 Let us continue our qualitative study of the wave equation and today we are going to look at what is called causality principle and finite speed of propagation a property which is exclusive to hyperbolic equations we will look at them through examples. This is these are the two sides of domain of dependence and domain of influence of the concepts. Causality principle, finite speed of propagation. Causality means cause and effect. What are the reasons in the past that are responsible for the current state? What will be the future events for which the current state is responsible for our influences? These questions were answered in lecture 5.2 using the explicit formulae for solutions to the Cauchy problem. In this lecture we attempt to answer the same questions without using the formulae for solutions. This kind of justifies the use of the word qualitative analysis in the title of this chapter because we are not using any quantitative formula for the solutions. The analysis presented in this lecture is a typical illustration of an a priori analysis. A priori means done before. Conclusions can be drawn on solutions despite zero knowledge on their existence. We may not be even knowing whether solution exists or not. Still we can conclude certain things about the solution of course if they exist that we do not know. So in this discussion once again like in lecture 5.2 we are going to switch off the non-homogeneous term and we have already explained reasons for that in lecture 5.2. So let us state in the form of a theorem. The hypothesis is let you from Rd cross 0 infinity to R be a classical solution of the Cauchy problem for homogeneous wave equation that is this is the problem where phi and psi are given of course in the Cauchy problem and use a solution to this Cauchy problem. Conclusion for X0 t0 a point in the space time the value of u of X0 t0 depends only on the values of phi and psi in the closure of this open ball with center X0 and radius ct0. Closure of this open ball is nothing but the closed ball with the same radius and center. Recall that B of X0, ct0 is the open ball with center at X0 in Rd and having a radius of ct0 lying in Rd. So maybe we just briefly write what that is. So those elements in Rd whose distance the Euclidean distance from the point X is less than ct0. This is the open ball and the closed unit ball we do not use but then let me introduce good notation is that those elements in Rd such that y minus X is less than or equal to ct0. Proof of this theorem it follows immediately from the formulae for solution to the Cauchy problem namely D'Alembert formula for D equal to 1, Poisson-Kirchhoff formulae for D equal to 2 and 3. This was done in lecture 5.2. A direct proof is presented in this lecture without using the explicit formulae for solutions. Of course we have to use something that may be some experience. Okay we prove the theorem for D equal to 1 only why is that because the proof for D equal to 2 and 3 are similar but for the obvious modification that needs to be done to the proof of D equal to 1. To avoid a lot of repetition we do not do the proof for D equal to 2 comma 3 but mention one important inequality there and how that can be derived just the idea we will present. So proof of Cauchy principle for D equal to 1. Let X0 t0 be a point and r cross 0 infinity. Consider the characteristic triangle which is a triangle formed by the characteristics lines through X0 t0 on the x axis that is this is the point X0 t0 this point is X0 minus ct0 is the point X0 plus ct0. This is called the characteristic triangle. What are these lines? This is X minus ct equal to X0 minus ct0 and this line is X plus ct equal to X0 plus ct0. Of course this line is nothing but t equal to 0 x axis. So fix a t such that t lies between 0 and t0 and draw the line t equal to t we will get a trapezium recall point X0 t0 X0 minus ct0 X0 plus ct0 draw this line t equal to t. So we get a trapezium which is formed by the intersection of all these lines that is this part let us call it f. So multiply the wave equation this is a trick the wave equation you have to multiply with ut you get this and re-organize the terms you get this if you expand this it will reduce to this. Now this is in a good shape because here derivative utt into ut is there u x x into ut is there whereas here some t derivative of some quantity x derivative some quantity is there it is like the divergence theorem we are in a good shape to apply divergence theorem therefore this is a good arrangement. So in fact as you see the divergence with x and t here of this quantity is 0 that is precisely this equation this equal to 0 means this. Now integrate this equation on the trapezium region that we have indicated on the previous slide you get this now we are ready to do integration by parts or apply Green's theorem and divergence will be converted to something else. So what we get is that this divergence of course it is equal to 0 is now converted to an integral on the boundary of f f is the trapezium region boundary of f is actually the trapezium and that this becomes the integrand dot n d sigma n is the normal outward unit normal to the points of boundary of f of course that varies from point to point on the boundary of f. So the boundary of the trapezium consists of four lines actually the boundary of the trapezium region is the trapezium itself it consists of four lines they are the base of the