 Welcome, we begin the study of heat equation starting from this lecture. In this lecture we are going to take a few preliminary steps into the solution of Cauchy problem for the heat equation. The outline of this lecture is as follows. First we introduce what is Cauchy problem for heat equation and take the first steps in solving the Cauchy problem for homogeneous heat equation. In the next lecture we are going to complete the study of the Cauchy problem for non-homogeneous heat equation. So what is Cauchy problem for heat equation? Given functions f, phi on appropriate domains, Cauchy problem is to solve u t – u xx equal to f of x, t, x belonging to r and t positive and u of x0 equal to phi of x for x in r. Notation C21 of r cross 0 infinity. In case you have seen this notation C21 somewhere else, you forget about that, it is not about that notation, it is a notation restricted to the heat equation study. So the function space C21 of r cross 0 infinity consists of all functions defined on the domain r cross 0 comma infinity taking values in real numbers such that u is continuous on r cross 0 infinity, u t, u x are continuous on r cross 0 infinity that is u is C1 of r cross 0 infinity, u xx is continuous on r cross 0 infinity. In C21, in this notation C21, 2 and 1 stand for the number of derivatives with respect to x and t variables respectively. As you see with respect to x, we have 2 derivatives that is why the 2 here and this 1 is for one derivative of u with respect to t. Let us define what is the meaning of solution to the Cauchy problem. Let u be a function such that u is C21 of r cross 0 infinity and that is intersection C of r cross closed 0 comma infinity. Observe here it is open 0 infinity, this is closed 0 comma infinity. u is said to be a solution to the Cauchy problem if u satisfies the heat equation u t minus u xx at every point x comma t is equal to f of x comma t for x in r and t positive that holds and u of x 0 is phi of x that also holds for every x in r. So the Cauchy problem u t minus u xx you may call this as heat operator h of u equal to f of x t and u x 0 equal to phi x on appropriate domains x belongs to r t positive here and x belongs to r here. May be solved in 2 steps. Step 1 solve the Cauchy problem with f identically equal to 0. Step 2 use Duhamel principle to get a solution to the non-homogeneous equation with 0 Cauchy data and superposition of the 2 solutions which are obtained in step 1 and 2 would then give solution to the Cauchy problem for the non-homogeneous equation that is solution to this problem. So Cauchy problem for homogeneous heat equation u t minus u xx equal to 0 and u x 0 equal to phi of x. So the first step obtain a general solution to this equation a solution which satisfies u x 0 equal to phi x that is the idea let us try that. Before that some exercises whenever you use a solution to u t minus u xx equal to 0 any derivative of u is also solution assuming that such derivatives exist okay and let a be positive define w of xt equal to u of ax comma a square t then w is also a solution. Fix y in r and look at a translate of u okay u of x minus y comma t call it w of xt that w is also a solution. Let us solve heat equation using a similarity transformation we will see what that means a key observation is follows for every a positive heat equation is invariant under the change of coordinates that means if you shift from xt to z comma tau the heat equation now will become w tau minus w zz equal to 0. Observe this z square by tau is x square by t. So we look for solutions to heat equation respecting the above symmetries. So let v be a function of one variable v of z okay this z is different from this z I should have called it v of some other variable s okay. So u of xt equal to v of x by root t. So take a function of one variable and look for u of xt as v of x by root. So x by root t is coming from this this equation x square by t equivalently x by root t. So substituting this ansatz u of xt equal to v of x by root t into heat equation gives us this equation very easy to check I am not going to prove this. Now such transformations are called similarity transformation which transformation this instead of looking for xt look for x by root t and they help in reducing the number of independent variables PDE became ODE in our example the heat equation became this ODE. So the equation which we have obtained on the earlier slide it can be solved explicitly. So its general solution is given by this v z equal to c 1 times 0 to z exponential minus s square by 4 ds plus c 2 where c 1 and c 2 are arbitrary constants. So therefore u of xt is v of x by root t. So put z equal to x by root t so that that is only here z is only here so it is only here. So this is an expression for u of xt. Now we want to make sure u of x 0 is phi x then we are done we have solved the Cauchy problem for the homogeneous heat equation. Let us see what is u of x 0? u of x 0 is limit of u of xt as t goes to 0 plus which is actually minus c 1 root pi plus c 2 if x is negative and if x is positive it is c 1 root pi plus c 2 please check this. Now u which is given here u of xt is smooth function for every t positive only at t equal to 0 there is some issue if t is positive it is an indefinite integral of course with this variable here of some nice function smooth function c infinity function. So therefore u is smooth there is no problem for u however u of x 0 has a