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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?

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    $\begingroup$ X and T definitely don't have dimensions of temperature separately, dimensional analysis will show you that $\endgroup$ – Triatticus Sep 11 '19 at 17:53
  • $\begingroup$ 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. $\endgroup$ – waltzingmonkey Sep 11 '19 at 18:04
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    $\begingroup$ @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. $\endgroup$ – probably_someone Sep 11 '19 at 18:39
  • $\begingroup$ Thanks, @probably_someone $\endgroup$ – waltzingmonkey Sep 12 '19 at 1:57
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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.

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

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

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