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Fully nonlinear first order PDEs (advanced topic)

We consider fully non-linear multidimensional equation \begin{equation} F(x,u,\nabla u)=0 \label{equ-2b.1} \end{equation} (we prefer such notations here) with $x=(x_1,\ldots,x_n)$, $\nabla u= (u_{x_1},\ldots, u_{x_n})$ and the initial condition \begin{equation} u|_\Sigma = g \label{equ-2b.2} \end{equation} where $\Sigma$ is a surface. If $F=u_{x_1}-f(x,u,u_{x_2},\ldots,u_{x_n})$ and $\Sigma=\{x_1=0\}$ then such problem has a local solution and it is unique. However we consider a general form under assumption \begin{equation} \sum_{1\le j\le n} F_{p_j} (x,u,p)\bigr|_{p=\nabla u} \nu_j \ne 0 \label{equ-2b.3} \end{equation} where $\nu=\nu(x)=(\nu_1,\ldots,\nu_n)$ is a normal to $\Sigma$ at point $x$.

Consider $p=\nabla u$ and consider a characteristic curve in $x$-space ($n$-dimensional) $\frac{dx_j}{dt}=F_{p_j}$ which is exactly (\ref{equ-2b.6}) below. Then by chain rule \begin{align} &\frac{dp_j}{dt}= \sum_k p_{j,x_k} \frac{dx_k}{dt}= \sum_k u_{x_jx_k} F_{p_k} \label{equ-2b.4}\\ &\frac{du}{dt}=\sum_k u_{x_k} \frac{dx_k}{dt}=\sum_k p_k F_{p_k}. \label{equ-2b.5} \end{align} The last equation is exactly (\ref{equ-2b.8}) below. To deal with (\ref{equ-2b.4}) we differentiate (\ref{equ-2b.1}) by $x_j$; by chain rule we get \begin{equation*} 0=\partial_{x_j} \bigl(F(x,u,\nabla u)\bigr) = F_{x_j}+ F_u u_{x_j} + \sum_k F_{p_k} p_{k,x_j}= F_{x_j}+ F_u u_{x_j} + \sum_k F_{p_k} u_{x_kx_j} \end{equation*} and therefore the r.h.e. in (\ref{equ-2b.4}) is equal to $ -F_{x_j}- F_u u_{x_j}$ and we arrive exactly to (\ref{equ-2b.7}) below.

So we have a system defining a characteristic trajectory which lives in $(2n+1)$-dimensional space: \begin{align} &\frac{dx_j}{dt}=F_{p_j}, \label{equ-2b.6}\\ &\frac{dp_j}{dt}=-F_{x_j}-F_u p_j ,\label{equ-2b.7} \\ &\frac{du}{dt}=\sum_{1\le j\le n} F_{p_j}p_j.\label{equ-2b.8} \end{align} Characteristic curve is $n$-dimensional $x$-projection. Condition (\ref{equ-2b.3}) simply means that characteristic curve is transversal (i. e. is not tangent) to $\Sigma$.

Therefore, to solve (\ref{equ-2b.1})-(\ref{equ-2b.2}) we

1. Find $\nabla\_\Sigma u=\nabla\_\Sigma g$ at $\Sigma$ (i.e. we find gradient of $u$ along $\Sigma$; if $\Sigma=\\{x\_1=0\\}$ then we just calculate $u\_{x\_2},\ldots,u\_{x\_n})$;
2. From (\ref{equ-2b.1}) we find the remaining normal component of $\nabla u$ at $\Sigma$; so we have $(n-1)$-dimensional surface $\Sigma^*=\\{(x,u,\nabla u), x\in \Sigma\\}$.
3. From each point of $\Sigma^*$ we issue a characteristic trajectory described by (\ref{equ-2b.6})-(\ref{equ-2b.8}). These trajectories together form $n$-dimensional hypesurface $\Lambda$ in $(2n+1)$-dimensional space.
4. Locally (near $t=0$) this surface $\Lambda$ has one-to-one $x$-projection and we can restore $u=u(x)$ (and $\nabla u =p(x)$).

However this property (d) is just local.

Remark. We have not proved directly that this construction always gives us a solution but if we know that solution exists then our arguments imply that it is unique and could be founds this way. Existence could be proven either directly or by some other arguments.

Remark.

1. Important for application case is when $F$ does not depend on $u$ (only on $x,\nabla u$) and (\ref{equ-2b.6})-(\ref{equ-2b.8}) become highly symmetrical with respect to $(x,p)$: \begin{align} &\frac{dx\_j}{dt}=F\_{p\_j}, \label{equ-2b.9}\\\\ &\frac{dp\_j}{dt}=-F\_{x\_j} , \label{equ-2b.10}\\\\ &\frac{du}{dt}=\sum\_{1\le j\\le n} p\_jF\_{p\_j} .\label{equ-2b.11} \end{align} This is so called *Hamiltonian system* with the *Hamiltonian* $F(x,p)$.
2. In this case we can drop $u$ from consideration and consider only $(x,p)$-projections of $\Sigma^*$ and $\Lambda$.