trapezium denoted by b given by the equation t equal to 0 a part of the characteristic denoted by k 1 given by the equation x plus ct equal to x naught plus ct naught upper part of the trapezium denoted by t t for top given by the equation t equal to t and a part of the characteristic denoted by k 2. So remember this is the point x 0 t 0 then we have drawn this. So this is x minus x naught minus ct naught x naught plus ct naught then we took this part. So this equation for this is t equal to t. So this we are calling a b for base this is k 1 p for top and k 2. So the integral is now on these lines. So we need to determine what is the normal to each of these sides. So base of the trapezium outward unit normally 0 minus 1 top the unit outward normally 0 1. So this is the base this is in the direction of negative t axis because these x axis this is t axis positive t axis direction. So this is in the negative direction so 0 minus 1 and here the top normally is in this direction outward unit normal it is in the direction of the positive t axis so 0 1. Then we have the sides of the trapezium which is k 1 and k 2 at this point because k 1 itself is straight line it is very easy it is constant on the direction is the same on all the points on k 1. Similarly the outward unit normal is in this direction and that is constant for all points on k 2. Therefore life is simpler so this integral on the boundary now we can split into 4 parts which is b union k 1 union t union k 2. So on b we have applied what is the normal similarly on k 1 k 2 1 t we have used the formula for the normal that we have written down on the previous slide we get this expression. So some of these 4 terms is 0. Now we do a trick the integral on k 1 can be expressed like this and this is an integral on k 1 of some non-negative quantity because of the presence of the square this is always greater than or equal to 0 and hence this integral is greater than or equal to 0 that is that is the property of d sigma non-negative functions internally will be non-negative similarly here remember this d sigma is nothing but the measure on the boundary which is coming from the domain trapezium. So this has all the nice properties if you integrate a non-negative function integral will be non-negative. So the integrals over k 1 and k 2 both of them are non-negative numbers okay. So here I have some of 4 quantities in that some of 2 of them is non-negative. So what can I say about the some of the other 2 it should be non-positive that means that these 2 terms together is less than or equal to 0 which means I have the integral of this quantity on the top is less than or equal to integral of this quantity on the bottom. If you notice from this inequality if u is 0 on the bottom u and ut are 0 on the bottom then this is 0 then they will be 0 on the top also this is the one which gives us uniqueness of solutions as we are going to see on the next slides. So this inequality is called domain of dependence inequality that means on the trapezium on the top portion the integral is less than or equal to integral on the bottom portion. So let u and v be solutions of the Cauchy problem defined w by the difference u and v that is w equal to u minus v note that w solves the wave equation due to linearity of the wave equation of course and the Cauchy data will be 0 because both u and v are solutions to the same Cauchy problem. Therefore both u and v will satisfy the same Cauchy data and hence the difference will be satisfying the 0 Cauchy data. Therefore w of x0 is 0 and wt of x0 is 0 on the bottom. Now we can apply the domain of dependence inequality and conclude that we have this right hand side has become 0 right hand side was the same integral on b that is 0 because on b both wt and wx are 0. Therefore this is what we have but if you look at this already non-negative quantity and we are saying that is less than or equal to 0. So this is always greater than or equal to 0. Therefore the only possibility is that integrand is 0 which means wt at what point x, t and wx at the point x, t is 0. Now t is arbitrary we will choose and capital T less than t0. Therefore we get wt of xt and wx of xt is 0 for every xt belonging to the characteristic triangle. This implies that w is a constant function but w is already 0 on b. Therefore it must be 0 everywhere. W is 0 everywhere is same as saying u is equal to v everywhere in the characteristic triangle. Therefore the solution is the same. u at x0, t0 will be same as v at x0, t0. So proved on the last slide w is a 0 function inside the characteristic triangle. In particular u of x0 equal to v of x0, t0 this finishes the proof of the theorem. This is another proof of uniqueness of solutions to the Cauchy problem. We already gave one proof of uniqueness earlier. Now this is another proof of uniqueness of solutions. One more proof we are going to see using the energy method. Pretty much the actors in that energy method we have already seen in the domain of dependence inequality. We will do that in the forthcoming lectures. Consequences of domain of dependence inequality. The above discussion shows that u of x0, t0 depends only on the values of the Cauchy data on the base of the characteristic triangle defined by x0, t0. Namely the interval on the x axis x0 minus ct0, x0 plus ct0. In other words the Cauchy data phi psi at a spatial point x0 can influence the solution only in the region enclosed by the two