We come now to some basic examples of variational problems in one dimension and all with a nice
geometric flavor, just to give you the feeling of what a classical typical problem in calculus of
variation should be. As a necessary preliminary knowledge we recall the fact that if a function
is a Sobolev W11 over some segment AB in the real line with values in Rn, it doesn't matter,
then it has a unique continuous representative over the closed segment. Actually this
representative is absolutely continuous, which is a stronger condition than simply continuity.
Now since of course W1P continuously embeds inside W11 is Jensen inequality, the same holds
for functions in W1P and so we can always identify such functions with their continuous
representatives in order to make sense of pointwise values, pointwise statements and the fundamental
theorem of calculus. So u of x minus u of y is equal to the integral from x to y of u'. This
fact is useful in order to introduce this notation for the set of Sobolev function with
fixed values on the endpoints of the interval AB. So W1P, Y1, Y2, AB with values in Rn is a set of
functions W in W1P, AB with values in Rn such that W of A is equal to Y1 and W of B is equal to Y2.
Basically we are fixing the boundary data. You see in one dimension everything works nicely
because we have these nice results on the existence of a continuous representative up
to the boundary and so the boundary values always make sense. In higher dimension is a little more
complicated. And then we introduce the space W1P per periodic where we put periodic boundary
conditions so we can ask that W of A is equal to W of B. This of course is simply the choice of Y2
equal to Y1 if you want to use the previous definition. So as already mentioned a classical
typical problem for calculus of variation with a clear geometric interpretation is the geodesic
problem. Find the shortest curve which connects two given points into Rn. Clearly a solution is
a straight line segment from one point to the other. There can be many variants. You can put
for instance obstacle. I don't know. You should find the shortest possible curve which does not
enter some particular region. Or you can post the problem in another setting. For instance you can
instead of putting an obstacle you can put a constraint. So ask that this curve must belong
to a manifold. So you take two points on a manifold and you look for the shortest possible
curves which connects these two points. By the way this is how you define the Riemannian
distance on a smooth manifold. Or even you can try to generalize this problem to non-Euclidean
space. And then by the way this is a property of a distance in Rn and in these more general
settings it becomes sometimes a definition of a distance. Whenever you're able to prove that this
minimization problem does indeed have a solution. Let us come back to the classical Euclidean case.
So clearly we represent these curves as the images of functions w from ab into Rn and the length then
it's given by the parametric length integral which is simply the integral from a to b of the
norm of w prime. The Euclidean norm of w prime. w belongs to w11 ab so it's continuous,
absolutely continuous, so it's well defined over ab. Here there is a little sketch of all the
possible curves and you see the segment of course. It's the shortest theorem. If we take a, b, y1,
y2 in R as above, I mean ab in R, y1, y2 in Rn, then if we have u in w11, w2,
w1, w2 from ab into Rn we have that u minimizes the function, the length function, the parametric
length integral in w11 from y1 to y2 ab with values in Rn which means l of u is a single then l of w
for every other w in the same space if and only if u is a monotone parametrization of the line
segment from y2, sorry from y1 to y2 which means u of t can be written as 1 minus tau of t y1 plus
tau of t y2 for t in ab and tau which belongs to w1101 ab. So it's a scalar sublet functions which
takes in a takes value 0 and in b takes value 1 which makes sense no? So if you put t equal to a
you get y1 if you get if you put t equal to b this term vanishes so you get y2.
Now one may say but wait the solution therefore is not unique. Indeed it's not. The point is that
the solution per se is not unique but the image of this function u of t is unique. It does not
depend on the choice of tau and it is the segment. This comes from the fact that we are actually
looking for a geometric object not for a particular way to parametrize this curve. That's the point.
Very well and
yeah. So in the two-dimensional cases so when the images are true so we'll reply in our geodesics
we notice that curves can be also seen as graphs of functions defined from some segment x1 x2 into r
and this allows to see this problem in another in this case equivalent way
by looking for the minimization of the non-parametric length integral. So basically in
this particular case there is a simpler way to write down the the functional before and
and then we have to find this minimum of course and again we have similar results also for the
non-parametric curves so if u in w11 y1 y2 from x1 or defined on the segment x1 x2
then u is a solution to a problem so minimizes l in w11 y1 y2 over the interval x1 x2
if and only if u is a fine so u of x is equal to y2 minus y1 over x2 minus x1 x
plus y1 x2 minus y2 x1 over x1 x2 minus x1 so it's this the segment which connects
the point x1 y1 and x2 y2 basically and you see this time the solution is unique because
we are not having a parametrization of the curve we have in this case the curve
they were watching the curve of the graph as the graph of a function so there is a unique way to do
it but this works only for n equal to true so it's just to point out the difference.
