 Welcome to this lecture. In this lecture we are going to discuss about the non-homogeneous wave equation how to solve it using a very general principle called Duhamel principle. So the outline for the lecture is as follows. First we recall certain things from the lecture 4.1 and then we state the Cauchy problem for the non-homogeneous wave equation. And we introduce Duhamel principle after giving a motivation for it. Then we apply Duhamel principle to wave equation we do for 1, 2 and 3 space dimensions. So let us recall from lecture 4.1 certain things about the Cauchy problem for non-homogeneous wave equation. So given functions phi psi and f Cauchy problem is to find a solution to the wave equation this is the Duhamel operation operator equal to f on Rd cross 0 infinity such that u of x 0 is phi x ut of x 0 equal to psi x. Phi and psi are generally referred to as Cauchy data and f is referred to as a source term. The Duhamel operation operator is a linear operator since Duhamel applied to u plus v is Duhamel acting on u plus Duhamel acting on v. Therefore a solution to the Cauchy problem which was stated on the earlier slide that is the non-homogeneous wave equation with the Cauchy data that may be obtained as sum of two functions v and v tilde where v is the solution to the Cauchy problem where the wave equation is without the source term. It is called homogeneous wave equation. When there is no source term it is referred as homogeneous wave equation and v tilde is a solution to the non-homogeneous wave equation but with 0 initial conditions. We will state more clearly on the next slide what the problem precisely is. So u equal to v plus v tilde and v satisfies this homogeneous wave equation and the Cauchy data and v tilde satisfies non-homogeneous wave equation and 0 Cauchy data or 0 initial conditions. So when you add v plus v tilde by the linearity of the dilambation operator, dilambation of u will be dilambation of v plus dilambation of v tilde that will be 0 plus f therefore f. So you get the non-homogeneous wave equation satisfied by u and then look at the first initial condition v x 0 is phi x, v tilde x 0 is 0 therefore u of x 0 is phi x. Similarly ut of x 0 is vt of x 0 plus v tilde t x 0 one of them is psi and the other one is 0 therefore the addition will be psi, the sum will be psi. So therefore u indeed satisfies the non-homogeneous wave equation and recall that we have already analyzed these problems. The homogeneous wave equation and the Cauchy problem for it we have analyzed already therefore it is enough that we analyze this or know how to solve this problem then we know how to solve the non-homogeneous Cauchy problem for the wave equation. So let us look at the Cauchy problem for non-homogeneous wave equation with homogeneous data or 0 initial conditions or 0 Cauchy data and we want to solve this Cauchy problem in dimensions 1, 2, 3 because we are discussing only wave equation in 3 dimensions 1, 2 and 3. These are the 3 dimensions in which we are discussing wave equation so far. So a general principle called Duhamel principle will be used. So what is Duhamel principle? It says solutions to non-homogeneous linear differential equations that is also known as equations with source terms may be obtained as a superposition of solutions to the associated homogeneous differential equations. In other words what it says this is a rough statement, precise statement we are going to see later. If you want to solve non-homogeneous equations it is enough to solve homogeneous equations that is what it says. Of course the initial data will be tailored using the source terms because after all we want to solve the non-homogeneous equation that means source term is given to us. So we have to use that and make or manufacture this homogeneous differential equations in the corresponding Cauchy problem. Essentially the initial data will be tailored using the source terms. Motivation for Duhamel principle we will give using initial value problem for a second order linear ODE. So initial value problem for a second order linear ODE is given by this d2y by dt square plus mu square y equal to ft y of 0 is y0, y dash 0 is y1. So in this f y0 y1 are given and we are supposed to find a solution to this ordinary differential equation satisfying these two initial conditions. If you notice this looks like wave equation because in wave equation we have a u2t term that is like d2u by dt square but instead of mu square y we have minus Laplacian y equal to f of xt. Naturally wave equation being in partial differential equation you have both x and t variables but one can view wave equation also as an ordinary differential equation. Here in ordinary differential equation y is a function of t given any t y of t will be a real number. In the wave equation given any t u of t will be a function of x that is the difference we will come back to this later. So first let us see how Duhamel principle works in this example. So the solution to the initial value problem one can explicitly solve and write down as this y of t is y0 cos mu t plus y1 sin mu t by mu plus 1 by mu integral 0 to t sin of mu into t minus tau into f tau dt. This is a formula one can verify we are already used to this in a course on ODE we would have got this solution. Now we want to see each and every