# Commutation of differentiation and averaging operations

I've been studying Turbulence, and there's a lot of averaging of differential equations involved. The books I've seen remark offhandedly that differentiation and averaging commute

for eg. < $$\frac{df}{dt}$$ > = $$\frac{d<f>}{dt}$$

Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation?

Stephen Tashi
I'm not sure I understand the notation. I would denote a temporal average of df/dt as the quantity <df/dt> so the braces include the dt, and the time derivative of some other kind of average of f as (d<f>/dt).

Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

$$\int \frac{d}{dt}(f(t))$$

$$= \frac{d}{dt} \int f(t) dt$$

I'm not sure I understand the notation. I would denote a temporal average of df/dt as the quantity <df/dt> so the braces include the dt, and the time derivative of some other kind of average of f as (d<f>/dt).

Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

$$\int \frac{d}{dt}(f(t))$$

$$= \frac{d}{dt} \int f(t) dt$$

That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involve independent variables. Is there a rigorous justification for doing it when both the differentiation and integration involve the same variable?

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

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lurflurf
Homework Helper
That is just the Leibniz rule. What can go wrong does not involve independent variable, but rather convergence. Two limits are being interchanged, so trouble may result. Often convergence is uniform justifying interchange, in other cases more subtile analysis is needed.

Stephen Tashi
That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involve independent variables.

That's a good point. What I wrote suggests an idea, but it doesn't really make sense to write
$$\frac{d}{dt} \int f(t) dt$$
since the integralion shouldn't leave a function that still depends on t. So you should double check what you said about <f> being a temporal average. Perhaps it is a spatial average.

Is there a rigorous justification for doing it when both the differentiation and integration involve the same variable?

Yes, there are conditions that say when one may do this. I can look them up. They aren't something that most people remember since functions describing physical phenomena are usually "well behaved".

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

LaTex on the forum looks crazy because of a problem with the pages showing a stale cache of LaTex expressions. (There are threads discussing this.) What you must do is to "preview" your page and then uses the browsers refresh or reload function after that.

lurflurf
Homework Helper
It would help if the definition given were given here. The usual temporal average does not remove temporal dependence. Often an overbar is used instead of a bracket to avoid confusion. For example
$$\int_{-\infty}^{\infty} f(x,s)w(s-t) ds$$
or
$$\frac{1}{2h}\int_{t-h}^{t+h} f(x,s) ds$$

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Here's my definition of temporal average -

<f(t,x)> = $$\stackrel{lim}{T\rightarrow\infty}$$ $$\frac{1}{T}$$ $$\int_{t_{0}}^{t_{0}+T} f(t,x) dt$$

And it's all making even less sense to me now. With this definition, <f(t,x)> is independent of t (as it should if the "average" is to make sense.).

So is

$$\frac{\partial<f>}{\partial t}$$ = < $$\frac{\partial f}{\partial t}$$ > ?

Shouldn't the L.H.S. be zero? I'm begiining to think there's something wrong in the problem I'm working on ....

lurflurf
Homework Helper
As you have written it those are not equal.
It would be more usuual and less nonsensical if it were
$$\frac{1}{T}\int_{t}^{t+T} f(s,x) ds$$