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\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Discussion 1 -- Diffusion on {\bf R}}
\centerline{February 1, 2001}
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\noindent
{\it Background}
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Before we continue on to the study of boundary value problems for
the wave equation in 1 space variable, we want to introduce the
other major PDE that we will focus on this term.  This other 
equation is often called the {\it heat equation} because
it serves as a model for {\it heat conduction} and was originally
introduced in that context.  To derive the equation, though, we will
consider a different but related situation where our physical intuition
may be somewhat more immediately applicable.  

Namely, let's consider a motionless liquid filling a long, straight, thin
tube with constant cross-sections (their exact
shape is not relevant, but you can think circles if that is helpful).  
Imagine that a chemical substance (e.g. a dye) is introduced
at some time and begins to spread out, or {\it diffuse} through the liquid.\footnote{${}^1$}{To be precise,
the physical mechanism for the spreading out is the essentially random
motions of the individual molecules of the liquid ``pushing the dye along''; 
the liquid as a whole is {\it not moving} -- there are no currents in the 
tube itself.}   To describe the diffusion process, we
consider the {\it concentration} 
of the dye.  {\it Note:} the hypothesis that the tube is {\it thin}
means that we will ignore possible variation of the concentration
on each cross-section of the tube perpendicular to the $x$ axis.
We assume that concentration is constant on each cross section, so 
it depends on only one space variable $x$ -- the longitudinal distance 
along the tube, and on time.  This means that the concentration
is a function $u(x,t)$, and the units of concentration are {\it mass per 
unit length} (volume of a section of the tube is proportional to its length).
The {\it key fact} about diffusion is that:
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\noindent
{\it The dye will tend to spread out from areas of high concentration
to areas of low concentration}
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\noindent
The precise version of this is an empirical relation (i.e. experimental 
observation) called {\it Fick's Law}:
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\noindent
(FL) {\it The amount of dye traveling to the right past $x = x_0$ per unit time 
at time $t_0$ is proportional to the derivative $u_x(x_0,t_0)$.}
Note that $u_x$ is the ``gradient'' of $u$ with respect to the 
spatial variable in this 1D setting.  Dye traveling to the {\it left}
corresponds to a negative amount flowing to the right.
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\noindent
Today, we want to use this information derive a PDE governing $u$, and 
consider some of its solutions.
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\noindent
{\it Discussion Questions}
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\item{A)} What does the function $u(x,0)$ represent in the 
diffusion set-up?  Same question for $u(0,t)$.  What does $u_t(x,t)$
represent?  What does $u_x(x,t)$ represent?  If we write Fick's
Law as an equation, what is the sign of the proportionality constant,
recalling that dye always flows from areas of higher concentration
to areas of lower concentration?
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\item{B)} Consider the section of the tube 
extending from $x$ to $x_1 = x + \Delta x$.
Show that the {\it total mass} of the dye {\it in that section} (which depends 
on $t$), is computed by the integral
$$M(t) = \int_{x}^{x_1} u(x,t)\ dx\leqno(1)$$
(Argue as in the derivation of integral formulas for areas, volumes, etc.:
Subdivide, approximate, recognize your approximation as a Riemann sum
for an integral, then ``let $n\to\infty$'' to get the exact mass.)
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\item{C)}  From B, it follows that the rate of change of $M$ with
respect to $t$ is given by 
$$M'(t) = {d\over dt}\int_{x}^{x_1} u(x,t)\ dx = \int_{x}^{x_1} u_t(x,t)\ dx\leqno(2)$$
(That is, to differentiate $M$ with respect to $t$, we can integrate the time
derivative of the concentration.)  We will assume that no dye
enters or leaves our section of the tube through its lateral surface.
Thus dye enters our section only by flowing in across the 
ends at $x$ and $x_1$.  Using this and (FL) (Fick's Law)
above, argue that 
$$M'(t) = {\rm flow\ in} - {\rm flow\ out} = ku_x(x_1,t) - ku_x(x,t)$$
(Be careful:  This equation {\it does not say that
the flow in is $ku_x(x_1,t)$, just that the difference of the flow in 
and the flow out is the difference on the right.}
What does $u_x > 0$ at a point mean?  What does $u_x < 0$
mean?  What sign does the proportionality factor $k$ have here?)
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\item{D)}  From (1) and (2),
$$\int_{x}^{x_1} u_t(x,t)\ dx = ku_x(x_1,t) - ku_x(x,t)$$
Divide both sides by $x_1 - x$ and let $x_1 \to x$ to derive
the {\it diffusion equation}:
$$u_t = ku_{xx}\leqno(3)$$
{\bf CAUTION:}  This equation is also a 2nd order linear PDE, but
it involves just the {\it first} derivative
of $u$ with respect to $t$, NOT $u_{tt}$ as in the wave equation.
This makes a large difference!
The techniques for solving the wave equation that we have studied
{\it do not apply} to this equation.  
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\item{E)} For today, we will consider solutions $u(x,t)$ of
(3) for all $x \in {\bf R}$, but just $t > 0$.
Show that each of the following functions is a {\it solution}
of the diffusion equation (3).  Also
answer each of the questions connected with the specific solutions.
\itemitem{1)} $u_1(x,t) = a x + b$ ($a$,$b$ constant).  What is "special"
about this type of solution?  (What happens to $u$ as
$t$ changes?)
\itemitem{2)}  $u_2(x,t) = x(x^2 + 6kt)$.  Is this a physically reasonable
concentration function for a diffusion situation?  If so, what situation does
it describe?
\itemitem{3)}  $u_3(x,t) = e^{-kt} \sin(x) + 2$ and $u_4(x,t) = e^{-kt} \cos(x) + 2$.
\itemitem{4)}  
$$u_5(x,t) = {1\over \sqrt{t}} e^{-x^2\over 4kt}$$
Note that $u_5$ is not defined at $t = 0$.  Plot $u_5(x,t)$ for $k = 1$ and 
several different $t$ (at least $t = 1/4,1/2,1,2$) and describe
what is happening as $t$ increases.  What do you think the graph
of $\lim_{t\to 0^+} u_5(x,t)$ looks like? 
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\noindent
{\it Assignment}
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Group writeups due in class Thursday, February 8.
\bye



