Partial Differential Equations/The heat equation

Fundamental solutions, Green's functions and Green's kernels Poisson's equation This chapter is about the heat equation, which looks like this: for some . Using distribution theory, we will prove an explicit solution formula (if is often enough differentiable), and we even prove a solution formula for the initial value problem. Green's kernel and solution 0 100% #FFFFFF Lemma 6.1: Proof: Taking the square root on both sides finishes the proof. 0 100% #FFFFFF Lemma 6.2: Proof: By lemma 6.1, . If we apply to this integration by substitution (theorem 5.5) with the diffeomorphism , we obtain and multiplying with Therefore, calculating the innermost integrals first and then pulling out the resulting constants, 0 100% #FFFFFF Theorem 6.3: The function is a Green's kernel for the heat equation. Proof: 1. We show that is locally integrable. Let a compact set, and let such that . We first show that the integral exists: By transformation of variables in the inner integral using the diffeomorphism , and lemma 6.2, we obtain: Therefore the integral exists. But since , where is the characteristic function of , the integral exists. Since was an arbitrary compact set, we thus have local integrability. 2. We calculate and (see exercise 1). 3. We show that Let be arbitrary. In this last step of the proof, we will only manipulate the term . If we choose and such that , we have even Using the dominated convergence theorem (theorem 5.1), we can rewrite the term again: , where is the characteristic function of . We split the limit term in half to manipulate each summand separately: The last integrals are taken over for . In this area and its boundary, is differentiable. Therefore, we are allowed to integrate by parts. In the last two manipulations, we used integration by parts where and exchanged the role of the function in theorem 5.4, and and exchanged the role of the vector field. In the latter manipulation, we did not apply theorem 5.4 directly, but instead with subtracted boundary term on both sides. Let's also integrate the other integral by parts. Now we add the two terms back together and see that The derivative calculations from above show that , which is why the last two integrals cancel and therefore Using that and with multi-dimensional integration by substitution with the diffeomorphism we obtain: Since is continuous (even smooth), we have Therefore 0 100% #FFFFFF Theorem 6.4: If is bounded, once continuously differentiable in the -variable and twice continuously differentiable in the -variable, then solves the heat equation Proof: 1. We show that is sufficiently often differentiable such that the equations are satisfied. 2. We invoke theorem 5.?, which states exactly that a convolution with a Green's kernel is a solution, provided that the convolution is sufficiently often differentiable (which we showed in part 1 of the proof). Initial Value Problem 0 100% #FFFFFF Definition 6.5: Let and be two functions. The spatial convolution of and is given by: 0 100% #FFFFFF Theorem and definition 6.6: Let be bounded, once continuously differentiable in the -variable and twice continuously differentiable in the -variable, and let be continuous and bounded. If we define , then the function is a continuous solution of the initial value problem for the heat equation, that is Note that if we do not require the solution to be continuous, we may just take any solution and just set it to at . Proof: 1. We show From theorem 7.4, we already know that solves Therefore, we have for , which is why would follow if This we shall now check. By definition of the spatial convolution, we have and By applying Leibniz' integral rule (see exercise 2) we find that for all . 2. We show that is continuous. It is clear that is continuous on , since all the first-order partial derivatives exist and are continuous (see exercise 2). It remains to be shown that is continuous on . To do so, we first note that for all Furthermore, due to the continuity of , we may choose for arbitrary and any a such that . From these last two observations, we may conclude: But due to integration by substitution using the diffeomorphism , we obtain which is why Since was arbitrary, continuity is proven. Exercises Sources Fundamental solutions, Green's functions and Green's kernels Poisson's equation