What is the meaning of the notation $F: \mathbb{R}^{n^k} \times \mathbb{R}^{n^{k-1}} \times\dotsb\times \mathbb{R} \times U\rightarrow \mathbb{R}$

Question

A $k^{th}$ order PDE is defined by $$F(D^ku(x),D^{k-1}u(x),\dotsc,Du(x),u(x),x)=0,$$ where $x$ is an element of $U$. I know that $\mathbb{R}^n$ is an $n$-dimensional real Euclidean space. I am not familiar with the notation used for $F$. I came across this definition in the PDE by Lawrence C Evans. Also, how is this a $k$th order PDE? Aren't there supposed to be more than one variable? Theres only $x$ here.

Answer 1

Let me just recall the notations:

• $U$: an open set in $\mathbb R^n$,
• $x = (x_1, \cdots, x_n)$ is in $U$,
• $u$ is a function $u: U \to \mathbb R$,
• for any $k = 1, 2, \cdots$, $D^k u(x)$ is a shorthand notations for $$u_{i_1\cdots i_k} = \frac{\partial ^k u }{\partial x_{i_1} \partial x_{i_2} \cdots \partial x_{i_k}} \ \ \ \ (\text{at } x\in U),$$ where $i_1, \cdots i_k$ are in $\{1, \cdots, n\}$. So for all $x\in U$ and $u$, $D^k u(x)$ is an element in $\mathbb R^{n^k}$.

A $k$-th order PDE is an equation of the form

$$F(D^ku(x),D^{k−1} u(x),…,Du(x),u(x),x)=0, $$

where $F$ depends on

• $x\in U$,
• $u(x)\in \mathbb R$,
• $Du (x) \in \mathbb R^n$, ...
• $D^ku (x) \in \mathbb R^{n^k}$.

For example, the Laplace equation

$$ \Delta u = \frac{\partial ^2 u }{\partial x_1^2} + \cdots + \frac{\partial ^2 u }{\partial x_n^2}=0$$

is a second order PDE. In this case $F (p_{ij}, p_i, u, x) = p_{11} + p_{22} + \cdots + p_{nn}$.

Answer 2

$D$ is the differential operator, $D^k$ is the $D$ applied $k$ times. In other words, $Du(x)=u'(x)$, $D^2u(x)=u''(x)$, and so on. This is a $k$th order PDE because derivatives up to the $k$th derivative $D^ku(x)=u^{(k)}(x)$ occur.