Partial Differential Equation - Mathemetics Video Lecture - Engineering Mathematics

we
welcome to the lecture series on
differential equations for undergraduate
students
today's topic is partial differential
equations like ordinary differential
equations partial differential equations
are also equations in which is to be
solved for unknown quantity and the
unknown quantity here is also a function
but unlike our noil differential
equations here the unknown function
could be of more than one variable and
the equation could be a relationship
between that unknown function and its
partial differentiation so we could
define that partial differential
equation as a partial differential
equation is an equation involving a
function U of several variables and its
partial derivatives with respect to
those independent variables so ours in
the ordinary differential equations we
were having a function unknown function
which we had denoted by Y and one
variable that we denoted by X here we
may have more than one independent
variable status now we'll denote those
independent variables as X Y and T so on
and the unknown function as you let us
say some examples of partial
differential equations say del u bar del
X is equal to Del u by Del T u times del
u by Del X minus del u by Del T is equal
to 0 ETA e to the power x times del u by
Del X plus 4 times del u by Del T is
equal to t1 upon X del u upon del X plus
del 2 u upon del X 2 minus del 2 u upon
Del T 2 is equal to X square del 2 u
over del X 2 minus e to the power 2 X
del 2 u upon Del T 2 is equal to u cube
so you see here is in all these
equations the unknown function is U in
the first equation this is U is a
function of two variables X
90 and we do have a relationship with
the partial derivatives with respect to
X and with respect to T similarly he
also here also this is also an unknown
function is of the two variables X and T
this last equation is also known as
one-dimensional wave equation when we do
have del u upon del T is equal to C
square times del 2 u over del X 2 this
is known as one-dimensional heat
equation another equation known as Del T
u upon del X 2 plus del 2 u upon Del Y 2
is equal to 0 you see here the
independent variables are X and Y and
the unknown function is U this is known
as the tool I mention Laplace equation
del 2 u over del X 2 plus del 2 u over
del Y 2 is equal to a function of x and
y this is known as two-dimensional
Poisson equation you see here is the
differences that is here is the right
hand side is 0 here is a function of x
and y this function is a known function
another example is Del 2 u upon Del T 2
is equal to C square times del 2 u upon
del X 2 plus del 2 u upon Del Y 2 this
is 2 dimensional wave equation is not
one dimension there is two dimensional
you see here the unknown function U is
of the three variables X Y and T and we
do have partial derivatives with respect
to T x and y then we do have del 2 u
over del X 2 plus del 2 u over del Y 2
plus del 2 u over del Z 2 is equal to 0
this is actually three dimensional
laplace equation here we do have three
variables X Y and in all these
equations wherever we have used the C
that was used as a constant that is the
constant and we had used the variables
as one variable as T another variables
as X Y and jet so this variable T is
referred as join and this X vultured
they are nothing but the Cartesian
coordinates and you see sometimes he
said is one-dimensional equation and the
x we said is two dimension or three
dimensional they are because of this
or many variables of this Cartesian
coordinates we are having as here we are
having two Cartesian coordinates x and y
we call it two-dimensional here we are
having is XY undred so we call it three
dimensional whenever we are having is
del UX plus del u del u del u X del u
over del X plus del u over del T is
equal to see kind of equation though we
had only X and T that is one dimensional
we are calling it all these equations
which I have shown you here as an
example they are special importance in
the partial differential equations that
is why here have just given a little bit
of insight of those kind of equations
let us understand some more basic terms
and concepts in the partial differential
equation it is similar to the ordinary
differential equations but some more
concepts you would be requiring so basic
concepts first is order as in the
ordinary differential equations we
define the order of a differential
equation as the highest order of the
derivatives so here also we say is the
order of the highest derivative is
called the order of the equation say for
example del u over del X plus del u over
del T is equal to 0 this has the
derivative of the first order only
similarly u times del u over del X minus
del u over del T again both the
derivatives of the first order only e to
the power X del u over del X plus 4
times del u over del T is equal to T
again he are having all the derivatives
occurring are of the first order only so
the highest order of the derivative
occurring or the first order and in the
second equation we do have the unknown
function U also but the derivative is
only of the first order so all these are
the examples of first order equations
