 Welcome to this lecture on initial boundary value problems for wave equation. We have already discussed this problem earlier. In this lecture we are going to see a method called separation of variables method for the solution of IBVPs. In lecture 4.9 the following IBVP was solved starting from first principles which is the homogeneous wave equation and phi and psi is the Cauchy data and 0 Dirichlet boundary conditions. We described separation of variable methods to solve this IBVP in this lecture. So first thing is keep only one nonzero initial data that is keep either phi or psi that means keep phi make psi 0 or make phi 0 and keep psi. So in this lecture we are going to consider with psi equal to 0 for obvious reasons because the equation is linear. Therefore if you solve this IBVP with psi equal to 0 and then one more IBVP with phi equal to 0 and add these two solutions you will get solution to this IBVP. So solution u to the IBVP may be obtained as u1 plus u2 where u1 solves the IBVP with psi equal to 0 and u2 solves the IBVP with phi identically equal to 0. The idea superposition works because the equation is linear and homogeneous. Therefore we assume that psi is identically equal to 0 in the rest of this discussion. Solving the IBVP with phi identically equal to 0 is completely similar and is left as an exercise. Of course in the computation that we are going to do we can keep the psi as it is and see where it is going to play a role where the psi will appear in our computations. I will point out that part. It is only to keep the calculations to minimum and theorems to you know reasonable length that I am considering psi to be identically equal to 0. So what are the main steps involved in separation of variables method? The IBVP features homogeneous wave equation and 0 Dirichlet boundary condition. This is to be kept in mind. Step one is we derive two families of ODE's from wave equation. How do we get that? Look for solutions of the wave equation in the separated form that is u of xt is not any arbitrary mix of xt but it is a product of a function of x and a function of t. Such solutions we are going to look for. They are known as solutions in the separated form. So the wave equation will give rise to two families of ODE's index by a single parameter lambda we are going to see that 1 for x and 1 for t. Now from solving PDEs we are reduced to solving for ODE's which are considerably easier than the PDEs. So the Dirichlet boundary conditions in the IBVP will yield boundary conditions for x and the initial conditions in the IBVP will yield initial condition for t. We will see exactly later. So step 2 is obtaining nonzero solutions to the two families of ODE's. So the BVP for x turns out to be an eigenvalue problem. Recall eigenvalue problems from your ODE courses. It turns out that only a countable number of BVPs from the family indexed by n and n will have nonzero solutions. We will see that. And for each of the eigenvalues we need to solve an initial value problem for t. So that means first we solve the problem for x and then solve for t. At the end of step 2 we have a countable number of nonzero functions x and x into t and t. Then the step 3 is to propose a formal solution to the initial boundary value problem as a superposition of the functions that we have obtained namely x and x into t and t. So superposition of this function is proposed as a formal solution. It reminds to check that the formal solution is indeed a solution. So far this is a method that is being described to obtain a solution and once you think you have a solution you must check that is a solution. So step 1 let us implement wave equation giving rise to an IVP and BVP. How it is going to come? Let us see. So separation of variables looks for solutions of this form. As we said uxt is a function of x into a function of t. We take this and substitute in the wave equation. uxt becomes x t double dash and similarly uxx will become x double dash into t. Therefore the wave equation becomes this. Now we divide this equation with xx into tt and then rearrange terms we get this. We get t double dash by t equal to c square x double dash by x. Now if you observe in this equation the left hand side is a function of t only because it depends only on t t of t t double dash of t. So it is a function of t only and rhs is a function of x only. So a function of t only equal to a function of x only. So the only possibility is that they are constant functions. So both functions are equal to a constant function. It means that there is a lambda real number such that this equal to lambda and this equal to lambda which is written as this. These are 2 equations. t double dash by t equal to lambda and c square x double dash by x equal to lambda. So one of the tasks is to determine the lambda which arise out of this separated solutions. So thus we get 2 ods which is x double dash minus lambda by c square x equal to 0 and t double dash minus lambda t equal to 0. Now we have boundary condition u of 0 t equal to 0. From that boundary condition we get x0 into tt equal to 0 for all t positive. In fact t greater than or equal to 0. So that means x of 0 must be 0. If x of 0 is not 0 t of t is identically equal to 0 function. Once t is identically equal to 0 function then xx into tt will be identically equal to 0 function. So we are not doing any great job in finding a 0 solution. So we are not interested in that. So we do not want to admit t to be identically