Maybe my questions are stupit, but I'm so unsure about some properties of the heat equation.

Suppose we have a compact and smooth manifold M (possibly with boundary), then in a lot of books they write the heat kernel as $K(x,y,t)=\sum\limits_{i=0}^{\infty}e^{-\lambda_i t}u_{i}(x)u_i(y)$, where $\lambda_1 < \lambda_2\leq\lambda_3...$ are the eigenvalues of the Laplace operator acting on functions and the $u_i(x)$ are the corresponding normalized eigenfunctions.

1) What is the exact meaning of the equation $K(x,y,t)=\sum\limits_{i=0}^{\infty}e^{-\lambda_i t}u_{i}(x)u_i(y)$? Why does the sum on the right hand side converge for all $x,y \in M$ and $t\in (0,\infty)?$

2) Sometimes in books they interchanging the sum with differentiation and integration respectively, i.e. they write something like: $$\int_M K(x,y,t) dy=\int_M \sum\limits_{i=0}^{\infty}e^{-\lambda_i t}u_{i}(x)u_i(y)dy = \sum\limits_{i=0}^{\infty}\int_M e^{-\lambda_i t}u_{i}(x)u_i(y)dy.$$

or $$ \frac{\partial }{\partial x} K(x,y,t) = \frac{\partial }{\partial x}\sum\limits_{i=0}^{\infty}e^{-\lambda_i t}u_{i}(x)u_i(y) = \sum\limits_{i=0}^{\infty}\frac{\partial }{\partial x} e^{-\lambda_i t}u_{i}(x)u_i(y)$$

What are the exact reasons which justify those interchanges? I wold be very glad if you can help me.

Best regards


1 Answer 1


The eigenfunctions $u_j$ of the Laplacian in $\Omega$, corresponding to the eigenvalues $\lambda_j$, constitute an orthogonal (and WLOG orthonormal) basis of $L^2(\Omega)$. Thus if $v_0(x)=v(x,0)$, where $u$ is a solution of the Heat Equation (with homogeneous boundary conditions) then $$ v_0=\sum_j \langle v_0,u_j\rangle u_j, $$ and as $\mathrm{e}^{-\lambda_j^2 t}\,u_j(x)$ is a solution of the Heat equation, then so is $$ \sum_j \langle v_0,u_j\rangle\mathrm{e}^{-\lambda_j^2 t}\,u_j(x)=\sum_j\int_{\Omega}\mathrm{e}^{-\lambda_j^2 t}\,u_j(x)\,\bar u_j(y)\,v_0(y)\,dy =\int_{\Omega} K(x,y,t)\,v_0(y)\,dy. $$


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