similarly if we see other examples del 2
over del T 2 is equal to C square times
del 2 u over del X 2 here the derivative
occurring is of the second order this is
also of second order del 2 u over
x2 plus del 2 u over del y 2 is equal to
0 here also the second-order derivative
that the highest order derivative is
occurring is the second order in this
also the highest order derivative which
is occurring this of the second order
you remember that this is our our
one-dimensional wave equation basis two
dimensional Laplace equation this is two
R I mention this is two dimensional
Poisson equation the highest order
derivative is occurring is of the second
order similarly in this one versus two
dimensional wave equation so we do have
all these equations the order of these
equations is the second order mixed the
term is linear partial differential
equation as in the case of ordinary
partial differential equations we set it
to be linear if the derivatives and the
unknown functions are occurring linearly
so similarly here also we are defining
linear partial differential equation the
partial differential equation is linear
if it is of the first degree in unknown
function and its partial derivative say
for example in our examples if we see
this example del us del u upon del X is
equal to Del u upon Del T we are having
is that the derivatives are occurring in
the first degree and linearly here also
the derivative that is second derivative
first derivative all are occurring
linearly the coefficient is C square C
is a constant here also del 2 u over del
T 2 is equal to C square del 2 u over
del X 2 the coefficients is C square
which is a constant so they are
occurring linearly in this equation you
see e to the power x del u over del x
plus 4 times del u over del T is equal
to T the coefficient of del u upon del X
is e to the power X it is not a function
or this is not containing you
similarly the coefficient of this is 4
on the right hand side this is T this is
you so we are having is that the unknown
function and its derivatives they are
occurring mean they are occurring
linearly so this is also linear here
also we see that derivative is the
function that the coefficient of the
derivatives del u over del X that is 1
by X the derivative of this coefficient
of this by 2 u over del X 2 that is the
1 del 2 u over del T 2 the coefficient
is minus 1 we are having is that the
coefficients are not having the function
of u are the other derivatives of
unknown function so this is also linear
so all these are examples of linear in
partial differential equations we'll
define some more kind of linear
equations one we call semi linear if all
the derivatives of U occur linearly with
coefficients depending only on X we had
seen in the last examples that is some
that the coefficients were depending on
X somewhere the coefficients were
depending on T so now if the
coefficients are depending only on X so
when I am calling it X it may have X Y
or depending upon if it is of
higher dimensional equation then this
equation is called semi linear let us
see some example of this one if I take
this example del 2 u over del X 2 plus
del 2 u over del y 2 is equal to f of X
Y this is semi linear here the code you
are having is that the coefficients we
are not talking about the other one we
are just talking about the coefficients
of the unknown function of its
derivative they are constant here here
also it is constant and what we are
having this that is here we are having
this at the right hand side that is the
constant function it is U and a function
of X and U we are saying is that
function is having u linearly and its
coefficient is depending only on X its
coefficients is are depending on this
may be a function of this one but the
coefficients would
only on X this is semi linear one
similarly we would define one more kind
of different partial differential
equation that we are calling quasi
linear partial differential equation
what it is if all highest order
derivatives whatever wave the second
order or third order first order highest
order we have to take in that equation
that occurs linearly with coefficients
depending on X U and lower order
derivatives of U then the equation is
called quasi linear see for example del
u over del T plus u times del u over del
X is equal to zero here up the unknown
function is U its highest derivative
over here is of the first order ifs
coefficient is depending on X U and
lower order derivatives because there is
first order so there is no lower order
derivative unknown function U is this so
here the derivative is here the
coefficient is U this we call quasi
linear differential equation in ordinary
differential equations we had only non
linear differential equations and there
we said is that is if the coefficients
are the constants are the coefficients
or the function of X there we have
discussed only one particular case of
coefficient of x and that we called
Cauchy equation there also we have done
something is called the homogeneous and
non homogeneous in the similar manner
here also we define this particular
equation is known as in use it bars our
equation nonlinear equation let us see
if it is not linear semi linear par
