 Welcome back. Today, we will start a new topic. So, in this chapter, we study two point boundary value problems. So, for short, we will denote it by b v p. So, in many important applications concerning differential equations, the data on the unknown function is given not at just one point, but many points. And we now discuss, restrict our discussion to two point boundary value problems. So, in contrast with initial value problems, which we have already studied. So, the data on the unknown function governing a differential equation are given just at one point, that is the initial value problem. And this is, when it is more than one point, the data is given at more than one point, then it is referred to as two point or more point boundary value problems. So, our discussion is restricted to two point boundary value problems. And let me start with some examples. So, just to see what are the difficulties, what are the issues. The first example is again u double dot plus u equal to 0. And again recall dot represents d by dt differentiation with respect to t variable, that is time variable and u the unknown function. So, this is second order equation. And for the initial value problem, we give data of u and u dot at one single point. Now, we are interested in two point boundary value problems. So, data is given at two points. So, for example, say u 0 is equal to 0 and u b, b is another point, say it is given as beta. So, we have to find a solution to the differential equation in this interval 0 b, 0 b, which satisfy the differential equation u double dot plus u equal to 0. And in addition to that, it also satisfies these two boundary conditions at 0 and b. So, since this is a linear problem, we know that the general solution is given by… So, that we have already learnt. So, u of t is c 1 sin t plus c 2 cos t for this linear equation. This is general solution. And if we want this to satisfy these two boundary conditions, so let us start with at u equal to t equal to 0. So, u 0 equal to 0 implies that you substitute the value t equal to 0 in this general solution and require that u 0 equal to 0, then you see that c 2 is 0. So, then this cos t term is not there. So, we have just therefore, u t is equal to… I will just now replace c 1 by just c sin t. So, c is a constant. And now, you have to determine that constant c is possible in order to satisfy the second boundary condition u b. So, let us calculate that thing. So, beta is equal to u b, this is equal to c sin b. Now, you see already some difficulties here. For example, if b is equal to n pi, then sin b is sin n pi and that is 0. n is an integer, in this case positive integer does not matter. Therefore, if beta is not 0, there is no solution. So, you just one observation when b equal to n pi, n is a positive integer, then we will not have a solution if beta is not equal to 0. So, that the other the boundary condition at t equal to b is not satisfied in this case. On the other hand, if b is not n pi, then sin b is not 0. And then, choosing c is equal to beta by sin b, we have a unique solution. You have a unique solution for the boundary value problem. So, even in the simplest case of a second order equation with constant coefficients, we may have a solution or we may not have a solution. So, let us continue with one more example. So, this is minus u double dot equal to f t, 0 less than t less than 1. And now, again we have 2.0 and 1. So, instead of u, I will give the data u double u dot 0 is equal to gamma 0 minus u dot 0 is equal to gamma 1. We will see why I am putting minus 1 in a minute gamma 1. So, this represents steady state heat conduction, heat flow in a rod. Physically that is the… So, rod is placed in the interval 0, 1, these two end points and these represent the heat flux. So, that is one physical interpretation. Now, let us… And f is the force interval, f is given to us. And now, let us integrate the differential equation. So, we have integral 0 to 1 minus u double dot t d t is equal to 0 to 1 f t d t. This is from the differential equation. And the left hand side, just we can integrate it. And just you find that this is minus u dot plus u dot 0 and that is gamma 1 plus gamma 1. So, hence we have this equality between the forcing term and the boundary conditions. So, necessarily the forcing term and the boundary conditions should satisfy this. Let me write it. So, 0 to 1 f t d t gamma 1 plus gamma 0. So, now, this relation will tell us which boundary conditions and which forcing terms are possible and which are not to have a solution to the BBB. So, for example, if f is identically equal to 1 and gamma 1 equal to gamma 0 is 0. So, more generally gamma 1 plus gamma 0 is 0. Then, this condition is not satisfied. Then, BBB has no solution because this condition is violated and the left side you get 1 and the right hand side you get 0. So, that is violated. So, there are no solutions. So, on the other hand, so if I keep the boundary condition same, now I take f t is equal to sin 2 pi x gamma 1 equal