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Citation: Guillermo Samperio-Ramos, J. Magdalena Santana-Casiano, Melchor González-Dávila, Sonia Ferreira, Manuel A. Coimbra. Variability in the organic ligands released by Emiliania huxleyi under simulated ocean acidification conditions[J]. AIMS Environmental Science, 2017, 4(6): 788-808. doi: 10.3934/environsci.2017.6.788
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In this paper we study systems of linear hyperbolic equations on a bounded interval, say,
∂t(υϖ)=(−C+00C−)∂x(υϖ),0<x<1,t>0, | (1a) |
Ξ(υ(0,t),ϖ(1,t),υ(1,t),ϖ(0,t))T=0,t>0, | (1b) |
υ(x,0)=˚υ(x),ϖ(x,0)=˚ϖ(x)0<x<1, | (1c) |
where
\begin{equation} \boldsymbol{\Xi}_{out} (\boldsymbol{{ \upsilon}}(0, t), \boldsymbol{{ \varpi}}(1, t))^T + \boldsymbol{\Xi}_{in}(\boldsymbol{{ \upsilon}}(1, t), \boldsymbol{{ \varpi}}(0, t))^T = 0, \quad t > 0. \end{equation} | (2) |
An important class of such problems arises from dynamical systems on metric graphs. Let
\begin{equation} {\partial}_t \boldsymbol{p}^j+ \mathcal{M}^j{\partial}_x\boldsymbol{p}^j = 0, \quad 0 < x < 1, \; t > 0, \; 1\leq j\leq m, \end{equation} | (3) |
where
Such problems have been a subject of intensive research, both from the dynamics on graphs, [1,10,5,4,11,17,19], and the 1-D hyperbolic systems, [7,21,15,14], points of view. However, there is hardly any overlap, as there seems to be little interest in the network interpretation of the results in the latter, while in the former the conditions on the Riemann invariants seem to be "difficult to adapt to the case of a network", [11,Section 3].
The main aim of this paper, as well as of the preceding one [2] is to bring together these two approaches. In [2] we have provided explicit formulae allowing for a systematic conversion of Kirchhoff's type network transmission conditions to (1b) in such a way that the resulting system(1) is well-posed. We also gave a proof of the well-posedness on any
To briefly describe the content of the paper, we observe that if the matrix
\widehat {\left(\widehat{\boldsymbol{\Xi}_{out}}\right)^T\widehat{\boldsymbol{\Xi}_{in}}} |
is the adjacency matrix of a line graph (where for a matrix
The main idea of this paper is similar to that of [3]. However, [3] dealt with first order problems with (2) solved with respect to the outgoing data. Here, we do not make this assumption and, while(1) technically is one-dimensional, having reconstructed
The paper is organized as follows. In Section 2 we briefly recall the notation and relevant results from [2]. Section 3 contains the main result of the paper. In Appendix we recall basic results on line graphs in the interpretation suitable for the considerations of the paper.
We consider a network represented by a finite, connected and simple (without loops and multiple edges) metric graph
\mathcal{F}^j = \left(\begin{array}{cc} f^j_{+, 1}&f^j_{-, 1}\\ f^j_{+, 2}&f^j_{-, 2} \end{array}\right), |
the diagonalizing matrix on each edge. The Riemann invariants
\begin{equation} \boldsymbol{u}^j = (\mathcal{F}^{j})^{-1} \boldsymbol{p}^j\quad\text{and}\quad \boldsymbol{p}^j = \binom{f^j_{+, 1} u^j_1 + f^j_{-, 1}u^j_2}{f^j_{+, 2}u^j_1 + f^j_{-, 2}u^j_2}. \end{equation} | (4) |
Then, we diagonalize (3) and, discarding lower order terms, we consider
\begin{equation} {\partial}_t\boldsymbol{u}^j = \mathcal{L}^j{\partial}_x\boldsymbol{{u}}^j = \left(\begin{array}{cc}- \lambda^j_+&0\\0&- \lambda^j_- \end{array}\right){\partial}_x\boldsymbol{{u}}^j, \end{equation} | (5) |
for each
Remark 1. We refer an interested reader to [7,Section 1.1] for a detailed construction of the Riemann invariants for a general
The most general linear local boundary conditions at
\begin{equation} \boldsymbol{\Phi}_{ \mathbf{v}} \boldsymbol{p}( \mathbf{v}) = 0, \end{equation} | (6) |
where
\begin{equation} \boldsymbol{\Phi}_{ \mathbf{v}} : = \left(\begin{array}{ccccc}\phi^{j_1}_{ \mathbf{v}, 1}& \varphi^{j_1}_{ \mathbf{v}, 1}&\ldots&\phi^{j_{|J_{ \mathbf{v}}|}}_{ \mathbf{v}, 1}& \varphi^{j_{|J_{ \mathbf{v}}|}}_{ \mathbf{v}, 1}\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ \phi^{j_1}_{ \mathbf{v}, k_{ \mathbf{v}}}& \varphi^{j_1}_{ \mathbf{v}, k_{ \mathbf{v}}}&\ldots&\phi^{j_{|J_{ \mathbf{v}}|}}_{ \mathbf{v}, k_{ \mathbf{v}}}& \varphi^{j_{|J_{ \mathbf{v}}|}}_{ \mathbf{v}, k_{ \mathbf{v}}}\end{array}\right), \ \end{equation} | (7) |
where
\begin{equation} \boldsymbol{\Psi}_{ \mathbf{v}} \boldsymbol{u}( \mathbf{v}) : = \boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}( \mathbf{v})\boldsymbol{u}( \mathbf{v}) = 0. \end{equation} | (8) |
For Riemann invariants, we can define their outgoing values at
Definition 2.1. Let
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Denote by
\begin{equation} k_{ \mathbf{v}} : = \sum\limits_{j\in J_{ \mathbf{v}}} (2(1-\alpha_j)l_j( \mathbf{v}) +\alpha_j ). \end{equation} | (9) |
Definition 2.2. We say that
● a sink if either
● a source if either
● a transient (or internal) vertex if it is neither a source nor a sink.
