 Welcome back. We will continue the discussions on the existence and uniqueness of solutions of initial value problem, existence and uniqueness of solution of initial value problem. In the previous lecture, we have seen several examples of initial value problems. Some are having infinitely many solutions and some are having no solutions and others having infinitely many solutions. So, now we will characterize under what condition an initial value problem will have unique solution like the problem posed by Hadamard. Once an initial value problem is given, under what condition the system has a solution and under what condition the solution is unique and conditions for which the solution continuously varying with respect to the initial condition. So, consider the initial value problem given by the first order differential equation dy by dx is equal to f of x comma y with the initial condition y at x 0 is y 0. So, f is a function. So, here f is a function which is defined on a domain D, a subset of D which is a subset of r 2 to r is a function. Two variables as it is defined on r 2 f is a function of two variables x and y with respect to the initial value problem. The first argument x that is a independent variable and y is a dependent variable which is of course the solution we are looking for. So, f is defined on a domain D. So, D is a domain in r 2. By domain we mean D is an open set in r 2 and also it is connected, it is a connected set. Any two given bonds in D that can be joined by a continuous curve which is completely inside D. So, D is a domain in r 2. So, our initial value problem given by dy by dx is equal to f of x y and y at x 0 is y 0. So, this is your x axis and this is your y axis and a point. So, this point is x 0, this point is y 0. So, initial point is x 0, y 0. We are looking for a function starting from this point and f is defined on the domain D and to make sure that first of all the solution, the initial value problem has a solution. We require certain conditions on f. So, let me first state the conditions and we will do the proof later. So, for existence, so if f of x y is continuous on D, then there exists a solution y of x to the initial value problem in some neighborhood, in some interval say x 0 to x 0 plus h. So, we will prove this existence results later. So, broadly speaking, if f is continuous in the given domain, the domain contains initial point x 0, y 0, then one can show that there exists a solution at least in some neighborhood that contains the initial point x 0, y 0. Now, uniqueness problem, so if f x y is lipschitz continuous with respect to the second argument, argument y on the domain, then the initial value problem has a unique solution. Rather, the solution of the initial value problem is unique if solution exists and if x y is lipschitz continuous with respect to the second argument that is y, then the solution is unique and the third problem is stability. So, if f of x y is continuous with respect to x and lipschitz continuous with respect to y, then the solution of the initial value problem varies continuously with respect to the initial condition y 0. So, these are the three major solution to the Hadamard problem. So, we will address each one of them separately and we will give sufficient proofs for each of them. See the existence problem, what we require is the continuity of f with respect to both x and y, continuous with respect to x and y. So, because f is defined on the domain D, which is a subset of R 2, if f is continuous with respect to x and y, then we can show that there exists a solution in some neighborhood containing the initial point x 0, y 0 and the second problem uniqueness is guaranteed if f is lipschitz continuous with respect to the second argument y and of course, continuity is required for the existence of a solution. So, continuity is also required for the existence of solution and uniqueness is guaranteed if f is lipschitz continuous with respect to the second argument. And the stability, the continuous dependence of the solution on the initial condition y 0 is also guaranteed if f is lipschitz continuous with respect to y and continuous with respect to x. And we will show it and for a lipschitz continuity, we have already seen the definition and examples of a lipschitz continuous function in the preliminaries. There we treated function of a single variable, a function of x only f of x and also we have seen sufficient conditions under which a function is lipschitz continuous and examples of lipschitz continuous functions and examples of functions which are not lipschitz continuous. Now, since we deal with the functions of two variables x and y, we will briefly deal with the lipschitz continuity of f with respect to the second argument y. So, definition 1 x y be a function which is defined on D which is a subset of R 2 x y plane and is mapping to R be a function not necessarily linear. So, could be a linear or non-linear function. Now, f is said to be lipschitz continuous with respect to y, there exist a constant called it alpha greater than 0 such that f of x y 1 minus f of x y 2 the absolute value of this is less than or equal to alpha times y 1 minus y 2 for all x y 1 and x y 2 x y 1 comma x y 2 are in the domain D. And if there exist a constant alpha of satisfying this condition you take the least of all such alphas and we call that alphas lipschitz constant. So, alpha is a