Selenia we call it del u over del X
square plus del u over del y whole
square is equal to C square you see the
derivative is Del u over del X and del u
over del Y and we are having its degree
two so they are not having a degree one
so this is nonlinear so
test example I was giving now as in the
ordinary differential equations here for
the linear partial differential
equations we again define the
homogeneous and non homogeneous so a
linear partial differential equation is
said to be homogeneous if each term of
the equation contains either the unknown
function R its derivatives and otherwise
the equation would be called
non-homogeneous what we mean by if you
do remember there is when we have used
in the Honorary differential equations
the our linear differential equations we
said is that all the derivatives on the
unknown function were occurring linearly
on the right hand side is 0 and if right
hand side was not 0 we called it to be
non-homogeneous the same thing is being
said here in the different manner that
is all that is each term is containing
either unknown function or its
derivative otherwise it is 0
so a homogeneous equation example del 2
u over del T 2 is equal to C square
times del 2 u over del X 2 so if I try
to write in that manner in which we used
to write the ordinary differential
equations we would write it as del 2 u
over del T 2 minus C square times del 2
u over del X 2 is equal to 0 so the
right hand side would be 0 similarly del
u over del T is equal to C square times
del 2 u over del X 2 that is each term
contains either the unknown function are
its derivatives we are not having any
function any term here which is not
containing U or its derivatives
non-homogeneous equation you see e to
the power x value over del X plus 4
times del u over del T is equal to Del u
over it should be del T is equal to T
now you see we are having this
containing the derivative of unknown
function containing the derivative of
unknown function but this is the
function this is the term which is not
contained neither containing you nor
it's any derivative so this is non
Oh genius equation similarly you see
here this term is containing the
derivative with respect to X this is
also containing the derivative with
respect to X this term is also
containing the derivative with respect
to T but this term is not containing any
of you or its derivatives first-order
second-order or with respect to X or T
anything so this is again equation which
is non-homogeneous now as well in the
ordinary differential equations we
define the solution we said is that
differential equations are in unknown
function u that is the unknown quantity
is a function U so here the solution
would also be a function U so which
function you would call the solution the
definitions is a solution of a partial
differential equation in some reason R
of the space of independent variables is
function that has all the partial
derivatives appearing in the equation in
some domain containing R and satisfies
the equation everywhere in our let's see
what we are saying is if there is an
equation partial differential equation
that means it will contain the
relationship between the unknown
function and its derivatives with
respect to certain coefficients so what
we say is a function which is continuous
and how all the partial derivatives
which are appearing in that equation
that is it must have all the partial
derivatives of that function and then
that function and its derivatives must
satisfy the equation everywhere in that
reason R then you would call this a
solution in that region R before seeing
is that is what is the solution let us
revisit our partial differentiation a
little bit so that whatever the terms
are whatever the partial derivatives we
would be involving in our
understanding of this partial
differential equation I am just giving
that little bit understanding over here
so let us see it with the help of
certain examples now if my function U is
of X and T where function is of X square
into sine T now we do know that we could
write its partial derivative with
respect to X that means we would treat T
as constant and the derivative of X is 2
times X and sine T that is as treated as
constant similarly if I want to find out
the partial derivative with respect to T
I would get it as X square into cos T
where this axis is being treated here as
a constant and the derivative of sine T
is nothing but the causative take to
another example if u is a function of X
square plus T Square whole square plus e
to the power x then what will be its
partial derivative with respect to X and
with respect to T so first let us take
with respect to X this was a function of
X and T so we will just treat it as a
chain rule or the function of function
so 2 times X square plus T Square and
then the derivative of this would be 2 X
because T is treated as a constant plus
this is the derivative of e to the power
X with respect to X is e to the power X
so what we do get is finally 4 times X
into X square plus T square plus e to
the power X now take the derivative with
respect to T and treat X as a constant
then del u over del T would be 2 times X
square plus T Square as a whole 1 and
