 Welcome to tutorial on initial boundary value problems for wave equation. In lecture 4.9 we have solved an initial boundary value problem with the Dirichlet boundary conditions. In this tutorial we consider some more problems where we will change the boundary condition to mixed boundary conditions, we change the domain from a finite interval to a semi-infinite interval and so on. So we are going to solve with 3 to 4 problems in this tutorial. The first problem is reduce the following initial boundary value problem with nonzero Dirichlet boundary conditions here u of 0 t is g t, u of l t is h t. This is the problem in the last class we considered with g equal to 0 h equal to 0. At that time we mentioned that this problem can be reduced to an i b v p with 0 Dirichlet boundary conditions. So the problem is how to do it that is what we are going to discuss. So trick is to find a function v such that it satisfies the boundary conditions v of 0 t is g t and v of l t is h t. Suppose you find such a function then if you will define a function w which is u minus v okay where u is a solution that we want to find and suppose we have found a v satisfying these conditions and if you look at u minus v what problem is w solves? w solves w t t minus c square w xx is equal to minus d'Alembertian acting on v because d'Alembertian acting on u is 0 because we want to solve homogeneous wave equation therefore minus d'Alembertian v d'Alembertian is a linear operator therefore it distributes over u minus v. Of course we have a new term but this is known term because if you know the function v you know the d'Alembertian of v. So it is a known function. So we have got a wave equation which we started with had no source terms but now we have a source terms and the initial displacement w of x 0 is u x 0 minus v x 0 but u x 0 should be phi x. So phi x minus v x 0. Similarly the initial velocity w by dou t of x 0 is psi x minus dou v by dou t at x 0. v is a known function so this is a known function this is also a known function and the source term is also a known function and we have 0 boundary conditions the Dirichlet boundary conditions have become 0 because u satisfies u should satisfy g u of 0 t is g but v of 0 t is already g therefore g minus g will be 0. So w of 0 t is 0 similarly w of l t if you plug in here u l t minus v l t u l t is supposed to be g h and v l t is h by construction therefore the difference is 0. So w l t is 0 ok. Now question is how do you find such a function v? So we must interpolate v the function v that should interpolate these 2 functions g t and h t. At x equal to 0 it should be g t and x equal to l it should be h t. So the simplest function we can think of is this is also called a linear interpolation if you are familiar with terminology in numerical analysis you will immediately follow this why it is called interpolation. So now look at v x t when I put x equal to 0 this term is not there because x is 0 what I get is g t. When I put x equal to l l by l is 1 so it is h minus g and a plus g therefore we get h. So this v satisfies what is required and thus we have converted our problem of initial boundary value problem to another initial boundary value problem with source term and the initial displacement and velocity have changed because of this v and the boundary conditions became 0 boundary conditions that is the advantage. So let us look at the problem 2 in lecture 4.9 using first principles we solved an initial boundary value problem with Dirichlet boundary conditions. Now the problem 2 that we are going to discuss now is about solving an IBVP with mixed boundary conditions we are still considering we are going to still consider finite interval 0 l in the lecture 4.9 we consider u 0 t and u l t being prescribed here we consider derivative of u is prescribed at one of the boundaries and the function itself is prescribed at the other boundary. Once again we want to solve this using first principles. So given phi which is c 2 0 1 psi which is c 1 0 1 find a solution to the homogeneous wave equation and initial conditions are as usual initial displacement is phi initial velocity is psi. Now comes to the boundary condition mixed boundary conditions because we are using derivative dou u by dou x at x equal to 0 that is given to be 0 and u is given at 1 x equal to 1 u of 1 t is given of course is given as 0. So 0 boundary conditions but the nature of the boundary conditions have changed derivative on the one boundary and the function on the other boundary is prescribed. Now what is solution from first principles the idea is that u of x t is given by f of x minus c t plus g of x plus c t. So the plan is to find what these functions are f and g of course we want to solve the initial boundary value problem which is given by phi and psi therefore f and g we want to get an expression in terms of phi and psi. Of course we have to use initial and boundary