 So, welcome to this lecture on wave equations. We are going to start our study of wave equation starting from this lecture onwards. In this lecture, we are going to derive an equation which governs the transverse vibrations of vibrating string. The outline for today's lecture is first we will derive wave equation as a mathematical model for transverse vibrations of strings. Then we present the Maxwell's prediction that light is an electromagnetic wave, it is based on wave equation actually. And we introduce the two problems that we are going to study for the wave equation. One of them is called Cauchy problem or initial value problem and the second problem is initial boundary value problems. These are the two kinds of problems that we are going to study for wave equation, mostly for one-dimensional wave equation. So, wave equation as a mathematical model for vibrating strings. So, this is the picture of the string. The string is lying along x axis at time t equal to 0. So, therefore, the position vector p of s 0 is s 0 0 that means x equal to s, y equal to 0, z equal to 0. At this point of time you can imagine that the string is infinite in length or finite in length that does not matter. So, consider a perfectly flexible elastic string lying along x axis at time t equal to 0. The elastic string is assumed to be one-dimensional, thus points on the string may be described by a real parameter s belongs to finite interval 0, l if the string is finite or s in r it is infinite string or even s in 0, infinity both are same actually as far as this modeling goes. But as far as the problems that are being will be posed s belongs to r is different from s belongs to 0, infinity that we will see in the discussion of initial boundary value problems. So, let p of s t it is a vector in r 3 denote the position of a point s on the string as we have introduced here points on the string are identified with s. So, it is a point s on the string x equal to s and at the time instant t. So, p of s t is given by x of s t, y of s t, z of s t and here we are writing a finite string. String lies along x axis means that p of s 0 is s 0 0. This is how we have identified points on the string at time t equal to 0 we said s 0 0 is represents points on the string. And when we say s is between 0 and l it means we are considering a finite string. So, goal is to derive an equation which governs the vibrations of this string. So, how do we do that? Pick a small segment of a string. Let us say for s between a and b. Of course, a and b itself is between 0 and l assume that the density when we say density the usually in the first course in physics density is mass density. But in mathematically modeling we will see there are lots of densities that is why we stress here it is a mass density of the string is given by a function rho of s between 0 and l. So, therefore if you want to consider the mass what is the mass of the string between a and b it is by integrating this density you get. Density has the dimensions of mass per unit volume and you are multiplying integrating over volume therefore you will get mass that is the understanding of this. So, whenever you have density if you integrate you get that quantity in this case you get the mass. So, Newton's second law it says that rate of change of momentum equals the net force acting on it. So, we would like to apply this Newton's second law. So, to apply Newton's second law we need to know what is this rate of change of momentum and what are the forces acting. So, momentum is related to the velocity. So, the instantaneous velocity of the string at the point here is dou p by dou t at s t these are the time instant t. Momentum of the string between s equal to a and s equal to b momentum is mass into velocity for mass we have a density mass density. So, we multiply the density with dou p by dou t and integrate we get the momentum of the string segment between s equal to a and s equal to b is the integral a to b rho of s dou p by dou t of s t ds. So, what is rate of change of momentum d by dt of this. So, d by dt of this quantity that is the rate of change of momentum. So, we have got one side of the Newton's second law. So, we need to see what is the other side namely the forces acting on the segment ab. This can further be expanded d by dt you can go inside and dou p by dou t become dou to p by dou t square. So, net force acting on the segment ab. So, there are two kinds of forces acting on the segment ab what are those first one is internal forces these are the tensile forces forces due to the tension in the string and second is external forces like gravity and many other forces. So, this is the picture of the vibrating string these are time instant t vibrating string at time instant t this is a position p s t is x s t y s t z s t. So, this is the string that we are this is a segment ab that we are considering. So, at b there is somebody who is pulling in this direction by the rest of the string and at a it is this side of the string which is pulling in this direction. So, tensile forces acting along tangential directions. So, therefore, the tangential direction at this point is dou p by dou s b t and dividing that with its length will give us a unit vector unit tangent vector in this direction is this. So, at this point a the it is in the opposite direction opposite means it is the other direction. So, that we model with a minus sign here otherwise it is exactly the tangential direction. So, let us see what are the internal forces which are acting on the string segment whatever we saw in the picture now we are going to write it down. So, tensile force at s equal to a this is the segment ab these are at one of the ends at s equal to a it is exerted by the part of the string from s equal to 0 to s equal to a it acts along tangential direction and is given by this is the tangential direction and