 Welcome back. Recall that in the previous lecture, we were constructing Green's function for linear second order boundary value problems. So, let me recall what we are trying to do. So, this second order equation t by d t by d t q t equal to f of t in the interval a b and we have taken the boundary conditions y a equal to y b equal to 0. So, p is a continuously differentiable function and a positive function in interval a b q and f are continuous functions. So, let me recall the procedure we adopted to construct the Green's function. So, we started with w 1, w 2, two linearly independent solutions of the homogeneous equation. So, that is f is identically 0 and w 1 a equal to 0. So, one of them satisfies boundary condition at one end and other one at other point a b. So, we also normalized w 1, w 2 so that the Ronskian of w 1, w 2 is equal to 1 by p t and these conditions the Green's function of the boundary value problem was defined by this expression w 1 s, w 2 t if a is less than or equal to s less than or equal to t and other way round is less than s less than and the solution required solution for the boundary value problem the solution y t is given by this integral a to b g t s and we verified that this y given by this integral satisfies both the boundary conditions as well as the differential equation. So, we will illustrate this one through an example simple example. So, example very simple one. So, y double dot minus y. So, in the interval 0 less than t less than 1 and again I take the boundary conditions. So, y vanishing at both the end points 0 and 1. So, in this so this is a second order equation with constant coefficients. So, if you study the homogenous equation so e exponential t exponential minus t are linearly independent solutions of the homogenous equation. And we have to construct two linearly independent solutions satisfying boundary conditions at different points and if you just look at the boundary condition here. So, you just w 1 t which is a linear combination of e to the t and e to the minus t and we want that to vanish at 0. So, one that easily found is sinh t. So, this is sign hyperbolic t. So, let me just you all know this thing, but let me just write that and you see that w 1 0 is 0. So, that is a solution of the homogenous equation and it satisfies this boundary condition. And similarly at t equal to 1 you just see that this fellow will do. So, again when you expand this this is a linear combination of e to the t and e to the minus t. Hence it is a solution of the homogenous equation and at t equal to 1 this is 0. So, now let us check whether that normalized or not. So, let us compute the Ronskian. So, Ronskian. So, you just w 1 t w dot t minus w 1 dot t and w 2. So, if you use these expressions and write the derivative this is simply minus c 1 c 2 and you use addition formula for the hyperbolic sign function what you find is sinh t 1. So, that is a constant. So, choose because we want this to be 1 Ronskian to be 1 by p t and p is 1 identically 1 here. So, choose c 1 c 2 such that c 1 c 2 is equal to minus 1 divided by sinh 1 sinh hyperbolic 1 and then the green function in this case. So, g t s is minus sinh s sinh 1 minus t by this normalizing thing sinh 1 if 0 less than s less than equal to t and minus sinh t into sinh hyperbolic 1 minus s divided by t is less than s less than 1 and the solution is just simply why t is equal to 0 to t 0 to 1 g of t s f of s d s. So, if you take a particular f for example, if for example, f is identically equal to 1. So, we can integrate this easily we find and so you can easily check that why t is equal to. So, let me write that 1 minus e to the minus 1 divided by e minus e to the minus 1 this is coming from sinh 1. So, e to the t plus e minus 1 by e e minus 1 e to the minus t and there is a minus 1. So, you can verify this. So, this satisfies the equation y double dot minus y equal to 1 with boundary conditions y 0 equal to y 1 equal to 0. So, as I remarked earlier so if we again change the boundary conditions the solution w 1, w 2 they change and accordingly the green function changes. So, green function is very much dependent not only on the equation, but also on the boundary condition. So, that you should keep in mind and we have a nice recipe in this case to construct the solutions of the boundary value problem via the green function. So, we will discuss the uniqueness little later. So, we have found a solution. The next question obviously is that whether it is unique or not. So, now we go to general second order equations and see how far we can carry whatever we have done in linear case. So, second order linear equation. So, let the second order general equation. So, I consider this non-linear equation. So, y double dot equal to f of t y. So, again this is