I've a little troubles in proving local uniqueness of solution for Cauchy problems concerning quasilinear PDE's. It's a little bit boring, but I tried to be as clear as possible.
Suppose $\Omega$ is an open and connected subset of $\mathbb{R}^2$ and let $a(x,y,z),b(x,y,z),c(x,y,z)$ scalar functions of class $C^1$ in $\Omega \times \mathbb{R}$. Let $I$ an open interval and $f=f(s)$, $g=g(s)$ and $h=h(s)$ be $C^1(I)$.
We want to prove local existence and uniqueness of a solution for the Cauchy problem $$\begin{cases} a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)\\u(f(s),g(s))=h(s)\qquad s\in I\end{cases},$$ under certain conditions using method of characteristic.
Consider for each fixed $s\in I$ the autonomous system of ODE's $$\begin{cases}\frac{d}{dt}x=a(x,y,z)\\\frac{d}{dt}y=b(x,y,z)\\\frac{d}{dt}z=c(x,y,z)\end{cases}$$ with initial conditions $$\begin{cases}x(0)=f(s)\\y(0)=g(s)\\z(0)=h(s)\end{cases}.$$
Because $a,b,c$ are $C^1$, then for every $s \in I$ I find a unique maximal and global solution $x=X(s,t),y=Y(s,t),z=Z(s,t)$ defined on an open interval $J_s$ containing $0$ such that $X(s,0)=f(s)$, $Y(s,0)=g(s)$, $Z(s,0)=h(s)$.
Now consider the function $$(s,t)\longrightarrow (X(s,t),Y(s,t),Z(s,t))\qquad [1] $$ for $s \in I$, $t \in J_s$.
Fix $s_0 \in I$ and let us reason in a neighborhood of $(s_0,0)$, trying to define a good domain for $[1]$ and then talk about invertibility.
Luckily we can choose an interval $J$ independent of s (on which IVPs for characteristic equations have solutions) provided we are willing to restrict ourselves to an interval $I_0$ containing $s = s_0 $ instead of the entire interval I. Thus we may assume that the domain of the vector-valued function given in $[1]$ is $I_0 × J$. Thanks to the differentiable dependence of solutions to initial value problems for ODEs, the vector-valued function given in $[1]$ is continuously differentiable. Thus we are interested in the invertibility, near $(s, t) = (s_0,0)$, of the function $[1]$. Note that at the point $(s, t) = (s_0,0)$, the Jacobian of the function in $[1]$ is given by $$J=\left|\begin{matrix} X_s(s_0,0) &X_t(s_0,0)\\Y_s(s_0,0) &Y_t(s_0,0)\end{matrix}\right|=\left|\begin{matrix} f'(s_0)& a(f(s_0),g(s_0),h(s_0))\\g'(s_0) &b(f(s_0),g(s_0),h(s_0))\end{matrix}\right|.$$ Provided that $J\neq 0$ we can use inverse function theorem and state that exist a neighbourhood $U$ containing $(s_0,0)$ and a neighbourhood $W$ containing $(f(s_0),g(s_0))$ such that the previous map considered from $U$ to $W$ is invertible. That is, we get two functions $S,T$ defined on $W$ such that $$s=S(x,y),\,\,\,t=T(x,y).$$ If we now define $$u(x,y):=Z(S(x,y),T(x,y))$$ then $u$ solves the Cauchy problem on $W$. In fact for every $s \in I$ such that $(f(s),g(s))\in W$ we have $$u(f(s),g(s))=Z(S(f(s),g(s)),T(f(s),g(s)))=Z(S(X(s,0),Y(s,0)),T(X(s,0),Y(s,0)))=Z(s,0)=h(s)$$and it's easy to see that $u$ solves the PDE by differentiating.
So a solution exist in a neighborhood $W$ of $(f(s_0),g(s_0))$.
Now come my problems, because I want to prove that $u$ is the unique solution on $W$. Following the analytical proof given by F. John - Partial Differential Equations-Springer US (1975), suppose that $u'$ is another solution of the Cauchy problem on $W$. Let $(x',y')\in W$.
