While proving the equivalence between the Dirichlet problem (i.e. the potential is known on the surface of every conductor) and the mixed problem (i.e. the potential is known on some conductors and the charge is known on the remaining ones) for conductors in stationary conditions I stumbled upon the following problem:

"Let us assume that we have already established that for an isolated system of conductors $\{c_j\}_{j=1,...,N}$ the following is true: $$ Q = CV $$ where $Q=(Q_j)_{j=1,...,N}$ and $V=(V_j)_{j=1,...,N}$ are the vector representing the potentials and the charges on the conductors and $C$ is known as the capacitance matrix and it is invertible. Now, suppose that, WLOG, $(Q_j)_{j=1,...,M}$ and $(V_j)_{j=M+1,...,N}$ for $M \in \{1,...,N\}$ are known and find the remaining components of both vectors."


This problem can be stated more generally in the form of a theorem:

"Let us consider the linear system $$Ax=y$$ where $A$ is an $n\times n$ invertible matrix with complex entries and thus $x,y$ are $n$-th tuples of complex numbers; let us also introduce the following notations: $$x_1 \equiv (x_j)_{j=1,...,m}, \hspace{2mm} x_2 \equiv (x_j)_{j=m+1,...,n}$$ $$y_1 \equiv (y_j)_{j=1,...,m}, \hspace{2mm} y_2 \equiv (y_j)_{j=m+1,...,n}$$ $$ A = \begin{pmatrix} B & C\\ D & E \end{pmatrix}, \hspace{2mm} B\in \Bbb C^{m\times m}, C\in \Bbb C^{m\times (n-m)}, D\in \Bbb C^{(n-m)\times m}, E\in \Bbb C^{(n-m)\times (n-m)} $$ where $m\in \{1,...n\}$. Now, suppose that $x_1,y_2$ are known and show that there exist unique $x_2,y_1$ such that the system is satisfied."


I tried to write out the multiplication of $A$ by $x$ by making use of the blocks in order to obtain, after proper rearrengement, a linear system where $(x_2,y_1)$ is the unknown vector and the rest is given. The result is:

$$ \begin{pmatrix} I_{m\times m} & C\\ O_{(n-m)\times m} & E \end{pmatrix} \begin{pmatrix} y_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} -Bx_1\\ y_2 - Dx_1 \end{pmatrix} $$ where $I$ and $O$ denote the identity and null matrices respectively, with the order being specified by the pedices. Unfortunately the new matrix of coefficients is not necessarily invertible; indeed there is a counterexample to the previously stated theorem by taking:

$$ A= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $$

because in this case we end up with $y_1 = x_2$. Nevertheless, there exists a sufficient condition for the theorem to be true: if the columns of the matrix obtained by adjoining $C, E$ vertically generate $\text{span}(e_j)_{j=m+1,...,n}$ (where $e_j$ is the $j$-th vector of the canonical basis of $\Bbb C^n$). In fact, in this case, the new matrix of coefficients becomes invertible. However this condition is not necessarily satisfied by the capacitance matrix.

I would like to know if there is a necessary and sufficient condition for the theorem to be true. Also, it may be worthwile to notice that the capacitance matrix is symmetric and has positive entries; in fact, I expect the theorem to be true for that matrix because the solution to the mixed can be proved to be unique indipendently from this result.


As pointed out in the comments by @Anne Bauval, $E$ being invertible is a necessary and sufficient condition. Indeed it is easy to see why it is sufficient, while the necessity follows from the fact that otherwise we could choose $y_2$ so that $Ex_2 = y_2 -Dx_1$ has no solution. The only thing that remains is verifying that the capacitance matrix satisfies this condition. I thought about it for some time now and I guess it could be true if the submatrix $E$ could be identified with the capacitance matrix of the system of last $(n-m)$ conductors, in which case it must be invertible.

  • $\begingroup$ The condition is: $E$ is invertible. $\endgroup$ Jul 19, 2023 at 23:39
  • $\begingroup$ @AnneBauval I guess you are right and this is the necessary and sufficient condition. Still, I do not see how this solves the problem for the capacitance matrix. $\endgroup$ Jul 20, 2023 at 7:19

1 Answer 1


Instead of subscripts, give each partition a distinct name $$\eqalign{ \def\c#1{\color{red}{#1}} \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \m{B&C\\D&E} \cdot \m{w\\x} &= \m{y\\z} \\ }$$ and solve for the unknown $x{\rm\;and\;}y$ partitions.

First, solve for $x$ $$\eqalign{ z &= Dw + Ex \\ x &= E^{-1}\LR{z-Dw} \\ }$$ then solve for $y$ $$\eqalign{ y &= Bw + Cx \\ &= Bw + CE^{-1}\LR{z-Dw} \\ }$$ The invertibility of $A$ and $E$ are related via the Schur complement
Mathematically, it is only necessary that $E$ be invertible to solve for $x{\rm\;and\;}y$.
The invertibility of $A$ is not required (but is enforced by the underlying physics).


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