WEBVTT - autoGenerated
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We come now to some basic examples of variational problems in one dimension and all with a nice
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geometric flavor, just to give you the feeling of what a classical typical problem in calculus of
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variation should be. As a necessary preliminary knowledge we recall the fact that if a function
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is a Sobolev W11 over some segment AB in the real line with values in Rn, it doesn't matter,
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then it has a unique continuous representative over the closed segment. Actually this
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representative is absolutely continuous, which is a stronger condition than simply continuity.
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Now since of course W1P continuously embeds inside W11 is Jensen inequality, the same holds
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for functions in W1P and so we can always identify such functions with their continuous
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representatives in order to make sense of pointwise values, pointwise statements and the fundamental
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theorem of calculus. So u of x minus u of y is equal to the integral from x to y of u'. This
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fact is useful in order to introduce this notation for the set of Sobolev function with
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fixed values on the endpoints of the interval AB. So W1P, Y1, Y2, AB with values in Rn is a set of
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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.
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Basically we are fixing the boundary data. You see in one dimension everything works nicely
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because we have these nice results on the existence of a continuous representative up
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to the boundary and so the boundary values always make sense. In higher dimension is a little more
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complicated. And then we introduce the space W1P per periodic where we put periodic boundary
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conditions so we can ask that W of A is equal to W of B. This of course is simply the choice of Y2
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equal to Y1 if you want to use the previous definition. So as already mentioned a classical
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typical problem for calculus of variation with a clear geometric interpretation is the geodesic
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problem. Find the shortest curve which connects two given points into Rn. Clearly a solution is
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a straight line segment from one point to the other. There can be many variants. You can put
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for instance obstacle. I don't know. You should find the shortest possible curve which does not
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enter some particular region. Or you can post the problem in another setting. For instance you can
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instead of putting an obstacle you can put a constraint. So ask that this curve must belong
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to a manifold. So you take two points on a manifold and you look for the shortest possible
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curves which connects these two points. By the way this is how you define the Riemannian
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distance on a smooth manifold. Or even you can try to generalize this problem to non-Euclidean
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space. And then by the way this is a property of a distance in Rn and in these more general
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settings it becomes sometimes a definition of a distance. Whenever you're able to prove that this
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minimization problem does indeed have a solution. Let us come back to the classical Euclidean case.
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So clearly we represent these curves as the images of functions w from ab into Rn and the length then
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it's given by the parametric length integral which is simply the integral from a to b of the
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norm of w prime. The Euclidean norm of w prime. w belongs to w11 ab so it's continuous,
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absolutely continuous, so it's well defined over ab. Here there is a little sketch of all the
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possible curves and you see the segment of course. It's the shortest theorem. If we take a, b, y1,
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y2 in R as above, I mean ab in R, y1, y2 in Rn, then if we have u in w11, w2,
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w1, w2 from ab into Rn we have that u minimizes the function, the length function, the parametric
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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
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for every other w in the same space if and only if u is a monotone parametrization of the line
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segment from y2, sorry from y1 to y2 which means u of t can be written as 1 minus tau of t y1 plus
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tau of t y2 for t in ab and tau which belongs to w1101 ab. So it's a scalar sublet functions which
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takes in a takes value 0 and in b takes value 1 which makes sense no? So if you put t equal to a
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you get y1 if you get if you put t equal to b this term vanishes so you get y2.
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Now one may say but wait the solution therefore is not unique. Indeed it's not. The point is that
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the solution per se is not unique but the image of this function u of t is unique. It does not
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depend on the choice of tau and it is the segment. This comes from the fact that we are actually
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looking for a geometric object not for a particular way to parametrize this curve. That's the point.
