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 $u(0,y)=g(y)$?
Best Answer
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}$.