 Welcome to this lecture. From this lecture onwards, we start solving wave equations. In today's lecture, we are going to solve wave equation in one space dimension. Here we are going to consider only homogeneous wave equation for the reasons that we explained in lecture 4.1. The solution is known as Dallambert formula. The outline is that we introduced the Cauchy problem for one-dimensional wave equation. And then we look at the canonical form for the one-dimensional wave equation. Using that, we obtain a solution which is known as Dallambert formula for the Cauchy problem, solution of the Cauchy problem. So, Cauchy problem for one-dimensional wave equation and the plan of action to solve it, we highlight the key points. So, the Cauchy problem is this. Given functions phi in c2 of r that is twice continuously differentiable and psi which is c1 of r, 1 times continuously differentiable on r, we need to solve this problem. This is the equation we write WE wave equation 1D for one space dimension. So, equation is u2t equal to c square uxx. We are interested in solving for x in r and t positive. We could also solve for t in r that we will discuss perhaps in a tutorial. So, u2t equal to c square uxx in x belongs to r, t belongs to 0 infinity. Then we have two conditions which are called initial conditions or Cauchy conditions that is ux0 is phi x, utx0 is psi x, phi and psi are given with this much smoothness. So, this is what is called a Cauchy problem for one-dimensional wave equation. So, what are the main steps in solving the Cauchy problem? The wave equation that is WE1D it is transformed into its canonical form. The new coordinate system is called system of characteristic coordinates. A general solution to the equation in the canonical form is obtained because the canonical form is very easy to solve. And going back to the xt coordinates the general solution to the wave equation is obtained as a function of x and t and solution or the Cauchy problem is then obtained by imposing the initial conditions IC1 and IC2 on the general solution that we obtained. And the solution that we finally get is what is called D'Alembert's formula. So, first point is to reduce our wave equation to canonical form. So, recall about change of variables. Suppose that we have a change of coordinates from xt to psi eta and vice versa given by psi equal to phi xt eta equal to psi xt and x equal to capital phi of psi eta t equal to capital psi of psi eta. A function u gets transformed into a function in terms of psi eta using these relations and vice versa. So, we are going to find an equation satisfied by w solve it and get u using the w that is the strategy. So, how does this wave equation change under change of variables? So, let us transform w e 1 d into psi eta coordinates. We are yet to find these coordinates we will find them. So, differentiate this relation u xt equal to w of phi xt psi xt with respect to x and t you get u x equal to w with respect to the first variable which is psi and whoever is here that is phi with respect to x okay plus w with respect to eta and psi with respect to x. This is the chain rule and similarly ut of course what we need is ut t and u xx so that we can go back and substitute in the equation. So, please do these computations on your own. So, u xx will turn out to be having this expression and ut t this expression. The argument is xt which we are suppressing here otherwise the equations will be in many more lines okay let us substitute for u x and u xx and ut t in the wave equation and we get an equation of this form okay x and t we have to substitute capital phi and capital psi. So, wave equation is a hyperbolic equation to find its canonical form we require that w xi xi and w eta eta their coefficients must be 0 what are they this is a coefficient of w xi xi this is for w eta eta we want that to be 0. But if you observe the equations are the same of course this is written for phi this is written for psi otherwise equation itself is the same. So, now for xi equal to phi and eta equal to psi to define a change of coordinates phi and psi I cannot take it to the same function in fact what I need is this Jacobian to be nonzero. So, I need solutions of this equation phi and psi of this equation with this property because that is when I know that there is this change of coordinates possible thanks to inverse function theorem. So, this equation can be factored into this if you multiply you will see that phi x phi t term will get cancelled. So, this if you multiply precisely this therefore what we do is that we find a phi satisfying one of these equations and psi the other equation okay. So, any one of them is 0 product is 0 that is the logic here. So, we do this and very easy to write down the general solutions