 Welcome to this lecture on maximum principle for heat equation. In this lecture, we are going to prove the maximum principle stated for bounded intervals in R. Of course, there is a version of the maximum principle for x belongs to R as well, which we are not going to discuss in this course. And then we give a few applications of maximum principles. One of them being justification that the formal solution obtained by the separation of variables method is indeed a solution to the initial boundary value problem. Recall that we have used separation of variables method to solve an initial boundary value problem in lecture 7.3. Recall notations R denotes the rectangle 0 L cross 0 t. This is a x space x varies in 0 L and the time varies in 0 to capital T. CH denotes the collection of all functions defined on the rectangle R taking values in the real numbers such that the function phi the first order derivatives with respect to x and t and the second order derivative of phi with respect to x, all of them belong to the space C of R closure that is their continuous functions on R closure. Definition of what is known as parabolic boundary, this is a subset of the boundary of the rectangle which plays a role in the maximum principle. So, the boundary of the rectangle R consists of the lines L1, L2, L3, L4, L3, L4 and this is the rectangle R. The parabolic boundary of the rectangle R which is denoted by boundary PRP for parabolic boundary of the rectangle. So, that is parabolic boundary of the rectangle is defined as union of L1, L2 and L3 that means L4 is not included, L4 is part of the boundary of the rectangle but is not included in the parabolic boundary. Here is the picture L1, L2, L3 this is dou PR and L4 is also a part of the boundary of rectangle R. Let us state now the maximum principle. Let U be a solution to the heat equation U t equal to U xx then the maximum value of U on R closure is achieved on the parabolic boundary dou PR. Proof of maximum principle it is an exercise in calculus exactly like the maximum principle for the Laplace equation or for the harmonic functions which we proved is also an exercise in calculus. So, since U is a continuous function on R closure, R closure is a compact set the maximum value of U is attained somewhere in R closure. We would like to show that this maximum is also attained on the parabolic boundary dou PR. In other words this is what we have R and dou PR consists of these 3 lines. So, we would like to show that the maximum is attained on either here or here or here. It may be attained in somewhere else also but the maximum principle does not say about that maximum principle says it is definitely achieved on dou PR. So, let M capital M and small M be defined by capital M is a maximum of U on R closure and small M is a maximum of U on the parabolic boundary. So, clearly M is less than or equal to M because dou PR is a subset of R closure therefore M is always less than or equal to capital M. The proof of the theorem will be complete if you prove that M less than M is not possible. In that case M equal to M will hold and that is precisely the conclusion of the maximum principle. So, we are going to show that M less than capital M is not possible. Let L4 star denote L4 without the n points, L4 minus 0 t comma L comma t because 0 t and L comma t are already in the parabolic boundary of R. So, we want to remove that from L4 and call it L4 star. Assume that M is strictly less than M holds. So, let X1 t1 be a point in R union L4 star be such that U of X1 t1 equal to capital M. Such X1 t1 exists because capital M is strictly bigger than small M. Capital M is a maximum of U on R closure while small M is a maximum of U on the parabolic boundary. Therefore, there will be a point X1 t1 which is essentially R closure minus the parabolic boundary. In other words R union L4 star such that U achieves the value M at that point X1 t1. Now, we are going to define a function V by V of Xt equal to U of Xt plus M minus M by 4L square into X minus X1 whole square. So, we are adding a term to U and what we are adding is always non-negative because M is strictly bigger than small m. So, this is positive X minus X1 square is always greater than equal to 0. So, for Xt in the parabolic boundary we have V of Xt less than or equal to small m because U of Xt is less than or equal to small m on the parabolic boundary plus M minus M by 4L square which is as it is from here. Now, X minus X1 whole square is less than or equal to L square and that equal to L square L square gets cancelled and we get M minus M by 4 and that is strictly less than capital M. So, that is V of Xt is strictly less than capital M whenever Xt belongs to the parabolic boundary of R. Further V of X1 comma T1 is U of X1 comma T1 because this term is 0 when X is equal to X1 and U of X1 T1 is capital M. Thus, the function V attains its maximum value say M dash on R union L4 star. So, let X2 comma T2 belonging to R union L4 star be such that V of X2 T2 is M dash. Note that X2 lies in the open interval 0 comma L. Note that if we X2 T2 belongs to R then we must have Vt of X2 T2 equal to 0. So, what are the possibilities for X2 T2 if it is in R union L4 star either it is in R or in L4 star. If it is in R then we must have Vt of X2 T2 equal to 0. And if X2 T2 is actually on L4 star then Vt of X2 T2 is greater than or equal to 0. Why? Because of this. So, if X2 T2 is here this is the situation 1 X2 T2 belongs to R then this is an interior point at which you have maximum that is why the first order derivatives are 0. On the other hand if it happens on L4 star this is L4 star then so X2 T2 is here. So, T