 Welcome back. We are trying to prove the Cauchy-Piano existence theorem, where we use only the continuity assumption on the function f. To prove the piano theorem, first we prove the existence of an approximate solution to the initial value problem. Let us recall the theorem we have stated, that is theorem 1. Consider an initial value problem d over dx with initial condition d over dx is f of x y with initial condition y at x 0 is y 0, where the function f is continuous on a rectangle defined by x minus x 0 is less than equal to a and y minus y 0 is less than equal to b. And we take some constants m is the maximum value of f in the rectangle and h is minimum of a and b by m. Then the conclusion of the theorem is that there exists for a given epsilon greater than 0, there exists an epsilon approximate solution to the initial value problem on the interval x minus x 0 less than equal to h. So, we will prove this theorem now. So, the proof is that for a given epsilon. So, let epsilon greater than 0 be given and let us consider the rectangle r. So, this is a rectangle r we defined, where this side is y minus b y 0 minus b and this side is y 0 plus b. This side of the rectangle is x 0 plus a and this side of the rectangle is x 0 minus a. And we are looking for a solution approximate solution on the interval x 0 to x 0 plus h. The given initial point is here x 0 y 0 is a given initial point. And now, we make polygons and join them to get an approximate shape of the solution curve. And approximate solution will be constructed and approximate solution. So, by approximate means epsilon approximate epsilon is given. Solution will be constructed for the interval x 0 x 0 plus h the right hand side to the given point x 0 and a similar construction will define it for the left hand side left interval that is for x 0 minus h to x 0. We will prove the epsilon approximate solution to this interval x 0 x 0 plus h. The approximate solution will consist of the epsilon approximate solution the solution will consist of. So, polygonal path starting from x 0 y 0 starting from the given point x 0 y 0 the initial point. So, it start from here this point x 0 y 0 there is a finite number of a finite number of straight line segments joined end to end to form the polygonal path that is a graph of the approximate solution. So, what we will do is since f of x y is uniformly continuous on r on the rectangle r rectangle r is a closed and bounded set in r 2 and f is continuous that gives a f is a uniformly continuous function on r. So, therefore, this implies that for every epsilon. So, we start up with an epsilon for that epsilon greater than 0 there exist a delta of course, the delta will be a function of epsilon such that f of x y minus f of say x tilde y tilde and this difference is less than epsilon whenever x minus x tilde is less than delta and y minus y tilde is less than delta. So, this follows from the continuity in particular uniform continuity f on r. Now, what we do is now we divide the interval divide the interval on which we are looking for a solution x 0 x 0 plus h this interval we divide into n parts into n parts say x 0 is a first point then write it x 1 less than x 2 less than x 3 etcetera x n that is the last point which is x 0 plus h and make the maximum of x k minus x k minus 1 for any k which is less than or equal to minimum of delta which is delta we define just above delta epsilon and delta epsilon y m. So, the procedure is as follows. So, from x 0 y 0 the given initial point construct a straight line a straight line with slope f of x 0 y 0 which the slope we get from the differential equation we are starting from the point x 0 y 0 and we know the slope of the solution at the point x 0 y 0 is f of x 0 y 0 from the differential equation as d y by d x is f of x y. So, you construct a straight line with a slope f of x 0 y 0 proceeding to the right of x 0 to the right of x 0 until it intersect the line x is equal to x 1. So, what we do is we start from the point x 0 y 0 and with slope f of x 0 y 0 draw a straight line let it touch the line x is equal to x 1 at this point. So, we got a line segment having slope f of x 0 y 0. So, let the line this line intersect the line x is equal to x 1 at some point say at the point x 1 y 1 x 1 is the point right to x 0 and y 1 is the point at which the straight line hits the line x is equal to x 1. Now, this line segment the line segment lies inside the triangular portion inside the triangular region bounded by the lines by the lines starting from x 0 y 0 with the slopes m is equal to a slope m and minus 1. See this