conditions for that and find out the compatibility condition that phi and psi must satisfy so that the f that we construct and the g that we construct are actually c 2 functions thereby u of x t is actually a classical solution to the given i b v p. Of course that is left as an exercise to you. We have done exactly a same exercise with different boundary conditions that is all we have analyzed what should be the compatibility conditions in that problem. Now it is similar you will see the even the wave that we are going to solve the problem is similar. So you should be able to do this exercise please do that. Let us start from this formula u x t equal to f of x minus c t plus g of x plus c t. The same computation that we did for Dirichlet boundary conditions exactly same conditions will come if you are looking at the what happens to f and g are from the initial conditions. How much of f and g are determined by initial conditions? We get the same expression I am not doing this computation because we have done this already twice. First of all we have done in deriving the Dallambert formula for the Cauchy problem on R. Secondly we have done in lecture 4.9 where we solved i b v p with Dirichlet boundary conditions. So exactly same thing you get the same expressions for f and g remember the validity. It is valid for xi between 0 and 1 and here eta between 0 and 1. So when we substitute in this formula we get the solution in terms of x and t which is this. Valid for xi between 0 and 1 xi is x minus c t and eta between 0 and 1 which is x plus c t. So in terms of x t the validity is this. So initial conditions determine the solution in some region and that region is the 00 region. If the 00 region in the diamond picture in fact the diamond picture is exactly same as what we saw in lecture 4.9 it is described like this 0 less than or equal to xi less than or equal to 1, 0 less than or equal to eta less than or equal to 1 in terms of xi eta in terms of x and t it is described as x minus c t less than or equal to 1 and x plus c t less than or equal to 1. Now let us draw the picture here this is the diamond picture we are talking about so on. So this is xi equal to 0 is xi equal to minus 1 these are the lines and this line is eta equal to 1 this line is eta equal to 2 and let us draw I mean exactly same picture we are going to write x and t this is how it goes the picture this is a line x minus this line is x minus c t is equal to 0 this line is x minus c t is equal to minus 1 this line is x plus c t is equal to 1 this is x plus c t is equal to 2 and so on and this is the region 00 this is the region 00. Now what do we get if we use boundary condition 1 now we have analyzed the information that we get from initial conditions so let us see what we will get if we use boundary condition 1 what is the boundary condition 1 that is dou u by dou x at 0 t equal to 0 this is the condition given to us. Now if we use this in this formula what is u x of 0 t it is f dash at minus c t I am differentiating and then substituting x equal to 0 so chain rule this equal to 0 so thus we have what do we have is f prime of minus zeta plus g prime of zeta equal to 0 for what zeta greater than or equal to 0 we have to always remember for what range of zeta this equation is valid. Now this I will write it as d by d zeta of minus f of minus zeta plus g of zeta equal to 0 if you differentiate with respect to zeta we exactly get this f prime of minus zeta and 1 minus sign will come that will make it plus so this is same as that now on integrating from 0 to zeta the above equation of course so it is a derivative equal to 0 so fundamental theorem calculus that will give you minus f of minus zeta plus g of zeta plus f of 0 minus g of 0 equal to 0 and we have given some reasons why we drop certain constants like this in both in the derivation of D'Alembert formula and also in the IBVP with Dirichlet boundary conditions in lecture 4.9. For the same reasons which I will not repeat we can drop that and what remains is f of minus zeta is equal to g of zeta valid for zeta greater than or equal to 0. So this is what we get from using the boundary condition 1 what does this mean f is determined see g is known in interval 0, 1 therefore this formula will give us there is a minus sign here this formula will give us the values of f on minus 1, 0 using values of g on which interval 0, 1 so this is the information we get from boundary condition 1. Let us see what we get from boundary condition 2 what is the second boundary condition it is u of 1 comma t equal to 0. So substituting the formula for u we get f of 1 minus ct plus g of 1 plus ct equal to 0 valid for all t greater than or equal to 0 here also t greater than or equal to 0. So this means that f of 1 minus zeta plus g of 1 plus zeta equal to 0 zeta greater than or equal to 0. So that implies g of eta it means I am setting eta equal to 1 plus zeta