then the derivative of this function
would be only 2t because X has been
treated as a constant and since it is a
constant so the derivative won't occur
here so what we get 40 times X square
plus T Square now this is a we have used
the here the chain rule of the ordinary
derivatives now for the partial
derivatives the chain will says that is
if u is a function
one of the two variables R and s where
both R and s are also the function of
two variables say X and T that is R is a
function of X and T and s is also a
function of X and T then if I want the
derivative of U with respect to X and D
Y U is actually a function of R and s
then the chain rule says that derivative
partial derivative of U with respect to
X is can be given as del u over del R
into del R over del X plus del u over
del s into del s over del X in the
ordinary differentiation what we had we
had only one variable and then though
that was a function of another variable
so we had it only do you / DJ done digit
by DX now here because it is of the two
variables so we are adding it up if it
is more than two variables again we
would go on this adding it up in that
manner similarly the partial derivative
with respect to T can be given as the
derivative of U with respect to R and
then derivative of R with respect to T
all of the partial ones because R is
also a function of X and T and then the
partial derivative of U with respect to
s that del u over del s and the partial
derivative of s with respect to T that
is del s over del T let us see one
example suppose U is sine of r plus s
square where your R is T times e to the
power X and s is x + T I want the
partial derivative of U with respect to
X and T so first the partial derivative
of U with respect to R del u over del R
that is cosine of R plus s square under
derivative of RS 1 s has been treated
constant if I want the derivative with
respect to s then it would be cosine R +
r SS square that is the derivative of
sine R + s the square and then the
derivative of s is cod that is - is
and would require actually Dell the
derivative of R and s with respect to T
and X so derivative of R del R over del
X treating X as the variable and T as a
constant T times e to the power X
similarly del R over del T would be e to
the power X T is constant is derivative
of T is 1 del s over del X would be 1
and del s over del T would be 1 so using
the change would del u over del X after
u over del R into del R by Del X plus
del u over del s into del s over del X
you would get cos R plus s square into T
to the power X plus 2 s cos R plus s
square into 1 that we can rewrite it in
this form so replacing my R and s bar
the function that is RS T times e to
power X and plus is X plus T so it is
this one similarly if I try to find it
out del u by Del T that would be del u
by Del R into del R by Del T plus del u
by Del s into del s by Del T now
substitute this value by Del R cos R
plus s square del R by Del T was e to
the power X del u by del s is 2 s cos r
plus s Square and del s by Del T was 1
so we would be getting this again
substituting this R and s square in the
terms of X and T would get this one so
this is what is your chain rule now
after this basic concepts revisiting let
us try to find out verification of
solution to the partial differential
equations that is what we will try to
see given certain partial differential
equation we will try to see if a given
function is solution of it or not that
is what why you call it solution if it
is having those partial derivatives
which are involving in
in that equation and if that unknown
function and its derivatives are
satisfying that equation so let us take
one example consider this first order
wave equation let us call the first
order wave equation C times del u over
del X plus del u over del T is equal to
0 where this constant C is the constant
and this is called the wave speed now we
see that is in ordinary differential
equations what we had find it out that
is there was certain function which we
said is that they are satisfying the
equation and we did find it out that
where the function was same with a
constant change here you see here
actually what will happen that is we
would have be having many functions
which would be satisfying now let us say
that is with the example and then we
would find it out let us see take the
function U as sine of X minus CT the
derivative so we do know that is sine is
a continuous and differentiable function
and the derivative with respect to X and
with respect to T both would be existing
so we have here the first order
derivatives so we do find out what is
Del u by Del X that would be cosine of X
minus CT and the derivative of X minus C
T with respect to X would be 1 and
derivative with respect to T that is
again cosine X minus CT and the
derivative of X minus CT with respect to
T would be minus C so it is Del u over
del T is minus C times cosine X minus CT
so now if you see is C times del u over
del X what it would be it is C times
cosine X minus CT plus del u over del T
that is minus C times cosine X minus CT
that gives me 0 that is the equation
satisfied that means this is the
solution to the equation C times Del u
over del X plus del u over del T is