in the interval a less than t less than b and now I take general linear boundary conditions. So, not able to do much with general boundary conditions. We consider so boundary conditions b c. So, a 0 y a minus a 1. So, there is no particular reason why I take minus, can also put plus no problem. So, a 1 y dot a is equal to alpha b 0 y b plus b 1 y dot b is equal to beta. So, in the previous case what we had was a 0 equal to 1 and a 1 equal to 0 and alpha equal to 0 and similarly b 0 equal to 1 b 1 equal to 0 and beta is 0. So, only the condition is that this a 0 a 1 should not vanish simultaneously. So, that we will put here and similarly other condition. So, they do not vanish simultaneously. One of them can be 0 and alpha beta are again given real numbers. So, now we will discuss this. So, let me again put some numbers here for this is 1 a and the boundary conditions 1 b and the whole thing together called problem 1. So, would like to discuss the existence of a solution to this problem 1. That means, we want a function y to be determined which is 2 times continuously differentiable and satisfies this given differential equation and these boundary conditions at the end points a and b. See unlike in the linear case we do not even know how to start because here there is no concept of the homogeneous equation because everything is clubbed here. So, I cannot separate the linear part and inhomogeneous part homogeneous part and inhomogeneous part and how to start. So, that is. So, we are already facing a problem we want to start way to start. A formal approach is the following. So, we start with an initial value problem start with I B P. So, this is B B P. So, this is we want to discuss and we start with an initial value problem. What is the initial value problem equation is the same. So, now, I use different unknown function. So, u double dot equal to f of t u dot. So, this is second order equation and now I impose conditions only at a. So, I will not bring in b because we are studying just some initial value problem. So, at one point we give the data what is that data. So, you just a 0 u a which is same as the condition at a in the B B P. And now I will have another independent condition at a only e 1 u dot a I put n s s and this s is at my choice. So, this. So, s is any real number for the time being and these two conditions are linearly independent linearly independent. What does that mean? You cannot obtain one condition from the other and in terms of the coefficients this means. So, that means the initial conditions independent that is this matrix you have a 0 a 0 minus a 1 c 0 minus c 1 is non singular that means you have this a 1 c 0 minus a 0 c 1 not 0. That is the meaning of that linearly the conditions are linearly independent. So, by fixing. So, since the second condition is at our choice by fixing c 0 c 1 we may assume. So, this is kind of normalization that a 1 c 0 minus a 0 c 1 is equal to 1. So, then you just forget it and we have that insularly problem. So, let me just again. So, this I V P. So, this let me call it. So, this is 2 a and suppose I V P has a solution defined. Suppose the I V P has a solution defined a less than or equal to t less than or equal to b. So, you want the solution to exist all the way up to b and perhaps one beyond b, but at least we want up to b. So, here we have to. So, we have to include we have to impose certain conditions to ensure this like global leafshedness and other things. So, we have now the solution for the. So, let me just write again u double dot equal to f of t u u dot and we have this a 0 u a minus a 1 u dot a equal to alpha and c 0 u a minus c 1 u dot a equal to s. s is at our freedom and we have normalized. So, that this a 1 c 0 minus a 0 c 1 is 1. So, these two conditions are linearly independent and we have a solution which exists for the entire interval a b. So, to emphasize the dependence on s. So, denote the solution by u t and you stress this s and s is coming from the second condition we are imposing at a and now you form this. So, consider the function phi of s. Now, I define this function. So, which is b 0 u b. So, u is the solution of the initial value problem. So, remember that plus b 1 u dot b s minus beta. So, if you remember the second condition boundary condition. So, it is of that form. So, I am just evaluating the solution of the initial value problem at t equal to b. So, this is nothing but the evaluation of the initial value solution of the initial value problem at t equal to b. And now this is a purely a function of s because everything b 0 b and beta and you now everything is fixed. So, this is just a function free from r to r. So, now that you have got a function defined from r to r. So, for every s I will have solution to the initial value problem and