Questions and Answers
Some questions asked during my office hours or by e-mail are of general
interest. Also some questions asked in class are too deep (or irrelevant)
to be answered in class, given the time limitation.
So I post here these questions with my answers
THEY ARE POSTED IN THE INVERSE CHRONOLOGICAL ORDER:
MOST RECENT QUESTIONS FIRST.
Q: What happens if we iterate Fourier transform?
A: You can easily figure this out. First, see what happens if we apply it
twice. The formula for the Fourier transform is almost the same as the
one for the inverse transform, so you almost recover your original function.
This gives F(F(f))(x)=f(-x) (Check!). So when we apply Fourier
transform four times, the result will be F(F(F(F(f))))(x)=f(x),
so we return to the same function we started with! This gives a complete answer
to the question: now we can easily figure out what happens if we apply
Fourier transform any number of times, the sequence has period four.
You surely know another mathematical operation with a similar property:
multiplication by i, the imaginary unit. Or geometrically, rotation by
90 degrees. This analogy is not accidental, of course, and we will explore
it a bit further in this course. This question suggests me a nice
examination problem: given some specific function, apply Fourier transform
2002 times to it, and write the answer. And of course, I will choose a function
whose FT is hard to find:-)
Q: Is the function defined by f(x)=1, when x is rational, and f(x)=0,
when x is irrational, integrable in the sense of Riemann?
A: NO. (I made a mistake when I answered this question during the lecture
on Febr. 1). In the definition of Riemann integral, you break the interval,
say [0,1] into finitely many sub-intervals, then you multiply the length
of each sub-interval by the value of the function at some point x of
this subinterval and add the results. The integral is the limit of these sums,
IF THIS LIMIT EXISTS. For our function the limit does not exist, because
we can take either a rational point or irrational on every interval,
so the sum can be made, for example, zero, or 1, no matter how small
our subintervals are.
The integral exists only in the sense of Lebesgue, and it is equal to
zero.
Q: Problem 1 on p. 7 says that the normal distribution with standard deviation=sqrt(time) satisfies the heat equation. Is this an accident?
A:
Of course, this is not an accident. (In fact this is one of the deepest questions I was ever asked by students:-) The simplest answer would be to refer to the books of Marc Kac (strongly recommended to all interested in mathematics.)
Coming to the essence of the question. Imagine a particle, which starts a
random walk (Brownian motion) at the point 0 on the x-axis. Let us look
at the position of this particle at the time time t>0. It is easy to understand
(if you ever had a probability course, like MA 519, for example, or higher),
that
this position will, be random, normally distributed, with average zero
(what else could it be?!) and standard deviation sqrt(t).
Now, what is heat? According to
the modern theories (kinetic theory of gases, statistical mechanics), heat
is the random motion of molecules. If you look at an individual molecule,
it performs something like a Brownian motion. The average kinetic energy of
molecules at some place is what we call temperature. These considerations make
at least plausible the connection between the random motion and the heat
equation. Actually this connection was subject to exploration by
mathematicians in XX century, including Mark Kac and many others. The
exact mathematical theorem that relates solutions of the heat equation
and is called Kac Formula. I also have to mention here the so-called
Feynman-Kac formula, which gives similar probabilistic interpretation
of solutions of Schrodinger's equation. The main difference between
the Schrodinger equation and the heat equation is the factor i, the square root
of -1. This difference can be removed by a simple change of the time variable.
From the point of view of physics this means that we consider
"pure imaginary time"!
Let me give a simpler example, related to the Laplace equation.
Imagine a bounded region D in the plane.
Some (say, positive) function f is defined on the boundary. Suppose a
random motion (Brownian motion) starts as a point (x,y) inside D.
If you don't know what Brownian motion exactly is, imagine a drunkard,
who is initially places at the point (x,y), and starts walking,
one step at a unit of time, but every step is taken in a random direction.
You may assume that there are 4 possible directions: N, E, S and W.
Also assume that the steps are very small, so s/he will make a lot of steps,
before s/he reaches the boundary of D. But s/he will eventually reach
the boundary for sure! (This is more or less evident, and also this is a
THEOREM.) Remember, in the very beginning, it was mentioned, that a positive function f was defined on this boundary. Now imagine that the drunkard gets some
money when s/he reaches the boundary for the first time (and then s/he stays there, to sleep or maybe, to buy another drink).
The amount of money is described by that positive function f on the boundary,
mentioned above. It depends on the point where the drunkard reaches the boundary
for the first time. The question is: what is the EXPECTATION of the total gain
of the drunkard in this experiment? To be sure: expectation is the average amount
of money s/he receives, if the experiment is repeated many times.
The answer depends on the starting point (x,y),
and of course, on the function f.
THEOREM. Let u(x,y) be the expectation of the gain of the drunkard. Then u
is the solution of the Laplace equation in the region D, with boundary condition
f.
This exciting theorem is discussed at length in the books of Mark Kac,
but the statement and proof in the most general setting is due to a
Japanese mathematician Kakutani. The problem of finding the solution
(it is unique) of the Laplace equation in a region, with given boundary
conditions is called the Dirichlet Problem.
The drunkard described above provides
a way to solve it numerically. Sometimes this way is the most effective one.