 Welcome, we resume our study of Cauchy problem for heat equation that we have started in lecture 7.1, we ended lecture 7.1 by defining what is known as heat kernel or the fundamental solution associated to heat equation. Using that we are going to express a solution of the Cauchy problem for both for homogeneous heat equation and for non-homogeneous heat equation. So outline of this lecture is as follows, we recall briefly recall the Cauchy problem for heat equation that we have discussed in the last lecture that is lecture 7.1 and then we solve the Cauchy problem for homogeneous heat equation and then go on to solve for the non-homogeneous heat equation. So Cauchy problem consists of solving the heat equation ut-uxx equal to f with the right hand side the non-homogeneous heat equation with the prescribed initial conditions u of x0 equal to phi of x. The functions f and phi are given and we have to find a function u which satisfies these two conditions. We have introduced this notation of c21 r cross 0 infinity c21 stands for two derivatives with respect to x and one derivative with respect to t and it consists of all the functions defined on the domain r cross 0 infinity such that u, ut, ux, u double x are all continuous on r cross 0 infinity. So solution to the Cauchy problem was defined as follows a function u which is c21 of r cross 0 infinity intersection continuous function on r cross closed 0 comma infinity is a solution to the Cauchy problem if it satisfies the heat equation with the right hand side f and ux0 equal to phi of x and the Cauchy problem for the non-homogeneous equation can be solved in two steps. First you solve the Cauchy problem for the homogeneous heat equation that is you set f equal to 0. So you solve ut-uxx equal to 0 ux0 equal to phi x. In second step we use Juhamel principle to get a solution to the non-homogeneous equation namely ut-uxx equal to f of xt with 0 Cauchy data that is u of x0 equal to 0 and superposition of the two solution that we obtain in step 1 and step 2 will give you a solution to the problem that we want to solve. So first we started looking at the homogeneous heat equation and the Cauchy problem for that. We observed that under this change of variables z equal to x tau equal to a square t the heat equation remains invariant and therefore we thought of this variable x by root t here. So we look for the solution u of xt as v of x by root t where v is a function of one variable. So when we substitute this ansatz into the heat equation we end up getting a second order ODE which can be totally solved and this is general solution of the ODE and therefore u of xt is expressed as this C1, C2 are constants. Then we thought we can get the solution to the Cauchy problem by specializing the C1, C2 we will get the solution but that was not to be the case and we have found out that u of x0 is given by minus C1 root pi plus C2 for x negative and C1 root pi plus C2 for x positive. So there is no way that we can solve the given Cauchy problem using this analysis that we have done so far and then we observed that this u has a jump assuming that C1 is nonzero if C1 is 0 then u is constant right u of x0 is C2 and u of xt is C2 so it is constant. So constant solutions are known to be solutions in any case they do not solve the Cauchy problem. So if C1 is nonzero there is a jump that we observed. The moment we observe a jump in the function the derivative will pick up a Dirac delta that observation was made. So in lecture 7.1 we observed that u of x, t would be useful to solve the Cauchy problem. What is ux of x, t? It is 1 by root 4 pi t exponential minus x square by 40 is what we have obtained. We briefly recall the reasons why ux of xt is expected to be useful in solving the Cauchy problem. So ux of xt has this expression where C1 is equal to 1 by root 4 pi which I have not written here. It is smooth function as for t positive but what happens to t goes to 0 that approximates Dirac delta function. The reasons were explained that the integral is 1 and as t goes to 0 we can observe that the graph of ux stippens and starts to concentrate at x equal to 0. Then we said that the concentration can be made to happen at any point y in R instead of 0 by a translation that is you look at ux of x minus y comma t instead of ux x comma t then ux of x minus y comma 0 is like delta y. So the collection ux of x minus y comma 0 indexed by y forms a basis in the sense of this equation. Therefore, we expected that the superposition of ux of x minus y comma t into phi y yields a solution to the Cauchy problem. This will be formulated as a theorem very shortly. So we define what is called heat kernel and the fundamental solution by this formula kxt equal to 1 by 2 root pi t exponential of minus x square by 40. This is in fact ux of x comma t that we have obtained. So LMA define the function k1 from r cross r cross 0 infinity to r by k1 of x by t equal to 1 by 2 root pi t exponential of minus mod x minus y square by 40. This is nothing but k of x minus y comma t. This is k of x minus y comma t. So then k1 has the following properties. What are they? k1 is a C infinity function in this domain very obvious because it is defined to exponential function and inside you have minus x minus y whole square by 40. So this is a polynomial divided polynomial and t is not 0 in the domain therefore, this is always a smooth function. So C infinity and it is positive function because it is defined through exponential. So this is always strictly greater than 0 and more importantly dou by dou t minus dou square by dou x square of k1 x y t is 0. Why is this true? What is k1 of xy comma t? It is k of x minus y comma t. Yes, but what is k of x minus y comma t? ux of x minus y comma t. What is u? u is solving heat equation that is how we have found u through v. So ux being a derivative of the solution it is still a solution that is why you get this one that dou by dou t minus dou 2 by dou x square k1 of x y t equals 0 and this is of course only a translation by y. So we are using here some of the exercises that we have stated in lecture 7.1. Whenever you use a solution u of x minus y comma t is a solution for every y whenever you use a solution