this is the direction that we have chosen minus because we are at the end point a the left end point in this interval and there is a number t minus at this is what is the tension and tensile force at s equal to b similarly is exerted by the part of the string from s equal to b to l and it acts along tangential direction and is given by this is the tangential direction at the point b and t plus bt. So, this minus and plus denote that this is something coming from the left side on the string and this is coming from the right side on the string we will soon see that they are the same at any point you do not have two values t minus a t and t plus a t no both are same we will see that soon. So, now how do we model external forces they are represented by the force density fst equal to f1 f2 f3 our current interest is not in modeling these external forces suppose you decide you want to do effects of gravity then you would like to incorporate explicitly how the gravity forces apply on the string. So, we are not interested in that we are generally taking any external force. So, we are not going to model we are not going to consider what are the external forces and what are the resultant f here we are not going to do that. So, net force acting on a string segment is here a to b rho s fst ds is a force this is mass density. So, this is the force and this a b and this is the internal forces. So, one is external one is internal. So, Newton's second law says that rate of change of momentum is equal to the net force acting on it. So, applying Newton's law. So, here we are treating the string segment as a particle. So, this is the rate of change of momentum and that equals the net force which is here. So, this is the equation that we have got so far. Now, suppose we tend b to a in other words we are trying to be as close to a particle as possible by taking this very small imagine a equal to b we are at indeed a particle. So, we tend b to a and what will happen to this integrals if inside integrants are reasonable this integral will be 0 this integral will also be 0 and when b goes to a this term goes to t plus a t dou p by dou s a t by modulus dou p by dou s a t and this anyway is a that equal to 0 and that will give us the t minus is same as t plus. Therefore, we may draw plus and minus in the equation here. So, let us drop that. So, after dropping this is the equation that we have now we would like to write this also as an integral this looks like some quantity evaluated b minus same quantity evaluated at a. So, this looks like integral from a to b of certain derivative of some quantity. So, that is exactly this here this is dou by dou s of this into ds a to b that will give you that this evaluated b minus this evaluated at a which is exactly this. So, we can write this now advantage is that all the terms are converted into integrals and we have this expression. Now, this expression equal to 0 this equation holds no matter what a and b is. Therefore, the integrand must be 0 we have discussed this kind of issues while dealing with conservation loss. So, since a and b are arbitrary in this equation the integrand must be 0. So, this is an equation that we have got where p we have a second derivative with respect to t this is the external force and p is here and then t there is something which is unknown t otherwise that is if t is known this will be an equation for p a second order equation in t. So, we obtained this equation using a general principle which is Newton's second Newton's second law and in the above equation tst is what the tension how the string is what material it is made up of it will depend on that. So, one needs to model or postulate the factors on which the tension depends and the manner in which it depends. So, we need to find a constitutive law if you recall the modeling that we did for the traffic problem this is exactly that this is like modeling the road. Tension is a function of the stretch factor this is an assumption this is an assumption. So, it is a function n of mod dou p by dou s st comma s and the above constitutive law along with this differential equation which we got that they model a vibrating string of course it is a non-linear equation in p. So, the model for a vibrating string turned out to be highly non-linear. So, we would like to understand using simpler equations where we can actually perhaps solve the problem. So, one looks for model which approximates this model of course you need to assume more conditions on the nature of vibrations then we can get more simplified model and then one has to see whether this model approximates the original model well this is how the approximations go in practice. So, in order to simplify the model we make a new assumption on the vibrations what is that assumption vibrations are small. So, how do we model this vibrations are small the x st y st z st at time t equal to 0 x was s and y and z are 0. So, we assume that it is a small change from there. So, s plus epsilon small x of s t this is epsilon y of s t epsilon z of s t and tension also we model like this t not a background tension t 0 plus a small variation of that epsilon t 1 st and f st is also a small force epsilon f st. Now, we want to go and substitute in the equation that we have obtained here. Therefore, we need to compute what dou p by dou s is. So, we will do those computations now. So, dou p by dou s square is derivative of this with respect to s is 1 plus epsilon dou x by dou s that is what is here and similarly we get the other two terms. So, rearranging the terms we get this expression. Now, we here we use this approximation 1 plus a square root is 1 plus half a plus order of a square. So, this is 1 plus this entire thing is a I have a square here. So, I want to compute dou p by dou s. Therefore, I need to take square root on the both the sides of this equation then I have