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Consider the above system. We will drive the Fick's law from it.

Let $\sigma(x,y)$ be the concentration inside the box centered at $(x,y)$. Then, (using some physical argument which I will skip in here). Let $j(x+dx/2,y)$ be the flux across the section located at $(x+dx/2,y)$.

$$j_x (x+dx/2, y) = D \frac{\sigma(x,y) - \sigma(x+dx,y)}{dx}$$

$$j_x (x-dx/2, y) = D \frac{\sigma(x-dx,y) - \sigma(x,y)}{dx}$$

Then the net flux along the x-axis

$$j_x (x-dx/2, y) + j_x (x+dx/2, y) = -D \frac{\sigma(x-dx,y) - \sigma(x+dx,y)}{dx}$$

In the limit $dx \to 0$,

$$2 j(x,y) = -D \partial_x \sigma $$

However, original Fick's law do not contain that factor 2. Why? What am I doing wrong?

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The right hand side of the finite difference

$$ j_x (x-dx/2, y) + j_x (x+dx/2, y) = -D \frac{\sigma(x-dx,y) - \sigma(x+dx,y)}{dx} $$

contains the difference between the concentration at points that are distant $2 dx$, but you divide by $dx$ only. Therefore, in the limit of $dx$ that goes to zero, you obtain $2 \partial_x \sigma$. This factor of 2 appears now on both sides of your equation and you can simplify it.

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