In the previous lecture, we were trying to prove the Picard's existence and uniqueness theorem and the proof was divided into 4 parts and we proved part A and part B. So, let me just recall the Picard's existence and uniqueness theorem. Let d be a domain in R 2 and f is a function from d to R, a real valued function satisfying the following conditions. f is continuous on d and f x y is Lipschitz continuous with respect to y with a Lipschitz constant alpha greater than 0 and x 0, y 0 the initial point of the initial value problem that is assumed to be an integer point in the domain d and we take 2 constants A greater than 0, B greater than 0 such that the rectangle defined by this is fully inside the domain d and we use a notation m is a maximum value of f in the rectangle which is attained because f is continuous and h is the minimum of A and B by A. Then the Picard's theorem is that the initial value problem has a unique solution in the interval x minus x 0 is less than or equal to h and the main idea in the Picard's theorem is we defined what is known as a Picard's iterative scheme, the iterands, we defined the Picard's iterands by phi n is equal to y 0 plus integral x 0 to x f of t phi n minus 1 t dt n varies from 1 to n is 1, 2, 3, etcetera. So, we get a sequence of functions. The main idea of the proof is we prove that this sequence of functions converges uniformly to a function phi in the interval x 0 or x 0 plus h and then we show that that limit function is a solution to the initial value problem and by uniqueness theorem which we have already proved where we used a Lipschitz condition, the solution is unique, the limit function is a solution and the solution is unique. So, we divided the proof into four parts. The first part is we have shown that phi n defined by the iterative scheme is well defined and phi n is a sequence of functions having continuous derivatives and phi n x for every n is inside the rectangle r and in part b, we found that the sequence of function phi n that satisfies and estimate phi n x minus phi n minus 1 x the absolute value of which is less than or equal to m by alpha into alpha h to the power n by n factorial for n going from 1, 2, 3, etcetera. And this happens on the interval x 0, x 0 plus h. Now, today we will prove part c and part d. So, part c what we want to prove is as n goes to infinity, we will prove that the sequence of function phi n that converges uniformly to a function phi on the interval x 0, x 0 plus h and part d, we will show that the limit function phi which is a limit of the sequence of function phi n that is nothing but the solution of the given initial value problem on the interval x 0, x 0 plus h. So, let us start with the proof of c. So, proof part c. So, in c, we want to prove that phi n converges, the sequence phi n converges uniformly to some function phi on the interval x 0, x 0 plus h and from note that from part b, we got an estimate for phi n. From part b, we got the inequality. So, part b, we got the inequality phi n x minus phi n minus 1 x. The absolute value of this is less than or equal to m by alpha times alpha h to the power n by n factorial. Now, we consider the series of positive constants. So, the right hand side, if you look at the right hand side and make a series of positive constants by using the right hand side, the series of positive constants on the RHS. So, that is this series m by alpha, alpha h to the power n by n factorial as n goes from 1 to infinity. So, which is m by alpha times alpha h by 1 factorial plus alpha h square by 2 factorial plus etcetera. So, this converges to m by alpha times e to the power alpha h minus 1. See, in this series, if you add 1 and if you subtract 1. So, this summation is e to the power alpha h and subtract 1, you get m by alpha times e to the power alpha h. So, this converges the right hand side term as a series that converges to this quantity. Now, we will consider the infinite series. So, consider the infinite series summation n goes to 1 to infinity phi n x minus phi n minus 1 x. So, this series, we are discussing the convergence of the series and what is the limit of the series. So, each term of the series is bounded by a positive constant, which we got in, we proved in part b. So, now by Weissach's m test. So, each series, each term phi n x minus phi n minus 1 x, this is less than or equal to m by alpha, which we proved in part b to the power n by n factorial. And since the series formed by the right hand side m by alpha, alpha h to the power n by n factorial converges. We now invoke the Weissach's m test, which we discussed in the preliminaries by Weissach's m test. The series n is equal to 1 to infinity phi n x minus phi n minus 1 x. This converges and it converges uniformly on the interval, the interval which we are concerned about is x 0, x 0 plus h. It converges on this. Now, if this series, infinite series converges, what is