Is there a relation between say the heat equation $u_t -\Delta u = f.$ and advection-diffusion equation $-\Delta u + c \cdot \nabla u = f$? I have heard several people use this argument in many talks and articles but this is particularly not clear to me.

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    $\begingroup$ I do not really understand what your question is - may be you can elaborate in a little more detail? The heat equation reads $u_t = \Delta u$ while the advection-diffusion equation is $u_t + c \nabla u = \Delta u$, so in a sense heat equation is advection-diffusion with $c=0$, but I assume this is not all you are asking? $\endgroup$ – Daniel Jul 28 '15 at 12:18
  • $\begingroup$ Thank you Daniel! I mean the relation between these equations $u_t - \Delta u= f$ and $-\Delta u + c \cdot \nabla u = f.$ $\endgroup$ – uli.xu Jul 28 '15 at 12:25
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    $\begingroup$ I am still not sure what you mean. You say you find this argument (not sure which) in articles, can you may be link an example of such an article? This might help to clarify your question. $\endgroup$ – Daniel Jul 28 '15 at 12:28
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    $\begingroup$ Where in the article is this argument used? $\endgroup$ – Daniel Jul 28 '15 at 12:56
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    $\begingroup$ OK, if you make no distinction between the time and space dimension, both equations can be written in a common form (using $\nabla_t := (\partial_t,\partial_{x_1},\dots,\partial_{x_d})^T$). I don't think the similarity goes much deeper than that, though. Full discretization of the space-time cylinder is a very specific situation, and I'd be surprised to see this argument made in any other set of circumstances. $\endgroup$ – Christian Clason Jul 28 '15 at 13:08

So after browsing the paper a bit, I think that the answer is essentially what Christian Clason stated in his comment.

It seems that the original question refers to the statement just above Equation (3) in the article linked by kwesi : There, the authors say that the advection-diffusion equation (Equation (1) in the paper)

$\frac{\partial c}{\partial t} + \sum_{i=1}^{d} u_i \frac{\partial c}{\partial x_i} - \sum_{i,j=1}^{d} D_{ij} \frac{\partial}{\partial x_{j}}\frac{\partial c}{\partial x_i} = 0$,

can be rewritten as

\begin{equation} \nabla \cdot \left( B c - A \nabla c \right) = 0 \end{equation}

with $A = \begin{pmatrix} 0 & 0 \\ 0 & \mathbf{D} \end{pmatrix}$ and $B = (1, \mathbf{u})$ when $\nabla$ is defined as the gradient in space and time, that is

$\nabla = \left( \frac{\partial}{\partial x_0}, \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_d} \right)$

with $x_0 \equiv t$. In this sense, the time-dependent heat equation

$c_t - \sum_{i,j=1}^{d} \frac{\partial}{\partial x_{j}}\frac{\partial c}{\partial x_i}$

can be written as an advection-diffusion problem in space-time

$\nabla \cdot \left( (1,\mathbf{0}) c - A \nabla c \right) = 0$.

The paper then proceeds to analyse space-time discontinuous Galerkin methods based on this formulation of the problem.

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    $\begingroup$ Thanks Daniel. I think Christian's comments were very good and your examples with $\mathbf{B}=(1,0)$ strengthens it. $\endgroup$ – uli.xu Jul 29 '15 at 12:55

Yes there is.

The heat equation has the same mathematical form as the diffusion part of the diffusion-advection equation. Both are Parabolic operators and behave very similarly.

For example, in the heat equation, temperature will "diffuse" from the hottest zone to the cooler zones. This is very similar to how a substance will "diffuse" from the area of highest concentration to the lowest concentration. The behavior is almost identical.

In many numerical schemes, you can always treat the diffusion part of the diffusion-advection equation separately from the advection part, so effectively the diffusion part can be treated as like a heat equation or viceversa.


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