equal to 0 therefore we conclude x of 0 is 0. Similarly we have one more boundary condition u of lt equal to 0 from which we get xl into tt equal to 0 therefore xl for the same reasons xl equal to 0. So we have got boundary conditions for x. What are those? x of 0 equal to 0 and x of l equal to 0. Therefore we have the ode that we have already obtained. Now we have got the boundary value problem. These are the boundary conditions. And the IVP for t is given by t double dash minus lambda t equal to 0 and t prime 0 equal to 0. Of course there is one more initial condition that I cannot apply that is namely u of x0 equal to 5x. We have the second one which is ut of x0 equal to 0. So in that if we substitute u equal to xx into tt we get t prime 0 equal to 0. So this is one initial condition for t. So if we had the function psi we cannot even write t dash of 0 equal to 0. If psi was present we cannot get this initial condition in which case we work with only this ode. We do not consider any initial value problem because we do not get one. Solve them and that features two constants being a second order ode. It will have two arbitrary constants. And then we consider the product solutions and in step 3 we propose a formal solution as superposition of these solutions. And then determine the constants that we have in coming from this ode using the two initial conditions that we have namely ux0 equal to 5x, utx0 equal to psi x. Since we are assuming psi x equal to 0 in this lecture we get this condition t prime 0 equal to 0 by using this condition. So this is the boundary value problem that we have for x x double dash minus lambda by c square x equal to 0 x of 0 equal to x of l equal to 0. Now the lambdas for which the BVP admits a non-zero solution remember we are interested in only non-zero solutions. Such lambdas are called eigenvalues and the corresponding non-zero solutions are called eigenfunctions. So let us start our search for eigenvalues and eigenfunctions. Note that a real number lambda can be 0 positive or negative. So we are going to make our divide our search into 3 parts. We will ask whether it is possible to get 0 eigenvalue, positive eigenvalues, negative eigenvalues. That is because the equation solution is like that. It is x double dash plus some constant times x equal to 0. So depending on that constant is 0 positive or negative the form of the solution changes. That is the reason for making this separate studies in the 3 cases. Let us look at lambda equal to 0. So the BVP is x double dash equal to 0, x of 0 equal to x of l equal to 0. General solution of the ODE x double dash equal to 0 is A x plus B where A and B are constants. Now we are going to use these boundary conditions and determine A and B. So applying the boundary conditions we get A and B to be 0. That is obvious because this is a straight line and these conditions are demanding that it should pass through 0 0 and L 0. That means the graph must be parallel to x axis which is the case only when it is a 0 function. So lambda equal to 0 is not an eigenvalue. So we do not bother about that. It is not interesting. Let us see whether it is possible to find eigenvalues which are positive, lambda positive. So once lambda is positive we can write it as lambda equal to mu square where mu is positive and the BVP becomes x double dash minus mu square by c square x equal to 0 and boundary conditions are x of 0 and x of l both of them are 0. Now the general solution of this ODE is given by A e power mu by c x plus B e power minus mu by c x. If you have not written lambda equal to mu square we would have here square root of lambda by c into x minus square root of lambda by c into x. Just to reduce the notation clutter we have supposed that lambda looks like mu square where mu is positive. So applying the boundary conditions we get A equal to B equal to 0. So please go through these computations and conclude by yourself that this is indeed the case that A and B both of them are 0. Thus lambda positive is not an eigenvalue. It cannot be eigenvalue. That means no eigenvalues are positive. So now let us search for negative eigenvalues whether such eigenvalues are there or not. So since lambda is negative we may write lambda equal to minus mu square and mu is positive where mu is positive. The BVP for x becomes x double dash plus mu square by c square x equal to 0 x of 0 and x of l both of them are 0. Now a general solution of this ODE is in terms of sine and cosine. So it is A cos mu by c x plus B sine mu by c x. Now we have to see whether there are A and B's for which these boundary conditions are satisfied. So nonzero A and B's. So applying the boundary conditions what we get is A equal to 0 because x of 0 if you put this is 0, sine 0 is 0. Therefore what you get is A times cos 0 which is 1 equal to 0. And when you put x equal to l this is the equation you get. Since A is already equal to 0 this relation is essentially B sine mu by c l equal to 0. So we want at least one of the constants AB to be nonzero but A is already 0. Therefore to have been on 0 we must have sine of mu by c l equal to 0. So sine of something is 0 means that guy is a multiple of pi. So mu n equal to c n pi by l, n belongs to n. So these are the solutions of this. Should n belongs to z is this a mistake n belongs to n or n belongs to z because solutions of sine of something equal to 0 means that has to be a multiple of pi, it can be an integral multiple of pi. Therefore should it