 In our studies of wave equation we have solved the Cauchy problem for the wave equation in dimensions 1, 2, 3 and also initial boundary value problem when D is equal to 1. We have proved uniqueness of solutions by at least 2 methods so far. In this lecture we are going to see a third method which is known as energy method. So that is the topic of this lecture is uniqueness by energy method. So first we look at Cauchy problem in Rd and we show that energy is conserved. We will define what is energy here. So Cauchy problem for homogeneous wave equation is we are in d space dimensions so square d delambarsion in d dimensions equal to 0 and this is Cauchy data phi and psi. Energy associate to the Cauchy problem for the wave equation is defined by E of t equal to this where u is the solution to the Cauchy problem. When is this integral meaningful? Because the integral is an Rd we have to ask is this integral meaningful? Is it a finite real number for each fixed t? Of course that would require that ut to be square integrable on Rd and norm grad u square equivalently all partial derivatives square of the all partial derivatives must be integrable on Rd. A simple condition which guarantees this is this for each fixed t u should have compact support. The function x going to u of xt is of compact support. Then this integral is really on a bounded set on a compact set and these are continuous functions they in fact u is a c2 function therefore ut is c1 and radu will also be c1. Therefore this integral there is no problem. They are definitely continuous functions and we are integrating on a compact set instead of Rd because of this hypothesis being satisfied. So then E of t makes sense. Of course this is in turn guaranteed when the Cauchy data itself is of compact support. We have seen this already. Cauchy data compact support solution to the homogeneous wave equation is also of compact support for each fixed t or we need to assume things like this for each fixed t this function should have some good decay properties so that these integrals will be finite. We will not elaborate more on this point. We are going to deal with this kind of assumptions in this lecture. So let the Cauchy data be compactly supported functions on Rd and let u be solution to the Cauchy problem for the homogeneous wave equation. Then the assertion is d by dt of this quantity which we call it energy is 0. That means energy is constant function of t. Energy depends on t right. It is independent of t and hence it is actually equal to energy at 0. What is energy at time 0? What is ut at x comma 0? It is psi and what is grad u? That is simply grad phi when time is 0. u is equal to phi. In other words, et is a constant function that is the energy is conserved. Let us prove this theorem. The proofs by energy method invariably go through the same first step which is to multiply the given equation with ut. This was also done exactly in the causality principle proof. Multiply the equation with ut then we rearrange you get this way. Now we are planning to integrate this on Rd. So exactly same equation on Rd. Now usually this term is actually energy right d by dt of the energy. So everything depends on this term now how the energy behaves. Here this is dou by dou xi is there. So we plan to do integration by parts in this term and then conclude things about the energy. As mentioned before, assumptions on the Cauchy data should be such that these integrals are meaningful. The first term on the LHS in this equation namely this term is 0. This is because the functions phi and psi are of compact support. That implies that the function x going to u of xt is also of compact support for each fixed t. In other words, the function x going to u of xt is identically equal to 0 for sufficiently large values of norm x. Let kt denote the support of the function x going to u of xt. So kt suggests it will depend on t. Of course, this function also depends on t. For each fixed t kt denotes the support of this function. Let R positive be such that kt is contained in this ball of radius R by 2 with center at the origin. Because this is a compact set, we can always find such an R. Of course, ball of radius R by 2 is contained in ball of radius R having the same center. Now the first term that we wanted to show is equal to 0. Look at this integral. It is an Rd because the support of x going to u of xt is compact. This integral is really on kt or let us say integral on this ball or maybe on this ball. So for convenience, we write it as the integral on the bigger ball. Now we are going to do integration by parts. Integration by parts in this integral will give you one domain integral and one boundary integral. The domain integral here will be 0 because it is dou by dou xi of c square u to u xi into 1. So when the dou by dou xi shifts to 1, it will be 0. So what we are left with will only be the surface integral or