once I obtain that solution of the initial value problem I compute this thing and call it phi of s. So, now ask the question. So, is there an s star such that phi of s star is 0 that is the question. So, remember this phi is a function from r to r. So, this is no guarantee that there is a root of that thing. So, in t there is if yes then the solution then the function y t. So, this is the definition. So, u s star is a solution of. So, this you remember this is solution of I v p. So, this we already know and certain conditions on the right hand side function f we always have a solution at this point uniqueness and other thing will not discuss. So, just there exist a solution and in addition to that if there is an s star satisfying this condition which is a root of the function phi then using that solution of the initial value problem we obtain a solution of the boundary value problem. So, in the literature this is called this is called shooting method and this is widely used in obtaining numerical solutions of the BVP. So, this is very important method and same thing can also be used for theoretical purposes as we are studying. So, if there are more solutions to this equation if there are more rows to this phi then we obtain as many a solutions for the boundary value problem. So, in general we cannot say that there is there could be no solution there could be only one solution there could be multiple solution and depending on that the boundary value problem will have a solution a unique solution or multiple solutions or no solutions. So, in order to sum up so just let me state a theorem. So, let me just put B the region consisting of all triplets t u 2. So, a is less than equal to t less than equal to b and u 1 u 2 in R. So, this is the region and assume. So, this is our right hand side m f. So, this is a function of t u 1 u 2 is globally lift is that is if we take f of t u 1 u 2 minus f of t v 1 v 2 is less than or equal to some m 1 u 1 minus v 1 plus m 2 u 2 minus v 2 for all t u 1 u 2 t v 1 v 2 in the region. So, let me not write that for all in R. So, then d v p has as many solutions as the as many solutions as the roots of the function pi. So, for every root of this function pi you get a solution of the boundary value problem. So, remember that what pi is. So, this one somewhat very strong condition and that ensures. So, this is just for the ensures that a solution of I v p exists for all t. So, if you have some other means of ensuring the same thing we do not need this strong condition. So, in fact, the let me give an example. So, an example somewhat difficult integration, but this doable. So, this is again second order equation. So, you let me just write y y double dot plus lambda e to the u is equal to 0 in 0 less than t less than 1 and y 0 equal to 0. Let us consider this b v p. So, if you follow this shooting method. So, what we should do is this consider the I v p u double dot plus lambda equal to 0 and one initial condition is same as the one coming from the b v p and second one we want linearly independent. So, it has to be in u dot equal to s, s is at our choice. So, as an exercise write down the solution in explicit. So, this can be done. So, this no problem with that. So, in fact, you can use the conservative nature of the equation. So, you can easily integrate once and you convert that into a first order equation and then using method of you separate the variables and integrate it. So, it is bit complicated, but doable. So, you can explicitly solve it and in the solution you see that the dependence of s very clearly in that. Now, you will check using that solution, you should using that solution check for which s, which s if there is any is not it is not guaranteed if there is any if there is any the solution u also satisfies the other boundary condition namely u of 1 equal to 0. And if and also you check check for the uniqueness whether there is one s or more s check for uniqueness. So, it is a good exercise. So, you will see how this shooting method is at work. So, why this example? So, I why I want to this example is this in this case this f of t u u dot very very simple it is just minus lambda u if you want to write it. And this is not globally lift is not globally, but yet the solution exists for all t in the interval 0 1. So, that is important. So, there are I mean that is one one was to state as a theorem one has to be very stringent. So, maybe you put lots of conditions, but in practice such stringent conditions may not be required. So, for example, this one. So, even without the the assumption of globally lift check you can write down the solution explicitly and you see that the solution exists for the entire interval 0 1. So, this is one example. And so, let me again continue why this shooting method. So, what