 So, we have discussed the inversion theorem, we have discussed the Parseval formula and we have discussed the fact that the Fourier transform maps L 2 onto L 2. So, now we use these ideas of Fourier transform to solve the initial value problem for the heat equation and later we will discuss the wave equation. So, what is the initial value problem for the heat equation? The heat equation in one space variable ut equal to uxx, u at x0 equal to f of x that is the first display that you see in the slide. Let us assume to begin with the initial datum f of x lies in the Schwarz space s and let us compute the Fourier transform with respect to the spatial variable x. In other words you treat the T as an auxiliary parameter and you take the Fourier transform with respect to x. So, take the Fourier transform of the heat equation it is it is d dt of u hat in the first term. In the second term remember that minus del 2 by del x square the Fourier transform is multiplication by chi square. So, when you Fourier transform the heat equation what you get is d dt of u hat of chi t plus chi square u hat equal to 0 that is an ordinary differential equation for u hat. Of course, it is a linear ODE, but how are you going to get the initial conditions? So, when time t equal to 0 what is u hat you simply take the initial data that is u at time t equal to 0 is f of x. So, take its Fourier transform you get u hat of chi 0 equal to f hat of chi. So, you got the ODE and you got its initial condition. So, you immediately solve this ODE u hat of chi t equal to c e to the power minus t chi square put t equal to 0 and remember that u hat of chi comma 0 is f hat of chi and. So, we get u hat of chi comma t is f hat of chi x of minus t chi square. Now x of minus t chi square is a Gaussian and it is the Fourier transform of another Gaussian. What is that function whose Fourier transform is x of minus t chi square either use inversion theorem or use the fact that the Fourier transform of a Gaussian is another Gaussian with a different parameter. So, observe that e to the power minus t chi square is a Fourier transform of capital G x t and the capital G x t is x of minus x square upon 4 t 1 upon root 4 pi t. This function G of x t that has been displayed on the slide is called the heat kernel and this heat kernel plays a very crucial role in the theory of probability. Probability, stochastic processes, diffusion processes, harmonic analysis, differential geometry in all these subjects the heat kernel plays a very fundamental role. So, now that we identified what function is it whose Fourier transform is this factor we can write u hat of chi t is f hat G hat chi t recall the convolution theorem the Fourier transform of the convolution is a product of the Fourier transforms and so the last equation can be written as u hat of chi t equal to G star f hat. So, now use the inversion theorem u of x t equal to 1 upon root 4 pi t integral minus infinity to infinity f of s x of minus x minus s the whole squared upon 4 t d s. We have got an integral representation for the solution of the initial value problem. We have solved the heat equation and obtained its solution as an integral. The integral representation as I said is very important because it enables us to make estimates easily. The formula was derived by assuming that the initial condition f of x was in the Schwarz space. Now you may ask what is the initial data is not in the Schwarz space what is the initial data is in L 2 for example, how do we cope with the problem? It is very simple observe that the integral there is an x of minus x minus s the whole squared upon 4 t this exponential factor is rapidly decreasing. So, I can allow the f of s to be in spaces like L 2. In fact, even if the function f of s grows exponentially fast if it is an function of exponential type the integral still makes sense. So, what you could do is you could take this integral representation you can allow the function f of x to be of exponential type the integral will converge very rapidly. So, you can differentiate under the integral sign and you can verify that the integral still defines the solution of a heat equation with this initial data f of x. So, this is an exercise for you to check that even if the function f of x is of exponential type the formula that we obtained does indeed furnish a solution of the initial value problem that we are trying to solve. Some more exercises for example, you take the initial condition to be x squared solve the heat equation take the initial condition to be cosine A x solve the heat equation what about the solution with initial condition sin A x. So, these are some standard initial conditions with respect to which you can solve the heat equation using the integral formula that we have obtained in the previous slide. Suppose the initial condition is a continuous function which is positive on the interval minus 1 1 you got a initial data which is positive on minus 1 1 and 0 outside the closed interval minus 1 1 then look at the integral that we obtained the integral the exponential factor is obviously positive and the datum is non-negative. So, f of x is non-negative. So, the integrand is non-negative and so the integral will be strictly positive. So, this integral will be strictly positive no matter how small t is because you are integrating over the whole space and integrating over the whole space and. So, what we notice is that if the initial condition is a continuous function which is positive on the open interval minus 1 1 and 0 outside the closed interval minus 1 1 then the solution is going to be positive at all the points x t no matter how large x is and how small t is. So, it means that the effect of initial heat distribution. So, the heat distribution is localized on the interval minus 1 1 and the heat distribution is 0 outside minus 1 1 and now I could take x to be very very large say 10 kilometers away and t to be very very small say a few nanoseconds, but you will see that u of