Method of Characteristics, PDE, Jacobian condition Q

[binbagsss]

Hi,

I'm looking at the Jacobian condition which is $J= a \frac{dy_{0}}{ds}-b\frac{dx_0}{ds}$
where the pde takes the form $c= a\frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y}$, where $a=\frac{\partial x}{\partial \tau }$, $b=\frac{\partial y}{\partial \tau }$, $c=\frac{\partial u}{\partial \tau}$

If $J=0$, there are two possibilities: zero solutions, or an infinite number of solutions.
The zero solution case corresponds to the initial curve not being a charecteristic, and the infinite solution case to the initial curve being a characteristic.

Question:

I have in my notes that there exsists a solution if $\frac{dx_{0}}{ ds}\frac{1}{a}= \frac{du_{0}}{ds}\frac{1}{c}$ holds and no solution if this does not hold.
So I conclude that this condition holding must be equivalnet to the intial curve being a characteristic. I'm just stuggling to see how it says this?

What I know:

$J$ is a conidtion ensuring the characteristics are not tangential to the curve where initial conditions are prescribed.

Any help in seeing how the 2 are equivalent, greatly appreciated.

 

[MisterX]

Let a solution to the the PDE be $f(\mathbf{x})$. Suppose a curve that is parameterized by $t$ that is $\mathbf{x}(t)$. So we have $f(\mathbf{x}(t))$
Suppose we have a PDE of the form
$$\sum_i a_i(\mathbf{x}) \frac{\partial f}{\partial x_i} = 0$$
By the chain rule
$$ \frac{df}{dt}= \sum_i \frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial t}$$
On the curves where $f$ is constant (the level curves of $f$)
$$ \frac{df}{dt}= \frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial t} = 0$$
Because of the given PDE the above will be true if $\frac{\partial x_i}{\partial t} = a_i(\mathbf{x})$
We call solutions to this autonomous system$\frac{\partial x_i}{\partial t} = a_i(\mathbf{x})$ "characteristics".
If our initial conditions "curve" is not ever parallel to the level curves, then we may thing of these characteristics as propagating out from the initial conditions to specify the value of $f$ at every point you can reach. This can potentially be the entire space.

However if this is not the case and the initial conditions are parallel to the characteristics then you may have an issue. Since we know the value of $f$ is constant, the initial conditions had better be constant there. If not there is an inconsistency (no solution). If the rules governing characteristics and the initial conditions agree, you may still be faced with a solution which is not unique.

This can be generalized to quasi-linear and non-linear first order PDEs, but all I am able to submit to you at this time is this simple case as I understand it. I know I still haven't fully answered your question.

 

[binbagsss]

Let a solution to the the PDE be $f(\mathbf{x})$. Suppose a curve that is parameterized by $t$ that is $\mathbf{x}(t)$. So we have $f(\mathbf{x}(t))$
Suppose we have a PDE of the form
$$\sum_i a_i(\mathbf{x}) \frac{\partial f}{\partial x_i} = 0$$
By the chain rule
$$ \frac{df}{dt}= \sum_i \frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial t}$$
On the curves where $f$ is constant (the level curves of $f$)
$$ \frac{df}{dt}= \frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial t} = 0$$
Because of the given PDE the above will be true if $\frac{\partial x_i}{\partial t} = a_i(\mathbf{x})$
We call solutions to this autonomous system$\frac{\partial x_i}{\partial t} = a_i(\mathbf{x})$ "characteristics".
If our initial conditions "curve" is not ever parallel to the level curves, then we may thing of these characteristics as propagating out from the initial conditions to specify the value of $f$ at every point you can reach. This can potentially be the entire space.

However if this is not the case and the initial conditions are parallel to the characteristics then you may have an issue. Since we know the value of $f$ is constant, the initial conditions had better be constant there. If not there is an inconsistency (no solution). If the rules governing characteristics and the initial conditions agree, you may still be faced with a solution which is not unique.

This can be generalized to quasi-linear and non-linear first order PDEs, but all I am able to submit to you at this time is this simple case as I understand it. I know I still haven't fully answered your question.

Thanks for your reply.

I think understand the majority of it. Although I fear I may be missing a few things as I don't fully undersand entirely why we need the characteristics to not be tangential to the curve where the initial conditions are prescribed.

From what I understand, when initial conditions are not parallel to the level curves there is no issue: $J ≠ 0$.

When they are parallel $J ≠ 0$ or $J=0$. If the initial conditions are constant on the characteristic we either get infinite or a unique solution, depending upon whether $J ≠ 0$ or $J=0$ intially. (Looking at the definition you gave for a characteristic, this means the initial data must be a characteristic curve).If not constant, no solution.

I'm a little confused about using the terms 'tangential' and 'parallel' because if tangential we have $J=0$ and in the above discussion we only have parallel sometimes when $J=0$ . So tangential to the intial curve must imply two things: 1)the initial curve and the characterisic are parallel 2) the initial curve is constant on the char

I think then, the next step in deriving the intial condition in my original post is to satisfy the 2 criteria listed just above.

Criteria 1: $a_{0}=k u_{0}$, $b_{0}=k u_{0}$. k some constant
Criteria 2: $\frac{ \partial u_{0}}{ \partial t } =0=\frac{ \partial u_{0}}{ \partial x }a + \frac{ \partial u_{0}}{ \partial y }b$

But these 2 equations didnt help me get to the condition

Have I understood correctly and are my thoughts on the right or wrong track?

Thanks in advance.

 

[binbagsss]

bump.

 

[MisterX]

I am not sure what is "s" $x_0$ or $y_0$

 

[binbagsss]

Apologies, where the initial curve is parameterised by $s$, defined by $x(t)=x_{0}(s)$, $y(t)=y_{0}(s)$, $u(t)=u_{0}(s)$ at $t=0$.