 Welcome back to the lectures on Existence and Uniqueness of Solutions of Initial Value Problems. We have discussed about the Lipschitz continuity of a function of two variables and also we proved in the last lecture, the Grownwald's Lemma, which will be used in proving the uniqueness of solution of an initial value problem. Let us recall the Grownwald's inequality, which we did yesterday. The Grownwald's inequality recalls Grownwald's inequality or Grownwald's lemma. So, let f x and g x be two real valued function. So, we two real valued function defined on some interval a, b such that both f x and g x are non-negative on the interval a, b. The inequality, then the inequality f x is less than or equal to c plus integral a to x f t g t d t with a constant k, k times, where c and k are positive constants implies then this inequality implies f x is less than or equal to c times e to the power integral a to x k g t d t for x in the interval a, b. So, this inequality says if whenever we have an inequality, f x is less than or equal to c plus integral a to b k times f t g t that implies that f x is less than or equal to c times this constant c times exponential of the integral a to x k into g t d t. The advantage of this inequality is that in the first inequality f x is coming on both sides and the second inequality gives a bound for the function f, the right hand side is independent of f. And a special case of this Grosvenor's inequality, we will use in proving the uniqueness theorem. So, that I state as a corollary, let f x be a real valued and non-negative function, continuous function defined on a, b. In the Grosvenor's inequality also, we require f x and g x be to real valued continuous function. And also continuity is required for giving a meaning to the integral. So, here the corollary let f x be a real valued and non-negative continuous function defined on a, b and k, b a constant. Then the inequality f x is less than or equal to integral a to a to x k times f t d t implies f x is equal to 0 on the interval a to b. So, this is a special case of Grosvenor's inequality where the constant c is 0, where the constant c is 0 and g t or g x is 1, but c we assume to be a strictly positive, it is strictly greater than 0, but that can be tackled. Let us see a proof of it, proof of corollary. Let epsilon greater than 0 be a given number, let epsilon be an arbitrary number given. Now, consider the inequality, consider the inequality f x is less than or equal to epsilon plus integral a to x k times f t d t, integral a to x k times f t d t. Now, f x is a continuous non-negative function and the k is already given k is a strictly positive constant and epsilon is strictly positive. Now, by applying Grosvenor's inequality on this, so by applying Grosvenor's lemma with, remember Grosvenor's lemma is f x is whenever f x is less than or equal to c plus integral a to x k into f t g t d t, we have the inequality f x is less than or equal to c times exponential of the integral a to x k into g t d t. See, if you apply this Grosvenor's inequality here with g x is equal to 1 and c is epsilon which is of course, greater than 0, then we get, then we get f of x is less than or equal to c times c is epsilon times e to the power integral k times integral a to x k times g is 1. So, this is k into d t which is equal to epsilon e to the power if we integrate x minus k times x minus a. So, therefore, this implies that this function, non-negative function f x is less than or equal to epsilon times e to the power k and x varies from a to b. So, the upper bound is b minus a. So, e to the power b minus a is a finite quantity and this is true for all epsilon any given epsilon. So, since epsilon is arbitrary and f x is greater than or equal to 0, we have f x is identically 0. So, this quantity is a finite quantity and this is true for all epsilon and f is non-negative. So, implies that f x is equal to literally 0. So, this form of Grosvenor's lemma we will apply in the uniqueness result. Let us consider the come to the uniqueness result. So, consider the initial value problem. So, d y by d x is equal to f of x comma y with the given initial condition y at x 0 is y 0. So, this is the initial value problem and we will prove that if f is Lipschitz continuous with respect to x, then the solution of this initial value problem is unique. So, by making use of the Lipschitz type, Lipschitz continuity condition, we will prove that the uniqueness of result and also we will invoke the Grosvenor's lemma. So, tools Lipschitz condition Lipschitz continuity of f with respect to y Grosvenor's lemma. So, we will also invoke a basic lemma for the initial value problem, a basic lemma for the initial value problem. Let me state and prove a basic lemma for the initial value problem. If y x is a solution of the initial value problem, d y by d x is equal to f