term closely and then we will see how Duhamel principle comes into play. So before that let us introduce this homogeneous equation. Remember Duhamel principle is something to do with the homogeneous equations now where the so we will make the equation homogeneous and the initial data has to be Tyler made using this source term. So with that view in mind let us discuss homogeneous ordinary differential equation with initial conditions where y of 0 is 0 that is initial position is 0 and initial velocity is y1, y prime 0 is y1. A solution to this we will denote it as S y1 of t. So S for solution y1 to remember that y prime 0 is y1 and solution for this ODE, ODE is in the background homogeneous and this map which maps given y1 a real number to this solution S y1 of t of this problem is called a source operator and of course we know the formula and that is given by y1 times sin mu t by mu. So S y1 of t is solution of this homogeneous ordinary differential equation and with y1 as the initial speed or initial velocity and initial displacement or initial position is 0. So in terms of the source operator the solution of the IVP is given by this yt equal to S y0 of t d by dt of that plus S y1 of t plus integral 0 to t S f tau t minus tau d tau. What is S f tau? It is a solution operator. What does it do? It maps f of tau as given as the initial velocity S f tau of t will be the solution of the given ODE. Let us define the integrand to be a function of t with the tau hanging around. So w of t and semicolon tau is S f tau of t minus tau. We will observe what happens to this w. What is that w satisfies? So where each tau positive the function w as a function of t is defined for t bigger than tau and it solves the ODE. So it is a solution to the homogeneous equation and it satisfies initial condition. When you put t equal to tau w will be 0 and wt of tau tau will be f tau. Remember that is how the source operator is defined. Source operator is defined like that. When t is equal to tau it is S f tau at 0. So that is why that is 0 because S f tau solves with 0 initial displacement and this is a velocity. So that is d by dt of this quantity at 0 will be f tau. Now if you observe this, this first two terms, some of the first two terms is a solution to the homogeneous ordinary differential equation and the Cauchy data is satisfied. That is y of 0 is y 0, y dash of 0 is y 1 is satisfied. Now this term essentially is solving the non-homogeneous ordinary differential equation with 0 initial conditions and how is it obtained as an integral? You can think integral is a certain sum therefore generalization of the sum therefore this is a superposition of the quantities inside. But what are the things inside? They are solution to the homogeneous equations. So this is the explanation or the motivation of Duhamel principle coming from initial value problem for a second order linear ODE. Now let us apply Duhamel principle to wave equation. So wave equation as a second order ODE, one may interpret the wave equation as an ODE in the variable t. In the case of ODE is for each fixed t, y of t is a real number or an element of Rd depending on whether we are dealing with a scalar equation or a system of equations. In the case of wave equation for each fixed t, u of xt is a function of x in Rd and thus takes value since space of functions defined on Rd. So let us compute the source operator in the context of wave equation and construct a solution to the Cauchy problem for non-homogeneous wave equation. So let us see what is the source operator for the wave equation. What is that Si of t for the ODE, right? For the wave equation Si of xt, what is the Si represents the initial velocity and initial displacement is 0 and you solve the homogeneous equation exactly as it was the case in ODE. So Si of xt let it denote the solution to this Cauchy problem for the homogeneous wave equation and 0 initial displacement and initial speed as Si. The source operator is well defined, we have to ask when is it well defined that means we have to ask when is that we have a classical solution to this. If D equal to 1, we need Si to be in C1 because we have a Dalambert formula which is applicable and then for D equal to 2 or 3, we need Si to be in C2 that is when the source operator is well defined. Now we expect the function, this is precisely that Sy is 0 dash, now it is y0 is phi and y1 is Si in the context of wave equation, this is Sy1 and this is integral 0 to t S f tau which we now wrote f subscript tau of t minus tau x dependence is there, so we write that. So this we expect to solve the non-homogeneous wave equation and the Cauchy data, so f tau of x stands for f of x, tau. So we expect that to solve this Cauchy problem, let us call this as a non-homogeneous Cauchy problem, so we have to check that. So in this formula, some of the first two terms on the RHS namely these two terms is a solution to the homogeneous wave equation and satisfies the initial conditions. What are those? Initial displacement is phi and initial velocity is Si, so these conditions are taken care by these two terms. Therefore what we expect is this term will be a solution to the non-homogeneous wave equation with a 0 initial data. Now remark on the first term, so we have to check that these are indeed solutions to the homogeneous