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Very well and
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yeah. So in the two-dimensional cases so when the images are true so we'll reply in our geodesics
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we notice that curves can be also seen as graphs of functions defined from some segment x1 x2 into r
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and this allows to see this problem in another in this case equivalent way
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by looking for the minimization of the non-parametric length integral. So basically in
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this particular case there is a simpler way to write down the the functional before and
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and then we have to find this minimum of course and again we have similar results also for the
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non-parametric curves so if u in w11 y1 y2 from x1 or defined on the segment x1 x2
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then u is a solution to a problem so minimizes l in w11 y1 y2 over the interval x1 x2
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if and only if u is a fine so u of x is equal to y2 minus y1 over x2 minus x1 x
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plus y1 x2 minus y2 x1 over x1 x2 minus x1 so it's this the segment which connects
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the point x1 y1 and x2 y2 basically and you see this time the solution is unique because
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we are not having a parametrization of the curve we have in this case the curve
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they were watching the curve of the graph as the graph of a function so there is a unique way to do
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it but this works only for n equal to true so it's just to point out the difference.
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Another incredibly relevant basic I mean classical problem ratio problem is the
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planar isoperimetric problem so this means we are looking for
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sorry we want to prove one of these three equivalent principle for simple closed curves
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in our show with length l and area a area of the region enclosed by the curve
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it's not difficult to show that these statements are equivalent let's check
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a that says that for fixed length l the largest possible area a enclosed by the curve of length
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l is l squared over 4 pi with equality if and only if the curve is a circle for fixed b for fixed
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area a the shortest possible length is l equal to square root of 4 pi a equality if and only if
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it's a circle and we have the isoperimetric inequality this is principle c that is the area
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of enclosed by the curve will be always less equal than l squared over 4 pi where l is length
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of the curve and the quality if and only if the curve is a circle so you see that a and b are
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equivalent and given a or b of course this inequality holds and given this condition
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we see that also these two are satisfied so we will discuss formulation b so you fix the area
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enclosed by a simple curve closed curve such that so it's parameterized by an element in w11
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periodic so the the initial and the end point coincide and there are no other self-intersections
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so the curve is simple this is the curve is simple we are looking for the minimization of the length
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of integral over all possible such possible curves with the same fixed area
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so so we take our curves to be parameterized i mean to be modelled by elements of w11 periodic
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from a b into r2 they include the oriented area oriented because it can be positive in it as a
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positive sign if the curves runs counterclockwise and negative if it runs clockwise can be proved to
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be equal to one of these three integrals so basically the integral from a to b of w1 w2 prime
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where w1 is w2 are the first and second components of the curve w this follows by
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geometric consideration of the divergence theorem and it would be a task of the exercises of course
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i'm not going to tell you how to prove it now accepting this factor what we are going to to
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do again is to minimize the length function so you see l of w is again the same for the first
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problem over the admissible class which is the class of a a which is the curves in w11 periodic
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from a b into r2 such that the area is constant and equal to i mean it's fixed to be equal to a
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for a positive because i mean just just not necessary but in a in an area usually it's
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positive but you see it's a sign oriented area so so we reach this theorem on the planarized
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overimetric problem with parametric curves that is if we have a b and a capital a as before
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then a function u in ua a sorry a a is a minimizer of l if and only if u is a simple
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parametrization of a circle so u of t can be written as some y0 plus r cos of tau theta sinus
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of tau theta for some y0 into r2 r positive tau in w11 of a b monotone with a positive first
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derivative such that tau of b minus tau of a is equal to two pi naturally actually one can take
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also a to b in r because it's an oriented area and the case with the negative a coincides with