lipschitz constant constant of f with respect to the argument y. Now, as in the single variable case if D is the all space R 2, if D is the entire space R 2, then we say that f is globally lipschitz continuous globally lipschitz with respect to y otherwise locally lipschitz. For existence and uniqueness for uniqueness result when a local lipschitz continuity is sufficient. Now, let us see an example say 1. So, consider a function f x y which is given by x plus 3 y where x and y they are on the ender x y is on the ender R 2. So, let us check the lipschitz continuity of f. So, f of x y 1 with respect to y f x comma y 2 which is x plus 3 y 1 minus x plus 3 y 2. So, which is equal to 3 y 1 minus 3 y 2 3 times y 1 minus y 2. Therefore, modulus of x y 1 minus f x y 2 is equal to 3 times y 1 minus y 2 for all x y 1 and x y 2 in R 2. So, this shows that f is lipschitz with a lipschitz constant 3. So, therefore, this function f given by f is equal to x plus 3 y is lipschitz with constant lipschitz constant alpha is equal to 3 and this lipschitz continuity is global because it is true for all points x y 1 and x y 2. So, therefore, it is a globally lipschitz with respect to y. Now, consider another example. So, consider the function of two variables x and y given by x square plus y square. So, this two variable function. So, let us check the lipschitz continuity. So, f of x y 1 the first point and f of x y 2 the second point these two points are arbitrary points in the x y plane that is in R 2 which is equal to x square plus y 1 square minus x square plus y 2 square. So, which is equal to y 1 square minus y 2 square which is expanded as y 1 plus y 2 into y 1 minus y 2. So, therefore, f of x y 1 minus f of x y 2 is equal to if is less than or equal to y 1 plus y 2 if we can bound times y 1 minus y 2. So, we know that this quantity y 1 plus y 2 as long as these points are arbitrary varying in the all R 2 is not bounded for all x y 1 and x y 2. Of course, if a y is bounded the y coordinate is bounded then this quantities can be bound by a constant. So, for example, if you take an infinite strip this is x axis and this is y axis if I take I vary y between say minus b and b minus b and b and if I take this infinite strip. So, if I call this as d. So, d is set of all x y in R 2 such that x is varying from minus infinity plus infinity and y is varying from minus b to plus b. So, on this t the given function says y 1 plus y 2 is bounded by 2 b. So, now y 1 plus y 2 is less than or equal to 2 b on d. So, therefore, we conclude that this function the function given by f x is equal to x square plus y square is Lipschitz continuous. If Lipschitz with respect to y with the constant the Lipschitz constant alpha is equal to 2 b. So, a function of 2 variables x and y and we have verified the Lipschitz continuity and this Lipschitz continuity is local. So, this is Lipschitz continuous and this function is Lipschitz and the Lipschitz continuity is local. So, this is locally Lipschitz with respect to y and that is on the domain we just defined d. So, on the domain defined by this set of 4 x y such that x is varying from minus infinity plus infinity and b is varying from minus b to plus b is an infinite strip parallel to x axis and we now see a function which is not Lipschitz. So, example of a non-Lipschitz function. So, example 3. So, consider a function f of x y a function of 2 variables which is given by x into square root of y. So, x into square root of y and suppose that this is defined on a domain where x varies from 0 to 1 and y varies from 0 to 1. So, the domain you are looking for domain is set of 4 x y in R 2 such that x is varying from 0 to 1 and y is also varying from 0 to 1. So, this is your domain. Now, let us check the Lipschitz continuity of this function on this domain. So, let us compute f of x y 1 minus f of x y 2. Now, we are looking for a condition like this f of x y 1 minus f of x y root value of this one is less than or equal to alpha times y 1 minus y 2 for all x y 1 in D and x y 2 also in D for all points x y 1 and x y 2 in D. We show that it is not possible for this function to have a condition like this. So, this is a question mark whether this condition is satisfied or not for any arbitrary point x y 1 and x y 2. Let us take two points x y 1 I am taking as 1 0 1 y arbitrary y and x y 2. Let us take that as 1 0 1 y where y is a value. So, where y varies between 0 and 1 y could be any value between 0 and 1 and x 2 y 2 is a 0 and x is 1. In both points the x coordinates are 1 1 y and 1 0. Let us compute. So, therefore, f of 1 y minus f of 1 0 this is equal to f of 1 y by definition of pure function. So, x is 1 and this is square root of y minus f of 1 0. So, y coordinate is 0 this is 0. So, this is equal to square root of y and y is positive. So, this I write as y is 1 by square root of y times this is. So, times y minus 0. So, if we can bound this quantity by a constant finite number alpha then it satisfies Lipschitz condition for this point. And we know that as y is approaching to 0 from right side of course, the quantity 1 y square root of y it goes to infinity showing that there does not exist an alpha satisfying a finite number alpha positive satisfying the condition if I call this condition as star satisfying star. So, as y approaches to 0 as this is x axis and this is y axis as y approaches to 0 then this quantity 1 1 upon square root of y that