 For wave equation we have so far discussed Cauchy problem. Cauchy problem is posed on rd cross 0 infinity that is x belongs to rd and t is positive. We have considered wave equation in 3 space dimensions namely d equal to 1, 2 and 3. In this lecture we are going to discuss wave equation when it is posed on not necessarily on rd but on a subset of rd. In this lecture we are going to restrict ourselves to d equal to 1. So we consider wave equation which is posed on sub intervals of r because the equation is posed on sub intervals of r they have boundaries and the problem that we are going to consider is what is known as initial boundary value problem and we are going to solve it using first principles in this lecture. So the outline of the lecture is as follows we state the initial boundary value problem with what are called Dirichlet boundary conditions in one space dimension that is in r and then we solve this IBVP using first principles I will explain what the first principle mean here. In the end we would like to express the solution obtained in the form of the Dalambert solution. Recall Dalambert formula gave us a solution to the Cauchy problem in r. Now we would like to have a similar formula even for an IBVP that is the goal here. So IBVP on a bounded interval in r with Dirichlet boundary conditions what is it? It is also called sometimes IBVP for a finite string. Remember the one dimensional wave equation we have modeled as a model for the transverse vibrations of a string. So since the string is finite it is called finite string here 0L that is the string position is 0 to L. So given functions phi which is in C2 on the interval 0L and psi which is C1 on the interval 0L please note the closed intervals here. Find a solution to the homogeneous wave equation it means the IBVP we are going to consider the homogeneous wave equation posed on the domain x in 0L T positive. With the initial conditions the initial displacement is phi x and initial velocity is psi x and Dirichlet boundary condition these are new compared to the Cauchy problem in r that is because we are considering x in the domain 0L. So it has two boundary points namely 0 x equal to 0 and x equal to L and that there we describe U of 0 T equal to 0 and U of L T equal to 0. In other words this is the domain we are considering 0L this is x this is T. Here we want that the wave equation to be satisfied homogeneous wave equation and this is T equal to 0 here we prescribe U x 0 and U T x 0 this is phi of x this is psi of x and here these are the boundary points this is x equal to 0 and x equal to L here we prescribe that U of 0 T is 0 U of L T equal to 0. We have introduced this initial boundary value problem in a tutorial in the lecture 4.3 earlier. So what to do with non-zero Dirichlet boundary conditions? The Dirichlet boundary conditions prescribe the values of the unknown function on the boundary. We are considering 0 Dirichlet boundary conditions an IBVP with non-zero Dirichlet boundary conditions may be transformed to an IBVP with 0 Dirichlet boundary conditions but we may have to pay a cost that will be introduced as source term and also modify the initial speed and velocity initial displacement and velocity. So the PDE would become non-homogeneous but source term of course consists of a known function so that is something good and we will discuss this question in a tutorial what is the question how to transform a problem with non-zero Dirichlet boundary conditions to a problem with 0 Dirichlet boundary conditions. So what about other boundary conditions? So IBVPs with other boundary conditions may also be considered Neumann conditions that is dou U by dou x 0 T and dou U by dou x at L T are prescribed. In Dirichlet we are prescribing U 0 T and U L T instead of U we prescribe dou U by dou x at the boundary points x equal to 0 and x equal to L at the boundary x equal to 0 is actually a line and x equal to L for all times they are prescribed positive times or Robin conditions which is a combination of Dirichlet and Neumann. So alpha times U 0 T plus beta times dou U by dou x at 0 T this is prescribed and alpha U L T minus beta dou U by dou x L T they are prescribed what are alpha beta they are real numbers alpha to be positive. So any mix of these types of boundary conditions may be considered in other words what we are saying is this is the domain that we are considering okay this is 0 this is L here we may have U of 0 T and here we may have dou U by dou x of 0 T of L T we may prescribe this we may also prescribe something like this alpha U 0 T plus beta dou U by dou x L 0 T here and here we prescribe for example U of L T. So any mix of conditions is also possible one can study finding solutions from first principles might become cumbersome for these other boundary conditions not so much for Neumann conditions but for Robin conditions it is going to be cumbersome given any boundary conditions due to linearity of the wave equation one may find a solution to IBVP as a superposition of solutions to simpler problems. So when we are when we have to solve an IBVP we should identify simpler problems out of them and then solve them and then superpose the solutions and that will be a solution thanks to the linearity of the wave equation or wave operator on IBVP for non-homogeneous equations and other domains. So IBVP for non-homogeneous equation may be solved using Duhamel principle exactly as we did for the Cauchy problem. We are studying the IBVP posed on an interval 0 L that is only for convenience one could also consider an interval AB it is similar one may ask what about infinite intervals or semi-infinite intervals that means zero intervals of the type 0 