 Welcome back. We have been discussing about the various aspects of existence, uniqueness and stability of initial value problems. Today, in this lecture, we will see the possibility of continuing the solution. So, what we have discussed so far is, given an initial value problems, say d by d x is equal to f of x y and initial condition y at x 0 is y 0 and the existence and uniqueness theorem that gives sufficient conditions on f, say f is f x y is continuous on some domain d in R 2 and f is lipchitz continuous with respect to y on d. Then the existence and uniqueness theorem guarantees that there exist a solution. So, there exist a unique solution y in some interval x minus x 0 is less than or equal to h, where h is a small quantity is a quantity defined by x, h is equal to minimum of a and b by f. What are these a and b? So, here a is defined a and b are the parameters, which we used to define a rectangle inside the domain d. So, rectangle inside the domain d is defined by set of whole x y such that x is less than or equal to say x minus a, x minus a initial point is less than or equal to a, y minus y 0 is less than or equal to b. So, these are the parameters a and b. They are chosen in such a way that this rectangle is inside the domain d and m, where m is a constant, which is the maximum of the modulus of the function value when x y in d or in r. So, the existence and uniqueness theorem guarantees that there exist a solution that solution is defined on an interval x minus x 0 is less than h. If h is a very small quantity, then the solution, which we obtain is very that is defined only on a small, very small interval. If we visualize it, say this is your x 0 point and this is y 0, you are looking for a solution and the solution is now defined some small interval x 0 plus h and x 0 minus h. So, the question, natural question, what about the solution outside this interval to the right side of x 0 plus h and to the left hand side of x 0 minus h. Can we continue? Can we extend the solution towards the right and towards the left? If it is possible, we say the solution can be continued and if it can be continued indefinitely that is for all x, then we say the solution exists for or a solution is defined for all x on the x axis. But if the system is non-linear, then the existence of solution on the all real axis, on the all x axis may not be possible always. But whereas, if the system is linear, then we may be able to extend or continue the solution to the entire x axis. So, in this lecture, we are going to discuss about the possibility of continuing the solution outside the domain of interval which is also the interval which was guaranteed by the existence theorem. So, to get a feel of it, let us consider an example. So, take an example that is a well-known example, which we have already solved. See the example. So, consider a differential equation d x d y by d x is y square is a non-linear differential equation and the initial condition given is y at 1 is minus 1. So, this x y plane, the solution the initial point y at 1, this is point 1, it is minus 1, this is the initial point 1 minus 1. Now, the existence theorem guarantees that there exists an interval on which this equation has a unique solution. So, let us verify these things. So, first of all our f f of x y in this example is y square, which is lipschitz on any bounded rectangle. So, let us define a rectangle r, r is given by set of all x y point in the x y plane such that x minus the initial point x minus 1 is less than equal to a and y minus 1, which is y plus 1, which is less than equal to b. So, obviously, f is lipschitz, lipschitz continuous on this rectangle. If it is lipschitz continuous on this rectangle and obviously, if it is continuous on r, so therefore, the existence and uniqueness theorem applies here and it says there exists a unique solution in a neighborhood x minus x 0, which is x minus 1 is less than equal to h. We are going to try to find what is h compute h such that x minus x is less than equal to h is a domain on which the solution is defined. So, in our case x 0 is x 0 is 1 that is x minus 1 is less than equal to h. So, domain in which solution is guaranteed by existence and uniqueness theorem that is right. So, what is h? h is defined as h is minimum of a and b by m. a and b are as given in the definition of r and m is maximum value of the function f of x y in the rectangle. So, x y belongs to the rectangle and since f is just y square, f x y is just y square. We know this f is y square and that will have maximum value or this point. So, let us load this one. This line is x is equal to x 0 plus h that is x 0 plus h is 1 plus a and this line is y is equal to minus 1 minus b and this line is minus 1 plus b and this line is x is equal to 1 minus a and we know that the maximum value of m is attained on this line. So, maximum value of y square is on this side when y is equal to minus 1 minus b. So, m is minus 1 minus b square which is same as b plus 1 square. So, we are looking for h which is minimum of a and b by m. m we got as b plus 1 to the whole square. So, we want to know which is minimum. So, what is the minimum and maximum? Let us compute what is the value of this. The maximum value of f b if I define f of b is equal to b by b plus 1 square, then f prime b with respect to b is 1 minus b by b plus 1 cube. So, maximum value for f b for b greater than 0 is b is equal to 1. So, maximum is obtained at when b is equal to 