characteristics starting from x0, 0 on the x axis which is given by xt in the space time domain such that x minus ct is less than or equal to x0 and x plus ct is greater than or equal to x0. Let us look at the causality principle for d greater than or equal to 2 and characteristic cone. What is a characteristic cone? The characteristic cone also called light cone at a point x0 t0 in space time is defined as the set xt in rd cross 0 infinity such that norm x minus x0 equal to ct times mod t minus t0 where norm x minus x0 is a Euclidean distance in rd between x and x0. The characteristic cone at a point x0 t0 together with its interior together with its interior this is only the surface this is with the interior we are considering now that is called solid light cone that is the solid light cone is this set. So this is the picture this you can imagine r2 or r3 it is easy to imagine r2, r3 you cannot really imagine because the picture will be in four dimensions but imagine this is r2 then this is what is called forward cone or future cone this is the backward cone or past cone this is vertex which is x0 t0. So the definition of a characteristic cone is consistent with the definition of a characteristic hyper surface which we have introduced in lecture 3.7. Recall that the analytic characterization for a hyper surface given by phi of xt equal to 0 to be a characteristic surface is at grad phi dot a grad phi equal to 0 where a is the diagonal matrix minus c square minus c square minus c square d times and 1 for the wave equation in d space dimensions. So this equation takes this form when we expand this gradient is in xt and what is a is a diagonal matrix described in the last slide. So when we do that we get this expression it is phi t square minus c square into norm grad phi square equal to 0. So this equation is nothing but phi t square minus c square mod grad phi square equal to 0 or you can put norm grad v square if you want or phi t is equal to plus or minus c mod grad phi. Note that this function which is here is a solution to the last equation you can substitute and check. Now what is phi of xt equal to 0 represent it is nothing but the characteristic cone through the point x0 t0. So why the characteristic cone is also called light cone characteristic cone is a union of all light rays that emanate from the point x0 t0 which travel at the speed c that is mod dx by dt equal to c this is the speed expression for the speed. In other words this set which we have here is nothing but union of these sets what is this this is a line right passing through the point x0 t0 because when t equal to t0 in the direction v. So take all this direction or the vectors with this length c and then this is precisely that. So both the sets are same that is why the characteristic cone is also called light cone speed is c okay t cross sections of light cone what they are each t cross section of the solid light cone will be a solid sphere if you omit solid here light cone will be sphere. Solid sphere means the entire interior is included that is for each fixer t the intersection of light cone with the hyper plane t equal to t is a d dimensional sphere lying in the hyper plane t equal to t that is sphere lying in d dimensions not d dimensional sphere sphere will be one dimension less if you consider solid sphere yeah. So let us not discuss that it is a sphere lying in rd that is what this sentence means when the t sections are projected in the space rd they are the spheres with center x0 and radius c times mod t minus t0 clearly as t goes to infinity the spheres are expanding. So what is the idea of the proof of causality principle in d greater than or equal to 2 as in the case of d equal to 1 the main idea is to multiply the wave equation with ut and then integrate it on some part of the solid light cone and then perform integration by parts. The region of integration was a trapezium shaped region for d equal to 1 which would now correspond to first term of the cone that is the reason why we used the notation f to denote the trapezium shaped region in d equal to 1 first term of the cone. So imagine this is the kind of cone okay so you cut this so you have this so this is the first term of the cone. So you have a bottom portion you have a top portion and you have a lateral portion. So multiply the homogeneous wave equation with ut and rearrange exactly in one dimension we have done so we get this and then integrate on the first term of the cone defined by t equal to 0 this is the bottom portion t equal to t is the top portion. Now proceeding exactly as in d equal to 1 the domain of dependence inequality we obtain that is exactly the same integral over t is less than or equal to integral over b. From here the uniqueness of solutions to Cauchy problem follows again okay you take u and v to be solutions even for the non-homogeneous Cauchy problem subtract u minus v call it as w that will satisfy the homogeneous wave equation with the homogeneous Cauchy data which means that integral on b is 0 on bottom that function and the derivative with respect to t will be 0 therefore we have this is 0 and this is true for every arbitrary t and therefore in the first term both of them coincide and hence even at the point x naught t naught same proof. So with these modifications the proof given for d equal to 1 goes through for d equal to 2 comma 3 also we will see one more proof of uniqueness using energy