line segment which we just drawn from x 0 towards the right of x 0 that is going up to x 1 and this line segment is inside the triangular portion this is a triangular portion triangular region. This is bounded by the straight line having slope minus m and another straight line having slope m and another straight line having slope minus m and also by the line x is equal to x 0 plus this line this line is x 0 plus h. So, it is a triangle formed by three lines and this approximate solution portion of the approximate solution starting from x 0 y 0 lies inside this triangular portion and why what is the guarantee that this line segment will lie inside the triangular portion. So, this follows. So, from the definition of h which is minimum of a and b by m and the fact the bound for f of x is h plus m f is bounded by m. Now, we define solutions on each of these sub intervals. Now, we define solutions we now define approximate solution each sub interval in the same way. So, in the first interval that is in the interval x 0 x 1 is the first interval there the solution is y of x is equal to starting from y 0 plus slope slope is f of x 0 y 0 and this is up to the and x varies up to x 1 this is from x minus x 0 this is a straight line equation of a straight line having slope f of x 0 y 0 and starting from the point x 0 y 0 this solution on the interval x 0 x 1 and similarly, so on the interval x 1 x 2 the solution is y of x is equal to and there we start from the last point of the previous interval which is y at x 1 which we obtained from the first a solution of the first interval y at x 1 plus the next slope is f of x 1 and the point there is y of x 1. So, this is a slope and x minus x 1. So, this is a solution defined on it is a straight line starting from the last point x 1 x 1 y 1 that is x 1 y at x 1 and similarly, if we keep on doing on the interval x k minus 1 x k on this interval. So, y x is given by the solution given by y of x k minus 1 plus the slope at y k minus 1 f of y x k minus 1 into x minus x k x k minus 1. So, this in general is a solution on any kth interval of 2 points x k minus 1 x k where k is equal to 1 2 3 etcetera. So, if we join all these solutions these are the solutions straight lines we obtained on each of the interval and if we join them and we get a function which is having a piecewise derivative piecewise continuous derivative. So, the function y obtained on the interval x 0 x 0 plus h is a function having piecewise continuous derivative. So, that is y is an element of c 1 p x 0 x 0 plus h. So, in the property of approximate solution we want the solution to be in c 1 p on those points at which the derivative are defined. Let us consider, so for any x and x tilde let us take 2 points x and x tilde in x 0 x 0 plus h. To study further properties of y let us take 2 points x tilde and x tilde x and x tilde in the interval x 0 and x 0 plus h. Without loss of generality let us assume that x c x tilde is an element in one of the sub intervals. So, it is in x i minus 1 x i and x is in x j minus 1 x j and assume that this is x i minus 1 this is x i. So, this is x j minus 1 this is x j and we take x from here. So, x tilde is somewhere in the interval between x i minus 1 and x i and x is from the interval x j minus 1 x j. So, consider the difference y of x minus y of x tilde y of x minus y of x tilde. This we can write this as y of x minus y of x j minus 1 write it to fresh y of x minus y of x tilde is equal to y of x minus y of x j minus 1 plus y of x j minus 1 adding and subtracting j minus 1 minus y of i plus y of x i minus let us consider y of x minus y of x tilde is equal to y of x minus y of x j minus 1 plus y of x j minus 1 minus y of x i plus y of x i minus y of x tilde if you group them. So, this is less than or equal to y of x minus y of j minus 1. So, remember y x j minus 1 is less than x plus x tilde plus y of x i minus y of x tilde and if you look at the properties of the solution the function y because y is a solution and y is a solution satisfied by the formula given that is the equation of a straight line. So, this is equal to less than or equal to m times x minus x j minus 1 plus x tilde. So, remember the form for y x is equal to y of x k minus 1 or in this case j minus 1 plus f of x j minus 1 y x j minus 1 y of x is y of x j minus 1 plus f of slope x j minus 1 y x j minus 1 into x minus x j minus 1 and we know that by definition this quantity. So, there is if you take the difference y x minus y x j minus 1 is less than or equal to the slope x j minus 1 y x j minus 1 into x minus x j minus 1 and this quantity is