2.2. First order PDEs (continued)

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

## 2.2. First order PDEs (continued)

### Multidimensional equations

Remark 1. Multidimensional equations (from linear to semilinear) \begin{equation} au_t + \sum_{j=1}^n b_j u_{x_j}=f(x_1,\ldots,x_n,t,u) \label{eq-2.2.1} \end{equation} and and even quasilinear (with $a=a(x_1,\ldots,x_n,t,u)$ and $b_j=b_j(x_1,\ldots,x_n,t,u)$) could be solved by the same methods.

Example 1. If $a=1$, $b_j=\const$ and $f=0$ the general solution of (\ref{eq-2.2.1}) is $u=\phi (x_1-b_1t,\ldots,x_n-b_nt)$ where $\phi$ is an arbitrary function of $n$ variables.

Example 2. Find the general solution of \begin{gather*} u_t + 2tu_x + xu_y=0. \end{gather*} Finding integral curves \begin{gather*} \frac{dt}{1}=\frac{dx}{2t}=\frac{dy}{x}=\frac{du}{0} \end{gather*} from the first equation we conclude that $x-t^2=C$ and then $dy= xdt = (C+t^2) dt \implies y- Ct -\frac{t^3}{3}=D$. Then $D= y -(x-t^2) t -\frac{t^3}{3}=y-xt +\frac{2t^3}{3}$.
Therefore \begin{gather*} u(x,y,t)=\phi(x-t^2, y-xt +\frac{2t^3}{3}) \end{gather*} is the general solution.

Example 3. Find the general solution of \begin{gather*} u_t + 2tu_x + xu_y=12 xtu. \end{gather*} Finding integral curves \begin{gather*} \frac{dt}{1}=\frac{dx}{2t}=\frac{dy}{x}=\frac{du}{12 xtu} \end{gather*} from Example 2 we know that $x=t^2+C$ and $y= Ct + \frac{t^3}{3}$. Then \begin{align*} \frac{du}{u}=x\,tdt=&(4t^3 +12 Ct)dt\implies \ln(u)= \bigl(t^4 +6Ct^2\bigr) +\ln(E)\\ & \implies u= E \exp \bigl(t^4 +6Ct^2\bigr)=\exp(-5t^4+ 6xt^2)E\\ & \implies u = \exp(-5t^4+ 6xt^2)\phi(x-t^2, y-xt +\frac{2t^3}{3}) \end{align*} is the general solution.

### Multidimensional non-linear equations

We consider fully non-linear multidimensional equation in $\mathbb{R}^{n}$ \begin{equation} F(x,u,\nabla u)=0 \label{eq-2.2.2} \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{eq-2.2.3} \end{equation} where $\Sigma$ is a hypersurface. 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{eq-2.2.4} \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{eq-2.2.7}) below. Then by the 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{eq-2.2.5}\\ &\frac{du}{dt}=\sum_k u_{x_k} \frac{dx_k}{dt}=\sum_k p_k F_{p_k}. \label{eq-2.2.6} \end{align} The last equation is exactly (\ref{eq-2.2.9}) below. To deal with (\ref{eq-2.2.5}) we differentiate (\ref{eq-2.2.3}) by $x_j$; by the chain rule we get \begin{multline*} 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{multline*} and therefore the r.h.e. in (\ref{eq-2.2.5}) is equal to $-F_{x_j}- F_u u_{x_j}$ and we arrive exactly to equation (\ref{eq-2.2.8}) 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{eq-2.2.7}\\ &\frac{dp_j}{dt}=-F_{x_j}-F_u p_j ,\label{eq-2.2.8} \\ &\frac{du}{dt}=\sum_{1\le j\le n} F_{p_j}p_j.\label{eq-2.2.9} \end{align} Characteristic curve is $n$-dimensional $x$-projection of the characteristic trajectory. Condition (\ref{eq-2.2.4}) simply means that characteristic curve is transversal (i. e. is not tangent) to $\Sigma$.

Therefore, to solve (\ref{eq-2.2.3})-(\ref{eq-2.2.4}) we need to

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{eq-2.2.3}) we find the remaining normal component of $\nabla u$ at $\Sigma$; so we have $(n-1)$-dimensional surface $\Sigma^*={(x,u,\nabla u), x\in \Sigma}$ in $(2n+1)$-dimensional space.
3. From each point of $\Sigma^*$ we issue a characteristic trajectory described by (\ref{eq-2.2.7})-(\ref{eq-2.2.9}). 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 $\nabla u =p(x)$).

Remark 2. However this property (4) is just local. Globally projection can stop be one-to-one, and points where it is not one-to-one even locally (that means, it is not one to one even in the small vicinity of such point) have a special role. Their projection to $x$-space are called caustics.

Remark 3. 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 4.

1. Important for application case is when $F$ does not depend on $u$ (only on $x,p=\nabla u$) and (\ref{eq-2.2.7})-(\ref{eq-2.2.9}) become highly symmetrical with respect to $(x,p)$: \begin{align} &\frac{dx_j}{dt}=F_{p_j}, \label{eq-2.2.10} \\ &\frac{dp_j}{dt}=-F_{x_j} , \label{eq-2.2.11}\\ &\frac{du}{dt}=\sum _{1\le j\le n} p _jF _{p_j} .\label{eq-2.2.12} \end{align} This is so called Hamiltonian system with the Hamiltonian $F(x,p)$.
2. In this case we can drop $u$ from the consideration and consider only (\ref{eq-2.2.10})--(\ref{eq-2.2.11}) describing $(x,p)$-projections of $\Sigma^*$ and $\Lambda$, finding $u$ afterwards from (\ref{eq-2.2.12}). Typical example: Hamiltonian mechanics, with the generalized coordinates $x$ and generalized momenta $p$ .