the limit of it? To what it is converging? For that, let us consider the partial sequence of partial sum. So, consider the sequence of partial sum of the above series. So, call it s n. s n x is the n partial sum plus if you add y 0 to it. So, y 0 plus the partial sum, n goes from 1 to, say i goes to 1 to n, i goes from 1 to n, phi i x minus 1 x. So, just if you expand this plus terms and minus terms, they cancel each other and this becomes, by definition this is your phi n x, which is defined by the Picard's iterative scheme. So, phi n x is the partial sum of the infinite series and since the infinite series converges uniformly on this interval, say the partial sum x n x, which is phi n x that converges the sequence of partial sums. So, if you suppress x, this converges uniformly to a limit function, say phi on the interval x 0, x 0 plus h. Therefore, this implies that the sequence of functions phi n defined by the Picard's iterative scheme converges uniformly the desired interval x 0, x 0 plus h and also from part a, which we have proved, each phi n is continuous on x 0, x 0 plus h. So, therefore, the sequence of functions converges uniformly to phi and each phi n is continuous. So, therefore, we can invoke the theorem, which we discussed in the preliminaries to conclude that the limit function phi itself is continuous. So, from part a, each phi n is continuous on and hence the limit function phi itself is continuous x 0, x 0 plus h. So, in conclusion, so the four we have, so conclusion is the sequence phi n converges to phi on x 0, x 0 plus h and phi is an element of the set of four continuous functions defined on x 0, x 0 plus h. Now, we will prove the next section that is part d, proof of part d. There we will show that this limit function phi, which we just got as a limit of the sequence of functions defined by the Picard's iterative scheme, is a solution to the initial value problem. So, to prove the limit function phi satisfies the initial value problem. Since each phi n x satisfies the estimate phi n x minus y 0 is less than or equal to b on the interval x 0, x 0 plus h, which we have proved in part a. Since each phi n x is inside the rectangle r or r 1, so we get phi x minus y 0 is less than or equal to b on x 0, x 0 plus h. So, this sequence phi n converges uniformly to phi on the interval. So, the four, the limiting function phi x satisfies phi x minus y 0 is less than or equal to b on this interval. And we also have the convergence phi n to phi as a uniform convergence. So, this converges uniformly on the interval x 0, x 0 plus h. We will prove that the function f x phi n x, this converges uniformly to f x phi x. So, uniformly x 0, x 0 plus h. So, how this is done? So, we have already proved that phi n converges to phi uniformly on x 0, x 0 plus h. And by using that, and f is given to be continuous and f is having nice properties, lipschitz continuity and continuity with respect to x and lipschitz continuity with respect to phi. So, therefore, if we find f of x phi n x minus f of x phi x, which is a limit function. So, this is less than or equal to by using the lipschitz continuity of f with respect to the second argument. It is alpha times phi n x minus phi x. So, uniform convergence of phi implies, uniform convergence of phi n implies that for every epsilon greater than 0, there exists a delta, there exists a positive number n and that n depends upon epsilon only such that n positive such that phi n x minus phi x, this difference is less than epsilon for all n greater than n epsilon. So, you can take this epsilon by alpha also, another epsilon just to get epsilon at the end. So, therefore, we find the estimate f of x phi n x minus f of x phi x, which is less than or equal to alpha times phi n x minus phi x. And uniform convergence of phi n implies that whenever x is for all n small n larger than the capital n, this can be made less than epsilon by epsilon by alpha. So, this is less than or equal to alpha times epsilon by alpha, which is epsilon for all n greater than capital n, which is a function of epsilon. So, for a given epsilon greater than 0, we could prove that there exists an n such that f of x phi n minus f of x phi x, this difference, the absolute value of the difference is made less than epsilon for all n greater than n of epsilon. So, this shows that f x phi n x converges to f x phi x uniformly x 0 x 0 plus h. Now, since f of x phi n x is continuous for each n for each n. So, this is a continuous for each n on the interval x 0 x 0 plus h. The limit function f t phi t function f x phi x is also continuous x 0 x 0 plus h. The convergence is uniform and for each n, the sequence, each term of the sequence is continuous and the sequence converges uniformly. So, therefore, the limit function is also continuous follows from the theorem we discussed in the preliminaries. So, therefore, phi of x is equal to limit of n goes to infinity phi n x, which is equal to by definition phi n is y 0 plus limit n goes to infinity integral x 0 to x f of t phi n t d t. Now, we invoke theorem 3 that we did in the preliminaries on the interchange of limit and integration of sequence of functions. So, since the convergence is uniform and each term f of t phi n x phi n t is continuous, we can interchange this limit and the integration. So, this is equal to y 0 plus integral x 0 to take the limit inside. Limit n goes to infinity f of t phi n t d t.  