square root of 1 plus something 1 plus something square root is given by 1 plus a by 2 plus order of a square. So, that will give us this expression. Please pause for a while and do all these computations by yourself make sure the computations are correct. So, please do it by yourself. Now, in view of this expression we need to compute this right this is what appears in the equation dou p by dou s by modulus of that. So, that will turn out to be a quantity of this type. Now, so this system now becomes this system. So, we have epsilon here there are epsilon square terms there are terms without epsilon. So, what we do now is equate the coefficients of epsilon from LHS and RHS of course LHS coefficient of epsilon is simply this these quantities rho dou 2 x by dou d square rho dou 2 by dou d square rho dou 2 z by dou d square and here the first term will be rho f 1. Let us look at the first equation. So, coefficient of epsilon I mean the one which multiplies epsilon is this and here it is this. Now, here we have to find out with the t naught no you do not get anything with epsilon you get 1. So, you get dou by dou s t 1 you get. So, this is the first equation similarly you can see that you get the other two equations. Now, if you observe this the second and third equation they look similar they describe the transverse vibrations of the string where the first one describes the longitudinal vibrations of the string. So, both look like this dou 2 u by dou t square you can put u is equal to y r z you will get these two equations equal to f plus t naught by rho dou 2 u by dou x square. So, this is called the one-dimensional wave equation. So, a small remark on the models for transverse vibrations of string. So, we followed the treatment presented in the book by Pinsky on partial differential equations and boundary value problems with applications. This is the title of the book by Pinsky for a good account of derivations of equations governing transverse vibrations for that matter vibrations of strings. See the article on the string equation of Narasimha by A. S. Vasudev Murthy in this book connected at infinity a selection of mathematics by Indians. It is a trim series book by published by Hindustan book agency that is where he describes an equation of Narasimha and this is actually the reference to the original paper of Narasimha derived a model for transverse vibrations of a string. The difference between various derivations of transverse vibrations of a string and these which I have quoted here is that while deriving transverse vibrations people assume that there are no longitudinal vibrations effects on that whereas these models take into that account and then further simplify. Therefore, you see you start with a very correct physical assumptions derive a model and then you know what assumptions you are making to get the equations that you get how the equations get simplified under more assumptions. So, that is the remark on this and discussion on this equation is what is there in this article by ASVM. Now, let us turn our attention to Maxwell's prediction that electromagnetic wave nature of light. So, these are what are called Maxwell's equations I will not read out. Is the electric field, B is magnetic field, mu naught and epsilon naught are permeability and permittivity respectively. So, divergence of B equal to 0, divergence of E equal to 0 and there are equations for curl. So, this is a first order system of equations. Now, applying curl to both sides of the first equation. So, we get curl E equal to minus dou by dou t of curl B. Now, we have a formula in the vector calculus curl curl is nothing but minus Laplacian this is dou 2 E by dou x square plus dou 2 E by dou y square dou 2 E by dou z square those are these are Laplacian and this is gradient of divergency into divergency. And the second equation in the Maxwell system which is for curl B that is the second equation we use these equations and we end up this gets simplified to this equation Laplacian E is equal to mu naught epsilon naught dou 2 E by dou t square naught this is system of 3 equations even E to E 3 of course they are all same. Now, when D is equal to 1 let us try if this equation has a solution of this type. So, now it is not dou Laplacian E it is simply dou 2 E by dou x square. So, we try a solution of this form substitute in this equation here wave velocity is mu and wavelength is lambda. Now, just compute what is dou 2 E by dou t square this is what we get sin becomes cos cos becomes minus sin that is why you get minus every time you differentiate with respect to t you pick up a minus nu by 2 pi by lambda. So, square similarly dou 2 E by dou x square you can compute go back and substitute in this equation you get that the velocity is 1 by root mu naught epsilon naught. Now, by the time Maxwell did the above computations the values of epsilon 0 mu 0 are known from experiments and this quantity 1 by root mu naught epsilon naught it is known to have the dimensions of velocity that is also known. But thanks to the computations of Maxwell with these known values of epsilon naught mu naught he obtained that this velocity is approximately equal to 3.107 into 10 to the power 8 meters per second which is close to the speed of light. So, Maxwell predicted that light is a form of electromagnetic radiation of course this was possible thanks to his computations using Maxwell's equations the computations led to a wave equation which admits waves moving with speed of light as its solutions. This highlights the importance of wave equation of course existence of electromagnetic radiation was later proved experimentally by Hertz. Now, let us discuss briefly the types of problems studied for wave equation the first problem is a Cauchy problem in full space it is also known as initial value problem. So, wave operator in d space variables. So, let us