just changing the names of the variables because now I am going to define g using the values of f on some interval. So when eta equal to 1 plus zeta what is minus zeta 1 minus eta. So therefore g of 1 plus zeta is minus f of 1 minus zeta but what is minus zeta 1 minus zeta so that is equal to minus f of 2 minus eta. Of course valid for what eta greater than or equal to 1 if you notice here 1 plus zeta zeta is always greater than or equal to 0 therefore 1 plus zeta is always greater than or equal to 1 and I am replacing 1 plus zeta with eta therefore eta is always greater than or equal to 1. So what do I have g of eta is equal to minus f of 2 minus eta when eta is greater than or equal to 1 this is the information coming from second boundary condition. So let us write down the information that we got from both of the both initial and boundary conditions. So what did it say first one is initial conditions it gave us f g are known on this interval 0 1 second thing is the bc 1 boundary condition 1 that gave us f values. So f known on minus 1 0 because it is expressed in terms of values of g on 0 1 that is why we got this. Now the last one which we got bc 2 f known on minus 1 comma 1 implied that g is known on 1 comma 3 and importantly f g satisfy some relations what is that f g satisfy f of minus zeta equal to g of zeta zeta is greater than or equal to 0 and g of eta is equal to minus f of 2 minus eta eta greater than or equal to 1. So let us call this as a call this as b now let us write one consequence of this let us call it consequence 1. So for xi less than or equal to minus 1 f of xi is equal to g of minus xi this is by a because if xi is less than or equal to minus 1 xi is negative and for negative things we know by a you can get this relation. But now xi is less than or equal to minus 1 means minus xi is greater than or equal to 1 when somebody is greater than or equal to 1 b is applicable. So g of anybody is minus f of 2 minus that so 2 minus eta but eta is minus xi here. So it is 2 plus xi. So what we have let us write down f xi is equal to minus f of 2 plus xi valid for xi less than or equal to minus 1. So this is a conclusion we get. So now f is known on minus 1 1 already this above relation will mean that f is known on minus infinity comma 1. So let us look at the second consequence. So let us briefly recall again what was a f of minus zeta equal to g of zeta if zeta is greater than or equal to 0 and b was g of eta equal to minus f of 2 minus eta of course eta is greater than or equal to 1. Now let eta is to be greater than or equal to 2 then what happens g of eta is equal to minus f of 2 minus eta by b is applicable. Now 2 minus eta is less than or equal to 0 because eta is bigger than or equal to 2. Therefore by a I get this equal to minus g of eta minus 2 by a. So therefore what we get g is known okay let us summarize what we get here g of eta is equal to minus g of eta minus 2 valid for eta greater than or equal to 2. From here it follows that g is known on 0 2 implies g is known on 0 infinity. As you remember from the last lecture and as well as from this picture this from this where it came from okay this is how things were right. Now f of xi is what is required and xi starts from 0 minus 1 and so on and of course this is xi equal to 1. So we need f values for xi's which are less than or equal to 1 and g's which are this is eta equal to 0 eta equal to 1 and so on. So we actually need the values of g from 0 to infinity and values of f from minus infinity 1 and that we have achieved by these are the two consequences that because of that we know the values of f and g they are determined. Now let us look at the solution in the region 1 0 let us briefly let us draw the picture. So this is the region 1 0 region is this we want to solve inside this. So x t belongs to region 1 0 what does that mean it says something about x minus t and x plus t this is x minus t equal to 0 this is x minus t equal to 1 it lies between that. So minus 1 less than or equal to x minus t less than 0 this is x minus t equal to 0 this is x minus t equal to minus 1 and what about x plus t this is x plus t equal to let us use blue color this is x plus t equal to 0 this is x plus t equal to 1. So this is the meaning of x t belongs to region 1 0 now what is u of x t it is f of x minus t by definition that we started with. Now x minus t is between minus 1 and 0 therefore f of x minus t x minus t is negative and x minus t is negative and we have determined the values f of minus zeta is g of zeta. So this is nothing but g of t minus x this stays as it is because x plus t is between 0 and 1 so g of g is known in 0 1. Now let us substitute the expressions for g that we know f and g in this case only g is relevant on 0 1 we know that expression for g using that what we get is phi of t minus x by 2 plus 1 by 2 1 by 2 into 0 to t minus x psi as ds this is g of t minus x g of x