L. Imposing Neumann boundary conditions

Neumann boundary conditions are handled much more simply than Dirichlet ones: the flux prescribed at the boundary immediately gives the vector $\overrightarrow{s}$ (nonzero only at the boundary nodes), which enters only the right-hand side of the system — the stiffness matrix $a^2 K$ need not be changed.

Stationary case

For the five-node example (the fluxes $g_0$ and $g_4$ are prescribed; the source on the end simplices is zero, $F_0$ ${} = F_4$ ${} = 0$ ) the modified right-hand side vector is:

\displaystyle \overrightarrow{P}

\displaystyle {} = \overrightarrow{F}

\displaystyle {} + a^2 \overrightarrow{s}

\displaystyle {} = \begin{pmatrix} a^2 g_0 \\ F_1 \\ F_2 \\ F_3 \\ a^2 g_4 \end{pmatrix}.

Non-stationary case

For the non-stationary problem the operator is $D$ ${} + \Delta t \cdot a^2 K$ , and the flux enters the right-hand side with the factor $\Delta t$ :

\displaystyle \left[D + \Delta t \cdot a^2 K\right] \cdot \overrightarrow{q}_n

\displaystyle {} = \Delta t \cdot \overrightarrow{F}

\displaystyle {} + D \cdot \overrightarrow{q}_{n-1}

\displaystyle {} + \Delta t \cdot a^2 \overrightarrow{s}.

The modified right-hand side vector:

\displaystyle \overrightarrow{P}

\displaystyle {} = \Delta t \cdot \overrightarrow{F}

\displaystyle {} + D \cdot \overrightarrow{q}_{n-1}

\displaystyle {} + \Delta t \cdot a^2 \overrightarrow{s}

\displaystyle {} = \Delta t \cdot \overrightarrow{F}

\displaystyle {} + D \cdot \overrightarrow{q}_{n-1}

\displaystyle {} + \begin{pmatrix} \Delta t \cdot a^2 g_0 \\ 0 \\ 0 \\ 0 \\ \Delta t \cdot a^2 g_4 \end{pmatrix}.

The operator $D$ ${} + \Delta t \cdot a^2 K$ is already nonsingular (the damping matrix $D$ is positive definite, $(D + \Delta t \cdot a^2 K)\,\overrightarrow{1}$ ${} = D \overrightarrow{1}$ ${} \neq 0$ ), so the non-stationary Neumann problem is solvable even without Dirichlet conditions.