Neumann boundary conditions are implemented by introducing ghost points outside the domain and then using the boundary conditions to eliminate the ghost points. For example, see this question.

I have a set of three equations in variables $(u,v,w)$ which are one dimensional functions of $x$, and I cannot see how to eliminate the ghost points for the equation for the $w$ variable.


The problem is something like this,

$ \frac{\partial u}{\partial t} = a_u\frac{\partial^2 u}{\partial x^2} + b_u\frac{\partial u}{\partial x} + f_u(x,u,v,w) \\ \frac{\partial v}{\partial t} = a_v\frac{\partial^2 v}{\partial x^2} + b_v\frac{\partial v}{\partial x} + f_v(x,u,v,w) \\ \frac{\partial w}{\partial t} = a_u\frac{\partial u}{\partial x} +a_v\frac{\partial v}{\partial x} + f_w(x,u,v,w) \\ $

The set of equations describe a two species advection-diffusion problem where the third equation couples to the other two. I am having problems applying Neumann boundary conditions to the third equation. Notice that the third equation does not contain a differential of $w$, but it does contain differentials of both $u$ and $v$.

In discretizated form (only showing the $w$ equation, where $\Delta t$ is the time step and $s_u$ and $s_v$ are the coefficients of the equations after discretization),

$ w_j^{n+1} - s_u\beta(u_{j+1}^{n+1} - u_{j-1}^{n+1}) - s_v\beta(v_{j+1}^{n+1} - v_{j-1}^{n+1}) = s_u(1-\beta)(u_{j+1}^{n} - u_{j-1}^{n}) + s_v(1-\beta)(v_{j+1}^{n} - v_{j-1}^{n}) + w_j^n + \Delta tf_{j}^n $

The problem

The above is a Crank-Nicolson discretization, but for the boundary points let's use an implicit scheme (setting $\beta=1$),

$ w_1^{n+1} - s_u(u_{2}^{n+1} - u_{0}^{n+1}) - s_v(v_{2}^{n+1} - v_{0}^{n+1}) = w_1^n + \Delta tf_{1}^n $

The ghost points, $u_0^{n+1}$ and $v_0^{n+1}$ can be eliminated, by using the boundary conditions for the $u$ and $v$ variables. However, applying Neumann boundary conditions to the $w$ variable gives,

$ \frac{\partial w}{\partial x} = \sigma |_{j=1} \\ \frac{1}{2 \Delta x} (w_{2}^{n+1} - w_{0}^{n+1} ) = \sigma $

The problem is that the boundary equation cannot be substituted into the equation for $j=1$ because it does not contain at term for $w_{0}^{n+1}$.


Do you have any suggestions on how I can apply Neumann boundary conditions to this equation?


1 Answer 1


As there are no spatial derivatives of $w$ in the PDE system, asserting Neumann boundary conditions on $w$ overdetermines the problem. Provided you have an appropriate initial condition for $w$, and thus for $w^0_j$, the discretized evolution equation for $w^{n+1}_j$ (including for $w^{n+1}_1$) you derived above fully determines $w$ for all time.

To put the above in slightly different terms, consider that, in your discretization, no boundary condition on $w$ is needed to evolve the solution forward in time. Given an initial condition $w^0_j$ defined on the boundary and interior, the evolution equations for $u$, $v$, and $w$ can all be solved without reference to values of $w$ across the boundary (i.e., ghost cells). The key is that, because there are no spatial derivatives of $w$, the ghost term $w^n_0$ does not appear in the discretized evolution equations for $u$, $v$, or $w$.

  • $\begingroup$ Maybe this is one of the places where physics meets maths... the boundary condition on $w$ are dependent, meaning that $dw/dx$ is completely determined by the values of $u$ and $v$ are the boundaries. I think this gets round the system being over-constrained because I am not enforcing this boundary to take a specific value. I am thinking of applying second order accurate one-sided differences instead of second-order two-sided difference for the boundary conditions. This doesn't introduce ghost points. Sounds reasonable? $\endgroup$
    – boyfarrell
    Commented May 13, 2013 at 4:58
  • $\begingroup$ @boyfarrell You cannot in general add Neumann boundary conditions (e.g., constraints on $dw/dx$ at the boundary) to a fully determined system without overdetermining that system. Consider the Poisson equation $-u_{xx} = f$ on $[0,1]$ with $u(0) = u(1) = 0$. This is a fully determined system. I cannot also stipulate that $u_x(0) = 1$ without accepting the possibility that no solutions exist. This is the same situation as your problem except no boundary condition is needed to begin with on $w$. $\endgroup$
    – Ben
    Commented May 13, 2013 at 16:05
  • $\begingroup$ Thanks for the interesting discussion, I will try as you suggest. $\endgroup$
    – boyfarrell
    Commented May 14, 2013 at 3:14

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