Let $k\geq 1$, $n\geq 2$ and let $U\subseteq \mathbb{R}^n$ be open and connected. Let $F:U\times \mathbb{R}\times\mathbb{R}^n\times \cdots \times \mathbb{R}^{n^k}\to \mathbb{R}$ a given function. A k-th order pde is an equation of the form $$F(x_1,\dots,x_n,u,u_{x_1},\dots,u_{x_n},u_{x_1 x_1}, u_{x_1 x_2},\dots,u_{\underbrace{x_n x_n\cdots x_n}_{k-\text{times}}})=0,\qquad [1]$$ where $u=u(x_1,\dots,x_n):U\to\mathbb{R}$ is an (unknown) function and $u_{x_i}:=\frac{\partial u}{\partial x_i},\,\forall i=1,\dots n$.

Definition A classical solution for $[1]$ is a function $\overline{u}=\overline{u}(x_1,\dots,x_n):U\to \mathbb{R}$ such that $\overline{u}\in C^k(U)$ and $\overline{u}$ satisfies $[1]$, ie for all $\underline{x}\in U$ $$F(\underline{x},u(\underline{x}),u_{x_1}(\underline{x}),\dots,u_{x_n}(\underline{x}),\dots,u_{x_n x_n\cdots x_n}(\underline{x}))=0.$$

I have some problems understanding the notion of solution for a pde like $[1]$+ some boundary or initial conditions. Supposing we have $$\begin{cases} F(x_1,\dots,x_n,u,u_{x_1},\dots,u_{x_n},u_{x_1 x_1}, u_{x_1 x_2},\dots,u_{{x_n x_n\cdots x_n}})=0,\qquad (x_1,\dots,x_n)\in U \qquad[1]\\ u(x_1,\dots,x_n)=g(x_1,\dots,x_n)\qquad \text{on the curve } \Gamma\subset \overline{U}\qquad\qquad\qquad[2] \end{cases}$$ where $\overline{U}$ is the closure of $U$ and $g$ is a given function. My question is the following:

What is the definition of (classical)solution for a problem like this? Do we have to require a function $\overline{u}\in C^k(U)\cap C^0(U\cup \Gamma)$ such that $\overline{u}$ satisfies $[1]$ and $[2]$? Or maybe $\overline{u}\in C^k(U)\cap C^0(\overline{U})$ or $\overline{u}\in C^k(\overline{U})$ ? I really don't understand what is, by definition, the domain of a solution $\overline{u}$ for a problem like this, and where do we have to require smootness.

Thanks in advance.

  • $\begingroup$ Normally it would be $C^k(\bar{U})\cap C^0(U\cup \Gamma)$; i.e., the minimal required. But sometimes you can easily prove more regularity, such as $u \in C^k(\bar{U})$, in which case authors might just take the latter for the definition. There is no hard and fast rule either way. $\endgroup$ – Jeff Oct 1 '18 at 3:26
  • $\begingroup$ Thank you so much Jeff, I think you probably want to write $C^k(U)\cap C^0(U\cup \Gamma)$ instead of $C^k(\overline{U})\cap C^0(U\cup \Gamma)$. $\endgroup$ – eleguitar Oct 3 '18 at 18:23
  • $\begingroup$ Yes, of course. I can't edit my comment now though. $\endgroup$ – Jeff Oct 5 '18 at 2:00

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