Another incredibly relevant basic I mean classical problem ratio problem is the
planar isoperimetric problem so this means we are looking for
sorry we want to prove one of these three equivalent principle for simple closed curves
in our show with length l and area a area of the region enclosed by the curve
it's not difficult to show that these statements are equivalent let's check
a that says that for fixed length l the largest possible area a enclosed by the curve of length
l is l squared over 4 pi with equality if and only if the curve is a circle for fixed b for fixed
area a the shortest possible length is l equal to square root of 4 pi a equality if and only if
it's a circle and we have the isoperimetric inequality this is principle c that is the area
of enclosed by the curve will be always less equal than l squared over 4 pi where l is length
of the curve and the quality if and only if the curve is a circle so you see that a and b are
equivalent and given a or b of course this inequality holds and given this condition
we see that also these two are satisfied so we will discuss formulation b so you fix the area
enclosed by a simple curve closed curve such that so it's parameterized by an element in w11
periodic so the the initial and the end point coincide and there are no other self-intersections
so the curve is simple this is the curve is simple we are looking for the minimization of the length
of integral over all possible such possible curves with the same fixed area
so so we take our curves to be parameterized i mean to be modelled by elements of w11 periodic
from a b into r2 they include the oriented area oriented because it can be positive in it as a
positive sign if the curves runs counterclockwise and negative if it runs clockwise can be proved to
be equal to one of these three integrals so basically the integral from a to b of w1 w2 prime
where w1 is w2 are the first and second components of the curve w this follows by
geometric consideration of the divergence theorem and it would be a task of the exercises of course
i'm not going to tell you how to prove it now accepting this factor what we are going to to
do again is to minimize the length function so you see l of w is again the same for the first
problem over the admissible class which is the class of a a which is the curves in w11 periodic
from a b into r2 such that the area is constant and equal to i mean it's fixed to be equal to a
for a positive because i mean just just not necessary but in a in an area usually it's
positive but you see it's a sign oriented area so so we reach this theorem on the planarized
overimetric problem with parametric curves that is if we have a b and a capital a as before
then a function u in ua a sorry a a is a minimizer of l if and only if u is a simple
parametrization of a circle so u of t can be written as some y0 plus r cos of tau theta sinus
of tau theta for some y0 into r2 r positive tau in w11 of a b monotone with a positive first
derivative such that tau of b minus tau of a is equal to two pi naturally actually one can take
also a to b in r because it's an oriented area and the case with the negative a coincides with
the case of the parametrization tau in which it is monotone but non-increasing so that tau prime
is negative we have a clockwise parametrization and so we have a negative oriented area
of course there are many many many variants also for this problem
well one of the possible variants is the so-called ditto problem in which the curves w are not
closed so it's not required that the starting and the end points coincide but they satisfy
some boundary condition for instance we ask that w the second component in a and in b is equal to
zero say so the sorry we are yeah okay we're here we are starting and finishing on the x-axis
for instance why is it called the ditto problem because of a very famous myth on the four
foundations of kartago the town in tunisia which was the the rival of ancient rome
this pro this myth says that queen ditto when she
ran away from her from a homeland in a in libya asked her to be to have a piece of land for her
and their followers to the king of the population of tunisia yarb and he replied she could have
as much as it could be contained in the skin of a nox type a bull and what she did was to cut
the skin in very very thin narrow pieces put them all together create a very long curve so that she
surrounded by this half circle a region nearby the sea so here the constraint was the sea which i
mean was not of course a straight line the costal shape but let's assume it for for simplicity and
so which is the area maximizing a curve for a given length half a circle see this is a
this is the formulation a of the isoparametric problem
okay then
so of course one could wonder if it's possible to deal with this kind of problem also in the
non-parametric framework so not not to work with the parameterization of some curve but directly
with the graph of some function but the problem is that the graph cannot be a closed curve because
functions don't take two values in a point however if you consider this kind of variance
in which basically one of the parts of the of the curve that's truly enclosed in area is
fixed so the rest can be seen as a graph here above the x-axis is a graph so we are again
brought to the minimization of the non-parametric length integral and that the constraint that
the area here contained is fixed and what is this area is the integral from x1 to x2 of w
if this is the graph of w this is all more familiar
so we define the admissible class in the usual way adding some boundary conditions so say here
it will be zero zero because w should be equal to zero and also be w choose