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the case of the parametrization tau in which it is monotone but non-increasing so that tau prime
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is negative we have a clockwise parametrization and so we have a negative oriented area
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of course there are many many many variants also for this problem
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well one of the possible variants is the so-called ditto problem in which the curves w are not
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closed so it's not required that the starting and the end points coincide but they satisfy
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some boundary condition for instance we ask that w the second component in a and in b is equal to
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zero say so the sorry we are yeah okay we're here we are starting and finishing on the x-axis
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for instance why is it called the ditto problem because of a very famous myth on the four
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foundations of kartago the town in tunisia which was the the rival of ancient rome
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this pro this myth says that queen ditto when she
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ran away from her from a homeland in a in libya asked her to be to have a piece of land for her
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and their followers to the king of the population of tunisia yarb and he replied she could have
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as much as it could be contained in the skin of a nox type a bull and what she did was to cut
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the skin in very very thin narrow pieces put them all together create a very long curve so that she
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surrounded by this half circle a region nearby the sea so here the constraint was the sea which i
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mean was not of course a straight line the costal shape but let's assume it for for simplicity and
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so which is the area maximizing a curve for a given length half a circle see this is a
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this is the formulation a of the isoparametric problem
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okay then
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so of course one could wonder if it's possible to deal with this kind of problem also in the
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non-parametric framework so not not to work with the parameterization of some curve but directly
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with the graph of some function but the problem is that the graph cannot be a closed curve because
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functions don't take two values in a point however if you consider this kind of variance
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in which basically one of the parts of the of the curve that's truly enclosed in area is
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fixed so the rest can be seen as a graph here above the x-axis is a graph so we are again
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brought to the minimization of the non-parametric length integral and that the constraint that
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the area here contained is fixed and what is this area is the integral from x1 to x2 of w
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if this is the graph of w this is all more familiar
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so we define the admissible class in the usual way adding some boundary conditions so say here
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it will be zero zero because w should be equal to zero and also be w choose
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no it's correct sorry it's correct both index one index two w should be zero
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now w is a scalar of course and we have a similar result with a little difference
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so if we take u in this admissible class for the non-parametric case of the planarized
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of parametric inequality then u is a minimizer if and only if the graph of u is a straight line
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or a circular arc from x1 x y2 sorry x1 y1 to x2 y2 in R2
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so first of all notice that this effect the fact that the solution must be either a line
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or a circular arc is telling us that in some cases there is no solution there is no minimizer
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because if the area is too large or too small you cannot find any segment or circular arc
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from x1 to x2 which encloses the correct area because see you must start from this point and
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arrive to this point here you can take i mean there will be a limit for the shape of circles
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both in one way or the other and there is exist only one segment
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um which connects these two points so say i don't know if a is very large so that
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um you will need to if a is too large you cannot possibly find this a
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a circular arc that could enclose the area and this result is telling you that
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the admissible class a tilde of a does not contain any minimum so that
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the minimization problem give us no solution the infimum of this functional exists of course
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because it's a number that you can find it of course negative number but it's not attained
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so it's not a minimum this is possible and this is the one of the simpler cases in which this can happen
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and then notice that if we choose y1 equal to y2 this is dido's problem of course but there is an
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additional rigidity sorry which is given by the fact that the starting points on the x axis are
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fixed instead if one would consider the dido problem but with the parametric version so we
