What is the meaning of the notation $F: \mathbb{R}^{n^k} \times \mathbb{R}^{n^{k-1}} \times\dotsb\times \mathbb{R} \times U\rightarrow \mathbb{R}$ 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.
 A: 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 $F (p_{ij}, p_i, u, x) = p_{11} + p_{22} + \cdots + p_{nn}$. 
A: $D$ 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.
