A Vector Calculus Identity for Characteristic Projections in PDEs

[Gregg]

In the notes it says that

\text{v}\cdot \nabla \text{u} = |\text{v}|\frac{du}{dl}

\text{v} = (a(x,y), b(x,y))

l is the arclength in the v-direction.

Why is this?

The LHS is the projection of v onto the gradient of u, the other thing is the magnitude of v, multiplied by the du/dl.

 

[LCKurtz]

In the notes it says that

\text{v}\cdot \nabla \text{u} = |\text{v}|\frac{du}{dl}

\text{v} = (a(x,y), b(x,y))

l is the arclength in the v-direction.

Why is this?

The LHS is the projection of v onto the gradient of u, the other thing is the magnitude of v, multiplied by the du/dl.

l is the arclength of what in the v direction? And you haven't told us what u is. Nor what v represents. What we have here is an example of "guess the question". Since I have a bit of time on my hands, I will expound a bit about a question I think might be relevant.

Let $u(x,y)$ be a scalar field, perhaps the temperature at $(x,y)$. Let a curve $C$ be given by $\vec R(t) = \langle x(t), y(t)\rangle$ represent the location of a moving particle. What if we want the rate of change of $u$ as the particle moves along $C$? Well we have$$
\frac{du}{ds}=\frac{du}{dt}\frac{dt}{ds}=
(u_x\frac {dx}{dt}+ u_y\frac{dy}{dt})\frac{dt}{ds}$$Multiply both sides by$$
\frac{ds}{dt}=\frac 1 {\frac{dt}{ds}}$$to get$$
\frac{du}{ds}\frac{ds}{dt}=u_x\frac {dx}{dt}+ u_y\frac{dy}{dt}=
\nabla u\cdot \frac {d\vec R}{dt}= \nabla u\cdot \vec V$$

where $\vec V$ is the velocity of the particle. Since $v = \frac{ds}{dt}=|\vec V|$ is its speed, this can be written$$
|\vec V|\frac{du}{ds}=\nabla u\cdot \vec V$$This looks a lot like your result, using $s$ instead of $l$ for arc length. There may be a more direct way of getting the result but, hey, I'm not even sure I worked the problem you were thinking about.

 

[Gregg]

It is the derivation for characteristic projections in PDEs. First order linear PDE with the form

$a(x,y) u_x + b(x,y) u_y + c(x,y) u = d(x,y)$

$v \cdot \nabla u + c u = d \Rightarrow |v| \frac{du}{dl} + cu = d$

So I have looked at what you wrote, you have parameterised it so that s is the position or arc length along some curve C?

 

[LCKurtz]

It is the derivation for characteristic projections in PDEs. First order linear PDE with the form

$a(x,y) u_x + b(x,y) u_y + c(x,y) u = d(x,y)$

$v \cdot \nabla u + c u = d \Rightarrow |v| \frac{du}{dl} + cu = d$

So I have looked at what you wrote, you have parameterised it so that s is the position or arc length along some curve C?

Yes. Now, I am not a PDE expert so I'm not going to comment directly on your PDE question. I do suspect, though, that if you look at and understand what I have done, it will likely apply to your context. Good luck with it.