to gamma 0 is 0. Then, certainly that relation is satisfied. Then, the relation is satisfied and we have many solutions. The solution again you can just take this f t equal to sin 2 pi t and solve the differential equation and you see that you have this 1 by 4 pi square. So, you can check that this is satisfied. The equation in terms so satisfies the boundary conditions. So, there is no problem there and A is arbitrary constant. So, again you see the solution is not unique. In one case, there is no solution and in another case, we have infinitely many solutions. So, these are the typical problems we face in the study of two point boundary value problems. So, with this let me now discuss a general theory first for linear equations. So, how to obtain a solution for a linear second order equation satisfying boundary conditions. So, here so for simplicity, so let me just consider y double dot plus alpha t y dot plus beta t y is equal to g of t. So, this is in A less than t less than b and for simplicity, let me take the boundary conditions as y A equal to 0 y B equal to 2. So, this is the problem here. So, we have this interval A B. So, we are to find a solution for this equation satisfying these two boundary conditions. So, we want a C 2 solution y is a C 2 function, it is twice continuously differential function satisfying this differential equation and these boundary conditions at the A and B. So, while dealing with linear equation, we have lots of tools already developed. So, we are going to make use of that. So, the theory for linear equation is somewhat easier. So, let me start with this. So, let me put some numbers. So, this is equation 1, this is equation 2. So, this alpha beta g are continuous functions. So, some minimum hypothesis defined on this closed interval A B. So, what would like to view this one as. So, if alpha beta are fixed. So, fixing alpha beta. So, what would like to do in analogy with first order equation? So, first order equation it was just one integration was involved and similar thing we expect even for the second order equation. So, there is this integration box. So, we will see what that integration is and here the input g and comes out solution, solution y. That is what would like to view this boundary value problem 1 and 2. So, we will simplify little bit, we will rewrite 1. So, this is for some simplification, rewrite 1 as this d by d t of some p t y dot plus q t y is equal to m. So, there is some advantage in this thing. So, let us exploit that thing. So, this is I call it 3. So, what is p? p is a positive function and p is also c 1 on A B. So, it is quite continuously differential function which is positive that is important. So, very easy to see that 1 implies 3. So, first 3 implies 1 that is easier part. So, 3 implies 1. So, if you expand this by this differentiating this product, you see that alpha is p dot by p and beta is q t by p and g is alpha. And for the converse, you have to just find what p is and p comes from this. So, put p t is equal to exponential A to t alpha s. And since this is an exponential, so p is always positive and it is differentiable because alpha is continuous and we are integrating it. So, that will produce a c 1 function. So, you can verify and with this p, you can calculate what q is and what f is. So, now we restrict this our attention to this 3 and again recall the boundary conditions y a they are not changed y a equal to j. So, this is remember that. So, let. So, a procedure. So, let u 1, u 2 be 2 linearly independent solutions of homogeneous equation 3. So, that is in 3 you take f 0 and then you and this existence of these two linearly independent solutions come from general theory of linear equations and we are sure that they are there. Then by using variation of constants formula, a general solution of 3 is given by. So, one part consists of solution of the homogeneous equation and that is a linear combination of u 1 and u 2. And the other one is coming from the particular integral and let me just write that. So, this is A to t f s W s. So, check this one from a bit complicated, but straight forward. So, u 1 s u 2 t minus u 1 t u 2 s d s. So, this is for the homogeneous equation and this is for the inhomogeneous part. So, this is general solution of 3. Now, we want this y to satisfy the boundary conditions y a equal to y b equal to 0. And when t is equal to a, this integral is no more there and one that integral is there when t equal to b. So, let us write those two equations. So, therefore, a u 1 a plus b u 2 a equal to 0 and a u 1 b plus b u 2 b is equal to a to b. So, let me not write the whole thing. So, you just copy from the previous line. Only thing is I am taking the other side. So, there will be a sign change. So, a b are constants. Now, the a and b satisfy these two equations linear equations and we hope to solve them provided this determinant is non-zero. The coefficient of this coefficient matrix, the determinant of the coefficient matrix is not 0. So, assuming here the determinant