1. So, f of 1 the maximum value is f of 1 is 1 by 1 plus 1, 1 by 1 plus 1 square is 1 by 4 and h. So, therefore, h is minimum of a and the maximum value for b is 1 by 4. Therefore, if a is greater than or equal to 1 by 4, if a is greater than or equal to 1 by 4, b by b plus 1 the whole square is less than or equal to a for all b positive. So, this implies that h is equal to b by b plus 1 the whole square which is less than or equal to 1 by 4. So, does not matter what value of a is. So, regardless the value of a of a. On the other hand, if a is strictly less than 1 by 4, if a is strictly less than 1 by 4, then by definition naturally h is strictly less than 4. Thus in any case the a is greater than 1 by 4 or a is less than 1 by 4 in any case the value of h we obtained is 1 by 4. So, there is now for b is equal to 1, a greater than or equal to 1 by 4 gives h is equal to minimum of a comma b by b plus 1 the whole square. So, which is minimum of a 1 by 4 which is equal to 1 by 4. So, thus the best. So, this tells us the best possible value of h value 4 h to have a solution. So, to have a solution is 1 by 4 to have a solution is 1 by 4. So, the tells us by the existence theorem by the existence theorem there exist solution. In fact, unique solution for x minus 1 is less than equal to 1 by 4. So, that the interval is 3 by 4 less than equal to s less than equal to 5 by 4 is the interval on which we have a solution. But this non-linear initial value problem we solved by using analytical method and we found that see if we solve this equation d over d x is equal to y square separating out the variable y square is equal to d x and integrating and this is 1 by minus 1 by y is equal to x plus c or y is equal to minus 1 by x plus c putting the initial condition y at 1 is minus 1 gives us c is equal to 0. So, therefore, the solution general solution which we get by analytical method is y x is equal to minus 1 by x is a solution. But this solution we know that this solution is defined solution is defined for all x between 0 and infinity. We look at the graph of it the solution. So, the given point is 1 this is minus 1 and the interval in which we got the solution is only this. So, 3 by 4 and this is 5 by 4. So, existence theorem guarantees that there is a solution only on this region that the solution is given by this portion. But look at the solution 1 minus 1 by x says the solution is defined between 0 and infinity. So, therefore, the question natural question is can we continue the solution the existence theorem guarantees that there is a solution in the interval 3 by 4 5 by 4 can we extend the solution to the right and also to the left to get a solution in a more large interval. The solution of the initial value problem we want to continue to the right and to the left this continuation this process is known as a continuation of the solution. So, we deal with this one. So, continuation of solution. So, consider the differential equation d y by d x is f of x y and y at x 0 is the initial condition y 0 and existence theorem implies that there exist a unique solution existence and uniqueness theorem implies that there exist a unique solution y on x minus x 0 less than equal to h. So, therefore, the solution exist for all x in the interval x 0 minus h to x 0 plus h. Now, the idea is to continue the solution to right and to left. So, we will deal with the continuation of solution to the right. The similar argument holds for continuation to the left. So, graphically we have the domain d and inside the domain we have a rectangle r and the initial point is inside the rectangle and existence theorem says that the solution exist to right and to left this is x 0 plus h. So, solution exist the point is x 0 and we have a solution. So, this is x 0 y 0. So, existence theorem says there exist a unique solution y let me just call it instead of y it is a solution say phi 0 phi 0 on this interval. So, phi 0 is defined phi 0 x is defined on x 0 minus h and x 0 plus h even at the n point x 0 plus h and if I call. So, let x 1 is equal to x 0 plus h that is this point and the value of phi 0 at x 1 the n point if you call us y 1. So, this point is x 1 y 1. So, x 1 is x 0 plus h and y 1 is the value of the solution at x 1. So, the point x 1 y 1 is a point inside the rectangle and the rectangle is inside the given domain d and the function satisfies all nice properties like continuity and a Lipschitz continuity on the all domain d. So, therefore, the question is now can we start the solution can we take the differential equation with a new initial point. So, consider the initial value problem with new initial condition y at x 1 is y 1. So, therefore, we are trying to start a solution we are going to start a solution from the new point x 1 y 1. Now, again apply the existence theorem that theorem says that we can find we can find another rectangle find another rectangle another rectangle such that x 1 y 1 is a point inside the rectangle and continue the solution or get a solution to a new point. So, this point is now solution on the first interval is we denote by phi 0 with the new point with the new initial point let phi 1 be a solution that is defined which is defined on x 1 less than equal to x less than equal to x 1 plus some h 1 h 1 is a similarly defined a constant for the new rectangle. So, now, because of the existence theorem we could find another solution phi 1 which starts from x 1 y 1 which was n point of the previous solution