PROFESSOR: Hi, everyone.
Welcome back.
In this problem, I'd like to
take a look at convolution
and Green's formula.
Specifically, for
part one, we're
just asked to compute
the convolution of t
with itself, t.
In the second part, we're asked
to compute the convolution of e
to the minus k*t with another
exponential, e to the a*t.
And then in part B, we're to use
the result from part A as well
as Green's formula to solve
x dot plus k*x equals e
to the a*t.
And specifically, we're
interested in the rest
initial condition
solution to this formula.
OK.
So I'll let you think
about this problem,
and I'll be back in a minute.
Hi, everyone.
Welcome back.
OK, so for part one,
we're asked to compute
the convolution of two
functions, t with itself,
and I'll just quickly write down
the formula for a convolution.
So if I have two
functions, f and g,
then the convolution is
defined as, in this case,
the integral from
zero to t of f,
so the function f,
evaluated at, we're
going to use tau as the
integration variable,
multiply by g, t
minus tau, d tau.
So notice how the
variable t appears
in two places in this formula.
It appears in one place
in the argument of g,
and it appears in the
bound of integration.
Meanwhile, tau is the variable
we're integrating over,
and tau appears in two places
as well, one in f and one in g.
So if we're interested
in t convolved with t,
we have the integral from zero
to t of tau times t minus tau.
So notice how f is
just t and g is t,
so when we insert tau and t
minus tau into the function,
we end up just getting tau
times t minus tau, d tau.
And just expanding this
out, we have t times tau
minus tau squared, d tau.
So in the integral,
the variable t always
appears just as a constant.
So in this case, we
have t, tau integrated
becomes t squared over 2.
When I integrate from zero
to t, tau squared integrates
to 1/3 t cubed,
and zero drops out
from the other end
of the integral.
So we end up with t
cubed, 1/2 minus 1/3,
which is equal to 1/6 t cubed.
So there's the answer
we're looking for.
For part B, or for
question two, again we
have another computation to do.
So we have more
integrals to work out.
And in this case,
it's two exponentials.
So we have zero to t.
The first one is going
to be e to the k*tau.
And the second one is e to
the a t minus tau, d tau.
So we can just
expand things out,
and I get e to the a*t times
e to the minus a plus k tau d
tau, zero to t.
And again, because t is just
a dummy variable in this
integral, we're not
integrating over t,
I can just think of e to
the a*t as a constant.
So I'm really just
integrating the function e
to the minus a plus k tau.
And when I integrate
this function,
I get 1 over minus a plus
k, e to the minus a plus k,
tau evaluated at the
bounds, zero and t.
So the negative sign here
just flips the bounds.
So we have a plus k.
Substituting in zero
just gives us 1,
and then substituting in t gives
us e to the minus a plus k*t.
And I can expand out,
multiply out, e to the a*t.
And when the dust settles,
we have e to the a*t minus e
to the minus k*t.
So this is just the computation
of the convolution e
to the minus k*t
and e to the a*t.
So note now if, for example,
k were to equal negative a,
we would have zero
in the denominator
and zero on the numerator.
So one way to compute
the special case when
k is equal to negative a
is to use L'Hopital's Rule
and differentiate the top
and the bottom, for example,
with respect to k.
Another alternative way of
computing the special case when
a is equal to negative k is
just to plug in e to the k,
and then just work
out this integral,
and you'll obtain
a different answer.
And now lastly, for
part B, we're interested
in the differential equation
x dot plus k*x equals e
to the a*t.
And we're interested in finding
the particular solution that
has rest initial conditions.
So for example, x of
zero is equal to zero.
And we want to use
Green's formula.
So just recall that there's
the impulse response
formula for the
weight function w,
which solves the same
differential equation,
but with the delta function
on the right-hand side.
And w, in this case, as
we've seen in lecture,
is actually e to the negative
k*t when t is bigger than zero,
and at zero when t
is less than zero.
So notice how e to the minus k*t
is exactly the function that we
convolved in part
A. Specifically,
we convolved it
with e to the a*t.
So for example, Green's formula
says that the particular
solution that has rest initial
conditions is going to be
the weight function convolved
with the right-hand side
of the differential
equation, e to the a*t.
In this case, on the
domain of integration
for the convolution, w is
just e to the minus k*t.
So we have e to the minus k*t
convolved with e to the a*t.
So Green's formula
gives us the solution
to this differential
equation, which
has rest initial conditions.
And specifically-- we've already
worked this out from part A,
so I can just write
down what the answer is.
It's 1 over a plus k, e to the
a*t, minus e to the minus k*t.
And just as a quick
check to make sure
that we've done
everything correctly,
we can plug in t equals zero.
And when t equals zero, we have
1 minus 1, which gives us zero.
So indeed, this
x is the solution
to this differential
equation, which
has rest initial conditions.
I'd just like to
conclude this problem.
Quick recap.
We worked out
several convolutions,
and specifically we were able to
use Green's formula in addition
to the convolution
that we worked out
to compute the solution
with rest initial conditions
to an ODE.
Moreover, Green's function
is very useful because notice
how we could have
computed the convolution
for any right-hand
side function here.
So in some sense, we're able to
generalize and find solutions
to differential equations which
have arbitrary right-hand side
forcings, not just sines and
cosines, periodic functions,
or other simple
functions which we've
been looking at in the past.
So I'd like to conclude here,
and I'll see you next time.
