 Welcome back. So, in the last class we were discussing boundary value problems for general second order equations. So, this is y double dot equal to f of t. So, this is the equation and the boundary condition. So, this is in a less than t less than b a 0 u a minus a 1 u dot a is equal to alpha b 0 u b plus b 1 u dot b equal to beta alpha beta are given real numbers and. So, this is 2 a boundary conditions 1 b. The condition was this a 0 and a 1 do not vanish simultaneously and similarly b 0 and b 1 do not vanish simultaneously. A formal approach to a solution of this boundary value problem 1 a and 1 b totally called 1 was through an initial value problem discussed this last time u dot equal to u double dot equal to f t u u dot call it 2 a and now we impose conditions on u and u dot only at t equal to a. So, this the first one is sorry for the this thing. So, this is not u this is y by t y dot u a minus a 1 u dot a equal to alpha. So, there is no change in the first condition and now we want a since its second order equation want to impose one more initial condition and this we want it to be independent of the first one. So, this is minus c 1 u dot a equal to s. S is a parameter at our choice and again a real number and this is 2 b and by independence what I mean is this a 1 c 0 we have fixed c 0 and c 1 such that this is 1. So, let u be a solution of I v p in the interval a b. So, last time we stated a theorem where some sufficient conditions were imposed on the right hand side function f. So, that the solution of the initial value problem exists for all t in this interval and with this u we form this real valued function p of s remember the second condition on u depends on s. So, we form this b 0. So, u we emphasize the dependence on s. So, we say that u at t and f s is in the dependence on s. So, this is u b s plus b 1 u dot b s minus beta. So, just notice that this phi of s is nothing, but the second boundary condition. So, if there exists. So, this is the conclusion we do last time. So, if there exists s star such that p of s star is 0 then y given by y of t is equal to u of t s star is a solution of d v t and this is called shooting method. So, in this approach we attack this bound value problem through an insular problem and this is a big if. So, this in general for every s star satisfying this phi s star equal to 0 that means the roots of the function phi we get a solution of the boundary value problem. So, last time I stated a uniqueness result. So, let me just again recall. So, theorem 1. So, let R be the region this t u 1 u 2 t is in the interval a b and u 1 u 2 in R. This condition looks stronger, but that is sufficient for the solution of the initial value problem to exist in the entire interval a b. So, suppose f is uniformly lifted in u 1 u 2 variables in R. So, that last time I stated that let me just state one more time u 1 u 2 minus f of t v 1 v 2 less than or equal to some m 1 u 1 minus v 1 plus m 2. Then d v p has as many solutions as the roots of. So, for every root of this function real valued function. So, remember this phi is from R to R. So, now we go to the second theorem some uniqueness result theorem 2 in addition to the hypothesis of theorem 1 assume. So, now some smoothness condition on f remember f is this f of t u 1 u 2 function of three variables t and u 1. So, this del f by del u 1 is positive and by del u 2 is bounded this is not norm just absolute value in R that is the reason in stated in theorem 1 and. So, a 0 b 0 a 0 a 1 have same sign and b 0 b 1 have same sign and a 0 and b 0 do not vanish simultaneously. So, I already assume that a 0 and a 1 do not vanish simultaneously and b 0 b 1 do not vanish simultaneously those are standing hypothesis on the boundary conditions. And in any addition to that now we are assuming that a 0 and a 1 have the both both have the same sign b 0 b 1 have the same sign and a 0 b 0 do not vanish simultaneously then the b v p has a unique solution. So, let me sketch a proof of this it is really simple and just use this one variable calculus, but little technical in detail. So, sufficient to show sufficient to show that phi has a unique root. So, if you just recall the conclusion of theorem 1 the b v p will have as many solutions as the roots of phi. So, if we show that phi has a unique root then b v p will have unique root. So, again let me recall. So, recall this phi of s is b 0 u b s plus b 1 u dot b s and u is the solution of the initial value problem. And since phi is a real valued function defined on real line. So, this is this is accomplished. So, we show phi has a unique root by showing that its derivative d phi by d s is bounded away from d phi by d s is bounded away that is d phi by d s bigger than or equal to c positive or d phi by d s less than or equal to minus is strictly less than 0 for all s. So, either it stays all the time all for all s it is for stays positive or it stays negative. So, just as an example if you take e s its derivative e is e to the s the derivative e to the s is bigger than or equal to 0 for all s in r, but it is not bounded away from 0. So, infimum e to the s s in r is 0. So, e to the s is not an example. So, once this is satisfied. So, you can show that. So, this is a simple calculus lemma. So, suppose phi from r