comma infinity that is much more simpler. So that is why we are not studying that instead we are studying a more complicated which is a finite interval 0 L case this is much simpler the methods that we are following in this lecture will apply and solution can be obtained in much simpler manner. So there are two methods that we are going to discuss to solve the initial boundary value problem. One is solution from first principles that is what we are going to discuss in this lecture. The starting point for this approach is the fact that solution to the homogeneous wave equation is of the type f of x minus ct plus g of x plus ct that means it is a superposition of a left moving wave and a right moving wave. Solution by the method of separation of variables that is another method this we are going to discuss in the next lecture and that is a very general more general method it does not assume a specific knowledge of the general solution or the problem of the wave equation. This is a very general method it is applicable for many linear equations we are going to see later on in the course that we will apply this for Laplace equation and also heat equation. So let us start discussion of the solution of IBVP from first principles. The main idea as explained before is that general solution to the equation wave equation homogeneous wave equation is a superposition of a right propagating wave and a left propagating wave that is uxt equal to f of x minus ct plus g of x plus ct. A general classical solution would be exactly this formula where f and g are c2 functions. So what is the action plan for us find expressions for fg of course we are solving IBVP. So therefore naturally the expressions for them will involve the data in the problem namely phi and psi how we get that we have to use the initial and boundary conditions and find out the compatibility conditions that phi and psi must satisfy. It is actually not enough that phi is a c2 of 0l and psi is c1 of 0l as mentioned earlier that is not enough they must satisfy some compatibility condition between them so that the f and g are c2 functions. So what are the goals in this find expressions for the functions f and g obtain a solution to the problem at each point of the strip 0l cross 0 infinity one sided strip. Usually strip is like this infinite thing but no we are starting we are stopping here so obtain solution here 0l cross 0 infinity express f and g in terms of phi psi find conditions under which f and g are c2 functions and finally present the solution in the form of D'Alembert formula. Now let us ask what is the information that we can get on f and g from initial conditions the initial condition ux0 equal phi x gives us fx plus gx equal to phi x always keep writing this domain for which this equality is valid very important if you ignore this we can easily make mistakes we have another initial condition which is ut of x0 equal to psi so on differentiating u of xt we did this earlier exactly the same computation we did this while deriving the D'Alembert formula. So ut dou u by dou t of xt is by chain rule f prime at the point x minus ct into derivative of x minus ct with respect to t that will give us minus c similarly g prime of x plus ct and derivative of x plus ct with respect to t is c so that is what exactly we have here when you put t equal to 0 it reduces to this. So integrating this last equation over the interval 0 to x will give us this expression minus fx plus gx equal to 1 by c integral 0 to x psi as ds minus f0 plus g0. So now we have two equations in fact linear equations for the fx and gx one equation is here one equation is here therefore we can solve for f and g. So solving the system of equations gives us fxi equal to this and g eta equal to this. I am using different notations here xi and eta because we know normally we use xi for x minus ct that is what we are going to substitute later to get a formula uxt and here we are going to put g of x plus ct that is why we are using eta fine. So when you write u of xt equal to f of x minus ct and plus g of x plus ct something will happen from here but this will get cancelled. This is f0 minus g0 by 2 this is exactly minus of that so when you add it will be 0 it is a constant. So finally what do we have initial conditions determine f and g only on the interval 0l f and g are determined only on the intervals for xi in 0l and eta in 0l and we dropped here the constant terms for the reasons that I have explained earlier. So when we substitute in the formula uxt equal to f of x minus ct plus g of x plus ct what we get is the D'Alembert's solution exactly D'Alembert's solution. What is this domain where xi is between 0 and l and eta is between 0 and l that is exactly same as all those xt for which x minus ct is between 0 and l and x plus ct is between 0 and l. So initial conditions determine solution by the D'Alembert's formula but this solution we obtain only for few xt not for all xt in the strip that we wanted to solve for it is only here in this region that means solution is determined in this region which is written here xt in 0l cross 0 infinity such that x minus ct lies between 0 and l and x plus ct lies between 0 and l. We have a figure on the next slide let us look at that and also the interpretation in terms of xi and eta okay xi between 0l eta between 0l look at this picture this is exactly this picture this is this is the region where the D'Alembert's formula gives us solution okay if you look at this this is x minus ct equal to 0 and if you just write a line here that is x minus ct equal to l so this is xi equal to 0 so this is xi equal to l so xi lies between 0 and l similarly eta eta lies between eta