 Welcome to this lecture on the Cauchy problem for wave equation in two space dimensions. We are going to use Hadamard method of descent to find solution to the Cauchy problem. The outline of the lecture is as follows. First we recall from lecture 4.5 where we have solved Cauchy problem for three space dimensions. Using that we will find a solution of Cauchy problem in two dimensions and towards the end of this lecture we will give a short tutorial on wave equation in higher dimensions where we are going to solve a few Cauchy problems. Let us recall from lecture 4.5 where we considered the Cauchy problem for three-dimensional wave equation. This box 3 stands for the wave equation in three dimensions equal to 0 that means homogeneous wave equation and this is the Cauchy data u of x0 is phi and ut of x0 is psi. Assuming that the Cauchy data satisfies phi is a C3 function and psi is a C2 function we have obtained an expression which is called Poisson-Kirchhoff formulae which represents a solution. We have obtained in fact four versions of the same formula. So, we derived f1 from there we have got f2 by expanding this derivative which gave rise to the first two terms. The third term is as it is in f1, f2 is the same and then these are the integrals on the unit sphere with center at the origin when we change back the variables we get an expression in terms of the sphere with center x radius Ct. So, that gave rise to another two formulae f3, f4. So, they are all called Poisson-Kirchhoff formulae. Now, we are going to use them to obtain a solution in two dimensions. The idea is called Hadamard method of descent. We are solving 3D problem first and then going to the 2D problem. So, it is a descent it is coming down from higher dimensions to lower dimensions. So, this is the two dimensional wave equations Cauchy problem where the equation is same Cauchy this is the homogeneous wave equation in two space variables x1 and x2 initial conditions are phi and psi respectively which are u and ut. And we also assume that this phi is smoothness is C3 and psi is smoothness is C2 on R2. So, phi and psi are given we want to solve a function we want to solve this wave equation and obtain a function which is a solution to this homogeneous wave equation and satisfying the Cauchy data which is given here ux0 is phi x, utx0 is psi x. So, Cauchy problem for wave equation in three dimensions was solved and solution is given by Poisson-Kirchhoff formulae. The idea behind Hadamard method of descent for solving Cauchy problem in 2D is solutions to 2D wave equation may also be thought of a solutions to wave equation in 3D only that these solutions do not depend on the third variable let us say x3. We use x1, x2 for the variables in R2 and x1, x2, x3 for the variables in R3. So, it does not depend on the third variable and solution u of x1, x2, x3, t is expressed now using Poisson-Kirchhoff formulae and it would not depend on x3 as a Cauchy data does not depend on x3 because we are starting with a Cauchy problem for the 2D problem 2D wave equation. So, u obtained as above solves the Cauchy problem this is the idea. So, one can reduce the integrals on spheres in 3D because the Poisson-Kirchhoff formulae in 3D feature integrals on spheres in R3. They can be reduced to integrals on disks in R2. So, this is the Cauchy problem that we start with for 2 dimensional wave equation as discussed before. Now, let us pose it for 3D wave equation. So, for that we need to add u x3, x3 that is when we get the 3D wave equation still homogeneous equation and u of x, 0 equal to phi x is given for x in R2. So, we simply put u of x, x3, 0 equal to phi x, x is x1, x2. So, this is the way we are writing we reserve this boldface x notation for x1, x2 which is an R2 as mentioned here. So, we join another variable x3 another coordinate xx3 is in R3. So, phi and psi if you notice this is a Cauchy problem for the 3D wave equation and the Cauchy data depends only on the first two variables. It does not depend on x3 because boldface x means x1, x2. Let us introduce a notation let us denote this sphere in R3 which center at x1, x2, 0 and radius ct that is set of all elements y in R3 which are at a distance ct from the point x1, x2, 0. Now, from the Poisson Kirchhoff formula F3 we get the expression for u of x1, x2, 0, t to be this. We are taking x3 equal to 0 because the solution does not depend on the x3 variable. So, therefore, I can take x3 equal to any point that is why I have taken 0 for convenience to compute these quantities and that let us call a function of x1, x2, t. Now, since phi is c3 of R2 and psi is c2 of R2 this formula gives a classical solution to the Cauchy problem in 3D because the phi and psi does not depend on x3 therefore trivially it is a function of