plus t is phi of x plus t by 2 plus 1 by 2 0 to x plus t psi as ds. Let us club the like terms so this is equal to phi of t minus x plus phi of x plus t by 2 plus 1 by 2 0 to t minus x psi plus 0 to x plus t psi psi of s ds. So let us analyze what is there in this brackets the integral in the brackets and we will then come back to this so we have determined the solution. Now we would like to express it as D'Alembert form that is why we would like to do a little more work. So let us look at these integrals that we have 0 to x plus t psi s ds 0 to t minus x psi plus t minus x 2 x plus t psi we can always write this. So what I am using here is something from the calculus we know a to b f is always equal to a to c f plus c to b f whether or not c belongs to the interval a b does not matter only thing is f should be defined in such intervals then we can always do this. So with that I get this expression therefore this term now becomes this I will substitute with this and then see what we get. So what we get is 0 to t minus x psi is also coming one more time here so 2 times. So let me write that what we get is phi of u of x t equal to phi of t minus x plus phi of t plus x by 2 plus 1 by 2 into 2 times integral 0 to t minus x psi s ds plus t minus x 2 x plus t psi s ds is what we get. Now here I would like to extend phi so let me write phi x and then I would like to have x minus t here t plus x is in 0 1 there is no need to extend so I just simply use phi here by 2 plus 1 by 2 and here what I would like to have is x minus t to x plus t psi extended then this is in the dilumbered form. Now if you look at this first term this suggests that we must define phi as an odd function in some interval no no sorry even function because phi of t minus x I want phi of x minus t no minus n here therefore even I want to define as even. So what we do is a small picture we will draw here 0 so we have x minus t is here in this region 1 comma 0 so minus 1 is here x minus t is here and then t minus x is here and then 1 is here and here we have x plus t this distance is same as this distance because x minus t and t minus x are equidistance from 0. So what we do now is that phi x we want to define on minus 1 0 whatever is needed only we do. So phi x of any x is equal to phi of minus x because minus x will be in 0 1 interval and there we know phi already it is given there so phi of minus x similarly psi x on this as psi x of x equal to psi of minus x. If we do this this term is taken care we got this equality now we have to worry about this why these two integrals put together is equal to this in this integral. So what is this 2 times t minus x 0 to t minus x psi as ds this is nothing but x minus t to x plus t psi extended because an even function and this interval x minus t x minus t comma x plus t is symmetric about 0 it becomes 2 times that therefore the one which is here is precisely integral x minus t to x plus t here psi extension now this is not x minus 2 to x plus t it is actually t minus x. So this is t minus x t minus x that is equal to this 2 times this and what is next is t minus x to x plus t. So if you combine you get this so therefore we have got the Dallambert form also. Let us solve the problem in the region 1 comma 1 so x t belongs to region 1 1 that means that minus 1 is less than equal to x minus t is less than 0 and 1 less than x plus t less than or equal to 2. So therefore u of x t is equal to f of x minus t plus g of x plus t because x minus t is between minus 1 and 0 f of zeta will be g of minus zeta so therefore it is g of t minus x now x plus t is between 1 and 2 and when you are more than 1 there is a formula for g in terms of f which is minus f of 2 minus x minus t this is what we have. So what did we use here we use 2 formulas what are they f of minus zeta equal to g of zeta if zeta is greater than or equal to 0 and g of eta equal to minus f of 2 minus eta if eta is greater than or equal to 1 we use this. Now using the expression for g we get u of x t equal to phi of t minus x by 2 plus 1 by 2 integral 0 to t minus x psi s ds this is the first term. Second one f of 2 minus x minus t that will give us phi of 2 minus x minus t by 2 plus 1 by 2 integral 0 to 2 minus x minus t psi s ds this is what we have. So now we need to define some extensions let us do once for all 0 is here minus so between 0 and 1 phi and psi are known. So between minus 1 and 0 so extend functions as even functions extend phi psi as even functions with respect to x equal to 0 in 1 comma 2 here you extend as odd functions extend phi psi as odd functions with respect to x equal to 1 if you do that you get this formula u x t is equal to phi extended of x minus t plus phi extended of x plus t by 2 plus 1 by 2 x minus t to x plus t psi extended of s ds this is what we have. In this part is clear so we just discuss why this equal to sum of these 2 integrals we are going to discuss that. So