Vamos a suponer $latex n=3$ para reducir el tamaño de las matrices.
Empezamos suponiendo que conocemos:
$latex frac{partial}{partial x}|_{0,0,}u, frac{partial}{partial x}|_{0,1}u, frac{partial}{partial x}|_{0,2}u$
$latex frac{partial}{partial y}|_{0,0}u, frac{partial}{partial y}|_{1,0}u$
$latex frac{partial}{partial y}|_{0,2}u, frac{partial}{partial y}|_{1,2}u$
$latex u|_{2,0}, u|_{2,1}, u|_{2,2}$
Discretizamos:
$latex frac{u_{-1,0}-2u_{0,0}+u_{1,0}}{h^2} + frac{u_{0,-1}-2u_{0,0}+u_{0,1}}{h^2} = f_{0,0}$
$latex frac{u_{-1,1}-2u_{0,1}+u_{1,1}}{h^2} + frac{u_{0,0}-2u_{0,1}+u_{0,2}}{h^2} = f_{0,1}$
$latex frac{u_{-1,2}-2u_{0,2}+u_{1,2}}{h^2} + frac{u_{0,1}-2u_{0,2}+u_{0,3}}{h^2} = f_{0,2}$
$latex frac{u_{0,0}-2u_{1,0}+u_{2,0}}{h^2} + frac{u_{1,-1}-2u_{1,0}+u_{1,1}}{h^2} = f_{1,0}$
$latex frac{u_{0,1}-2u_{1,1}+u_{2,1}}{h^2} + frac{u_{1,0}-2u_{1,1}+u_{1,2}}{h^2} = f_{1,1}$
$latex frac{u_{0,2}-2u_{1,2}+u_{2,2}}{h^2} + frac{u_{1,1}-2u_{1,2}+u_{1,3}}{h^2} = f_{1,2}$
En las fronteras, sabemos que:
$latex frac{u_{1,0}-u_{-1,0}}{2h} = frac{partial}{partial x}|_{0,0}u Leftrightarrow u_{-1,0}=u_{1,0}-2h frac{partial}{partial x}|_{0,0}u$
$latex frac{u_{1,1}-u_{-1,1}}{2h} = frac{partial}{partial x}|_{0,1}u Leftrightarrow u_{-1,1}=u_{1,1}-2h frac{partial}{partial x}|_{0,1}u$
$latex frac{u_{1,2}-u_{-1,2}}{2h} = frac{partial}{partial x}|_{0,2}u Leftrightarrow u_{-1,2}=u_{1,2}-2h frac{partial}{partial x}|_{0,2}u$
$latex frac{u_{0,1}-u_{0,-1}}{2h} = frac{partial}{partial y}|_{0,0}u Leftrightarrow u_{0,-1}=u_{0,1}-2h frac{partial}{partial y}|_{0,0}u$
$latex frac{u_{1,1}-u_{1,-1}}{2h} = frac{partial}{partial y}|_{1,0}u Leftrightarrow u_{1,-1}=u_{1,1}-2h frac{partial}{partial y}|_{1,0}u$
$latex frac{u_{0,3}-u_{0,1}}{2h} = frac{partial}{partial y}|_{0,2}u Leftrightarrow u_{0,3}=u_{0,1}+2h frac{partial}{partial y}|_{0,2}u$
$latex frac{u_{1,3}-u_{1,1}}{2h} = frac{partial}{partial y}|_{1,2}u Leftrightarrow u_{1,3}=u_{1,1}+2h frac{partial}{partial y}|_{1,2}u$
La matriz queda:
$latex left(
begin{array}{ccc|ccc}
-4 & 2 & 0 & 2 & 0 & 0 \
1 & -4 & 1 & 0 & 2 & 0 \
0 & 2 & -4 & 0 & 0 & 2 \ hline
1 & 0 & 0 & -4 & 2 & 0 \
0 & 1 & 0 & 1 & -4 & 1 \
0 & 0 & 1 & 0 & 2 & -4
end{array}
right)$
Simetrizable como:
$latex left(
begin{array}{ccc|ccc}
-1 & frac{1}{2} & 0 & frac{1}{2} & 0 & 0 \
frac{1}{2} & -2 & frac{1}{2} & 0 & 1 & 0 \
0 & frac{1}{2} & -1 & 0 & 0 & frac{1}{2} \ hline
frac{1}{2} & 0 & 0 & -2 & 1 & 0 \
0 & 1 & 0 & 1 & -4 & 1 \
0 & 0 & frac{1}{2} & 0 & 1 & -2
end{array}
right)$
Tenemos $latex 6$ ecuaciones con $latex 6$ incognitas y la matriz tiene rango $latex 6$, por lo que la solución es única.