c3 R3 and psi is also c2 of R3. So, let us compute this integral which is appearing here this integral and this integral is exactly same because anyway we have no idea what phi and psi are. So, it is a general phi let us simplify this integral which is posed on this sphere to something which is on a disc that is what we are going to do now. So, let S plus denote the upper hemisphere since the integrand is independent of the y3 variable integral on the whole sphere is 2 times integral on the upper hemisphere. Note that the upper hemisphere is a graph of this function which is defined on x1, x2 center CT radius disc to R and H of y1, y2 is square root of c2, t2 minus x1 minus y1 square minus x2 minus y2 square where D of x1, x2, ct denotes the open disc with center at x1, x2 and having radius CT. Therefore, the surface measure D sigma on S plus which is given as S plus is given by y3 equal to H of y1, y2. This is what we mean by saying S plus is the graph of this function H is given by this. This is the fact from multivariable calculus. D sigma is equal to square root of 1 plus norm grad H square dy1 dy2. So, that means D sigma is equal to CT by square root of c square t square minus x1 minus y1 square minus x2 minus y2 square dy1 dy2. After plugging in this value of norm grad H square we get this. Now substituting in this formula, this formula reduces to this formula or it changes to this formula. This formula is posed on a particular sphere in R3. Now this is posed on a disc in R2 which is what we like because we are saying the solution at the point x1, x2 at time t is given by this formula. So, this formula is also called Poisson Kirchhoff formulae. We have to remember that this is in two dimensions. So, if you compute the differentiation here dou by dou t here we get this. Rewriting the integral in the above equation using change of variable we get it on the disc of radius 1 center at the origin. Please do these computations by yourself particularly this change of variables very simple. So, all the three formulae represent solutions to Cauchy problem for wave equation in 2D. As in the case of 3D the formulae are still known as Poisson Kirchhoff formulae and these findings we state as a result. So, theorem let phi be C3 and R2 and psi be C2 function on R2. Then a classical solution to the Cauchy problem is given by this formula Poisson Kirchhoff formulae. So, now how do I check this? It is a matter of verification. So, that is left as an exercise to you. So, let us now begin a short tutorial on wave equation in higher dimensions. The problem 1. So, this is a wave equation we are taking C equal to 1. So, wave equation in 3 space variable homogeneous equation we take ux0 equal to 0 ut to be 1 in the inside the unit ball up to the boundary and 0 outside the norm x bigger than 1 it will be 0. Of course, immediate worry is the data phi is the C3 function yes it is a C infinity function in fact fine but psi is not C2 it is actually discontinuous. So, as before we did for the wave equation in 1D do not bother about this that this is not a smooth function as required for a classical solution being given by Poisson Kirchhoff formula go ahead and compute the solution within quotes because now we are not sure that it is a classical solution but still go ahead and compute that is what we are doing. So, since phi is 0 the first term drops out what you have is only this. So, therefore it is this. So, we need to compute what this integral is for the given psi. Since the function psi is identically equal to 0 outside this ball or the closure of the ball uxt the integral reduces to integral on sxt intersection v01. Now, if the sphere is completely inside the ball. So, let us write this ball this is a ball in 3D of radius 1 suppose my x is here and this is the sphere this is the sphere S of xt. So, if the sphere is completely inside the ball that is a situation here then what happens what about this intersection sxt intersection v01 it is simply sxt. Therefore, uxt is 1 by 4 by t integral over sxt of function psi is 1 on the ball. Therefore, what we get is the surface area of sxt which is 4 by t square we are in R3 sxt has 4 by t square is a surface area. So, therefore uxt equal to t and this situation occurs if and only if t is less than 1 minus norm x just a rough picture 0, 1 this is x this x and the radius is t this distance is t. So, you can just draw this line. So, this distance is norm x this distance is t that is a farthest point. So, t plus norm x should be less than 1 therefore t is less than 1 minus norm x. Now, let us consider the second situation sphere is completely outside the ball when will how will that happen you have this ball let us use a different color. So, I am x here. So, this is one way that the sphere