x minus t to x plus t of psi extended is x minus t okay before that maybe it is worth drawing a picture 0 here x minus t is here of course minus 1 is here this is symmetrically placed here this length is same as this length and then we have a 1 here and we have x plus t here and here we have 2 minus x minus t this distance is same as this distance okay. So x minus t to t minus x psi extended plus t minus x to 2 minus x minus t of psi no extended because we are in the interval 0 1 therefore psi extended coincides with psi ds plus 2 minus x minus t into x plus t psi extended. Now here the first integral is 0 sorry the first integral is actually twice the integral from 0 to t minus x because psi extended function psi was extended as a even function due to even extension that is why now this term is 0 because psi is odd about odd function about x equal to 1 that is how we have extended to the functions to the right side of 1 has odd functions up to this interval 2 therefore that is 0 and this is an interval which is symmetric about 1 that is what we saw the distances are same from 1 and therefore integral is 0. So this is the reason why we have the Dallambert form formula further questions related to problem 2 obtain solution in the region Mn we have obtained only in the regions 1 0 and 1 1 so get a formula for Mn also and find conditions on phi and psi which ensures that the solution obtain is a classical solution to the IBVP and express the solution in the Dallambert form we have done this in the two regions that we considered do it for a general region Mn as well. Now let us look at the third problem 3a because the same problem has two three parts so as a first part what we are going to do is we are going to consider IBVP on a semi infinite interval in lecture 4.9 we said this is much more simpler than a bounded interval we will see why it is much more simpler. So given functions phi in C2 of 0 infinity and psi which is C1 of 0 infinity find a solution to the homogeneous wave equation and initial conditions as before the initial displacement is a phi and initial velocity is psi now there is only one boundary because the domain is 0, infinity there is a boundary only at x equal to 0. So the boundary condition is again once again we consider Dirichlet boundary condition u of 0 t equal to 0. Now proceeding as in the last problem we note that the initial conditions determine f and g only on this interval 0 infinity no surprise again it so these are the expressions for f and g exactly the same as before. So substituting in this formula we have this expression which is a Dallambert form of the solution and it is valid in this region x minus t is greater than or equal to 0 x plus t is greater than or equal to 0 in this region. So this region let us see how it looks. So now what we have is picture is like this x here t here x minus t C is playing a role this is x minus t equal to 0 here x is always greater than t here x is always less than t. So we have determined the region x minus t greater than or equal to 0 and x plus t greater than or equal to 0 actually is this region let us call it as 1. So in the region 1 initial conditions determine the solution. So what remains to do is to find the solution in the region 2 the region 1 looks like the 1 0 that we had earlier. So essentially there are only 2 regions now whereas if you are in a finite interval case it had in many infinitely many regions. Let us see what is the information that we get from the boundary condition. So we have only one boundary condition which is u of 0 t equal to 0 t greater than or equal to 0. So what we get is f of minus Ct plus g of Ct equal to 0 t greater than or equal to 0. So that means f of xi is equal to minus g of minus xi for all xi less than or equal to 0. So this is the information. So therefore u of x t in region 2 yeah this is a original formula f of x minus t plus g of x plus t but now that becomes because x minus t is negative using this relation this is minus g of t minus x plus g of x plus t. We can substitute the expression for phi or for g in terms of phi and psi that will give us minus phi of t minus x by 2 minus half 0 to t minus x psi s ds plus g of x plus t is phi of x plus t by 2 plus 1 by 2 integral 0 to x plus t psi s ds. So this is nothing but phi extended of x minus t plus phi of x plus t by 2 plus 1 by 2 integral x minus t to x plus t psi extended of s ds. If we want this, if we want u x t equal to this we have to tell how we have to extend. There is no need to extend the other side because this side phi and psi are given. So we have to get only this side and this suggests phi extension of x minus should be equal to minus phi of t minus x that means extend as odd functions extend phi psi as odd functions with respect to 0. So we have this. So please check the validity of this equation. Now a small question which is an exercise understand the solution in