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PROFESSOR: So as a
more technical aside,

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let's analyze more
carefully the problem

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of release of viral load from
a drop by process of diffusion.

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So here, again, I
sketch a droplet,

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which would typically be an
aerosol droplet in the size

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range of, let's say, microns.

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And the virion of
interest has a size

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that is much smaller than that
on the order of, let's say,

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100 nanometers.

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And this white path is showing
how such a virion would

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go from its initial position R,
let's say, as a radio position,

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to the boundary.

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Now, the general
problem of finding

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the expected first passage time
from the point inside a domain

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to a boundary is a
classical problem

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in the theory of stochastic
processes and random walks.

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And it has the following
representation.

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So the mean are
expected first passage

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time from a point to
an absorbing surface

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solves the following problem.

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Is the Laplacian of that
time with a minus sign

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is 1 over D where D
is the diffusivity.

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And so this is the DV
that I described before,

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the diffusivity of the virion.

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And the boundary condition
for this equation

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is that it's an absorbing
boundary, basically.

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So when the virion gets
to the surface it's gone.

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And that's when the
stochastic process finishes.

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So it's tau equals 0 on
radius capital R, which

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is the radius of the droplets.

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So that's on the boundary.

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So in the case of a
spherical drop, then

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we can write this equation
in spherical coordinates.

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So that's minus 1 over
r-squared, r derivative

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of r-squared D
tau DR. And that's

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equal to 1 over capital D.

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And then again, our
boundary condition

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is the tau of capital
R is equal to zero.

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Another boundary condition we might
mention is that D tau DR at R

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equals 0 is zero.

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That's a symmetry
boundary condition.

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So when you're right in
the middle, basically,

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there should be no--

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there's no favorite direction
for the diffusion process.

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And so, therefore, the
derivative this time

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with respect to R must
be 0 at the center

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because it's a symmetry
boundary condition.

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OK, so we can now go ahead
and solve this problem.

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Let me put the r-squared on
their side and use primes

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with no derivatives.

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So let's write this
as r-squared tau

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prime, prime equals
minus r-squared over d.

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And if I now
integrate both sides,

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I get r-squared
tau prime equals--

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and then here we
get a minus r-cubed

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over 3D plus a constant.

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And that constant immigration
according to the symmetry

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boundary condition has to
be 0, because tau prime is

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0 at r equals 0.

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So we can simplify
this and write

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tau prime is minus r over 3D.

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And so then I can
integrate again.

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And I found that tau of r is--

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well, I integrate this,
I get r squared over 2

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is integral of r.

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So that gives that
an r-squared over 6.

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So there will be a 6
D and minus r-squared.

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And there'll be a
constant integration.

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And in order to satisfy this
boundary condition of vanishing

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a capital R, you
can see that I can

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write the constant this way.

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And I'll write it as
capital R squared over 6D

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is a constant of integration.

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So basically, the profile of
the main first passage time

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is essentially a parabola.

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So as a function of distance R
from the center of the droplet,

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you have this shape to the
mean transmission time.

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And the maximum here,
the maximum value

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is tau 0 is the maximum value.

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And that is r-squared over 60.

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So that's if you happen
to be unlucky and right

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in the middle.

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That's the longest
you would expect

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to take from a particular part.

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Now on the other hand,
as I've sketched here,

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in a typical droplet, if
the virions are randomly

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distributed, some of them,
like this guy over here,

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happen to be very
close to the surface.

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So you're not going to have
to wait this long for them

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to escape.

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So now we can ask the question,
what is the average escape

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time over all the initial
positions of the virus,

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or virion, assuming that
the virions are uniformly

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distributed at random in
the initial condition.

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So if we do that, then
we're solving for the--

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that's all [INAUDIBLE],,
so I'll put it over here.

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The average escape time or
first passage time for the virus

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will be tau bar.

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Well, what I'll do is
I'll integrate over

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all the positions and
then divide by the volume

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because a uniform distribution.

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So I'll write the
integral over the volume

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as 4 pi, the solid angle times
integral from 0 to capital

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R of tau of r r-squared DR. And
I'll divide by the total volume

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4/3 pi capital r-cubed.

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OK, so if we do
this integral here,

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so when we plug-in
this, first of all,

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we can see that
we get, obviously,

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for the constant term,
r-squared over 60,

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we just have a ratio of volume
over volume, which is one.

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So we get r squared
over 6D is the first term,

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just integrating that constant.

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But then it's 1 minus.

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And then instead
of r-squared, you

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have r-squared times
this r-squared.

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It says r to fourth.

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So you integrate that.

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You get r to the fifth over 5.

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And so then you get
1/5 over this 1/3 here.

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And that gives you 3/5.

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And so that gives, when
I subtract one minus 3/5,

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I get 2/5.

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And 2 over 6 is 1/3.

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And so this ends up
being r-squared over 15D

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as we had previously quoted.

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So it's worth going through
the calculation just to see.

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This number 15 is an
order of magnitude.

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It's larger than 10.

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So while we normally
estimate diffusion times

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to be of order length
squared divided by D,

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so r-squared over
d, this calculation

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shows that the actual
average time is much smaller

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than that by a factor of 15.

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And that's precisely because
the virus virions sometimes

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find themselves initially
near the surface.

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So they get out more easily.

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So you don't have to
wait for diffusion

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across the entire drop.

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That sets the overall scale.

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But the average
drop diffusion time

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is actually less than that.