the boundary integral. So you have this integral because dou by dou xi, you get the ith component of an outward normal nu i d sigma. Now if you see summation over i equal to 1 to d of u xi nu i is nothing but the normal derivative dou by dou nu. And this is equal to 0. Why is that? In fact, we need not write this step. I am writing this because for a future use, we are going to use it later on in this lecture. Why is this 0? Because u xi is 0. Why is u xi 0? Because see u is supported in kt and kt is contained in the ball of radius r by 2 and this integral that we have is on the sphere s of 0, r. In particular u is 0 in this annular region. Therefore, all the derivatives will be 0 here. u xi s will be 0 and hence this term will be 0 because u xi on the sphere is 0 and hence we have this equality. So thus we get the second term which reminds in the equation equal to 0. But second term is nothing but d by dt of et. So d by dt of et equal to 0. That means e is a constant function. That means energy is conserved. So e of t equal to e of 0. But what is e of 0? It is this expression. I have to put t equal to 0 in the definition of e of t. But I know what these are. That is u t is psi and u f x 0 is phi x therefore, grad u will be grad phi. So therefore, energy for all times positive is actually equal to the energy when t equal to 0 which is given by Cauchy data. So now uniqueness of solutions to Cauchy problem. So let us consider a non-homogeneous equation with the Cauchy data. We want to show it has a unique classical solution. So what is this general strategy for showing any problem has a unique solution is let u and v are solutions, consider the difference and show that the difference is 0. So we would like to see what is the equation that u minus v solves. u satisfies this equation v satisfies exactly same equation instead of u I have a v here equal to f v x 0 is phi x v t x 0 is psi x. When I subtract the operator is linear here the dilumination operator is linear these conditions are also linear in u and u t. So w will satisfy homogeneous wave equation with a 0 Cauchy data but we proved E of t equal to E of 0 for homogeneous wave equation. Therefore, we have to see what is E of 0 here E of 0 is given in terms of psi and phi which is 0. Therefore, for this problem E of 0 is 0 and therefore E of t is 0 for all t by what is E of t it is this expression this expression equal to 0 integrand is always non-negative it is sum of 2 non-negative quantities. So that is 0 if and only if each of the terms is 0 that means w t is 0 and grad w is 0 this implies w is a constant function and w is a constant function and it should be 0 function because it is 0 at time t equal to 0. So we are assuming the solutions to be classical solutions therefore w of xt equal to 0 for all xt in rd cross 0 infinity. In other words u of xt equal to v of xt that means we have shown uniqueness to the Cauchy problem. Let us look at the second problem which is called equi partition of energy it means energy is partition into 2 equal parts let us see what what are the 2 parts. Let phi and psi have compact support with the usual regularity assumptions and let u be a solution to the homogeneous wave equation with phi and psi as the Cauchy data. Show that there exists a capital T there is a time t such that for all times after that t greater than or equal to t we have the kinetic energy which is given by half u t square equal to potential energy which is half u x square energy is sum of kinetic energy and potential energy and what we are showing here is that there is a time capital T after which kinetic energy equals potential energy that means energy is equally partitioned equi partition of energy. How do we show this we will use we know the formula for the solution we compute u t we compute u x substitute here and see what we get. So this is the formula called a Dallambert formula which gives solution to the Cauchy problem for the homogeneous wave equation. Let us compute the derivatives of u let us compute u t first I have to differentiate phi and then differentiate x minus t with respect to t which will give me a minus sign and again differentiate phi differentiate t with respect to t therefore you have 1 plus 1 by 2. Now here we need to differentiate here x and t both are in the limits of the integration so we have to use what is called a Leibniz rule for differentiation of integrals. So this is as we discussed earlier it is a combination of fundamental theorem calculus and chain rule. So psi of x plus t into derivative of this with respect to t which is 1 minus psi at this point x minus t into derivative of this quantity x minus t with respect to t which is minus 1 therefore you get a plus here and this quantity this is u t similarly you can compute u x here you see that phi prime plus psi x plus t is there here phi prime plus psi x plus t. So let us separate those terms