wave equation, we are going to do that and they satisfy the Cauchy data, we will check that. So now let us concentrate on the first term, it is dou by dou t of S phi of x, t. It is a derivative with respect to t of what? S phi of x, t. What is S phi of x, t? It is a solution to the wave equation by definition, solution to the wave equation, homogeneous wave equation, initial displacement 0, initial velocity phi x. Now what we have the first term is a time derivative of a solution to the homogeneous wave equation. Therefore it is itself a solution, one can check that. If you have a homogeneous wave equation and let us say u is a solution to that then dou u by dou t is also a solution, dou k u by dou t k is also a solution. We have already mentioned this in one of the tutorials earlier, otherwise you can just check. So it is a solution that is not doubt. So it follows from derivative of a solution is a solution for the homogeneous wave equations. Now we have to check the initial data. So this first term satisfies initial displacement phi and initial velocity to be 0. So let us check initial displacement that is dou by dou t S phi of x, t at t equal to 0. We want to check this is phi x. This is by definition because what is S phi? S phi is solving homogeneous wave equation with initial velocity as phi. This is exactly that dou by dou t of that is initial velocity. So that is phi. So this is by definition. Now we have to check this condition. Initial velocity is it 0 or not? That means we have to take this term, differentiate this with respect to t. That means essentially dou 2 by dou t square of S phi and check that that is 0. So this is what we have to check. Dou 2 by dou t square of S phi of x, t at t equal to 0 compute what it is. By the equation because it satisfies homogeneous wave equation. So u t, t equal to c square Laplace in u. So that is what I have written. But now I will evaluate at t equal to 0. That is why both sides I put t equal to 0. Now what is Laplacian? It is a differential operator with respect to x. What am I doing here? I want to take the value at t equal to 0. Therefore, I can as well take the evaluate at t equal to 0 and then take the derivative with respect to the other variables, space variables or take the derivatives and then take t equal to 0. Both are same because the evaluation I am going to make is for t and the derivative that I have in the front is Laplacian that is with respect to x. Therefore, when I do that I get this. But what is S phi of x 0? By definition of S phi the initial displacement it is of the solution. So therefore that is 0. Therefore it is 0. Now if you notice instead of Laplacian you could have had any operator that does not depend on t. This is an interesting and important observation. So the Duhamel principle application is not limited to wave equation only. You can apply u t, t equal to any operator L of u as long as L does not involve the variable t. So therefore we check the first term satisfies homogeneous wave equation and it satisfies the initial conditions u of x 0 equal to phi, u t of x 0 equal to 0. Now let us move to the second one which is more straight forward because just the definition. The second term is by design or definition is a solution to homogeneous wave equation with initial displacement 0 and initial velocity as psi. That is the definition. Therefore the sum of the first two terms is going to solve homogeneous wave equation because both of them are solutions to the homogeneous wave equation and initial conditions will be now u x 0 equal to phi and u dx 0 equal to psi done. Now we have to concentrate on the third term now. What about the third term? So the integral term satisfies 0 initial conditions because if you check t equal to 0 integral collapses there is no integral so it will be 0 and check that the derivative is also 0 that I leave it as exercise very easy to check. So what remains to really check is that it satisfies the non-homogeneous wave equation. Then we would have proved that this is a solution to the non-homogeneous Cauchy problem. Let us do this checking. So first we have to compute the derivatives. If you want to check this solution to the wave equation you have to compute two time derivatives on two x derivatives. Let us compute the first derivative dou by dou t of this. Now we are going to use Leibniz rule. Whenever you have integrals which depend on the variable with respect to which you are differentiating there is a rule how to differentiate that particular integral. I am going to follow that. So therefore you get this derivative equal to the inside thing evaluated at tau equal to t that will give you this because tau becomes t Sf t x comma when tau equals t it is 0. So you get this plus you are going to differentiate inside the integral. So 0 to t remains as it is and dou by dou t of the inside quantity. So we have this. What is Sf t of x comma 0? It is 0 by definition that is how the source operator is defined and I have not changed the integrand as it is. So one at a time. Now let us differentiate once more. So we use Leibniz rule for differentiation of integrals here and it is actually a consequence of fundamental theorem of calculus and chain rule. Sf t of x 0 is 0 by definition of the source operator. So first derivative of the integral term is this. Now we will compute the second derivative. So this is being carried forward from the previous slide. This is the first derivative of the integral term. Therefore second derivative becomes first derivative of the first derivative. Now exactly again by the Leibniz rule this will be the integrand evaluated at tau equal to t plus differentiate inside with respect to t. This quantity is precisely f of x t. Once again from the definition of the source operator. Now here we see the dou 2 by dou t square of some solution. Sf tau after all is a solution to the wave equation. Therefore dou 2 by dou t square can be written in terms of the Laplacian. So it is c square Laplacian of whatever is here and Laplacian comes outside and we have this. Therefore we have proved that this integral term satisfies dou 2 by dou t square minus c square Laplacian equal to f x t. Take this term to the LHS then precisely we have this. That is integral term satisfies non-homogeneous wave equation. And rest of this lecture is devoted to writing down the source operator explicitly in each of the dimensions 1 to 3. And obtain explicit formulae of solutions to the non-homogeneous Cauchy problems. Before that let us give a remark on the integrand in the particular solution. Yeah we may call this thing as a particular solution. This is a nomenclature that we use even in ODE's. It is one particular solution that means it is one solution of the non-homogeneous equation. So as before we already introduced this in the context of ODE's. So integrand you call it as w of x t, tau then the function w satisfies homogeneous wave equation and it satisfies the initial conditions when t equal to tau w is 0 and the derivative of w when t equal to tau is f of x tau. So in other words you see this is a superposition of solutions to the homogeneous equation with initial data which is Tyler made using the source terms. This explains Duhamel principle. So let us obtain explicit expressions. We have already solved the homogeneous wave equation on Cauchy problem. So it is just a matter of writing down what source operator is. So by Dalamert formula the source operator Si is this 1 by 2c integral x minus c2 to x plus ct Si sds. So solution to the non-homogeneous wave equation is this dou by dou t of s phi plus Si plus the integral term 0 to t Si of some t minus tau extra dt which is there on the previous slide. Okay we can simplify this further and we get this expression because here it looks a very complicated expression. We can use the Leibniz rule and get this expression. So if you notice the first of two terms is precisely the Dalamert formula that means when f is 0 phi and Si that is solution of the homogeneous wave equation and the Cauchy problem for that. And this is the new term which we have added to get solution to the non-homogeneous Cauchy problem. Now of course we have to ask this question when is it a classical solution? What are the assumptions needed? Of course if you look at the formula it is meaningful. This is meaningful if phi is let us say continuous integral we need therefore Si should be continuous again some integral so f should be continuous. But that is not enough for classical solution we require phi in C2, Si in C1 we have already noted down this when we discussed the homogeneous problem for the one-dimensional wave equation. Now what about this? Remember this f this term how did we get? The f went into initial velocity in the definition of source operator therefore we need some first derivative of f to be continuous. So f should be continuous so that this makes sense in addition you need first derivative of f with respect to x to be continuous. That is what is needed to make sure that the source operator is well defined with this as the initial velocity done. Now the comment on the domain of integration in this integral what is this? It is a triangular region formed by x axis on the two characteristics which are passing through the point x t that is if you look at this take a point this is the point x, t. So this is a line which are going through this is x minus yeah we should have used x0 t0 then this is x minus ct equal to x0 minus ct0 that is the line and this line is x plus ct equal to x0 plus ct0 and this is x axis. So this is that triangular region if you see this is where we plot x and t here. So here this formula you just read with x0 t0 that is what I am reading x0 t0 this is also t0. So this point is t0 you fix any thing in between let us say t equal to tau okay then you are going from this point to this point this is precisely the two points here this and this okay. So this triangle is called characteristic triangle. Now let us move on to the two space dimensions by Poisson Kirchhoff formula the source operator is given by this formula and now this is the expression dou by dou t of s phi plus s psi plus that integral. So we get this formula using a change of variable it is convenient to write it on disc of radius one with centre zero the homogeneous equation part. Again same question under what hypothesis it is a classical solution no need to see look at the formula phi integral involved so phi continues grad phi integral is there therefore grad v continuous psi continuous f continuous we can put such conditions but then we want a solution for which we