We denote the sets of sources, sinks and transient vertices by
We observe that if
A typical example of (8) is Kirchhoff's law that requires that the total inflow rate into a vertex must equal the total outflow rate from it. Its precise formulation depends on the context, we refer to [8,Chapter 18] for a detailed description in the context of flows in networks. Since it provides only one equation, in general it is not sufficient to ensure the well-posedness of the problem. So, we introduce the following definition.
Definition 2.3. We say that
To realize the requirement that the outgoing values should be determined by the incoming ones, we have to analyze the structure of
\begin{equation} \{1, \ldots, m\} = : J_1\cup J_2\cup J_0, \end{equation} | (10) |
where
J_{ \mathbf{v}} : = J_{ \mathbf{v}, 1}\cup J_{ \mathbf{v}, 2}\cup J_{ \mathbf{v}, 0}. |
We also consider another partition
\begin{equation} k_{ \mathbf{v}} = \sum\limits_{j\in J^0_{ \mathbf{v}}} \alpha_j + \sum\limits_{j\in J^1_{ \mathbf{v}}} (2-\alpha_j) = |J_{ \mathbf{v}, 1}|+ 2(|J^0_{ \mathbf{v}}\cap J_{ \mathbf{v}, 2}| + |J^1_{ \mathbf{v}}\cap J_{ \mathbf{v}, 0}|). \end{equation} | (11) |
Then, by [2,Lemma 3.6],
We introduce the block diagonal matrix
\begin{equation} \mathcal{{\tilde{{F}}}}_{out}( \mathbf{v}) = {\rm diag}\{\mathcal{{\tilde{{F}}}}_{out}^j( \mathbf{v})\}_{j \in J_{ \mathbf{v}}}, \end{equation} | (12) |
where
\mathcal{{\tilde{F}}}_{out}^j( \mathbf{v}) = \left\{\begin{array} {ccc} \left(\begin{array}{cc}0&0\\0&0\end{array}\right)&\text{if} &j\in (J_{ \mathbf{v}, 0} \cap J_{ \mathbf{v}}^0)\cup (J_{ \mathbf{v}, 2}\cap J_{ \mathbf{v}}^1), \\ \left(\begin{array}{cc}f^j_{+, 1}(l_j( \mathbf{v}))&f^j_{-, 1}(l_j( \mathbf{v}))\\f^j_{+, 2}(l_j( \mathbf{v}))&f^j_{-, 2}(l_j( \mathbf{v}))\end{array}\right)&\text{if}& j\in (J_{ \mathbf{v}, 0} \cap J_{ \mathbf{v}}^1)\cup (J_{ \mathbf{v}, 2}\cap J_{ \mathbf{v}}^0), \\ \left(\begin{array}{cc}f^j_{+, 1}(0)&0\\f^j_{+, 2}(0)&0\end{array}\right)&\text{if} &j\in J_{ \mathbf{v}, 1} \cap J_{ \mathbf{v}}^0, \\ \left(\begin{array}{cc}0&f^j_{-, 1}(1)\\0&f^j_{-, 2}(1)\end{array}\right)&\text{if}& j\in J_{ \mathbf{v}, 1} \cap J_{ \mathbf{v}}^1. \end{array} \right. |
Further, by
In a similar way, we extract from
\widetilde{\boldsymbol{u}}^j_{out}( \mathbf{v}) = \left\{\begin{array} {ccc} (0, 0)^T&\text{if} &j\in (J_{ \mathbf{v}, 0} \cap J_{ \mathbf{v}}^0)\cup (J_{ \mathbf{v}, 2}\cap J_{ \mathbf{v}}^1), \\ (u^j_1(l_j( \mathbf{v})), u^j_2(l_j( \mathbf{v})))^T&\text{if}& j\in (J_{ \mathbf{v}, 0} \cap J_{ \mathbf{v}}^1)\cup (J_{ \mathbf{v}, 2}\cap J_{ \mathbf{v}}^0), \\ (u^j_1(0), 0)^T&\text{if} &j\in J_{ \mathbf{v}, 1} \cap J_{ \mathbf{v}}^0, \\ (0, u^j_2(1))^T&\text{if}& j\in J_{ \mathbf{v}, 1} \cap J_{ \mathbf{v}}^1, \end{array} \right. |
and
Proposition 1. [2,Proposition 3.8] The boundary system (8) at
\begin{equation} \boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}_{out}( \mathbf{v}) \boldsymbol{u}_{out}( \mathbf{v}) + \boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}_{in}( \mathbf{v})\boldsymbol{u}_{in}( \mathbf{v}) = 0 \end{equation} | (13) |
and hence it uniquely determines the outgoing values of
