The following is paraphrased from Introduction to integral equations with applications by Abdul Jerri:

Let $F$ be a functional with domain $D$. Start with $F(x,t,u(t))$. If $F$ has a continuous partial derivative $\partial F / \partial u$ in $D$, then we can use the mean-value-theorem:

For any $u_1(t)$ and $u_2(t)$ in $D$, there is an $\eta(t)$ between them, $u_1(t) < \eta(t) < u_2(t)$, such that $\displaystyle\frac{F(x,t,u_1(t)) - F(x,t,u_2(t))}{u_1(t) - u_2(t)} = \frac{\partial F}{\partial u}(x, t, \eta(t))$.

I have two questions:

  1. What precisely is meant by the notation $\displaystyle\frac{\partial F}{\partial u}(x, t, \eta(t))$? (specifically, how is $\displaystyle\frac{\partial F}{\partial u}$ defined?)

  2. How can I generalise this result to suit a functional of the form $F(x, t_1, t_2, u(t_1), u(t_2)$) (where $F$ takes values in $\mathbb R$ and $u_1, u_2 : \mathbb R \to \mathbb R$, if this makes any difference)?

Any explanations, insight or references to relevant sources of information would be greatly appreciated.

  • $\begingroup$ For 1. I think that it's the derivative with respect to the third argument. $\endgroup$ Jan 26, 2012 at 14:38
  • $\begingroup$ @DavideGiraudo What does it mean to take the derivative of a functional? (is this the same as the 'functional derivative': en.wikipedia.org/wiki/Functional_derivative?) $\endgroup$
    – Matt
    Jan 26, 2012 at 14:40
  • $\begingroup$ Depends on what is meant by a Functional at this point. From your statement, since $F$ depends on $x,t$ also, it appears to me that $F$ is merely a mapping $\mathbb{R}^3 \to \mathbb{R}$ for which you insert as the third variable $u$ something that is actually a function of $t$. In which case $\partial F/ \partial u$ is nothing more than the usual partial derivative with respect to the third coordinate of the function $F$. $\endgroup$ Jan 26, 2012 at 15:01
  • $\begingroup$ It is most likely not a functional derivative, which would usually be written $DF$ or $dF$ for Frechet or Gateaux derivatives, or $\delta F / \delta u$ in physics notation for functional derivatives, and furthermore by the equation given, $\partial F/ \partial u(x,t,\eta(t))$ takes scalar values, whereas the functional derivative should take value in the dual space to some infinite dimensional vector space. $\endgroup$ Jan 26, 2012 at 15:05
  • $\begingroup$ @WillieWong Ok, that makes sense, thanks. Any ideas about a generalisation? $\endgroup$
    – Matt
    Jan 26, 2012 at 15:12

1 Answer 1


In the first place you are given a real-valued function $F\colon (x,t,u)\mapsto F(x,t,u)$ defined in a certain domain $D$ of $(x,t,u)$-space. Here $u$ is an independent variable as are $x$ and $t$. At any point $p:=(x,t,u)\in D$ this $F$ has partial derivatives $F_x(p)$, $F_t(p)$, and $F_u(p)$ which can also be addressed by means of the notation ${\partial f\over\partial x}(p)$.

In addition you are considering functions $u(\cdot)\colon \ t\mapsto u(t)$ where it is guaranteed that all resulting points $\bigl(x,t,u(t)\bigr)$ are lying in $D$. In this case the original $F\colon\ D\to{\mathbb R}$ produces for each such function $t\mapsto u(t)$ a new function depending on ($x$ and) $t$, given by $$(x,t)\mapsto F\bigl(x,t,u(t)\bigr)\ .$$ That's why $F$ is called a functional now: $F$ transforms one function of $t$ into some other function of $t$. (This has nothing to do with "functionals" in the realm of linear algebra.)

Let's look at the mean value theorem under these circumstances. When the segment connecting the points $(x,t,u_1), (x,t,u_2)\in D$, $\ u_1<u_2$, is contained in $D$ then the ordinary mean value theorem for one variable ($u$ here) guarantees the existence of an $\eta\in\ ]u_1,u_2[\ $ such that $${F(x,t,u_2)-F(x,t,u_1)\over u_2-u_1}\ =\ {\partial F\over \partial u}(x,t,\eta)\ .$$ It follows that given two functions $t\mapsto u_i(t)$ as described there is a function $t\mapsto\eta(t)$ such that your large equation holds.

In the second situation there seems to be only one function $u(\cdot)$ involved, so the partial derivative of $F$ with respect to the third variable does not enter. Before I elaborate more you should indicate what kind of formula you expect to hold. Note that the mean value theorem guaranteeing a $\tau$ with $u(t_2)-u(t_1)=u'(\tau)(t_2-t_1)$ does not hold for vector-valued functions $u(\cdot)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .