The difference between the nonlocal and local conditions problems

Question

In some problems involving ordinary differential equations, subsidiary conditions are imposed locally. In some other cases, nonlocal conditions are imposed.

In this paper: Existence and uniqueness of a classical solution to a functional-differential abstract nonlocal Cauchy problem Byszewski studied this form of functional-differential nonlocal problem:

$(1)\left\{\begin{matrix} u'(t)=f(t,u(t),u(a(t))),\:\:t\in I \\ u(t_0)+\sum_{k=1}^{p}c_ku(t_k)=x_0 \end{matrix}\right.$

With $I:=[t_0,t_0+T], t_0<t_1<...<t_p\leq t_0+T, T>0$ and $f:I\times E^2\rightarrow E \:$ and $\:a:I\rightarrow I \:$are given functions satisfying some assumptions; $E$ is a Banach space with norm $\:\left \| . \right \|; x_0\in E, c_k\neq 0 \:\:(k=1,...,p)\: p \in \mathbb N$.

And here, in the classical Robin problem: $$u''(t) + f(t,u(t),u'(t)) = 0$$

With local conditions: $u(0)= 0$ and $u'(1) = 0.$

Or

With nonlocal conditions: $u(0)= 0$ and $u(1) = u(\eta)\;\:\eta\in(0,1)$

My question is:

-When we say that The conditions of functional-differential problem are local or nonlocal?

-In which situation we impose local or nonlocal conditions?

Thank you!

Answer 1

This is not an answer but rather an extended comment.

The derivative in its ordinary sense is local: for any neighborhood of $x$, if $f \equiv g$ on that neighborhood and $f^{(n)}(x)$ exists then $g^{(n)}(x)$ exists and $f^{(n)}(x) = g^{(n)}(x)$. So, equations of the form $$ x^{(n)}(t) = F(t, x(t), x'(t), \dots, x^{(n-1)}(t)) $$ or $$ \frac{\partial u}{\partial t}(t, x) = \frac{\partial^2 u}{\partial x^2}(t,x) + F(t, x, u(t, x)) $$ are referred to as local.

Generally, nonlocal denotes that in an equation in question there is something that does not belong to the above category. For instance, we can replace in an ODE the derivative by a derivative of fractional order: the latter does not have the locality property as described in the first paragraph.

Equations of the form $$ x'(t) = F(t, x(t), x(a(t))), $$ where $a$ is a given function, are, to the best of my knowledge, (almost) never called nonlocal: the standard name appears to be (functional) (ordinary) differential equations with deviating argument (retarded or delayed if $a(t) < t$, and advanced if $a(t) > t$).

The boundary conditions considered in Byszewski's paper look like "usual" multipoint boundary conditions.

It seems that the author just chose to call problems considered by him functional-differential nonlocal problems. I think it would be proper to ask him directly: [email protected].

Answer 2

I would like to give another example, a very classical one. Consider a smooth open subset $\Omega$ of $\mathbb R^n$ and let $T$ be a smooth vector field tangential to the boundary $\partial\Omega$. The Oblique Derivative Problem is $$ \begin{cases} ∆ u=0\quad\text{in $\Omega$,}\\ Tu+\alpha \frac{\partial u}{\partial \nu}=g\quad\text{on $\partial\Omega$,} \end{cases} \tag{1}$$ where $\alpha$ is a given function on the boundary. When $T=0$ and $\alpha =1$, you recognize the Neumann problem. If you add a zeroth-order term $\beta$ to $T$ and make $T=0$, $\alpha =0, \beta=1$, you find the Dirichlet problem. Now you consider the Green operator $G$ you have solving uniquely the Dirichlet problem $$ ∆ Gw=0, \quad u_{\vert\partial \Omega}=w, $$ so that $(1)$ is equivalent to $$ Aw:=Tw+\alpha \frac{\partial Gw}{\partial \nu}=g, \tag{2}$$ on $\partial\Omega$. The operator $A$ is a first-order pseudo-differential operator on the boundary and, if $\alpha$ is not identically vanishing, is not a differential operator. It is not difficult to see that if $\alpha$ is non-vanishing, the operator $A$ is elliptic. If you assume that at points where $\alpha$ vanishes, the vector $T$ is non-vanishing, then it implies that $A$ is principal type. Of course $A$ is a non-local operator, namely $$ \text{support } A u\not\subset\text{support } u $$ in general. Most boundary problems are in fact non-local problems on the boundary.