En el segundo caso, suponemos que todas las fronteras son Neumann:
$latex frac{partial}{partial x}|_{0,0}u, frac{partial}{partial x}|_{0,1}u, frac{partial}{partial x}|_{0,2}u$
$latex frac{partial}{partial y}|_{0,0}u, frac{partial}{partial y}|_{1,0}u, frac{partial}{partial y}|_{2,0}u$
$latex frac{partial}{partial y}|_{0,2}u, frac{partial}{partial y}|_{1,2}u, frac{partial}{partial y}|_{2,2}u$
$latex frac{partial}{partial x}|_{2,0}u, frac{partial}{partial x}|_{2,1}u, frac{partial}{partial x}|_{2,2}u$
Si discretizamos:
$latex frac{u_{-1,0}-2u_{0,0}+u_{1,0}}{h^2} + frac{u_{0,-1}-2u_{0,0}+u_{0,1}}{h^2} = f_{0,0}$
$latex frac{u_{-1,1}-2u_{0,1}+u_{1,1}}{h^2} + frac{u_{0,0}-2u_{0,1}+u_{0,2}}{h^2} = f_{0,1}$
$latex frac{u_{-1,2}-2u_{0,2}+u_{1,2}}{h^2} + frac{u_{0,1}-2u_{0,2}+u_{0,3}}{h^2} = f_{0,2}$
$latex frac{u_{0,0}-2u_{1,0}+u_{2,0}}{h^2} + frac{u_{1,-1}-2u_{1,0}+u_{1,1}}{h^2} = f_{1,0}$
$latex frac{u_{0,1}-2u_{1,1}+u_{2,1}}{h^2} + frac{u_{1,0}-2u_{1,1}+u_{1,2}}{h^2} = f_{1,1}$
$latex frac{u_{0,2}-2u_{1,2}+u_{2,2}}{h^2} + frac{u_{1,1}-2u_{1,2}+u_{1,3}}{h^2} = f_{1,2}$
$latex frac{u_{1,0}-2u_{2,0}+u_{3,0}}{h^2} + frac{u_{2,-1}-2u_{2,0}+u_{2,1}}{h^2} = f_{2,0}$
$latex frac{u_{1,1}-2u_{2,1}+u_{3,1}}{h^2} + frac{u_{2,0}-2u_{2,1}+u_{2,2}}{h^2} = f_{2,1}$
$latex frac{u_{1,2}-2u_{2,2}+u_{3,2}}{h^2} + frac{u_{2,1}-2u_{2,2}+u_{2,3}}{h^2} = f_{2,2}$
En las fronteras, sabemos que:
$latex frac{u_{1,0}-u_{-1,0}}{2h} = frac{partial}{partial x}|_{0,0}u Leftrightarrow u_{-1,0}=u_{1,0}-2h frac{partial}{partial x}|_{0,0}u$
$latex frac{u_{1,1}-u_{-1,1}}{2h} = frac{partial}{partial x}|_{0,1}u Leftrightarrow u_{-1,1}=u_{1,1}-2h frac{partial}{partial x}|_{0,1}u$
$latex frac{u_{1,2}-u_{-1,2}}{2h} = frac{partial}{partial x}|_{0,2}u Leftrightarrow u_{-1,2}=u_{1,2}-2h frac{partial}{partial x}|_{0,2}u$
$latex frac{u_{0,1}-u_{0,-1}}{2h} = frac{partial}{partial y}|_{0,0}u Leftrightarrow u_{0,-1}=u_{0,1}-2h frac{partial}{partial y}|_{0,0}u$
$latex frac{u_{1,1}-u_{1,-1}}{2h} = frac{partial}{partial y}|_{1,0}u Leftrightarrow u_{1,-1}=u_{1,1}-2h frac{partial}{partial y}|_{1,0}u$
