So, the last exercise, instead, was more about the calculations, mostly, just to get, I mean,
to make you experiment some real way of solving by hand a somehow elementary case of the heat
equation on a bounded domain, in particular, here, in one dimension, and you were given
also some diffusion constant k, not equal to 1, but it's not so relevant, and you're working on a
bounded domain, so the segment zero L for some L positive, and you were given the initial band
value problem, so ut equal to k uxx in zero L times zero plus infinity, ux0 is equal to phi of x,
the initial condition, u0t is equal to zero, and the derivative in x evaluated at x equal L is equal
to zero for every t positive, and your task was to use the separation of variables method in order
to find the solution, which means, we should select only the solutions of the form u of xt equal to v
of x f of t, we plug this function inside the equation to find the derivative in time, of course,
it was only on f derivative in x only on v, so we get v of x f prime of t equal to k v second of x
f of t, which means f prime of f of t equal to k v second over v, and this means that the f prime
over f and v second over v are actually constant, because here the right hand side depends only on x,
the left hand side is only on t, and so if I take a derivative in t, you get zero and vice versa, so
there must exist some lambda such that f prime of f is equal to lambda f of t, and v second of x is
equal to lambda over k v of x, particularly will be the solution to this eigenvalue problem, you may
say how do you get the condition v of zero equal to zero, v prime of l equal to zero from here, of
course, and from this definition, it's not difficult to show. The first thing to notice is that lambda
must be negative, if we want to find a non-zero solution, I mean, it's admissible, but it's not
very interesting, and also it does not, it cannot possibly solve, help us in solving the initial
boundary variable problem given by this exercise. Indeed, if lambda is zero, well, then v of x is
equal to a x plus b, it's linear, but then the condition v of zero equal to zero means that b must
be zero, v prime of l equal to zero means that a must be zero, and so v must be identically zero.
If instead lambda is positive, then v of x will be equal to a times e to the beta x plus b e to the
minus beta x, where beta is the square root of lambda over k, and the initial condition will give
us that a plus b must be zero, and beta a e to the beta l minus b e to the minus beta m must be zero,
but since b is equal to minus a, here there is actually written, beta a e to the beta l plus
e to the minus beta l is equal to zero, but this is non-negative, so this is impossible,
so the only possibility is that a must be zero, so v is identically zero. So, excluding all these
cases, lambda should be negative, and so v of x can be written as a cos beta x plus b sinus beta x,
where beta is the square root of absolute value of lambda over k, and a and b satisfy the initial
condition, which means a must be zero, and b, well, b must satisfy this condition over here,
so in other words a must be zero, and here we have b beta cos beta l equal to zero,
and now we see that b also will be zero, unless beta l is equal to 2 m plus 1 pi over 2 for some
m in n zero. Therefore, the eigenvalues, the values of lambda for which we find non-zero solution
are minus k 2 m plus 1 squared pi squared over 4 l squared, because they correspond to choosing beta
equal to m m plus 1 over 2 pi over l. Therefore, the eigenfunctions are some v m divided by b m
sinus of m plus 1 over 2 pi over l x. On the other hand, of course, the equation f prime of t equal
to lambda m f of t give us f m of t e to the, equal to e to the lambda m t, sometimes some constant,
and so the general solution u is given as a series of a function v m f m, so we write it in this way,
and notice that it is quite relevant to all this discussion and the fact that lambda must be
negative, because otherwise here if lambda would be positive or zero, here we would get not this
very nice decay in term in t. I mean, if lambda would be positive, here this stuff will grow
exponentially and that would be really a problem. So, for some constant c m here, which of course
is the multiplication of that, but we still have the initial condition in time, u of x zero must be
phi of x. Therefore, this constant c m must be the Fourier coefficients of phi, so if you want,
you can write it explicitly like this, or you can even recall that the integral from zero to l of
sine squared m plus one half pi over l x is simply m l, but I mean, that's not so relevant.
The purpose of the exercise, I accept everything, and okay, now one may say,
how can you say that, I mean, in order to claim this, you're basically saying that phi can be
written only as a sum, as a series of sinus, with in particular sinus of this form, which means
sinus of odd number pi over two l x. Why? Why are we excluding the cosinus and why are we excluding
sinus of m pi over l x? Why? Because of the initial condition, sorry, because of the fact that
u zero t is equal to zero, u x l t is equal to zero. This implies that if we extend the phi,
if we want to, now here I'm not very rigorous, but the point is that if we want to see phi
written using the Fourier series, so as an element of the completion of the space,
of the linear space generated by the family of sinus of pi m pi l x,
m pi over two l x, and cosinus m pi over two l x, two l x, then the functions in this family,
we must pick the functions in this family that satisfy these two conditions, and these are only
this family of sinus. Or, if you want to see it in another way, one can extend the function phi
in another way, over minus l l in this way here, and then to all r by periodicity, phi is odd,
and don't notice that this is coherent with the fact that in the origin must be zero,
odd functions are zero in the origin, so this tells us that we have an expansion only with
sin functions, and then the condition on the first derivative, basically saying the first derivative is
even, because the function is odd, so the first derivative must be even, m must vanish in the
external points, this means that cosinus of pi m over two must be zero, but this happens if and only
m is odd. And so, of all these series, we are selecting only the odd values of m, and we are done.
I mean, this was the degree, the level of rigor that was required, so I accepted this.
So thank you very much for your attention, and I guess we shall see each other online
after the Christmas vacations. Bye bye.