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Helmholtz Diffusion equation with reaction term: $$ k\Delta u + u = f ~ \text{in} ~\Omega $$ $$ \nabla u \cdot \mathbf{n} = 0 ~ \text{in} ~\partial \Omega $$

For sufficiently small $k$ (relative to the mesh), there can be oscillations (with Lagrange elements) that return a negative $u$ in some regions, specially when $f$ is a step function. I tried to apply a DG method with the Symmetric Interior Penalty formulation, and I am still seeing these negative values. Is this possible to happen? I was told that DG methods preserve positivity and cannot return negative values. Is there any way to preserve positivity in DG methods for elliptic equations?

Preserve positivity: If $f(x)>0 ~\forall x\in\Omega$, then $u(x)>0 ~\forall x\in\Omega$

Edit again: What if the boundary conditions are of zero flux? Edited the problem.

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  • $\begingroup$ What do you mean by preserving positivity? Do you mean that $u(x) > 0\ \forall x \in \Omega$? $\endgroup$
    – nicoguaro ♦
    Jul 20, 2017 at 14:32
  • $\begingroup$ Yes, I edited my question for clarity. $\endgroup$
    – balborian
    Jul 20, 2017 at 15:15
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    $\begingroup$ I'm not sure that is correct. If we take the 1D version of your equation $k \frac{d^2u}{dx^2} + u =f$, and take $f=k=1$, the solution is: $u(x) = A \sin(x) + B \cos(x) + 1$. This is positive for some values of $A$ and $B$, but not always. $\endgroup$
    – nicoguaro ♦
    Jul 20, 2017 at 16:21
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    $\begingroup$ Correct -- there is no reason why the solution should be positive. In fact, for particular values of $k$ (the inverse of the eigenvalues of the Laplacian), the equation is singular and there is no unique solution any more. $\endgroup$
    – Wolfgang Bangerth
    Jul 20, 2017 at 20:50
  • $\begingroup$ Does the situation change if there is zero flux in the boundary? $\endgroup$
    – balborian
    Jul 22, 2017 at 17:19

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