Solution domain of PDEs

Question

Suppose we have a PDE in the following general form $$ F\left(x,y,\ldots, U, \frac{\partial U}{\partial x},\frac{\partial U}{\partial y},\ldots,\frac{\partial^2 U}{\partial x^2},\frac{\partial^2 U}{\partial x \partial y}, \frac{\partial^2 U}{\partial y^2}, \ldots\right)=0 $$ where $F$ is a given function of the independent variables $x, y , \ldots$, of the unknown function $U$ and of a finite number of its partial derivatives.

My question. Why the sought for numerical or analytical solutions of this PDE are customarily assumed to be defined on $(0,1)$? Why the independent variables are customarily not assumed to belong to $(1,2)$ or to any other interval $(a,b)$, where $a,b \in \Bbb N$?

An example.

Answer 1

It is just a matter of notation: it could have been any interval and the solution method would have been the same.

A qualitative explanation.

Since the general solution of a PDE of the second or higher order can be found only in very few cases, and even in those cases it is almost always useless, scientists and practitioners alike prefer to consider special problems for PDEs, for example the following one $$ \begin{cases} F(\boldsymbol{x},\nabla^l \boldsymbol{U})=0 & \boldsymbol{x}\in \Omega\\ B(\boldsymbol{U})= \boldsymbol{f} & \boldsymbol{x}\in \Sigma\subseteq{\partial\Omega} \end{cases} $$ where

• $F(\boldsymbol{x},\nabla^l \boldsymbol{U})$ is a synthetic notation for the general partial differential operator above written by using the higher order gradient $\nabla^l$, $l\ge 2$,
• $B$ is a boundary operator i.e. an operator which is defined on a (possibly proper) subset $\Sigma$ of the boundary $ \partial\Omega $ of the domain $\Omega$ where a solution $\boldsymbol{U}$ of the problem is required to be found,
• $\boldsymbol{f}$ is a function defined on $\partial\Omega$.

Now, provided the used methods for solution (numerical or analytic) are adequate, it does not matter what is the exact structure of the domain. Therefore, in example problems, the unit interval (or more generally the unit cube) is a standard choice.

In the example above, we have that $\Omega=[0,1]\times[0,1]\times\Bbb R_{\ge 0}$ possibly because, by making this choice, the form of the solution to the posed problem (a mixed problem) is slightly simpler.