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!

  • $\begingroup$ Thanks for asking this, I have also been wondering how I might write the advection-diffusion equation in a conservative form. The one thing that has been stumping me is how to consider the boundary conditions. It seems like a natural way to invoke conservation would be to require periodic boundary conditions, but that certainly wouldn't be a realistic condition for certain types of problems. $\endgroup$
    – Loonuh
    Commented Nov 23, 2014 at 23:50
  • $\begingroup$ (if you remove the time derivative then) it seems that you are trying to get at the concept known as a Green's function. $\endgroup$ Commented Nov 24, 2014 at 6:05

2 Answers 2


Your description "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" seems to indicate a slight misunderstanding of derivatives. Since the term on the left of your equation ($\partial_t u$) is a time derivative, the "small unit of time" must be infinitesimally small. Therefore only points infinitesimally far from $x$ need to be considered and you would end up right back at the advection equation with a spatial derivative.

On a related note, differential equations coming from conservation laws can also be written in integral form, which is typically the way they are derived. Any decent book on fluid dynamics will include a derivation of the Euler and Navier-Stokes equations starting from the conservation laws to write the integral form and then derive the differential form. For the advection equation this goes as follows:

Assuming no sources or sinks of $u$, conservation of $u$ says that the amount of $u$ between $x_1$ and $x_2$ at time $t_2$ is given by the amount at $t_1$ plus the flux into and out of the domain at the boundaries during the time from $t_1$ to $t_2$: $$ \int_{x_1}^{x_2} u(x,t_2) dx = \int_{x_1}^{x_2} u(x,t_1) dx + \int_{t_1}^{t_2} \left[ v(x_1,t)u(x_1,t) - v(x_2,t)u(x_2,t) \right] dt. $$ Rearranging and dividing both sides by $(t_2-t_1)$ gives $$ \int_{x_1}^{x_2} \dfrac{u(x,t_2) - u(x,t_1)}{t_2-t_1} dx = \dfrac{1}{t_2-t_1} \int_{t_1}^{t_2} \left[ v(x_1,t)u(x_1,t) - v(x_2,t)u(x_2,t) \right] dt. $$ Then, taking the limit as $t_1\to t_2$ (assuming $u$ is continuous in time) gives $$ \int_{x_1}^{x_2}\partial_t u(x,t) dx = v(x_1,t)u(x_1,t) - v(x_2,t)u(x_2,t). $$ Finally, dividing by $(x_2-x_1)$ and taking the limit as $x_1\to x_2$ gives the advection equation $$ \partial_t u(x,t) = -\partial_x(v(x,t)u(x,t)). $$


Whilst Doug's answer seems to be strictly correct, I've been informed that it is possible to re-write the spatial derivative as an integral if one uses a Dirac delta function with $\delta(x-y)$. I'm not sure how useful this is for any practical applications of the advection equation, but I find it interesting nevertheless. The 'trick' is to use $h(y,x)=v(y)\partial_{y}[\delta(x-y)]$. Integration by parts then gives

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

$= \int_y \partial_{y}[\delta(x-y)] v(y) u(y,t) dy $

$= \delta(x-y) v(y) u(y,t) - \int_y \delta(x-y) \partial_{y} [v(y) u(y,t)] dy $

$= - \partial_{y} [v(y) u(y,t)]. $


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