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Let $p$ be a function of the form $\mathbb{R}^2 \to \mathbb{R}$.

How do i find the derivatives of the following expressions with respect to p(x,y) :

a. $\int_\mathbb{R^2} p$

b. $ \dfrac{\partial p}{\partial x}$

I am not sure if the question makes any mathematical sense. I am trying to solve a problem in which i have to differentiate w.r.t functions.

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I doubt you really have to differentiate with respect to a function. Looking at the tags, I'd say you might have stumbled on the unfortunate notation of calculus of variations: you have to differentiate a function of many variables F(x,y,y') with respect to the second or third parameters. They might look like functions, but they still are just variables. You could always just rewrite this F(u,v,w). Then you differentiate with respect to "v" or "w". You could also do the composition F(x,y(x),y'(x)) for example, and differentiate that with respect to x, and not get the same answer. –  Vhailor Jun 4 '11 at 3:34
I think in my case, I might have to differentiate w.r.t to a function, please see the actual problem that i am trying to solve : math.stackexchange.com/questions/43110/… –  AnkurVijay Jun 4 '11 at 3:44
@Vhailor Okay, i think I get your point to some extent. Are you suggesting that my real problem is of the following nature $$ \dfrac{\partial}{\partial z} \int_x\int_y z dx dy$$ where z is some function of x,y. –  AnkurVijay Jun 4 '11 at 3:55
@Vhailor: It's not true that "they might look like functions, but they still are just variables". Calculus of variations is fundamentally different from calculus with real variables. First, you'd have to consider the functions as uncountably many variables at uncountably many points, not just one variable, but more importantly, there's a link between $y$ and $y'$ that you can't just ignore. This leads to the Euler-Lagrange equations, which don't simply set a derivative of a function with respect to its arguments to zero. –  joriki Jun 4 '11 at 5:33
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2 Answers

up vote 2 down vote accepted

"Calculus of variations" is the correct buzzword, you may also look for "analysis in infinite dimensions" or, more generally, "nonlinear functional analysis".

Your question can be made mathematically precise, one of the best (but somewhat sophisticated) sources is still:

  • Richard S. Hamilton: The Inverse Function Theorem of Nash and Moser (Bulletin (New Series) of the American Mathematical Society Volume 7, Number 1, July 1982)

First, the functions that you have to differentiate are "functions of functions", that is the variables are members of function spaces. The paper I cited treats calculus on such function spaces that are "Fréchet spaces", which is sufficiently general for many applications.

To make your question precise, we'll have to add an assumption about the function space we are talking about, but I'll leave that implicit, i.e. to you :-)

(Hint: a good choice would be the Schwartz space of smooth rapidly decreasing functions, see Wikipedia.)

Your first "function of functions" is e.g. the mapping $$ p \mapsto \int_{\mathbb{R}^2} p(x, y) d x d y $$ where I assume that the integral is with respect to the Lesbegue measure and our function space is such that this integral exists and is finite. Let's call this mapping $F$.

Now, the definition of a directional derivative is just the same as in finite dimensions, let $t$ be a real parameter and $q$ be another function, then $$ DF(p, q) := \lim_{t \to 0} \frac{1}{t} (F(p + t q) - F(p)) $$ which is the directional derivative (which is synonymous to "partial derivative") of $F$ (at p) into the direction $q$. If this limit exists for all functions $q$ (and in a neighborhood of p), and is jointly continuous in the variables $p, q$, then we say that $F$ is continuously differentiable (at p). While the function $F$ that we differentiated was a function of "one variable", the derivative $DF$ is a function of two variables, i.e. on the product space.

Ok, I think you'll be able to calculate $DF(p, q)$ for both the examples you cited, if not, come again :-)

Edit: Addendum, a simple example, let's say that $S(\mathbb{R})$ is the space of Schwartz functions on $\mathbb{R}$ and define a function $F$ via: $$ F : S(\mathbb{R}) \to S(\mathbb{R}) $$ $$ f \mapsto f' := \frac{d}{d x} f $$ Now let's see what $F(p ,q)$ may be, if it exists. Since we have $$ F(p + t q) = \frac{d}{d x} (p + t q) = p' + t q' $$ because the derivative is a linear operator (hint hint!), we get $$ DF(p, q) := \lim_{t \to 0} \frac{1}{t} (F(p + t q) - F(p)) = \lim_{t \to 0} \frac{1}{t} (p' + t q' - p') = \lim_{t \to 0} \frac{1}{t} (t q') = q' $$ Ergo, the limit exists and is independent from the first argument and linear in the second.

But we can do a nonlinear example, too, define: $$ F : S(\mathbb{R}) \to S(\mathbb{R}) $$ $$ f \mapsto f^2 $$ Then we get $$ DF(p, q) := \lim_{t \to 0} \frac{1}{t} (F(p + t q) - F(p)) = \lim_{t \to 0} \frac{1}{t} (p^2 + 2 t p q + t^2 q^2 - p^2) = 2 p q $$ As you can see, the calculus in infinite dimensions is a little bit different from the one in finite dimensions :-)

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Tim, I understand the point about functions of functions. However I dont know how to find the limits on F which is a mapping. If you can explain via some example ( not necessarily the ones i wish to evaluate), it'd be great. –  AnkurVijay Jun 4 '11 at 20:05
@AnkurVj: No problem, will do. –  Tim van Beek Jun 4 '11 at 21:06
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You should take a look at the calculus of variations. You can use this to solve optimization problems for integrals including unknown functions and their derivatives. The solution is slightly more complicated than just taking derivatives, but on a general level the variation of a functional with respect to a function is somewhat analogous to the derivative of a function with respect to a variable.

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