blows up. So, therefore, Lipschitz condition is not possible. So, the conclusion is the function f x y given by x square root of y is not Lipschitz on the given domain D. So, therefore, this function is not Lipschitz continuous and not I mean locally Lipschitz continuous on D. So, therefore, always to check Lipschitz continuity by this method may not be that very straightforward. Now, as in the case of one variable there are sufficient conditions which ensure the Lipschitz continuity of a function of two variables. So, now, I will state a theorem that ensures or guarantees the Lipschitz continuity of a function of two variables with respect to y. So, theorem says sufficient condition to guarantee Lipschitz continuity f which is a function of two variable with respect to the second variable y. So, we deal with sufficient conditions. So, we state this in the form of a theorem. So, theorem 1 let D be a closed let D be a closed domain in R 2 that is a domain is a connected set an open connected set which is also closed. So, we look for a closed domain in R 2 such that the line joining the line joining any points in D lies entirely within D. So, we take a domain a closed domain such that any two points the line joining them should lie completely inside D. Suppose that the supremum suppose that f x y suppose that f x y is differentiable is differentiable with respect to y the supremum of the partial derivative del f by del y the supremum of the absolute value of the partial derivative of f with respect to y where x and y lies in the domain D the supremum is attained and which is equal to alpha is a finite number. If the partial derivatives are bounded in this sense then the conclusion of the theorem is then f of x y which is from of course, from D to R is Lipschitz continuous with respect to the second argument with respect to y on the domain D with the Lipschitz constant alpha. So, the sufficient condition that guarantees the Lipschitz continuity of a function f is that. So, function f x y if f is differentiable with respect to the second argument y and if the partial derivatives with respect to y is bounded or supremum of the absolute value of the partial derivative is a finite number in the domain then the function f itself is Lipschitz with respect to y. The proof is very simple I will just give the proof, the proof is just by using the mean value theorem. So, let x 1, x y 1 and x y 2 to be 2 points in D that x y 1, x y 2 be 2 points in D. Now by Lagrange know mean value theorem by the mean value theorem by the mean value theorem we have there exist a number psi between y 1 and y 2 such that f of x y 1 minus f of x y 2 is equal to y 1 minus y 2 times del f by del y at the point x psi. So, this psi is a point between y 1 and y 2 that is guaranteed that is assured because we assume that in the domain D any the line joining any 2 points that lies entirely inside. So, therefore, psi is a point between y 1 and y 2 which is totally inside the domain D. Now take the absolute value. So, this implies the absolute value of f of x y 1 minus f of x y 2 which is obviously less than or equal to supremum of del f by del y at x i times y 1 minus y 2 for all x y 1 and x y 2 inside D. And by hypothesis this is bounded by this is alpha. So, this is less than or equal to alpha times y 1 minus y 2 for all x y 1 and x y 2 in D. So, proving that f x y is elliptics with respect to y with elliptics constant alpha. So, therefore, if a function is differentiable it is easy to check the elliptics continuity if you can find the bound for a partial derivative with respect to y. So, let us take the example a similar example which we considered. So, let f of x y is equal to x plus y square where x varies from minus infinity to plus infinity and y is bounded by b. So, absolute value of y is less than or equal to b. So, obviously del f by del y the partial derivative of f with respect to y is 2 y and del f by del y is equal to 2 y is equal to 2 times y and y is bounded by b. So, therefore, this is less than or equal to 2 times b. So, this implies that f of x y given by x plus y square is elliptics on if we say this is a domain. If domain D is given by this on the domain D with alpha is equal to 2 b and at this junction we know that the conditions in the theorem is not a necessary condition to for a elliptics continuity. It is just a sufficient condition. If the partial derivatives are bounded, then the function is elliptics continuous. There is a strong sufficient condition, but there are functions which are elliptics continuous, but at the same time not satisfying the hypothesis of the above theorem. So, we give a counter example. So, consider a function f of x y is given by x into absolute value of y, where the domain is which is defined on a domain where x y such that mod of x is less than or equal to a and y is between minus infinity plus infinity. So, also an infinite strip where y varies. So, from minus infinity to plus infinity x is bounded between minus a to a. So, this is a domain. So, if you look check for elliptics continuity f of x y 1 minus f of x y 2, where x y 1 and x y 2 are two points in the domain which is equal to x absolute value of y 1 minus x absolute value of y 2 and if you take the absolute value of this. So, this is less than or equal to mod of x into absolute value of y 1 minus y 2 and as x is bounded by a. So, this is a times absolute value of y 1 minus y 