 Welcome. In this lecture, we are going to discuss certain concepts associated to solutions of heat equation and heat equation itself. One of them is infinite speed of propagation and then we look at the energy associated to the heat equation and using energy we will show that initial boundary value problem that we have considered in earlier lectures has a unique solution. Then we go on to state a couple of examples which illustrate that heat equation is kind of irreversible that means we cannot consider t to be negative unlike the wave equation. So, this backward problem for the heat equation is not well posed. So, the outline is we start with infinite speed of propagation we will show that the solutions to heat equation have this property. Then we show uniqueness for IBVP via energy method and then we finally conclude by showing two examples which illustrate that backward heat equation Cauchy problem for it is not well posed. Recall from lecture 7.2 where we have established this theorem. Whenever phi is a continuous and bounded function if you define u of xt by this formula then u is a solution to the heat equation ut equal to uxx for t positive and satisfies the initial condition u of x0 equal to phi x. In other words, this formula represents a solution to the Cauchy problem for homogeneous heat equation. So, the solution has the following domain of dependence property like we have discussed domain of dependence and domain of influence for wave equation we are going to discuss the same for heat equation. So, solution at any point xt depends on the values of the initial data phi x at all x in R that is visible from this integral all the values of phi are used to find u of xt at any point x, t. Thus domain of dependence of solution at any point xt is the entire real line because all the values of phi matter phi of y as y belongs to R they matter in determining the solution at any point x and at any time t positive. The domain of influence of any point on x axis is r cross 0 infinity for the same reason this shows that the information from Cauchy data reaches all points instantly because Cauchy data is given at t equal to 0. Whereas, u of x comma t for whatever may be the x all the values of phi matter in this they influence the solution at every point. So, with infinite speed this is unlike the wave equation where an initial disturbance takes some time to reach other points. But here it reaches instantly with infinite speed this suggests that heat equation may not be suitable to study physical phenomena. Let us look at the formula again assume that the Cauchy data phi is also compactly supported function now and it is a nonzero function and such that phi is greater than or equal to 0 for every x in R that is a non negative compactly supported continuous function. From the formula for solution u of xt is strictly positive because integrants are greater than or equal to 0. But it is strictly greater than 0 somewhere because phi is nonzero function and hence this integration after the integration we get strictly positive quantity. So, u of xt is strictly greater than 0 for every x in R in particular outside the support of phi also and t positive. So, it is like lamp lit at one point heat reaches everywhere. There is a story in the literature where a person is trying to cross a river where the water is very chill and suddenly he notices a lamp is lit at the other bank and then he starts to feel the warmth. This happens in literature and in perhaps in mind not in the physical world. So, this is another instance illustrating infinite speed of propagation property of the heat equation. This suggests again that heat equation may not be suitable to study physical phenomena. So, models trying to remedy infinite speed of propagation. So, people have created models which do not have this infinite speed of propagation they were proposed in literature. You may consult the book by Olber, the book of Olber partial differential equations for a nonlinear diffusion equation that exhibits finite speed of propagation. There are some fractional order partial differential equations they are also proposed and they do not have infinite speed of propagation maybe you are happy because it has finite speed of propagation. But the physical relevance is rarely discussed. On the other hand, heat equation is suited for long time studies of a t large and has shortcomings for small times. So, let us now discuss uniqueness for IBVP via energy method. So, an energy associated to heat equation. So, we have to get what we want to declare as energy. So, multiply the heat equation u t minus u x x equal to 0 with u and integrate over the interval 0 L that gives us 0 to L u u t dx equal to 0 to L u u x x dx. Note that the LHS of the above equation namely this can be written as this. So, integrating by parts on the RHS of this equation we get these terms. So therefore, what we have is d by dt of 0 to L half u square dx equal to minus 0 to L u x u x dx plus u u x at the point 0 and L. So, define energy to be half 0 to L u square of x dx. This is non-increasing if u x 0 to L is less than or equal to 0 because its derivative which is here this is non-positive and this is also non-positive then d by dt of the energy is non-positive that means energy is non-increasing. For