I am working my way through a textbook (Richard Haberman fourth edition) on the heat equation as an example of applied partial differential equations. I am not familiar with the concept of a separation constant an it keeps coming up in the derivations. Forgive me I am a neuroscience major not a math major.

For instance I am on chapter two, we are discussing Laplace's Equation for heat flow in a rectangular surface. We are given this equation $$\frac{1}{h}\frac{\partial^2 h}{\partial x^2}=-\frac{1}{\phi}\frac{\partial^2 u}{\partial y^2}=\lambda, $$

where \lambda is the eigenvalue or separation constant of this gradient. I understand an eigenvalue in the context of linear algebra (which I understand well enough) and I am willing to accept that functions are infinitely indexed vectors but I am still confused as to how I can just pull that separation constant out of the air. What conditions need to be met to make this assumption?

Edit: Here is the page in my text this is taken from, maybe there is relevant information I am not including. enter image description here


1 Answer 1


The point is that

$$f(x) = \frac{1}{h }\frac{\partial^2 h}{\partial x^2}$$

is independent of $y$, while

$$g(y)=-\frac{1}{\phi}\frac{\partial^2 u}{\partial y^2} $$

is independent of $x$. So you are in a situation where

$$ f(x) = g(y), \ \ \ \text{for all }x, y.$$

This implies that $f, g$ are both constant functions. For example, choose $y=0$, then $f(x) = g(0)$ for all $x$. So $f(x)$ is a constant function. Similar for $g$.

Thus $f(x) = g(y) = \lambda$.

  • $\begingroup$ Sorry by constant functions do you mean their sum equals zero? $\endgroup$ Aug 17, 2020 at 4:58
  • 1
    $\begingroup$ @AngusCampbell No. I mean a function which takes the same value for all $x$ (e.g. $f(x) = 10$) $\endgroup$ Aug 17, 2020 at 4:59
  • $\begingroup$ I don't know if that assumption is correct, In the text it says h(x) oscillates and phi(y) is composed of exponential. They are definitely not constants unless this is the trivial case where they all equivalent to 0. $\endgroup$ Aug 17, 2020 at 18:55
  • $\begingroup$ @AngusCampbell It is not that $h$ is constant, but $\frac{1}{h} \frac{\partial ^2 h}{\partial x^2}$ is $\endgroup$ Aug 17, 2020 at 19:03
  • $\begingroup$ ohhhh that makes sense. $\endgroup$ Aug 17, 2020 at 19:17

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