region 2 that we obtained in terms of reflections. Something like this I will just indicate but I will not do it. So this is the region 2. So take a point here go like that that is all right I mean then if you come here shift to the other one that will be l in terms of this. Now the question is the solution that we obtained is it a classical solution. So let us write down the solution that we obtained once more what we got is f of psi is equal to half phi of psi minus half 0 to psi psi s ds if psi is greater than or equal to 0 and minus half of phi of minus psi minus half of 0 to minus psi psi s ds if psi is less than 0 and for g g of eta equal to half phi eta plus 1 by 2 integral 0 to eta psi s ds. So is u of xt equal to f of x minus t plus g of x plus t where f and g are given as above a classical solution that is the question or when will it be a classical solution that means we are indirectly asking for conditions on phi and c whether they need to satisfy any conditions extra conditions on phi and c which we call sometimes compatibility conditions we are looking for such things such conditions are necessary or is it automatic when will be a classical solution that is the question. Now let us observe that as far as g is concerned there is no doubt g is c 2 of 0 infinity and we do not require the values of g for negative values. So that is fine this is easy second one is about f f is c 2 of r we would like to have that because x minus c t takes all the values in r. Now f is in c 2 of r that is doubtful at some points at only one point psi equal to 0 in terms of xt it is on the line x equal to t there is some doubt otherwise there is no problem for this function is nicely defined nicely defined only at the interface psi equal to 0 there could be some issues. So we will analyze that on the later okay let us analyze that. So f is continuous continuous at psi equal to 0 if and only if phi of 0 equal to 0 and f is c 1 at psi equal to 0 so basically to conclude this what you have to do is we have a split formula for f so pass to the limit on both sides as psi goes to 0 you get something like phi of 0 by 2 equal to minus phi 0 by 2 some such thing therefore you will get phi 0 equal to 0. So please do the computation the next one I am going to do f is c 1 if and only if limit of the derivative I am going to take directly the derivative and derivative is phi dash of psi by 2 minus half psi psi this limit is equal to limit on the other side psi going to 0 minus phi dash of minus psi by 2 plus half psi of minus psi. So that is if and only if phi dash of 0 by 2 minus half psi 0 equal to phi dash of 0 by 2 plus half of psi of 0. So you see that this cancels and what we get is psi of 0 must be 0. So f is c 1 at psi equal to 0 if and only if psi of 0 is 0 so we got phi of 0 equal to 0 psi of 0 equal to 0. Now we have to still ask is it c 2 that will put one more condition. So f is c 2 at psi equal to 0 if and only if I am taking the secondary derivative in the formula for f on both sides of psi less than 0 and psi greater than 0 and that gives me one side this is the limit of the second derivative this limit should be same as the limit from the other side. So that is if and only if phi double dash at 0 by 2 minus half psi dash of 0 equal to minus phi double dash of 0 by 2 minus half psi dash of 0. So now this goes off therefore even only if phi double dash of 0 equal to 0. So we have got 3 conditions phi of 0 psi of 0 must be 0 no conditions on first derivatives second derivatives which mean 0. So these are the compatibility conditions if they are satisfied then what we have got is indeed a classical solution and we have expression in terms of phi and psi as well as in terms of f f itself is expressed in terms of phi and psi. Let us move on to problem b 3 b using Duhamel principle find u of 1 comma 2 where u is a solution to ut t minus uxx equal to x square t. So we have a source term now initial conditions we are taking as 0 phi is 0 psi is 0 boundary conditions we take Dirichlet boundary condition as before. So we want to solve this problem non-homogeneous wave equation with a source term given by x square t and 0 Cauchy data 0 boundary data we want to solve this we want to use Duhamel principle. For Duhamel principle what we need is a source operator. So what is the source operator definition it is the one which maps psi maps to s psi. So s psi of x comma t is a solution psi should come as initial velocity everything else should be 0. So ut t minus uxx homogeneous wave equation equal to 0 x positive t positive fine and u of x comma 0 equal to 0 x is equal to 0 ut of x comma 0 equal to psi of x is 0 and we have a boundary condition that should be satisfied. So given psi find the solution that is called s psi. So recall that the source operator is well defined if psi should be c 1 function and psi of 0 must be 0 this is part of the compatibility condition there