 Welcome to this lecture on initial boundary value problems for heat equation. We are going to use method of separation of variables to solve the initial boundary value problem. So what is an initial boundary value problem? We will define and we will define what is the meaning of its solution. So the initial boundary value problem for heat equation consists of solving the heat equation u t equal to u x x posed on the domain x belonging to the interval 0 l and t belonging to the interval 0 comma capital T where capital T is a fixed number. u of 0 t equal to g 1 of t valid for t between 0 and capital T. So this is one of the boundary conditions. The second boundary condition is u of l comma t that is also prescribed to be g 3 of t. And then the initial condition u x 0 equal to g 2 of x where x is between 0 and l. So where g 1, g 2, g 3 are given functions. So we are given these three functions g 1, g 2, g 3. We are supposed to find a function which solves this equation and also satisfying these three conditions. So this is x equal to 0, x equal to l. So this is a boundary. So we are prescribed u of 0 t here and here u of l comma t and this is u of x comma 0. Of course we have put this is the t axis. So t equal to t here. We are not prescribing any condition on this. If we prescribe a condition on this line as well then it will be a boundary value problem. We are going to see in a future lecture that boundary value problem for heat equation is not well posed. So that is why we consider only initial boundary value problem. So initial condition you have x 0 or time t equal to 0 and these are the boundary conditions which are given. Of course we can consider other kinds of boundary conditions. Here we have considered what are called Dirichlet boundary conditions. We can consider other types of boundary conditions exactly like we did for the wave equation. So what is the meaning of a solution to the initial boundary value problem? Let R denote the rectangle 0 l cross 0 t. Let C h, h for heat denote the collection of all functions phi defined on this rectangle taking values in real numbers such that the functions phi, the first order derivative is phi x and phi t and second order derivative with respect to x phi xx these are all continuous on R closure. In fact we do not require it to be continuous on R closure because the conditions are prescribed only on this on this on this. We do not require it to be continuous here. A function V belongs to C h is said to be a solution to the initial boundary value problem and R if V satisfies the heat equation and satisfies the 3 conditions. The 2 boundary conditions V 0 t equal to G 1 t and V l t equal to G 3 t and the initial condition. A remark since the equation U t equal to U xx is linear and homogeneous the principle of superposition holds for its solutions. Thus a solution to the given I B V P may be obtained by a superposition of solutions of 3 I B V P's where each one of these I B V P's feature exactly one of the 3 functions G 1, G 2, G 3 and other 2 functions are 0 functions. We are going to describe separation of variables methods when G 1 and G 3 are equal to 0. So, given a function phi defined on interval 0 l find a solution to the homogeneous heat equation U t equal to U xx for x in 0 l and t positive initial condition U x 0 equal to phi x for x between 0 and l and the boundary conditions which are directly boundary conditions we are considering U of 0 t is 0, U of l t is 0. In other words what we have done is we have set U equal to 0 as a boundary condition here and we are considering only this to be general function. So, this is just one of those 3 I B V P's that we mentioned earlier. If you notice here we have t positive of course we can consider t in the interval 0 comma capital T for some capital T positive that also we can consider. So, as a consequence if you want to define what is the solution of this it must be in CH for each capital T positive. We are going to discuss more about the solution of this in a future lecture. So, we describe a separation of variables method to solve this I B V P. So, what are the main steps involved in separation of variables method? The I B V P features homogeneous heat equation and 0 Dirichlet boundary conditions. So, step 1 2 families of ODE is obtained from heat equation. How do we get that? We look for solutions to heat equation in the separated form u x t equal to x x into t of t. The heat equation will give rise to 2 families of ODE is indexed by a single parameter lambda one for x and another for t that is one family of ODE is for x and one family of ODE is for t. The Dirichlet boundary conditions in the I B V P will yield boundary conditions for x. The step 2 is obtaining non-zero solutions to the 2 families of ODE's. The B V P for x turns out to be what is known as eigenvalue problem. So, it turns out that only a countable number of B V P's from the family indexed by lambda n n belongs to n will have non-zero solutions. For each of these eigenvalues we need to then solve for t at the end of step 2 we have a countable number of non-zero functions x and x into t and t. Step 3 proposing a formal solution as a superposition of this non-zero function that we have obtained. So, superposition of the function x and x t and t is proposed as a formal solution to the I B V P and then it remains to check whether this formal solution is indeed a solution what are the conditions that are needed on the initial data file. So, step 1 heat equation gives rise to 2 ODE's and one of them there is a boundary value problem. So, method