equal to 0 now you see what it is
solution how it is looking up
if I take CT too busy we have taken X
minus CT so CT I have taken zero are you
could say that is C I am taking as one
and T as zero then this would bigger so
this function is solution to the given
equation if I take T is equal to 1 RT is
equal to PI by 4 and getting is that the
solution is little bit changing towards
the right shifting towards the right the
function is same but it is shifting
towards the right you see at T is equal
to PI by 2 it is shifting this 1/3 PI by
4 it is again more shifting this is
shifting let us see that is what we have
find it out we are finding it out that
at T is equal to 0 this is the function
at T is equal to PI by 4 this is the
function at T is equal to PI by 2 this
is the function at T is equal to PI by 3
PI by 4 this is the function at T is
equal to PI this is the function so what
we are having is that is my function the
shape of the function is same but it is
shifting towards its right as we are
changing our CT so either you could say
is that T we are keeping fix and see we
are changing or you could says rather it
should be that is C is been fixed on T
changing that is T as I said is we are
treating it as a time so with the time
this wave is shifting towards right now
let us take another function U as 1
minus X minus C T whole square in the
range when X minus C T whole squared is
between 0 and 1 and it is 0 otherwise
now we see that this is also satisfying
our wave equation see what will be del u
over del X this function is actually we
find it out that is is continuous but
may find out that del u over del X in
this region it is minus 2 ty
sex - city derivative of X is 1 and in
this region it is 0 similarly the
derivative with respect to T would be
the derivative of CT would be minus C
actually so would be getting 2 times CX
minus T in this reason and zero so now
if I consider our wave equation that was
C times del u over del X plus del u over
del T VC is that is what it is if I
multiply it with the C these things are
just C times del u over del X is same as
minus del u over del T in this reason
this is same on in this reason that is
when X minus CT is greater than 1 both
are 0 so they are satisfied it for all X
that means this is also a function this
function is also satisfying our wave
equation let us see what this function
is this function is this one when I am
taking is C is equal to 0 that is from
minus 1 to plus 1 this is a wave are
this kind of curve and then so if I move
this C 2 to 1 what I get is that
function is shifted from 0 to 2 if I
move it 2 to 1 2 3 it is moved with 2 to
4 now you see the shape was not changing
the shape was exactly the same from 0 to
1 but the function is shifting towards
the right you see this is with 0 1 2 3 &
4 so what we are having is the shape is
same the only thing is that the wave
that is this wave which it has been
forming that is C but it is shifting as
with respect to time this is shifting
towards its right now let us treat
another one now here what I am taking is
this function f of X minus C T where F
is any arbitrary function I am NOT
giving any form of if in previous two
examples I had given that if I sign and
then if I have given as a particular
function now here it is f is a general
function now let us see whether if I
take treat this as a general function
just continuous and has partial
derivative with respect to X and T will
it satisfy our wave equation so use this
chain rule we get value by Del X as f
dash X minus CT and del u over del T as
minus C times f - f - means is that is I
am taking is the derivative with respect
to Jade that is X minus CT now what we
are getting is del u over del X is f - Z
and del u over del T is minus C times F
- straight that is if I take C times F -
n that is same as minus del u over del T
so this equation is satisfied what I was
requiring is that is f is continuous and
has partial derivatives are
differentiable with respect to both this
one so this is also a solution what it
says in our differential equations if
you do remember we were used to find out
a solution which involves arbitrary
constants now here what we are getting
is I am finding out a solution which is
involving actually arbitrary function
that says this up I do have in partial
differential equations my solutions main
wall arbitrary functions and how to
determine this unique solution does this
partial differential equation does not
have unique solution because we said is
an ordinary differential equations we do
had unique solution that bogus general
solution then there was unique solutions
we said is that there would be unique
solution which we could find out from
the general solution that is given
particular values to the arbitrary
constants if we had used certain site
conditions what those cost side
conditions we said is that there were
initial conditions initial conditions
means if it is a first order equation we
said is initially that is from where the
function is starting our
if it is second order we had to initial
conditions first was that is from where
the function is a start in the second
verse that maybe that has at the first
derivative is a initial point so we are
