Writing PDE in terms of x and y

[Carla1985]

Hi all,

I am hoping someone can help me understand a PDE. I am reading a paper and am trying to follow the math. My experience with PDEs is limited though and I am not sure I am understanding it all correctly. I have 3 coupled PDEs, for $n$, $f$ and $c$, that are written in general form, and I would like to write them in 2d (in terms of x and y directions). The equations for $f$ and $c$ are fairly straightforward, but I am having some trouble with the one for $n$:

$$\frac{\partial n}{\partial t} = D^2 \nabla n - \nabla \cdot (\chi(c) n \nabla c) - \rho \nabla \cdot (n \nabla c) $$

$D$, and $\rho$ are constants. The first term on the RHS confuses me most as I thought $\nabla$ means gradient, so would return a vector of length 2? The second term I think expands to

$$\frac{\partial}{\partial x}\left(\chi(c) n \frac{\partial c}{\partial x}\right) + \frac{\partial}{\partial y}\left(\chi(c) n \frac{\partial c}{\partial y}\right)$$

is this correct? Thank you very much for your help, Carla.

 

[I like Serena]

Hi Carla,

What you write is correct.
The nabla operator applied to a scalar function of (x,y) returns indeed a vector of length 2.
And the dot product of the nabla operator is indeed what you wrote.

 

[topsquark]

I would double-check the equation for \(\displaystyle \partial n / \partial t\). My guess is that the first term should be \(\displaystyle \nabla ^2 n\).

-Dan

 

[I like Serena]

I would double-check the equation for \(\displaystyle \partial n / \partial t\). My guess is that the first term should be \(\displaystyle \nabla ^2 n\).

Indeed. It could also have been written as $\Delta n$ with the Laplace operator, which is the same as $\nabla^2 n$.