\begin{equation} \boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}_{out}( \mathbf{v})\quad \mathit{\text{is nonsingular}}. \end{equation} | (14) |
In this case,
\begin{equation} \boldsymbol{u}_{out}( \mathbf{v}) = - (\boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}_{out}( \mathbf{v}))^{-1} \boldsymbol{\Phi}_{ \mathbf{v}} \mathcal{F}_{in}( \mathbf{v})\boldsymbol{u}_{in}( \mathbf{v}). \end{equation} | (15) |
To pass from (3) with Kirchhoff's boundary conditions at each vertex
\begin{equation} \boldsymbol{\Psi}' \gamma \boldsymbol{u} = 0. \end{equation} | (16) |
We note that the function values that are incoming at
\begin{equation} \boldsymbol{\Psi}^{out} \gamma \boldsymbol{u}_{out} + \boldsymbol{\Psi}^{in}\gamma\boldsymbol{u}_{in} = 0, \end{equation} | (17) |
where
Using the adopted parametrization and the formalism of Definition 2.1, we only need to distinguish between functions describing the flow from
\begin{equation} \begin{split} \boldsymbol{{\upsilon}} &: = \left((u^j_1)_{j\in J_1\cup J_2}, (u^j_2)_{j\in J_2}\right) = ( \upsilon_j)_{j\in J^+}, \\ \boldsymbol{{ \varpi}}& : = \left((u^j_1)_{j\in J_0}, (u^j_2)_{j\in J_1\cup J_0}\right) = ( \varpi_j)_{j\in J^-}, \end{split} \end{equation} | (18) |
where
In this way, we converted
Using this construction, the second order hyperbolic problem (3), (17) was transformed into first order system (1) with (17) written in the form (2). However, it is clear that (1) can be formulated with an arbitrary matrix
how to characterize matrices \boldsymbol{\Xi} that arise from \boldsymbol{\Psi} so that (1) describes a network dynamics?
For a graph
\boldsymbol{{\Psi}}_{ \mathbf{v}}^{out} = (\psi_{ \mathbf{v}, i}^j)_{1\leq i\leq k_{ \mathbf{v}}, j\in \boldsymbol{J}_{ \mathbf{v}}^-}, \qquad \boldsymbol{{\Psi}}_{ \mathbf{v}}^{in} = (\psi_{ \mathbf{v}, i}^j)_{1\leq i\leq k_{ \mathbf{v}}, j\in \boldsymbol{J}_{ \mathbf{v}}^+}. |
Since no outgoing value should be missing, we adopt the following
Assumption 1. No column or row of
These matrices provide some insight into how the arcs are connected by the flow which is an additional feature, superimposed on the geometric structure of the incoming and outgoing arcs at the vertex. In principle, these two structures do not have to be the same, that is, it may happen that the substance flowing from
Definition 3.1. Let
Using this idea, we define a connectivity matrix
\mathsf{c}_{ \mathbf{v}, lj} = \left\{\begin{array}{lcl} 1&\text{if}& \boldsymbol{{\varepsilon}}^j\;\text{flow connects to}\; \boldsymbol{{\varepsilon}}^l, \\ 0&&\text{otherwise}. \end{array} \right. |
Remark 2. We observe that
● the above definition implies that for
●
For an arbitrary matrix
Lemma 3.2. If
\begin{equation} \mathsf{C}_{ \mathbf{v}} = \widehat{\left(\widehat{\boldsymbol{{\Psi}}^{out}_{ \mathbf{v}}}\right)^T \widehat{\boldsymbol{{\Psi}}^{in}_{ \mathbf{v}}}}. \end{equation} | (19) |
Proof. Denote
\sum\limits_{r = 1}^{k_{ \mathbf{v}}} \hat{\psi}_{ \mathbf{v}, r}^i\hat{\psi}^j_{ \mathbf{v}, r} \neq 0. |
This occurs if and only if there is
Let
Definition 3.3. Let
As before, we construct a connectivity matrix
\begin{equation} \mathsf{c}_{ \mathbf{v}, ij} = \left\{\begin{array}{lcl} 1&\text{if}& \boldsymbol{{\varepsilon}}^j\;\text{and}\; \boldsymbol{{\varepsilon}}^i\;\text{are flow connected}, \\ 0&&\text{otherwise}. \end{array} \right. \end{equation} | (20) |
