Heating a long cylinder: steady states 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 further away, 'nearer to the ends' of the cylinder is still 'felt' by $u$ on $D \times \{ 0 \}$, but 'much less'.
Question. How does one establish estimates that quantify the 'waning influence' of $\varphi$ on $u$ with the height?  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?
Edit. Apparently there are estimates with exponentially decaying weight $w(t) = \mathrm{e}^{-C \lvert t \rvert}$, perhaps something like
\begin{equation}
\lvert u(\cdot,0) \rvert_{L^2(D)}
\leq C \lvert \mathrm{e}^{-C \lvert t \rvert} \varphi\rvert_{L^2(\partial D \times (-L,L))}.
\end{equation}
Although I am inclined to believe these bounds by fiat, I am ultimately most interested in the arguments used to derive them (or indeed a reference to such arguments). This wasn't explicit in the wording of the earlier version of the question, so I've amended it to clarify this.
 A: To separate contribution of the "ends" and the lateral surface, write $w=u+v$, where $u$ has zero
boundary conditions on the lateral surface, and $v$ is zero
on the ends.
The estimate $|u|\leq Ce^{-kt}\|\phi\|$ is true, where $k$
is the smallest eigenvalue of the Laplacan for $D$, and $t$
is the distance to the "ends" of the cylinder, and the constant $C$ depends on the norm used. This generalizes the similar estimate of Carleman for dimension 2.
Some references are:
H. KELLER, Sur la croissance des fonctions harmoniques s'annulant sur la frontiere d'un
domaine non bornd. C. R. Acad. Sci. Paris 231 (1950), 266-267.
A. DINGHAS, Das Denjoy-Carlemansche Problem fur harmonische Funktionen in $E^n$. Det.
Kgl. Norske Videns. Selsk. skr. (1962) No. 7, 12 pp.
A. HUBER, Ober Wachstumseigenschafien gewisser Klassen yon subharmonischen Funktionen.
Comment. Math. Helv. 26 (1952), 81-116.
They all considered domains more general than cylinders, and worried about precise estimates. But for a straight cylinder, the stated estimate follows from very general compactness arguments, if you do not worry about the constant $C$. The key fact is that in the infinite cylinder
($D\times R$) all positive harmonic functions with zero boundary conditions  are those with separated variables, so
they have the form $u(x)(ae^{-kt}+be^{kt})$ where $x$ is the coordinate in $D$, and $u$ is a positive eigenfunction.
A: $\newcommand{\R}{\mathbb R}$Sorry for being too sketchy in the following answer, time permitting, I'll try to expand.

Step 0. Some more-or-less classical potential theory. Let $D$ be an open set in $\R^d$ (with $d \geqslant 3$ for simplicity), and assume that $D$ is sufficiently regular (for example, Lipschitz). Let $p_t^D(x, \xi)$ be the heat kernel in $D$: the fundamental solution of the heat equation $$\frac{\partial u}{\partial t}(t, x) = \Delta u(t, x) $$ with $u(t, x) = 0$ when $x \in \partial D$. In other words, $$u(t, x) = \int_D p_t^D(x, \xi) f(\xi) d\xi$$ is the (unique) solution of the heat equation with initial condition $u(0, x) = f(x)$.
The Green function in $D$ can be defined by the time integral of the heat kernel:
$$G_D(x, \xi) = \int_0^\infty p_t^D(x, \xi) dt .$$
Note that this is always finite (except when $x = \xi$, of course), because $p_t^D(x, \xi) \leqslant p_t^{\R^d}(x, \xi)$, and the integral of $p_t^{\R^d}(x, \xi)$ is the Newtonian potential kernel $c_d |x - \xi|^{2 - d}$. Furthermore, $G_D(x, \xi)$ is zero on the boundary (this is not quite obvious, though), and — formally — we have $$ \Delta_x G_D(x, \xi) = \int_0^\infty \Delta_x p_t^D(x, \xi) dt = \int_0^\infty \frac{\partial p_t}{\partial t}(x, \xi) dt = 0 - \delta_\xi(x) ,$$ so that $G_D(x, \xi)$ is the fundamental solution for the Poisson problem $$ \Delta u(x) = -f(x) $$ in $D$, with $u(x) = 0$ for $x \in \partial D$. And indeed, one can rigorously prove that if $$u(x) = \int_D G_D(x, \xi) f(\xi) d\xi$$ for, say, continuous and bounded $f$, then indeed $\Delta u = -f$ in $D$ and $u = 0$ on $\partial D$.
Finally, if $D$ is regular enough ($C^{1,1}$ is the usual condition), then the Dirichlet problem in $D$: $$\Delta u(x) = 0$$ with Dirichlet boundary condition $u(x) = f(x)$ for $x \in \partial D$ can be solved using the Poisson kernel: $$u(x) = \int_{\partial D} f(\xi) P_D(x, \xi) \sigma(d\xi),$$ where $\sigma$ is the surface measure on $\partial D$ and the Poisson kernel $P_D(x, \xi)$ is the boundary derivative of the Green function: $$P_D(x, \xi) = \frac{\partial G_D(x, \cdot)}{\partial \nu}(\xi) = \lim_{s \to 0^+} \frac{G_D(x, \xi + s \nu)}{s} \, ,$$ where $\nu$ is the inward normal vector at $\xi$. This follows relatively easily from the divergence theorem (or Green's identities).
By the way, for a general open set $D$, the solution of the Dirichlet problem is given in terms of the harmonic measure: $$ u(x) = \int_{\partial D} f(\xi) P_D(x, d\xi) ,$$
which is again closely related to the Green function, but this is a completely different story.