bounded by this is bounded by m. So, therefore, we get the first time is less than or equal to m times x minus x j minus 1 and similarly the other times are coming as a solution from the each interval we can compute it to get this is also m times x i minus x tilde. Now, use of fact that this x 1, x 2, x 3 they are all in the increasing order. So, therefore, this absolute value we can get rid of that this is less than or equal to m times x minus x j minus 1 plus x j minus 1 minus etcetera minus plus x j x i minus x tilde and we will be able to cancel out these terms to obtain m times absolute value of x minus x tilde. So, what we have shown is the absolute value of y of x minus y of x tilde where x and x tilde are from anywhere in the interval x 0, x 0 plus h is less than or equal to a constant m times x minus x tilde. So, in particular now, so you call this inequality as star we will be using it again. So, in particular let x tilde is equal to x 0 the first initial point x 0 then what we have is y of x minus y of x 0 which is less than or equal to m times x minus x 0. So, that is absolute value of y of x minus y of x 0 is y 0 is less than or equal to m times x minus x 0 is bounded by h and which is again bounded by b by definition of m by definition of h this is bounded by m. So, therefore, what is the conclusion the conclusion is. So, we obtained y x for any x minus y 0 is less than or equal to b. So, therefore, this implies that x y of x this is a point inside r for x in the interval x 0, x 0 plus h. So, hence f of x y of x is defined y of x is a approximate solution we defined we constructed and f of x y x is defined for all x for x in the interval x 0, x 0 plus h. Now, we consider an x in the interval say x k minus 1 x k a particular interval. So, we get also we know that in this interval the solution y x is given y. So, we have this interval the solution is y x is equal to y of x k minus 1 plus f of x k minus 1 the slope f of x k minus 1 times x minus x k minus 1. This is a equation of the straight line or the solution defined in the interval x k minus 1 x k. Obviously, f prime x is a slope is f of x k minus 1 y of x k minus 1 which we will use later. So, let us see what is y of x minus y of x k minus 1 here y of x the solution minus y of x k minus 1 the solution at the starting point. So, this is less than or equal to if you make use of this equation y of x minus y of x k minus 1 is bounded by the bound of f which is bounded by m. So, this is m times x minus x k minus 1 which is of course, less than by definition of our partition m into delta y m which is delta. So, that delta is a function of epsilon. Now, if we take the difference of the derivative y prime x minus f of x y x. So, in this interval if you find the difference of this solution the derivative of the solution with the right hand side of the differential equation y prime x minus f of x y x. In fact, this is going to be the difference between the solution. So, which is equal to we have seen that y prime x is f of x k f of x k y x k minus 1. So, this is f of x k minus 1 y x k minus 1 minus f of x y of x and from the uniform continuity of f we have seen that. So, this is less than or equal to epsilon by the uniform of f. So, whenever x minus x k k minus 1 and x the difference between x k minus 1 x is less than delta and y x k minus 1 and y x less than delta this difference is less than epsilon by the uniform continuity of f. So, what does it say? So, this implies that if y now satisfies all the properties of an epsilon approximate solution. So, this implies that y x. So, y x is an epsilon approximate solution. So, y is in c 1 p and y prime minus y this is less than epsilon for all points at which the derivatives are defined. So, that is not satisfied only at a finite number of points at which the derivatives having gem type discontinuity. Now, with these background we are going to prove the celebrated theorem of Cauchy and Piano. So, Cauchy Piano theorem on the existence of solution Cauchy Piano existence theorem. So, in the previous theorem under the same conditions of the previous theorem we are going to show that the initial value problem is going to have a solution. So, the statement of the theorem is let f x y be continuous on the rectangle r which is defined as set of whole x y such that x minus x 0 is less than or equal to a y minus y 0 is less than or equal to b. Then there exists a solution to the initial value problem d y by d x is equal to f of x y y at x 0 is y 0. In the interval x minus x 0 is less than or equal to h where h is minimum of a b by m with the m