to r be such that d phi by d s is bounded away from 0. So, then there exist unique s star such that phi of s star is 0. So, it is a very simple thing. So, it is just it is not hard to see that. Now, coming back to the so proof. So, you have to establish that d phi by d s is bounded away from 0. So, for this we introduce something. So, introduce xi t is d u by d s. So, I am taking partial differential equation derivative with respect to s remember s appears in the initial condition. So, now I am varying that s. So, I am considering u as a function of s initial condition it appears in the initial condition. And now look at the equation again for the u. So, look at recall that. So, u double dot equal to f of t u u dot. So, you know implicitly depends on this s. So, we will differentiate this equation with respect to s. So, differentiate with respect to s. So, this is this is what we learn in general theory the continuous dependence on the of the solution of the initial value problem with respect to initial data. So, if you do that thing then you get xi double dot equal to. So, let me just write it p t xi dot plus q t xi. So, remember this is xi the derivative of the solution of the initial value problem with respect to s. So, what is p? So, where p t just coming from this you are differentiating the right hand side with respect to s. And so you are using just the implicit differentiation. So, this is what you get. So, d f by d u dot. So, now everything let me write once then it is understood and q t is just t does not depend on s. So, do not have to worry about that and same thing. So, this is called variation equation. So, this plays an important role now variation equation. And remember why this xi is important if you go look at the expression for phi, phi we are analyzing this phi phi is nothing but u b s b 0 plus b 1 u dot b s. And we are trying to show the derivative of phi is bounded away from 0. So, what is the derivative of phi? Phi is nothing but this derivative of u with respect to s and that is precisely xi. So, that is why this role of xi is important. So, now you see that from hypothesis this q is positive. By hypothesis q is positive and p is bounded. So, we are going to make use of this. So, let me again rewrite here. So, xi double dot equal to p t xi dot. So, you keep on recalling this. So, this let me call it 3 a and similarly if you now differentiate the initial conditions to a what you get is xi of a is equal to you do that thing. So, let me just write what you get is by differentiating initial conditions to a with respect to s. So, everything with respect to s we obtain a 0 xi a minus a 1 xi dot a this is 0 because this is alpha there. So, alpha does not depend on s. So, that is we get that one. So, c 0 xi a minus c 1 xi dot a and there is an s here. So, when I differentiate with respect to s we will get 1 and by our choice of c 0 you immediately see that xi a if you solve these two equations you get a 1 and xi dot a is equal to a 0. So, let me call this as 3 b. Now, just concentrate on this. So, remember a 0 a 1 they have the same sign and they do not vanish simultaneously. So, let us fix. So, assume. So, the other case is similar assume a 0 is non negative and a 1 is non negative and look at the conditions 3 b. So, if a 1 is positive then xi a is positive and this implies xi t is positive in a small labour rule in this interval a 2 a epsilon. So, if a 1 is positive there is no problem there and if a 1 is 0 then a 0 is positive by our choice and look at the second condition xi dot a equal to a 0 a 0 is positive. So, xi is again increasing. So, again that implies xi t is in some a 2 a epsilon. This epsilon may be different. So, this may be different. What we are trying to say here is xi remains positive with our this assumption important. If we assume the both are negative then we will have negative thing here, but just for definiteness let us fix the signs of a 0 and a 1. So, that is what we get. So, therefore, xi t is positive for t in a a epsilon for some epsilon positive. So, claim. So, this is again very simple calculus argument xi t is positive. For all t in a epsilon. So, up to b the xi remains positive. So, only at a it may be 0 or it may be positive. So, that is the claim. So, once we have this thing it is the proof follows very easily and this is also very simple argument. So, suppose not a calculus argument suppose not. So, what does that mean? So, there exists some t star in a b such that xi t star is less than or equal to 0. So, may be better draw some diagram. So, there is a here, there is t star there, and there is b here. So, we already know that xi is positive. So, positive in some small neighborhood that is important is positive there. So, it may be 0 here and then it comes. So, looking at the picture. So, therefore, xi has a positive maximum it could be just relative maximum no problem in a t star. And we want to rule out this a t star is already ruled out because xi t star is less than or equal to 0. So, it cannot be a positive maximum. So, we want to rule out this positive maximum at a. So, how to do that thing? So, if a 0 is positive then xi dot a is a 0 positive. So, positive maximum cannot occur at a. And if a 0 is 0 then look at xi double dot a from the equation. So, this