of separation of variables looks for solutions of the form u x t equal to x x into t t of course x in this finite interval 0 L and t positive. So, substituting this ansatz in the heat equation gives us this equation and dividing both sides of this equation with x x and t t and rearranging terms will give us t prime by t equal to x double dash by x. So, in this equation t prime by t equal to x double dash by x the LHS is a function of t only and RHS is a function of x only. Such an equation can hold if and only if both the functions are identically equal to a constant function. It means that there exists lambda in R a real number lambda such that t prime by t equal to x double dash by x equal to that lambda. One of the tasks is to find all possible lambdas which are coming from the separated solutions nonzero separated solutions. So, this equation which we have got on the last slide gives rise to 2 ODE's 1 for t and 1 for x what are they x double dash minus lambda x equal to 0 and t prime minus lambda t equal to 0. It is not surprising the heat equation has 2 derivatives with respect to x variable. So, therefore the equation satisfied by capital X is second order and the equation satisfied by capital T is first order because the heat equation features only the first order derivative with respect to t. Now, we have 2 boundary conditions using them we will get a boundary value problem for x. So, using the boundary condition u of 0 t equal to 0 what we get is x of 0 into t of t equal to 0 for every t. So, this can mean that either x of 0 is a 0 or t is identically equal to 0 function. We are not interested in t of t identically equal to 0. Therefore, x of 0 is 0. If t of t is identically equal to 0 we get nothing because as we saw in step 3 we are going to propose as superposition of x n x and t n t as solutions. If t n t is 0 for some n it does not make sense because it is 0 there is no term like that. Therefore, we cannot admit t to be 0 and hence we conclude x of 0 is 0. Similarly, using the boundary condition u of l t equal to 0 we get x of l equal to 0. So, we got 2 conditions for x x of 0 is 0 x of l is 0. So, we have the boundary value problem for x given by x double dash minus lambda x equal to 0 and the values of x at x equal to 0 and x equal to l are 0. So, the ODE for t is t prime minus lambda t equal to 0 we do not get any condition for t unlike what we have got in the wave equation case. So, finding nonzero solutions to BVP for x that is a step 2. So, this is the boundary value problem we are interested in solving and we are looking for nonzero solutions of this boundary value problem. Of course, as you see x identically equal to 0 is a solution for every lambda but we are not interested in that. So, the lambdas for which the BVP admits a nonzero solution are called Eigen values and the corresponding nonzero solutions are called Eigen functions. So, let us start our search for Eigen values and Eigen functions. Note that the lambda which is a real number it can be 0 positive or negative. Therefore, our search we will conduct in 3 steps lambda equal to 0, lambda positive, lambda negative. Why is that? It is because of this equation that we have x double dash minus lambda x equal to 0. If lambda is positive we can write the solution in terms of exponentials, general solution. If lambda is negative the general solution is in terms of the sine and cosine functions and of course, if lambda equal to 0 x is A x plus B x of x equal to A x plus B. So, the nature or the form of the solution changes depending on whether lambda is 0 positive or negative that is the reason why we are going to divide our search for Eigen values and Eigen functions into 3 steps. So, let us take lambda equal to 0, then the boundary value problem for x becomes x double dash equal to 0 and x of 0 equal to x of L equal to 0. General solution of x double dash equal to 0 is A x plus B for some constants A and B real numbers. But now we are going to look for those solutions which satisfy these boundary conditions. That means what we get is A equal to B equal to 0 because if you see look at this the graph of this is a straight line and that is supposed to pass through 0 comma 0 as well as L comma 0 and the only straight line which passes through 0 comma 0 as well as L comma 0 is x of x equal to 0, y equal to 0. So, A equal to B equal to 0 the line is the x axis itself. So, this lambda equal to 0 is not an Eigen value. What about lambda positive? Are there positive Eigen values? So, because lambda is positive we can always write down as lambda equal to mu square it is in the interest of non complicated notations where mu is positive equation is x double dash minus mu square x equal to 0 boundary condition remains the same. Now general solution of the ODE x double dash minus mu square x equal to 0 is combination of exponentials. So, A e power mu x plus B e power minus mu x. Imagine if you are not set lambda equal to mu square we would have had here square root of lambda instead of a simple looking mu. That is the reason we always do like this. The moment something is positive we write it as mu square for mu positive. So, now we have to check if there are solutions which satisfy these boundary conditions and nonzero functions. So, applying the boundary conditions x of 0 equal to x of L equal to 0 it is easy to see that we get A equal to B equal to 0 it means no nonzero solutions exist for this boundary value problem. So, no positive number is an eigenvalue. So, now what remains is to look for eigenvalues which are