of sxt does not intersect it is outside completely outside or another way is I have a x here it could be this. So, in this case also it does not intersect the ball, ball is here the sphere is this blue color thing. So, it does not intersect then that case uxt is 0 because intersection is empty set. So, uxt is 0 the integral is taken and domain is empty set. So, integral is 0 and this situation occurs either if t is less than equal to norm x minus 1 or when t is bigger than 1 plus norm x. So, please deduce this from the picture draw the picture and you can deduce this. Now, something else is remaining what remains is to analyze the case where the sphere and the ball intersect partially what we have considered is that never intersects are total intersects. So, let xt be such that a part of the sphere and the ball intersect and this happens in two situations for x non-zero we have to be careful with x equal to 0. So, let us consider for x non-zero t is greater than or equal to 1 minus norm x that is one situation coming from the first picture coming from the later pictures it is this 0 less than or equal to norm x minus 1 less than or equal to t t less than or equal to 1 plus norm x. Now, combining the last two sets of inequalities the nonemptive intersection is possible if and only if this happens. Modulus of norm x minus 1 should be less than or equal to t t less than or equal to 1 plus norm x this is a situation. So, what about uxt for such xt there are two things every time. We have to find out what is u of xt and what is the range of xt for which that uxt is valid. The domain of the integral in this formula sxt intersection b01 namely this it is called spherical cap of course, on the intersection psi is till 1 therefore, the computation is becoming very easy. So, what we get is uxt equal to 1 by 4 pi t into this surface area of the spherical cap and surface area of the spherical cap is known to be given by this formula. Therefore, the solution we write down in this form this is coming from the first picture this is coming from the second picture this is from the other case for x non-zero. You see why we put x non-zero because norm x is coming in the denominator that is natural it is coming. So, at x equal to 0 u of 0 1 t equal to 1 is a problematic point we will discuss this example again later when we are going to discuss how the confined disturbances propagate are propagated by wave equations in various dimensions. This is an example which will be very useful to understand that. So, there is some problem at t equal to 1 if you recall the Cauchy data has a trouble at sphere 1 it is discontinuous across norm x equal to 1 that finally leads to this trouble you have 0 1 equal to 0 and something else happens as t goes to 1 at the point x equal to 0 which we will discuss later. So, even though the Cauchy data is discontinuous we do have a classical solution in some regions identify what are those some of them. This is the same philosophy we applied in the case of Burgers equation. There also we worked with discontinuous initial conditions we found the formula then we realized that there are regions where it is actually solution to the Burgers equation and regions where it is a solution to the Cauchy problem for the Burgers equation the same thing here. So, please identify some of them. Let us move on to problem 2 where we have now changed the psi in the previous problem to phi now. Now, phi is 1 norm x less than equal to 1 0 if norm x greater than 1 but psi is 0 again as before do not worry about the smoothness of the Cauchy data go ahead and compute this Poisson formulae now reduces to this earlier it was given by the second term now it is even given by the first term. But if you notice who is inside this dou by dou t this integral term is precisely same as before. In fact, the function phi now is the same as psi in problem 1. So, we know what this is therefore it is a matter of just finding the derivative and get solution to this Cauchy problem. So, and we get this this is a solution in problem 1 now in problem 2 this is a solution there will be troubles at these boundary points of these domains of validity and that is where the there are possible singularities for the solutions. Otherwise in the interior of these ranges it must be fine and these are the solution. Now, let us look at problem 3 which is again a homogeneous 3 dimensional wave equation with c equal to 1 where we have a specific nice initial conditions phi x is x 1 square plus x 2 square it means this does not depend once again on x 3 and psi is 0 of course so it does not depend on any variable it is a constant. So, because phi does not depend on x 3 we have 2 ways of solving this problem 1 using Poisson Kirchhoff formulae in 3 space dimension second