Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
$k$ has units, too! –  Rahul Aug 1 '12 at 5:45
1  
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
add comment

2 Answers 2

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!

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.