Set $s'=S(x',y')$ and consider the characteristic curve $\Gamma$ that solves $$\begin{cases}\frac{d}{dt}x=a(x,y,z)\\\frac{d}{dt}y=b(x,y,z)\\\frac{d}{dt}z=c(x,y,z)\end{cases}$$ with initial conditions $$\begin{cases}x(0)=f(s')\\y(0)=g(s')\\z(0)=h(s')\end{cases}.$$
Because $u$ and $u'$ solve the Cauchy problem, their corresponding integral surfaces both passes through the point $(f(s'),g(s'),h(s'))$ as the characteristic curve $\Gamma$ does for $t=0$. So the integral surfaces must contain the part of $\Gamma$ whose projection on $xy$ plane is contained in $W$. In particular for $t'=T(x',y')$ we have $$u'(x',y')=u'(X(s',t'),Y(s',t'))=Z(s',t')=Z(S(x',y'),T(x',y'))=u(x',y')$$ by definition of $u$.
Here is my question: even if we now that $(s',t')\in U$, who ensure me that $(s',0)\in U$ too, and so I'm sure that both integral surfaces have the point $(f(s'),g(s'),h(s'))$ in common? I mean, am I sure that if a point $(\overline{x},\overline{y})\in W$ and $s=S(\overline{x},\overline{y})$ then $(f(s),g(s))\in W$?
I think I have to consider a neighborhood of $(s_0,0)$ contained in $U$ that is a rectangle to be sure that the previous hold: in this way every selected characteristic will pass through the space initial curve $(f(s),g(s),h(s))$ and so the problem is solved. I apologies for all this words for a problem that is probably trivial and is not about PDE's!!
Thanks in advance.
Let $\boldsymbol{x}:=(x,y,z)$, $\boldsymbol{F}(\boldsymbol{x}% ):=(a(\boldsymbol{x}),b(\boldsymbol{x}),z(\boldsymbol{x}))$, $\boldsymbol{\gamma}(s):=(f(s),g(s),h(s))$, and $\boldsymbol{x}_{0}% :=\boldsymbol{\gamma}(s_{0})$. I assume that $\boldsymbol{F}$ is defined on an open set $U$ of $\mathbb{R}^{3}$ which contains $\boldsymbol{x}_{0}$. Since $\boldsymbol{F}$ is of class $C^{1}$, there exist a closed cube $Q(\boldsymbol{x}_{0},r)$ centered at $\boldsymbol{x}_{0}$ and of side-length $r$ which is contained in $U$ and two constants $M>0$ and $L>0$ such that \begin{align*} \Vert\boldsymbol{F}(\boldsymbol{x})\Vert & \leq M\quad\text{for all }\boldsymbol{x}\in Q(\boldsymbol{x}_{0},r),\\ \Vert\boldsymbol{F}(\boldsymbol{x}_{1})-\boldsymbol{F}(\boldsymbol{x}% _{2})\Vert & \leq L\Vert\boldsymbol{x}_{1}-\boldsymbol{x}_{2}\Vert \quad\text{for all }\boldsymbol{x}_{1},\boldsymbol{x}_{2}\in Q(\boldsymbol{x}% _{0},r). \end{align*} Since $f$, $g$, $h$ are continuous, there exist $\delta>0$ such that $$ |f(s)-f(s_{0})|\leq\frac{r}{4},\quad|g(s)-g(s_{0})|\leq\frac{r}{4}% ,\quad|h(s)-h(s_{0})|\leq\frac{r}{4}% $$ for all $s\in\lbrack s_{0}-\delta,s_{0}+\delta]$. Hence, $\boldsymbol{\gamma }(s)\in Q(\boldsymbol{x}_{0},r/2)$.