x t is going to be positive. So, x could be 10 kilometers and t could be a few nanoseconds and. So, it means that as soon as time becomes positive the effect of the heat distribution which is localized on minus 1 1 is felt everywhere in space as soon as t is positive. So, how is this physically possible? This can be described by saying that the effect of initial heat distribution is instantaneously propagated throughout space this is obviously not a physically tenable situation it will take some time for the heat distribution to propagate. The heat equation nevertheless explains physical phenomenon accurately because although t is very small and x is large the value of u of x t that you get from the integral is going to be very very small though in real terms it appears to be untenable it is a good approximation the heat equation is a very good approximation to real physical phenomenon despite the fact that the speed of propagation is infinite. Now, let us take another differential equation which we saw using Fourier transforms this time it is not a partial differential equation it is an ordinary differential equation it is called the airy's equation airy George Biddle airy studied this function the that bears his name in connection with his researches on the intensity of light in a neighborhood of a caustic. You must see Watson's treatise on the theory of Bessel's function page 188 the work dates back to 1838 before commencing on the discussion of airy's function here is a pointer to the interesting life of Sir George Biddle airy another interesting account that I read long ago was by the astronomer Patrick Moore unfortunately I am not able to locate the exact book at the moment but airy's equation is this equation 4.15 y double prime x minus one third x y equal to 0 that is the airy's differential equation now I also mentioned the word caustic what is a caustic now I do not know whether you have observed this suppose for example you have a bowl which is made of metal a metallic bowl like a stainless steel bowl for example and you pour milk in the stainless steel bowl and there is light coming from far such as sunlight or say from a from a tube light what you will see on the surface of the milk are two curves are two bright curves that you see those are caustics those are envelopes of these rays so in geometrical optics you you talk about the rays of light and these rays envelop a certain curve and that curve is a caustic so near a caustic airy found that the intensity of light must be infinite so what it shows is that geometrical optics fails near a caustic but that is all physics it is optics let us not get into that let us look at Fourier transforms and its use in the study of equation 4.15 before taking up the airy's equation let us discuss a few things that you have studied in your elementary calculus courses of how to handle this integral 0 to infinity sin x dx upon x 4.16 that is a display in the slide that you see this integral converges as we know of course we also know its value it is pi by 2 you can use Cauchy's theorem to calculate the value but the value is not important at the moment what we need to look at is how to prove the convergence of this integral 4.16 let us break up the integral into a sum of two integrals 0 to pi by 2 sin x dx upon x plus integral from 1 to infinity sin x dx upon x the integral from 0 to pi by 2 is an innocent looking integral because sin x upon x is continuous on 0 to pi by 2 you define it to be 1 at the origin so the first of these two integrals is a proper Riemann integral of a continuous function on a bounded interval the second integral that you see integral from 1 to infinity sin x dx upon x is what we need to show converges okay so let us turn to that how do you deal with this integral 1 to infinity sin x dx upon x you first write it as limit as R tends to infinity integral from 1 to R sin x dx upon x we must examine whether this limit exists or not let us write sin x as d dx of Cauchy's with a negative sign thrown in and let us perform an integration by parts of course there will be two terms in one term the derivative will shift from Cauchy's to the other factor that will give you Cauchy's dx upon x squared the negative sign will go away because of integration by parts and it will reappear because of the derivative on the 1 upon x factor and the boundary terms are cosine 1 minus cosine r upon r now cosine 1 is an innocent constant as r goes to infinity cosine r upon r goes to 0 and we are left with limit as R tends to infinity integral from 1 to R Cauchy's dx upon x squared but Cauchy's dx upon x squared is a convergent integral on 1 to infinity so limit as R tends to infinity integral from 1 to R Cauchy's dx upon x squared is simply Cauchy's dx upon x squared integral from 1 to infinity so we have seen how to handle convergence issues of this conditionally convergent integral now let us take another example before we take up the Aries function this integral will be an integral which is similar to the one that we are going to encounter so what we are going to encounter is a more complicated integral but let us take a similar one which is slightly easier so consider this integral 0 to infinity cosine of x squared dx this is called the Fresnel integrals integral 0 to infinity cos x squared dx and integral from 0 to infinity sin of x squared dx these two integrals are called Fresnel integrals they also appear in optics in connection with the theory of Fresnel diffraction I am sure you have seen how to calculate these integrals using Cauchy's theorem in complex analysis and you also know how to find the value of this root pi by 2 root 2 is the answer but again the exact value of the integral is not important at the moment what is important is how to understand whether this integral converges or diverges in fact it converges conditionally again the integral from 0 to 1 is standard proper Riemann integral we need to discuss the convergence on the interval 1 to infinity