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How do we determine the units used in a differential equation? Yes, in theory a PDE has nothing to do with units, but I'm interested in this question from a modeling point of view. By units, I mean the following. In an ordinary differential equation, finding the correct units seems relative straightforward. For example, if we have a function $u : [0,T] (seconds) -> \mathbb{R} (meters)$, we know that taking the derivative with respect to time gives velocity, $(meters)/(seconds)$. Taking two derivatives with respect to time gives acceleration, $(meters)/(seconds)^2$. Therefore, when we write an ODE $$ \frac{\partial^2}{\partial t^2} u = f, $$ we know that $f$ should have units $(meters)/(seconds)^2$. In a PDE, this same trick doesn't seem to work. For example, say we have a function $u : [0,T] (seconds) \times \Omega (meters^2) \rightarrow \mathbb{R} (celsius)$ where $\Omega \subseteq \mathbb{R}^2$. Then, we write the heat equation $$ \frac{\partial}{\partial t} u - k \frac{\partial^2}{\partial x^2} u - k\frac{\partial^2}{\partial y^2} u = f $$ or more simply as $$ \frac{\partial}{\partial t} u - k\Delta u = f. $$ Now, using the above trick, the term $\frac{\partial}{\partial t} u$ has units $(celsius)/(seconds)$. However, the term $\Delta u$ has units $(celsius)/(meters^2)$. In this context, it doesn't make sense to add the two terms. It also doesn't give clear insight into what the units of the forcing function $f$ need to be. As such, what are the correct units for the heat equation and what's the general rule for establishing units for an arbitrary PDE?

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$k$ has units, too! – Rahul Aug 1 '12 at 5:45
Specifically, in this case it has to have units $\mathrm m^2/\mathrm s$ for the units to match. – Rahul Aug 1 '12 at 5:46
You mixed up the numerators and denominators in all the units up to "in a PDE". For instance, acceleration has units meters per second squared. – joriki Aug 1 '12 at 6:04
You're right, I flipped the units in the ODE part of the post. It's now been fixed. – wyer33 Aug 1 '12 at 6:09
up vote 0 down vote accepted

$k$ has units $[distance^2 / time]$, so $$\frac{\partial}{\partial t}u - k\Delta u = \left[\frac{temp}{time} - \frac{distance^2}{time} \frac{temp}{distance^2}\right] = \left[\frac{temp}{time}\right] $$

You are already performing the dimensional analysis correctly!

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I think that instead of units, you should study the scaling behaviour of PDEs. For example, what is the effect of simultaneously scaling $t$ and $x$ in the heat equation? And noting that if you scale $x$ with $a$ and $t$ with $a^2$, the equation is unchanged (except for the RHS, if nonzero). Scale invariant solutions are particular interesting and useful when they exist – one classic example being the fundamental solution of the heat equation.

The connection between scaling and the units of physics is via Buckingham's Π theorem.

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