be n belongs to z. So it is a common thing to get confused at this point. It happens to everybody when they are solving first few problems involving the separation of variables method. No it is n belongs to n not n belongs to z this is not correct because if n belongs to z that means a negative number you take mu n becomes negative. But you see when we are looking for lambda less than 0 we suppose that lambda equal to minus mu square and mu is positive. Therefore mu n cannot be negative. Therefore n belongs to n is correct. Keep this in mind be very careful at such points. So let us see what we have got so far about the boundary value problem for x. The eigenvalues and corresponding eigenfunctions are indexed by natural numbers. They are actually countable. So any countable set can be indexed by natural numbers. So lambda n equal to minus c square n square pi square by l square and x n of x is sine n pi by l x. So we have got now some numbers lambda n. Now for each of such n we have to solve the initial value problem for t. So solve this ODE for t with lambda equal to lambda n for each n. The initial condition ut of x 0 as we already discussed gives us a condition t prime 0 equal to 0. Therefore solution of t double dash minus lambda n t equal to 0 with t prime 0 equal to 0 gives us t n equal to b cosine of n pi c by l into t. So now we are in a position to propose a formal series solution to the IVVP using superposition principle. So that is summation n equal to 1 to infinity b n sine n pi by l x cos n pi c by l into t. So this is precisely x n x into t n t. So we are considering superposition of x n x t n t with coefficient p n summation n equal to 1 to infinity. Now we do not use equality right away here that is because we do not know whether this series converges or not. So the unknown coefficients what are these b n's that also we should have an idea and they will be determined using the initial condition u x 0 equal phi x. Recall that we have considered psi equal to 0. If you have not considered psi equal to 0 what we will have here is sine n pi by l x into b n cosine of n pi c by l t plus c n sine n pi c by l into t. Then we have determined both b n and c n using the two initial conditions one is this u x 0 is phi x and second one is u t x 0 equal to psi x. Because we assume psi equal to 0 we only have one term here we do not have an extra term. That information was already used that namely that psi equal to 0 was already used in obtaining solution for t. Now using the initial condition u x 0 equal to phi x what we have is put t equal to 0 cosine 0 is 1. So phi x is b n sine n pi by l x. So this gives us an idea what should be the coefficient b n is. Later on we have to prove everything rigorously. This is only for guessing what should the b n be. So b n are the Fourier sine coefficients of the function phi. That means phi when we express as a Fourier series it should have only sine terms that means cosine terms must be missing and b n's will be the coefficients of sine terms. Cosine terms must be missing tells us that function should be kind of odd function we will see that. So extend the function phi to the interval phi is given only on 0 l right. So extend it to minus l gamma l as an odd function with respect to x equal to 0. Let the extended function be still denoted by phi then the Fourier series of phi takes this form phi x equal to summation n equal to 1 to infinity b n sine n pi by l x. Note here there are no cosine terms here that is because phi is an odd function. The coefficients corresponding to cosine function cosine terms will be 0 and what is b n? It is given by this expression. Formally if you want to get you can easily get multiply both sides with sine k pi by l x integrate between 0 to l you will get such an expression for b k. Now formal solution to the IBVP is now given by this. So I have just substituted the values of b n. So this is still a proposed candidate solution to the IBVP that is why we still use this approximation symbol just to make sure that we have not checked things. So when is this formal solution which is defined above is indeed a solution. We have an answer on the next slide that is a theorem. Let phi be C4 4 times continuously differentiable and be such that phi 0 phi l phi double dash 0 phi double dash l all of them are 0. Let psi be identically equal to 0. So what we want is phi smooth. These are the compatibility conditions but that we already encountered when we solved the same problem using first principles and we here we are assuming psi equal to 0. Then the function defined by this formal infinite series and that is a solution to the IBVP. Let us prove this theorem. We need to show that the infinite series converges and the resultant function that we get is twice continuously differentiable function. Let us call it u itself. So infinite series converges defines a function u of xt and that function is C2 function and then it solves the wave equation and it is continuous up to the boundary of 0 l cross 0 t so that the boundary conditions are meaningful namely u of 0 t and u of l t and initial conditions are meaningful that is u of x 0 is phi x. We also need ut of x 0 equal to 0 that means the derivative of u should also be continuous to 0 0 l cross closed 0 t. So once we prove the convergence of the infinite series we will replace the symbol that approximate symbol with equality. So let us check by formally differentiating the infinite series we get ut