I am interested in non-linear PDEs which can be split into different cases depending on the value taken by the underlying variables $-$ see equation \eqref{eq:pde} below.
More concretely, let us consider the unknown function $v=v(t,x)$ where $(t,x)\in\mathbb{R}_+\times\mathbb{R}^2$. We consider a subset $D\subset\mathbb{R}$, parametrized by some functional which might be exogenous or endogenous to the PDE (see examples below). Let $f$ and $g$ be two functions. Then $v$ satisfies the following PDE: \begin{align} \begin{aligned} &v=f(t,x,v_t,v_x,v_{xx})\mathbb{1}_{x\in D}+g(t,x,v_t,v_x,v_{xx})\mathbb{1}_{x\in D^c} \end{aligned}\label{eq:pde}\tag{1} \end{align}
Namely, the PDE for $v$ is different in $D$ and $D^c=\mathbb{R}\backslash D$:
- One example with "endogenous" parametrization of the domain $D$ is: \begin{align} D&=\{(t,x):v_x>0\}\\[5pt] v&=\left(\alpha v_t+\beta v_{xx}+\gamma v_x\right)1_{\{x\in D\}}+\left(\alpha v_t+\beta v_{xx}-\gamma v_x\right)1_{\{x\in D^c\}} \end{align} where the PDE can also be rewritten: \begin{align} v&=\alpha v_t+\beta v_{xx}+\gamma\max\{v_x,-v_x\} \end{align}
- An example with "exogenous" parametrization is: $D=\{(t,x):h(t,x)\in E\}$ where $h$ is some function taking values in $\mathbb{R}$ and $E$ some subset of $\mathbb{R}$.
Note that in any case, equation \eqref{eq:pde} can be "split" as follows: \begin{align} \label{eq:pde-1}\tag{2} &v=f(t,x,v_t,v_x,v_{xx}), \quad x\in D \\[5pt] \label{eq:pde-2}\tag{3} &v=g(t,x,v_t,v_x,v_{xx}), \quad x\notin D \end{align}
Is there a set of conditions under which we can solve PDEs $\eqref{eq:pde-1}$ and $\eqref{eq:pde-2}$ separately, then "stick" them together using indicator functions over the domain $D$ to obtain the solution to the initial non-linear PDE $\eqref{eq:pde}$? That is, there exists functions $F$ and $G$ which solve $\eqref{eq:pde-1}$ and $\eqref{eq:pde-2}$ respectively such that: \begin{align} v(t,x)=F(t,x)\mathbb{1}_{x\in D}+G(t,x)\mathbb{1}_{x\in D^c} \end{align}
Intuitively, if $f$ and $g$ are well-defined at the boundary $\partial D$ of the domain, and if the limits of $f$ and $g$ when $x$ approaches the boundary are the same, I would think that we can follow the procedure above. Does this make sense? Any helpful resource to look up? Or am I fully wrong and missing something?
