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Vector calculus identity

  1. Dec 3, 2012 #1
    In the notes it says that

    [itex] \text{v}\cdot \nabla \text{u} = |\text{v}|\frac{du}{dl} [/itex]

    [itex] \text{v} = (a(x,y), b(x,y)) [/itex]

    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.
     
  2. jcsd
  3. Dec 3, 2012 #2

    LCKurtz

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    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.
     
    Last edited: Dec 3, 2012
  4. Dec 3, 2012 #3
    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?
     
  5. Dec 3, 2012 #4

    LCKurtz

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    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.
     
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