is this right this is the T direction. So, there is a maximum at this n point T2 in other words T2 is actually capital T. So, this point is X2 comma capital T if it belongs to L4 star. And hence partial derivative with respect to T is greater than or equal to 0 and this follows from the difference quotients. In either of the 2 cases we have Vt to be greater than or equal to 0. So, since V is given by this formula Vt of X2 T2 that means we are going to differentiate with respect to T we get ut of XT T2 and this does not depend on T. So, its T derivative is 0. Now, U is a solution to the heat equation. So, ut of X2 T2 is UXX of X2 T2. But what is UXX of X2 T2? We can compute from here in terms of V. UXX will be VXX minus the second derivative with respect to X of this term which is here. Now observe this is a non-negative quantity in fact a positive quantity you are subtracting something from VXX. So, this quantity is strictly less than VXX. This we have anyway 0 less than or equal to Vt of X2 T2 we proved on the last slide. And now that Vt of X2 T2 is strictly less than VXX of X2 T2. So, this is what we proved on the last slide. But VXX of X2 T2 is less than or equal to 0. Why? Because V attains maximum at X2 T2 and this leads to contradiction because VXX is strictly positive on one hand and less than or equal to 0 on the other hand and that is a contradiction. Therefore, M less than capital M is not possible and this completes the proof of maximum principle. As a corollary we can reduce minimum principle. Let U be a solution to the heat equation. Then the minimum value of U on our closure is actually attained on the parabolic boundary. This follows from the maximum principle that we have just proved we have to apply this to the function V equal to minus U. If U is a solution to heat equation minus U is also solution to the heat equation. Therefore, we can apply the maximum principle for V. Note that the maximum principle proved here is like the weak maximum principle that we have proved for Laplace equation. A strong maximum principle similar to that for harmony function which we have proved as a consequence of the mean value property it also holds in the context of heat equation. We are not going to discuss that we are going to just state it. So, strong maximum principle if U is a solution of the heat equation, suppose that the maximum value of U on our closure is achieved at a point in the rectangle then it must be constant on the rectangle. For its proof please consult books on partial differential equations for example, by Di Benedetto or Evans. The maximum principle that I have mentioned earlier which is stated for X belongs to R is available in this book Benedetto. Now as an application of maximum principle we are going to show the uniqueness of solutions to initial boundary value problems. Recall the IBVP that we have considered for the heat equation U t equal to U xx, U 0 t is G 1 of t, U L t is G 3 of t, U X 0 is G 2 of x where G 1, G 2, G 3 are given functions. And R denotes the rectangle 0 L cross 0 t, C H denotes the collection of functions having this property. A function V in C H is set to be a solution to the IBVP if V satisfies the heat equation and the initial and boundary conditions. So, uniqueness for IBVP, so the initial boundary value problem given here has at most one solution. The proof of uniqueness there is a standard procedure which is to assume that U 1 and U 2 are solutions of the IBVP, consider the difference and show that the difference is 0. So, we want to show U 1 equal to U 2, so consider the difference U 1 minus U 2 call it W. And W satisfies the IBVP for the heat equation with 0 initial boundary data. So, applying maximum and minimum principles to W, we conclude that W attains both its maximum and minimum on the parabolic boundary, but W is 0 on the parabolic boundary. This is precisely the data on the parabolic boundary 0. Therefore, W is identically equal to 0 and this proves the uniqueness. Now, as another application, we are going to see that the formal solution for IBVP is indeed a solution. Remember, the formal solution was obtained using separation of variables method in lecture 7.3. So, in lecture 7.3, we described separation of variables method to solve the following initial boundary value problem. Given a function phi, look at the heat equation posed on this domain and initial condition is phi of x and Dirichlet boundary conditions 0, 0 boundary conditions. And the following formal solution was derived. Uxt is given by this infinite series. We have obtained these coefficients as coefficients in the Fourier sine series for the function phi of x. Using maximum principle, we are going to establish that the formal solution is indeed a solution. In fact, we use maximum principle to establish that the initial conditions are taken by this function. The fact that this defines a function which is twice differentiable with respect to x and once differentiable with respect to t, mainly follows from this factor which is here, the exponential factor. We are going to see the proof. So, let us state it as a theorem. Let phi be continuous function and such that the Fourier series, in fact Fourier sine series of phi converges uniformly to phi and phi of 0 equal to phi of l equal to 0. Then the function defined by this formal series expansion, it is indeed a solution to initial boundary value problem. So, we have to check that this series defines a function which is 2 times differentiable with respect to x, 1 times differentiable with respect to t and then that function actually satisfies the heat equation and the initial and boundary conditions