So, therefore, this is equal to y 0 plus integral x 0 to x f of t phi t d t. So, your left hand side is phi of x. So, therefore, the limit function takes the form phi x is equal to y 0 plus integral x 0 to x f of t phi t d t. Now, if you recall, so, therefore, phi of x is equal to y 0 plus integral x 0 to x f of t phi of t d t. So, limit function of the sequence phi n which is the sequence obtained from the Picard's iterative scheme. Now, if you invoke from the basic lemma, basic lemma, any function satisfying this integral equation. So, any function satisfying the integral equation has to satisfy the initial value problem. The function satisfies initial value problem. So, therefore, the Picard's iterance converges uniformly to the solution of the initial value problem. And this solution is unique. The uniqueness follows from the uniqueness theorem which we proved. So, this proves the existence of a solution to the IVP. Now, uniqueness for uniqueness what we require is the Lipschitz continuity which is already assumed the theorem. So, now, the uniqueness of solution follows from the uniqueness theorem proved earlier. So, this completes the proof of Picard's existence and uniqueness theorem. And a few things to remark, although Lipschitz continuity was used in the above theorem to establish existence result, it is possible to establish existence theorem just by assuming only continuity assumptions on f. However, to establish uniqueness of solution, one need to use condition like Lipschitz continuity of f with respect to y or some other condition weaker or stronger existence. So, for existence of solution to the initial value problem, let us recall d y by d x is equal to f of x y, y at x 0 is y 0. For existence of solution to the initial value problem, only continuity condition only continuity condition on f is sufficient. Only continuity condition is necessary. Only continuity condition is, but for uniqueness we need stronger condition than continuity. Say for example, Lipschitz type for example, Lipschitz condition. And one can also use a weak version of Lipschitz type condition to ensure the uniqueness. So, if we, that is a remark 1. So, remark 2.  This is a Lipschitz condition, but this is replaced by times L n of 1 by y 1 minus y 2. This is a weaker condition than the Lipschitz condition for all x y 1 and x y 2 on the domain or the rectangle. A weaker version of Lipschitz condition is sufficient to ensure uniqueness, ensure uniqueness of solution to the initial value problem. But still, continuity condition is enough to improve the existence. Now, we look into the Piano, Cauchy Piano theorem on existence of solution that requires only continuity condition on the function f. So, now we look into the Cauchy Piano, Cauchy Piano existence theorem. Now, we look into we state and prove Cauchy Piano existence theorem for the initial value problem where the function f is continuous on a domain D, continuous on a domain D in R 2, but not Lipschitz continuous with respect to the second argument that is y. To prove the existence theorem, we first define what is called an epsilon approximate solution for the initial value problem. And just by using continuity on f, subsequently we define a sequence of approximate solutions. We define a sequence of approximate solutions for the initial value problem and we show that the sequence of epsilon approximate solution that is uniformly bounded and equicontinuous. So, once we have a uniformly bounded and equicontinuous sequence of functions, we make use of Arzela Ascoli theorem to extract a subsequence of the sequence that converges uniformly to f function. And later, we prove that that limit function is a solution to the initial value problem. That is all idea of Cauchy Piano existence theorem. Cauchy Piano existence theorem, we first define what is known as an epsilon approximate solution to the initial value problem. So, definition is called an epsilon approximate solution. So, consider the initial value problem which is d y by d x equal to f of x y with initial condition y at x 0 is y 0. This is the initial value problem where the function f x y is a real function, real valued function defined on a domain called as d and epsilon approximate solution of an epsilon approximate solution of the initial value problem I v p on an interval called it I which is a set of 4 x is that absolute value of x minus x 0 is less than