is just this one u 1 u 2 b minus u 1 b u 2 a not 0 we can solve a b. So, now you write a and b from and now you say and uniquely. So, find a b substitute. So, let us go back and see what this one. So, let us call this some equation I have put 3 there. So, let me put 4. So, after solving for a and b from those equations you have solved now a b from these two linear equations in form. So, I will let me just write the end result. So, it is bit involved, but very straight forward. So, let me just write that b f s. I forgot to tell what w s is. Let me complete that thing b u 2 s minus u 1 s u 2 b. You also check that I might have done some mistake, but you can verify yourself these are just linear equations. So, there is no problem in verification and this is where that determinant comes u 1 a u 2 b minus u 1 b u 2 a d s and as usual there is other part there is no change there. So, let me just a t and let me just put as it is. So, what is w s? Let me just say what w s is. So, w s r to w t in the Ronskian to add t and since both these they are linearly independent solutions of the homogeneous equation. So, this is by definition u 1 t u 2 dot t u 1 dot t and since they satisfy the same homogeneous equation you can easily check that this is actually given by. So, this we already done earlier for generally linear systems. So, it is constant by p t and remember we have assumed p t is bigger than 0. And looking at this expression let us go back again. So, this complicated let me call it 5 this complicated expression. So, the first integral is a to b and second one is only from a to t. So, we can split this integral into two parts. So, y t is something a to t plus t to b. So, remember this because we are going to imitate this. So, looking at this expressions in the integrand. So, it is convenient to. So, looking at the integrand integrands in 5 it is convenient to introduce the following. Introduce the following functions. So, namely w 1 t is equal to u 1 a u 1 t u 2 a and w 2 t at the other end this is at one end t equal to a and this one is u 1 b u 2 t minus u 1 t. And w 1 and w 2 being linear combinations of u 1 and u 2 these are just constants u 1 a u 2 a they are just constants here. So, they are linear combinations of u 1 and u 2. So, therefore, w 1 w 2 are also solutions of the homogeneous equation 3. So, right hand side is 0 and moreover if you look at these expressions. So, moreover and this is the advantage we take now moreover w 1 a is equal to 0 and w 2 b is 0. So, they are the solutions of the homogeneous equation and each one of satisfying one boundary condition. So, w 1 satisfy the boundary condition at a and w 2 satisfy the boundary condition at p. And now you just forget whatever calculations we did and start afresh from this w 1 w 2 that is just to get a motivation how to obtain this w 1 w 2. So, now let us start with so afresh. So, let w 1 w 2 be two linearly independent solutions of the homogeneous equation 3. We have to remember that homogeneous equation 3 satisfying w 1 a equal to 0 and w 2 b equal to 0. So, if you have a more general condition this is what we have to do. You have to find two linearly independent solutions of the homogeneous equation 3 satisfying one of them satisfies boundary condition at one point and another one at another point. So, for simplicity we have taken this simple boundary condition here that namely y a equal to y b equal to 0 and w 1 satisfy the boundary condition at 1 point a and w 2 satisfy the boundary condition at b. Now, define so g of t s it has two parts. So, let me draw a picture later. So, this is w 1 s w 2 t if a is less than or equal to s less than equal to t and other way around w 1 t w 2 s t is less than s less than equal to. So, if you take the square a b. So, this is the line t equal to s in one portion this is s less than equal to t and here t is less than that is a. So, this is the first portion and that is what is happening and define and put and little comment later just look at the definition of g. So, g has two parts one in a t. So, this integral splits into two parts a to t and t to p similar to the one we did for t and we expect this y 2 satisfy the differential equation and also the boundary condition. And this is the integral box we are talking at the beginning and. So, see this integral box is just manufactured with the help of these two linearly independent solution nothing else and that comes from the just homogenous equation and now we put this input f the right hand side and we obtain corresponding solution. So, this is what we have obtained and now we will verify that y t indeed satisfies the differential equation 3 and also the boundary conditions. So, this is the in the next 10 minutes we will do that thing. So, so cling under suitable normalization we will see what this normalization is y satisfies d e 3 differential equation 3 and boundary conditions. So, it is very easy to verify. So, remember y t is given by. So, let me just write that. So, a w 1 is w 2 t f s d s plus t 2 