of these equations because this is the equation with constant coefficients. So, phi is a function of x minus ct alone and psi is a function of x plus ct alone. So, please check for yourself we did this in the first order partial differential equations chapter. So, for simplicity let us take phi equal to x minus ct psi equal to x plus ct. Now there is a definition these are lines x minus ct equal to constant and x plus ct equal to constant they are called characteristics of the one dimensional wave equation. The new coordinate system psi equal to x minus ct eta equal to x plus ct is called characteristic coordinate system. Note that characteristics consist of two families of straight lines having slopes plus or minus 1 by c one has 1 by c the other family has minus 1 by c each family consist of parallel lines. So, when c equal to 1 the families are actually orthogonal families. The characteristic coordinate system is an anticlockwise rotation of the xt coordinate system. We will see in the next slide we have a picture. So, this is our original x and t coordinate system and this is psi and eta psi equal to x minus t eta equal to x plus t they are orthogonal because c equal to 1. Look at this point p this point p in terms of xt coordinate is given by x equal to minus 0.5 t equal to minus 0.5 in terms of psi eta it is lying on this line x equal to 3. So, psi is 0 and eta is minus 1. In this coordinate system given by psi equal to x minus ct eta equal to x plus ct the wave equation becomes this we already noted down. But now we have insisted a must be 0. Similarly, c is 0 and b turns out to be minus 2c square the computation will give c is 0 d is also 0 is also 0. So, therefore the equation that we have is minus 4c square w psi eta is 0. Of course, c is non-zero we can just cancel this and you get w psi eta is 0. This is a canonical form of the homogeneous wave equation. Now we can integrate this with respect to eta first and then psi next or even the other order is also fine and we get this expression. If you integrate with respect to eta first then what you have is w psi is a function of psi alone and then w eta will turn out with this w psi eta is f psi plus g eta. So, note that the function w is c2 on r cross r if and only if the capital F is c2 and capital G is c2. So, w is c2 function if and only if f and g are c2 functions. So, in terms of the xt coordinates the general solution of the wave equation is given by go back and substitute for xi equal to x minus ct and eta equal to x plus ct. So, we get this where f is c2 and g is c2. So, we proved the following result classical solution to wave equation. Let f belongs to c2 and g belongs to c2 the function you define by uxt equal to f of x minus ct plus g of x plus ct is a classical solution of the homogeneous wave equation. No Cauchy problem yet we have just solved the wave equation homogeneous wave equation. Now we impose initial conditions on this and we determine what f and g should be in terms of the given functions phi and psi. Before that let us do a geometric interpretation as a wave propagation. Let us fix a time instant t equal to t naught the graph of f of x plus ct naught is precisely graph of f of x shifted by ct naught units to the left. Thus f of x plus ct represents a backward moving or left moving wave with speed c. Similarly, the graph of g of x minus ct naught is precisely the graph of g of x that means shape is not changing but it is getting translated by ct naught units to the right side. Thus g of x minus ct represents a forward moving or right moving wave with speed c. Thus any solution of the wave equation is a superposition of a wave which is moving forward and another moving backward. Now let us derive the D'Alembert formula for solution of Cauchy problem. The general solution of the wave equation is u of xt equal to f of x minus ct plus g of x plus ct. Now we have to use our given Cauchy conditions u of x0 equal to phi x. So I put t equal to 0 in this equation I get phi x is equal to fx plus gx. So this is the equation that I have for f and g. Now we have another initial condition that is ut x0 is psi x. So differentiate this equation with respect to t and then put t equal to 0. So if you differentiate this with respect to t what you get is f prime into x minus ct into minus c plus g prime at x plus ct into c and when you put t equal to 0 what you have is minus cf prime x plus cg prime x that is equal to psi x therefore you have this. So we have two relations connecting f and g in terms of given phi and psi but this equation has f dash and g dash. So it is better to get it off the derivative and that is done by integrating. Integrate both sides of this equation so that the left hand side becomes a combination of f and g and here already