Differentiation of infinite sum in the solution to heat equation

Question

Let $\Omega\subset\Bbb R^d$ be a bounded domain and suppose that $\partial \Omega$ is $C^\infty$. Suppose that you are given $v_0=\sum_{j}a_ju_j$, where each $u_j\in H_0^1(\Omega)$ is an eigenvector for $-\Delta$ with associated eigenvalue $\lambda_j$. I'd like to verify that $$ v(x,t)=\sum_{j} e^{-\lambda_jt}c_ju_j $$ is a solution of the heat equation $$ \begin{cases} \partial_t v - \Delta v=0 &\text{ in }\Omega\times(0,\infty)\\ v=0&\text{ in }\partial\Omega\times(0,\infty) \\ v(x,0)=v_0 &\text{ in }\Omega.\end{cases}\label{E}\tag{E} $$ My question is, how to manage the derivative in $t$? For the derivatives in $x$, I guess I can look at the weak form of problem \eqref{E} and use that $\nabla$ is continuous on $H_0^1(\Omega)$ to differentiate under the sum. Is there an inequality that allows us to manage the derivative in $t$?

Answer 1

To differentiate an infinite sum, you can use the theorem of derivation under the integral, see https://en.wikipedia.org/wiki/Leibniz_integral_rule in the section "Measure theory statement". Indeed, you can see a sum as an integral with respect to the counting measure.

What you have to verify then is that the absolute values of the time derivatives of $t \mapsto e^{-\lambda_jt}a_ju_j$ are bounded by some non negative $\theta_j$ that don't depend on the time such that $\sum_{j \geqslant 0} \theta_j < +\infty$.

Indeed, for all $|\partial_te^{-\lambda_jt}a_ju_j| = \lambda_je^{-\lambda_jt}|a_ju_j| \leqslant \lambda_j|a_ju_j|$ (remember that the $\lambda_j$ are non negative because $-\Delta$ is self-adjoint and positive semi-definite). Therefore, your constants $a_j$ must be such that for all $x \in \Omega$, $\sum_{j \geqslant 0} \lambda_j|a_ju_j(x)| < +\infty$. In this case, everything works perfectly.

Notice that it you only assume that $\sum_{j \geqslant 0} |a_ju_j(x)| < +\infty$, then for all $t \geqslant t_0 > 0$, you have $\sum_{j \geqslant 0} \lambda_je^{-\lambda_jt_0}|a_ju_j(x)| < +\infty$ because $\lambda_je^{-\lambda_jt_0} \leqslant 1$ for $j$ large enough (remeber that $\lambda_j \rightarrow +\infty$) thus $(x,t) \mapsto v(x,t)$ admits derivatives with respect to $t$ for all $x$ and all $t \geqslant t_0 > 0$, hence for all positive time, but a priori not at $t = 0$.