2. So, this shows that the function f given by x into mod y is elliptics with elliptics constant alpha is equal to a. However, this function does not have a partial derivative with respect to y as it is not differentiable with respect to y at 0. So, therefore, the conditions in the theorem is just necessary just a sufficient. So, the conclusion the condition in the theorem is just sufficient, but not necessary to guarantee elliptics continuity of f with respect to y. So, therefore, a given differential equation we can check the continuity of f with respect to x and y and also check the elliptics continuity of f with respect to y to address the question of existence, uniqueness and the continuous dependence of the solution with respect to the initial condition and also note that elliptics continuity is a stronger notion than continuity. Any elliptics continuous function is continuous, but less than the differentiability. For proving the uniqueness result, we employ a very important lemma which is known as Gronwald's lemma. So, this will be invoked to prove the uniqueness of solution of an initial value problem. So, let me state the Gronwald's lemma. Suppose that f x and g x continuous real valued functions with a condition that f of x is greater than or equal to 0 and g of x is also greater than or equal to 0 on an interval call it a b. So, if f x and g x are two continuous real valued functions both non negative and defined on an interval a b, then if we have an inequality that is f x is less than or equal to some constant c plus k times integral a to x g s f s d s, where c and k are non negative constants. Then the conclusion is f of x is less than or equal to c times exponential integral a to x g s d s. So, this inequality is known as a Gronwald's inequality. If f x is less than or equal to c plus k times integral a to x g s f s d s, then f x has to be bounded by f x is less than or equal to c times e to the power integral a to x g x d s. In the first inequality f x is there on both sides left hand side and the right hand side and the conclusion is f x can if this is the case, then f x can be bounded by a function which is independent of f. The proof of this inequality is, so what we have is we have f x is less than or equal to c plus k times integral a to x g s f s d s, this we have. And if I take the arches and denote this by g x, so let g x is equal to c plus k integral a to x g s f s d s. So, this implies that, so thus what we have is f of x is less than or equal to g x by definition. It is a hypothesis because g x is your arches, so f of x is less than or equal to g x. And also g of a, if you evaluate g at a integral a to a which is a 0 which is just c. And if you differentiate g, g prime x is equal to if you use Leibniz formula and differentiate this integral, so first part is 0, then k times, this is k times g x f x. So, therefore, g prime x is, so therefore, we get g prime x is equal to k g x f x. And this f x is less than or equal to g x, so this implies that this is less than or equal to k times g x capital G x. In other words, what we have is g prime x is less than or equal to k small g x into capital G x. And if you divide this by g x, so g prime x divided by g x is less than or equal to k small g x. Now, integrating this, integrating over the interval a x, we have that integral of g prime x by which is g prime is called g s by g of s d s integral a to x is less than or equal to integral a to x k times g of s d s by using the integral of derivative by the function which is nothing but the logarithm ln of g s and evaluated at these two points a and x is less than or equal to integral of a to x k g s d s. So, since g of s at a is c, this is ln of g x divided by c less than or equal to integral a to x k g s d s. And taking exponential on both sides, we get g x is less than or equal to c times e to the power integral a to x k g s d s. And from our hypothesis, we know that f x is less than or equal to, we know that f x is less than or equal to g x. So, therefore, f x is, so this implies f x is less than or equal to c times exponential integral a to x k g s d s. So, this proves a ground walls inequality. And as a remark, in the ground walls inequality, I can say as a corollary, see as a corollary in the ground walls inequality, if c is equal to 0 in ground walls inequality, then f of x is less than or equal to k times integral a to x g s f s is d s. This implies that f of x is equal to 0. So, that proof follows very easy, easily because if we take for every epsilon, we take an epsilon greater or every epsilon greater than 0, the ground walls inequality holds. And if that holds, then you get epsilon times k integral a to x g s f s. So, that makes that f x to be 0, since f x is non-negative. So, this result we will be using in proving the uniqueness result, the ground walls inequality. Now, we have built up necessary tools to prove the uniqueness of an initial value problem, if the function f is slipschitz continuous with respect to y. So, that the uniqueness will be proved in the next theorem. So, therefore, two things we have seen that one is, if the function is continuous with respect to x and y, then there exists a solution to the initial value problem. If the function is slipschitz with respect to y, then the solution is unique and the continuous dependence will be shown later. We will see the proof of this in the subsequent lectures.