negative. So, since lambda is negative we can write down lambda to be minus mu square where mu is positive. So, the BVP for x becomes x double dash plus mu square x equal to 0 and the boundary condition is x 0 equal to x L equal to 0. So, general solutions of the ODE x double dash plus mu square x equal to 0 is x of x equal to A cos mu x plus B sin mu x. Once again if you are not set lambda equal to minus mu square what we would have had here is square root of minus lambda x which is confusing that is why we choose to write lambda equal to minus mu square. Now, applying the boundary conditions x of 0 equal to x of L equal to 0 what do we get? Suppose we use x of 0 equal to 0 then what we get when I set x equal to 0 A plus 0 equal to 0 that means A is 0. When I put x of L equal to 0 I get this condition but note A is already 0. So, essentially the condition is B sin mu L equal to 0. I do not want B to be 0 so the condition is sin mu L equal to 0. So, mu must satisfy sin mu L equal to 0 that is what we are going to see on the next slide. So, since we are interested in non-zero solutions to the boundary value problem at least one of the constant A B should be non-zero but we already have A equal to 0. Therefore, to have been on 0 we must have sin mu L equal to 0 which means mu n equal to n pi by L where n is a natural number. So, whenever mu n is of the form n pi by L for some natural number n sin of mu n L is 0. So, let us summarize what we have got on eigenvalues and eigenfunctions for the boundary value problem for x. The eigenvalues and corresponding eigenfunctions are indexed by n belongs to n, lambda n is minus n square pi square by L square remember lambda equal to minus m square lambda is eigenvalue. So, lambda equal to minus mu square and x n x is sin mu x. So, x n x is sin mu is n pi by L x. Now, we need to solve the ODE for t that is a step 2 that is very simple. So, solve the ODE for t with lambda equal to lambda n for each n in n and call it t n. What is the equation? t dash minus lambda n t equal to 0 that is t dash equal to lambda n t. So, answers will be in terms of the exponential. So, it is a constant times exponential of minus n square pi square by L square into t. These are the solutions. Now, let us propose a formal solution to the initial boundary value problem. So, we propose a formal solution using superposition principle as this this please note this is not really superposition principle this is infinite superposition that is why I have put them in quotes. So, u x t is proposed to be this this is x n x this is t n t and we take a combination of them and propose that as a solution to u x t. Of course, each one of these terms solves heat equation that is why we believe that the sum will also solve, but we have infinite sum. So, some justification is required. Since we are not doing the justification currently, we just put this symbol. Now, what are the coefficients b n they need to be determined but we have one condition that we have not yet used which is u x 0 equal to phi x. So, put t equal to 0 you get u x 0 equal to phi x when you put t equal to 0 this term will not be there because this will be 1. So, what we have is phi x equal to summation n equal to 1 to infinity b n sin n pi by L into x. So, thus b n are the Fourier sin coefficients of the function phi. See phi of x is given in terms of only pure sin series that is why this is called Fourier sin series for the function phi and b n's are the coefficients. So, we need to determine them. So, extend the function phi to the interval minus L comma L as an odd function with respect to x equal to 0 because phi is given on the interval 0 L. So, let the extended function be still denoted by phi then the Fourier series of phi it takes this form because now the extended function is odd function it will not feature cosine times the Fourier series will not feature cosine terms it features only the sin terms. And the coefficient b n is given by 2 by L integral 0 to L phi x sin of n pi by L x dx. We already discussed this in the context of wave equation. So, a formal solution to the IVVP is given by this the only difference is we have substituted what are the coefficients b n. The coefficients b n are the Fourier sin coefficients for the function phi. So, when is this formal solution defined above is indeed a classical solution. So, answer is on the next slide. So, theorem let phi be such that the Fourier series of phi converges uniformly to phi. In fact, the Fourier sin 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 is a solution to IVVP. If you recall we have defined the solution to IVVP only when t is restricted to finite intervals 0, t. But here the IVVP we have posed is for all t positive. So, what it means is that this function belongs to C H for every capital T fixed that is the meaning of the solution. A comment on the uniform convergence of Fourier series because is what we have assumed in the theorem. There are conditions under which the Fourier sin series of phi converges uniformly to phi in the interval minus L to L or equivalently on 0 L. Once such set of conditions is that phi is continuous and the integral 0 to L phi dash square dx is finite. And of course, phi of 0 equal to phi of L equal to 0 that is also required. So, let us summarize what we read in this lecture. IVVP for heat equation was solved using separation of variables method. A formal solution to IVVP was proposed. In a future lecture, we will prove that the formal solution obtained by separation of variables method is indeed a classical solution. Thank you.