is no compatibility condition and phi because phi is already 0. So it satisfies all the compatibility conditions. So now what is s x s psi of x t what is an expression for this it is 1 by 2 t minus x to x plus t psi s ds if x is less than t that means it is in the region 2 x t is in the region 2 if x t is in the region 1 it is this is in the region 1 this is the expression for s psi. Now what do we need in the Duhamel principle we need we need to find what is the source operator corresponding to f tau. So what we need for the Duhamel s f tau of x t and that is nothing but half integral t minus x to x plus t f of s comma tau ds x less than t and half x minus t to x plus t f of s tau ds x greater than or equal to t this is s f tau then u the solution at x t the non-homogeneous equation solution is given as a superposition of these s f tau s f tau x t minus tau t tau and what we want to compute is u of 1 comma 2 that is what we are asked to find. So therefore t equal to 2 0 2 s f tau x equal to 1 t equal to 2 so that is what it is what is s f tau 1 comma 2 minus tau you can substitute and get that formula s f tau 1 comma 2 minus tau is equal to half t minus x that is 2 minus tau minus 1 there is a 1 minus 2 minus tau minus 1 is 1 minus tau this is 3 minus tau f of s tau ds this happens if 1 is less than 2 minus tau that is tau is less than 1 and other one is half integral tau minus 1 to 3 minus tau of f of s tau ds this is for 1 greater than or equal to 2 minus tau which is tau greater than or equal to 1. So therefore u of 1 comma 2 is equal to it is an integral from 0 to 2 but we are now split that into 0 to 1 plus 1 to 1 to 2 0 to 1 half 1 minus tau 2 3 minus tau our f of x t is x square t therefore the integrand is s square tau ds and then d tau plus integral from 1 to 2 of 1 by 2 tau minus 1 to 3 minus tau s square tau ds d tau. Let us call this term as A the first term and this as B and we will compute them separately what is A and B. So A is half I have brought out half to the front integral 0 to 1 then tau is here then 1 minus tau to 3 minus tau s square ds then d tau. That is nothing but s square integral will be a cube by 3 that 3 comes out and it becomes 1 by 6 into 0 to 1 tau into 3 minus tau cube minus 1 minus tau cube d tau. This after computation becomes 13 by 2 therefore 13 by 12 that is what it becomes. Now the B you can compute B and that value comes out to be 23 by 30 it is a matter of integration so please do it therefore u of 1 comma 2 is 13 by 12 plus 23 by 13 on simplification this becomes 1 1 1 by 60. So this is the answer we have done this. Let us look at the problem 3 c here we need to find u of 1 2 for the same non-homogeneous equation but there is a Cauchy data here now sin x other one is 0 the initial velocity to richly boundary condition. So this we solve using superposition principle again. Now before we do anything let us note that phi x equal to sin x right this is the initial displacement this satisfies the required compatibility conditions that we found so that we have a classical solution and psi is 0 therefore it satisfies psi of 0 equal to 0. So compatibility conditions are satisfied we are in classical solution. So solution that we obtain is a classical solution. Let V and W solve this is the dilumbaration of V equal to 0 V of x 0 is sin x V t of x 0 is 0 V of 0 t is 0 and W satisfies the non-homogeneous equation x square t x positive t positive and rest of the conditions are 0 conditions W of x 0 W t of x 0 are 0 and W of 0 t is also 0. Then u is equal to V plus W solves the problem that we posed in 3C so it is a given IVVP in problem 3C. So therefore if you want u of 1 comma 2 that is nothing but V of 1 comma 2 plus W of 1 comma 2 but W of 1 comma 2 we already computed. So that is this plus 1 1 1 by 60. So we have to simply compute V of 1 comma 2 in other words we have to compute the solution of the homogeneous wave equation with initial displacement as sin x initial velocity 0 and boundary conditions 0 that we already obtained a formula for the solution. We simply use that formula. So V of 1 comma 2 x equal to 1 t equal to 2 so clearly x is less than t so we are in the region 2 so we have to apply that formula. So that is given by minus phi of 1 plus phi of 3 by 2 psi is 0 so there is no other term and this is nothing but sin 3 minus sin 1 by 2 therefore u of 1 comma 2 is V of 1 comma 2 plus W of 1 comma 2. Okay let us summarize what we did in this tutorial. We starting from first principles we solved an IVVP on a bounded interval with mixed boundary conditions. Starting from first principles again we solved an IVVP on a semi-infinite interval with Dirichlet boundary conditions. Applied Duhamel principle and obtained solution to an IVVP with source terms. Introduced a trick that is a problem 1 which converts an IVVP with non-zero Dirichlet boundary conditions to 0 Dirichlet boundary conditions IVVP. Thank you.