GILBERT STRANG: OK,
this is the video about two neat functions--
the step function and its derivative
the delta function. So if I can just introduce
you to those functions and show you that
they're very natural inputs to a
differential equation. They happen all the
time in real life. And so we need to understand
how to compute these formulas and compute with them. OK, so the first one is the step
function and it's-- I'll call it h after its inventor who was
an engineer named Heaviside, started with an H.
And the step function, let me write the formula. h of t is 0 for t negative and
1 for t greater or equal to 0. OK. So, that's the step function. It just has two values
and it has a jump. You could say jump
function also. Jump function, step function. All right. And notice I've also graphed
the shifted step function. What happens to any
function including this one if I change from t, which
jumps at 0, to h of t minus t? If I put in t minus some fixed
number t as the variable, then the jump happens. So the jump will
happen when this is 0. Step functions
jump when that's 0. And that's 0 at t equal to t. So the jump in dotted line. So the shifted step function
will just shift over. That's the complete
effect of changing from t to t minus a capital T, is
just to shift the whole thing by capital T. OK. So you keep your eye on the
standard step function, which jumps at t equals 0. It jumps by 1. And take its derivative. So what's the derivative
of this step function? Well, the function is 0 along
there, so the derivative is 0. The function is
constant along here, so the derivative is again 0. It's just at this one
point everything happens. So now this is the
delta function. The delta function runs
along at 0, continues at 0, but at t at 0, the
whole thing explodes. The derivative is infinite. You see an infinite slope there. And the point is infinity is
not a sufficiently precise word to tell you exactly
what's happening. So we don't have really--
this graph of a delta function is not fully satisfactory. It's perfect for all the
uninteresting boring part. But at the moment of
truth, when something happens in an instant,
we need to say more. We need to say more,
not just its infinite. And again, if it's shifted,
then the infinite slope happens at t equal a
capital T. So the infinity is just shifted over. And that'd be the
delta function there. So this is what I would use. If that was the source term
in my differential equation, what would that mean? If this was the q of
t in the differential equation reflecting
input at different times, that function would say no input
except at one moment and one instant, capital T. At
that instant of time, you put 1 in, over
in an instant. And remember, that
otherwise q of t has been a continuous input. Put in $1.00 per year
over the whole year. This one puts in
$1.00 at one moment. But of course, you see that
that's really what we do. So, you see that that's
a function we need to do things in an instant. And as I took the example of a
golf club hitting a golf ball, well, it's not quite 0 time. But it's so close to zero time
that the two are connected. And then the ball takes off. And so a simple model,
a workable model is to say it happens in 0
time with a delta function. So I really want to
use delta functions. And they're not
difficult to use. They're just not quite
perfect for calculus because the derivative
of the step function is not quite
legitimate at the jump. OK. But what you can do, the part
of calculus that works correctly is integration. Integration tends to
make things smoother. The delta function--
sorry, the step function is the integral of
the delta function. Right? We're going in the
opposite direction. We take derivatives,
we get craziness. If we take integrals
to go from delta-- so the integral of the
delta is the step function. And that's really how you
know a delta function. That's the math way to
describe more exactly than this arrow that just fires
off what the delta function is doing. So the key property
of the delta function is to know what
it's integral is. The integral of
the delta function is the total deposit
over, let's say, it started-- time could have
started even at minus infinity, and it could go on
forever to plus infinity. So that's the total
deposit, the total input coming from this
source term delta of t. And what is the answer? Well, the integral of delta
should be the step function. The step function
out in infinity is 1. Back at minus infinity it's 0. Do you see what I'm saying here? This would be h of t evaluated
between t equal minus infinity and plus
infinity because those are the limits of integration. And what do I get? At plus infinity the
step function is 1. This is 0. So I get 1. And everybody catches
on to that key fact that the total integral of
the delta function is 1. Again, you only made the
deposit at one moment, but that deposit
was a full dollar. And that, adding up all
deposits is just that $1.00. So, that's the integral