of x, y and y at x 0 is y 0, then y satisfies the following Volterra integral equation, Volterra integral equation given by y of x is equal to y 0 plus integral x 0 to x f of t y of t d t. So, this is the integral equation. We denote this by I e and the initial value problem is I v p. So, the basic lemma is if y is a solution of the initial value problem, then y satisfies the Volterra integral equation y x is equal to y 0 plus integral x 0 to x f of t y t d t. Conversely, if y is a solution of the integral equation I e and y belongs to the class of all continuously differentiable function defined on some interval x 0 x 1 with x 1 is some number greater than x 0. So, if it is a solution of the integral equation, which is continuously differentiable on the interval x 0 x 1, then y satisfies the initial value problem. It means y is also a solution of the initial value problem. So, one way we are converting the problem of solution of the differential equation into the problem of the solution of an integral equation. This integral equation I e, the solution of the integral equation can be treated separately. The solution of an integral equation need not be differentiable. In case it is differentiable, we can show that it is a solution of the initial value problem. So, the proof of this basic lemma, so what we have is the differential equation d over dx is equal to f of x y. Now, integrating with respect to x over an interval x 0 to x we get, so integral x 0 to x dy dx, dx is equal to integral x 0 to x f of, we change the variable to t and y is also a function of x y of t dt and this is y at x minus y at x 0 is equal to integral x 0 to x f of t y t dt. We know that the initial condition y at x 0 is y 0. Therefore, I am pushing this to the right hand side, y of x is equal to y 0 plus integral x 0 to x f of t y of t dt. So, this is an integral equation, volt around integral equation. Therefore, if y is a solution to the initial value problem, then y satisfies the integral equation. Now, we will prove the other way. Suppose that y is a solution to this integral equation or y satisfies this integral equation and y is also continuously differentiable, then we can. So, suppose y is in c 1 x 0 x x 1 and y satisfies i e, then differentiating with respect to x, differentiating with respect to x we obtain. So, dy by dx is equal to derivative of y 0 is 0 plus now dy by dx of integral x 0 to x f of t y of t dt using the Leibniz rule for differentiating an integral. So, we get this turns out to be f of x y of x. So, the derivative of this integral is f of x y of x. So, therefore, we get dy by dx is equal to f of x y of x. So, therefore, the function satisfies the differential equation and it is very easy to verify that y at x 0 for the integral equation. See in this integral equation, if you look at the integral equation and if you substitute, if you replace x by x 0, if you replace x by x 0, then this is integral x 0 to x 0 that is vanishes and y at x 0 is y 0. So, obviously, y at x 0 is y 0. So, this implies that y satisfies the initial value problem. Now, as I pointed out a remark, so one can study solutions of the integral equation without differentiability assumption, without differentiability assumptions on y, but only with continuity assumption. So, such solutions, such solutions are known as mild solution, solutions or weak solutions of the initial value problem. So, if we do not require the differentiability condition on y, then the solutions of the integral equation, what are the integral equations just defined are known as mild solutions or weak solutions. Now, we are ready to prove the uniqueness result. So, theorem, we now state and prove the uniqueness theorem for the initial value problem. Suppose that f of x y is continuous, f of x y is continuous and lift is continuous with respect to y on a rectangle r, which is in r 2, where r is defined as r is a rectangle set of fold x y such that x minus x 0 is less than equal to a y minus y 0 is less than equal to b for some constants a and b positive. So, we assume that suppose that f is a continuous and lift is continuous with respect to y on a rectangle r and the rectangle is defined this way with lift is constant alpha, then the initial value problem d y by d x is equal to f of x y y at x 0 is y 0, then the initial value problem and you say then the solution then the solution of the initial of the initial value problem is unique. If it has solution, then the solution is unique, but because of the continuity with continuity is assumed with respect to both x and y because of the continuity there exists a solution and here our major emphasis is on the uniqueness and since f is lift is continuous with respect to y, we are going to prove that the solution is going to be unique. So, if solution exists then the solution is unique, it cannot have more than one