having is that in ordinary differential
equation we had initial conditions for
higher order in had differential
equations say for in it harder we had n
minus one one ass derivative we had got
the initial conditions for all that is n
initial conditions whatever we had
learned there in the ordinary
differential equations that is other
than the initial conditions we were
having the boundary conditions as well
when the function was defined on a
certain finite trees and we said is that
on both the boundaries we had certain
conditions imposed upon on in the
boundary that those problems he called
boundary value problems and we had seen
that is in the boundary value problem
sometimes the solution was existing
sometimes the solution was not existing
now in the similar manner here to
determine this unknown function if we
require some sight conditions so we call
them are some additional side conditions
they could be of the form of initial
conditions or they could be of the form
of boundary conditions are sometimes the
combination of both and so we are
calling them as in the under
differential equation if you do remember
we were calling the initial conditions
if the we are solving a problem we
called it initial value problems here
also we would call them initial value
problem with boundary conditions imposed
upon we call them boundary value problem
and when we are having the combination
of both that is my partial differential
equation may be having both kind of
sight conditions initial conditions as
well as boundary conditions and we call
them initial and boundary value problem
or similar are simply initial boundary
value problems now with this
if you remember in the boundary value
problem in ordinary differential
equations we had find it out with
certain boundary conditions the solution
is existing and with certain boundary
conditions the solution was not existing
while in initial value problems you find
it out that is although the solution was
existing and that solution was unique
here on cane with these site conditions
whatever the conditions that is the data
is given that is the function at initial
value is some particular value or some
particular function actually here it
would come out we are saying is that
this values and is called data in this
partial differential equation and with
respect to those data if for some
initial value problem or boundary value
problem our initial boundary value
problem if unique solution is existing
because in boundary value problems in
the ordinary differential equations we
had find out that all the ten solution
was not existing so here what would say
if with respect to these site conditions
whatever the site conditions are that we
are calling the data with respect to
that data if for any partial
differential equation the unique
solution is existing they will call that
problem as well first problem otherwise
it is called ill-posed problem now on so
come back to our example we had find out
that is we were talking about the first
order wave equation and we had find out
that is a general function f of X minus
C T could be solution of that equation
now we want to find out unique solution
so let us impose first the initial
condition what initial condition here it
would be we say at time T is equal to 0
the solution should be of the two
variables U as xn T so we are saying is
at T is equal to 0 u X is a function of
X
now here F could be of any particular
function now in general I'm telling
suppose F is a known function then you
XT we had find out that the solution is
f of X minus CT now if I use T is equal
to zero what it would give it would give
me that is at T is equal to zero here I
would get UX zero and here if I put T is
equal to zero I would get capital f of X
now this capital f of X is same as my
initial condition says is U at X 0 is
equal to F X that says is this capital F
X must be this known function f X so
what we are saying is that this initial
condition with sit that function u X is
a function of X the known function my
solution is actually turns out to be
that function f itself so what would be
my function my solution my solution
would be f of X minus CT that is the
initial condition ur would give the
unique solution with reference to that
initial condition that is function moon
function so say for example I take FX s
e to the power minus X square then what
will be my solution my solution u X T
would be e to the power minus X minus C
T whole square let us see is it a
solution to this our equation if I take
curves derivative with respect to X I
would be getting it as e to the power
minus X minus CT square into minus 2
times X minus C T and the derivative
with respect to T would be e to the
power minus X minus CT square into 2 C
times X minus CT so you can just check
that this is satisfying other cell
actually we had already shown that it is
satisfying with any F so whatever this F
I am keeping it it would satisfy this
solution if I put T is equal to 0 or
would get here e to the power minus X
square that is my initial condition is
also satisfied
how this function is looking like you
see here again I have taken two
different values of CT when CT is zero
this is the this is the function when CT