equal to l is here this line is eta equal to l and a parallel line here would be eta equal to 0 okay common part of that is precisely this triangular region this triangle that is where we have the D'Alembert's solution so the initial conditions determine the solution in this triangular region so this picture is sometimes called diamond picture strip as a union of diamonds in terms of xi eta coordinates where the characteristic lines are xi equal to constant and eta equal to constant in terms of xt coordinate there x minus ct equal to constant and x plus ct equal to constant observe that the values of f are needed on minus infinity 0 and g are needed on 0 infinity why is that let us look go back to the picture okay this picture we obtain as you see x minus ct equal to 0 x minus ct equal to minus l minus 2l and so on minus 3l and so on therefore the values of f are needed because we have an expression f of x minus ct values are needed only from this l to infinity minus infinity here similarly if you look at g what do we need is here this is x plus ct equal to 0 x plus ct equal to l here x plus ct equal to 2l 3l and so on therefore since we have g of x plus ct we need that g is defined on 0 comma infinity that is all we need so values of f are not needed for positive real numbers other after l and the values of g are not needed for negative real numbers and the various regions in this picture the boundaries are the lines the characteristic lines now if you see the 00 we have not used boundary conditions that means boundary conditions do not influence the solution in the region 00 so fix in x not in 0l the information from the boundary x equal to 0 reaches the point x not at time x not by c example let us draw this we will draw the picture at the end and the information from the boundary x equal to l reaches the point x 0 at time because the distance to the boundary is l minus x not speed is c therefore l minus x not by c is the time taken thus information from neither other boundaries reaches the point x not for all times t which is less than or equal to minimum of both of them minimum of these two times information does not reach x not and the region x 0 consists of such points x t okay this is this is the point 0 this is the point x 0 this is 0 this is l so the distance is x 0 speed is c therefore the time is x not by c okay this distance is l minus x not therefore the time taken is is here l minus x not by c of course in this picture as you see x not is very close to 0 than l that is why this time is less if you are taken a y 0 here this will be the corresponding time y 0 by c and this information from here will be here this time in any case if you take the minimum of the times up to here the information does not reach and this is true of every point in x t okay set of all those x t is this the times and the spatial points they are the ones which are in the region 0 0 now let us see what information we can get using the boundary conditions one of the boundary conditions is u 0 t equal to 0 so what we get is f of minus c t plus g of c t equal to 0 because x equal to 0 we get this equation now I observe here that f c t c is always positive speed t is positive so c t is positive minus c t is negative so f can be made sense for a negative real number provided you know the g for the corresponding positive real number c t what do I know about g g we have already determined in the interval 0 l therefore now f can be given meaning or defined or f is determined in the interval minus l to 0 because g is known on 0 to l that is the idea we are going to use so we write f of zeta equal to minus g of minus zeta for zeta less than equal to this is of course true for every zeta but only thing is that I know the formula for g only when this minus zeta is in 0 l that is that there are two aspects here one this is a relation satisfied by f and g this is true for every negative zeta there is no doubt here this is true but if do you know the expression for f that can be done only when zeta is in minus l to 0 because g is known only in 0 to l now let us use other boundary condition u of l t equal to 0 substituting in the formula for you we get f of l minus c t plus g of l plus c t is 0 for t greater than or equal to 0 so this condition how do we see you have l here this is a point l plus c t that means you travel c t to the right of l and you travel to the left side of l that is l minus c t l plus c t so f at l minus c t plus g at l plus c t is 0 that means f at l minus c t is minus of g at l plus c t see had it been the same function imagine it was some function f of l minus c t plus f of l plus c t equal to 0 imagine this is something different from what we are considering here then this actually means what f of l minus c t is equal to minus of f of l plus c t what does this mean it means that the values at l plus l you travel c t distance to the right side and to the left side the values are tied like this it just means imagine l equal to 0 what is it f of minus c t equal to minus f of c t it means f is odd about 0 so and this condition is what is called f is odd about the point l but here it is not the same f it is different f and g so this will give us f of l plus zeta equal to minus g of l minus zeta for zeta less than or equal to 0 because c t can be any positive real number therefore this zeta can be any negative real number our goal is to find expression for f and g right so these are the information that we got this we got from the one boundary condition this we got from the another boundary condition now let us compute for zeta less than or equal to l f of zeta minus 2 l that is equal to minus g of 2 l minus zeta this is actually