$latex frac{u_{2,1}-u_{2,-1}}{2h} = frac{partial}{partial y}|_{2,0}u Leftrightarrow u_{2,-1}=u_{2,1}-2h frac{partial}{partial y}|_{2,0}u$
$latex frac{u_{0,3}-u_{0,1}}{2h} = frac{partial}{partial y}|_{0,2}u Leftrightarrow u_{0,3}=u_{0,1}+2h frac{partial}{partial y}|_{0,2}u$
$latex frac{u_{1,3}-u_{1,1}}{2h} = frac{partial}{partial y}|_{1,2}u Leftrightarrow u_{1,3}=u_{1,1}+2h frac{partial}{partial y}|_{1,2}u$
$latex frac{u_{2,3}-u_{2,1}}{2h} = frac{partial}{partial y}|_{2,2}u Leftrightarrow u_{2,3}=u_{2,1}+2h frac{partial}{partial y}|_{2,2}u$
$latex frac{u_{3,0}-u_{1,0}}{2h} = frac{partial}{partial x}|_{2,0}u Leftrightarrow u_{3,0}=u_{1,0}+2h frac{partial}{partial x}|_{2,0}u$
$latex frac{u_{3,1}-u_{1,1}}{2h} = frac{partial}{partial x}|_{2,1}u Leftrightarrow u_{3,1}=u_{1,1}+2h frac{partial}{partial x}|_{2,1}u$
$latex frac{u_{3,2}-u_{1,2}}{2h} = frac{partial}{partial x}|_{2,2}u Leftrightarrow u_{3,2}=u_{1,2}+2h frac{partial}{partial x}|_{2,2}u$
La matriz, por tanto, queda:
$latex text{A6}=left(
begin{array}{ccc|ccc|ccc}
-4 & 2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \
1 & -4 & 1 & 0 & 2 & 0 & 0 & 0 & 0 \
0 & 2 & -4 & 0 & 0 & 2 & 0 & 0 & 0 \ hline
1 & 0 & 0 & -4 & 2 & 0 & 1 & 0 & 0 \
0 & 1 & 0 & 1 & -4 & 1 & 0 & 1 & 0 \
0 & 0 & 1 & 0 & 2 & -4 & 0 & 0 & 1 \ hline
0 & 0 & 0 & 2 & 0 & 0 & -4 & 2 & 0 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & -4 & 1 \
0 & 0 & 0 & 0 & 0 & 2 & 0 & 2 & -4
end{array}
right)$
Simetrizable como:
$latex text{A6s}=left(
begin{array}{ccc|ccc|ccc}
-1 & 1/2 & 0 & 1/2 & 0 & 0 & 0 & 0 & 0 \
1/2 & -2 & 1/2 & 0 & 1 & 0 & 0 & 0 & 0 \
0 & 1/2 & -1 & 0 & 0 & 1/2 & 0 & 0 & 0 \ hline
1/2 & 0 & 0 & -2 & 1 & 0 & 1/2 & 0 & 0 \
0 & 1 & 0 & 1 & -4 & 1 & 0 & 1 & 0 \
0 & 0 & 1/2 & 0 & 1 & -2 & 0 & 0 & 1/2 \ hline
0 & 0 & 0 & 1/2 & 0 & 0 & -1 & 1/2 & 0 \
0 & 0 & 0 & 0 & 1 & 0 & 1/2 & -2 & 1/2 \
0 & 0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 & -1
end{array}
right)$
En este caso, tenemos $latex 9$ ecuaciones con $latex 9$ incognitas pero la matriz tiene rango $latex 8$, por lo que tenemos infinitas soluciones. Hay que conservar.