is go back to the equation. So, xi double dot a is a 0 positive. So, this is a 0 positive is equal to p t xi dot a plus q t xi a. And now we are assuming a 0 is 0. So, that means, this is 0 and this is q t xi a and xi a is a 1. So, this is strictly positive. So, and at a maximum the second derivative cannot be positive it can only be less than or equal to 0. So, therefore, again the positive maximum cannot occur at a. So, that means, what the positive maximum occurs at an interior point t 0. Let us go back. So, therefore, the positive maximum occurs at t call it t 0 or t 1 or t 1 in a t star. So, at an interior point. So, this is an interior it is not either a or t star. So, then we know that when a maximum occurs at an interior point. So, therefore, we have this xi of t 1 is strictly positive xi dot t 1 is 0 because it is a maximum. And one more condition this xi double dot t 1 is less than I put this is from calculus. But again you go back to the equation variation equation. So, xi dot t 1 is equal to p t xi dot p t 1 xi dot t 1 plus q t 1 xi t 1 and xi t 1 is and again this is 0 and this is positive. So, whole thing is positive, but at a maximum xi dot xi double dot t 1 is negative. So, this is a contradiction. So, therefore, our claim is true. So, therefore, what is our claim xi t is positive for all t up to b. So, that is important. So, this is the first step and the remaining steps are very straight forward. So, now, look at again go back to the look at the variation equation xi double dot equal to p t xi dot plus q t xi. So, everything at t. So, let me just write that. And just now, we have proved that this xi t is positive up to b and q is positive by assuming that this is a contradiction. So, this one is positive for all t in a. So, therefore, we have this inequality we can put greater than or just. And fortunately, we can integrate an inequality. We cannot differentiate an inequality, but we can certainly integrate an inequality. So, integrate this and preserving the sign of inequality integrate. You will not do directly. So, you have to multiply by exponential 0 to t. Let me write p s multiplied by this and then this is the integrating factor. So, multiply and integrate. So, what you obtain is. So, xi dot t is bigger than a 0. This is coming from the condition xi double dot a dot a equal to not 0 here a. Our point is this is exponential p s t s not p s. So, let me use some other variable may be eta. So, this is true for all t in a. And now, this integrate one more time. So, integrate one more time to obtain. So, finally, we have xi t is bigger than a 1 plus a 0. So, there is a double integral. So, let me write it first and then. So, a 2 t let me write a 1 plus a 1 plus a 1 plus a 1 plus a 1 plus a 2 d eta and then exponential as it is a 2 eta p eta 1. So, there is a double integral here. Now, you use the hypothesis. So, use the hypothesis mod p is less than equal to m. So, we are just using one sided. So, p is bigger than equal to minus m. m is some positive constant. And then if you plug in this hypothesis on p. So, we obtain xi dot t is bigger than a 0 e to the minus m t minus a and xi t bigger than a 1 plus a 0. So, we have to do some integration there. Let me just write it e to the m t minus a divided by m. In the second inequality there are double integral. So, if you do that double integral, you get this. And this is valid for all t in a v. So, remember again. So, you have to just remember that you have to do some integration. So, you have to do some integration. So, you have to do some integration. So, you have to remember that thing. So, recall phi s is b 0 u b s plus b 1 u dot b s. So, therefore, d phi by d s is just b 0 d u by d s at b s plus b 1 d u dot by d s at b s. And by our notation this is nothing, but xi of b s plus b 1 xi dot at b s. And our aim is to show that d phi by d s is bounded away from 0 that you remember that. So, now, we have got the expressions for xi and xi dot for all t. So, in particular for t equal to b and let us see whether they are bounded away from 0. So, that is our next task. So, if a 0. So, look again let me go back there. Just look at this, look at here. So, both xi dot b is greater than or equal to 0 and xi b since there is a 1 there and a 0 and a 1 are not simultaneously 0 xi b is always bounded away from 0. There is no problem about xi b. This is always bounded away from 0 and this in general not that is just I want to. So, note that xi b is let me write y b let me write t is bounded away from 0 for all t in a b. Of course, we are just interested in the final point b. The same is true for xi dot t if a 0 is strictly positive. So, the trouble occurs when a 0 is 0 and now again look at the conditions. So, if a 0 is 0 then b 0 is not 0. So, this is one of the hypothesis. In the second theorem we want this a 0 b 0 should not vanish simultaneously and also b 0 also b 0 b 1 have the same sign. If a 0 is 0 b 0 is not 0 and b 0 and b 1 have the same sign and again now look at phi s recall this b 0 u b s plus b 1. So, our only concern now is what happens if a 0 is 0. So, if a 0 is 0 then b 0 is not 0 and b 0 and b 1 they have same sign. So, if b 1 is 0 that is no problem if b 1 is not 0 b 1 and b 0 have same sign. And we are already seen that this derivative of u with