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I have encountered the following linear PDE in the context of reaction-diffusion processes \begin{equation} \partial_t (\nabla^2f) - \nabla\cdot \left( \nabla (\nabla^2 f) - m^2 \, \nabla f \right) = 0, \end{equation} for $(x,y,z) \in [0,L]\times \mathbb{R} \times \mathbb{R}$, where $f$ is a potential field and $m$ a constant (which represents a screening term). I am looking for the general form of the solution (especially at long times) with Neumann boundary conditions, \begin{equation} \partial_x \left( \nabla^2 f - m^2 f\right) \big\rvert_{x=0,L} = 0. \end{equation}

Update

This PDE is taken from a simple model for a charged system, for which the electric potential $f$ is related to the charge density $\rho$ via the Poisson equation $-\nabla^2 f = \rho$. With appropriate boundary conditions on $f$ and treating $\rho$ as known, there is a unique solution for $f$ in terms of $\rho$ which can be written as ($r=(x,y,z)$) \begin{equation} f(r,t) = \int_{x'=0}^L\int\int d^3 r' \, K(r-r') \rho(r',t). \end{equation} $K$ is a known, nonlocal kernel (think of the Coulomb kernel).

The dynamics of the charge distribution $\rho$ is then given by a reaction-diffusion form \begin{align} \partial_t \rho = \nabla^2 \rho - m^2 \rho. \end{align} The boundary condition is what makes it somewhat nontrivial. For no-flux boundaries, it is given by \begin{align} \left(\partial_x \rho - m^2 \partial_x f\right) \big\rvert_{x=0,L} = 0. \end{align} We can either express this in terms of $\rho$ only, where it takes a nonlocal integral form that contains $K$; alternatively, I thought rewriting everything in terms of $f$ would make this more tractable - but then of course we need to deal with a higher-order PDE.

Question 1
Is this PDE with the given boundary condition, either written in terms of $\rho$ or in terms of $f$, well-posed? Assuming a specific initial condition is also given, is there a unique solution to this dynamics, or does it require more (less?) boundary conditions?

Question 2
Is there a way to obtain the general form of the function $\rho = \rho(x,y,z, t) $. I can, for example, guess a form given by $\rho = \sum_n g_n (y,z,t) \, \cos(n\pi x/L)$ with some functions $g_n$ (to be determined by further boundary conditions); but is this the most general form, or could there be other solutions which cannot be written like this (for instance solutions that cannot be obtained from a separation of the variables)?

Question 3
For the case that $m L \gg 1$ a naive approximation is to discard the Laplacian term in the boundary condition and write the equation as \begin{equation} \partial_x (-m^2 f) \big\rvert_{x=0,L} \approx 0, \end{equation} which implies a same cosine expansion but with $g_n = g_n (t)$. Does this represent a sensible approximation? I guess this would amounts to some singular perturbation, but I have no clue about how one should proceed with this.

Saïd M
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  • You write a couple of times $f=f(x,y,z)$, but your PDE has $t$ as a variable. Do you mean $f=f(x,y,z,t)$? – JackT Jun 06 '21 at 23:40
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    Your equation is $\nabla^2(\partial_t f+\nabla^2 f-m^2f) =0$ but usually reaction diffusion would have $\partial_t f-\nabla^2 f$... – Bob Terrell Jun 06 '21 at 23:41
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    @JackT yes, $t$ is a variable too. I corrected this. – Saïd M Jun 07 '21 at 07:40
  • @SaMaSo The term $\nabla (\nabla^2 f) - m^2 , \nabla f$ adds a $3$-tensor ($\nabla^3 f$) to a vector ($\nabla f$)? – Nicolas Jun 07 '21 at 07:50
  • @BobTerrell this is for a charged system, and the electric potential and charged density are related by a Poisson equation, $-\nabla^2 f = \rho$. The whole equation can be written in the usual reaction-diffusion form in terms of $\rho$, but the BC explicitly includes $f$ and that's why I decided to write everything in terms of $f$. – Saïd M Jun 07 '21 at 08:27
  • @Nicolas I'm not sure if I understand your comment correctly, but $\nabla^2 f$ is a scalar. – Saïd M Jun 07 '21 at 08:29
  • @SaMaSo So I do not understand what is $\nabla^2$. Could you define it please? – Nicolas Jun 07 '21 at 08:48
  • @Nicolas I believe it is just the Laplacian – JackT Jun 07 '21 at 09:15
  • @Nicolas it is indeed just the Laplacian $\nabla^2 f = \partial_x^2 f + \partial_y^2 f + \partial_z^2 f$. – Saïd M Jun 07 '21 at 09:31

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