equal to a. So, is a function the approximate solution epsilon approximate solution is a function call it phi defined on I such that the following properties are satisfied. First x for every x in the interval I is in the given domain d x on I. Second property, so phi is c 1 class this on I this is one time continuously differentiable phi is phi has continuous first derivative except possibly for a finite set call it s of points on I where the derivative phi prime may have finite discontinuity or symbol discontinuity have symbol discontinuity means a kind of jump discontinuity. And third condition for approximate solution is the difference between phi prime minus f of x phi x phi prime minus f of x phi x is less than epsilon for all of the x in the interval I except for the points on this finite set s. So, for a given initial value problem initial value problem we define an epsilon approximate solution. So, phi is said to be a function phi is said to be a function phi on an interval I is a function given by x minus x is less than or equal to a is said to be an approximate solution epsilon approximate solution. If for this properties x phi x is in the domain and phi is continuously differentiable except on a set of finite number of points. And the difference between phi prime and f of x phi x this is a difference there are this is made smaller than epsilon on the interval I minus s. So, what we do now is we will prove that under continuity assumption on f just by continuity assumption on f with respect to both the terms of x and y there exist epsilon approximate solution to this initial value problem. So, we prove the following theorem and one note is so the solution phi or any function phi which is an element of c i continuous functions defined on the interval I satisfying the satisfying the second property which we just have seen the property to this property that is phi is in c 1 on I except possibly for a finite set s of points on I where phi prime may have symbol discontinuity. So, if any function satisfying the second property phi is said to be is said to have piece wise continuous set to have piece wise continuous derivative on the interval I and is denoted by and is denoted by phi is in the set the class. C 1 p I a class of functions having piece wise continuous derivatives. So, we state the theorem so theorem 1. So, this theorem in this theorem we prove that under continuity assumptions on f with respect to x and y there exist epsilon approximate solutions for the function to the initial value problem. Consider the initial value problem d y by d x is equal to f of x y with initial condition y at x 0 is y 0. So, initial value problem suppose that f of x y is continuous on the rectangle r given by r is set of 4 point x y such that x minus x minus x 0 is less than or equal to a and y minus y 0 is less than or equal to b and let m be a constant m is equal to maximum of the function f of x y maximum of the function m and x y lies on r and h is a constant defined by h is a constant defined by minimum of a b by m. So, then the conclusion of the theorem is then given epsilon greater than 0 then given epsilon greater than 0 there exist epsilon approximate epsilon approximate solution to the initial value problem the interval x minus x 0 less than or equal to h. Theorem does not say anything about the uniqueness theorem says that there exist an epsilon approximate solution to the initial value problem under the continuity assumption on f. And then the solution is the constant m since f is continuous on the rectangle is a closed set inside r 2. So, therefore bounded and closed a compact set and the maximum is attained this m is defined and h is the minimum of a and b by m depends upon the value of m. So, we will prove this theorem the whole idea of this theorem the proof of the theorem proof. So, we take let epsilon greater than 0 be given then. So, this is x y plane and this is a rectangle and say this is a point x 0 y 0 x 0 y 0 and say this line is x is equal to x 0 x 0 x 0 x 0 x 0 x 0 plus a if h is smaller than a this is x is equal to x 0 plus h and we divide the interval x 0 to x 0 plus h into n n parts. So, the first point is x 0 x 0 y 0 then x 1 plus x 2 x 3 etcetera. Then we define approximate solution starting from x 0 y 0. We will approximate the solution by straight lines the idea is starting from x 0 at x 0 I know what is a slope of the solution. Slope of the solution is f of x 0 y 0 I make a straight line then from that straight line meets a line x is equal to x 1 at some point from there I again find the slope and I make line segments and join the line segments to get a polygon and that polygon is an approximate solution as the mesh the difference of x i and x i plus 1 is very small then the difference of the actual solution and the approximate solution can be made small. We will do the details of the proof in the next lecture. Thank you.