method later on. So what are the consequences of domain of dependence inequality the solid backward cone is called the past history of the vertex x naught t naught. So if this is x naught t naught the past cone we said is this right so this is the past and this will be the future of t naught in other words the Cauchy data at a spatial point x naught can influence the solution only in the future cone with vertex at x comma 0 which is forward solid light cone emanating from x naught comma 0. The figure on the next slide depicts past and future cones located at a point x naught t naught. So we have already seen this picture. So finite speed of propagation finite speed of propagation is a common feature for wave equation in all dimensions we are going to study the speed of propagation of the Cauchy data thus the homogeneous wave equation is studied. We illustrate finite speed of propagation using examples for d equal to 1 to 3. So let us consider d equal to 1 let the initial data phi and psi be 0 outside this interval minus 3 comma minus 2 therefore u of 0 0 is 0 x equal to 0 t equal to 0 that is actually phi of 0 0 is not in this interval therefore phi is 0. Let us now study the behavior of u of 0 t that means I am standing at x naught equal to 0 I want to study what happens for t positive. For t positive such that minus ct is bigger than minus 2 that is t less than 2 by c u of 0 t will be 0 we will see this in a picture it will be very easy. So that is the information at t equal to 0 has not reached the point x naught equal to 0 till this time t1 equal to 1 by c from which time the information will be received at this point. Thus it took a time of 2 by c to travel a distance of 2 why the distance of 2 this interval is at a distance of 2 from the point x naught equal to 0 and thus the speed is c u of 0 t remains 0 till t equal to 2 and illustration of this example is given on the next slide. So here we are standing at this point x naught equal to 0 right now at time t equal to 0 because this is t equal to 0 the information is only here in this interval minus 3 comma 2 outside that phi and psi are 0 therefore u is 0 here let us consider this instant time instant 1 then also if you see from our formulae this does not intersect the interval minus 3 2 therefore u is still 0 here when you go to 2 that is when you pick up some information from here from the interval it has reached this point 0 at time t equal to 2 because it is hitting this point possibly phi is nonzero here who knows but because the support of phi and psi is 0 it will be 0 only here phi of 0 phi of minus 2 psi of minus 2 will be 0. So till time 2 you will not reach the moment you cross time 2 2 plus something this time then definitely you are intersecting this piece this side you get nothing because this side anyway phi and psi are 0. So you may pick up some information from here that means information from this interval minus 3 comma minus 2 where the support of phi psi lies is reaching the point x not equal to 0 which is at a distance of 2 distance is 2 it takes time 2 time 2 units. So this example is with c equal to 1 so speed is 1 distance is 2 so you take 2 units of time to reach information so after 2 the information starts coming for example if you are at 3 as you see here domain of dependence for u of 0 comma 3 contains minus 3 comma minus 2 intersection non-empty in fact in this example it contains. Now if you are at this point 4 time 4 at this point you see that this is the interval definitely in this you have the support of phi and psi so information has reached there. So in d equal to 2 comma 3 let us look at consider a ball b of 0 r center 0 origin at the origin and radius r it let it denote open ball of radius r with center at origin support the suppose that the Cauchy data is supported inside this ball we know that u of x t depends on the values of the Cauchy data on this b this is a closed ball x c t intersection b 0 r only this is non-empty we have non-zero solution otherwise it will be 0. So if the above intersection is empty u will be 0 in other words for each fixed t positive the support of the function x going to u of x t is contained in union over y in this ball of radius r center origin of the closed ball with center as y and radius c t which is nothing but this ball ball center 0 radius r plus c t. Thus for each fixed t the support of the solution is a compact set if the Cauchy data is compactly supported we have observed this in d equal to 1 and illustrated with the picture also in other words the support spreads with finite speed let y be not in this ball of radius r with center 0 what will happen then not only u of y comma 0 is 0 this is the initial time right phi and psi are concentrated inside b 0 or outside that phi and psi are 0 y is a point which is outside that therefore u of y 0 is what it is phi of y and y is not in this ball therefore this is 0 but also u of y t will continue to be 0 for all times up to norm y minus norm y minus r by c this is referred to as finite speed of propagation so what we have done in this lecture is we have derived the domain of dependence in equality for d equal to 1 for d equal to 2 3 we just give the idea so using domain of dependence in equality we proved uniqueness of solutions to the Cauchy problem the second proof of uniqueness and we revisited domains of dependence and influence with past future cause effect points of view thank you