need higher kind of higher smoothness phi should be c3 psi should be c2 and f should be continuous and also gradient should be continuous with respect to x and the second derivative should also be continuous with respect to f reason is exactly the same as we gave for d equal to 1 this f is going to sit in according to the Duhamel principle this f the source term will go to the initial velocity and for initial velocity we need c2 s that is why this condition let us move on to three space dimensions again from Poisson Kirchhoff formula in three dimensions s psi has this expression and therefore by Duhamel principle u has this expression now which values of f are needed to determine u at the point x0 t0 that is a question we ask so we write the expression for u of x0 t0 of course f is not coming here but if you ask what values of psi and phi are needed it is clear that we need the values on the sphere of radius ct0 with centre x0 what are the values of f which are needed that is what we are asking here so set of all y tau size that tau is here y is here and tau between 0 and t0 so they are needed on this set now what is y is here y minus x0 norm is c times t0 minus tau c into t0 minus tau exactly that this is the sphere y should belong to that and tau should belong to 0 to t0 that is exactly what I have written here but what is this set this is a subset of r4 r3 cross 0 infinity this is the backward cone with vertex at x0 t0 we know the cone in three dimensions right so this is something similar we have to imagine in four dimensions I will show you a picture in other words the value of u at x0 t0 depends on the sources coming from the backward cone with vertex at x0 t0 so figure of a backward cone is on the next slide it is here this is a point x0 t0 what we are used to when it is r2 in the plane and you have a cone this is called backward cone or past cone is called forward cone or future cone we will discuss about this in a future lecture more on this but this is what is needed the values of f are needed on this cone therefore since y is in the sphere that means that norm y minus x is c into t minus tau the formula may be written like this now t minus norm y minus x by c and norm y minus x earlier it was written t minus tau but now that we can convert in terms of y and x and if you just combine what you get here that is precisely the ball of radius ct with center at x so the ball is written as a lot of spheres which are union take the union or yeah so that is what you have for each fixed tau you have a sphere and then take the union as tau varies from 0 to t so you get this so again the same question when is this a classical solution to the Cauchy problem of course formula is meaningful under some set of assumptions on phi psi and f but if you want a classical solution you need phi to be c3 psi to be c2 and f to be here continuous f a gradient to be continuous and second order to be continuous on r3 cross 0 infinity so the value of f of yt for any y in this ball of x comma ct does not play any role in determining the solution at the point xt on the other hand the values of f of yt at the retarded time t which is t minus norm y minus x by c that is what they place a role if you see the formula and the situation is totally different for d equal to 1 I will request you to check what is that difference from 3d in 1d and this integrand is called a retarded potential because instead of t we have t minus norm y minus x by c so justify the word retarded time after you learn about Newtonian potentials in the context of Laplace equation which will be done later on. So let us solve an example solve a problem Cauchy problem non-homogeneous Cauchy problem I have taken simple initial data because I want the computations to be simpler. Of course we have already said this is a formula f is a smooth function no problem so this will represent a classical solution because phi is smooth psi is 0 and f is definitely smooth you can compute directly by plugging in these values we do not do that what we do is we apply Duhamel principle directly. So for the given Cauchy problem psi is 0 and therefore the solution in terms of the source operator is given like this now we have to find the source operator and that is given by this by the Dallambert formula which we can compute that is xt s phi of s phi of x, t is xt. Now let us compute what is s f tau by definition it is this given by this formula and compute what is f tau of s that is s into e power tau compute these integrals you get t minus tau into x e power tau. Now find this integral because this is what will give you solution to the non-homogeneous wave equation and that is given by this integral on simplification we get this formula x e power t minus xt minus x. Now we have to add these two okay dou by dou t of s phi is required s phi was xt so dou by dou t will be simply x plus this we will plug in from the last slide solution is x into e power t minus t. So we have introduced Duhamel principle applied it to obtain solutions to non-homogeneous question problems we observed that the Duhamel principle works for any operators or the form ut t minus lu for wave equation l is the Laplacian for what operators it will work is the operator real can be any differential operator not involving the variable t in it. Thank you.