Note that, contrary to an internal vertex, here the connectivity matrix is symmetric. We also do not stipulate that
Lemma 3.4. If
\begin{equation} \mathsf{C}_{ \mathbf{v}} = \widehat{\left(\widehat{\boldsymbol{{\Psi}}^{out}_{ \mathbf{v}}}\right)^T \widehat{\boldsymbol{{\Psi}}^{out}_{ \mathbf{v}}}}. \end{equation} | (21) |
Proof. As before, let
\sum\limits_{r = 1}^{k_{ \mathbf{v}}} \hat{\psi}_{ \mathbf{v}, r}^i\hat{\psi}^j_{ \mathbf{v}, r} \neq 0. |
Certainly, by Assumption 1,
We adopt an assumption that the structure of flow connectivity is the same as of the geometry at the vertex. Thus, if
Assumption 2. For all
\mathsf{C}_{ \mathbf{v}} = \boldsymbol{1}_{ \mathbf{v}} = \left(\begin{array}{cccc}1&1&\ldots&1\\ \vdots&\vdots&\vdots&\vdots\\ 1&1&\ldots&1 \end{array}\right). |
We observe that the dimension of
If
Assumption 3. For all
Remark 3. Assumption 3 is weaker than requiring each two arcs from
Proposition 2. Let
\begin{equation} \boldsymbol{\Psi}_{ \mathbf{v}} \boldsymbol{u}( \mathbf{v}) = 0, \end{equation} | (22) |
contains a Kirchhoff's condition
\begin{equation} \sum\limits_{j \in J_{ \mathbf{v}}} (\psi^j_{ \mathbf{v}, r} u^j_1( \mathbf{v}) + \psi^j_{ \mathbf{v}, r} u^j_2( \mathbf{v})) = 0, \end{equation} | (23) |
with
Proof. Condition (23) ensures that each entry of the
Example 1. Consider the model of [20], analysed in the framework of our approach in [2,Example 5.12], i.e.,
\begin{equation} {\partial}_tp^{j}_1 + K^j {\partial}_x p^j_2 = 0, \quad {\partial}_tp^{j}_2 + L^j {\partial}_x p^j_1 = 0, \end{equation} | (24) |
for
\boldsymbol{p_1}( \mathbf{v}) \in X_{ \mathbf{v}}, \quad T_{ \mathbf{v}}\boldsymbol{p_2}( \mathbf{v}) \in X^\perp_{ \mathbf{v}}, |
that is, denoting
\begin{equation} \sum\limits_{j\in J_{ \mathbf{v}}}\phi^j_r p^j_1( \mathbf{v}) = 0, \quad r \in I_2, \quad \sum\limits_{j\in J_{ \mathbf{v}}} \varphi^j_r \nu^j( \mathbf{v})p^j_2( \mathbf{v}) = 0, \qquad r \in I_1, \end{equation} | (25) |
where
\begin{align*} p^r_1(0) & = u^r_1(0)+ u^r_2(0) = 0, \quad r = n_{ \mathbf{v}}+1, \ldots, |J_{ \mathbf{v}}|, \\ p^r_2(0) & = u^r_1(0)- u^r_2(0) = 0, \quad r = 1, \ldots, n_{ \mathbf{v}}. \end{align*} |
Thus
On the other hand, the Kirchhoff condition,
\begin{equation} \sum\limits_{j\in J_{ \mathbf{v}}} \nu^{j}( \mathbf{v}) p^{j}_2( \mathbf{v}) = 0, \end{equation} | (26) |
see [20,Eqn (4)], satisfies the assumption of Proposition 2, as we have
\begin{align*} 0& = \sum\limits_{j\in J_{ \mathbf{v}}}\nu^j( \mathbf{v})p^j_2( \mathbf{v}) = \sum\limits_{j\in J_{ \mathbf{v}}}\nu^j( \mathbf{v}) (f^j_{+, 2}( \mathbf{v})u^j_1( \mathbf{v}) +f^j_{-, 2}( \mathbf{v})u^j_2( \mathbf{v})) \\& = \sum\limits_{j\in J_{ \mathbf{v}}}\nu^j( \mathbf{v})\sqrt{K^jL^j} (u^j_1( \mathbf{v}) -u^j_2( \mathbf{v})) \\ & = -\sum\limits_{j \in J^0_{ \mathbf{v}}} \sqrt{K^jL^j}u^j_1(0) -\sum\limits_{j\in J^1_{ \mathbf{v}}} \sqrt{K^jL^j}u^j_2(1) \\ &\phantom{x}+\sum\limits_{j \in J^1_{ \mathbf{v}}} \sqrt{K^jL^j}u^j_1(1) +\sum\limits_{j\in J^0_{ \mathbf{v}}} \sqrt{K^jL^j}u^j_2(0), \end{align*} |
where we used [2,Eqn 5.2]. Hence, by Proposition 2, Assumption 2 is satisfied.