no it's correct sorry it's correct both index one index two w should be zero
now w is a scalar of course and we have a similar result with a little difference
so if we take u in this admissible class for the non-parametric case of the planarized
of parametric inequality then u is a minimizer if and only if the graph of u is a straight line
or a circular arc from x1 x y2 sorry x1 y1 to x2 y2 in R2
so first of all notice that this effect the fact that the solution must be either a line
or a circular arc is telling us that in some cases there is no solution there is no minimizer
because if the area is too large or too small you cannot find any segment or circular arc
from x1 to x2 which encloses the correct area because see you must start from this point and
arrive to this point here you can take i mean there will be a limit for the shape of circles
both in one way or the other and there is exist only one segment
um which connects these two points so say i don't know if a is very large so that
um you will need to if a is too large you cannot possibly find this a
a circular arc that could enclose the area and this result is telling you that
the admissible class a tilde of a does not contain any minimum so that
the minimization problem give us no solution the infimum of this functional exists of course
because it's a number that you can find it of course negative number but it's not attained
so it's not a minimum this is possible and this is the one of the simpler cases in which this can happen
and then notice that if we choose y1 equal to y2 this is dido's problem of course but there is an
additional rigidity sorry which is given by the fact that the starting points on the x axis are
fixed instead if one would consider the dido problem but with the parametric version so we
consider parametric curves then yes the second component of this curve should be zero should
always start from zero and try from zero with respect to the y component yes but
there will be no no constraint on the values of the x component of the curve so it could move
back or forward and so this will help ensuring the existence of solutions more in general
sorry finally we come to the brachistochrome problem so the brachistochrome problem is very
nice and it is a it has a nice difference with respect to the other two problems
while the the first two problems came from very ancient times i mean one is the basic
problem of geometry which is the shortest path between two segments i mean for the ancient greeks
it was obvious it was the segment two points sorry and the second instead is mentioned in
the ancient mythology of greeks and romans so quite all they would say the brachistochrome
problem is a typical problem of the age of well the beginning of calculus the time of
Newton Leibniz and the brothers Bernoulli indeed it was posed by the brother Bernoulli one of the
two um particular Johan Bernoulli it was posed this problem in 1696
why i'm saying this because it's the problem to determine the shape of a curve which with
given end points so x1 and x2 and x1 and y1 and x2 y2 in R2 still in the plane such that
an idealized point mass so a body with some non-negative mass i mean a positive mass
but idealized in the sense we assume it's a without dimension of course is a physical
simplification which starts from x1 y1 with zero speed and then slides on the curve
under the influence of gravity and without any friction and reaches x2 y2 in the shortest time
so it's a problem that you see comes directly from the work of Newton from the idea of
the existence of gravity without this incredible discovery of Newton for the time it was
groundbreaking of course even now it's one of the greatest accomplishments in the in the history
of physical sciences this problem could not exist that the mere idea of this problem could not be
possible no so the funny thing is that Johan Bernoulli posed this problem in 1696 to i mean
and received the solutions from the brilliant most brilliant minds of the time from from his
brother Jacob from Leibniz from the L'Hôpital and from Newton himself however Newton himself
Newton at the time was also not anymore a professor he was working if i'm not wrong already for the
either for the Royal Academy of Sciences or the already for the for the for the tax office
of the government because when he was old he stopped working in physics and became an officer
of the interior he dealt with the taxes tax evasion and was brilliant also in that work
anyway Newton solved the problem it is told in one night and then sent the solution
sent a solution to Johan Bernoulli without signing it but it is reported that Johan Bernoulli
wrote that he recognized the lion from his claw mark
hey that's that's really nice historical anecdote about mathematics but let's not now come to
the mathematical model for this problem because you see the formulation of this problem is
exquisitely physical i mean while the previous two of course were i mean were not so difficult
to visualize so one could say that also those problems had a very clear concrete interpretation
they also had a very simple geometrical modeling simple i mean for our standards not also for
the ancient ones not so much i mean the interviews but the notion of a curve of an area of a segment
were kind of known at least in the naive sense here instead we have to translate these
these new concepts from the well the the dynamic which was just born into the language of
mathematics so first of all without loss of generality we need to assume that Y2 is less
equal than Y1 because because the the point is falling and cannot fall upwards it falls down
