Lecture 55

Welcome back, in the last lecture we had discussed the Fourier transform on higher dimension as and we got Planck's rule as well as the inversion formula. So, in today's lecture we are going to see some of the applications of Fourier transform related to PDE and also later on we will see its application to other problems emerging from science and engineering. To start with on R n define the Laplacian eta is equal to summation i from 1 to n del square by del x i square. So, now consider the equation delta of u is equal to 1 by c square del square u by del t square. So, for dimension n equal to 3 this determines the behavior of electromagnetic wave in the vacuum. Now for convenience, so let us take consider c to be equal to 1 and then the problem what we are interested is delta u is equal to del square u by del t square on R n cross 0 infinity and u of x 0. So, this is on the upper half plane we are considering when we come to the boundary of the upper half plane that is t is equal to 0 it is given by f of x this is on R n and then this is the second order differential equation PDE. So, this is x 0 is equal to g of x and for our suitable use of Fourier analysis let us take let us assume both f and g to be in Schwarz space it is an ideal situation I mean. So, we are assuming to perform to convince ourselves how to solve this equation ok. This is called a Cauchy problem for wave equation. So, this is a wave equation and this is for wave equation ok. So, then like what we had seen in one dimensional case, now if we are taking the Fourier transform on both the sides of the equation by taking Fourier transform we get delta u hat xi comma t Fourier transform we are taking only on the space variable. So, this is equal to del square u by del t square hat at xi t and this what we know we have computed this that this is going to be minus 4 pi square mod xi square then u hat of xi comma t and this on to compute the Fourier transform on the right hand side we have seen that the operator del by del t and del square by del t square they are going to commute with the Fourier transform because we can we are assuming nice condition we can take out the derivative outside the integral. So, now what we get is that this is del square by del t square and then this is u hat xi t. So, this now has is an ODE second order ODE and then this will imply that u hat of xi comma t is equal 2 pi mod xi t plus B xi sin 2 pi mod xi t. Now, what we know is that we need to find out the constants A xi and B xi and what is known to us is that u hat xi at 0 this is equal to integral over R n u of x comma 0 e to the power minus 2 pi i xi dot x dx which is nothing, but this is f of x this is f hat of xi. So, u hat xi at 0 this is equal to A of xi and t is equal to 0 sin is 0. So, this is nothing, but what we have observed is f hat of xi. Similarly, if we want to take del by del t of u hat xi at 0 this is again what we are going to get this is G hat at xi and now u hat of xi is this. So, this is going to be A xi minus sin 2 pi mod xi into 0 and then 2 pi mod xi plus B xi cos 2 pi mod xi into 0 into 2 pi mod xi. So, this will imply 2 pi mod xi into B xi this is equal to G hat at xi. So, B xi is equal to G hat by 2 pi. So, now, if we plug in the value then we get u hat of xi comma t this is equal to f hat of xi cos 2 pi mod xi t plus G B xi is G hat of xi by 2 pi mod xi. So, this is sin 2 pi mod xi t divided by 2 pi mod xi this is what is if u hat. So, now, by the these are all f and G are nice class of functions what we have taken and therefore, by Fourier we can write x at t this is equal to integral over R n f hat of xi cos 2 pi mod xi t plus G hat of xi sin and e to the power 2 pi i xi dot x this is what is going to be the solution of the equation wave equation we had considered and obviously, u of x comma 0 if you take it then this is f hat of xi this guy is going to vanish. So, this is f of x u t of x comma 0 is nothing, but by the Fourier inversion this is G of x. So, this is the solution as a matter of fact one can show that the solution is unique, but in the present lecture we are not going to address that problem. So, now, the next one we will consider is the heat equation on the higher dimension. So, consider the time dependent heat equation this is equal to delta of u is equal to del u by del t on R n cross 0 infinity and u of x comma 0 this is equal to f of x and we as before we are assuming that f is going to be in the Schwarz class. So, again we play the same game taking Fourier transform we get that minus 4 pi square mod xi square u hat of xi comma t this is equal to and this is del by del t of u hat of xi t again this is a ODE first order ODE. So, we know precisely what this is going to be. So, this will imply that u hat of xi comma t this is equal to some constant and then e to the power minus 4 pi square mod xi square hence what we have got that u hat at xi 0 this like before we can get that this is f hat of xi. So, this will imply that u hat xi t this is equal to f hat xi is replaced by f hat of xi into e to the power minus 4 pi square. So, this will imply now if we define H t of x this is equal to integral on R n e to the power minus 4 pi square mod xi square t and e to the power 2 pi i xi dot x dx. So, d xi then we know that this is how to compute the Fourier transform of the Gaussian this is nothing, but the dilated Gaussian. So, this is 1 by 4 pi t to the power n by 2 because 1 by square root of 4 pi t what we need to deal with and this is minus mod of x square by 4 t. So, this is the usual heat kernel this is what exactly what we had got in dimension 1 2. So, u of x t this is nothing, but f convolution of H t at x and now next if we want to compute steady state heat equation this is nothing, but del of u plus del square u by del t square this is equal to 0 and you take u of x 0 this is equal to f of x