Step 1. First, consider the Poisson problem in $D \times \R$, with boundary data given by $f : \partial D \times \R \to \R$ (let us denote the boundary data by $f$ rather than $\varphi$, which we will need for the eigenfunctions; here $f$ is an arbitrary bounded and continuous function). The solution is given by the harmonic measure, which, due to translation invariance of the problem, is translation invariant itself:
$$ \begin{aligned} u(x, y) & = \int_{\partial D \times \R} f(\xi, \eta) P_{D \times \R}(x, y, d\xi d\eta) \\ & = \int_{\partial D \times \R} f(\xi, y + \eta) P_{D \times \R}(x, 0, d\xi d\eta) \end{aligned} $$
for an appropriate measure $P_{D \times \R}(x, y, d\xi d\eta)$. If $D$ is nice enough — say $C^{1,1}$ — then $P_{D \times \R}(x, y, d\xi d\eta)$ has a density function $P_{D \times \R}(x, y, \xi, \eta)$ with respect to the surface measure $\sigma(d\xi) d\eta$, and this density is called the Poisson kernel. Thus,
$$ u(x, y) = \int_{\R} \int_{\partial D} f(\xi, \eta) P_{D \times \R}(x, y, \xi, \eta) \sigma(d\xi) d\eta . $$
Translation invariance means that $P_{D \times \R}(x, y, \xi, \eta) = P_{D \times \R}(x, 0, \xi, \eta - y)$.

Step 2. How fast does $P_{D \times \R}(x, 0, \xi, \eta)$ decay with $|\eta|$? In my comment to the question, I sketched a probabilistic argument which shows exponential decay. Here is a more analytic (but still potential-theoretic) version of the same argument.
Let $p_t^D(x, \xi)$ be the heat kernel in $D$, and $p_t^{D \times \R}(x, y, \xi, \eta)$ be the heat kernel in $D \times \R$. Thus,
$$ p_t^{D \times \R}(x, y, \xi, \eta) = p_t^D(x, \xi) (4 \pi t)^{-1/2} e^{-(\eta - y)^2 / (4t)} . $$
Again if $D$ is nice enough ($C^{1,1}$ is more than enough, Lipschitz is already fine), than $p_t^D$ is known to be intrinsically ultracontractive. In particular,
$$ p_t^D(x, \xi) \approx C e^{-\lambda_1 t} \varphi_1(x) \varphi_1(\xi) $$
for $t > 1$. Here $\approx$ means that the ratio is bounded from above and below by positive constants.
Using an estimate $0 \leqslant p_t^D(x, \xi) \leqslant p_t^{\R^2}(x, \xi)$ (where $p_t^{\R^2}$ is the usual Gauss–Weierstrass kernel) for $t < 1$ and intrinsic ultracontractiviety for $t > 1$, by direct integration, we find the following estimate of the Green function, valid when $|\eta|$ is large enough (here I omit the details):
$$\begin{aligned} G_{D \times \R}(x, 0, \xi, \eta) & = \int_0^\infty p_t^{D \times \R}(x, 0, \xi, \eta) dt \\ & \approx \varphi_1(x) \varphi_1(\xi) \int_0^\infty e^{-\lambda_1 t} t^{-1/2} e^{-\eta^2 / (4 t)} dt \\ & \approx \varphi_1(x) \varphi_1(\xi) e^{-\sqrt{\lambda_1} |\eta|} . \end{aligned} $$
Now the Poisson kernel is the normal derivative of the Green function. Thus, if $D$ is a $C^{1,1}$ set,
$$ P_{D \times \R}(x, 0, \xi, \eta) \approx \varphi_1(x) e^{-\sqrt{\lambda_1} |\eta|} . $$
This may look as if we "differentiate both sides of an inequality", but it is not the case: since the Green function is zero on the boundary, the normal derivative reduces to a simple limit of $G_D(x, 0, \xi + s \nu, \eta) / s$, where $\nu$ is the inward normal vector at $\xi$.

Step 3. Now it remains to translate this into a result on $D \times (-L, L)$ with zero Neumann boundary condition on the bases. This, however, is pretty standard: if $u$ is the solution of the problem on $D \times (-L, L)$, then the function $v$ given by
$$ v(x, y + 4 n L) = u(x, y) , \qquad v(x, y + 2 n L) = u(x, -y) $$
whenever $x \in D$, $y \in (-L, L)$ and $n \in \mathbb Z$, is a solution of the corresponding Poisson problem in $D \times \R$. Using the Poisson representation for $v$, we find that
$$ u(x, y) = \int_{(-L, L)} \int_{\partial D} f(\xi, \eta) \sum_{n = -\infty}^\infty (P_{D \times \R}(x, y, \xi, \eta + 4 n L) + P_{D \times \R}(x, y, \xi, -\eta + 2 n L)) \sigma(d\xi) d\eta . $$
By using the estimate for the Poisson kernel found above, we easily see that again
$$ u(x, 0) \approx \varphi_1(x) \int_{(-L, L)} \int_{\partial D} f(\xi, \eta) e^{-\sqrt{\lambda_1} \eta} \sigma(d\xi) d\eta $$
uniformly in $L$ large enough and $f$.