is equal to one this is the function on
when CT is equal to two this is third
one is the function and CT is equal to
three this is the function again we are
getting is the shape of the function is
same that is this wave is same the thing
is that is it is changing towards or
shifting towards right so this is a
solution so solution what we are getting
this that solution is unique solution we
can get for one-dimensional wave
equation with initial condition where
initial condition is being posed at the
time T is equal to 0 the unknown
function takes form of a known function
so the solution would be in the form of
that known function because the shape of
the function is not changing the
function remains same only thing is that
with time T it is changing its place
that is it is shifting towards the right
here now let us come to the first order
partial differential equation first
order partial differential equation I
take a simplest of example of first
order partial differential equation del
u over del T plus C u times del u over
del X is equal to 0 C u that is the
coefficient of del u over del X it may
be a function of U also now if this C of
U is just a constant that is it is not
involving any unknown function or any X
T it is just a constant let us say C
then this equation is the equation just
now we have seen the one-dimensional
wave equation del u over del T plus C
times del u over del X is equal to 0 we
see that this is first order
because the order is one only it's
linear the derivatives are occurring
linearly the coefficient is constant
right hand side is zero that is we are
not having an e term which is not
involving you are its derivative so it
is homogeneous linear partial
differential equations we had already
seen that we had called it first order
wave equation with our examples for the
solutions we had seen that all the
solutions the in general actually we had
find out the solution f of X minus CT
and if I use this F to be different
values of the function now different
functions we had seen we had always find
it out that the solution of this is
giving is that is a graph of a function
which is shifting towards a and all
those examples we had that it was
shifting towards its right that is it is
changing the place only and we that is
why we do call it the transport equation
also because it is transporting the data
so you see is that is why we are calling
transporting say if we do have this
equation we see is that is if the
function is my e to the power minus X
square it is shifting towards right that
is it is transporting the function
towards its right because at the speed
what we are saying is it is transporting
at the speed C with this C speed it is
transporting towards or whatever it is C
now here see all the time I have taken
it to be positive value so it is
shifting towards the right now
how to find out the solution of this
first-order partial differential
equations we require to know that is
what is the method before knowing this
method we had seen only one example of
first-order wave equation let us
straight one more example and see that
is thus this also has more than one
function as its solution you know we see
is that this is now not and this is also
an homogeneous equation linear one we do
have del u over del T plus C times del u
over del X minus U is equal to 0 now
verify that function e to the power
minus T f of X minus C T is a solution
of this equation where this F again you
see this is an arbitrary function so I
am NOT treating it as a general I am
treating now with the arbitrary function
so the form of the solution we are
taking treat taking here is e to the
power minus T f of X minus C let us try
to see we have been given the function
as e to the power minus T f of X minus
CT we require the derivative with
respect to X and with respect to T so
first find out the derivative of this
with respect to X and T so where this
dash is denoting the derivative with
respect to X minus CT so del u over del
X would be e to the power minus T times
F dash X minus CT because the derivative
of X minus C T with respect to X would
be 1 del u over del T would be e to the
power minus T first this function and
then this f of X minus CT asks us then
the derivative of this with respect to T
is that C times e to the power minus T F
dash X minus CT dash means the
derivative with respect to X minus CT
now if substitute in the given equation
the equation was value over del T plus C
times del u over del X minus u is equal
to 0 what we have got that e to the
power minus T f of X minus CT minus C
times
power minus T F dash X minus CT this is
what is my del u over del T plus C times
this one C times e to the power minus T
of - X minus CT minus u U is my e to the
power minus TF times X minus CT now you
see this is cancelling this one this is
cancelling this one so I am getting it
is equal to zero so that was satisfying
the equation again what we have got is
actually that F here is a in this one F
is arbitrary let us see one more example
let us consider this semi linear
equation but the coefficient is X x
times del u over del X plus del u upon
del T is equal to 0 for X positive now
we are verifying that is f of X times e
