Consider a long cylinder $C = D \times (-L,L) \subset \mathbf{R}^3$, with heat applied to its horizontal boundary according to $\varphi$ and perfectly insulated ends. The steady state $u: C \to \mathbf{R}$ representing the temperature is the solution of the system \begin{equation} \begin{cases} \Delta u = 0 \quad \text{in $C$} \\ u = \varphi \quad \text{on $\partial D \times (-L,L)$} \\ \frac{\partial u}{\partial \nu} = 0 \quad \text{ on $D \times \{-L,L\}$}. \end{cases} \end{equation}

In heuristic terms, 'most important' for the values of $u$ on $D \times \{ 0 \}$ are the boundary values with similar height. Heat applied 'near the ends' of the cylinder are still 'felt' by $u$ on $D \times \{ 0 \}$, but 'much less'.

Question. I am looking for estimates that quantify the 'waning influence' of $\varphi$ on $u$ with the height. Does somebody have a tip? Maybe this would be in terms of a weight function $w: (-L,L) \to \mathbf{R}$, say $w(t) = \lvert L - t \rvert$ or something stronger?

Comment. Here is a back-of-the-envelope calculation inspired by the comment below; this falls a bit short of what I am asking but perhaps it points in the right direction.

Let $\Delta'$ be the Laplacian on $D$. Let $v$ be the solution of the equation \begin{equation} \begin{cases} \Delta' v = 0 \quad \text{in $D$} \\ v = \varphi \quad \text{on $\partial D$}. \end{cases} \end{equation} We can then decompose $u(\cdot,0) - v = \sum_{i \geq 1} a_i \phi_i$ in terms of the eigenfunctions $(\phi_i \mid i \geq 1)$ of $\Delta'$ on $D$, with associated eigenvalues $0 < \lambda_1 < \lambda_2 < \cdots$.

Extend $v$ to $C$ by setting $V(x,t) = v(x)$ and extend also $\Phi_i(x,t) = \mathrm{e}^{-\lambda_i t} \phi_i(x).$ Then $u$ and $V + \sum_{i \geq 1} a_i \Phi_i$ are harmonic in $C$ with $u = V + \sum_{i \geq 1} a_i \Phi_i$ on the slice $D \times \{ 0 \}$, so maybe (?) there could be an estimate like \begin{align} \lvert u(\cdot,t) \rvert_{L^2} &\leq C ( \lvert V \rvert_{L^2} + \sum_{i \geq 1} \lvert a_i \Phi_i \rvert_{L^2} ) \\ &\leq C ( \lvert v \rvert_{L^2} + \mathrm{e}^{-\lambda_1 t} \sum_{i \geq 1} \lvert a_i \rvert ) \\ &= C ( \lvert v \rvert_{L^2} + \mathrm{e}^{-\lambda_1 t} \lvert u(\cdot,0) - v \rvert_{L^2} ). \end{align} Then I guess one could flip this argument around, to instead estimate $u(\cdot,0)$ in terms of $u(\cdot,L/2)$ say, but I can't quite conclude this line.