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.
Best Answer
"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:
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 :-)