to the power minus T is a solution where
F is again any arbitrary function let us
try to see it we have been given that
the function as f of X into in power
minus T so this we will take as set and
use the derivative with respect to X and
T partial derivative respect to X it
would be f dash X time times e to power
minus T and derivative of x times in
power minus T with respect to X would be
e to the power minus T dash means the
derivative with respect to this whole
function as one variable similarly del u
over del T is x times e to the power
minus T F dash X of e to power minus T
now substitute this in the given
equation equation was X x value over del
X plus del u over del T so del u upon
del X X times this x times e to power
minus T F dash X time is power minus T
this is same as here this is with
respect to T so it should do with minus
sign actually then e to power minus T so
it should be with minus sign so we do
get it with minus is 0 that is
is also satisfying the equation so we
had God that is not only that first
order wave equation what we had
different kind of equations all are the
first order equations we had find out
that is the function f was arbitrary in
all those things but one more
interesting thing did you observe here
when we had first order wave equation my
argument that was X minus C T that is
unknown function f but the argument was
X minus C T and when I have taken one
other one there I was having is that my
argument was different and here also we
were having some multiplication that is
e to the power minus T and sometimes
something like that one when I have took
this example of a simile near equation I
have got that F is of course unknown but
this form this argument is x times e to
the power minus T what we had observed
that in all these examples the function
remains unknown and we had find out that
with initial condition in the wave
equations we had seen that that initial
condition was deciding what should be my
if if initially at time T is equal to 0
I give that the solution is taking this
particular form of the function then my
function is being decided from there but
the argument is changing now it says is
that in ordinary differential equations
what we had we had that my solution was
involving arbitrary constants so we try
to find out whether solution in general
with arbitrary constants and then we
using those side conditions we try to
find out the value of those arbitrary
constants and we call them the unique
solution and the solution with arbitrary
constants we call them the general
solution here what we are having is we
are having this that is a solution is
involving orbitary function
in these three examples we had observed
that argument of the function is
changing now the question comes is how
to find out this argument is there any
method which says is seeing the equation
can I obtain the argument R what should
be the argument of the function of
course arbitrary function because the
general solution we must get in the
terms of arbitrary function but that
arbitrary function is with respect to
some other functions are the arguments
are also functions can a determine those
functions are those arguments so we have
learnt today what is called partial
differential equation certain basic
concepts definition of partial
differential equations certain concepts
which we are using in the partial
differential equations they are more
like on a differential equations but
somewhere we had changed according to
the partial derivatives we had learned
some more concepts or more terms other
than partial differential equations we
call some semi linear equations and
quasi linear equations we had learnt
that solution of partial differential
equations are having unknown function as
an in the form of arbitrary function we
had seen that in ordinary differential
equation one function was defining R is
giving the solution of differential
equation in that function we can only
change with a constant but here what we
had find it out that the function itself
could be arbitrary we had seen that with
certain site conditions as in the
ordinary differential equations we call
them initial value problems or boundary
value problems here also we can use
certain site conditions to determine the
function uniquely and those we called
initial value problems boundary value
problems or we could call them initial
and boundary value problems of course
will come later on on all these
particular things but the first thing
which we had learned in the first are
partial differential equations that the
function has to be arbitrary that has to
be determined from the site conditions
what we do get that argument of that
function is changing as the equation is
changing so we would get the solution of
the partial differential equation that
is are try to find out the methods to
find out the solution of partial
differential equations as first method
we would learn about determining the
argument of the arbitrary function so in
the next lecture we'll go how to solve
the first-order partial differential
equations so today would be closing the
lecture here next lecture we'll go for
the solution of first-order partial
differential equations thank you