 Welcome back to the existence and uniqueness theorem of initial value problems. In the last lecture, we have seen that an initial value problem has a unique solution if the function satisfies a Lipschitz continuity condition with respect to the dependent variable y. So, we deal with the initial value problem. So, our initial value problem d over d x is equal to f of x y and the initial condition is y at x 0 is y 0. So, where f is a function which is defined on a domain which is a subset of r 2 to r and x 0 y 0 is a point an interior point inside d is d and x 0 this is point y 0. So, x 0 y 0 is the initial point initial condition and we now prove the existence of solution and uniqueness of solution under Lipschitz condition with respect to y and continuity condition with respect to x. So, Picard's theorem gives both existence and uniqueness and uniqueness part we have already proved and in this theorem I will concentrate only on the existence part. So, let me state the Picard's theorem d be a domain in r 2 f from d to r be a real valued function real function satisfying the following conditions satisfying the following conditions. First one is f is continuous f is continuous on d with respect to both the argument that is meaning of the all domain d and f of x y is Lipschitz continuous with respect to y on d with the constant with the Lipschitz constant alpha which is a constant positive constant and let x 0 y 0 be an interior point an interior point on d and let a and b be constants such that the rectangle are defined by set of all x y such that x minus x 0 is less than or equal to a and y minus y 0 is less than or equal to b this rectangle is inside the domain d and let m be a constant capital m be a constant defined by maximum of the function f x y where x y is in d this maximum exists because f is continuous. So, and we take x y to vary in the rectangle r rectangle is a closed set in inside d. So, therefore, this maximum exists and let us define another constant h is equal to minimum of a and b y m then the initial value problem has a unique solution y the interval x minus x 0 is less than or equal to h. So, this is a Picard's accessions and uniqueness theorem it gives both accessions and uniqueness the conditions we assume is f is a continuous on d and f x y is a Lipschitz with respect to y on d with a Lipschitz constant alpha and the initial point x 0 y 0 is an interior point on d and we take a rectangle which is inside the domain d is a closed rectangle inside the domain such that two constants are defined m is the maximum of the value function in the rectangle r and h is a minimum of a and b y m. So, if I look at so this is domain d and inside the domain we define a rectangle. So, this point is x 0 y 0 such the point x 0 y 0 and this is this side is x is equal to x 0 plus a and this side x is equal to x 0 minus a and this side is y is equal to y 0 plus b and this side y is equal to y 0 minus b. So, we now prove the theorem proof. So, the proof has more technical details. So, since r is a closed rectangle r is a closed rectangle. So, inside the domain d f satisfies all properties mentioned inside d inside r. Now, there are two situations two situations one is if a is less than b by m. So, remember our definition of the constant h which is a minimum of a b by a. So, h is defined as minimum of a b by m. So, in case if a is less than b by m then h is equal to a and if in that case we have a full rectangle the same rectangle h is equal to a and if a is not if a b by m if b by m is less than a then h is equal to b by m that case h is a number which is smaller than a. So, that case we will have another rectangle. So, we will have another rectangle. So, this line is x is equal to x 0 minus h. So, this line is x is equal to x 0 plus h. So, we have two rectangles one this you call it r 1 and the first rectangle is r or we can write r is set of all x y such that x minus x 0 is less than equal to a y minus y 0 is less than equal to b and r 1 is set of all x y x minus x 0 is less than equal to h y minus y 0 the bound for y is the same b. So, if a is less than b by m then these two rectangles coincide in this case r 1 is same as r if a b by m if a is greater than b by m in this case we will see that r 1 comes inside r. So, because existence and uniqueness theorem says that it has a solution the solution starting from x 0 y 0 is a solution and the solution exists in the interval x minus x 0 solution exists in the interval x minus x 0 less than equal to h. So, solution existence solution is on r 1. So, depending upon the value of a if a is less than b by m then this the solution exists for the larger rectangle if a is greater than b by m then the solution exists on a smaller rectangle r 1. So, we prove the theorem by the method of successive approximation we