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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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00:12:00,340 --> 00:12:05,030
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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00:12:34,960 --> 00:12:43,360
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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00:12:48,570 --> 00:12:50,510
That's the picture
of the solution.
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The graph of the solution.
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It has this 0 at
the left boundary.
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00:12:57,260 --> 00:13:00,390
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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00:13:05,690 --> 00:13:11,810
And I have made it
continuous at x equal A.
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The bar is not coming apart.
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00:13:13,480 --> 00:13:17,890
So that this solution
runs into that solution.
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That's good.
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That's one more condition.
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00:13:20,860 --> 00:13:23,990
But I need one further,
one final, condition.
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00:13:23,990 --> 00:13:27,830
And somehow I have to
use the delta function.
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00:13:27,830 --> 00:13:29,780
And what does the
delta function tell me?
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00:13:29,780 --> 00:13:35,110
I'm just going to go give you
the answer here, rather than
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a theory of delta functions.
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That equation.
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So you see what my solution is.
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It's a broken line
with a change of slope.
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It's a ramp.
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It has a corner.
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All those words describe
functions like this.
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So I have some slope going
up here, and some slope--
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and let me tell you.
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I'll tell you what
those slopes are.
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00:14:01,320 --> 00:14:04,200
I'll tell you what those
slopes are in this.
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So I'll tell you the answer
and then we'll check.
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00:14:08,330 --> 00:14:16,210
So that C turns out to be 1
minus A. So in this region,
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00:14:16,210 --> 00:14:20,190
I have 1 minus A times x.
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In that region.
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00:14:21,520 --> 00:14:28,130
And in this region, below,
so that's stretching.
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00:14:28,130 --> 00:14:30,830
The fact that it's
positive displacement
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means it's stretching.
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00:14:32,660 --> 00:14:34,790
Now this part is going
to be in compression,
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with that negative slope.
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And I think in this region
it's 1 minus x times A, which
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00:14:43,740 --> 00:14:45,840
will be coming from there.
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00:14:45,840 --> 00:14:50,870
So there is my solution.
259
00:14:50,870 --> 00:14:56,670
Because of the delta function,
I need a two part solution.
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00:14:56,670 --> 00:14:59,320
To the left of the delta
function, the point load.
261
00:14:59,320 --> 00:15:01,510
And to the right
of the point load.
262
00:15:01,510 --> 00:15:04,540
And then we could
check that at the load,
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00:15:04,540 --> 00:15:09,190
x equal A. This is
1 minus A times A.
264
00:15:09,190 --> 00:15:12,840
This is 1 minus A
times A. They do meet.
265
00:15:12,840 --> 00:15:16,800
And now comes this
mysterious fourth condition
266
00:15:16,800 --> 00:15:18,680
about the slopes.
267
00:15:18,680 --> 00:15:21,040
The slope drops by 1.
268
00:15:21,040 --> 00:15:24,940
Here the slope is
1 minus A. That's
269
00:15:24,940 --> 00:15:28,070
1 minus A is the slope there.
270
00:15:28,070 --> 00:15:33,540
And here the slope is minus
A. You see minus x times A,
271
00:15:33,540 --> 00:15:37,270
so the derivative is minus A.
272
00:15:37,270 --> 00:15:42,240
So it was 1 minus A. The
1 dropped away and left me
273
00:15:42,240 --> 00:15:46,930
with minus A. That's what
the solution looks like.
274
00:15:46,930 --> 00:15:51,100
And now I have to say
one word about why
275
00:15:51,100 --> 00:15:53,460
did the slope drop by 1.
276
00:15:53,460 --> 00:15:57,790
The slope dropped by 1,
from 1 minus A to minus A.
277
00:15:57,790 --> 00:16:01,120
And that has to come
from this delta function.
278
00:16:01,120 --> 00:16:04,250
And of course you remember
about the delta function.
279
00:16:04,250 --> 00:16:08,010
The key point is if when you
integrate the delta function,
280
00:16:08,010 --> 00:16:09,370
you get 1.
281
00:16:09,370 --> 00:16:13,110
So when I integrate
this equation, I get a 1
282
00:16:13,110 --> 00:16:15,540
on the right hand
side from the delta.
283
00:16:15,540 --> 00:16:18,850
And on the left hand
side, I'm integrating
284
00:16:18,850 --> 00:16:22,470
the second derivative, so
I get the first derivative.
285
00:16:22,470 --> 00:16:23,400
Great.
286
00:16:23,400 --> 00:16:26,310
The first derivative
at the end point,
287
00:16:26,310 --> 00:16:29,000
minus the first derivative
at the start point,
288
00:16:29,000 --> 00:16:31,030
should be the 1.
289
00:16:31,030 --> 00:16:33,510
And that's the drop of 1.
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00:16:33,510 --> 00:16:39,180
I'll do a full-scale
job with delta functions
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00:16:39,180 --> 00:16:41,450
in another video.
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00:16:41,450 --> 00:16:43,820
I want to keep this
one under control.
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00:16:43,820 --> 00:16:47,580
We're seeing the new idea
is boundary conditions,
294
00:16:47,580 --> 00:16:53,600
and here we're seeing a
delta function equation
295
00:16:53,600 --> 00:16:57,190
in this boundary value problem.
296
00:16:57,190 --> 00:16:58,960
Thank you.