being m is equal to the maximum of the function f x y where x y is in the rectangles. So, for all x y rectangle if m is a bound is a maximum of it then there exists a solution in the interval starting from the point x 0 y 0. So, let us quickly see the proof of it by using the ideas from the previous proof, previous theorem proof. Now, with epsilon n to be 1 by n where n is equal to 1, 2, 3, etcetera these small sequence of positive numbers. Now, for each epsilon n for each epsilon n there exists epsilon approximate solution called it y n x to the initial value problem. So, this is true because of our previous theorem for every epsilon n there exists an epsilon n approximate solution which we denote by y n to the initial value problem. Now, by theorem the previous theorem 1, the theorem we have shown that y x and y n x the approximate solution y n x minus y 0 is less than or equal to b or y n x is bounded by y 0 plus b. So, right hand side is independent of n. So, therefore the family of solution. So, therefore this implies that y n x the solution this sequence of solution is uniformly bounded. So, this is a uniformly bounded sequence of functions again from the previous theorem from the previous theorem. In fact, from the inequality from star if you look into the previous theorem we have shown that this is star this inequality. So, from star that y minus y x minus y x tilde is less than or equal to m x minus x tilde for all x and x tilde in the interval x 0, x 0 plus h. So, therefore again from star we have that y n x minus y n x tilde is less than or equal to m times x minus x tilde where x and x tilde are numbers from x 0, x 0 plus h. So, what does it say? It says that. So, this implies that the sequence y n is an equicontinuous. So, equicontinuous functions. So, therefore the approximate solutions epsilon n approximate solutions which are denoted by y n's. So, y n's are uniformly bounded and equicontinuous. Now, we invoke Arcella-Oscoli theorem. We here make use of Arcella-Oscoli theorem. So, thus by Arcella-Oscoli theorem there exist a subsequence of y n, a subsequence y n say y n k. So, there exist a subsequence of y n k such that y n k converges to some function y uniformly x 0, x 0 plus h. So, invoking Arcella-Oscoli theorem we get a sequence, a subsequence of the constructed sequence y n that converges uniformly to some function y in the interval x 0, x 0 plus h. So, this implies that y is continuous. So, again by using the theorem we discussed in the preliminaries a sequence of function converges uniformly and each y n k is continuous, then the limit function is also continuous. And y of x minus y of x tilde is less than or equal to m of m times x minus x tilde. Now, our task is to prove that this limit function y is a solution to the initial value problem. So, we now prove that the limit function y satisfies the initial value problem. So, for that consider the error term. So, consider the term denoted by delta n k of x which is equal to y n k which is a subsequence we extracted from the sequence prime minus f of x y n k x. So, this difference is this error if y prime n k exist, otherwise we take this 0, otherwise that is otherwise case is only for finitely many the many points. So, in other words this is a y n k y n k prime is equal to f of x y n k x plus delta n k x. Now, integrating over x 0 x we get. So, integrating we get y n k x k x is equal to y 0 plus integral x 0 to x f of t y n k t plus delta n k t d t. And we know that y n k converges to y uniformly. So, this implies that as f is continuous f of t y n k t converges f of t y t uniformly. So, uniformly as f is continuous f is uniformly continuous. So, this implies that the integral x 0 to x f of t y n k t d t converges to integral x 0 to x f of t y t d t. Further we have delta n k which is by definition is less than epsilon n which is 1 by n and passing to the limit this tends to 0. So, we prove that we prove that the limit of star is y of x is equal to y 0 plus integral x 0 to x f of t y of t d t. And obviously, y at x 0 is y 0 and by basic lemma y has to satisfy. So, by basic lemma y satisfies the initial value problem. So, therefore, the limit of this convergence of sequence y n k is a solution to the initial value problem. So, this proves that solution exist thus the limit function y n is a solution to the initial value problem. So, in this lecture we have proved that if we have an initial value problem d over d x is equal f of x y and f is continuous with respect to x and y then there exist a solution to the initial value problem. The uniqueness is not guaranteed.