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Suppose I have a 1D advection equation in conservation (divergence) form

$\partial_t u(x,t) = -\partial_x [v(x)u(x,t)],$

where $u$ is a conserved quantity in space, and $v$ gives the velocity of the flow of mass in space. I want to know if it is possible to re-write this in a way that describes the rate of change of $u(x,t)$ as a function of incoming mass from all other points in space (let us call these points $y$). In other words, can I write an equation of the form

$\partial_t u(x,t) = \int_y u(y,t) h(y,x) dy,$

where the function $h(y,x)$ gives the proportion of the mass $u(y,t)$ at position $y$ that moves to position $x$ in space in a small unit of time?

I arrived at the advection equation via the divergence theorem, which is why I can write $\partial_t u(x,t)$ in terms of the rate of change in the space at the same point $x$, but intuitively it seems to me there should be another possible formulation like I have described. I suppose the mass is only entering point $x$ from the $y$ points that are already 'very close' to $x$, which is why it is an advective equation rather than a diffusive equation, but I should still be able to sum over all the $y$ points from where the mass is coming from, shouldn't I!?

As I'm a novice in terms of the mathematics and the applications of the advection equation, I would like to know if the question I have asked is well-posed but also whether or not this is a typical problem people think about in applications? Thanks!

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  • $\begingroup$ certainly: $h(y,x)=v(y)\partial_y \delta(x-y)$ $\endgroup$
    – Carlo Beenakker
    Commented Nov 23, 2014 at 21:15
  • $\begingroup$ Thanks Carlo, but please could you clarify a bit more how you arrived at that? Your answer looks intuitively appealing, but when I try to solve for $h(y,x)$ myself I arrive at an ugly answer! I've not found any examples where people use the form I'm looking for, is there a reason do you think? Thanks for the help. $\endgroup$
    – Michael Andrew Bentley
    Commented Nov 23, 2014 at 23:14

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for $h(y,x)=\nu(y)\partial_y\delta(x-y)$ one has, upon partial integration:

$$\int_{-\infty}^\infty u(y,t)h(y,x)\,dy=-\int_{-\infty}^\infty \delta(x-y)\partial_y[\nu(y)u(y,t)]\,dy=-\partial_x [\nu(x)u(x,t)]$$

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  • $\begingroup$ I just had to learn about the Dirac delta function! This is amazing, thank you so much for your help. $\endgroup$
    – Michael Andrew Bentley
    Commented Nov 24, 2014 at 11:15
  • $\begingroup$ please could I just clarify a few things? Firstly, when I do integration by parts I get the first term as $v(y)u(y,t)\delta (x-y)$. For every $x$ not equal to $y$ this is zero and the term disappears, but at $x=y$ it is $\inf$, so why can we ignore it? Secondly, how do you know to use the delta function in defining $h(y,x)$, was it an 'ansatz' or is it something obvious that I am missing? Sorry for my ignorance, it's just I'm a biologist by training and this is all new to me. $\endgroup$
    – Michael Andrew Bentley
    Commented Nov 24, 2014 at 14:34
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    $\begingroup$ I'm not sure what you mean by "the first term", probably you mean the boundary term of a partial integration? Then you have to send $y$ to $\pm\infty$, and the boundary term vanishes. The answer to the second question is "yes, it's obvious you need a delta function" because you are trying to write a local relation in a nonlocal form, and the delta function restores locality. $\endgroup$
    – Carlo Beenakker
    Commented Nov 24, 2014 at 14:40

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