rearrange when we rearrange u t has this expression and u x has this separate this expression. If you notice the first two terms in both the expressions are the same second two terms this second term and this second term are also same but further so it looks like a plus b this looks like a minus b it is interesting to keep this analogy in mind because now we are going to ask when is this kinetic energy equal to potential energy. Half is always equal to half you cancel half u t square and u x square u t square as I mentioned a plus b whole square a minus b whole square. So a plus b whole square equal to a minus b whole square if and only if a b is 0 therefore when you substitute u t inside this and u x inside this the a square term b square term will get cancelled because it is the same on both sides here a square is this on the left side on the right side in the expansion for u x square you get a square here so they get cancelled because they are same similarly b is same so b square gets cancelled. So what remains is integral product of this quantity into this quantity equal to 0 please do this computation by yourself pause the slide here and then do the computation. Now we have assumed phi and psi have compact supports right so let them be contained in an interval a b any compact set you can always pull it inside some interval a b closed and bound interval. So let a b is b such that supports of phi and psi are contained in a b it means outside a b phi and psi are 0. Now the integrand on the LHS is identically equal to 0 because see one way of assuring that integral is 0 is to assure that integrand is 0. So that is why we are interested in making the integrand equal to 0. Whenever t is such that either x minus t is not in a b or x plus t is not in a b imagine x minus t is not in a b what will happen x minus t is not in a b therefore psi will be 0 phi dash will also be 0 therefore this term is 0 therefore this integral is 0. Similarly if x plus t is not in a b this first quantity in quantity in the first brackets is 0 it does not matter what this is product will be 0 and hence integral will be 0. So therefore the integrand of this integral and hence the integral itself is 0 whenever t is such that x minus t or x plus t one of them at least is outside the interval a b. So therefore we ask the opposite question suppose both of them are there in this interval a b then what can we say x minus t x plus t interval is lying in interval a b so we should be able to say something a here b here x minus t x plus t length of this interval is 2t length of the interval a b is b minus a therefore what should happen is 2t should be less than equal to b minus a. Suppose t is such that 2t is bigger than b minus a what does it mean both x minus t and x plus t cannot lie in the interval a b that means at least one of them is outside the interval a b and hence we have this integral equal to 0 as a consequence ke equals p now we ask that question. So therefore we choose t equal to b minus a then 2t will be bigger than b minus a therefore x minus t and x plus t both of them cannot lie simultaneously in the interval a b for any t bigger than or equal to t if one lies other cannot lie because both of them lie it means 2t is less than or equal to b minus a but I have chosen here 2 capital t is bigger than b minus a therefore 2 small t 2 times small t is greater than or equal to 2 times capital t and that is bigger than b minus a. So both of them cannot lie in interval a b whenever t is bigger than or equal to t and hence what we wanted the integrand is ideally equal to 0 and hence integral is 0. Therefore we have ke equal to pe for all t bigger than or equal to capital t. Now let us look at another proof of uniqueness for IBVP show that the IBVP with non-homogeneous wave equation non-homogeneous Cauchy data non-zero Cauchy data non-zero boundary conditions Dirichlet boundary conditions this has utmost one classical solution strategy is same let you and we be solutions take the difference take the difference and find out the equation satisfied by the difference. Therefore due to the linearity of the operator and the conditions that we have u, ut and u and u therefore the given problem has unique solution or at most one classical solution we had to be very careful here we are saying at most one classical solution we are not saying it has a solution that should be proved separately of course we have proved it. So it has utmost one classical solution if and only if this problem with homogeneous wave equation and 0 Cauchy data 0 boundary conditions has only trivial solution trivial solution is 0 solution of course we know that 0 u equal to 0 is a solution to this that we know but what we have to show is that is the only solution. Then uniqueness for the non-homogeneous initial boundary value problem follows. As I mentioned earlier energy method proceeds like this by multiplying