of this function at T is
equal to 0 with respect to T the
derivative with respect to T at T is
equal to 0 should also be 0 so let us
try to satisfy these initial conditions
they says that is 1 when I am putting T
is equal to 0 the
cos 0 is 1 and sine 0 is 0 so I would be
getting only from here the only term a
in I'm sine n PI X so what we would be
getting is summation n is running from n
to 1 to infinity a n times sine n PI x
over L Plus u P X comma T that should be
0 that I'm putting on the right hand
side so it will satisfy nishal
conditions that is if I find out this n
such that this is happening no what does
it says is that I should take my
whatever with that particular solution
that particular solution at T is equal
to 0 whatever we are getting my ends
should be actually the coefficients of
Fourier sine series of that function and
similarly when I take the derivative
with respect to T here I would be
getting is minus n PI over L n times
sine n PI T over L and plus n PI over L
n star sine n PI T over L that says is
that n PI over L would also come over
here that n star times n PI over L sine
n PI x over L should be equal to the
derivative of this partial this is a
particular solution with respect to T at
T is equal to C so what it says is that
I should get the solution as using the
Fourier series I could get that is
whatever this particular solution I do
have that evaluated at T is equal to 0
what function I do get that should be a
function of X we do get it in the terms
of the Fourier series and similarly we
should get over here in the N stars we
could get from the other initial
condition now let us do one example
suppose in this particular one I do take
my particular solution is x square plus
3 by 2 T Square now this would satisfy
our equation because our equation CSS
del 2 u over del X 2 minus del 2 u over
del R del 2 u over del T 2 minus del 2 u
over del X 2
is equal to one so the second derivative
with respect to T will give me here only
not three by two is in the case that is
three and the second derivative with
respect to X will give me two only so
three minus 2 would be 1 so this is
actually a solution of our
non-homogeneous differential partial
differential equation satisfying that
one so this is a particular solution now
in this particular solution if I put T
is equal to 0 what I do say get is let a
and I should get as the Fourier
coefficient of Fourier sine series are
minus X square if I take the derivative
with respect to T I would get it as 3t
and at t is equal to 0 that would be 0
that says this n starts I would be
getting it actually n PI over L times n
star would be 0 are a and stars would be
0 so what would be getting the solution
we would be getting the solution as
summation n is running from 1 to
infinity a and cos n PI T over L times
sine n PI x over L plus X square plus 3
by 2 T Square where ants are being given
as the Fourier coefficients
after Fourier sine series of minus X
square that is a and should be integral
0 to L minus X square sine n PI x over L
DX so that is how we are solving this
boundary value problem also as one
example let us move to one more
practical problem which says is that in
heat conditions we have already done
this heat equations that is this
equation which we call the Laplace
equation del 2 u over del X 2 plus del 2
u over del Y 2 is equal to 0 this we
have our turnaround in the first lecture
and second lecture we had already find
it out that this is an elliptic equation
it is not having any characteristic
moreover we are knowing is that this is
not well posed Cauchy problem is not
well posed that says it if I use any
initial conditions or boundary
conditions then on any characteristic
curves it's not necessary that that
me a unique solution that is it is not
the cost problem for such equations is
not well forced an important application
of fergus's coming from this
steady-state two-dimensional sheet flow
that is heat equation - damn this is a
two dimensional heat equation del 2 u
over del X 2 plus del 2 u over del y 2
is equal to Del u over del T if we say
is that from here this heat is heat flow
is steady state that means it is
independent of time if it is independent
of time that says del u over del T would
be 0 and then this toda Invention 'l
heat equation would be change into our
laplace equation this is one of a
particular importance let us so let us
see this is what is this elliptic
equation because elliptic equation this
Laplace equation is elliptic and from
here the Cauchy problem does not is not
well posed that is we may not get unique
solution so how do how to find out the
solution of this let us talk about this
two dimensional laplace equation if it
is on some reason that is it is defined
under some reason on the two dimensional
X while under some reason since the
Cauchy problem is not well posed what we
say is that boundary problems boundary
value problems that is what the boundary
for the kind of sign heat conditions we
can have after using the different side
condition if it is on the some reason
our if it is working then we can have
different kind of pose different kind of
boundary conditions and from those
boundary conditions we do have we name
them different problems one is that we
could put one boundary condition as my U
is given on the boundary of this reason
that is see this is then it is called
the District problem if it is given that
is because this Cauchy problem is not
well posed what it says is that its
normal derivative with respect to T may
not be having the unique solution that
says the other problem which will give
me the solution that is called a Neumann
in which we do have that normal
derivative UN is given on this boundary
see are we may have mixed kind of
problems in which on some part of this
see the U is given and on the rest of
the part of the see this un is given so
let us just talk about this literate
problem that is we are talking about the
boundary value problem with Laplace
equation where boundary conditions are
given for the U on the boundary of the
reason so for simplicity let us take the
first one that is directly from the one
dimensional heat equation if I am just
changing it to two dimensional heat
equation with initial with the del u
over del T is zero they steady state one
so let's say the rectangular reason so I
am just considering the digital problem
and rectangular reason so this is the
reason in which we are talking about
this reason and the boundary conditions
we are having is that is on the three
sides of the boundary we are putting
that is U is zero and on the upper side
of the boundary we do have that it is f