introduce wave equation of course we have derived wave equation in one space dimension using strings but you can write wave operator in d space variables which is nothing but utt minus c square into Laplacian u Laplacian u is precisely this dou 2 u by dou x 1 square plus dou 2 u by dou x 2 square up to dou 2 u by dou x d square this is the operator wave operator. This is called the d dimensional wave operator when we say refer to the word dimension what we mean here is a space dimension x 1 up to x d. So, the equation for the string was a one-dimensional wave operator. So, this square d this is a notation used square d stands for this operator it is also known as D'Alembertian operator. So, what is Cauchy problem for wave equation? So, given functions phi psi and f suitably Cauchy problem is to find a solution to this equation what is this equation this is a wave operator equal to f and you are given u x 0 to be phi and ut of x 0 to be psi. So, f phi psi are supplied find a solution you satisfying all the three equations the boldface x stands for x 1 x 2 x d which is an r d and c is positive number. So, we need to define what is the meaning of a solution? Solution how it should be it should be a function it should be as many times differentiable as you see the derivatives here two derivatives. So, I want the function to be two times continuously differentiable and I want these conditions to be satisfied these two conditions and that is it but if you see here this equation is valid in r d cross 0 infinity. So, the function must be defined an r d cross open 0 infinity but here I am asking the value at 0 should make sense that means you should be continuous on r d cross closer to 0 comma infinity and similarly ut must be continuous on r d cross closer to 0 comma infinity. So, that this makes sense and then we can ask it should be equal to the given psi that is all. So, a function is said to be a classical solution to the Cauchy problem if the function is twice continuously differentiable on its domain and these equality holds at every point in r d cross 0 infinity and it is continuous on this domain note here I have included a closer to 0 whereas here we have only open 0. So, both are different conditions this does not imply this continuity on the open set does not define that does not mean it is continuous up to the its closure. So, that u x 0 is now meaningful and then the equality u x 0 equal to phi x holds for every x in r d. Similarly, the ut should be continuous on this domain so that ut of any x comma 0 makes sense and then it is equal to psi x that holds. We can modify by replacing infinity with a capital T what do you mean by classical solution on a domain r d cross 0 comma t. Observation is that the delambition operator is a linear operator that means you apply square d to u plus v you get square d u plus square d v. There is a solution u to the Cauchy problem which we have stated earlier on the previous slides that may be obtained as sum of two things v plus v tilde what are the properties? v solves the Cauchy problem for homogeneous wave equation that means the right hand side f is 0 but phi and psi are there and v tilde solves non-homogeneous wave equation that means f is there but 0 initial conditions that means phi and psi are 0. So, that is precisely this v solves this problem notice here the right hand side in the equation is switched off it becomes 0 whereas here f is present but the initial conditions switched off the Cauchy conditions. If you add v plus v tilde because the linearity square d of v plus v tilde will be f because it will be 0 plus f that will be f and x 0 will be phi plus 0 that will be phi and v t will be v plus v tilde of t will be psi plus 0 which is psi. So, therefore plan of action for solving the Cauchy problem is solve it for the homogeneous wave equation first that is what we will do for d equal to 1 to 3 we derive actually the formula which are known by various names and we solve the non-homogeneous wave equation with 0 Cauchy data for that we use a general method called Duhamel principle. The Duhamel principle tells us that non-homogeneous equations can be solved using solutions of homogeneous equations and the second type of problem is initial boundary value problems. This is applicable when the equation is posed not x in r d but x in omega omega is a boundary domain r d you want the equations to be satisfied on omega that there could be a right hand side f here no problem there could be a right hand side f here. Similarly, there could be u t x 0 equal to psi x this is just an example of a IVVP. So, these are the initial condition the first two conditions here are initial conditions equation initial condition and this is a boundary condition u equal to 0 on the boundary of omega for all times. So, these are the initial boundary value problems and we will also solve initial boundary value problem when omega is a subset of r 1 an interval. Of course, one could consider more general boundary conditions than u for example, you could prescribe the Neumann derivative of u on the boundary or you can prescribe a combination of u and Neumann derivative. Neumann derivative is nothing but the normal derivative dou u by dou n grad u dot n on the boundary. So, IVVPs are more complicated than Cauchy problems which are posed in full space due to the presence of the boundary omega. So, let us summarize what we did. We derived a mathematical model for vibrating strings it was a nonlinear model under the assumption of small vibrations the nonlinear model reduced to linear wave equation and we discussed the computations which led Maxwell to predict the electromagnetic wave nature of the light and we introduced the Cauchy problem associated to wave equation outlined a plan of action for solving it and we introduced initial boundary value problem associated to wave equation. Thank you.