respect to s is bounded away from 0 and this is just infimum could be 0, but in this case we have b 0 not 0. So, therefore, again d phi by d s d phi by d s is bounded away 0 even when a 0 is 0. So, let me again recall this. So, if a 0 is 0 by our hypothesis b 0 is not 0 and b 0 and b 1 have the same sign and that is important because the b 0 and b 1 they let me there is a beta here that does not matter. So, there is a beta there b 0 and when I take derivative with respect to s that beta anyhow goes away. So, important that b 0 and b 1 have the same sign and b 0 is not 0. So, when I take the derivative it again bound I already know that I already know that xi xi is bounded away from 0 I already know that I already know that the only problem with respect to xi dot xi dot and that is now taken care of by this additional hypothesis. So, remember this is additional hypothesis. So, when a 0 is 0 we required that b 0 different from 0 and we also required that b 0 b 1 have the same sign and that assures that d phi by d s is again bounded away from 0 even when a 0 0. So, therefore, d phi by d s bounded away from 0 in all the cases and this completes the problem. So, this very simple calculus one dimensional calculus tools very nice proof. Of course, one can always question this whether they are necessary or they can be weakened and that is we can always look into the literature and see if weaker hypothesis are possible. So, again let me recall this is the one. So, again last time I stated this example hope you have worked out. So, again let me recall that. So, this y double dot plus lambda E y equal to 0. So, y 0 equal to 0 y 1 equal to 0. So, this lambda is positive. So, I approach this boundary value problem via this initial value. So, this is b v p. So, this is I v p. So, u dot u double dot plus lambda E u equal to 0. So, you put one this initial condition same thing and now this is an independent initial condition I put a parameter s. So, last time I mentioned that this can be solved explicitly a solution of I v p can be found in explicit form. So, do it. So, it is similar to the one we did in case of conservative equations. So, this is also in conservative equation explicit form. So, using that explicit form obviously, it contains this parameter s. So, next question is is there an s star. So, you call this solution as u t s is there an s star such that u at 1 s star is 0 because that is the condition we want at t equal to 1. So, that is the question. So, if s if s then y of t is equal to u of t s star is a solution of b v. So, do this thing and also do this and also discuss uniqueness. So, why I am stating this example is that this e to the u in this e to the u is not globally lift. So, this condition is not really necessary. What we want is this solution to the initial value problem to exist for the interval in question. In this situation we have just 0 1. So, we are satisfied if a solution of the initial value problem exists for all t in this interval. We are not bothered whether it blows up outside that thing. So, there are I mean the possibilities of weaker hypothesis. So, with that thing I come to an end of this simple discussion on boundary value problems. So, we first discussed the linear boundary value problems in which we have the advantage of fundamental solutions from the linear system. But in the case of non-linear system our approach was through this IVP and you can now refer literature and even see more general problems involving this boundary values. With that brief discussion on the boundary value problems for second order equations we come to the end of this course on ordinary differential equations. We hope that you these listeners have enjoyed the course. The topics of the course were mainly based on the syllabus of ODE courses in most of our universities and other institutions. As you might have noticed we have omitted some important topics such as regular and singular storm value theory and corresponding Eigen value problem or spectral problems and also systems with periodic coefficients and the corresponding flow k theory. At the same time we have added as a trade off from discussion on qualitative theory of autonomous systems and shooting method for the boundary value problem of a general second order equations. The qualitative theory for autonomous systems is an important topic not only in physical and engineering sciences, but also in mathematical biology especially mathematical epidemiology where many problems have been modeled as first order autonomous ODE's. The subject field of ODE is quite vast and it is not at all possible to include all the topics in a single course to satisfy the various users of the subject. We however hope that those who have followed the course meticulously will be able to read and understand the cited differences further topics of their interest with our energy levels permitting and time permitting. We certainly hope to return with another course on ODE in future comprising of storm level theory, flow k theory and some advanced topics in qualitative theory and perhaps some topics in non-autonomous systems also. We would like to certainly hear from you regarding the contents and usefulness of this course. Your constructive criticisms are always welcome. Thank you.