# Is separation of variables in the heat equation dimensionally consistent?

This may be a trivial question but is about the statement that the function $$U(x,t)$$ in the heat equation may be expressed in the form $$X(x)\cdot T(t)$$. It's that $$X$$ and $$T$$ both are functions which have outputs of dimension temperature. So does $$U$$.

Are the two sides of the equation dimensionally consistent?

• X and T definitely don't have dimensions of temperature separately, dimensional analysis will show you that – Triatticus Sep 11 '19 at 17:53
• Can you elaborate? X is a function takes position as input and outputs the temperature there? T takes in time and outputs the temperature at that time? Edit: what I wrote above doesn't make sense to me but then I wonder what the dimensions of X and T are. – waltzingmonkey Sep 11 '19 at 18:04
• @waltzingmonkey The division of the units is entirely up to you, as you're the one who is defining $X$ and $T$. The important part is that $X$ is a function only of space, and $T$ is a function only of time. My personal strategy is to non-dimensionalize both $X$ and $T$, so that both are simply pure functions of dimensionless parameters, and have all of the units carried by a constant, like $U(x,t)=kX(x/L)T(t/\tau)$, where $L$ and $\tau$ are arbitrary length and time scales. – probably_someone Sep 11 '19 at 18:39
• Thanks, @probably_someone – waltzingmonkey Sep 12 '19 at 1:57

The 2D heat equation $$\alpha \frac{\partial^2 U}{\partial x^2} = \frac{\partial U}{\partial t}$$ and the equations that result from separation of variables, $$T' = - \lambda \alpha T \qquad X'' = -\lambda X$$ are all linear in the functions to be solved for. This means that it is dimensionally consistent to assign them any units we please; all that is required for dimensional consistency is that $$[\lambda] = \text{m}^{-2}$$ and $$[\alpha] = \text{m}^2/\text{s}$$.

In particular, there is no requirement that either $$T$$ or $$X$$ have units of temperature. All that is required is that $$U = TX$$ has units of temperature—and even then, that's only on physical grounds. The first equation would be perfectly consistent if $$U$$ had dimensions of kilograms or coulombs.

• Thank you! That does clarify. – waltzingmonkey Sep 12 '19 at 2:06

It will be consistent if the product $$X\times T$$ has the same units as $$U$$. In general, a function does not hold the same units as its arguments. For example the function:

$$f(x)=e^{-x^2/\sigma^2}$$

Have no dimensions even though $$x$$ and $$\sigma$$ could be expressed in some length units.

• Thank you, @fgoudra. That helps. – waltzingmonkey Sep 12 '19 at 2:05