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 $u(0,y)=g(y)$?

1 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}$.