Oh, yeah, what's up guys, this is Chris Pipe, my friends call me Big C. I'm back again guys. Today we got some more featured on freedom. That's right guys. We try and select freedom partners from the comment section to feature on freedom and we get you more views, we try and get you more subscribers. Send some people your way to check out your channel. Now, as I say in all my videos guys, the only way to get in on this is to submit your video links in the comment section below. When you submit a link it goes in and it waits, it gets pending approval. I go in, I approve your comment and then I look at all the videos and then I select one or two or three or four and then I feature you on this channel here. So that's how it is guys, submit your videos. If you've submitted videos in the past and you haven't been selected yet, feel free to resubmit them. Not a big deal guys, I got lots to go through. With that out of the way, we've got three partners today that we're gonna feature. So let's just skip the four-point and get right into this. The first guy is Boba Vaughn. Let's talk about Boba Vaughn. All right guys, Boba Vaughn. This guy does some really awesome stuff on his channel. His edits are funny. He has a Witcher 3 montage. The game's only been out for like five or six days and he's already got a full montage out and it's hilarious and it's well edited. He's got great song choices and just a good time guys. So let's watch, let's cue the clip and then we'll talk a little bit more about Boba Vaughn. This is the Witcher 3 Wild Hunt funny moments. Enjoy. So there's a monster hunting Noel is there. Looks like we're gonna have to sort him in. Roachy baby, where are you? Come on Roachy baby. Dude, has been for a good 20 years. She's so old, dude. Let it be, honey. Has been for a good 20 years. It's awful frightening, awful. I meant what does it look like? Can we just go in? Oh, sh**. Go on John. I don't mean to do that. What? Roda. Holy f**k. How am I talking underwater? Drink some oceans. Back one man. Hey guys, I don't know man. That was pretty funny to me. If you, I recommend you go to his channel, check it out. He's doing a Witcher playthrough I believe. So it's probably worth it if you're on the fence about getting that game. A, you should probably get it and B, you should check out his channel. It's really good. All right guys, the second guy is New Spirit. Yes, New Spirit does the Grand Theft Auto V funny montage and he also edits like a champion. I can't, these two guys, they both send me messages and they're just awesome. It's like they went to the same editing school but hilarious stuff guys. He does a GTA V montage, like I just said. All funny parts. Definitely check out his channel. He also plays some weird dating game called Kitty Powers. Yeah, yeah, let's check. Kitty Powers. Hold on, before I talk more about New Spirit, let's watch some Kitty Powers. Ooh, hello kittens. I'm Kitty Powers, matchmaker extraordinaire. As we all know, the past of true love is long and hard. Which is why I've created my new dating simulator. Every client deserves their happy ending and it's your job to give them one. Yeah guys, he plays Kitty Powers but I guess, hey man, that looks like a fun game. But joking aside guys, he has a great channel. He's a Grand Theft Auto V montage, lots of other things there. Click on his link below.  You should do that, it's worth it. It's stuff you've seen today. He's a finger style guitarist and I figured A, his stuff is awesome.  How cool is that? Like a champ and guys, he can really pick guitar. In fact, he plays a lot of the more recent songs that have come out. He plays covers where he creates his own arrangement with his own finger style and what else can I say? He's awesome, if you like guitar, if you like acoustic guitar, this guy's channel is the one to check. He's got 3,600 some odd subscribers already. He's growing fast. Check it out. I've left him for last because I want you guys to see the whole song that he played, one of his songs and this is Hozier's Take Me to Church. All right guys, big C, check it out. I'll be back more with at least one or two more videos. Kling!


Shoot down the temperature like so keep your distance. Yes, sir Let's stay focused watch your sectors There's the communications outpost Could be the missing troopers, let's check it out Let's go six All right noble team spread out watch the approach Looks clear from this angle Distress beacons coming from just south of your commander. We're close Eyes peeled Round the beacon make out any ID negative, but it's military Why are we not seeing explosives residue noble three can you confirm any EX residue in the area? Plasma maybe there's a lot of blood on the ground All right, noble looks like there's nothing here. Let's move on Smoke at the next structure noble team you have permission to engage but be selective. We don't need a telegraph or presence Noble six moving to the house. I'm right behind you. I'm seeing heat seeks in the structure ahead They're not rebels their farmers look at them ask them what they're doing here mid-crash the kids But never cut up in Holly hiding sir. Take enough idea. I sombsidot. He's a child. I got She called his ass neighbors were attacked last night He had screens gunfire Stop around sunrise Says something in the fields Killed his son some thing Copy that get them back inside Osmont on Bethlehem getting noble team double-timing We've got military casualties two of the missing troopers It looks like they were interrogated It's messy see anything Noble to move up to the West we're about to be flanked Contact contact Here we go