are met. So, this is the formal solution that was proposed as a consequence of separation of variables method. Let us denote these coefficients by Bn. The series converges uniformly for xt belonging to 0l cross t0 comma t and of course, phi is integrable because we are assuming phi is continuous. So, Bn has this expression 2 by l 0 to l phi x sine n pi by l x dx. So, mod Bn is less than or equal to 2 by l modulus goes inside the integral. So, mod phi into mod sine n pi x by l and that is less than or equal to 2 by l 0 to l mod phi s dx which is a finite number because phi is continuous it is bounded it is integrable whatever reasons you want to give. So, therefore mod Bn e power minus n square pi square t by l square sine n pi x by l what is this? This is the nth term in the series that we have. This quantity now is less than or equal to the constant c for mod Bn times this is a non-negative quantity. So, it stays as it is and of course, modulus of the sine is less than or equal to 1. So, we have this estimate not only this because we are considering on this domain t0 comma t. Therefore, this exponential is dominated by this. So, this is what we proved on the last slide and this series summation nth term is this. This is convergent follows from the ratio test. We conclude that the series converges uniformly for x t in 0 l cross t 0 t. So, whenever you have uniform convergence of the infinite series it defines a continuous function. So, since t0 is arbitrary we conclude that u is continuous on 0 l cross open 0 close t. Now, differentiability. So, we have proved that series converges and it defines a continuous function on the rectangle of interest. Now, we are going to show that u t, u x and u xx exist and they are continuous on this domain. This follows from the fact that the series can be differentiated term by term once with respect to t and twice with respect to x. In fact, much more that we are going to see in a remark soon after finishing this proof. So, since the proofs are similar we are going to give the proof for u t just one derivative. Note that the series given here this we have obtained after differentiating with respect to t once is uniformly convergent for x in 0 l and t in t 0 comma t. It follows once again from the ratio test and the convergence of this series exactly the same estimates. So, you have to estimate modulus of this and you will get this with and a constant time step. And yeah this is precisely the inequality I was talking about we have this and this series as stated here converges and hence this converges uniformly in t 0 comma t for every t 0 and hence it defines a continuous function for every t 0. Therefore, it defines a continuous function on open 0 comma t as well 0 l cross open 0 comma t. So, having justified the differentiation of the infinite series in the formal solution it is easy to check that you satisfies the heat equation on this domain 0 l cross 0 t. It reminds to show that u satisfies the initial boundary conditions. So, in order to show that u is continuous on 0 l cross 0 t we show that the sequence of partial sums is uniformly Cauchy in 0 l cross 0 t. So, let the nth partial sum be denoted by S n of x t which is given by this. And for m greater than or equal to k let w k m be defined as Sm minus sk. Therefore, w k m has this expression each term in this finite sum actually solves heat equation. So, w k m is a solution to the heat equation and it satisfies this boundary data when x equal to 0 or x equal to l let us go back when x equal to 0 sin 0 is 0. So, therefore, this is 0 when x equal to l what we have is sin n pi that is once again 0 therefore, we have 0. So, these conditions are satisfied and then w k m of x 0 is simply this. So, applying maximum principle to the function w k m which is a solution to the heat equation and the and on the parabolic boundary it takes 0 and 2 parts of it and on the third part of it it is this. So, applying maximum principle to the function w k m we get that the supremum or the maximum is less than or equal to the maximum here w k m x 0. So, note that this is partial sums of a convergent uniformly convergent series. Therefore, this is uniformly Cauchy sequence this is Sm minus sk that is why it is a uniformly Cauchy sequence. Since the Fourier series for phi is assumed to converge uniformly to phi on the interval 0 l thus the function u is continuous on 0 l cross 0 t and satisfies the initial condition u x 0 equal to phi x remark the smoothing effect. Note from the above proof that the Fourier series was proved to be differentiable with respect to t by proving that the series resulting from term by term differentiation is uniformly convergent which of course followed from the presence of exponentially decaying term here. By similar argument it follows that the function defined by the series is infinitely differentiable with respect to x and t in the domain 0 l cross 0 t. Thus a solution of heat equation belongs to c infinity or an interior of a rectangle. Even when u of x 0 equal phi x is not phi may be just continuous but u is c infinity in the rectangle. This is described as a regularizing effect or smoothing effect of the heat equation. Let us summarize what we did in this lecture. Weak form of the maximum principle for IBVP of heat equation was presented and the strong form of the maximum principle was stated. And the following applications of the weak form of maximum principle were presented. Uniqueness for the IBVP for heat equation was proved and we have proved that the formal solution to IBVP which was obtained by method of separation of variables is indeed a solution. Thank you.