# What's the physical meaning of the integral of the squared temperature in a one-dimensional rod, used to prove uniqueness of soln. in heat eqn.?

Before getting to the question, some background. Let $u(x,t)$ be the temperature in a laterally insulated rod of length $L$, at position $x$ and time $t$. The temperature satisfies the heat equation

$\partial_t u = c \, \partial^2_x u$,

where $c > 0$, and let us assume $u(x,0) = f(x)=$ some given initial temperature distribution and assume $u(0,t) = 0$ and $u(L,t) = 0$ for all $t$. Thus, the ends of the rods are held at zero temperature.

To prove uniqueness of the solution, assume $u_1$ and $u_2$ satisfy the above conditions and let $u = u_1 - u_2$, which satisfies $\partial_t u = c\, \partial^2_x u$, $u(x,0) = 0$ and $u(0,t) = 0$ and $u(L,t) = 0$. The way everyone proves uniqueness (without using maximum principles) is to consider the integral

$V(t) = \int_0^L u(x,t)^2 dx$.

One then shows that $dV/dt \leq 0$ (this is easy). Thus, $V(t)$ is nondecreasing and hence $0 \leq V(t) \leq V(0)$ for all $t$. Since $V(0) = 0$, we get $V(t) = 0$ for all $t$ and hence $u_1 = u_2$.

Now, my question: What does $V(t)$ represent physically???

Not a single reference has ever explained what $V(t)$ represents, they simply introduce $V(t)$ "out of a hat", then show $dV/dt \leq 0$ without ever explaining what $V(t)$ physically represents. Thus, it seems the authors consider $V(t)$ as simply a "trick", but there should be some physical meaning of it.

What is the physical meaning of $V(t)$?

• Why should there be physical meaning to a mathematical trick (assuming it is just a mathematical trick)? – Kyle Kanos Jan 5 '15 at 17:56
• There might not be, but behind almost every mathematical "trick" I've ever seen (related to physics), there is some physical intuition that guides the person to make the trick. I'm hoping there is such intuition here! – Curiosity Jan 5 '15 at 18:05
• The diffusion equation is very similar to the Schroedinger equation for a particle with no potential, and the situation of the 1D rod with 0 temperature at the ends is much like a particle in an infinite square well. In the case of the particle, $V(t)$ tells you the probability that the particle is in the box. The boundary conditions mean the particle is trapped so $\dot{V}(t)=0$. Similarly, here $V(t)$ may give you information about the rate heat escapes the rod. With the ends fixed to 0, no heat escapes. Possibly this is the physical reasoning motivating this trick. – user27118 Jan 5 '15 at 21:10
• It would appear as if this is one way of defining a temperature potential. Your problem is isentropic so a potential is possible to find. Typically for any quantity $u$, $u = \partial \phi / \partial x$ where $\phi$ is the potential function. Specifically, potential functions appear any time the Laplacian vanishes -- which it does here given that your final solution says $\partial_t u = 0$. $V$ is the potential function in this instance. – tpg2114 Jan 6 '15 at 5:10
• If I were more certain I would have posted that as an answer. I'm not 100% positive on it, but it's very similar to how we deal with potential flow equations (as is the final resulting equation). – tpg2114 Jan 6 '15 at 5:14