and it is extended up to x 1 plus h 1 that is a maximum interval guaranteed by the existence theorem. So, we got two solutions phi 0 and phi 1. So, therefore, we have two solutions and if I combine these two solutions into one solution that phi x I defined phi x is a solution which is phi 0 x for x varies from x 0 minus h and its upper limit is x 0 plus h which we denoted by x 1 and the second solution is phi 1 x. So, phi 1 x is defined from x 1 is less than equal to x which is going to x 1 plus some h 1. Now, this combined solution this is now phi is a solution phi is a solution defined on the interval x 0 minus h to x 1 plus h. If we write down the formula for phi 0 phi 0 x is equal to by the basic lemma y 0 plus that was the initial condition plus integral x 0 to x f of t phi 0 t d t and phi 1 x is equal to starts from y 1 plus integral x 1 to x f of t phi 1 t d t. And phi t if you combine these two this is phi x is equal to the solution starting from y 0 and y 0 plus integral x 0 to x f of t phi t d t. So, this phi this phi t is defined by phi x here. So, therefore, we have extended the solution to. So, we thus continued the solution from x 0 minus h to x 0 plus h to a larger interval x 0 minus h to the x 1 plus h. Now, what we can do is we can repeat this process. Now, take x 1 plus h 1 as a new point if I call this point as x 2 x 1 plus h 1 as a new point x 2 and the value of the function phi 1 at x 2 if I call as y 2. So, I got a new initial condition x 2 y 2. Again, if x 2 y 2 is inside the domain and f of x 2 y 2 is defined it has got all properties since f is defined on d and the conditions of the existence theorem are satisfied. So, therefore, you can find an h. So, continue further to get the interval of definition enlarged. So, we continue this process. So, continue this process further. So, to get larger intervals like the end point is x 0 minus h, the right end point is x n plus h n where x n and h n is obtained at each step. So, repeating this indefinitely thus repeating the process indefinitely. So, this repeating this process indefinitely on both sides. So, what I have shown is only on the right hand side the same process can be done on the left hand side on both left and right. We continue to get we continue the solution to successively longer intervals. So, if I call those intervals are a n and b n where x 0 minus h x 0 plus h is the one which we started which we denote by a 0 b 0 and the newer one a 1 b 1 continued on both sides a 1 b 1 then which is a subset of the next one a 2 b 2 and so on to a n b n. And if take the limit of this a n let a is equal to limit a n n goes to infinity and b is equal to limit n goes to infinity b n then we obtain a largest open interval a less than x less than b over which the solution phi d y by d x is equal to f of x y such that the initial condition phi at x 0 is y 0 is defined. So, if you can find such a larger interval a less than x less than b on which the solution of this initial value problem is defined. Now, in this process there are two possibilities. So, two possibilities. So, one is this a is minus infinity and b is plus infinity. So, and in this case we say case solution is defined solution is defined for x is between minus infinity plus infinity. So, this is one of the possibilities that solution can be extended both to right and to left. So, that the solution is defined on the n r x axis. Now, the second possibility is either a is finite b is finite so or both are finite or both are finite. So, a is finite number or b is a finite number or both are finite or one is infinite the other one is finite. So, we remark that even if f is even if f is continuous and this lipschitz continuous on every bounded domain d if f is continuous and lipschitz continuous on every bounded domain d we cannot conclude the case one above. So, f could be continuous and lipschitz continuous on every bounded domain d still we may not be able to conclude that the solution is defined for the n r x axis. Taking an example the same example of d y by d x is y square the same example if we if we choose with an initial condition say y at 1 is minus 1. So, we have seen that the solution exists y x which is minus 1 by x the existence theorem says that the solution exists only in the interval 3 by 5 and 3 by 4 and 5 by 4. But we can now further extend to left and right that the solution is defined between 0 and infinity. So, what we have here is in the above two possibilities a is equal to 0 and b is infinity. So, remember f of x y which is equal to y square is lipschitz every bounded domain every bounded domain. But note that f is not lipschitz on an unbounded domain if f is globally lipschitz then there is a possibility of extending it to the entire the solution can be extended to the entire x axis. But here the non-linear function is not globally lipschitz solely lipschitz on bounded domains. So, if we lipschitz on an unbounded domain then we can expect larger opening interval over which the solution of the initial value problem is defined. So, we state this in the form of a theorem and which we will not prove it the proof follows if we use a successive approximation method theorem. So, let f x y be continuous. Let f x y be continuous in the unbounded domain denote this by d which is equal to set of 4 x y in R 2 such that a is less than x less than b x is bounded between a and b and minus infinity less than y less than infinity. It is an infinite strip where x is