equal to this you can see just by formal differentiation. So that means we are not saying that the infinite series converges or it can be differentiated term by term all such questions we are not answering we formally differentiate that means we pretend everything is alright and differentiate. So I use ut because you denoted that infinite expansion infinite series so ut will denote this similarly utt ux and uxx if you notice whenever you differentiate once you pick up n here in the expression for you there was no n it was just bn whenever you differentiate one derivative you get n here also n if you differentiate two times you get n square and n square. If you look at these terms here sin terms are bounded by 1 so essentially if you want to prove the convergence by dominating the series with a convergent series you need that summation mod n b n converges here you need mod n square bn converges so that in fact gives rise to our assumption why we have assumed that phi is c4 we will see those computations. So all the four series converge uniformly by comparison test that I have just mentioned if we have the following estimate for bn n square bn summation n square bn modulus bn less than equal to summation 1 by n square and I know summation 1 by n square is finite it converges so this is one sufficient condition. So once the series converges there is no doubted that utt equal to c square uxx holds we can see from the series that we have computed on the earlier slides. So therefore formal solution is indeed a solution of the wave equation. Now how do I get this decay estimate on bn bn recall this is a formula for bn now I write the sign term as d by dx of cos so I get minus 2 by n pi 0 to l phi x d by dx of this cos. So this means I am planning to use integration by parts so d by dx will be shifted to phi I get phi dash x here and cosine term here that gave me nn you see 1 n came on differentiating ones and there will be boundary terms which is this it is 0 because phi of 0 and phi of l are assumed to be 0. Now let us do this integration by parts once more namely we write cos as once again d by dx of sin and that we pick up n square here and again integration by parts gives you this boundary terms once again are 0 why the boundary term is 0 that is an exercise please see that earlier we got it from the compatibility conditions now we do not get it from compatibility condition but we get it from this sin term when x equal to l r 0 this is going to be a multiple of pi and sin is 0 at those points that is why this is 0 now we have got this repeat this 2 more times you get this n power 4. So therefore n square mod bn will be less than or equal to constant by n square therefore we get the desired dk estimate for bm. So the smoother the function is the faster the decay is for the coefficients so c4 gave us 1 by n power 4. Now initial and boundary conditions due to assumptions on phi the Fourier sine series of phi converges to phi and hence the initial condition is satisfied and other initial condition can also be checked. So at this point I would like to mention that the proof of theorem we have given is quite sketchy the intention was to show that the formal solution that we have time using separation of variables method or classical solutions only if we assume sufficiently smooth Cauchy data the actual theorems you can find in books on Fourier series for example there is a book by G. B. Falland on Fourier analysis and applications there you will find such theorems. Let us recall from lecture 4.9 we have proved the following existence and uniqueness theorem following first principles for the same problem same IVVP of course we considered even psi term there UTX0 was not assumed to be 0. In this lecture we assumed psi equal to 0. Now if you see phi c2 in the theorem in this lecture we proved the same theorem of existence but we needed higher smoothness so what is the use? Let us see a few disadvantages of separation of variables method. Solution by first principles gave classical solution under much milder assumptions on phi. In separation of variables method how much can we really conclude with those assumptions phi is in c2. As you have seen the proof we needed n square mod Bn to be less than equal to 1 by n square so that the series for second order derivatives converges. Now if phi is only c2 Bn's are only like mod Bn less than or equal to constant by n square that means the series for the second order derivative that we have obtained formally differentiating the formal solution may not converge. So we cannot make any conclusion about the convergence of the series that we obtained for UXX or UTT because the Fourier coefficients there were n square Bn. If we see to we do not know if summation n square mod Bn is a convergent series or not. So for physically relevant problems phi will not be even differentiable. Phi will typically be a piecewise linear function which is continuous. So therefore none of these two results are really applicable so we do not have really worry that separation of variables method required much more smoothness. Finally if you see both of them are useless as far as classical solution is concerned. Thus there is a need to generalize the very restrictive notion of the classical solution. We describe some ways to define notions of generalized solutions. For completeness sake it is out of the scope of the current course. We will do this later on. So formal series may not define a classical solution