solution if f is lift is continuous with respect to y. We will prove it, the proof we invoke the ground words lemma. So, let us assume that it has got two solutions, let y of x z of x be two solutions that y and z be two solutions of the initial value problem defined on some interval x 0, x 1. So, thus by the basic lemma, by the basic lemma, basic lemma says any solution that is satisfying the initial value problem will also satisfy the integral equation. So, basic lemma says a solution satisfying the initial value problem, a solution satisfying the initial value problem will also satisfy the integral equation. So, therefore, we are using the basic lemma. So, thus by the basic lemma y x is equal to y 0 plus integral x 0 to x f of t y of t dt and similarly, z is a solution. So, z x is equal to y 0 plus integral x 0 to x f of t z t dt is also a solution. Now, subtracting one from the other one, so subtracting we get y x minus z x is equal to y 0 and y 0 get cancelled and the integration domain of integration is common. So, integral x 0 to x f of t y t minus f of t z t dt. Now, if you take the absolute value y of x minus z of x is less than or equal to integral x 0 to x absolute value f of t y t minus f of t z t dt. Now, we have assumed that if we slip just continuous with respect to the second argument y. So, if we slip just continuous with respect to the second argument, so therefore, we can use a slip just condition slip just continuity. So, this is less than or equal to 0. So, by Lipschitz continuous alpha times integral x 0 to x y of t minus z of t dt this is by the Lipschitz continuity by Lipschitz continuity of f. So, now we have an inequality that absolute value of y x minus z x is less than or equal to alpha times integral x 0 to x absolute value of y t minus z t dt. So, now, if we use the corollary of Gronwald's lemma see the corollary of Gronwald's lemma, if you have a situation like this if f x is less than or equal to integral a to x. So, if f x is less than or equal to integral a to x k f t dt then f x has to be 0. Here, the condition shown f is f is non negative continuous and k is a positive constant. So, if this the conclusion is f t is equal to 0. So, if we come to our inequality, so now by using Gronwald's. So, we have obtained that absolute value of y of x minus z of x is less than or equal to integral x 0 to x alpha times y of t minus z of t dt. So, applying Gronwald's inequality Gronwald's lemma with. So, Gronwald's lemma we use a corollary f of x is absolute value of y of x minus z of x and k is equal to alpha and g x the corollary is 1 and c is 0. So, we get f of x is literally 0. So, that is y x minus z x is equal to 0 for x in the interval x 0 to x 1. So, the conclusion is, so this implies that. So, we started with two solutions y and z and we have come to a conclusion that the difference between these two solutions for all x is 0. So, that is y of x is equal to z of x proving the uniqueness of solution. So, the solution is unique. Now, we will consider a few examples. So, example with example 1 d over dx is equal to y plus exponential 2 x with initial condition y at 0 is 2. So, look at this symbol initial value problem. See, note that it is a linear differential equation. It is a linear differential equation and non-homogeneous. So, linear and non-homogeneous differential equation and this equation is first order. So, sufficiently theory has been already been developed and you have seen how to solve this equation by using the method using the integrating factor techniques and all. So, let us analyze the solution, existence of solution, uniqueness of solution. Here, the function f of x y, the right hand side is of this equation is y plus e to the power 2 x and this is continuous. So, with respect to x and y, it is a linear in y and it is an exponential function. In x, it is a continuous function in the all r 2. So, f is continuous and what about the Lipschitz continuity? So, f of x y 1 minus f of x y 2, which is equal to y 1 minus y 2. So, it is linear. So, therefore, it is obviously Lipschitz continuous. It is Lipschitz continuous. Here, the Lipschitz constant alpha is 1. So, it satisfies all the conditions of the uniqueness theorem. So, the function is continuous in x and y and the function is Lipschitz continuous with respect to y with the Lipschitz constant 1. So, therefore, by the uniqueness theorem, the IVP, this initial value problem has a unique solution. This linear problem has a unique solution. Remember that not all linear problems, linear initial value problems are having unique solutions. We have seen examples and further we will see examples. And by using the integrating factor and the method which you have already seen, we have seen that a solution of this equation is given by y of x is equal to e to power x by applying the initial conditions e to the power x plus e