WEBVTT

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GILBERT STRANG: OK.

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Well my problem today
is a little different.

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Because I don't have
two initial conditions,

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as we normally have for a
second-order differential

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equation.

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Instead, I have two
boundary conditions.

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So let me show you the equation.

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So I'm changing t
to x because I'm

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thinking of this as a problem
in space rather than in time.

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So there's the
second derivative.

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The minus sign is
for convenience.

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This is the load.

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But here's the new thing.

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I'm on an interval 0 to 1.

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And at 0-- let me take 0 for
the two boundary conditions.

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So my solution somehow
does something like this.

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Maybe up and back down.

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So it's 0 there, 0
there, and in between it

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solves the
differential equation.

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Not a big difference,
but you'll see

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that it's an entirely
new type of problem.

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OK.

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As far as the solution
to the equation goes,

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there is nothing enormously new.

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I still have a y particular.

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A particular solution
that solves the equation.

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And then I still have the y
null, the homogeneous solution,

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any solution that solves
the equation with 0

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on the right hand side.

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And in this example--
this is especially

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simple-- the null equation would
be second derivative equal 0.

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And those are the
functions, linear functions,

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that have second
derivative equal zero.

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So there's the general solution.

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And now I have to put in,
not the initial conditions,

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but the boundary conditions.

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OK.

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So I substitute x equal 0.

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And I substitute x
equal 1 into this.

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I have to find y particular.

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I'll do two examples.

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I'll do two examples.

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But the general
principle is to get

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these numbers, these
constants like C1 and C2,

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from the boundary conditions.

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I'll put in x equal zero.

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And then I'll have y of 0,
which is y particular at 0,

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still defined, plus
C times 0, plus d.

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That's the solution at the left
end, which is supposed to be 0.

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And then at the
right end, end, I

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have whatever this
particular solution is at 1,

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plus now I'm putting
in x equal 1.

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I'm just plugging in x
equals 0, and then x equal 1.

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And x equal 1, I have C plus D.
C plus D. And that gives me 0.

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The two 0's come
from there and there.

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OK.

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Two equations.

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They give me C and
D. So I'm all solved.

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Once I know how-- I
know how to proceed once

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I find a particular solution.

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So I'll just do two examples.

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They'll have two
particular solutions.

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And they are the most important
examples in applications.

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So let me start with
the first example.

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So my first example is going
to be the equation minus D

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second y, Dx squared, equal 1.

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That will be my load.

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f of x is going to be 1.

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So I'm looking for a particular
solution to that equation.

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And of course I
can find a function

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whose second derivative
is 1, or maybe minus 1.

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My function will be-- well, if
I want the second derivative

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to be 1, then probably 1/2 x
squared is the right thing.

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And that would give me a minus.

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So I think I have a
minus 1/2 x squared.

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That solves the equation.

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And now I have the Cx
plus D. The homogeneous,

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the null solution.

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And now I plug in.

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And again, I'm always
taking y of 0 to be 0,

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and also y of 1 to be 0.

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Boundary conditions.

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Again, boundary condition,
not initial condition.

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OK.

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Plug in x equal 0.

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At x equals 0, what do I learn?

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x equal 0.

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That's 0, that's 0, so
I learn that D is 0.

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At x equal 1, what do I learn?

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This is minus 1/2.

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D is 0 now.

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And x is 1.

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So I think we learn
C is plus 1/2.

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OK with that?

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At x equal 1, I'm
supposed to get 0

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from the boundary condition.

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So I have minus 1/2,
plus 1/2, plus 0.

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I do get 0.

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This is good.

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So this answer is-- Cx, then,
is 1/2 x minus 1/2 x squared.

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That's it.

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That's my solution.

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That function is 0 at both ends,
and it solves the differential

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equation.

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So that's a simple example.

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And maybe I can give
you an application.

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Suppose I have a rod.

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Here's a bar.

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And those lines that I
put at the top and bottom

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are the ones that give me
the boundary conditions.

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And I have a weight.

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A weight of 1.

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Maybe the bar itself.

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It gives-- elastic force.

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Gravity will pull,
displace, the bar downwards

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because of its weight.

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It's elastic.

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And this function
gives me the solution,

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gives me the distribution.

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If I go down a distance
x, then that tells me

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that this part of the
bar, originally at x,

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will move down by
an additional y.

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Moves.

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So this is now at x plus y of x.

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And that's the y.

