0
$\begingroup$

So, I know I'm missing something simple, but I can't find a way to solve the Laplacian with the boundary conditions I've got down.

The Problem:

"Consider the semi-infinite plate sketched below with thickness 2b. The temperature at the base (x = 0) is constant at $T_0$. Heat is transferred on both sides of the plate to a surrounding temperature $T_{\infty}$. The conductivity of the plate is k and the heat transfer coefficient for both sides is h. There is no heat generation. As $x \rightarrow \infty $, the temperature is finite. Solve for steady state temperature."

Below is a sketch of the problem

https://i.stack.imgur.com/6MVti.png

So far I've managed to define my PDE

$$\frac {\partial^2T}{\partial y^2} + \frac {\partial^2T}{\partial x^2} = 0$$

and boundary conditions

\begin{align} T(0,y) &= T_0 \\ T(\infty , y) &< \infty \\ \frac{\partial T}{\partial y}(x,b) &= - \frac hk [T_b(x) - T_{\infty}] \\ &= f(x) \\ \frac{\partial T}{\partial y}(x,0) &= 0 \end{align}

And I know I'm going to use my separation of variables, but I run into trouble when I try to solve my boundary conditions. I can't re-figure my boundary conditions in a way that gives me a known eigenvalue (generally of the form $\lambda = \frac{n\pi}b$) without giving me some $\sin(\infty)$ kind of a condition. I've tried for a few hours to find boundary conditions that work, but at this point I'm thinking I've either forgotten a simple trick, or I never learned how to deal with Laplacian boundary conditions approaching infinity. Any help would be appreciated.

Also, this is my first post and I'm sure I've left out some important detail or committed some faux pas. Please let me know if I've committed any breaches of etiquette, or if there's something I've left out.

$\endgroup$
0
$\begingroup$

This is because you are considering the equation on the half-line. Your domain is essentially $\Omega=[0,\infty)\times [0,y]$

If you look at the heat equation and wave equation on a bounded domain, you will see the solutions are typically given as Fourier Series, but on unbounded domains, the solutions look a lot different. You cannot use boundary methods and separation of variables to solve the PDE here.

Generally speaking, we want solutions $u(x,y)$ which "vanish at infinity" (tend towards $0$ as $||u(x,y)||_{L^2}\rightarrow \infty)$. Functions which are smooth and have compact support (denoted by $C_{c}^{\infty}(\Omega)$) include those that vanish at infinity $C_{0}(\Omega)$. These conditions imposed on the solutions are desirable because they make physical sense. The temperature/potential in a rod is not going to shoot off to infinity, it's going to decrease with time until it reaches a steady-state.

Now to actually solve the PDE, I don't think there is a slick and easy way to do it. Solutions to Laplace's equation in domains which aren't on a disk $D^n$ tend to be ad-hoc and very specific to the domain. Polynomials wouldn't work because they do not vanish at infinity. I would try to look for solutions which use sines and cosines. Alternatively, you could look for a harmonic function on the domain $[0,\infty)\times [0,y]$ (you don't necessarily need to work in the complex plane for this). Start by looking for qualitative solutions and then try to get a concrete, quantitative one.

$\endgroup$
  • $\begingroup$ Could you clarify what you mean by "look for solutions"? Do you mean guessing a solution like $T(x,y) = \sum A_n sin(\lambda y) e^{- \lambda x}$ and applying boundary conditions? As an aside, the professor who wrote this added a note that this problem should be solvable by separation of variables, but he's known for accidentally writing problems that aren't solvable or don't give real answers. $\endgroup$ – jtnew Apr 30 '17 at 14:07
  • $\begingroup$ Yes, something of that flavor would probably work. The professor likely made a mistake, boundary conditions do not work when your domain is unbounded, otherwise, the heat kernel, D'Alembert's solution, etc. all would be pointless, and they do not coincide with the Fourier series solution. There is a reason we have solutions for different domains. However, the condition that the solution is finite as $lim_{x \rightarrow \infty}$ is finite is exactly a condition commonly imposed on all solutions to PDEs on unbounded domains. You don't want the solution to "blow-up". $\endgroup$ – Kernel_Dirichlet Apr 30 '17 at 14:40
  • $\begingroup$ It is likely the PDE is written in bad form, $\infty$ is clearly not a defined value, so it makes no sense to think of it is a boundary in the normal sense, but imposing a condition at infinity where the solution either vanishes, or at least is finite (no blow-up), is valid and common practice. Separation of Variables may work here, but you in general have to be very careful as this is not a "true" boundary. $\endgroup$ – Kernel_Dirichlet Apr 30 '17 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.