Example 2. Let us consider the linearized Saint-Venant system,
\begin{equation} {\partial}_tp^j_1 = -V^j {\partial}_x p^j_1 - H^j{\partial}_x p^j_2, \quad {\partial}_t p^j_2 = -g {\partial}_x p^j_1 -V^j {\partial}_x p^j_2, \end{equation} | (27) |
see [2,Example 1.2], assuming that on each edge we have
\begin{equation} \binom{p^j_1}{p^j_2} = \binom{f^j_{+, 1} u^j_1 + f^j_{-, 1}u^j_2}{f^j_{+, 2}u^j_1 + f^j_{-, 2}u^j_2} = \binom{H^j u^j_1+ H^j u^j_2}{\sqrt{gH^j}u^j_1 -\sqrt{gH^j}u^j_2}. \end{equation} | (28) |
We use the flow structure of [11,Example 5.1], shown in Fig. 1, and focus on
p^j_1(0) = p^1_1(1), \quad p^j_2(0) = p_2^1(1), \qquad j = 2, \ldots, N. |
In terms of the Riemann invariants, they can be written as
\begin{align*} &\left(\begin{array}{ccccccc}H^2&H^2&0&0&\ldots&0&0\\\sqrt{gH^2}&-\sqrt{gH^2}&0&0&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\ldots&H^N&H^N\\0&0&0&0&\ldots&\sqrt{gH^N}&-\sqrt{gH^N}\end{array}\right)\left(\begin{array}{c}u_1^2(0)\\u_2^2(0)\\\vdots \\ u_1^N(0)\\u_2^N(0)\end{array}\right)\\ &\phantom{xxxx} = \left(\begin{array}{cc}H^1&H^1\\\sqrt{gH^1}&-\sqrt{gH^1}\\\vdots\\ H^1&H^1\\\sqrt{gH^1}&-\sqrt{gH^1}\end{array}\right)\left(\begin{array}{c}u_1^1(1)\\u_2^1(1)\end{array}\right) \end{align*} |
and it is clear that Assumption 2 is satisfied.
For a matrix
\begin{equation} A = (\boldsymbol{a}^c_{j})_{1\leq j\leq m} = (\boldsymbol{a}^r_{i})_{1\leq i\leq n}, \end{equation} | (29) |
that is, we represent the matrix as a row vector of its columns or a column vector of its rows. In particular, we write
\begin{align*} \boldsymbol{\Xi}_{out} & = (\xi^{out}_{ij})_{1\leq i\leq 2m, 1\leq j\leq 2m} = (\boldsymbol{\xi}^{out, c}_{j})_{1\leq j\leq 2m} = (\boldsymbol{\xi}^{out, r}_{i})_{1\leq i\leq 2m}, \\\boldsymbol{\Xi}_{in} & = (\xi^{in}_{ij})_{1\leq i\leq 2m, 1\leq j\leq 2m} = (\boldsymbol{\xi}^{in, c}_{j})_{1\leq j\leq 2m} = (\boldsymbol{\xi}^{in, r}_{i})_{1\leq i\leq 2m}. \end{align*} |
For any vector
Definition 3.5. We say that the problem (1) is graph realizable if there is a graph
Before we formulate the main theorem, we need to introduce some notation. Let us recall that we consider the boundary system (2), i.e.,
\boldsymbol{\Xi}_{out}(( \upsilon_j(0, t))_{j\in J^+}, ( \varpi_j(1, t))_{j\in J^-}) = - \boldsymbol{\Xi}_{in}(( \upsilon_j(1, t))_{j\in J^+}, ( \varpi_j(0, t))_{j\in J^-}). |
Let us emphasize that in this notation, the column indices on the left and right hand side correspond to the values of the same function. To shorten notation, let us renumber them as
In the second step we will determine additional assumptions that allow
Since we do not want (2) to be under- or over-determined, we adopt
Assumption 4. For all
\boldsymbol{\xi}^{out, c}_j\neq 0\quad\mathit{\text{and}}\quad \boldsymbol{\xi}^{out, r}_j \neq 0. |
Our strategy is to treat
Assumption 5. The matrix
\mathsf{A}: = \widehat {\left(\widehat{\boldsymbol{\Xi}_{out}}\right)^T\widehat{\boldsymbol{\Xi}_{in}}} |
is the adjacency matrix of the line graph of a multi digraph.
For
I : = \{i\in \{1, \ldots, 2m\};\; \boldsymbol{\xi}^{in, r}_i = 0\} |
and adopt
Assumption 6. For all
\mathit{\text{supp}}\; \boldsymbol{\xi}^{out, r}_i \subset V^{out}_j. |
In the next proposition we shall show that
Proposition 3. If Assumptions 4, 5 and 6 are satisfied, then the sets
\begin{equation} \{\{\mathcal{V_i}\}_{1\leq i\leq n}, \mathcal{V_S}\}, \end{equation} | (30) |
where
\begin{align} \mathcal{V_S} & = \{i \in \{1, \ldots, 2m\};\; \mathrm{supp\;} \boldsymbol{\xi}^{out, r}_i \subset V^{out}_{M'}\}, \end{align} | (31) |
\begin{align} \mathcal{V_i} & = \bigcup\limits_{s \in V^{out}_i}\mathrm{supp\;} \boldsymbol{{\xi}}^{out, c}_s, \;1\leq i\leq n, \end{align} | (32) |
form a partition of the row indices of both
Since the proof is quite long, we first present its outline.