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And that is 0 at the
bottom, 0 at the top,

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and positive in between.

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OK.

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That was a pretty quick
description of an application.

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And more important, a pretty
quick solution to the problem.

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Can I do a second example
that won't be quite as easy?

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OK.

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So again, my equation
is going to be

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minus the second
derivative equals a load.

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But now it will be a point load.

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A point load.

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That's a point
load at x equal A.

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This is my friend,
the delta function.

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The delta function,
you remember,

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is 0, except at that one
point where this is 0.

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This is 0 at the
point x equal A.

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In my little picture
of a physical problem,

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now I don't have any
weight in the bar.

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The bar is thin.

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Weightless.

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But I'm putting on, at the point
x equal A, right at this point,

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I'm attaching a weight.

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So this distance is x equal
A. Here's my weight, my load,

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hanging at this point.

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So I can see what will happen.

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That load hanging
down there will

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stretch the part above
the bar, above the load,

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and compress the
part below the load.

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So it's a point load.

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Very important application.

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OK.

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Now I have this
equation to solve.

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OK.

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I can solve it on the
one side of A, x equal A.

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And I can solve it
on the other side

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of x equal A. Let me do that.

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For x less than A, I have
minus the second derivative.

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And what's the delta
function for x below A,

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on the left side of the spike?

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0.

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And x on the right side
of the load, again, 0.

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And what are the solutions
to the null equation?

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y is Cx plus D on the left
side of the load, there.

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And now here it may have
some different constants.

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y equals, what shall
I say, E x plus F,

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on the right side of the load.

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And now I've got
four numbers to find.

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C, D, E, and F.
And what do I know?

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I know two boundary conditions.

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Always I know that y of 0 is 0,
from fixing the top of the bar.

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So y of 0 equal 0.

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And when I put in x equal
0, that will tell me D is 0.

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And then also y of 1 is 0.

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And that will be on
this side of the load.

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So when I put in x equal 1, that
will tell me that E plus F is

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0, at x equal 1.

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So it tells me that
F is minus E, right?

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So what do I know now?

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D is gone.

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0.

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F is minus E. So can I just
change this to F is minus E.

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So I had E x minus
E. E times x minus 1

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takes care of that
boundary condition.

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At x equal 1, it's gone.

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OK.

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But I still have two,
C and E, to find.

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So what are my two further
conditions at the jump?

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So far I'm on the left of the
jump, the spike, the impulse,

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the delta function.

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And on the right of it.

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But now I've got to say, what's
happening at the impulse?

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At the delta function.

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Or at the point load.

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OK.

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Well, what's happening there?

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I need two equations.

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I've still got C and E to find.

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So my first equation
is that at that load,

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the bar is not going
to break apart.

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It's just going to be stretched
above and compressed below.

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But it is not going
to break apart.

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So at the load, at
which is x equal A.

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So now I'm ready for x equal A.

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OK.

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What happens at x equal A?

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That's the same as that.

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Let me draw a picture
of the solution, here.

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Here is x.

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This is x.

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Here's y.

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Here is x equal 0.

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Here is x equal 1.

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I see a linear function.

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Cx up to the point
x equal A. And here

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I have a linear function
coming back to 0.

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You see?

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That's the picture
of the solution.

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The graph of the solution.

00:12:52.130 --> 00:12:57.260
It has this 0 at
the left boundary.

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It has 0 at the right boundary.

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It has, in between, it
is Cx in the x minus E.

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And I have made it
continuous at x equal A.

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The bar is not coming apart.

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So that this solution
runs into that solution.

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That's good.

00:13:19.550 --> 00:13:20.860
That's one more condition.

00:13:20.860 --> 00:13:23.990
But I need one further,
one final, condition.

00:13:23.990 --> 00:13:27.830
And somehow I have to
use the delta function.

00:13:27.830 --> 00:13:29.780
And what does the
delta function tell me?

00:13:29.780 --> 00:13:35.110
I'm just going to go give you
the answer here, rather than

00:13:35.110 --> 00:13:38.900
a theory of delta functions.

00:13:38.900 --> 00:13:40.270
That equation.

00:13:40.270 --> 00:13:44.405
So you see what my solution is.

00:13:44.405 --> 00:13:48.250
It's a broken line
with a change of slope.

00:13:48.250 --> 00:13:49.430
It's a ramp.

00:13:49.430 --> 00:13:50.950
It has a corner.