to the power 2 x is the only solution. This solution is found by using the method we studied earlier. And by the uniqueness theorem, this is only solution. And now, let us consider another example. So, example 2. So, let us consider a non-linear initial value problem. So, consider a non-linear initial value problem given by d y by d x is equal to x into sin y, where this function is defined on for x on a domain. Domain is defined by set of 4 x y, where x is bounded by 2 and y is free. So, y is varying from minus infinity plus infinity and x is bounded by 2. So, let the initial condition be y at 0 is 1. We will analyze the existence and uniqueness. So, first let us write down the function. The right hand side function f of x y is x into sin y is continuous on d. And we will check whether this function is Lipschitz continuous with respect to the second argument y. To check the Lipschitz continuity of f with respect to y, we have stated and proved a sufficient condition. If the partial derivative of f with respect to y is bounded in the given domain, then the function is Lipschitz continuous with respect to y in that domain. So, the partial derivative, so del f by del y is equal to x cos y. And if you take the bound, say you take supremum of this one for all x y in d. So, this is even maximum is enough. So, maximum is attained. So, x y in d, so this is equal to cos y is bounded by 1 and x is bounded by 2. So, the supremum is 2. Therefore, this implies that x sin y is Lipschitz continuous on d with respect to y with Lipschitz constant alpha is equal to 2. Therefore, this function satisfies all the conditions of the uniqueness result. So, f satisfies conditions of uniqueness theorem. Therefore, the conclusion is, so d over d x is equal to x sin y with y at 0 is 1 has a unique solution. A unique solution starting from the given point is 0 1. But the existence part as I have already mentioned, if the function is continuous with respect to x and y, then existence is guaranteed in some interval starting from the given initial point 0 1. Now, what we have proved or observed is the uniqueness is guaranteed because of the Lipschitz continuity property of f with respect to y. Now, we will see an example where we have more than one solution. That example we had already seen, a linear equation. So, I will take an example. So, example 3. So, consider the initial value problem. Of course, this is a linear one d over d x is equal to 2 by x into y and the initial condition y at 0 is 0. So, in our earlier sessions, we have obtained its general solution and also its solution satisfying the initial condition. So, note that y is equal to c into x square is a solution for the initial value problem, every value of c. Therefore, it has got, we have seen that it has got infinitely many solutions. Let us just check y or how we compare this with the uniqueness result. So, in this case f of x y is 2 by x into y on an interval say 0 to some number say 0 to 2, 0 is a point given and del f y del y obviously, it exists, it is 2 by x. But if we take the supremum of del f y del y when x and y are here, depends only on x, when x is in the interval say 0, 1 which is sup. So, this does not exist. We know that this blows up. So, therefore, the function f is not Lipschitz continuous with respect to y. See at x is equal to 0, you see the singularity is therefore, this differential equation as x goes to 0, 2 upon x blows up. So, if it is not Lipschitz continuous with respect to y, so therefore, there is no surprise why this equation does not have a unique solution. So, the uniqueness theorem does not apply here. So, uniqueness theorem does not apply here and we have several other examples which we dealt with. One example which can be done as an exercise so example 4. So, consider the initial value problem d over dx is equal to 3 y to the power 2 by 3. The initial condition y at 0 is 0. We have seen in the earlier lecture that this differential equation has got infinitely many solutions including y x is equal to x cube. So, it has got many solutions and in this case also if we look at the function f of x y is equal to 3 y 2 by 3. So, it is a continuous with respect to both the arguments, but it is not Lipschitz with respect to y on any domain containing the point 0, 0. So, not Lipschitz on any domain containing 0, 0. So, therefore, the uniqueness is not assured by the uniqueness theorem. So, given a differential equation we can make out whether the equation has a unique solution provided the function f is Lipschitz continuous with respect to y. That is one of the sufficient conditions. Again remember it is not necessary. There can be a function which does not satisfy Lipschitz continuity, but still thus it can have a unique solution. And we will prove the existence theorem in the coming lectures. So, by various existence theorems we will see.