Step 1. Reconstruct a multi digraph
Step 2. Identify the rows of
Step 3. Associate other rows of
Step 4. Associate remaining rows with vertices and construct a possible partition of the row indices.
Step 5. Check that the constructed partition has the required properties.
Proof. Step 1. By Assumption 5,
Let us recall that the entry
\mathsf{a}_{ij} = \widehat{\widehat{\boldsymbol{{\xi}}^{out, c}_i}\cdot \widehat{\boldsymbol{{\xi}}^{in, c}_j}} |
and if a row
Step 2. To determine the rows in
\mathcal{V_S} = \{i \in \{1, \ldots, 2m\};\; \text{supp}\; \boldsymbol{\xi}^{out, r}_i \subset V^{out}_{M'}\}. |
For any
Step 3. Now, consider the indices
Step 4. Next, we associate the remaining rows in
\mathcal{V}^{out}_i = \bigcup\limits_{s \in V^{out}_i} \text{supp}\; \boldsymbol{{\xi}}^{out, c}_s, \qquad \mathcal{V}^{in}_j = \bigcup\limits_{q \in V^{in}_j} \text{supp}\; \boldsymbol{{\xi}}^{in, c}_q. |
We first observe that if
\begin{equation} \mathcal{V}^{out}_i\setminus \{s \in \mathcal{V}^{out}_i;\; \boldsymbol{\xi}^{in, r}_s = 0\} = \mathcal{V}^{in}_j. \end{equation} | (33) |
Indeed, let
Step 5. We easily check that this partition satisfies the conditions of the proposition. We have already checked this for
We note that (30) does not contain rows corresponding to sinks and they must be added following the rules described in Appendix A. With such an extension, we consider the multi digraph
\begin{equation} \{\{\mathcal{V_i}\}_{1\leq i\leq n}, \mathcal{V_S}, \mathcal{V_Z}\}, \quad \{\{V^{out}_i\}_{1\leq i\leq n}, V^{out}_{M'}, \emptyset\}, \quad \{\{V^{in}_{j_i}\}_{1\leq i\leq n}, \emptyset, V^{in}_{N'}\}, \end{equation} | (34) |
where the association
\begin{equation} \begin{split} &\boldsymbol{\Xi}^i_{out}(( \upsilon_j(0, t))_{j\in J^+\cap V^{out}_i}, ( \varpi_j(1, t))_{j\in J^-\cap V^{out}_i})\\& \phantom{xxx} = - \boldsymbol{\Xi}^i_{in}(( \upsilon_j(1, t))_{j\in J^+\cap V^{in}_{j_i}}, ( \varpi_j(0, t))_{j\in J^-\cap V^{in}_{j_i}}), \quad 1\leq i\leq n, \\ &\boldsymbol{\Xi}^S_{out}(( \upsilon_j(0, t))_{j\in J^+\cap V^{out}_{M'}}, ( \varpi_j(1, t))_{j\in J^-\cap V^{out}_{M'}}) = 0. \end{split} \end{equation} | (35) |
This system can be seen as a Kirchhoff system on the multi digraph
Let
If we grouped all sources into one node, as before Proposition 5, then, by Lemma 3.4, the flow connectivity in this source would be given by
\mathsf{C}_{ \mathbf{v}}: = \widehat{\left(\widehat{\boldsymbol{\Xi}_{out}^{S}}\right)^T\widehat{\boldsymbol{\Xi}_{out}^{S}}}. |
However, such a matrix would not necessarily satisfy Assumption 3. Thus, we separate the arcs into non-communicating groups, each determining a source satisfying Assumption 3. For this, by simultaneous permutations of rows and columns,
\begin{equation} \boldsymbol{\Xi}^S = {\rm diag}\{ \boldsymbol{\Xi}^S_{i}\}_{1\leq i\leq k}, \end{equation} | (36) |
where
\xi^{out, r}_{S_i j} = \left\{\begin{array}{lcl} 1 &\text{if}& j\in V^{out}_{S_i}, \\0&\text{otherwise}.&\end{array}\right. |
For the sinks, it is simpler as there is no constraining information from (2). We have columns with indices in
\begin{equation} V^{in}_{\mathcal{V_i}} = \{j\in V^{in}_{N'};\; \text{supp}\; \boldsymbol{\xi}^{out, c}_j\cap \mathcal{V_i}\neq \emptyset\}, \quad i = 1, \ldots, n, S_1, \ldots, S_k. \end{equation} | (37) |
For each
\begin{equation} V^{in}_{\mathcal{V_i}} = V^{in}_{i, {1}}\cup\ldots\cup V^{in}_{i, {l_i}}, \quad i = 1, \ldots, n, S_1, \ldots, S_k, \end{equation} | (38) |
where
\xi^{in, r}_{\{i, {l}\}, q} = \left\{\begin{array}{lcl} 1 &\text{if}& q\in V^{in}_{i, {l}}, \\0&\text{otherwise}.&\end{array}\right. |