ok. Now again by the same method what do we get is that we get u hat of xi comma t plus del square by del t square u hat xi comma t this is equal to 0 and we know that then this is again an ODE we know how to solve it this is equal to a of xi then e to the power. So, this is minus 4 pi square mod xi square e to the power minus 2 pi mod xi. So, in order to find the value of u hat of xi 0 as before is f hat of xi now. So, u if we define P t of x such that is a kernel which is denoted by some dependent on some constant dimensional constant which is comes through the gamma function by t by t square plus mod x square to the power n plus 1 by 2 that is the analog of the Poisson kernel in the higher dimension then one can show that P t hat of xi this is equal to e to the power minus 2 pi mod xi t. Then u of x comma t this is equal to f convolution of P t at x that is what is going to be the solution of the steady state heat equation or for that matter the Dirichlet problem on the upper half plane. Now, I would like to indicate how does one arrive at this in the higher dimension which is a little tricky. So, I will not give the full details, but you can find out what is going on while computing this Poisson kernels are Fourier transform ok. So, computation of Fourier transform Poisson kernel ok. So, if we take f to be a nice class function with that means, which has a moderate decay in f is a function from R to C with a moderate decay, then there is a subordination principle which works then this is minus infinity to f of x dx this is equal to minus infinity to infinity f of x minus 1 over x dx ok. So, this is a little the trick is involved. So, now let us take RHS this is I can write it minus infinity to 0 f of x minus 1 by x dx plus 0 to infinity. Now, if I make a change a variable x going to minus of x what I will I am going to get is that this is 0 to infinity f of 1 over x minus of x dx plus 0 to infinity f of x minus 1 x 1 over x dx. So, now, we can see that x is a non negative number. Therefore, I can make the substitution substitute x is equal to e to the power t then what do we get here? Now, this becomes minus infinity to infinity f of e to the power minus t minus e to the power t then dx is e to the power t dt plus here 0 can be attained at minus infinity to infinity f of e to the power t minus e to the power minus t e to the power t dt. Now, in this if I make a change of variable then what we are going to get is minus infinity to infinity f of e to the power t minus e to the power minus t into e to the power minus t dt and here. So, if I take then this is e to the power minus t plus e to the power t then we have got dt and this is equal to minus infinity to infinity f of 2 sin hyperbolic t and this is 2 cos hyperbolic t dt. So, now, if we make a change of variable 2 sin hyperbolic t is equal to x if we take then obviously, what we are going to get that this is minus infinity to infinity f of x dx. So, this is the subordination principle what we have got. So, now, if I take in particular consider f of x is equal to e to the power minus t x square. So, therefore, what one is going to get is minus infinity to infinity e to the power minus t x square dx this is equal to minus infinity to infinity e to the power minus t x minus 1 by x whole square dx and which we know that left hand side is square root of pi by square root of t and then the right hand side is e to the power this is minus 2 t e to the power 2 t is going to come out minus infinity to infinity e to the power minus t x square and then minus t by x square dx. So, making a change of variable what we get and bring it to the other side e to the power minus of 2 t this is equal to square root of t by square root of pi and then this is look at this is an even function. So, therefore, I can make it 2 times 0 to infinity e to the power minus t of x square minus t by x square dx. Now, which making a change of variable as t x square is equal to y what we are going to get that this is 1 by square root of pi 0 to infinity e to the power minus y e to the power minus t square by y dy by square root of y just with a change of variable if you are taking t of x square is equal to y. And let t is equal to pi mod x then what we have is that e to the power minus 2 pi mod x is equal to 1 by square root of pi 0 to infinity e to the power minus of y and then e to the power minus pi square mod x square by y dy by square root of y. Now, we want to take the Fourier transform of that as you can see. So, if you take the Fourier transform e to the power minus 2 pi mod x this hat xi this is going to be 1 by square root of pi integral 0 to infinity e to the power minus of y then I am changing the order of the integration e to the power minus pi square mod x square by y and e to the power minus 2 pi i xi dot x and this is d dx and then this is dy by square root of y. As this we know how to compute the Fourier transform and then by using the gamma function this is we will get a constant and this one can is that this is P 1 hat at xi some constant which is going to depend on dimension. So, this precisely is going to be the this is P 1 of xi because 1 over 1 plus mod xi square is going to come into here with the gamma function and then that is what is going to give us the identity. So, this is the sketch I mean usually it is proved to the subordination formula. So, hence what we got the steady state heat equation on the upper half plane which means we are given a function on the boundary and we are extending it to be a harmonic function on the upper half plane and that solution is going to give by F convolution with the Poisson kernel. So, we will see some further application as well as what can be the possibilities of we will discuss little bit about the Fourier series in the higher dimension in the next lecture and then some more applications of that too. Thank you. .