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consider parametric curves then yes the second component of this curve should be zero should
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always start from zero and try from zero with respect to the y component yes but
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there will be no no constraint on the values of the x component of the curve so it could move
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back or forward and so this will help ensuring the existence of solutions more in general
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sorry finally we come to the brachistochrome problem so the brachistochrome problem is very
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nice and it is a it has a nice difference with respect to the other two problems
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while the the first two problems came from very ancient times i mean one is the basic
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problem of geometry which is the shortest path between two segments i mean for the ancient greeks
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it was obvious it was the segment two points sorry and the second instead is mentioned in
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the ancient mythology of greeks and romans so quite all they would say the brachistochrome
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problem is a typical problem of the age of well the beginning of calculus the time of
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Newton Leibniz and the brothers Bernoulli indeed it was posed by the brother Bernoulli one of the
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two um particular Johan Bernoulli it was posed this problem in 1696
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why i'm saying this because it's the problem to determine the shape of a curve which with
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given end points so x1 and x2 and x1 and y1 and x2 y2 in R2 still in the plane such that
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an idealized point mass so a body with some non-negative mass i mean a positive mass
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but idealized in the sense we assume it's a without dimension of course is a physical
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simplification which starts from x1 y1 with zero speed and then slides on the curve
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under the influence of gravity and without any friction and reaches x2 y2 in the shortest time
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so it's a problem that you see comes directly from the work of Newton from the idea of
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the existence of gravity without this incredible discovery of Newton for the time it was
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groundbreaking of course even now it's one of the greatest accomplishments in the in the history
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of physical sciences this problem could not exist that the mere idea of this problem could not be
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possible no so the funny thing is that Johan Bernoulli posed this problem in 1696 to i mean
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and received the solutions from the brilliant most brilliant minds of the time from from his
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brother Jacob from Leibniz from the L'HÃ´pital and from Newton himself however Newton himself
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Newton at the time was also not anymore a professor he was working if i'm not wrong already for the
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either for the Royal Academy of Sciences or the already for the for the for the tax office
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of the government because when he was old he stopped working in physics and became an officer
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of the interior he dealt with the taxes tax evasion and was brilliant also in that work
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anyway Newton solved the problem it is told in one night and then sent the solution
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sent a solution to Johan Bernoulli without signing it but it is reported that Johan Bernoulli
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wrote that he recognized the lion from his claw mark
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hey that's that's really nice historical anecdote about mathematics but let's not now come to
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the mathematical model for this problem because you see the formulation of this problem is
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exquisitely physical i mean while the previous two of course were i mean were not so difficult
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to visualize so one could say that also those problems had a very clear concrete interpretation
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they also had a very simple geometrical modeling simple i mean for our standards not also for
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the ancient ones not so much i mean the interviews but the notion of a curve of an area of a segment
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were kind of known at least in the naive sense here instead we have to translate these
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these new concepts from the well the the dynamic which was just born into the language of
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mathematics so first of all without loss of generality we need to assume that Y2 is less
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equal than Y1 because because the the point is falling and cannot fall upwards it falls down
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or if you want because of the conservation of energy principle i mean which is telling you this
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then you can assume that the X1 is less than X2 why because if they're equal
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and the the point is falling in a vertical line the solution is a segment trivial if X1 is larger
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then you reflect clearly it does not it's not relevant whether the point is going like this
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or like this the shape will be simply reflected so we can assume we are in this kind of situation
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now we want to model this curve and for doing so we select a C1 function Q
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from some interval AB into X1 X2 so this models this line here Q of A is equal to X1 and Q of B