Remark 4. We expect
Then, as in Remark 5, the incoming and outgoing incidence matrices are
A^+ = \left(\begin{array}{c} \mathsf{A}^+\\\boldsymbol{0}\\\boldsymbol{\xi}^{in, r}\end{array}\right), \qquad A^- = \left(\begin{array}{c} \mathsf{A}^-\\\boldsymbol{\xi}^{out, r}\\\boldsymbol{0}\end{array}\right) |
which, by a suitable permutation of columns moving the sources and the sinks to the last positions, can be written as
(39) |
respectively. Both matrices have
(40) |
where the dimensions of the blocks in the first row are, respectively,
Consider a nonzero pair
\begin{equation} \begin{split} &(a_{ij}\mapsto \{k^{ij}_1, \ldots, k^{ij}_h\}, a_{ji}\mapsto \{k^{ji}_1, \ldots, k^{ji}_e\}) \\ & = (a_{ij}\mapsto \text{supp}\; \boldsymbol{a}^{+, r}_i\cap \text{supp}\; \boldsymbol{a}^{-, r}_j, a_{ji}\mapsto \text{supp}\; \boldsymbol{a}^{+, r}_j\cap \text{supp}\; \boldsymbol{a}^{-, r}_i)\end{split} \end{equation} | (41) |
of columns of
Theorem 3.6. System (2) is graph realizable with generalized Kirchhoff's conditions satisfying Assumptions 1 and 2 for
1. for any
\begin{equation} (2, 0), \; (1, 1), \; (0, 2)\; \mathit{\text{or}}\; (0, 0); \end{equation} | (42) |
2. if
\begin{array}{l} if\;({a_{ij}}, {a_{ji}}) = (2, 0)\;or\;(0, 2), \;\;\;then\;k, l \in {J^ + }\;or\;k, l \in {J^ - }\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;and\;{c_k} \ne {c_l}, \\ if\;({a_{ij}}, {a_{ji}}) = (1, 1), \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;then\;k \in {J^ + }\;and\;l \in {J^ - }\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;k \in {J^ - }\;and\;l \in {J^ + }. \end{array} | (43) |
Proof. Necessity. Let us consider the Kirchhoff system (17). By construction, both matrices
\tilde A = (\tilde a_{ij})_{1\leq i, j\leq 2m} = \widehat{(\widehat{\boldsymbol{\Psi}^{out}})^T\widehat{\boldsymbol{\Psi}^{in}}} |
is block diagonal with blocks of the form
\boldsymbol{\Xi}_{out} = \boldsymbol{\Psi}^{out}P, \qquad \boldsymbol{\Xi}_{in} = \boldsymbol{\Psi}^{in}Q, |
where
A = (a_{ij})_{1\leq i, j\leq 2m} : = \widehat{(\widehat{\boldsymbol{\Xi}_{out}})^T\widehat{\boldsymbol{\Xi}_{in}}} = \widehat{(\widehat{\boldsymbol{\Psi}^{out}}P)^T\widehat{\boldsymbol{\Psi}^{in}}Q} = P^T\tilde AQ |
is a matrix where the indices
Sufficiency. Given (2), we have flows
Now, (42) ensures that there are no loops at vertices and that between any two vertices there are either two arcs or none. If
Finally, the assumption
Example 3. Let us consider the system
\begin{equation} \begin{split} {\partial}_t \upsilon_{j}+c_j{\partial}_x \upsilon_j & = 0, \quad 1\leq j\leq 4, \\ {\partial}_t \varpi_{j}-c_j{\partial}_x \varpi_j & = 0, \quad 5\leq j\leq 6, \end{split} \end{equation} | (44) |
where
\begin{equation} \left(\!\!\begin{array}{cccccc}0&1&1&0&0&0\\ 1&0&0&0&0&0\\ 1&1&0&0&0&0\\ 0&0&1&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\end{array}\!\! \right)\left(\!\!\!\begin{array}{c} \upsilon_1(0)\\ \upsilon_2(0)\\ \upsilon_3(0)\\ \upsilon_4(0)\\ \varpi_5(1)\\ \varpi_6(1)\end{array}\!\!\!\right) = \left(\!\!\begin{array}{cccccc}0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 1&1&0&0&0&1\\ 0&0&1&1&1&0\end{array}\!\! \right)\left(\!\!\!\begin{array}{c} \upsilon_1(1)\\ \upsilon_2(1)\\ \upsilon_3(1)\\ \upsilon_4(1)\\ \varpi_5(0)\\ \varpi_6(0)\end{array}\!\!\!\right). \end{equation} | (45) |
Thus,
A = \widehat{(\widehat{\boldsymbol{\Xi}_{out}})^T\widehat{\boldsymbol{\Xi}_{in}}} = \left(\begin{array}{cccccc}0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 1&1&0&0&0&1\\ 0&0&1&1&1&0\end{array} \right). |