X so let us post the problem we do have
del 2 u over del X 2 plus del 2 u over
del Y 2 is equal to 0 with boundary
conditions that you 0 Y is 0 u 0 Y is 0
u a Y is 0 and u X comma 0 is 0 and UX
comma B is FX let's see how to solve it
so just go by the separation of variable
method that is using the product
function FX on Gy we do get del 2 u over
del X 2 as f double dash XG y del 2 u
over del Y 2 as FX times T double dash Y
now substituting this in the given
equation del 2 u over del X 2 plus del 2
u over del Y 2 is equal to 0 and
dividing it by F into G what we do get F
Double Dash X upon FX is equal to minus
G double dash Y upon Gy since left hand
side is its on right hand side is y if
there has to be equal they must be
constant so let us say that is constant
is minus K what we are getting is 2 or D
are
ordinary differential equation one is f
double dash plus K F is equal to zero
and another is Z Double Dash minus KD is
equal to zero and here we do have the
boundary conditions if we change we do
get that F 0 is 0 and FA 0 here the
boundary condition will be having is C 0
is C so now we do have two boundary
value problems which is on the ordinary
differential equations for this boundary
value problem we do know that the eigen
values and eigen functions are that is
eigen values are in Phi over a whole
square and eigen functions there's a
solution if an x is sine n pi x over a
now second ordinal differential equation
is this one now if this is the eigen
values then this K would be writing in
just in Phi sin Phi over a whole a
square times T so this is 1 for this the
solution is actually a and e to the
power and Phi Y upon a plus B an e to
the power minus n Phi by f1 a now if I
put J is equal to 0 G at 0 is 0 this
give me a n plus BN is equal to 0 RB n
is equal to minus n so we do get the
solution as n times e to the power n PI
Y upon a minus e to the power minus n
Phi Y upon a that says is in the
hyperbolic function to a n sine
hyperbolic and Phi Y upon a so what we
have got the solution U and X comma Y as
n star sine hyperbolic and Phi Y upon l
into sine n pi x over a this 2n I am
writing this n star so we do have seen
that with this solution is satisfying
the boundary conditions on the three
sides the solution we have got n star
sine hyperbolic n PI Y upon a sine Phi n
PI x over a this is satisfying the
boundary conditions on the three sides
now let us see that is whether it would
satisfy the boundary condition on the
fourth side or not certainly single
solution we cannot find out this one so
we use this superposition principle and
lit this solution u XY as summation n is
running from 1 to infinity u n X comma Y
where u n x comma bar is this
so that says is numb we want to satisfy
this boundary condition the last one
that is U at X comma B is f X so u X
comma B if I am just using this UN X Y
over here if I put our Y is equal to B
what I would be getting is a list our
sine hyperbolic in 5y up in Phi B upon a
into sine n PI x over a is equal to FX
now from here if we do see what this
says is that this bracket this
coefficient in a star sine hyperbolic n
PI B over a this must be actually the
coefficient of Fourier sine series of FX
now if X is a function which is defined
from 0 to a at this point so now we have
to have odd expansion of this and it for
your expansion for the odd extension of
FX from 0 to a 2 minus a 2 plus C so now
we do have is n star using that it just
gives us 2 upon that is actually a n is
star times sine hyperbolic n pi b upon
air should be 2 upon L 2 a point a so
that whole thing I have taken this 2
upon their sine hyperbolic and Phi B
upon a integral 0 to a FX sine n pi x
over a DX using the Euler's formula for
the Fourier coefficients of odd
expansion of Fourier expansion for this
FX so we get the solution of this our
boundary value problem on the where this
boundary conditions we do have this one
in the rectangular reason ask this one
so now this is what we have got laplace
equation with boundary conditions we
have got now if I take non-homogeneous
laplace equation we do know that this is
called Poisson equation del 2 u over del
X 2 plus del 2 u over del Y 2 is equal
to F X Y
how to go for the solution normally we
are having is this again we do know that
this is elliptic equation again we do
know because this is elliptic equation
the Cauchy problem for such equation is
not well posed that says this we have to
choose the initial conditions are the
site conditions in such a manner so that
we do get some solutions unique
solutions normal condition is that is
whenever we are using non homogeneous
equations we use homogeneous initial
conditions are with homogeneous
equations we use non homogeneous initial
conditions we do have is that many times
homogeneous equations we do have that
the solution is existing just now we
have done this Laplace equation also
that homogeneous equation and this
boundary conditions we could say that we
could treat as initial conditions as
well also we could find out a solution
so if the problem is non-homogeneous we
can change it to the homogeneous one and
take the particular solution of this non
homogeneous one and add it up and find
out the using the initial conditions we
could find out the solution are we are
having is homogeneous equation with
non-homogeneous initial conditions that
already we have done so if you do have
non-homogeneous equation with
non-homogeneous initial conditions we
just go ahead with the homogeneous
equation with non-homogeneous initial
conditions and then out of the solution
of a particular solution of
non-homogeneous equation so in this
Poisson equation also we can define the
destroyed problem as well as the Neumann
problem in the destroyed problem it says
is that this you should be initial
condition should be GX for the boundary
on the Neumann problem says is that we
do have the normal derivative normal
derivative is defined on this one so
either we take this non-homogeneous the
Poisson equation with GX and HX as 0 and
R we do have this
here as deal on them GX and HX all
assets on all the cases as we have done
in the laplace equation similar manner
and in the wave equations we can do it
in the Poisson equation acid so today we
had learned the one-dimensional wave
equation in the non-homogeneous one
using homogeneous initial conditions
then using non-homogeneous initial
conditions then to find out a boundary
value problem using homogeneous initial
conditions and so on then we have gone
to the Laplace equation using the
boundary conditions and we had learned
about the Poisson equations as well so
that is all in today's lecture thank you