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I am studying the method of separation of variables for the heat equation. I am struggling to understand why the separation constant takes on a positive or negative value. For instance if we apply the method of separation of variables to this equation. $$\frac{\partial u}{\partial t}= \frac{k}{r} \frac{\partial}{\partial r} (r \frac{\partial u}{\partial r} ) $$

looking for product solutions $u(r,t)=\phi(r)h(t)$ then dividing by $\phi h $ yields... $$\frac{1}{kh}\frac{dh}{dt}=\frac{1}{r \phi} \frac{d}{dr} (r \frac{d \phi}{dr})=-\lambda$$

But the following problem ends up with a positive separation constant. $$\frac{\partial u}{\partial t}= k \frac{\partial^4}{\partial x^4} $$

looking for product solutions $u(x,t)=\phi(x)h(t)$ then dividing by $k \phi h $ yields... $$\frac{1}{kh}\frac{dh}{dt}=\frac{1}{\phi} \frac{\partial^4}{\partial x^4}=\lambda$$

Why does the first problem end up with a negative constant and the second gets a positive one?

Edit: Does it have something to do with this question Why can we assume the separation constant??

Angus Campbell
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    You can put $-\lambda$ or $\lambda$. There's not problem. –  Aug 25 '20 at 20:18
  • If the time interval is $[0,+\infty),$ I think in both cases we should have a negative constant, otherwise $h(t)=e^{k\lambda t}$ diverges in time – Vincenzo Tibullo Aug 25 '20 at 22:56
  • It's not over [0,+infinity], we are evaluating at equilibrium over x from 0 to L. So either time =0 or dt=0 This is a boundary value problem, it's leading up to calculating initial conditions (where t=0) and/or equilibrium, where u(x,0) = f(x) and dt=0, so we can effectively describe u(x,t) as u(x). I'm not sure if that makes any sense. This problem is from Richard Haberman 4th eddition Applied Partial Differential Equations. – Angus Campbell Aug 26 '20 at 22:58

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