# Local uniqueness of solution for quasi linear PDE

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!!

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 $$0, it follows that for all $$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$$.