if the Cauchy data and boundary data are not sufficiently smooth and compatible with each other. If you are happy with only the convergence of the formal series that means UXT equal to the infinite series the series converges we are very happy with it we do not ask questions about the derivatives. Then we could work with Cauchy data which is piecewise smooth. Now that series make sense or meaningful let us declare them as generalized solutions. These are the general ideas which go behind in generalizing the notions of solutions. In fact this may be a good compromise. Let us see from computational point of view where does separation of variables method stand. Imagine the infinite series that we obtained as a formal solution that we proposed it converges. Is it possible to compute? Except in very rare circumstances exact analytical solution is not available which can be easily computed. Here we have an infinite series. Thus a solution can only be computed approximately anywhere whatever may be the expression that you have for the solution except in rarest of the rarest cases you cannot compute exactly. Therefore one choice of an approximate solution is to truncate the infinite series to finitely minute terms. Note that finite sum always solves homogeneous wave equation exactly and boundary conditions are also satisfied. Only initial conditions are satisfied in an approximate sense. Be happy with that. So therefore some people take this point of view that yes I know that I have an infinite series. I know it does not converge also as long as it converges for the function may not be for the derivatives just the function it converges infinite series converges. Then I want to compute a solution only right. Of course notion of solution I have to generalize but I know a function which can be computed which is a candidate for solution and by truncating the infinite sum I am computing it approximately that is one point of view. Of course one has to do an error analysis in that case. Now separation of variables method in higher dimensions what are the new complications? So problem as before is the same given a sufficiently smooth function defined on a bounded domain in RD find a solution to still homogeneous wave equation initial conditions where I have taken psi equal to 0 already and Dirichlet boundary conditions for x belongs to boundary of omega you prescribe E of xt to be 0 homogeneous boundary condition. Perfect setting for separation of variables method the way we are seeing it. So try for the separated solution substitute in the wave equation you get this and then divide both divide with x, x, t, t rearrange you get this. Now we get left hand side is a function of t here right hand side is a function of x therefore they must be constant functions thus giving rise to a ODE and a PDE. So PDE for x and ODE for this and we have a boundary condition which is U of xt equal to 0 for all x in boundary of omega. Therefore we get here a boundary condition for x, x equal to 0 on the boundary of omega and here we get an initial condition for t as before. So this is the boundary value problem we have to solve. So earlier the wave equation gave rise to two ODE when we are considering one space dimension and ODE is we could easily solve. Now here it is a linear equation once again but solving this is not easy we do not know. So unless the domain is very special like a square imagine omega is a square once again we apply a separation of variables method and reduce that to two ODE's and do something so it can be done. But we are not going to discuss how to solve such problems in this course just wanted to show how separation of variable method looks like in higher dimensions. So separation of variables method has an Eigen function expansion method. Remember we considered homogeneous wave equation with homogeneous boundary conditions, non-homogeneous conditions and boundary data need to be handled separately. Of course you for example we have some tools like Duhamel principle. An alternative point of view is to see separation of variables method as an Eigen function expansion method and the following reference of a book explores this point of view. Of course this is out of scope of the current course. The reference is a book by Cain and Meyer, separation of variables for partial differential equations and Eigen function approach. So let us summarize. Separation of variables method was used to solve an IBVP with Dirichlet boundary conditions. IBVPs with other boundary conditions may also be solved following the procedure described here. We will take it up in the tutorial. Proofs of convergence of the formal series are difficult and even impossible without smooth and compatible data. Thus you may restrict yourself to finding formal series solutions to IBVPs in this course. Question, separation of variables method works whenever the domain is of a cross product type. We have seen 0 L cross 0 infinity, X of X T of T, X belongs to 0 L, T belongs to 0 infinity, X belongs to omega, T belongs to 0 infinity, A cross V, X belongs to A, T belongs to B or X belongs to A, Y belongs to B. A disk in R2, is it of this type? Disk in R2, is it of this type? Imagine you have a unit disk even. Does it look like a cross product of two domains? Obviously not. But in some other coordinate system it does look like. Find out what is that coordinate system in which it does look like a cross product type, perhaps after removing one point. Thank you.