# 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 $$$$\begin{cases} \Delta u = 0 \quad \text{in C} \\ u = \varphi \quad \text{on \partial D \times (-L,L)} \\ \frac{\partial u}{\partial \nu} = 0 \quad \text{ on D \times \{-L,L\}}. \end{cases}$$$$

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 $$$$\lvert u(\cdot,0) \rvert_{L^2(D)} \leq C \lvert \mathrm{e}^{-C \lvert t \rvert} \varphi\rvert_{L^2(\partial D \times (-L,L))}.$$$$ 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.

• Much stronger: the density function of the exit time from $D$ (of the 2-D Brownian) motion decays exponentially fast, so the harmonic measure in the infinite case also decays exponentially fast. By periodization, the same is true for the reflecting boundary. That is, the "weight" associated with $\partial D \times \{x\}$ is roughly $\exp(-\sqrt{\lambda_1} |x|) / \sqrt{4 \lambda_1}$, where $\lambda_1$ is the smallest eigenvalue of $\Delta$ in $D$ with Dirichlet boundary conditions. Sep 26 at 14:14
• @MateuszKwaśnicki Cool, that's very neat! So something like $\lvert u \rvert \leq C \lvert \mathrm{e}^{-Ct} \varphi \rvert$? Do you have any idea whether bounds like this can be obtained without resorting to Brownian motion? I'm just asking because I was hoping to use this setting to gain a better grasp of a similar, but non-linear problem. Sep 26 at 14:24
• Yes, something of that kind, although I thought about pointwise bounds. I tried to give some details in an answer below. $L^2$ bounds might actually be much easier to establish, as the problem becomes essentially 1-D. Sep 26 at 23:02

$$\newcommand{\R}{\mathbb R}$$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 $$f$$.

• Thanks for taking the time to write up your answer! I may have some questions once I've managed to read through it - I hope that's OK. Sep 26 at 23:19
• Sure, of course! Sep 27 at 6:40
• Sorry it took me so long to get back to you. There are a couple of things I was hoping you could clarify. The main one is the expression of the Green's function as an integral of the heat kernel. Is this always true, maybe provided the integral is convergent? Perhaps on every regular domain $\Omega$ - here the cylinder? Is there a heuristic explaining it? It seemed a bit weird to me because I thought the Green's function ought to be defined on $\Omega \times \Omega$, but the heat kernel on $\Omega \times \partial \Omega$. [...] Oct 2 at 17:14
• [...] Two other, smaller points. When you speak of 'the' solution of the Poisson problem in $D \times \mathbf{R}$, you mean the unique periodic solution, right? Because there should be other, non-periodic solutions, no? The last point: when you use the bounds (above and below) for the Green's function to get an estimate for the Poisson kernel, are you skipping over some details? It looks a bit like differentiating an inequality. Oct 2 at 17:18
• I expanded the answer a bit to address your comments. Roughtly: (1) Yes, the Green function is always the time integral of the heat kernel in dimensions three and above; and the same is true in dimensions one and two, as long as the complement of your domain is non-polar. In fact, this is true for much more general operators than the Laplace operator. (2) The heat kernel is defined on $\Omega \times \Omega$, too! (In fact: on $(0 \infty) \times \Omega \times \Omega$ if you include time as a variable.) [...] Oct 2 at 19:43

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.

• If I understand correctly, this answer is kind of orthogonal to the question. :-) You seem to be assuming zero boundary condition on the side of the cylinder $\partial D \times [-L,L]$ and Dirichlet boundary condition prescribed by $\phi$ on the bases. The original question asks for Dirichlet boundary condition given by $\varphi$ on the side, and Neumann boundary condition on the bases. That said, solutions of both problems are pretty much the same. Sep 26 at 17:28
• @Mateusz Kwasnicki: the question was that the "ends" are "not felt", or "felt little". I quantified it. Sep 26 at 19:16
• @AlexandreEremenko: Sure thing! This is what I meant by "solutions of both problems are pretty much the same". Sep 26 at 23:03