minus of the inside thing so I am using the first equation now just this rewriting of the same thing l minus zeta minus l now I am going to use the second equation I need the argument to be negative right this argument is negative yes because zeta is less than or equal to l zeta minus l is negative so minus g of l minus any negative quantity is given by f of l plus that negative quantity okay done by second equation but what is this f of zeta so f of zeta minus 2 l equal to f of zeta for every zeta less than or equal to l what does that mean we have l here okay take any zeta and take zeta minus 2 l the values of capital F are the same it means it is a periodic function on the left side of l so if you know the value of capital F on this interval minus l comma l you know the values everywhere every negative number thanks to this relation okay these follows just from the boundary conditions so if f is known on minus l comma l then f is known on minus infinity to l now for zeta greater than or equal to 0 let us show similar thing about g g of zeta plus 2 l is minus f of minus 2 l minus zeta by first equation and that is equal to minus of f of minus zeta because we already showed the periodicity which is g of zeta therefore if g is known on the interval 0 to l then it is known everywhere 0 to infinity now let us define f on minus l comma 0 using the boundary conditions we have already noted how to define this g is known on 0 l therefore f is known on minus l to 0 by this formula therefore now f is known is determined on minus l comma l therefore on minus infinity comma l now let us do the other one for g g is already known on 0 l let us define it on l comma 2 l so that g will automatically determined due to the periodicity to 0 to infinity now f is known on minus l comma 0 so we are going to use the second condition because this condition is saying I have l here 0 here minus l here I want to define in this region right up to 2 l in this region I want to define therefore we from the second equation we get this g eta equal to minus f of 2 l minus eta for l less than eta less than or equal to 2 l g is known on 0 to l therefore it is known on 0 infinity so please stop here and convince yourself about these relations various relation that we have derived so the information that we get and using both initial and boundary conditions as follows using initial conditions we got this formula for f on the interval 0 l using the boundary conditions we have extended not extended we have determined f on the interval minus l comma 0 this is just the formula for g of minus into minus minus g of minus i and the periodicity on g similarly for initially we have determined on interval 0 l then using the second boundary condition we have got this expression on l comma 2 l and we have periodicity so now we need to write the formula we have determined f and g everywhere so now we are going to write the formula for the solution so take any point x t in 0 l cross 0 infinity now if you take somebody in that diamond picture a point in the diamond picture it is going to lie in some region it is going to some region means what it is going to lie between xi equal to some number and xi equal to that number plus 1 similarly eta equal to a number and eta equal to that number plus 1 so that is what we are setting up the notation here so x minus c t will belong to some interval of this type in fact unique m is unique in n union 0 minus m l comma minus m minus 1 into l similarly eta x plus c t that belongs to an interval of this type for some unique m so this is actually determining the region m n we have marked the region 1 1 1 2 in that picture so a general region m n is characterized by this okay for example I take 2 comma 3 okay this is my m n m is 2 right so it lies between what xi equal to minus l to xi equal to minus 2 l yeah it is here xi equal to minus l to xi equal to minus 2 l what about eta this point I have taken so it is going to lie between here which is cut eta equal to 4 l and eta equal to 3 l okay and this is the region that we get finally so let xi belongs to let x t belongs to 0 l cross 0 infinity let m n be such that xi is in this interval x minus it is in this interval x plus it is in this interval that is the region m n now we need to write f of x minus c t right but as such we have formula for f only on minus l l rest is by periodicity so we are going to use periodicity and bring this value x minus c t into the interval minus l comma l similarly x plus c t we are going to bring it to 0 comma 2 l by some translations so by the periodicity of f f of x minus c t has this formula if m is even it is f of x minus c t plus ml if m is odd it is m minus 1 l I request you to do this computations by yourself similarly for g we have an expression okay so if m is even x minus c t plus ml is in 0 l now we have to be careful because f and g are f is known on minus l comma l formulas are different on minus l to 0 and 0 to l similarly for g formula is different on 0 l and l to 2 l that is why we are making this more finer classification if m is even then x minus c t plus ml is in 0 l if it is odd this is the one which f uses so this will be in minus l 0 and I know the formula for f on that so therefore f of x minus c t is given by f of this if m is even and f of this if m is odd similarly for g we can write down we get this formula now this is something that you must verify by yourself if m is even and n is even because I have to write now u of x t equal to f of x minus c t plus g of x plus c t but the values of f and g depend on whether m and n are even and odd so I have two cases for m and two cases for n so in overall I will have four cases so this