or if you want because of the conservation of energy principle i mean which is telling you this
then you can assume that the X1 is less than X2 why because if they're equal
and the the point is falling in a vertical line the solution is a segment trivial if X1 is larger
then you reflect clearly it does not it's not relevant whether the point is going like this
or like this the shape will be simply reflected so we can assume we are in this kind of situation
now we want to model this curve and for doing so we select a C1 function Q
from some interval AB into X1 X2 so this models this line here Q of A is equal to X1 and Q of B
is equal to X2 of course and for the second coordinate we choose a function W which depends
on Q for some functions W which will belong to W1 1 or Y1 by 2 of course over X1 X2 so our curve
gamma is given by gamma T equal to QT W of Q of T and it goes from AB into R2 the derivative gamma
prime is given by Q prime and W prime Q of T Q prime of T so that is the norm of this first
derivative is the square root of 1 plus W prime of Q of T squared Q prime of T you may wonder why
did not report the absolute value we can assume without loss of generality that Q prime is larger
equal than zero otherwise the point would move backwards horizontally but we assume that X2 is
larger than X1 so we're moving forward and not backwards actually we can even assume this is
strictly positive but we'll see it in a moment now we exploit the conservation of energy principle
and again remember the time 1696 it was just discovered I mean for me it's very fascinating
so the sum of the kinetic and potential energy is constant in time which means that the function
E of T which is the mass divided by 2 times the velocity squared so the norm of the velocity of
the vector of the derivative vector squared plus gravitational acceleration mass and the height
of the point mass which is the second coordinate of the of the W of QT this is constant
on T in AB which means E of T is equal to AE of A but in A remember with zero speed at starting
point the speed in the horizontal direction is zero in particular Q prime of A is zero so this
term vanishes and it remains G mass W of Q of A Q of A is X1 W of X1 is well one by our choice
of parameterization of course therefore we simplify the mass because it's irrelevant
and we get this equation over here by the way from this equation we see nice things first of all
we see that W must be always strictly less than Y1 and you may tell me but of course the point is
falling down it cannot go up yes sure this is the way to deduce it from math do something
obvious from physics that the point is not falling upwards also without loss of generality
we may assume that Q prime is strictly positive in the open segment because in A it is zero
why this because otherwise the point would stop and remember we are looking for the curves on
which the points reaches the end point x2 y2 in the shortest time so it makes sense to look only
of the of the functions that allow a fastest possible movement so if the velocity which is
modulated by the first derivative becomes zero at the same point but it's not the fastest possible
path you will you will agree so Q is a c1 function of a reb with Q prime strictly positive
therefore it has a continuous inverse we should call it tau from x1 to x2 into ab c1
over x1 x2 not an ab sorry and tau prime is positive and of course it represents
the time as a function of the first coordinate of the position vectors but then simple roots of
calculus the first derivative of tau is one over the first derivative of Q therefore the equation
over here can be rewritten this way one plus y prime squared over two tau prime squared plus
g of w equal to g of y1 on the interval x1 x2 basically we compose everything with the
we choose t to be tau of Q of t and so we arrive here therefore we can find that tau prime is equal
to the square root of one plus norm of y prime squared over two g times y1 minus w and
and plus because tau prime is positive here so there is no ambiguity into the square root
very well now we take the integral from x1 to x2 to have finally the total time which is the total
time from that the point x to go from x1 y1 to x2 y2 of course the final time minus the initial time
which is the integral from x1 to x2 of tau prime which is the integral from x1 to x2 of one over
square root of two g square root of one plus y prime squared over sorry w prime squared over y1
minus w therefore the back histogram problem consists in minimization of the total time
functional which is this integral i mean of course and up to this constant which is irrelevant for w in
w11 y1 y2 over the interval x1 x2 we can also add the condition that t of w is plus infinity if w
is larger equal than y1 on a set of positive Lebesgue measure in this way we impose we implicitly ask
that w of x is always less than than than y1 which makes sense because the point is falling down
and not flying theorem if x1 is less than x2 and y2 is less equal than y1 inner then u in w11 of
y2 y1 y2 over the interval x1 x2 is a minimizer of t if and only if the graph is a cycloidal arc
with upward point in cusp at x1 x2 but no interior cusp so basically it is this kind of figure
the cusp here is given by the fact that the derivative should be plus infinity
make it enough straight and here is that no
in other words the graph of u has this nice expression for some parameter r and phi 2 which
should solve of course the boundary conditions of course we are putting radix1 and y1 here so
that when phi is equal to zero this term vanish and then when phi is equal to phi 2 we should get
10 points and with this conclude the first introductory lecture thank you for your attention