b w 1 t w 2 s f s d s. Just remember this every time we have to do that thing and looking at this expression it is very easy to see that. So, clearly y a y a y 0 because w 1 a 0 a 0 and y b 0 because w 2 b 0. You see the way we have chosen the two linearly independent solutions of the homogenous equation they are helping us in satisfying the boundary conditions. And since we are assuming w 1 w 2 are also linearly independent. So, since w 1 w 2 are linearly independent their Ranskian is this we already seen is constant divided by p t and the normalization. So, we normalize. So, normalize means you just multiply w 1 w 2 by constants normalize w 1 w 2. So, that the Ranskian is 1 by p t. So, we want this constant to be 1 and that can always be achieved by multiplication by some constants. So, there is no problem there. So, this is the normalization we are talking about. So, now we directly verify that y satisfies 3. So, again let me recall what that 3 is the differential equation. So, it is d by d t of p t d y by d t plus q t y is equal to f t and w 1 w 2 satisfy the homogenous equation. So, that means d d t p t w i dot d by d w i by d t plus q t w i equal to 0. So, i equal to. So, remember that they satisfy this. They are linearly independent and their Ranskian is just 1 by p t. So, that also you remember. Now, y is given by that again let me not write it. So, just go back here. So, y is given by this. Remember that thing and now we will just get this. So, differentiate. So, you have to be bit careful there because the integral limits also involve p. So, you have to just be bit careful there. If you already done it differentiate y. So, y is given by. So, let me just write it. So, a 2 t plus t 2 p recall that. So, d y by d t is equal to. So, let me just write it a 2 t w 1 s w 2 dot t because t is only here f of s d s. And since this t is there in the limit. So, you have to also differentiate that. So, that we get. So, w 1 t w 2 t f t and if you do for the other integral similarly. So, now there is t in w 1. So, w 1 dot t w 2 s f s d s and now t is in the lower limit. So, get a minus here. So, w 1 t w t and this and this cancels. So, and now you are looking at equation 3. Now, multiply by p t. So, therefore, p t d y by d t is equal to a 2 t. Now, just p t just goes inside there that is all. So, there is no problem there that is it. And now again we differentiate this. So, d by d t p t d y by d t and this is similar to what we did in the previous step. So, just we have now a 2 t. So, w 1 s d by d t of p t d w 2 by d t f as it is. And now because of the presence of t in the integral limit. So, we have this p t w 1 t w 2 dot t f t and similarly from the other integral t to b. Now, it is d by d t of p t w 1 dot w 2 s f s d s. And now I have minus p t w 1 dot t w 2 t and f t. And remember w 1 w 2 they satisfy the homogeneous equation 3 and this is. So, simply this let me write here this is minus q t w 1 and similarly that one is minus q t w 2 t. So, and q t that nothing to do with the integration variable s. So, that q t just simply comes out and what is remaining is just minus q t and if you look at the expression this is just y of t. So, the integrals are taken care of now that portion is coming here and now about what about this one. Now, you just simplify them. So, you get p t w 1 t w dot t minus w 1 dot t w 2 t f t. And this nothing but the Ranskian of w 1 w 2 and we have normalized. So, that the Ranskian is equal to 1 by p t and this p t p t goes away. So, you have just plus f t. So, indeed. So, therefore, y indeed satisfies equation 3. So, I already checked that this y has already satisfied already and already seen. So, therefore, therefore, y is indeed a solution. So, in fact we have just remaining 5 minutes let me just say what we have done. So, we have started with just let me recall. So, w 1 w 2 w 1 w 2 are two linearly independent solutions of homogeneous equation 3 with. So, we have normalized Ranskian is equal to 1 by p t also. So, w 1 a equal to 0 and w 2 b equal to 0. So, if there are more general boundary conditions. So, you have to choose w 1 w 2 appropriately and then we define this g. So, it has a name. So, this is w 1 s w 2 t if a less than or equal to s less than or equal to p and other way around. So, g is called Green's function bound to value problem. So, it is very much dependent on the boundary values that is very important. So, if you change the boundary values if you take more general thing then the green function also changes and this g then gives us the solution of the boundary value problem. So, next time we will consider some examples of construction of this green functions and you are seeing that we are taking the advantage of the linear system and existence of linearly independent solutions and more generally the fundamental matrix if you are dealing with a first order linear system and that is going to help us in construction of the green function and which will eventually give solutions of our boundary value problem. Thank you.