is expression for you this is expression in terms of phi and psi m even n odd we get this and this is a final formula this is for m odd and n even note the final formula is not in the Dallambert form because Dallambert wants only phi here but there is a minus here and x minus c t here but something else is here so we will do that later so let us see m odd and n odd again mixing the formula for f f is this g is this so this is the expression now I am going to substitute the values of g and f that I know and that becomes this is the formula so now the question is are they classical solutions we obtained an expression for f and g and for u in the last four slides are they classical solutions are the functions f and g is c 2 functions since f and g are expressed in terms of phi and psi answers will depend on phi and psi so we are assuming that phi is c 2 of closed interval 0 l and psi is c 1 of 0 l is this good enough when is f a c 2 function this is what we know about f there is no doubt about smoothness of f when you are not at these points like 0 l minus l and integral multiples of l because inside functions are nice this is nice so it will be a smooth function so the doubt is only at points which are like l minus l 0 2 l minus 2 l etc because that is where we have breaks in the formula so f is continuous at psi equal to 0 that means let us see here psi greater than equal to 0 so you pass to limit as psi goes to 0 what you get this integral term goes off what you get is half phi of 0 and here from here you get half minus half phi of 0 both have to be same if f is continuous which means phi of 0 is 0 now these are all similar considerations f of minus l equal to f of l if and only if phi l is 0 please answer these questions for yourself f is differentiable at psi equal to 0 if and only if psi of 0 is 0 f dash of minus l is f dash of l why do we even have this because it comes from here this holds for f of course it will hold for f dash f double dash and so on and that is true if and only if psi l is 0 f is twice differentiable at psi equal to 0 if and only if psi double dash of 0 is 0 similarly f double dash at minus l is same as f double dash l if if and only if psi double dash of l is 0 so f is a c 2 function if and only if the following happens phi and psi satisfies the following compatibility conditions these are called compatibility conditions phi at the end point 0 and l are 0 psi is also 0 at the end point 0 and l so is psi double dash under the above conditions and phi and psi the function g is also c 2 so we have a classical solution if phi satisfies these two conditions and the these compatibility conditions so let us summarize our discussions in the form of a theorem it is done on the next slide the existence and uniqueness theorem let phi and psi have this smoothness further assume the following compatibility conditions these three sets of compatibility conditions the IBVP has a unique classical solution this is IBVP remember it is a devilishly boundary conditions if you are considering no I mean boundary conditions the compatibility conditions will change so let us get the solution in the dilumbered form that is another goal that we had for that what we have to do is phi and psi which are defined only on 0 comma l we need to extend to r because we need to take phi of x minus ct psi of x plus ct and so on they should make sense that means phi and psi should be defined for every real number yeah let phi 0 and psi 0 denote the extensions of the functions phi and psi respectively to the interval minus l 0 as odd functions with respect to 0 and then as 2 l periodic functions to r so phi 0 of x is minus phi of minus x psi 0 of x is minus psi of minus x this is what is defining or extending the function phi and psi as odd function to the interval minus l comma 0 after that we can extend it to r then u of xt has the dilumbered form this is exactly what we like right phi 0 of x minus ct plus phi 0 of x plus ct by 2 plus 1 by 2 c x minus c to x plus ct psi 0 of s ds this is exactly how the dilumbered formula looks for the Cauchy problem in 1d but now not phi and psi but in terms of phi 0 and psi 0 that is the only change let us look at an example of solution in the region 1 0 1 0 means m is 1 n is 0 so this formula which I have taken from the slide where the solution m is odd n is even I get this expression and this is a formula I get now in terms of the extended functions it becomes this because ct minus x will be in minus l 0 and we have extended phi 0 as an odd function therefore this quantity is precisely phi 0 of x minus ct this is between 0 l therefore there is no change it is the same and this integral becomes this integral that is left as an exercise for you though I will be doing on some other domain later on so find the value of u of 1 by 2 3 by 2 where u solves the ibvp given here where I have given psi equal to x into 1 minus x phi is 0 as usual the Dirichlet boundary conditions which are 0 here so solution for m odd n odd will apply here because we have to look at c is 1 so x minus t is half minus 3 by 2 there is minus 1 x plus t is 2 so m equal to 1 n equal to 1 that is that is what you have explained the reason I have written here the formula is precisely this which is here and it is a matter of computing this integral minus 1 by 6. Now let us get the solution using Dallambert form because on the last slide we have used the formula which is coming from this slide which gives solution for m odd n odd. So now we need to extend our function phi 0 and psi 0 but phi 0 is 0 no need to extend so this is the formula so we need to know what is psi 0 now this formula is half minus 1 to 2 psi 0 now psi 0 is a odd