the given equation with the suitable multiplier in the context of wave equation it is ut so on rearranging the terms as before we get this. So integrating the last equality over 0l we get this quantity. Now the first term is equal to 0 that is because ut of 0t into ux of 0t that is what will come once you do integration by parts in one variable we do not call it it is simply fundamental theorem of calculus. So maybe the second form so whenever you have derivative with respect to x and dx integrand evaluated at the upper limit minus integrand evaluated at the lower limit that is what will be the answer of this integral of course there is a minus sign that we will take care later. So ut ux values at l and at 0 that is what we have written here ut of 0t ux is 0t both of them are 0 because we know that u of 0t and u of lt are 0 because of the Dirichlet boundary conditions therefore ut of 0t is 0 and ut of lt is 0 therefore this product is 0 therefore this term will reduce to 0 or in fact in the same proof you can observe that this will be 0 then this term will not be there this term will be 0 but you do not require that for that for this thing to happen you do not require this Dirichlet boundary conditions. So even if you have Neumann boundary conditions you have ux 0t and ux lt is given to be 0 therefore this product is 0 even in that case it means we are now actually showing the uniqueness of solutions even to the Neumann boundary value problem. So what we have is dou by dou t of this equal to this and we have shown that equal to 0 and as before now this is dou by dou t of this quantity is 0 therefore this quantity is constant and this quantity is constant and that constant has to be 0 because when time t equal to 0 the initial energy is 0 because we are working with a 0 Cauchy data therefore this will be 0 as a consequence ut of xt is 0 ux of xt is 0 for all x and t and hence u is identically equal to 0 it means u is constant and that constant has to be 0 because u is already 0 on at t equal to 0 as well as on the boundaries if you are dealing with Dirichlet problem. So u of xt equal to 0 hence we conclude that the given IBVP has a unique solution. Let us look at an IBVP in D dimensions so homogeneous wave equation this is the Cauchy data but with some mixed condition like this on the boundary of omega. So let you be a solution to IBVP show that the energy defined by this formula decreases how do you show u of t is a decreasing function actually what we mean is u of t is a decreasing function we show that d by dt of ut is less than or equal to 0. So as before multiply the equation with ut rearranging the terms we will give you this integrate over omega this is what we like d by dt of ut so always we are doing anything is with this term do integration by parts this becomes c square ut and we already saw u xi into nu i summation i equal to 1 to d is dou u by dou n in one of the earlier problems we have demonstrated how this comes so this is what it is plus that term so from this equation what we get keep d by dt of ut on one side take the other thing to the other side. Now here we will use the boundary condition un and ut there is a boundary condition which connects them which after substituting that expression we get this. Now if you notice this integrand is greater than or equal to 0 d sigma has this property that it integrates non-negative function it gives a non-negative number c square is positive b is positive therefore the quantity on the RHS is negative less than or equal to 0 that means we have shown d by dt of et is less than or equal to 0. So e of t is a decreasing function of t we can also conclude uniqueness of the ibbp in this case that is left to you as an exercise. Now let us look at another equation now we have not just wave equation but we have some extra terms in the equation like this plus ut same question problem and these are the boundary conditions so initial boundary value problem for this equation. Here energy is a decreasing function of t as before multiply the given equation with ut rearranging your terms I am indicating in the red color the new terms that we never had earlier are new in this problem. Integrate and settle with this term see what happens once again it is 0 because of Dirichlet boundary conditions I am not explaining more because we have gained NF experience in deciding when is this integral 0. So therefore what I have is d by dt of et equal to minus 0 to l ut square which is less than or equal to 0. Therefore e of t is a decreasing function of now let us look at a semi-linear wave equation further also we show that there can be only at most one classical solution. So this is the delimbation right hand side is ut minus u cube if you observe ut in the previous problem it was on the other side now ut is on this side you try to follow the same proof forget about this u cube let this be removed okay remove u cube term and solve as before and see what you can show that is an exercise to you whether energy still decreases or not okay let us discuss this problem now we will show it has at most one classical solution starting point is the same let u and v be solutions consider the difference and see what is the problem that W satisfies whenever there is linear terms here after subtracting the two equations for u and v you get W tt W xx here W t here here you will get minus u cube plus v cube here you get W x0 W t W 0 t W lt everything is fine except for this term these are equation satisfied by W please pause the video make sure you get the computations correct computations. So multiply with W t equation might have changed but since we are in the context of wave equation you always multiply with W t that will give you this is the wave equation part this is the right hand side part integrate okay so this you deal with the condition that you have on W and these are the new terms let us see how to handle this will be your d by dt of E of t first time again is 0 because the Dirichlet boundary conditions so what we have is this let us do some estimation on the right hand side term which is here that u cube minus y v cube I wrote it as u minus v into u square plus v square plus u v so that u minus v became W that is why you have a W here this is a very simple estimation okay here the k will depend on explicitly u and v are known right depends on u and v so you take a bound for u and v on 0 l cross 0 t maybe so you have these terms now here a b is less than or equal to a square plus b square by 2 that is what I have used and then if you separate you get this now summarizing a last set this is the LHS RHS one term was this second term gave rise to a W t and W t square and W square W t square I have mixed with this W t square so W square will be the new term that you will see now okay at last we have an inequality featuring the function W and its derivatives but the problem here is that left hand side you have W t W x right hand side you have W t and W so this W is not there on the LHS because I would like to see it as d by dt of some quantity is less than or equal to some constant times the same quantity if I want to do that it asks me that maybe have a W here or I have a W x instead of W here so W on RHS maybe could be converted into a W x we will let us go ahead with that W x t can be written as this because this is by fundamental theorem it is W of x t minus W of 0 t which is 0 W of 0 t is 0 therefore this is simply W x t therefore modulus is less than or equal to integral modulus and then that can be further written as this into this this is what is called Cauchy Schwarz inequality integral f g integral mod f g is less than or equal to integral mod f square power half into integral mod g square power half so here f is this g is 1 so you get this now we know x is less than or equal to L therefore okay let us take the square first so this 1 by 2 will go away root x becomes x because x is less than or equal to L you get this and this is a non negative integrand 0 to x and 0 to L the relation is 0 to x is always less than or equal to the integral on 0 L therefore this inequality we have and integrate both sides with respect to 0 on the interval 0 L with respect to x and you get this so therefore this inequality which we had now becomes this now we are happy because the same W t W x on both the sides denoting E of t equal to this what we have is dou by dou t of E of t is less than or equal to k tilde times this is like E of t maybe I think there is a c square missing but that can be absorbed into this k tilde I can give c square here it is not a problem so it looks like this d E by dt less than or equal to m into E t okay now we will use grown walls inequality which says if d E by dt is less than or equal to m times E then the solution E of t is less than or equal to E 0 into E power m t okay E of 0 into m power E power m t is a solution of d by dt equal to m into E power t d E by dt equal to m E t this is a solution but because of the inequality we get inequality here that is what is grown walls inequality says now E of 0 is 0 therefore E of t is less than or equal to 0 but by definition E of t is always greater than or equal to 0 therefore E of t is 0 now it follows that W 0 therefore solution is unique to the IBVP let us summarize yet another proof of uniqueness of solutions to the Cauchy problem in full space RD was presented proof of uniqueness of solutions to IBVPs with a variety of boundary conditions was presented in fact Dirichlet and Neumann clearly the most complicated of all the problems presented in this lecture is the one for semi-linear wave equation we had to apply Cauchy-Schwarz inequality grown walls inequality and inequality connecting square integrals of W and Wx which is known as Poincare inequality in advanced theory of PDEs thank you.