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is equal to X2 of course and for the second coordinate we choose a function W which depends
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on Q for some functions W which will belong to W1 1 or Y1 by 2 of course over X1 X2 so our curve
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gamma is given by gamma T equal to QT W of Q of T and it goes from AB into R2 the derivative gamma
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prime is given by Q prime and W prime Q of T Q prime of T so that is the norm of this first
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derivative is the square root of 1 plus W prime of Q of T squared Q prime of T you may wonder why
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did not report the absolute value we can assume without loss of generality that Q prime is larger
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equal than zero otherwise the point would move backwards horizontally but we assume that X2 is
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larger than X1 so we're moving forward and not backwards actually we can even assume this is
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strictly positive but we'll see it in a moment now we exploit the conservation of energy principle
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and again remember the time 1696 it was just discovered I mean for me it's very fascinating
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so the sum of the kinetic and potential energy is constant in time which means that the function
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E of T which is the mass divided by 2 times the velocity squared so the norm of the velocity of
00:26:03.000 --> 00:26:11.000
the vector of the derivative vector squared plus gravitational acceleration mass and the height
00:26:11.000 --> 00:26:19.000
of the point mass which is the second coordinate of the of the W of QT this is constant
00:26:19.000 --> 00:26:29.000
on T in AB which means E of T is equal to AE of A but in A remember with zero speed at starting
00:26:29.000 --> 00:26:35.000
point the speed in the horizontal direction is zero in particular Q prime of A is zero so this
00:26:35.000 --> 00:26:46.000
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
00:26:46.000 --> 00:26:51.000
of parameterization of course therefore we simplify the mass because it's irrelevant
00:26:52.000 --> 00:26:58.000
and we get this equation over here by the way from this equation we see nice things first of all
00:26:58.000 --> 00:27:08.000
we see that W must be always strictly less than Y1 and you may tell me but of course the point is
00:27:08.000 --> 00:27:15.000
falling down it cannot go up yes sure this is the way to deduce it from math do something
00:27:15.000 --> 00:27:22.000
obvious from physics that the point is not falling upwards also without loss of generality
00:27:22.000 --> 00:27:28.000
we may assume that Q prime is strictly positive in the open segment because in A it is zero
00:27:30.000 --> 00:27:38.000
why this because otherwise the point would stop and remember we are looking for the curves on
00:27:38.000 --> 00:27:47.000
which the points reaches the end point x2 y2 in the shortest time so it makes sense to look only
00:27:47.000 --> 00:27:55.000
of the of the functions that allow a fastest possible movement so if the velocity which is
00:27:56.000 --> 00:28:01.000
modulated by the first derivative becomes zero at the same point but it's not the fastest possible
00:28:02.000 --> 00:28:11.000
path you will you will agree so Q is a c1 function of a reb with Q prime strictly positive
00:28:11.000 --> 00:28:17.000
therefore it has a continuous inverse we should call it tau from x1 to x2 into ab c1
00:28:18.000 --> 00:28:27.000
over x1 x2 not an ab sorry and tau prime is positive and of course it represents
00:28:27.000 --> 00:28:34.000
the time as a function of the first coordinate of the position vectors but then simple roots of
00:28:34.000 --> 00:28:40.000
calculus the first derivative of tau is one over the first derivative of Q therefore the equation
00:28:40.000 --> 00:28:47.000
over here can be rewritten this way one plus y prime squared over two tau prime squared plus
00:28:47.000 --> 00:28:53.000
g of w equal to g of y1 on the interval x1 x2 basically we compose everything with the
00:28:54.000 --> 00:29:03.000
we choose t to be tau of Q of t and so we arrive here therefore we can find that tau prime is equal
00:29:03.000 --> 00:29:11.000
to the square root of one plus norm of y prime squared over two g times y1 minus w and
00:29:12.000 --> 00:29:19.000
and plus because tau prime is positive here so there is no ambiguity into the square root
00:29:20.000 --> 00:29:26.000
very well now we take the integral from x1 to x2 to have finally the total time which is the total
00:29:26.000 --> 00:29:35.000
time from that the point x to go from x1 y1 to x2 y2 of course the final time minus the initial time
00:29:35.000 --> 00:29:41.000
which is the integral from x1 to x2 of tau prime which is the integral from x1 to x2 of one over
00:29:41.000 --> 00:29:49.000
square root of two g square root of one plus y prime squared over sorry w prime squared over y1
00:29:49.000 --> 00:29:56.000
minus w therefore the back histogram problem consists in minimization of the total time
00:29:56.000 --> 00:30:02.000
functional which is this integral i mean of course and up to this constant which is irrelevant for w in
00:30:02.000 --> 00:30:13.000
w11 y1 y2 over the interval x1 x2 we can also add the condition that t of w is plus infinity if w
00:30:13.000 --> 00:30:20.000
is larger equal than y1 on a set of positive Lebesgue measure in this way we impose we implicitly ask
00:30:20.000 --> 00:30:27.000
that w of x is always less than than than y1 which makes sense because the point is falling down
00:30:27.000 --> 00:30:37.000
and not flying theorem if x1 is less than x2 and y2 is less equal than y1 inner then u in w11 of
00:30:38.000 --> 00:30:48.000
y2 y1 y2 over the interval x1 x2 is a minimizer of t if and only if the graph is a cycloidal arc
00:30:48.000 --> 00:30:56.000
with upward point in cusp at x1 x2 but no interior cusp so basically it is this kind of figure
00:30:58.000 --> 00:31:02.000
the cusp here is given by the fact that the derivative should be plus infinity
00:31:03.000 --> 00:31:06.000
make it enough straight and here is that no
00:31:09.000 --> 00:31:15.000
in other words the graph of u has this nice expression for some parameter r and phi 2 which
00:31:15.000 --> 00:31:22.000
should solve of course the boundary conditions of course we are putting radix1 and y1 here so
00:31:22.000 --> 00:31:29.000
that when phi is equal to zero this term vanish and then when phi is equal to phi 2 we should get
00:31:29.000 --> 00:31:39.000
10 points and with this conclude the first introductory lecture thank you for your attention