one is using in 2 space dimension we do both because psi is 0 the Poisson Kirchhoff formulae of 1 reduces to this one. So, what we have to do is really to compute this integral. So, let us compute this 3 dimensional integral phi of x plus t nu is the first coordinate of x plus t nu square plus second coordinate of x plus t nu square which is here. Let us expand this it will give us x 1 square plus x 2 square plus t square nu 1 square t square nu 2 square which is here plus the 2 a b terms that is 2 t x 1 nu 1 2 t x 2 nu 2 is here. Now, here x 1 square plus x 2 square does not depend on nu so it comes out as a constant and with the multiplied with the surface area of the unit sphere. Here t square also comes out but there is a nu 1 square plus nu 2 square remember that is not 1 because here nu 1 square plus nu 2 square plus nu 3 square is 1. Therefore, we have to do some computation let us see how this term split. Yeah as I mentioned earlier this comes out as a constant times the surface area of this which is 4 pi. This is done the second term and third term are still kept together one at a time. Yeah now we have separated them into 2 terms in fact 3 terms 2 t x 1 nu 1 is here x 2 nu 2 is here. Now, we need to compute this integral this integral and this integral. It looks like this and this integrals will behave similarly and this integral we have to compute again. Due to symmetry of the sphere this is 0 because the amount of time this nu 1 is positive is also negative on the unit sphere that is the rotational symmetry because of that this is 0 and this is 0. Please convince yourself about this. Now, once again due to the symmetry what we know is this nu 1 square integral is same as integral of nu 2 square is same as integral nu 3 square. Let us call all of them to be i. Now, we are going to play a trick 3 i is nothing but this summation earlier we noticed that nu 1 square plus nu 2 square may not be 1 is not 1 in general except at few points. But now nu 1 square plus nu 2 square nu 3 square is here which we know is 1. Therefore, what this integral evaluates to just surface area of the unit sphere which is 4. Therefore, i is 4 pi by 3. Therefore, we know this integral it is 4 point x 1 square plus x 2 square plus t square 8 pi by 3 because there are 2 terms. So, 8 pi by 3. Therefore, we have computed the integral that we wrote multiplied with this t. So, this is the expression we have on finding the derivative we get answer to be x 1 square plus x 2 square plus 2 t square. So, this is the solution one can easily check now that this is a solution to the given Cauchy problem. And now we are going to discuss about a second method which I have outlined at the beginning. And realizing that the phi dip does not depend on x 3 variable we think that a solution to this can be obtained using the expression for 2D wave equation solutions. Of course, psi does not depend on x 3 because 0. So, this is the formula u of x 1 x 2 t. Later on we are going to write it as u of x 1 x 2 x 3 t solution does not depend on x 3. Now, let us do a change of variable z equal to y minus x pi t. So, then this y becomes x plus t z from here and dy 1 dy 2 that is a measure that becomes t square dz 1 dz 2. And here I have substituted and therefore the denominator changes and the domain which is a disk radius ct center x now becomes radius 1 center 0. Now, we have to compute this integral and then substitute. So, after substituting the formula for phi we get this. So, let us compute the integral which is here. I have written d for d disk of radius 1 with center origin just for notational convenience otherwise this integral you see how much space is coming. So, it will fill up the spaces to avoid that I have done d. So, d stands for the disk of radius 1 center of the origin. So, this on expansion numerator we will get this. Now, exactly as before we are going to compute these integrals which I leave it as exercise. This in both integrals are 0 using polar coordinates show that this integral is 2 pi and this integral is 4 pi by 3. Now, we are going to go and substitute in the expression for the solution. Please do these computations on your own. Once you compute integrals nothing is left everything is easy. So, we get of course the same solution and it does not depend on x 3. Because the problem is posed as a solution to the 3D wave equation we have to write u of x 1, x 2, x 3, comma t equal to x 1 square plus x 2 square plus 2D square. Let us summarize using Hadamard's method of descent solution to Cauchy problem for wave equation in 2D is found and we solve 3 Cauchy problems. We used very nice functions as Cauchy data to make computations very easy. We can only imagine the scale of difficulties in dealing with more general Cauchy data. Thank you.