 Welcome to tutorial on heat equation. In this tutorial, we are going to solve a few problems associated to heat equation. The first problem is a problem in calculus which is integral of e power minus x square is root pi. We have used this information in our analysis of heat equation that is why I decided to do this proof. This involves a small trick I will explain that. So integral of e power minus x square on R is equal to 2 times integral on 0 to infinity of the same integral e power minus x square because it is an even function. Therefore, it is enough to compute 0 to infinity e power minus x square dx. Then the answer we are interested will be 2 times i. So to compute i, we look at i square. i square is i into i that is this quantity into the same quantity because I am writing the 2 times I have used x in one integral and y in the other integral. Anyway, this is so called dummy variable. So I can use any variable I want. Now what I do is that this quantity I take it inside. Actually, I am assuming that this is finite quantity and that is why everything is fine. So I take this inside and I get this. Now this is nothing but 0 to infinity 0 to infinity e power minus x square into e power minus y square is e power minus x square minus y square dx dy. And that is equal to in the polar coordinates R equal to square root of x square plus y square. So therefore minus x square minus y square becomes minus R square and the area element dx dy becomes R dr d theta. Now we have to find out the limits for the integration variables. Look at this integral. This integral is on 0 infinity cross 0 infinity that is the first quadrant. So the first quadrant the radius goes from 0 to infinity. But the part of the circle which is in the first quadrant corresponds to theta equal to 0 to theta equal to pi by 2. For example, these are the circles I am integrating on. So the angle is 0 to pi by 2 here. Pi by 2 is the angle here. So I square equal to this. This is what we have seen on the last slide. Now it is very easy to integrate. Put s equal to R square then inside integral with respect to R becomes with respect to s in this form. And this is easily integrable e power minus s integral is e power minus s by minus 1 limits 0 to infinity. So you get 1 from there. So total is half. So integral with respect to theta from 0 to pi by 2 of 1 by 2 that is pi by 4. So I square is pi by 4. Therefore I is root pi by 2. Therefore this integral which is 2 times I is root pi. This is a very simple trick that we use in calculus. To compute an integral of one variable we go through computation of integral in 2 variables. Luckily because of the form of the integrand we can convert that into polar coordinates and we could easily compute. Now let us solve the Cauchy problem for homogeneous heat equation. Of course we know a certain formula for the solution of the homogeneous heat equation and the corresponding Cauchy problem with a phi of x. We have an integral. But that integral is very integrable only in very special cases where we are very lucky namely you see pi of x we have taken e power 3x because the kernel the heat kernel is also exponential type. So therefore this exponential should helps us. That is why we are able to integrate. Otherwise if it is some arbitrary function of x which is continuous and bounded we do know that it gives rise to a solution. But then that is it. We cannot really compute. We have to be really lucky to be able to compute. So without worrying about uniqueness issues because we know that Cauchy problem solution is not unique. It is unique only when you are looking at special classes. Let us not worry about that and simply compute this integral and see what we get. So we will compute a solution. Maybe that is the solution that is the only solution in some class or maybe there are more solutions we do not know at because this may not belong to the class for which uniqueness theorem holds. In fact we have not discussed much about uniqueness theorems for which I have just given you the reference of book by Debenedetto on PDEs. You can look at your leisure time. But as far as this course is concerned we simply want to compute this integral with phi of y equal to e power 3y. Can we do it or not? So this formula we have derived in lecture 7.2. So now this is exponential into exponential therefore it is exponential of this plus this. So I have written down exactly that. So this is same as this. So I am only working in the power of this e exponential e to the power something. So that power only I am manipulating. Now I express that like this so called completion of squares. The advantage now is that I am going to set y minus x minus 60 by something equal to a some variable. So I am going to put y minus x minus 60 by root 40 as p. So then this term the first term by 40 that simply becomes p square e power minus p square and what remains is 12 x t by 40 36 t square by 40. So this is e power minus p square and what remains here is 12 x t by 40 is 3x 36 t square by 40 is 90 and dy 1 by root 40 is your dp from here. Dp is 1 by root 40 dy. So this is exactly same as this. Now if you see exponential of 3x plus 90 does not depend on p. So it just comes outside and what we have is 1 by root pi into integral over r of e power minus p square dp. We just found out this is root pi. So root pi by root pi that is 1. So what you have is exponential of 3x plus 90. So we are very lucky