 We have been discussing about the accessions and the uniqueness of solutions of initial value problems. We have so far proved the Picard's accessions and uniqueness theorem and also Cauchy Piano accessions theorem. These proofs were involved and we used many notions from real analysis to prove the accessions and the uniqueness part. And if we make use of the functional analytic technique, especially the fixed point theory, then this accessions and uniqueness of the solution of an initial value problem can be very easily established without much effort. In this respect, the Banach contraction principle can be applied to derive the accessions and uniqueness of solution of an initial value problem provided the function in the initial value problem is good in the sense that satisfies Lipchitz's condition. And also, the Banach fixed point theorem will give us a way to compute the solution of the initial value problem. And the preliminaries we have already seen, the fixed point theorem Banach contraction principle and also a generalization of the Banach contraction principle. And let me just recall the Banach contraction principle if X is a normed linear space and then the space is complete with respect to the norm and we call this is a complete normed linear space. Let X be a complete normed linear space, a complete normed linear space is called a Banach space and T is an operator from X to X such that T is an operator, this could be a no linear operator. So, T is called a contraction if T x minus T y, the norm of T x minus T y is less than equal to alpha times norm of x minus y for all x y in x and for some alpha strictly less than 1. Of course, this has to be a no negative number. If a T is a contraction, example of, see example if f x is a function defined by half x is a linear function. So, it follows easily that f is a contraction, f is a contraction, f is slipchitz continuous and the slipchitz constant is less than 1 then is called a contraction, f is slipchitz with a slipchitz constant half, so here alpha is equal to half. Another no linear example is say f x is 1 by 4 sin x and we have already seen that sin function is slipchitz continuous. So, therefore, 1 by sin x is also slipchitz continuous with slipchitz constant 1 by 4, so this implies that f is a contraction on R, R is a complete no linear space, a Banach space. Now, the Banach fixed point theorem says, Banach fixed point theorem says or known as Banach also known as Banach contraction principle, if a T from a Banach space x to x is a contraction, then T has a unique fixed point say x star that means T of x star is same as x star, point x star is a fixed point. Further, if T is a contraction then T gives a unique fixed point, the same time it gives a computational algorithm to find the fixed point. Further, the sequence x n defined by x n plus 1 is equal to or x n is equal to T of x n minus 1, so x 0 arbitrary, so n is 1, 2, 3 etcetera. The sequence x n defined by x n is equal to T x n converges to the unique fixed point, to the unique fixed point T star of T. So, this is a computational algorithm we will see later that, this is nothing but the Picard's iterands, which we defined earlier. So, just example of a fixed point, see if I have a function f x is equal to x square, so obviously x is equal to 0 that f of x is equal to 0 square that is 0, x is equal to 0 and x is equal to 1 are fixed point f x is equal to x square. See geometrically, if you look into the graph x square and fixed point is in one dimensional case, the point of intersection of the function y is equal to x and the given function, so there are two fixed points, so 0 and 1. So, this gives you the fixed point, two fixed points are there and f x is equal to x, all points are fixed points, f x is equal to x cube, so you can find out the fixed point, they are the point of intersection of the graph of the curve with the diagonal line. Now, let us come back to the initial value problem, so the initial value problem, so we have the initial value problem d y by d x is equal to f of x y with initial condition y at x 0 is equal to y 0. Now, we state and prove the existence and uniqueness theorem for this initial value problem and the proof of the theorem will be based on Banach conduction principle. So, let me state the theorem, theorem is, so let f x y be a continuous function defined on a domain, so let f be Lipschitz continuous on d, Lipschitz continuous on with respect to y, Lipschitz continuous with respect to y on d. Then of course, other conditions like x 0 y 0 is an integer point of d, then there exist a unique solution to the initial value problem. So, this is our initial value problem on an interval x minus x 0 less than or equal to h, where h is minimum of a b by m and m is maximum of f x y x y is in some rectangle r which is inside d. So, r is a rectangle defined as earlier x y such that x is x minus x 0 is less than equal to a y minus y 0 is less than equal to b. For some a and b such that this r is inside the domain d. So, the theorem is, if f is continuous is a continuous function on d and f is Lipschitz continuous with respect to y on d, then there exist a unique solution to the initial value problem on an interval x minus x 0 is less than or equal to h, where h is defined by this, where a and b are parameters of a rectangle which is inside the domain. And that is alpha b the Lipschitz constant of f with respect to y. So, further we have so further the unique solution can be computed from the successive approximation y of n minus 1 x is equal to y 0 plus integral x 0 to x f of t, y n of t, d t, n is equal to 0, 1, 2, etcetera. So, this iterative scheme if you