jump discontinuity but in any case there is no way to obtain a solution to our Cauchy problem we want u of x 0 equal to phi x how will I get phi x from this this is just piecewise constant function with some jump discontinuity if c 1 is non-zero if c 1 is 0 of course it is just a constant function. So we need to exploit this of varying c 1 if possible and try to get solution but there is no way u of x 0 is not going to be phi of x. So what can we do for solving Cauchy problem now let us observe this for compactly supported function psi look at u of x 0 into psi dash of x of course this integral can be split into minus infinity 0 plus 0 to infinity I have not done anything the integration domain is split into 2 domains I am going to substitute the value of u of x 0 for x negative we have u of x 0 this for x positive u of x 0 is this now we can do integration right this is a constant so minus infinity to 0 of psi dash of x is psi of 0 and here this is constant so 0 to infinity of psi dash of x is minus psi of 0 psi is having compact support so the infinity does not matter only 0 matters if you simplify you get this minus 2 c 1 root pi psi of 0 if they function u of x 0 were smooth and not like this what we would have had is integral u of x 0 psi dash of x equal to minus u x of x comma 0 psi of x dx so we obtained on the last slide this expression for smooth functions u of x 0 we would get this so in some sense this and this are related loosely roughly speaking roughly speaking u of x comma 0 is like 2 c 1 root pi delta 0 this is the Dirac delta Dirac delta when you integrate loosely speaking when you integrate against some function will give you the value of the function at the point 0 that is why you have psi of 0 by a translation we can make this happen at any y so u of x minus y comma 0 will be delta y instead of delta 0 so from the observations on the last slide it appears that u x of x comma t would be useful to solve Cauchy problem u x of x comma t of course is a smooth function there is no problem for t positive what happens at t equal to 0 it looks like the Dirac delta function they are related with that so choose c 1 so that this integral is 1 that will give us c 1 to be 1 by 2 root pi so therefore u x of x comma t is 1 by 2 root pi t into exponential minus x square by 4 t this is going to play a crucial role in the solution of the Cauchy problem for homogeneous heat equation so as observed this function is smooth for t positive and what happens to it as t goes to 0 we will see some picture on the next slide it approximates this Dirac delta distribution in case you do not know what this is simply ignore what I have said the reason is that they have it has integral 1 for every t integral 1 with respect to x and as t goes to 0 plus one can observe that the graph steepens and starts to concentrate at x equal to 0 look at the picture on the next slide so here u x of x comma t graphs are drawn for various times t equal to 2 is in blue see here these are 1 t equal to 1 is in this red color and yellow is t equal to 0.25 you see it is getting steep it steepens around 0 at the same time the values are tending to 0 at the other places not becoming 0 but then getting closer and closer to 0 as t goes to 0. So, t equal to 0.1 is in this here in magenta color see it steepens because area it has to maintain to be 1 area under this curve and the this is shrinking where u is big value it is becoming smaller and smaller and therefore it is becoming bigger and bigger near x equal to 0. So, it is this u x of x comma t is concentrating near x equal to 0 as t is going to 0. So, one can make u x of x comma 0 to concentrate around any y by a translation we already observed that. So, you consider u x of x minus y comma t instead of u x of x comma t u x of x minus y comma 0 will be like delta y the collection u x of x minus y comma 0 indexed by y forms a basis roughly speaking in the sense that this when you integrate you will get phi of y. Thus we expect that the superposition of u x of x minus y comma t into phi of y yields a solution to the Cauchy problem. If I can get my phi of y using u x of x minus y comma 0 I hope that I can get my u of x t with phi as the initial data using u x of x minus y comma t that is a hope we have. This will be formulated as a theorem in the next lecture. I would like to give an analogy with the case of the Laplace operator. We looked at symmetries of the Laplace operator and motivated by them we tried special solutions that is radial solutions and that gave us fundamental solution for the Laplace equation that is done in lecture 6.2. In the case of heat equation also we looked for solutions which respected symmetry properties that heat equation enjoys. For Laplace equation the fundamental solution picked up Dirac delta function in the space of course there is no time in Laplace case. For heat equation a similar idea resulted in picking up Dirac delta at initial time and that is why we will be able to solve the initial value problem. We will continue this in the next lecture. Let us define what is called heat kernel that is 1 by 2 root pi t exponential minus x square by 4 t k of x t this is also known as heat kernel or fundamental solution we can call fundamental solution. Of course we need to verify that the fundamental solution defined above has the properties that are expected of a fundamental solution. We will take it up in the next lecture. Thank you.