ux of x comma t is also a solution. So these assertions follow from the formula k1. I have already explained you. Some more properties of k1. Integral of k1 is 1. For any delta positive limit t goes to 0 plus integral over mod y minus x strictly greater than delta of k1 x y td by is 0 uniformly for x in i. So proof of 4 in the lemma. So setting z equal to y minus x by 2 root t we have integral of k1 equal to this quantity and that is equal to 1 because this integral is a standard integral. This integral is actually root pi. So proof of 5 introducing a change of variable exactly as in the proof of 4 we get integral of k1 on this domain to be this integral. Now because this integral is root pi and that is its finite it is integrable as t goes to 0 this quantity is limit of this is 0. So this completes the proof of 5. So solution to Cauchy problem for homogeneous heat equation. This is the Cauchy problem for the homogeneous heat equation. It can be obtained using the fundamental solution or the heat kernel or the k1. k1 of x y t is nothing but k of x minus y comma t and that is the content of the following result. Theorem. Let phi be a continuous and bounded function. Define function u by this integral substituting the expression for k1 it is this. So u of x t is proposed to be this integral. Then u is a c infinity function and r cross 0 infinity, u is a solution to the heat equation ut minus u x x equal to 0. If we extend the function by setting u of x 0 equal to phi x then it is a continuous function on r cross closed 0 comma infinity. This says that the initial condition is realized by this function which is only defined by t positive. It can be extended continuously up to t equal to 0 that is why the closed 0 here and in a continuous manner that is why c here and it assumes its value phi x. So the integral in the definition of u which is here converges uniformly and absolutely indeed modulus of u of x t if you look modulus of this integral or modulus of this it is a positive function. So mod phi y and phi is bounded therefore that comes out as the supremum and what remains is 1 by root 4 pi t integral exponential of this quantity which is 1. So if phi is bounded that is why supremum is finite then u is also bounded. So checking that u is a solution to heat equation is easy once suitable differentiability properties for u are established that is assertion 1 is established. Of course, assertion 1 is c infinity what we need here is only that u is in c to 1. So proof of 1 let us show u x exist remaining derivatives u x, u t, u xx, u x t, u t, t and all higher order derivatives showing that they exist is similar. So we will show that u x of x t exists at every point in r cross 0 infinity. So formally differentiating we get this if it is differentiable u of x t is differentiable with respect to x then u x of x t must be this that is the idea. So what we will now show is this integral converges absolutely and uniformly and hence differentiation on the integral sign is valid and then we have this expression that is idea. So note that this expression I am going to differentiate with respect to x and write down so is this. Now we do a change of variable x minus y by root t equal to p then this integral will turn out to be this integral. Now this is less than or equal to this phi comes out as its supremum rest stay as it is and you have a mod p instead of p and that this integral is finite. So therefore the integral in this converges uniformly and absolutely and this justifies the differentiation under integral sign. Thus u is differentiable with respect to x once using similar computations and ideas we can conclude that u is c infinity this completes the proof of 1. Now what remains is the proof of 3 that is the initial condition is achieved that is u of x comma 0 makes sense and that is phi of x. So this delta we will choose later in view of integral being 1 of k 1 we can write this uxt is the first term here integral k 1 phi by dy phi of xi xi is some point okay phi of xi into integral k 1 integral k 1 is 1 you multiply that with phi xi so this equal to this because of this. Now that is less than or equal to mod y minus x less than delta in fact this integral we write it on mod y minus x and delta and mod y minus x greater than delta and then use triangle inequality. So in this inequality there are 2 integrals on the RHS one is this one is this. So the first integral can be made small using continuity of phi at the point xi and the second integral goes to 0 as t goes to 0 by lemma. So we will choose delta so that mod phi m minus phi xi is less than epsilon by 2 whenever y minus xi is less than 2 delta this is precisely consequence of the continuity of the function phi at the point xi. Thus we have this integral less than or equal to this integral because this is a bigger this is contained in this mod y minus x less than delta is contained in mod y minus z less than 2 delta and now by continuity where this is valid this quantity is less than epsilon by 2 is valid therefore we get this and this integral on some set is always less than equal to integral and bigger set which we know to be 1. So this is actually equal to epsilon by 2 I have written less than it should be equal so what we have got is this less than epsilon by 2. Now by phi of the lemma there exists t naught so that for t less than t naught we have this quantity is less than epsilon by 2. Combining the two inequalities that we got one is here and one is on the last slide what we get is that whenever mod x minus z is less than delta and t is less than t naught modulus of u x t minus phi xi is less than equal to epsilon. So thus u is continuous at points of x axis and u x 0 equal to phi x for each x in r. What about uniqueness? Note that the theorem that we have stated is the existence theorem we are proposing a formula and then we are saying that formula gives rise to a solution. So essentially it is in existence by exhibition of the solution. So in general uniqueness is not expected for Cauchy problem posed in x in r. So taken off example illustrates non uniqueness of solutions. So what is taken off example it is concerning the Cauchy problem u t equal to u x x that is a homogeneous