## Is it a functional differential equation [closed]

definition A functional differential equation is a differential equation in which the derivative $u'(t)$ of an unknown function u has a value at t that is related to u as a function of some other function at t. A general first-order functional differential equation is therefore given by $u'(t)=f(t,y(t),y(u(t)))$

I have not have the classes about functional. I think it is but I am not sure. So please give me a answer. Thank you.

Let $F(u)$ be a functional ,$f(u)=0$ is its Euler-Lagrange equation. We need to calculate the $u$ which minimize the $F$. So, as usually, we have the equation as follows: $$\frac{\partial u}{\partial t}=-f(u)$$ Through iteration, we get the final u (the original $u_{0}$ is arbitrary,ps1): $$\frac{u^{n+1}-u^{n}}{\triangle t}=-f(u)$$ $$\frac{u^{n+1}\left(x,y\right)-u^{n}\left(x,y\right)}{\triangle t}=-f\left(u\right)=\iint H(u(a,b))K(g\left(x,y\right)-g\left(a,b\right))dadb \qquad (1)$$ $$u,g:R^{2}\rightarrow R$$

$$\begin{eqnarray*} K:\, R & \rightarrow & R\ K(a) & = & \begin{cases} 1 & a=0\ 0 & otherwise \end{cases} \end{eqnarray*} \begin{eqnarray*} H:\, R & \rightarrow & R\ H(a) & = & \begin{cases} 1 & a\geq0\ 0 & otherwise \end{cases} \end{eqnarray*}$$

In the right side of equation 1, there is no derivative. And in the left side of equation 2, is $\partial u/\partial t$ a derivative according to t? Is equation 1 a functional differential equation?

1.how did you arrive at that formula with the double integral? To a specific point (x,y), for example x=1,y=6, then the equation becomes:$$\frac{u^{n+1}\left(1,6\right)-u^{n}\left(1,6\right)}{\triangle t}=-f\left(u\right)=\iint H(u(a,b))K(g\left(1,6\right)-g\left(a,b\right))dadb \qquad (2)$$ In the right side of the equation(2), we get a real number,so we could update $u^{n+1}\left(1,6\right)$. To all points in our plane coordinate, we use the same way to update every point.
2. how do you actually want to use this expression? I want to update values of u，I mean calculate the function u by iteration. $R^{2}\rightarrow R$

ps1:$u_{0}$ I initialize $u_{0}$ at random. In my experiments, it makes no difference to get the final results. I tested this, the final results were almost same. So I believe the difference between two results is very samll and can be neglect. As a result, I did not calculate the error. So in here, I can not say the results are exactly same. They are almost same to human's eyes.

This question is better suited to math.stackexchange.com. Equations of the form $\frac{du}{dt}=f(u(t))$ are called autonomous ODE, not functional differential equations. – Nilima Nigam Jul 25 2011 at 4:22
@Nilima Nigam，Thank you. But according to wiki, "In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of their derivatives with respect to that variable.A simple example is Newton's second law of motion, which leads to the differential equation" . However, in my question, $$u,g:R^{2}\rightarrow R$$. Since I need to calculate $u$ by iteration process, I introduce a variable $t$,please see equation (1). So I do not think it is a ODE. – math Jul 25 2011 at 9:23
@math: voting to close. You should re-post it to Math.SE: I don't remember seeing it the last time around. Anyway, Read your definition again. A functional DE is of the form $y'(t) = f(t, y(t), y(u(t)))$. Note that on the LHS it is $y'$ not $u'$. (You copied it down incorrectly from the link.) What you wrote is, indeed, as Nilima stated, just an ODE (though maybe one that takes value in a topological vector space; which sometimes can be expressed as PDEs). What you have is a functional DE only in the trivial sense that $u(t) = t$. – Willie Wong Jul 25 2011 at 14:33