recall this is the Piccarts known as a Piccarts iterates. So, same sequence of functions which we constructed in Piccarts iteration, y 0 function is just y second start from any arbitrary y 0 n is y 0. The proof of this theorem we will state by using Banach conduction principle. And also we note that in the Banach conduction principle if a t is a contraction, then t has a unique space point. If t n is a contraction, if t is an operator from a Banach space x to another Banach space x, if t n is a contraction for the sum n greater than or equal to 1, then also then t has a unique fixed point. So, this is known as a generalized Banach conduction principle, generalized conduction principle. So, we will be using this generalized Banach conduction principle to establish the proof of existence and uniqueness of solution of the initial value problem. So, let us come to the proof. So, by the basic lemma, so from the basic lemma we stated and proved earlier basic lemma, the solubility of the initial value problem follows if the following integral equation is solvable. The integral equation from the basic lemma y s is equal to y 0 plus integral x 0 to x f of t y of t d t. So, we have proved that if you want to solve the initial value problem, it is enough to solve this integral equation. This is known as a mild solution of the initial value problem. So, therefore, we are looking for a solution of this integral equation. Let us define an operator. So, let us define first a function. Let x be a function c a b, c set of all not a b, this is our interval we are looking for a solution from x 0 to c 0 to x 1. For some x 1 greater than x 0, set of four continuous functions defined on the interval x 0, x 1 and it is easy to show that this x is a but a x space. So, if we define a norm, so define a norm for x in x, norm of x is supremum or maximum of x of t, where t varies between x 0, x 1, it is called the soup norm or maximum. So, x is the set of all continuous functions of supremum and maximum. They are the same in this interval x 0, x 1 closed and bounded interval. So, therefore, x is a normed linear space, x with this norm is a normed linear space and also it is complete. Every Cauchy sequence in x with respect to this norm converges to a limit inside. So, therefore, it is a complete normed linear space, it is a complete normed linear space, that is a Banach space, it is a Banach space with soup norm. Now, we define an operator, remember our integral equation is y x is equal to y 0 plus integral x 0 to x f of t y of t d t. Now, I define an operator, define a mapping or an operator call it t from c x 0 x 1 to itself by t of y is a function at x is equal to y 0 plus the right hand side of your equation quality equation 1, right hand side of equation 1, y 0 plus integral x 0 to x f of t y of t d t. So, f is a non-linear function and so, therefore, t is a non-linear operator from the Banach space c x c to c. Now, if a t has a fixed point, if t has a fixed point that is, if you can find some y such that y is equal to t y, then y is equal to t y, y is equal to t y for some y, then that is nothing but your integral equation. So, therefore, the solubility of the integral equation 1 is equal into the existence of a fixed point for the operator t. So, therefore, my objective is to show that t has a fixed point, if t has a fixed point that is, there exist that is, there exist a y such that y is equal to, there exist a y such that y is equal to t y, then the fixed point y is a solution to the integral equation integral equation 1. If t has two fixed points, then integral equation has two fixed point, two solutions. If t has a unique fixed point, then the integral equation has a unique fixed point. If the integral equation has a unique solution, then the initial value problem has a unique solution. So, there is an equivalence between the solution of the initial value problem and the integral equation. If t has a unique fixed point, then the integral equation 1 has a unique solution. So, we are going to show that this operator t has a unique fixed point and we will show that t n is a contraction for some n greater than or equal to 1. So, we will prove that t n is a contraction, some sufficiently large n greater than 1. So, how do we go about it? So, let us define. So, by definition t of y of x, t of y of x is y 0 plus integral x 0 to x f of t y t d t. This is the definition of the operator t. So, let us find what is t y 1 of x minus t y 2 of x. If y 1 and y 2 are two functions in the Benak space, let y 1 and y 2 are two points C x 0, x 1, two continuous functions, then we take this difference and the absolute value. This is absolute value of integral x 0 to x, y 0 and y 0 will get cancelled and what we have is f of t y 1 t minus f of t y 2 t d t and this is less than or equal to and also by using the fact that f is Lipschitz continuous with respect to the second argument y. So, therefore, this is less than or equal to alpha times integral x 0 to x y 1 t minus y 2 t d t. So, this is one relation we will be calling, call it equation number 2. Now, this is also greater than or equal to alpha times integral x 0 to x. If I take the max of or sup of sup over t, t in the interval x 0 x 1 y 1 t minus y 2 of t can take this is greater than or equal to d t and this quantity is our norm of y 1 minus y 2. So, this is norm of sup norm y 1 minus y 2. So, therefore, this is less than or equal to alpha times integral x 0 to x norm of y 1 minus y 2 d t, which is equal to if you integrate it. Now, norm of y 1 minus y 2 is a constant. So, alpha times x minus x 0 into norm of y 1 minus y 2. So, if I take now t square. So, my aim is to find an n such that t n is a contraction. Now, if this constant alpha into x minus x 0, if I can bound this by a number which is less than 1, then t is a contraction. But this I cannot have unless I put some restriction on the Lipschitz constant alpha. So, but I do not have any restriction on the Lipschitz constant alpha. So, what I do is I take a composition, I take a t square. So, t square y 1. So, t square y 1 of x minus t square y 2 of x, which is nothing but the composition of t with the t 2 times. So, this is t of t y 1 of x minus t of t y 2 of x. So, this is less than or equal to if I use 2, this is less than or equal to alpha times integral x 0 to x t of y 1 of t of t minus t of y 2 of t d t. Now, using the inequality, call it this one. So, 3, if I use 3, t y 1 minus t y 2 absolute value t y 1 minus t y 2 from this equation number 3, I get this is less than or equal to alpha square times integral x 0 to x. So, you call it t minus x 0. So, x minus x 0 becomes t minus x 0 times norm of y 1 minus y 2 d t. This I can integrate to get alpha square by integral of t minus x 0 with respect to t is a t minus x 0 square by 2. So, alpha square by 2, so t minus that is evaluating from x 0 to x. So, x 0 minus x 0 is 0. So, therefore, this is x minus x 0 square into norm of y 1 minus y 2. So, this is times norm of y 1 minus y 2. So, conclusion if I take the supremum of this quantity over x varying on the interval x 0 x 1, then that becomes a norm. So, norm of t square y 1 minus t square y 2 is less than or equal to if I take maximum over there, I get alpha square into x 1 minus x 1 is a maximum x 1 minus x 0 square by 2 factorial into y 1 minus y 2. So, I can continue this procedure. So, if I continue this procedure to take one more times t composition, so t cube y 1 minus t cube y 2 just by integrating that integration is a process which is making the right hand side a good number. So, t cube y 1 minus t cube y 2 which you can show that this is less than or equal to alpha cube by x 1 alpha cube into x 1 minus x 0 cube by 3 factorial times y 1 minus y 2. So, if you continue like this t n y 1 minus t n y 2. So, t n y 1 minus t n y 2 can show that this is less than or equal to alpha to the power n x 1 minus x 0 to the power n by n factorial into y 1 minus y 2. So, what does it say? If my n is sufficiently large and x 1 minus x 0 is a finite quantity and alpha is ellipsis constant and the denominator I have n factorial this can be made. So, can be made less than 1 strictly less than 1 if n is sufficiently large. So, therefore, without any additional conditions on the ellipsis constant of f I can show that or we have seen that t n is a contraction. So, this implies that t n is a contraction and hence t n is a contraction for n large for n large and hence t has a unique fixed point by the generalized Banach conduction principle. So, therefore, this Banach conduction principle this proof gives us both existence and uniqueness. So, t n is a contraction means t has a unique fixed point. The unique fixed point happens to be the unique solution of the integral equation that amounts to be the unique solution of the initial value problem. And further the computation of the fixed point as we have seen x n plus 1 is equal to t of x n. So, starting from any arbitrary x 0 that gives us the or a sequence x n a sequence x n that converges to the unique solution of the initial value converges to the unique fixed point. So, in our context, so y of n plus 1 is equal to t of y n and y 0 arbitrary by the definition of t. So, y n plus 1 is y n plus 1 is a function of x is equal to y 0 plus integral x 0 to x f of t y n t d t and you choose the initial function y 0 t y 0 x as your y 0 y 0 x is your initial condition y 0. In fact, this iterative scheme if you compare with the Picard's existence and uniqueness theorem this is a Picard's iterative scheme. Let us see one example how the initial condition is or how the solution is computed by using this Banach fixed point theorem is illustrated in the following example. So, example let us take a very simple system a linear d y by d x is equal to y where y at 0 is 1 which we all know the solution. So, linear equation the solution y x is e to the power x we know the solution a priori. Here what is f? f of x y is y which is lectures and if it is continuous with respect to both the variables and also it slips with respect to the second argument and it satisfies all the properties. So, let us define the iterates. So, y 0 y 0 x is a initial function initial condition 1 and y n plus 1 x is equal to y 0 plus integral 0 to x here f n is y n y n t d t. Therefore, what is y 1? y 1 x is equal to y 0 is 1 plus integral 0 to x y 0 t d t and y 0 is a 1. So, this is equal to 1 plus integral 0 to x 1 d t. So, which is equal to 1 plus x and y 2 x is 1 plus integral 0 to x is y 1 t d t which is 1 plus integral 0 to x y 1 is 1 plus x. So, this is 1 plus t d t. So, which is equal to we get 1 plus x plus x square by 2. And similarly, if you find y 3 x is not difficult to show that y 3 x is 1 plus x plus x square by 2 plus x cube by 6 and so on. And you can see y n x is summation x to the power m n by n factorial y n is x to the power m by m factorial where m goes from 0 to n. So, this as n goes to infinity this goes to e to the power x as we expected. So, therefore, the computation of the fixed point by using the Banach conduction principle that amounts to be the Picard's iteration. And by using that one can compute iteratively the solution of a initial value problem provided if those good conditions of Lipschitz continuity etcetera are already satisfied in the equation. So, this portion completes the existence and