prove the theorem by successive approximation of the Picard's successive approximation of the Picard's iterands denoted by phi 1 x phi 2 x phi 3 x etcetera this iterates this Picard's iterands are defined on the interval x minus x 0 less than equal to h is defined on x minus x 0 is less than equal to h and are defined by phi 1 x. So, phi 1 is equal to y 0 plus integral x 0 to x f of t y 0 which is the initial condition that y 0 d t is a constant function d t and phi 2 x is y 0 plus integral x 0 to x f of t phi 1 t d t etcetera and phi n t. So, phi n of x is y 0 plus integral x 0 to x f of t phi n minus 1 t d t. So, called this as equation 1. So, we prove the existence of solution solution to the initial value problem on one side of the interval x 0 x 0 plus h. We will prove the existence on one side of the point x 0 to x 0 plus h similar arguments similar arguments hold for x 0 minus h to x 0. So, therefore, we prove only on one side and uniqueness is already been proved. Uniqueness of solution follows from Euler's theorem theorem which we proved in the previous lecture. Now, we divide the proof we divide the proof into four parts. So, part a part b part c and part d. So, part a is in part a we will prove that the functions the functions phi n is the sequence of function phi n defined by equation 1 is well defined that is and it is also continuous and it is called it is a part is well defined and b is phi n of phi n's have continuous derivatives continuous derivatives c it obeys an estimate phi n x minus y 0 the difference between the initial point and all and phi n x. This is less than or equal to b means all phi n's are within the rectangle b or x 0 x 0 plus h the interval on which we are proving the existence d when we evaluate the function f at this phi n's phi n x. So, this is well defined. So, this is part a in part a we will prove that the Picard's iterands defined by equation 1. So, defined by equation 1 this phi n x this equation this is well defined and phi n's have continuous derivatives on the interval x 0 x 0 plus h and phi n obeys an estimate phi n x minus y 0 is less than equal to b on this interval and f of x phi n x is also defined that is part a we will prove it and part b. So, part b is the functions phi n x the sequence of functions we defined by 1 satisfy the following inequality following estimate that is absolute value of phi n x minus phi n minus 1 x is bounded by m by alpha m is the maximum value of the function m is the maximum value of the function on the rectangle and alpha is ellipses constant times alpha h to the power n divided by n factorial. This happens for all x in the interval x 0 x 0 plus h. So, the Picard's iterands defines a sequence of function that sequence of function satisfies this estimate. Now, part 3 that is a part c is we will prove that as n goes to infinity as n goes to infinity the sequence of functions phi n this converges uniformly. So, this sequence of functions converges uniformly to a continuous function call it phi a continuous function phi on the interval x 0 x 0 plus h. Now, the fourth part part b of the proof is that we will prove that the limit function which is obtained in part c phi satisfies the limit function phi satisfies the given initial value problem given on the interval x 0 x 0 plus h. So, in short if you prove all these four parts part a b c and d then we obtain a limit function phi which happens to be the solution of the initial value problem that proves the existence of solution to the initial value problem and uniqueness is already been proved in the uniqueness theorem. So, let us prove part by part. So, proof of part a. So, this we will we prove this by mathematical induction prove this by mathematical induction as we will assume that the result is true for n minus 1 and then we will prove that it is also true for n then we will check this is correct for n is equal to 1 and then by the method of mathematical induction we conclude that this is true for all n. So, assume that phi n minus 1 x exist. So, phi n minus 1 x exist and it has continuous derivatives on the interval x 0 x 0 plus h and it satisfies the estimate and it satisfies phi n minus 1 x minus y 0 is less than or equal to b for all x in the interval x 0 x 0 plus h. Now, after assuming this we are going to show that this is these properties are also true for phi n if you assume that these properties are true for phi n minus 1 we are going to show that this property is true for n phi n. So, this implies. So, this above conditions implies this implies that x phi n minus 1 x the point x phi n minus 1 x is in the rectangle r 1 because you have this bound phi phi n minus 1 x minus y 0 is less than or equal to b. So, that shows that it and that x is in the x is in the interval x 0 x 0 plus h. So, therefore, the point x phi n minus 