00:13:50.950 --> 00:13:54.230
All those words describe
functions like this.

00:13:54.230 --> 00:13:57.810
So I have some slope going
up here, and some slope--

00:13:57.810 --> 00:13:58.830
and let me tell you.

00:13:58.830 --> 00:14:01.320
I'll tell you what
those slopes are.

00:14:01.320 --> 00:14:04.200
I'll tell you what those
slopes are in this.

00:14:04.200 --> 00:14:08.330
So I'll tell you the answer
and then we'll check.

00:14:08.330 --> 00:14:16.210
So that C turns out to be 1
minus A. So in this region,

00:14:16.210 --> 00:14:20.190
I have 1 minus A times x.

00:14:20.190 --> 00:14:21.520
In that region.

00:14:21.520 --> 00:14:28.130
And in this region, below,
so that's stretching.

00:14:28.130 --> 00:14:30.830
The fact that it's
positive displacement

00:14:30.830 --> 00:14:32.660
means it's stretching.

00:14:32.660 --> 00:14:34.790
Now this part is going
to be in compression,

00:14:34.790 --> 00:14:36.260
with that negative slope.

00:14:36.260 --> 00:14:43.740
And I think in this region
it's 1 minus x times A, which

00:14:43.740 --> 00:14:45.840
will be coming from there.

00:14:45.840 --> 00:14:50.870
So there is my solution.

00:14:50.870 --> 00:14:56.670
Because of the delta function,
I need a two part solution.

00:14:56.670 --> 00:14:59.320
To the left of the delta
function, the point load.

00:14:59.320 --> 00:15:01.510
And to the right
of the point load.

00:15:01.510 --> 00:15:04.540
And then we could
check that at the load,

00:15:04.540 --> 00:15:09.190
x equal A. This is
1 minus A times A.

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This is 1 minus A
times A. They do meet.

00:15:12.840 --> 00:15:16.800
And now comes this
mysterious fourth condition

00:15:16.800 --> 00:15:18.680
about the slopes.

00:15:18.680 --> 00:15:21.040
The slope drops by 1.

00:15:21.040 --> 00:15:24.940
Here the slope is
1 minus A. That's

00:15:24.940 --> 00:15:28.070
1 minus A is the slope there.

00:15:28.070 --> 00:15:33.540
And here the slope is minus
A. You see minus x times A,

00:15:33.540 --> 00:15:37.270
so the derivative is minus A.

00:15:37.270 --> 00:15:42.240
So it was 1 minus A. The
1 dropped away and left me

00:15:42.240 --> 00:15:46.930
with minus A. That's what
the solution looks like.

00:15:46.930 --> 00:15:51.100
And now I have to say
one word about why

00:15:51.100 --> 00:15:53.460
did the slope drop by 1.

00:15:53.460 --> 00:15:57.790
The slope dropped by 1,
from 1 minus A to minus A.

00:15:57.790 --> 00:16:01.120
And that has to come
from this delta function.

00:16:01.120 --> 00:16:04.250
And of course you remember
about the delta function.

00:16:04.250 --> 00:16:08.010
The key point is if when you
integrate the delta function,

00:16:08.010 --> 00:16:09.370
you get 1.

00:16:09.370 --> 00:16:13.110
So when I integrate
this equation, I get a 1

00:16:13.110 --> 00:16:15.540
on the right hand
side from the delta.

00:16:15.540 --> 00:16:18.850
And on the left hand
side, I'm integrating

00:16:18.850 --> 00:16:22.470
the second derivative, so
I get the first derivative.

00:16:22.470 --> 00:16:23.400
Great.

00:16:23.400 --> 00:16:26.310
The first derivative
at the end point,

00:16:26.310 --> 00:16:29.000
minus the first derivative
at the start point,

00:16:29.000 --> 00:16:31.030
should be the 1.

00:16:31.030 --> 00:16:33.510
And that's the drop of 1.

00:16:33.510 --> 00:16:39.180
I'll do a full-scale
job with delta functions

00:16:39.180 --> 00:16:41.450
in another video.

00:16:41.450 --> 00:16:43.820
I want to keep this
one under control.

00:16:43.820 --> 00:16:47.580
We're seeing the new idea
is boundary conditions,

00:16:47.580 --> 00:16:53.600
and here we're seeing a
delta function equation

00:16:53.600 --> 00:16:57.190
in this boundary value problem.

00:16:57.190 --> 00:16:58.960
Thank you.