heat equation and 0 initial data u x 0 equal to 0. Of course u x t identically equal to 0 is obviously a solution to this but unfortunately there is one more solution. The following function which is given as a series is also solution. What is G? G is defined by this formula. Now the analysis to show that this solution is not easy but it can be understood. It is elementary but not easy. So details on taken off example may be found in this book by Fritz John on partial differential equations or Helwig partial differential equation and many more books which has this. Now we state a fact without proof. So Cauchy problem admits only one solution when we are looking for solutions belonging to special classes of function that is uniqueness can be regained but we need to put conditions on the kind of function that we are looking at or admitting as solutions. So like bounded solutions. So if you are looking only for bounded solutions then the solution is unique or solutions have a controlled growth of this type then also solution is unique. In our case we are considering the bounded Cauchy data and therefore the solution is bounded which we have constructed that solution is bounded. Therefore solution is unique. So solution is unique. Reference you can see this book Dependent RTO PDE or you can look at again Fritz John and Helwig or Evans. So in the case of taken off example bounded solution is a zero solution. So that is the only so bounded solution. The series solution is not bounded solution. So now let us discuss how to solve Cauchy problem for non-homogeneous CT equation using Juhamel principle. We have already seen Juhamel principle when we were discussing wave equation and how we obtain solution to the non-homogeneous wave equation and the corresponding initial value problem or initial boundary value problems. So it is a very general principle we will apply that and obtain a solution to the non-homogeneous CT equation and corresponding Cauchy problem. The Cauchy problem for the non-homogeneous CT equation is ut-uxx equal to f of xt and u of x0 equal to phi of x. We are going to apply Juhamel principle to obtain a solution to the Cauchy problem. Juhamel principle expresses the solution of the non-homogeneous equation in terms of solutions of the Cauchy problem for homogeneous heat equation and that passes through what is known as source operator. Source operator for heat equation let S phi of x, t denote the solution to the Cauchy problem for the homogeneous heat equation with initial data given as phi x. Now the question is it well defined? Answer is yes because given a function phi which is bounded and continuous there is exactly one bounded solution to this Cauchy problem. So S phi denotes that solution. So we expect the function defined by this where f tau of x is f of x, tau to solve the Cauchy problem for the non-homogeneous heat equation. A comment on the terms that we see on the RHS of this equation the proposed solution. The first term solves homogeneous heat equation by definition and at t equal to 0 it takes a value phi of x by the definition of the source operator. The second term solves the non-homogeneous heat equation with zero Cauchy data. So remark on the first term once again the first term on the RHS is by design or by the construction of the source operator is a solution to the homogeneous heat equation and satisfies the initial conditions ux0 equal to phi x for x in R. Remark on the second term on the RHS in this formula the integral term satisfies zero initial conditions t equal to 0 you get 0 very easy to check. The integral term satisfies the non-homogeneous heat equation and we proceed to check this. So we have to compute the derivatives of the integral term. So what is dou by dou t of that that is sft of x, 0 plus 0 to t derivative goes inside the integral sign. Sft of x, 0 is f t of x by definition of the source operator plus we have this. So we have f of xt plus this term. So we use Leibniz rule for differentiation of integrals which is a consequence of fundamental theorem calculus and chain rule. So this is what we proved on the last slide for the first derivative of the integral term. Now since s of tau of x, t minus tau is a solution to the homogeneous heat equation it is time derivative equals second derivative with respect to x variable. Therefore we have this equal to this dou by dou t is dou 2 by dou x square. So bring dou 2 by dou x square outside this integral that means we are assuming that the differentiation and integral commute with each other and we get this. So that means the integral term satisfies the non-homogeneous heat equation. Who guarantees that all the computations that we have done are valid? We need to assume that the functions f and v are good. So remark on the integrand in the particular solution this one 0 to t Sf tau xt minus tau. So define w of xt tau to be the integrand of this. Then w satisfies homogeneous heat equation in this domain x belongs to Rt bigger than tau and it satisfies the initial conditions w of x tau comma tau is f of x tau for x in R. Thus the integral term in the Duhamel formula has this expression 0 to t w. W are solutions to the homogeneous heat equation with this as the initial data. Initial data is coming from the source term that is a general idea in Duhamel principle. So the source operator is given by this because this is the expression for the solution of the homogeneous heat equation with Cauchy data phi. Therefore u of xt is now given by this which is a solution to the non-homogeneous heat equation. We state a theorem without proof. If f, f t, f x, f double x are all continuous unbounded in this domain R cross 0 to t for every t positive then and of course phi is bounded and continuous then u that we have obtained using the Duhamel principle is indeed a solution. Defends a classical solution C21 R cross 0 infinity and its continuous up to t equal to 0 and it realizes the initial condition that part we have already checked. So we do not discuss its proof. It involves justifying the formal computations we made earlier namely which involved interchanging the differentiation and the integration that justification using the hypothesis on f. Thank you.