1 x is in r 1. Now, also if this point is in r 1 we have we can evaluate the function and we can evaluate this function at this point. So, f of x phi n minus 1 x is defined and phi n minus 1 is continuous and f itself is continuous is defined and is continuous with respect to x on x 0 x 0 plus h. So, further we have further if we evaluate the function f x phi n minus 1 since that is in the rectangle r 1 this can be evaluated for the function f because f is defined on r 1 and this is going to and we have seen this will have a bound less than or equal to m by hypothesis. So, maximum value of f on r is m that is the maximum value of it. So, this all x 0 x 0 plus h and all this discussion helps us to look into the integral. So, consider the function phi n x which is defined as y 0 plus integral x 0 to x f of t phi n minus 1 t d t. Now, we have seen that the function f is this function f is continuous with respect to x and it is well defined and therefore, the properties mentioned above helps us to conclude that. So, phi n x exist. So, y 0 plus integral x 0 to x and for the continuous function f this integral exist and has continuous derivative since f is continuous with respect to the second argument if it is continuous with respect to the second argument. So, therefore, this integral exist and that can be differentiated to get a continuous derivative on the interval x 0 x 0 plus h. Now, also consider also find the estimate phi n x minus y 0 the absolute value of phi n x minus y 0 by definition this is absolute value of integral x 0 to x f of t phi n t d t and this is less than or equal to integral x 0 to x absolute value of phi t phi n minus 1 phi n minus 1 above phi n minus 1 t d t and we have seen that f of t phi n minus 1 t is bounded by a constant m. So, therefore, this is less than or equal to integral x 0 to x m d t which is integral x 0 to x can straight away integrate this one and get m into x minus x 0 and x minus x 0 is less than or equal to h. So, this is m of less than or equal to h and m h is obviously, less than or equal to b by definition of h. So, h is minimum of a b by a. So, this all arguments helps us to conclude that x phi n x this point this lies in the rectangle. So, x phi n x lies in the rectangle r 1 and hence f at a f evaluated at x phi n x is defined since f is continuous f is continuous phi n is continuous and this f of x phi n x is continuous is defined and continuous on the interval x 0 x 0 plus h. So, what is it says it says that it says that all properties. So, says that this phi n is also satisfying all the properties we assumed in part a. Now, with assumption that if phi n minus 1 satisfies the properties then phi n also satisfies the properties. Now, let us check for the case when n is equal to 1. So, when n is equal to 1 case. So, we have phi 1 x which is equal to by definition y 0 plus integral x 0 to x f of t y 0 d t. So, obviously, phi 1 is well defined obviously, phi 1 is defined and f is continuous and y 0 is a constant function which is continuous. So, therefore, f is continuous and therefore, this phi 1 is defined and has continuous derivative on x 0 x 0 plus h and also if you find the estimate. So, also the bound phi 1 x minus y 0 by definition this is less than equal to integral x 0 to x f of t y 0 d t and f is bounded by m. So, this is less than equal to m into x minus x 0 which is less than equal to m h and which is bounded by b. So, obviously, phi 1 also satisfies these properties. So, therefore, this implies that x phi 1 x is in r 1 and hence f of x phi 1 x is continuous x 0 x 0 plus h. So, therefore, this implies that the properties are true for n is equal to 1. So, what we have proved in part a is the properties of phi n like they are all defined and phi n are having continuous derivatives and when phi n is put into f that is also continuous and phi n x phi n are in the rectangle these all properties are true for n is equal to n minus 1 that is what we assumed and from that we have proved that it is also true for the case n and it is also true for n is equal to 1 thus by the method of mathematical induction. So, by the method of mathematical induction phi n. So, phi n of that is a sequence of sequence of induction phi n sequence of functions defined in one by the Picard's iterands that possesses all desired properties or desired properties in the interval x 0 x 0 plus h. So, hence part a. So, part a of the proof is established by mathematical induction. Now, we look into part b proof part b what we are looking for we are looking for an example for the Picard's iterands. So, we again we prove this again by mathematical induction we prove this also by mathematical induction so assume that is true for n minus 1 assume that the estimate is true for n minus 1 that absolute value of phi n minus 1 x minus phi n