varying between a and b and y is varying in the entire y axis on the entire y axis. So, let f of x y satisfy lipschitz continuity lipschitz continuity on d then a solution call it phi solution phi of d y by d x is equal to f of x y with the phi at x 0 is y 0 the initial condition is defined on the entire open interval defined on the entire open interval a less than x less than b. So, if this is the case when the function f is continuous and lipschitz continuous on a given bounded domain is not bounded with respect to x and is lipschitz continuous on an unbounded domain of y axis then the initial value problem has a solution on the entire open interval x between a and b and in particular if a is minus infinity and b is infinity then the solution phi is defined or x that is x is varying between minus infinity plus infinity. So, here the solution exists globally. So, if therefore, if f is lipschitz on an unbounded domain then one can expect a solution on the entire real x axis. So, what we have discussed in this lecture so far is the existence and uniqueness result guarantees a small interval on which a solution exists and the solution is unique and we have seen that the solution can be extended towards the right and also towards the left provided the function f is good smooth. So, if the function is lipschitz continuous on an unbounded so globally lipschitz continuous then the solution can be extended further on an opening interval on which the function is assumed to be continuous on lipschitz otherwise still we have to have a small interval x minus x 0 is less than h. Now, let us look at one example consider an example. So, d over d x is equal to y square the same example at this time the initial condition is y at minus 1 is 1. So, the initial condition is given y at minus 1 is 1. So, y at minus 1 is 1 and we know that the solution is y of x which is equal to minus 1 by x plus c and putting the initial condition we get c is equal to 0 and the solution is y is equal to minus 1 by x and if you look at the interval see that solution exists. So, this solution exists for if you try to continue the solution further see the solution exists for minus 1 to 0. So, at 0 that blows up. So, therefore, the solution cannot be continued beyond this point. So, up to 0 it cannot be extended solution does not exist it cannot be continued to minus 1 to 1 minus 1 to 0 it cannot even include 0. That is natural because it is not violating the previous theorem previous theorem says if the function is lipschitz continuous on the unbounded domains on globally lipschitz continuous, but here if it is not f of x y which is y square is not globally lipschitz it is not globally lipschitz continuous. So, therefore, we cannot expect that the solution is solution can be continued further. So, we stop the discussion of continuation of a solution at this point and now we come back to system of equation come back to initial value problem which are not scalar, but a vector differential equation. So, we spend some time in dealing with the existence of solution existence and uniqueness of solution of vector differential equation. So, vector differential equations or is a system of differential equation or system of differential equations. So, for example, if you have say d x 1 by d t is equal to f 1 of x 1 x 2 and d t d x 2 by d t is equal to f 2 of x 1 x 2 t with initial conditions x 1 at t 0 is x 1 x 0 0 1 say and x 2 at t 0 is x 0 2. See if you have two equations coupled equations then what can we say about the existence and uniqueness of solution of this equation. So, in general we can have an n equations. So, if you have n equations so d x 1 so we will write in a vector form. So, d y d t of x 1 x 2 up to x n which is equal to f 1 x 1 x 2 up to x n t f 2 x 1 x 2 up to x n t and f n x 1 x 2 x 3 up to x n t you have n equations with the initial conditions x 1 at t 0 x 2 at t 0 and so on x n at t 0 is x 0 1 x 0 2 etcetera x 0 n. So, this is where each f i is a function from r n cross r to r n is a function not necessarily linear and if you denote this vector this x y just x x t and all these components this into say f of x t t then this equation can be written in the form d x by d t is equal to f of x t t with initial condition x at t 0 is x 0 x 0 is this quantity this is x 0 and this is x at t 0. So, the question under what condition this initial value problem has a solution and solution is unique and we have the stability of solution etcetera all these things can be similarly studied along the same line. See here the conditions are so our system is d x by d t is equal to f of let me write this t comma x this also a notation conventionally and x at t 0 is x 0. So, for each time would say t 0 t 1 is a time interval then x of t is in r n and f is from f is now r cross r n to r n and x 0 is in r n and we can show that if this function f is Lipschitz continuous with respect to the second argument f is Lipschitz then this implies that there exist a unique solution. So, if the vector value function the vector function f is Lipschitz with respect to the second argument x then this system has a unique solution and the solution can be continued and the solution is stable with respect to the initial condition x 0 and all these things can be studied and analyzed in a same manner as we have done for scalar case. So, with this we finish the existence uniqueness continuation and continuous dependence of solutions of initial value problem. Thank you.