K. Imposing Dirichlet boundary conditions

There is a simple and widely used method for imposing Dirichlet conditions — direct substitution of the known boundary values into the system (the row-and-column elimination method). Suppose a node $i$ carries the Dirichlet condition $q_i$​${} = g_i$, where $g_i$ is the known boundary value. Then on the system $A \overrightarrow{q}$​${} = \overrightarrow{P}$ we perform the following operations:

1. Right-hand side modification: For every row $j$​${} \neq i$ subtract the contribution of the known value $q_i$ from the right-hand side: $p_j :$​${} = p_j$​${} - A_{ji} \cdot g_i$.
2. Zeroing the row and column: Replace the $i$-th row and $i$-th column of the matrix $A$ with zeros, except for the diagonal element: $A_{ij} :$​${} = 0, \quad A_{ji} :$​${} = 0, \quad j$​${} \neq i$.
3. Setting the diagonal element: Set the diagonal element to one: $A_{ii} :$​${} = 1$.
4. Setting the boundary value: Replace the $i$-th component of the right-hand side with the boundary value: $p_i :$​${} = g_i$.

After these operations the $i$-th equation of the system takes the form $1 \cdot q_i$​${} = g_i$, which guarantees that the boundary condition holds.

Let us return to our one-dimensional example with five nodes $q_0, q_1, q_2, q_3, q_4$. Suppose Dirichlet conditions are prescribed at the boundaries:

Stationary case

The original system has the form:

The matrix $a^2 K$ is full (symmetric and tridiagonal); boundary conditions are imposed by replacing the boundary rows.

Denote $A$​${} = a^2 K$ and $\overrightarrow{P}$​${} = \overrightarrow{F}$. We apply the direct method for the nodes $q_0$ and $q_4$.

Step 1: Right-hand side modification for the interior nodes

For the node $q_1$ (index 1):

For the node $q_3$ (index 3):

Steps 2–4: Modifying the matrix and right-hand side for the boundary nodes

For the node $q_0$ (index 0):

• Zero the first row: $A_{01} :$​${} = 0$
• Zero the first column: $A_{10} :$​${} = 0$
• Set the diagonal: $A_{00} :$​${} = 1$
• Set the right-hand side: $p_0 :$​${} = g_0$

For the node $q_4$ (index 4):

• Zero the last row: $A_{43} :$​${} = 0$
• Zero the last column: $A_{34} :$​${} = 0$
• Set the diagonal: $A_{44} :$​${} = 1$
• Set the right-hand side: $p_4 :$​${} = g_4$

In the matrix $a^2 K$ the boundary rows and columns (nodes $q_0$ and $q_4$) are zeroed and the diagonal is set to one; the known values are moved to the right-hand side.

The modified right-hand side vector:

Note: since the source-density function is identically zero on the simplices $s_1$ and $s_4$, we have $F_0$​${} = 0$ and $F_4$​${} = 0$. However, the components $F_1, F_2, F_3$ may be nonzero.

Non-stationary case

For the non-stationary case the system has the form:

Denote $A$​${} = D$​${} + \Delta t \cdot a^2 K$ and $\overrightarrow{P}$​${} = \Delta t \cdot \overrightarrow{F}$​${} + D \cdot \overrightarrow{q}_{n-1}$. We apply the same direct method.

Right-hand side modification for the interior nodes

For the node $q_1$:

For the node $q_3$:

As in the stationary case, the boundary rows and columns of $D$​${} + \Delta t\, a^2 K$ are zeroed and the diagonal is set to one, while $q_0$​${} = g_0$ and $q_4$​${} = g_4$ go to the right-hand side.