I have one combination of f and g linear combinations. So these are system of two linear equations we will get from there we can determine f and g that is a strategy. So let us integrate this equation between 0 to x. So you get f plus g minus f plus g from here because once you integrate from 0 to x this is actually minus c times f prime minus g prime. So that is f minus g whole dash. So if integrate you get f minus g at x minus f minus g at 0 the 0 terms have been transferred this side and here you have integral 0 to x psi s ds which is here and then you divide with a c. So you get 1 by c please do this computation. So we have the following two equations for f and g this is coming from i c 1 f plus g equal to phi this is coming from i c 2 after integration minus f plus g equal to this. Now if you add up you get an expression for g if you subtract you get an expression for f. So we can solve this system of linear equations and obtain f equal to this expression and g equal to the expression. Now we take these expressions for f and g and substitute in the solution u which is given in terms of f and g. So this is the expression that we end up with u x t equal to phi of x minus c t plus phi of x plus c t by 2 plus 1 by 2 c integral x minus c 2 to x plus c t. This term this is a constant this is a minus of that constant so it actually got cancelled. So this is known as the Alambert formula for the solution of homogeneous wave equation and the Cauchy problem where phi is u x 0 psi is u t x 0. So phi is u x 0 so u x 0 is called initial displacement psi is u t x 0 that is the velocity initial velocity. So let phi belongs to c 2 of r and psi belongs to c 1 of r the function defined by the Alambert formula is a classical solution to the Cauchy problem this we have to check right what do we have to check u is a c 2 function why is it a c 2 function because the chain rules. We have phi is a c 2 function this is a c infinity function x minus c t similarly x plus c t show that phi of x minus c t is c 2 phi of x plus c t is c 2 as a function of x and t and psi c 1 but we are integrating therefore we gain one extra derivative that is why this term is c 2 you can use Leibniz rule to get the derivatives and then check that u is indeed a solution to the homogeneous wave equation and putting t equal to 0 is very easy put t equal to 0 this integral is from x to x so this 0 this term is 0 and what we get is phi x plus phi x by 2 so we get phi x differentiate with respect to t this equation and substitute t equal to 0 you should get psi x so it is a very straightforward computation so I leave it as an exercise for you to do it please write down in detail why this u is c 2 and why it satisfies the homogeneous wave equation and the two initial conditions in fact if you observe closely we proved that any solution of the Cauchy problem is given by D'Alembert formula thus we established that the Cauchy problem has a unique solution and is given by D'Alembert formula so go through the steps and convince yourself what did we do we are given homogeneous wave equation that we reduce to canal form there is no compromise there canonical form a c 2 solution if and only if it is given by f of xi plus g of zeta and f and g has to be c 2 no compromise there after that we went back and substituted so everywhere there is no loss of information therefore the uniqueness is there in the proof so in so please convince yourself that we have indeed proved uniqueness and in case you are still not convinced we will give another proof later on a couple of proofs later on for the uniqueness let us summarize what we did in this lecture wave equation in any space dimension is already in a good canonical form u t t minus c square Laplace in u equal to 0 you cannot ask for anything better than that however in one space dimension the wave equation has a simpler canonical form namely w xi eta equal to 0 and that played a major role in deriving D'Alembert formula on the other hand for wave equation higher space dimensions that is dimension 2 onwards it is not clear if there are such simpler canonical forms this aspect we discussed in the chapter 3 so it is not clear if there are such simpler canonical form which help immensely in obtaining a solution to Cauchy problem so for higher dimensional wave equations what we do we adopt a totally different approach we use what is called a method of spherical means we will introduce that and that helps in solving Cauchy problem for wave equation in 3d first three dimensions not 2d but 2 3d as we shall see D'Alembert formula also plays a role in the analysis so using spherical means we kind of reduce the Cauchy problem in 3d to some wave equation in 1d and that is where we can use the D'Alembert formula for the solution thank you