Introduction to Boundary Value Problems Video Lecture - Civil Engineering (CE)

welcome to an intro lesson to boundary
value problems the objectives of this
video are to define a boundary value
problem and also to solve a boundary
value problem given a general solution
to a differential equation notice in
these examples we will be given the
general solution to the differential
equation a boundary value problem is
just a special initial value problem so
for review a differential equation that
has given conditions allows us to find
the specific function that satisfies a
given de rather than just the family of
functions and these types of problems
are called initial value problems if the
conditions are given at more than one
point and the differential equation is
of order two or higher it is called a
boundary value problem abbreviated by
bvp a bvp problem can have none one or
many solutions let's take a look at our
first example we want to find a solution
to the given bvp problem the second
derivative of Y with respects to X minus
y equals 0 Y of 0 equals 0 Y of 1 equals
1 if we know that Y of x equals C sub 1
times e to the X plus C sub 2 times e to
the negative x is a general solution to
the differential equation so given the
general solution here and these two
conditions we want to find a particular
solution so if Y of 0 is equal to 0 now
if we substitute 0 for X into this
function here the function value must be
0 so C sub 1 times e to the 0 plus C sub
2 times e to the negative 0 which is
still just 0 must equal 0 so C sub 1
plus c sub 2 must equal 0
which means that C sub 1 must equal
negative C sub 2 or vice-versa so
they're opposites of one another so when
to keep track of this
now we'll use the fact that we know why
of one equals one
well if Y of 1 equals 1 then C sub 1 e
to the 1 plus C sub 2 e to the negative
1 must equal 1 again we're trying to
find the values of C sub 1 and C sub 2
notice how this equation still has two
unknowns but we do know that C sub 1 is
equal to negative C sub 2
so let's form our substitution here for
C sub 1 by replacing it with negative C
sub 2 so negative C sub 2 times E Plus C
sub 2 times e to the negative 1 must
equal 1 let's go ahead and factor out
our common factor of C sub 2 that's
going to leave us with negative e plus e
to the negative 1 which I'm going to
write as e to the negative 1 - e must
equal 1 we'll go ahead and divide by
this quantity here to solve for C sub 2
this simplifies to 1 so now we know that
C sub 2 is equal to 1 divided by the
quantity e to the negative 1 minus E and
since C sub 1 and C sub 2 are opposites
we also know that C sub 1 would be equal
to negative 1 all over e to the negative
1 minus E remember this is all the
information we need to solve this bvp
problem we now know C sub 1 and we now
know C sub 2 so let's go ahead and give
our solution on the next slide so Y of X
is equal to C sub 1 times e to the X or
negative 1 all over e to the negative 1
- e times e to the X plus C sub 2 which
is positive 1 all over e to the negative
1 - e times e to the negative x notice
how we do have a common denominator here
so we can write Y of X as this would be
negative e to the X plus e to the
negative X I'm going to go Ed and put
the positive term first so we have e to
the negative x minus e to the X all over
our common denominator of e to the
negative 1 - e this would be a
particularly the bvp problem now if we
did want to check this solution in the
differential equation we want to write
this in a slightly different form we
could write this as Y of x equals notice
how our denominator is a constant so we
could factor out 1 over e to the
negative 1 - e this would be times a
quantity e to the negative x minus e to
the X it might be easier to find the
derivative in this form rather than the
previous form now let's take a look at a
second example here we are given that y
double prime plus 4y equals 0 where Y of
0 equals 1 y prime of PI over 2 equals 2
and that Y of x equals C sub 1 cosine 2
X + C sub 2 sine 2
is a general solution to the given
differential equation so again we'll use
this general solution with these two
initial conditions to solve this bvp
problem since Y of 0 equals 1 we can
replace X with 0 in this function here
and know the function value is positive
1 so C sub 1 times cosine of 2 times 0
which is 0 plus C sub 2 times the sine
of 2 times 0 or 0 must equal positive 1
well the sine of 0 is equal to 0 and the
cosine of 0 is equal to 1 so this tells
us that C sub 1 must equal 1 now that we
know that C sub 1 is equal to 1 we can
rewrite Y of X as Y of x equals just
cosine 2 X since C sub 1 is equal to 1
plus c sub 2 sine 2 X now let's go ahead
and find the derivative of this so we
can use this second initial condition so
Y prime of X or Y prime can be equal to
the derivative of cosine 2 X which is
going to be negative sine 2x times to do
the chain rule so we have negative 2
sine 2 X plus the derivative of C sub 2
sine 2 X that will be C sub 2 times
cosine 2x times 2 or 2 C sub 2 cosine 2
X and now we use the fact that Y prime
of PI over 2 equals 2 to find C sub 2
well substitute PI over 2 for X and know
the function value must be equal to 2 so
we'll have negative 2 sine of 2 times pi
over 2 is just PI plus 2 C sub 2 times
the cosine of 2 times pi over 2 which is
just PI and this must equal positive 2
well sine pi is equal to 0 cosine of pi
is equal to negative 1 so here we're
going to have negative 2 C sub 2 equals
2
divide both sides by negative two we
have C sub 2 equals negative 1 now that
we know C sub 1 and C sub 2 we can solve
this bvp problem let's go ahead and do
that in the next slide again we found
that C sub 1 was equal to 1 and C sub 2
is equal to negative 1 and now we can
perform the substitution into our
general solution so Y of X is equal to
cosine 2x plus negative 1 times sine 2x
or just minus sine 2x ok I hope you
found these two examples helpful
thank you for watching
you