Jacobian and PDE I am wondering how to compute the Jacobian in order to know if a given PDE satisfying an initial condition has a unique solution or not. 
If I consider the PDE, $u_x=1$, satisfying the initial condition $u(x,0)=h(x)$ then the parametric solution is $(x(t,s),y(t,s),u(t,s))=(t+s,0,t+h(s)).$ The book gave that the Jacobian is $$J=\left[
  \begin{array}{ c c }
     1 & 0 \\
     1 & 0
  \end{array} \right]=0.
$$
How were those values obtained?
Also, what will be the Jacobian for the initial condition $u(0,y)=g(y)$?
 A: I'm writing this answer, due to no replies. This would be my approachment, and we can can discuss it if you like.
Let's say we have the following PDE:
$1\cdot u_x+ 0\cdot u_y=1$
We write $x,y,u$ as functions of $(t,s)$, such that $x=x(t,s), y=y(t,s), u=u(t,s)$. In this form, $t$ is the variable that parametrizes the curve and $s$ is the variable which indicates the position of the particular trajectory on the initial curve.
According to the method of characteristics, we take:
$\begin{cases}
\dfrac{dx}{dt}=1, & \dfrac {dy}{dt}=0, & \dfrac{du}{dt}=1\\\\
I.Cs:\text{ on }  t=0: x(0,s)=s, & y(0,s)=0, & u(0,s)=h(s)
\end{cases}$ 
Thus, we have:
$\begin{array}[t]{l}
x(t,s)=t+s\\
y(t,s)=0\\
u(t,s)=t+h(s)
\end{array}$
Consider the transformation: 
$x(t,s)=t+s$
$y(t,s)=0$
Now, the Jacobian matrix is defined as follows: $ J\big(x,y\big)(t,s)=\begin{pmatrix} \dfrac {\partial x }{ \partial t} & \dfrac{ \partial x }{\partial s} \\\\ \dfrac {\partial y }{ \partial t}& \dfrac{ \partial y}{\partial s}\end{pmatrix}=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$.
This is different to what you claim, due to many people define Jacobian matrix as the transpose matrix of the one I mentioned. If you do so, you will have the result, as mentioned in the book.
In the second case, I think it goes like this:
$\begin{cases}
\dfrac{dx}{dt}=1, & \dfrac {dy}{dt}=0, & \dfrac{du}{dt}=1\\\\
I.Cs:\text{ on }  t=0: x(0,s)=0, & y(0,s)=s, & u(0,s)=g(s)
\end{cases}$ 
Thus, we take:
$\begin{array}[t]{l}
x(t,s)=t\\
y(t,s)=s\\
u(t,s)=t+g(s)
\end{array}$
The Jacobian matrix is defined as follows: $ J\big(x,y\big)(t,s)=\begin{pmatrix} \dfrac {\partial x }{ \partial t} & \dfrac{ \partial x }{\partial s} \\\\ \dfrac {\partial y }{ \partial t}& \dfrac{ \partial y}{\partial s}\end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