only so let us concentrate on the integral from 1 to infinity cosine x squared dx where we make the change of variables x squared equal to u when you put x squared equal to u I get the integral from 1 to infinity cos u du upon 2 root u and now the 2 is of course an innocent constant cosine u can be written as d du of sin u and again I should integrate my parts as we did in the previous example exercise show that the integral j converges but not absolutely it is a conditionally convergent integral so conditionally convergent integrals can be pretty delicate and we must now learn to handle those we would like to subject Aries equation to the Fourier transform but the main question is whether the equation has a solution which is Fourier transformable for example does it have a solution in the Schwarz class does it have a solution in L1 do we have a right to take the Fourier transform but before taking up these technical issues let us proceed formally since Aries equation does not contain a y prime term the Ronskian of any two linearly independent solutions is going to be a constant and a non-zero constant so if one solution decays at infinity along with its derivative in other words if you have one solution in the Schwarz class then the second solution must actually grow rapidly at infinity so only one solution at most can be in the Schwarz class up to multiples of course so if at all we can solve this equation using Fourier transforms if you do get a solution that solution will be the unique solution in the Schwarz class but let us set aside the question of existence let us try to formally proceed and try to get the solution once we get the solution then we can hopefully work backwards and try to differentiate under the integral sign and check whether it satisfies the Aries equation or not then we can be assured that the solution that we have got is indeed a bona fide solution so let us proceed formally let us take the Fourier transform of equation 4.15 when you take the Fourier transform of y double prime it will be minus chi squared y hat and when you take the Fourier transform of x y it is going to be 1 upon i d d chi of y hat and the resulting ODE is 4.16 the resulting ODE is chi squared y hat minus 1 upon 3 i d d chi of y hat equal to 0 the advantage is that 4.16 is a first order linear ODE which we can immediately solve and we can get a solution up to constant y hat equal to x of i chi cube now this x of i chi cube is certainly not in l 1 let alone Schwarz class it is not even in l 1 so but nevertheless let us continue arguing formally and apply blindly the inversion theorem y of x equal to integral over r x puff i chi cube plus i x chi d chi the 1 upon 2 pi factor has been ignored because my differential equation is a homogeneous linear ODE so I can ignore multiplicative constants and I can write this integral 4.17 as y of x equal to integral over r cosine x chi plus chi q d chi because the sin integral will be 0 it is an odd function so now from a purely heuristic considerations ignoring all issues of convergence etc we have arrived at this equation 4.18 the problem is that this whole derivation leading to 4.18 is suspect for several reasons first we did not know that there is a solution which is the Schwarz class or even in l 1 secondly this the Fourier transform was e to the power i chi cube it is not clear whether the inversion theorem is applicable and many other similar issues so anyway we have this equation 4.18 and let us see what we can do with it first is to inquire whether 4.18 converges at all it is a conditionally convergent integrals and we could try to prove that 4.18 satisfies the differential equation by hoping to differentiate under the integral sign suppose a differentiate under the integral sign with respect to x what is going to happen I am going to get sin x chi plus chi cube with a negative sign but the derivative of whatever is inside namely I am going to pick a factor of chi when I differentiate it twice remember the Aries equation is second order od I am going to get cos x chi plus chi cube and then going to be a factor of chi squared it gets worse earlier the cosine was actually bounded now I have chi squared cos x chi plus chi cube so directly differentiating under the integral sign and checking whether 4.18 is a solution of Aries equation is going to be problematic and convergence of 4.18 has to be established it is a conditionally convergent integral now let us now make a change of variables remember our problem integral cos x squared dx from 1 to infinity how did we try that integral we put x squared equal to u similarly here we will put x chi plus chi cube equal to s let us look at this function x chi plus chi cube that is a monotone increasing function it is not monotone on the entire real line it will be monotone increasing only on r infinity where r is sufficiently large when you make a change of variables the change of variables must be a invertible map so monotone increasing is essential so it is going to be monotone increasing on some interval r infinity and we work on this interval r to infinity so let us look at this integral y of x from r to infinity cos s ds upon x plus 3 chi squared because ds will be equal to x plus 3 chi squared d chi so d chi will be ds upon x plus 3 chi squared so the integral transforms to 4.19 the integral from 0 to r is an innocent looking integral it is a proper Riemann integral like your cosine x squared dx from 0 to 1 or sin x upon x from 0 to pi by 2 those are proper Riemann integrals so the improper part is from r to infinity now clearly as s goes to infinity in this equation as s goes to infinity this chi of s must go to infinity so now let us integrate by parts in this integral 4.19 and we are led to a simple calculation will tell you that you get 4.20 you get chi upon x plus 3 chi squared the whole cube into sin s ds you can write down the boundary terms arising from integration by parts I am suppressing that I think we will stop here okay