minus 2 x is less than or equal to m alpha to the power n minus 2 divided by n minus 1 factorial into x minus x 0 to the power n minus 1 for x in the interval x 0 x 0 x 0 plus h. We assume that the estimate is true this inequality is true for n minus 2 so called this inequality as 2. Now, by using this inequality we find the estimate for phi n. So, then phi n x minus phi n minus 1 x this by definition of the iterands is integral x 0 to x f of t phi n minus 1 t minus f of t phi n minus 2 t d t. So, by definition of the function phi n now by part a by part a. So, what do we have in part a we have that phi n t phi n x minus in part a we have shown that phi n x minus y 0 phi n x minus y 0 this is less than or equal to b for all n and x in the interval x 0 x 0 plus h which we have established in part a. So, hence what we have is the point x phi n minus 1 x and x phi n minus 2 x these 2 both the points are in the rectangle r 1 for x of course, in the interval x 0 x 0 plus h. Since they are in r 1 f satisfies all the nice properties in r 1 including the Lipschitz continuity properties. So, therefore, by Lipschitz continuity by Lipschitz continuity of f. So, if you apply the Lipschitz continuity of f we have. So, on this difference if we apply Lipschitz continuity over here we have. So, absolute value of phi n x minus phi n minus 1 x which is less than or equal to alpha times alpha is a Lipschitz constant integral x 0 to x phi n minus 1 t minus phi n minus 2 t d t. So, here obviously, this is less than or equal to integral x 0 to x f of t phi n minus 1 t minus f t phi n minus 2 t d t. And applying the Lipschitz continuity and the Lipschitz constant is alpha. So, the 4 alpha will come out is alpha times integral x 0 to x 1 phi n minus 1 t minus phi n minus 2 t. Now, by the assumption. So, therefore, what is our left hand side? Our left hand side is phi n x minus phi n minus 1 x is less than or equal to what we have is alpha times x 0 to x absolute value of phi n minus 1 t minus phi n minus 2 t d t. Now, by using our assumption by 2, by 2 is our assumption. So, we assume this is our 2. We assume that 4 n minus 1 case this we have. If we assume this, then we get this is alpha times this alpha times integral x 0 to x. So, putting the values m of alpha n minus 2 n minus 2 by n minus 1 factorial and t minus x 0 to the power n minus 1 d t. If we integrate it with respect to t, so this is less than or equal to you get m into alpha I can take outside alpha to the power n minus 1 by n minus 1 factorial. Then integrating t minus x 0, we get t minus x 0 to the power n by n and you have to evaluate it at the point x 0 and x. So, this gives you that when x is equal to when t is equal to x 0, this vanishes and t is equal to x get x minus x 0. So, this is m into alpha to the power n minus 1 by n factorial n minus 1 into n factorial into x minus x 0 to the power n. So, with a just an adjustment of constants if I divide by alpha m by alpha into alpha to the power n by n factorial into x minus x 0 is always less than or equal to h. So, this is h to the power n. So, therefore, this is less than or equal to m by alpha into alpha h to the power n by n factorial. So, as x minus x 0 is less than or equal to h. So, therefore, this implies that the inequality is true for n. We assume that the inequality is true for n minus 1, then we could prove that the inequality is true for n. Now, for the case, so let n is equal to 1, check for the case n is equal to 1. When n is equal to 1, we have that phi 1 x minus y 0 is equal to integral less than or equal to integral x 0 to x f of t y 0 d t, t y 0 d t and f of t y 0 is less than or equal to m. So, therefore, this is less than or equal to m into x minus x 0, which is again less than or equal to m of h. So, it is true for the case alpha is equal. So, n is equal to 1. So, therefore, it is true. So, for alpha is equal to 1 case, it is true. So, therefore, by mathematical induction, the inequality is true for all n. So, therefore, this proves. So, as we have seen, the total proof of the Picard's existence and uniqueness theorem is divided into four parts. Part A gives some nice properties of the sequence phi n and part B gives very useful estimate for the sequence of functions phi n. And both these parts are proved. Now, part C and D, in part C, we will prove that the sequence phi n that converges uniformly to a continuous function. And part D, we will prove that that limiting limit function is a solution to the initial value problem. So, these two parts we will prove in the next theorem, next lecture. So, let us just recall. Part A gives nice properties on the sequence of functions phi n, which are the Picard's iterates. Now, part B gives a good estimate for the Picard's iterates phi n. Now, remaining part, we will prove in the next theorem. Bye.