that we could actually compute the integral. Now let us move on to the problem 3. For t positive let r be defined by this rectangle 0 pi cross 0 t. Let u be a solution to this problem which is initial bountiful problem homogeneous heat equation boundary conditions are 0 initial condition is sin square x. Show that 0 is less than or equal to u of x t is less than or equal to e power minus t sin x holds for x t in the rectangle. Hint is to use maximum principle because maximum principle allows us to compare certain things. So therefore the maximum principle using that is a hint here. So what we have to show is actually two inequalities. We have to show 0 is less than or equal to u x t and u x t is less than or equal to e power minus t into sin x. So let us prove the first one that is u x t is greater than or equal to 0. It follows from minimum principle. By minimum principle we know that minimum of u on r closure is actually minimum of u on the parabolic boundary of r. What is the parabolic boundary of r? This is r so parabolic boundary is this union of these three lines and u is 0 here, u is 0 here and u is sin square x here that is what is given to us. So minimum of u on the parabolic boundary is 0 because of this because minimum of sin square x on this 0 to pi is 0. So minimum is 0. Therefore minimum of u on r closure is 0 which means u is greater than or equal to 0 on r closure in particular on r. So this completes the proof of the first inequality. Now let us look at the second inequality. So we want to prove this. See if you want to apply maximum principle you can only apply that to solutions of heat equation. So therefore we naturally ask this question whether this is a solution of heat equation. Yes we have already seen when we were solving heat equation using separation of variables method that these are the things which are coming as the terms in the series. So is this a solution? Yes it is a solution. So what we do now is we define new function v which is u minus e power minus t sin x. This satisfies vt minus vxx equal to 0 because u is already a solution to the homogenous heat equation. This also solves homogenous heat equation. Therefore this equation being linear and homogeneous sum of two solutions or the difference of two solutions is a solution still. So v is a solution to this and what is that we want to show v of xt is less than or equal to 0. So showing this is same as showing v is less than or equal to 0. So in fact we will look at the problem that v satisfies. So homogenous heat equation v is a solution that we have already checked on r and v of 0t and v of pi t are also 0. v of x0 equal to sin square x minus sin x. Please check this it is very simple. So I am not going to check here. So this is the problem that v satisfies. Now by maximum principle applied to this v, maximum of v on our closure is nothing but maximum of v on the parabolic boundary. So therefore we will look at what is the maximum of v on the parabolic boundary. Parabolic boundary consists of three lines. One is the line x equal to 0, x equal to pi on which v is 0 and the other line is the x axis part between 0 to pi. So we need to look at what happens to this sin square x minus sin x. Sin square x minus sin x is always less than or equal to 0 on this interval. Why? Because sin x lies between 0 to 1 that is the reason we will see the details. So maximum of v on this is 0 this is because v of x0 is sin square x minus sin x that is always less than or equal to 0 and v of 0t and v of pi t both of them are 0. Therefore v of x t is less than or equal to 0 for every x t in r that implies u of x t is less than or equal to e power minus t sin x. This is what we wanted to show. So let us look at problem number 4. For t positive let RT be defined as this rectangle x between 0 to L and t between 0 to t equal to t. So this is what is called RT. So for example if you take t dash another line t equal to t dash this entire region will be RT dash. So let u be a solution to the homogenous heat equation in r cross 0 infinity and the boundary condition at x equal to 0 and x equal to L are 0 and u of x0 equal to phi x. Let m of t denote the maximum value of u on the rectangle RT and closure RT closure. Show that m of t is a constant function of t. So let t be less than t dash and RT is a subset of RT dash. RT dash is a bigger rectangle. Therefore RT closure also is contained in RT dash closure. Therefore m of t is less than or equal to m of t dash. m of t after all is a maximum of u on a smaller set. m of t dash is a maximum on a bigger set. Therefore this holds. So by maximum principle what is m of t? It is a maximum of u on the parabolic boundary of RT. Similarly m of t dash is maximum value of u on the parabolic boundary of RT dash. Parabolic boundary of RT is a subset of parabolic boundary of RT dash. u on both these sets takes the same values. Let us have just have a look at this picture. This is RT, this is RT dash. So this part, this part, this part is the parabolic boundary of RT. What is the parabolic boundary of RT dash? It is bigger than this. What is the new portion? It is only this much. And what happens to u there? u is 0. That is the boundary condition. Therefore and u is already 0 here. So u does not take any new value so that maximum increases. So therefore maximum is the same. So m of t equal to m of t dash. A similar statement can be made for the minimum also. Thank you.