example, if u satisfies u of 0 t equal to 0 and u of L t equal to 0 these are the kind of conditions that we have in initial boundary problem then actually this is 0. So, what we have is d by dt of E t is less than or equal to 0. As a consequence E of t is less than or equal to E of 0 for t positive. So, using energy method we established that solutions to the following IBVP are unique. The IBVP is u t equal to u x x plus f of x t x in 0 to L and t positive u of 0 t is g 1 t u of L t is g 3 t u of x 0 is g 2 x. So, these two are boundary condition this is initial condition. So, this is initial boundary value problem for non-homogeneous heat equation f g 1 g 2 g 3 are known functions. In fact, I am taking k equal to 1 here. So, let u and v be solutions to the IBVP on the previous slide on the strip 0 L cross 0 infinity then u equal to v on S. Let us see the proof of the theorem. How do we show uniqueness? We consider the difference u minus v and show that that is 0. So, that leads us to defining w equal to u minus v. Then we look at the problem solved by W. W solves homogenous heat equation because u and v solve non-homogeneous heat equation. Therefore, W solves homogenous heat equation and the 0 initial boundary conditions on the strip. Note that E of t is a non-increasing function that is E of t is less than or equal to E of 0. If you remember this we have derived under the assumption that you have 0 t and you have L t are 0 and in this case W of 0 t W of L t are 0 therefore, this is valid. So, the energy associated to W has this property. But at time t equal to 0 the energy is 0 half u square. So, if it is W it is half W square half integral W square sorry in half integral W square and W is 0. Therefore, E of 0 is 0. That means E of t is 0 and hence W of x t is 0 on the strip. Thus, IBVP has unique solution. Now, let us discuss backward heat equation and show through two examples that the Cauchy problem for it is ill-posed. The example one non-existence of solutions. Cauchy problem for backward heat equation let t be positive. Now, it is exactly the heat equation here u t minus u x x equal to 0 x in R. But now t belongs to minus t comma 0. That means I am considering this equation for negative t and u of x 0 equal to phi x for x in R. So, that is why this is called backward because you have prescribed t equal to 0 and you are interested in solving the heat equation for t less than 0. That is why it is called backward heat equation. Assume that phi is continuous and bounded function. The above problem is not solvable in the class of bounded functions. So, that means there is no solution. If a problem is not well posed, it can happen because of three reasons either there are no solutions or uniqueness is violated or there is no stability of solutions. So, here we are showing about non-existence of solution. That is the reason why this problem is ill-posed. So, what is the reason? Suppose the above problem is solvable for every choice of phi which is bounded and continuous. Then phi of x is given by this formula. We have derived this formula. We have shown that in the class of bounded and continuous functions the solutions are unique. So, this I am looking at u which is a solution to the backward problem and so that is why this is the condition like the initial condition there. Then phi of x is at a later time. This is a t equal to minus t. This is a time t equal to 0. So, it is given in terms of this integral. So, these holes. So, once these holes we can see that the RHS is a smooth function. There is no doubt about that. It is a smooth function and the last equation cannot hold if phi is only continuous because if this is smooth and phi is not even differentiable then this cannot hold right. Therefore, if phi is only continuous this equation cannot hold and hence our supposition that there is a solution for every choice of phi which is bounded continuous is not true. Therefore, the Cauchy problem for backward heat equation need not have solutions. So, second example is about instability of solutions. Look at this function sequence of functions u n of x t equal to 1 by n sin n x into e power minus n square t. Each of them solves the homogenous heat equation for every x and t x in R t in R. Look at u n of x comma 0. If we put t equal to 0 this is 1 therefore you have 1 by n sin n x. So, u n of x 0 is 1 by n sin n x and that goes to 0 as n goes to infinity in fact uniformly. Let us look at u n of x comma minus 1. Minus 1 is just one instance for 1 negative time t. Instead of writing let t not be negative less than 0 we have chosen t not equal to minus 1. Then this u n of x comma minus 1 what we get is 1 by n sin n x e power n square. Now as n goes to infinity this term 1 by n of course goes to 0 but e power n square goes to infinity. So, this is going to infinity and sin n x will be oscillating between minus 1 and 1. So, therefore u n of x comma minus 1 keeps oscillating towards plus infinity and minus infinity. In any case it is not going to 0 except for x being integral multiples of pi this will never go to 0. So, initial conditions are going to 0 but at a time t equal to minus 1 it does not go anywhere. The solutions to the Cauchy problem for backward heat equation does not have stability property. Thank you.