function therefore if you integrate between minus 1 to 1 it will be 0 therefore this integral is essentially from 1 to 2 and 1 to 2 how did we define psi by this manner then please do this computations by yourself you get the same of course you get the same answer now let us illustrate the solution in the region 1 comma 2 where we see the connection between the Dallambert form solution and reflections. So region 1 2 is algebraically given by these inequalities and solution is given by this this you can write from the slide where m is odd and n is even you get this. Now look at this picture here I am in this region 1 2 I take a point p I want to find the u at x naught t naught p is x naught t naught it is in the region 1 2 of course region 1 2 is characterized by this inequalities that you know so what I am saying here is how do I get this solution do I have to do every time the m even n odd find which m which n etc how it comes look at this point now through this point there are two characteristics which pass through one is this pink color which is moving to the right side and this is the violet color which is moving to the left side as t increases but what we have to do is we have to do backward thing. So take the point p we want to come towards x axis because that is where our initial data is there so therefore follow this characteristic and come back you will hit the bound boundary when you hit the boundary change over to the characteristic line from the other family and you hit this point this computation can be easily done after all these are straight lines and you know this equation x equal to 0 is x equal to L. So you come here and just one reflection you do you come here now if you come from this side okay come from this side you hit this and then switch over to the other character family characteristic passing through that point and then you hit again this boundary and again you switch over here and call it R okay. So this is L and this is R okay and exactly the coordinates are identified what is R what is L now this is a formula that we have written down on the previous slide no the formula in the region 1, 2 is given by this now this is just for a placement where what is happening so minus L, 0, L, 2, L, 3, L. Now x naught plus CT naught lies between 2 L 3 L so it is somewhere here and x naught minus CT naught is between minus L and 0 so it is here but what we have is CT naught minus x naught which is coming this side and this we are translating by 2 L so that we fit into 0, 2, L interval that is where we know the function f and g is right f g is known on 0, 2, L so this is x plus CT so g is responsible we come here and what is L CT naught minus x naught what is R is this point okay why are we doing this picture is because we want to get the dialomert formula in dialomert formula what is integral that is between x naught minus CT naught to x naught plus CT naught so let us look at this is integral we have this integral is nothing but minus of this just switching the limits you get this now this is the integral that appears in a dialomert formula that I am writing as between x naught minus CT naught to CT naught minus x naught that is from here to here and again from here to here and this piece is repeated so I subtract but this piece is 0 because this odd function and this is a symmetric interval around 0 so this is 0 here it is odd as well as periodic therefore the this is an integral on length 2 L interval so it will be 0 so what remains is just this in other words what we have is precisely the dialomert form so this integral we got instead of L 2 R we got x naught minus CT naught x naught plus CT now this is phi of x naught minus phi naught of x naught minus CT naught and this by periodicity is phi naught of x naught plus CT naught okay interpretation of the solution using reflection of waves let x t be a point in the region M n then the formula is this the L and R exactly as I suggested we have to come to L and R but that is multiplied with minus 1 power M minus 1 power N respectively and this is 1 by 2 C integral L 2 R L and R obtained by following characteristic lines backwards as described earlier let us also do one more point and try let me take a point here okay so first thing is go like this of course these are characteristic lines which is a family of parallel lines so not too much so this point you can compute the coordinates this is L and let us do the R okay then R okay so this here M n our M n is 2 2 how many reflections we made one reflection here and two reflections similarly here one reflection for this side and two reflections so this is actually the number of reflections that we are going to do you can identify this L and R and the formula as before is applicable what is the formula minus 1 power M in this case it is 2 into phi at this point L plus minus 1 power again N is 2 phi at R by 2 plus 1 by 2 C integral L 2 R psi s ds this formula holds is a very easy way to determine the solution so let us summarize what we did we consider the IBVP with Dirichlet boundary conditions it is solved from so called first principles we are calling it first principle because we have obtained to that the beginning we have obtained the solution to the homogeneous wave equation by transforming into the system of characteristic coordinates where the equation read W psi eta equal to 0 and solution was a function of xi plus a function of eta that is why we are calling this first principles and the same problem may also be solved using the ideas presented in tutorial which is called lecture 4.3 there we did problem one as an application of that problem to solution of IBVP this IBVP can also be solved thank you.