Let consider a general linear first order PDE in $xy$ space, ie $$a(x,y)u_x+b(x,y)u_y=c(x,y)u+d(x,y),\qquad (x,y)\in U$$ where $U$ is an open and connected subset of $\mathbb{R}^2$ and $a,b,c,d \in C^1(U)$ are given functions. Suppose also there is a Cauchy data $$u(f(s),g(s))=h(s),\,\quad s \in I$$ where $(f(s),g(s))$ is the parametric representation of the curve $\gamma_0$ contained in $U$. Here I is an interval of $\mathbb{R}$ and $f,g,h \in C^1(I)$.
By theory of method of characteristic I know that for each $s \in I $ $$\begin{cases} \frac{d}{dt}x(t;s)=a(x(t;s),y(t;s))\\ \frac{d}{dt}y(t;s)=b(x(t;s),y(t;s)) \end{cases}+\begin{cases} x(0;s)=f(s)\\ y(0;s)=g(s) \end{cases}, $$ gives us the unique characteristic curve which intersect $\gamma_0$ in $(f(s),g(s))$. Along this characteristic curve my initial value problem degenerate in $$\begin{cases} \frac{d}{dt}u(x(t;s),y(t;s))=c(x(t;s),y(t;s))u(x(t;s),y(t;s))+d(x(t;s),y(t;s))\\ u(x(0;s),y(0;s))=h(s) \end{cases}.$$ By theorem of existence and uniqueness of solution for first order ODE, we know that the previous has a unique solution.
So, if a solution of the original Cauchy problem exists, it is univocally determined on $(x(t;s),y(t;s))$.
Now my problem arises because my teacher says that we have to check if $(x(t;s),y(t;s))$ intersect $\gamma_0$ in another point except $(f(s),g(s))$, because if this is true then we would have an ODE with two initial condition (which except very special cases yields to contradiction).
I don't really understand this, because even if there exist $\overline{s}\neq s$ such that $(f(\overline{s}),g(\overline{s}))\in (x(t;s),y(t;s))$ then, by the first step, I can re-write the characteristic curve $(x(t;s),y(t;s))$ as $(x(t;\overline{s}),y(t;\overline{s}))$ such that $x(0;\overline{s})=f(\overline{s}), y(0;\overline{s})=g(\overline{s})$; in particular any (eventual) solution $u$ of the initial problem will satisfy $$\begin{cases} \frac{d}{dt}u(x(t;\overline{s}),y(t;\overline{s}))=c(x(t;\overline{s}),y(t;\overline{s}))u(x(t;\overline{s}),y(t;\overline{s}))+d(x(t;\overline{s}),y(t;\overline{s}))\\ u(x(0;\overline{s}),y(0;\overline{s}))=h(\overline{s}) \end{cases}$$ but I really don't find any general contradiction in this. This is certainly a contradiction if $c=d=0$ and $h(s)\neq h(\overline{s})$, but I really don't understand why I always find an ODE with two initial conditions.
Anyone can give me a hint? Thank you so much.
Let us illustrate with a caricatural simple example. We consider the linear advection equation $a=b=1$, $c=d=0$, where the initial data $h(s)$ is located on the unit circle $\gamma_0:(\cos s, \sin s)$ parametrized by $s \in I = [-\frac{1}{4}\pi,\frac{7}{4}\pi]$. The characteristics are straight parallel lines, which gives the following picture of the characteristic curves in the $x$-$y$ plane:
Since $u$ is constant along characteristics, we must have $h(\frac{3}{4}\pi-s) = h(\frac{3}{4}\pi+s)$ for all $s$ in $[0,\pi]$. If this symmetry property is not true, the problem cannot be solved in an unambiguous manner. Therefore, if characteristics and boundary data cross twice, the problem cannot be solved for arbitrary $h$, but it may only be solved in particular cases.