uniqueness of solution of initial value problem. So, we post three problems initially the Hadamard well postness problem. A model is well post if solution exists the existence problem and the solution is unique the uniqueness problem and the solution is stable. Stable the sense if the solution changes continuously with respect to the initial condition then the solution is stable. All these three properties of three conditions of Hadamard the one is left out is the stability the third condition. We now look into the stability aspect of the solution of the initial value problem. So, stability of solution with respect to initial condition consider 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. And if I change y 0 to say y 0 tilde if with y 0 I have a solution say y x what will be the corresponding solution of y 0 tilde what is the relationship between if y 0 is changed to y 0 tilde and what happens to the solution y and what will be the relationship or how much close if y 0 and y 0 tilde are very close how close y and y tilde the corresponding solutions that is our question. In other words it comes practically in many applications because in many experimental data we the initial data we take from the instrument from a device. While taking the reading from the instrument either the instrument may not be that very accurate or the person who is taking the observation is not very keenly taking there may be a possibility of little bit error in the initial data y 0. If a slight error in the y 0 makes a slight error a corresponding small error in the solution then the solution is reliable then we say the solution is stable if a small change in the initial data results in a small change in the solution then the solution is stable. A small change in very small change in the initial data results in a drastic difference in the solution then that solution is not reliable is not believable. Therefore, we want to make sure that the solution is stable in this sense. So, mathematically it is nothing but the continuity of the solution with respect to the initial data. So, stability is the continuity of solution with respect to initial data y 0. Again, if our function f our function f is nice if f is continuous with respect to x and y and f is lipschitz continuous with respect to y then the stability is also guaranteed. So, I state and prove that as a theorem if the differential equation is in the differential equation f if f is continuous with respect to x and lipschitz continuous with respect to y then the system the solution is continuous changing continuously with respect to the initial data or the system is stable. So, I state and prove the theorem. So, theorem is suppose that f of x y is continuous on D a subset of R 2 and lipschitz continuous with respect to y on D then the solution of the initial value problem d y by d x is equal to f of x y y at x 0 is y 0 is stable or solution changes continuously with respect to the initial data. So, the initial data y 0. So, proof is straight forward by using ground mass inequality proof. Let y 0 and y 0 tilde be to initial and let. So, since f is a lipschitz it has a solution it has a unique solution for every x 0 every y 0 since y with y 0 there exists a unique solution and y 0 tilde also there is a unique solution. So, because of the lipschitz continuity. So, if we choose a two initial conditions y 0 and y 0 tilde are two unique solutions collect y x and y tilde x be the corresponding solutions of IVP. So, such solution exist and such solutions are unique. So, therefore, what we have is corresponding to y 0 we have a solution y x is equal to y 0 plus integral x 0 to x f of t y t d t and with the y 0 tilde the solution is y tilde x is equal to y 0 tilde plus integral x 0 to x f of t y tilde t d t. So, there are two solutions. If I find the difference say y x minus y tilde and I take the absolute value which is less than or equal to taking the difference y 0 minus y 0 tilde plus integral x 0 to x and since f is lipschitz continuous with respect to the second argument one alpha comes out. So, this is y t minus y tilde t d t by lipschitz continuity f with respect to y. Now, using by Gronwall's inequality by Gronwall's inequality we can see left hand side and right hand side both you have y x minus y tilde. So, that can be bounded by another function on one side that is y x minus y tilde x is less than or equal to y 0 minus y 0 tilde into exponential alpha times integral x 0 to x d t and this can be bounded this is less than or equal to y 0 minus y 0 tilde e to the power alpha into x minus x 0 and x minus x 0 this can be bounded by h or if you take x 1. So, that is h less than or equal to y 0 minus y 0 tilde into e to the power alpha x 1 minus x 0. So, what does it say? It says that y x minus y tilde x is less than or equal to y 0 minus y 0 tilde e to the power alpha x 1 minus x 0 which is a finite quantity. So, whenever y 0, so for every small change in y 0 the difference between y 0 and y 0 tilde is small then the corresponding difference between y x and y tilde is also small. So, for every epsilon greater than 0 there existed delta that delta is given by epsilon by e to the power alpha x 1 minus x 0. If I choose this delta then it follows that this so y is continuous with respect to y 0. So, now it is obvious that the solution is also continuous with the initial data. So, therefore, if a function if that initial value problem the function satisfies a elliptical type condition then the solution is also stable with respect to the initial data. So, with this we finish the stability aspect of the solution. We have seen the existence, the uniqueness and stability. Bye.