is happening shooting method? So, geometric picture. So, here. So, this is a this is b. So, for let me just for take for example, simple thing u a equal to 0 u b equal to 0 just for the illustration. So, what we are doing is so starting here, but with a velocity. So, u dot a is equal to s u dot a is equal to s. So, we are really we are shooting some particle from here and it may just go somewhere depending on s at b. And if I choose another s, it may go somewhere. So, would like to determine an s for which we eventually go there. So, that is why it is called shooting method. So, by changing the initial velocity in this particular setup of boundary conditions we are looking for a particular velocity with which if we shoot we are reaching the destination namely u b equal to 0. So, that is the idea. So, that is why it is called shooting method. So, again let me go back to the analytical description. So, we started with let me again go back to y dot equal to f t y y dot and a 0 plus b 1 y dot b is equal to beta. So, this is the b b p call it 1 a 1 b and the corresponding initial value problem. So, u double dot equal to f of t u u dot and a 0 u a minus a 1 u dot a is equal to alpha and c 0 u a minus c 1 u equal to s and we are normalizing. So, a 1 c 0 minus a 0 c 1 is 1. So, normalizing that and we are calling the solution you stretching the importance and the dependence of s. And then we found this function coming from the second boundary condition b 0 u b s plus b 1 u dot b s minus beta and whenever there is a root of. So, if phi s star is 0 then this y t defined by. So, this is definition u of t s star solves b b p y and if there are more roots then we will also have more solutions for the b b problem and this follows from uniqueness of the initial value problem. So, the important thing is this. So, we are stated a theorem using the global leafshedness of f. So, we have existence and uniqueness. So, later on we also need continuous dependence on the solution we are going to differentiate with respect to s and as of now. So, existence and uniqueness results of i v p are giving us the solutions of the boundary value problem provided this function phi has roots. If this has no roots then by this method we are not able to get produce any solutions of the b v p. So, that is. So, you can imagine if you go to higher order equations or even first order systems. So, this phi say remember here the phi is from r to r and soon we will state a result where existence of roots is guaranteed conditions on phi. So, if you go to higher order equations or more general first order systems of first order equations then this phi will be a mapping from some Euclidean space to some other Euclidean space and the study of its zeroes existence of its zeroes becomes more complicated and that is where you need the tools from non-linear analysis. In this one dimensional case we can do all that thing just using one dimensional calculus and that is what we are going to do. So, let me just state a result and we will discuss more detail next time. So, just I will just write a theorem in addition to the hypothesis hypothesis of global liftage etcetera. So, we have also several conditions on the coefficients etcetera. Assume so this is some extra assumptions on the non-linear right hand side. So, del f by so remember f of t u 1 u 2. So, that is the right hand function. So, del f 0 and del f by u 2 bounded. So, a 0 a 1 non-negative product b 0 b 1 non-negative and a 0 and b 0 do not vanish simultaneously. So, this is the condition. So, this means a 0 a 1 are of the same sign and similarly b 0 b 1 and a 0 b 0 do not vanish simultaneously. So, this is the condition. Then b v p has a unique solution. So, we will prove this theorem because it is quite interesting bit technical, but it uses only one dimensional calculus. So, the idea of the proof is let me just state that idea of the proof. So, look at this phi again from r to r and show that phi has a unique root that is phi of s star and there is no other s for which phi vanishes. So, even in one dimensional case we have of course several conditions. So, one of the conditions to have such a result in one dimensional case you to show that this phi is a function of s. So, you show that d phi by d s next time we will do that d phi by d s is bounded away from 0. So, this is one sufficient condition under which this is true and what we are aiming at these conditions additional conditions on f and also on the coefficients. Ensure this that is what we use one dimensional calculus using this hypothesis to show that this d phi by d s is bounded away from 0 and then the calculus lemma will give us the unique root satisfying this thing and that will prove that b v p has a unique solution. So, we will see next time. Thank you.