Taking $0<T\leq\min\{r/(4M),1/(2L\}\}$, it follows that for all $s\in\lbrack s_{0}-\delta,s_{0}+\delta]$ the Cauchy problem \begin{align*} \frac{d\boldsymbol{x}}{d\tau}(\tau) & =\boldsymbol{f}(\boldsymbol{x}% (\tau)),\\ \boldsymbol{x}(0) & =\boldsymbol{\gamma}(s), \end{align*} has a unique solution (this is just by Banach's fixed point theorem). We claim that $\boldsymbol{x}(\tau)\in Q(\boldsymbol{x}_{0},r)$ for all $\tau\in \lbrack-T,T]$. Indeed, $\boldsymbol{x}(0)=\boldsymbol{\gamma}(s)\in Q(\boldsymbol{x}_{0},r/2)$, and so by continuity we have that for $\tau$ very small $\boldsymbol{x}(\tau)\in Q(\boldsymbol{x}_{0},r)$. But as long as $\boldsymbol{x}(t)\in Q(\boldsymbol{x}_{0},r)$, we have that $$ \boldsymbol{x}(\tau)=\boldsymbol{x}(0)+\int_{0}^{\tau}\boldsymbol{f}% (\boldsymbol{x}(t))\,dt $$ and so \begin{align*} |x(\tau)-f(s_{0})| & \leq|x(\tau)-f(s)|+|f(s)-f(s_{0})|\\ & \leq\int_{0}^{\tau}|a(\boldsymbol{x}(t))|\,dt+|f(s)-f(s_{0})|\\ & \leq MT+\frac{r}{4}\leq\frac{r}{2}% \end{align*} by the choice of $M$, and similarly for $y$ and $z$. Hence, $\boldsymbol{x}% (\tau)\in Q(\boldsymbol{x}_{0},r)$ for all $\tau\in\lbrack-T,T]$.
Going back to your proof, take $U$ and take $W$ so small that \begin{align*} U & \subseteq[s_{0}-\delta,s_{0}+\delta]\times\lbrack-T,T],\\ W & \subseteq[f(s_{0})-r/2,f(s_{0})+r/2]\times\lbrack g(s_{0})-r/2,g(s_{0}% )+r/2]. \end{align*}
Let $\bar{u}=\bar{u}(x,y)$ be any solution of your Cauchy problem in $W$. Given $(\bar{x},\bar{y})\in W$, you have that $(s^{\prime},t^{\prime})\in U$. In particular, $s^{\prime}\in\lbrack s_{0}-\delta,s_{0}+\delta]$. Consider the following system of ODE \begin{align*} \frac{d\bar{x}}{d\tau}(\tau) & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar {x}(\tau),\bar{y}(\tau))),\\ \frac{d\bar{y}}{d\tau}(\tau) & =b(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar {x}(\tau),\bar{y}(\tau))), \end{align*} with the initial conditions$$ \bar{x}(0)=f(s^{\prime}),\quad\bar{y}(0)=g(s^{\prime}). $$ Define $\bar{w}(\tau):=\bar{u}(\bar{x}(\tau),\bar{y}(\tau))$. Then as you showed \begin{align*} \frac{d\bar{w}}{d\tau}(\tau) & =\frac{\partial\bar{u}}{\partial x}(\bar {x}(\tau),\bar{y}(\tau))\frac{d\bar{x}}{d\tau}(\tau)+\frac{\partial\bar{u}% }{\partial y}(\bar{x}(\tau),\bar{y}(\tau))\frac{d\bar{y}}{d\tau}(\tau)\\ & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau )))\frac{\partial\bar{u}}{\partial x}(\bar{x}(\tau),\bar{y}(\tau))+b(\bar {x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau)))\frac {\partial\bar{u}}{\partial y}(\bar{x}(\tau),\bar{y}(\tau))\\ & =c(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau))) \end{align*} and $$ \bar{w}(0)=\bar{u}(\bar{x}(0),\bar{y}(0))=\bar{u}(f(s^{\prime}),g(s^{\prime }))=h(s^{\prime}). $$ Hence, $(\bar{x}(\tau),\bar{y}(\tau),\bar{w}(\tau))$ is a solution of the Cauchy problem \begin{align*} \frac{d\bar{x}}{d\tau}(\tau) & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{w}% (\tau)),\\ \frac{d\bar{y}}{d\tau}(\tau) & =b(\bar{x}(\tau),\bar{y}(\tau),\bar{w}% (\tau)),\\ \frac{d\bar{w}}{d\tau}(\tau) & =c(\bar{x}(\tau),\bar{y}(\tau),\bar{w}(\tau)), \end{align*} with initial data $$ \bar{x}(0)=f(s^{\prime}),\quad\bar{y}(0)=g(s^{\prime}),\quad\bar {w}(0)=h(s^{\prime}). $$ But by uniqueness $\bar{x}(t)=X(s^{\prime},t)$, $\bar{y}(t)=Y(s^{\prime},t)$ and $\bar{w}(t)=Z(s^{\prime},t)$ for all $t$.