of the delta function. Now actually, to use
delta functions I need to give you a slight
generalization of that. So as I say, delta
functions are really known-- we don't like to
take their derivative. The derivative of
a delta function is a truly crazy function. It shoots up to
infinity and then it shoots down the
minus infinity, the slope of that arrow. But it's integrals that we want. So now let me integrate
from minus infinity to infinity my
delta function times any other function,
say f of t dt. That's something we'll
need to be able to compute. What's the right
integral for that? And again, delta
is doing everything at one moment at t equals 0. At that moment t equals 0,
at that moment when t is 0 and that's the only place any
action is happening, f of t is f of 0. It's whatever value it has
at that point t equals 0. And that's the answer. f of 0. So if f of t was the
constant function 1, then we're back to
our integral up there. If that's just 1, I'm
integrating delta of t. My function is 1, I get 1. But if that function is,
suppose that function is sine t. What's the integral of
delta of t times sine t dt? Well, sine t happens to
disappear just at the moment when the delta function is
ready to turn on at t equals 0. So the integral of delta of
t sine t is sine of 0 is 0. You have one term turned on,
but the other term turned off. So nothing happened altogether. Whereas the integral of
delta t e to the t-- yeah, tell me that one. The integral of
delta t e to the t dt is-- well e to the t is
doing all sorts of stuff for all time. But the delta function
is 0 all that time, except at t equals 0. So, the integral of
delta t e to the t dt would be 1 because at
that moment, t equals 0, the only important moment
would be e to the t function is e to the 0, and it's just 1. Let me ask you for
another example. The integral of minus infinity
to infinity of delta-- let me use the shifted
delta e to the t dt. Can you compute that integral. Well again, that function
is 0 almost all the time. The only time that impulse, the
moment that impulse hits is t equals capital T.
At that moment, this is equal to e
to the capital T. And that's all that matters. OK. So now, let me use a delta
function as the source term in our differential equation. So we are seeing one
last time one more-- I still call it a
nice function, even though it's not legitimately
a function at all, the delta. But let me solve
the equation dy dt equals ay plus the
delta function turned on at capital T. And let
me start it from 0. So I don't make an initial
deposit to my account. I don't make any deposit at all,
except at one moment t equal capital T. And in that moment,
I deposit $1.00 because delta-- this is the unit delta. If I was depositing $10,
I would make it 10 delta. OK. So we know what the solution
is from a deposit of $1.00 made at one time, t equal to capital
T. What is the solution? y of t, we have 0 up to t
equal to T. Nothing whatever has happened. And at capital T time t, in
goes the $1.00 and it grows. It grows so that it
grows over the remaining time e to the t
minus capital T. This is for t larger than T. t larger
than or equal I could say. When t and capital T are
equal, that's e to the 0. That's our $1.00 just gone in. When t minus capital T is a year
later, our dollar is worth e. When t minus capital T, when
it's been in there for a year, that $1.00 has
increased to-- well, that was if the interest
rate was 100% you may feel. You'd be fortunate to get that. But let's suppose you do. At 100% interest,
after one year, you might say, well, my
money just doubled because I got the interest
equaled the original. So I got twice it. But not true because that
money went in-- was growing. Interest was being
added, compounded through the whole year
so that after one year, starting with 1, you
have e at a is 100%. Oh, OK. My formula isn't incorrect
here because I had an a here and it belongs here. So let me fix that. It's e to the a t minus T.
That's the growth factor. That's the growth
factor up to time t starting from the
earlier time capital T. So you see that
we were able just to write down the solution
to the differential equation even though it's entirely new
or different or non-standard input. The step function input--
so we're finding here the impulse response. That's a very, very
important concept in engineering, the
impulse response, the response to an impulse. And for second order
differential equations, this is going to be-- it's
really a crucial function in the subject. So this is the
response to an impulse. It's the impulse response
from our standard first order equation that we've
been dealing with. Now we've got just
to remember one more step is still linear would
be to allow the interest rate to change. That's one lecture,
the next one. And then we get
non-linear equations. So that's what's coming. But here is delta functions
for the first time and not for the last time. Thank you.