Thus, there is a multi digraph
\boldsymbol{\Xi}^S_{out} = \left(\begin{array}{cccc}0&1&1&0\\ 1&0&0&0\\ 1&1&0&0\\ 0&0&1&1\end{array}\right) \quad\text{and so}\quad \boldsymbol{\Xi}^S = \widehat{\left(\widehat{\boldsymbol{\Xi}_{out}^{S}}\right)^T\widehat{\boldsymbol{\Xi}_{out}^{S}}} = \left(\begin{array}{cccc}1&1&0&0\\ 1&1&1&0\\ 0&1&1&1\\ 0&0&1&1\end{array}\right). |
This matrix is irreducible and thus we have one source. Therefore
A^+ = \left(\begin{array}{cccccc}1&1&0&0&0&1\\ 0&0&1&1&1&0\\ 0&0&0&0&0&0\end{array}\right), \qquad A^- = \left(\begin{array}{cccccc}0&0&0&0&1&0\\ 0&0&0&0&0&1\\ 1&1&1&1&0&0\end{array}\right) |
and consequently
Further,
\begin{align*} \text{supp}\; \boldsymbol{a}^{+, r}_1& = \{1, 2, 6\}, \; \text{supp}\; \boldsymbol{a}^{+, r}_2 = \{3, 4, 5\}, \\ \text{supp}\; \boldsymbol{a}^{-, r}_1& = \{5\}, \; \text{supp}\; \boldsymbol{a}^{-, r}_2 = \{6\}, \; \text{supp}\; \boldsymbol{a}^{-, r}_3 = \{1, 2, 3, 4\}, \end{align*} |
hence, by (41),
(a_{12}\mapsto \{6\}, a_{21}\mapsto \{5\}), \quad (a_{13}\mapsto \{1, 2\}, a_{31}\mapsto \emptyset), \quad (a_{23}\mapsto \{3, 4\}, a_{32}\mapsto \emptyset). |
To reconstruct
Consider a small modification of (44), (45),
\begin{equation} \begin{split} {\partial}_t \upsilon_{j}+c_j{\partial}_x \upsilon_j & = 0, \quad 1\leq j\leq 5, \\ {\partial}_t \varpi_{6}-c_6{\partial}_x \varpi_6 & = 0, \end{split} \end{equation} | (46) |
\begin{equation} \begin{split} \upsilon_5(0)- \upsilon_1(1)- \upsilon_2(1)- \varpi_6(0)& = 0, \\ \varpi_6(1)- \upsilon_3(1)- \upsilon_4(1)- \upsilon_5(1)& = 0. \end{split} \end{equation} | (47) |
The matrices
\begin{equation} \begin{array}{c} {\partial}_t u^1_1+c_1{\partial}_x u^1_1 = 0, \\ {\partial}_t u^1_2+c_2{\partial}_x u^1_2 = 0, \end{array}\;\begin{array}{c} {\partial}_t u^2_1+c_5{\partial}_x u^2_1 = 0, \\ {\partial}_t u^2_2-c_6{\partial}_x u^2_2 = 0, \end{array}\; \begin{array}{c}{\partial}_t u^3_1+c_3{\partial}_x u^3_1 = 0, \\{\partial}_t u^3_2+c_4{\partial}_x u^3_2 = 0, \end{array} \end{equation} | (48) |
with boundary conditions at
(49) |
Consider a digraph
Proposition 4. [6,Thm. 2.4.1] A binary matrix
For our analysis, it is important to understand the reconstruction of
\mathsf{A} = (\mathsf{{a}}_{ij})_{1\leq i, j\leq m} = (\boldsymbol{{a}}^c_j)_{1\leq j\leq m} = (\boldsymbol{{a}}^r_i)_{1\leq i\leq m}. |
If for some
Using the adjacency matrix of a line digraph, we cannot determine how many sources or sinks the original graph could have without additional information. We can lump all potential sources and sinks into one source and one sink, we can have as many sinks and sources as there are zero columns and rows, respectively, or we can subdivide the arcs into some intermediate arrangement. We describe a construction with one source and one sink and indicate its possible variants.
We introduce
\begin{equation} \begin{split} M: = \left\{\begin{array}{lcl} M' &\text{if}& V^{out}_{M'} = \{j;\; \boldsymbol{a}^r_j \neq 0\}, \\ M'-1 &\text{if} & V^{out}_{M'} = \{j;\; \boldsymbol{a}^r_j = 0\}, \end{array}\right.\\ N: = \left\{\begin{array}{lcl} N' &\text{if}& V^{in}_{N'} = \{j;\; \boldsymbol{a}^c_j \neq 0\}, \\ N'-1 &\text{if} & V^{in}_{N'} = \{j;\; \boldsymbol{a}^c_j = 0\}.\end{array}\right.\end{split} \end{equation} | (A.1) |
Thus, we see that the number of internal (or transient) vertices, that is, which are neither sources nor sinks is
\begin{equation} \mathbf{v}_j = \{V^{in}_j, V^{out}_i\}, \qquad a_{i_pj_r} = 1\;\text{for some/any }i_p \in V^{out}_i, j_r \in V^{in}_j. \end{equation} | (A.2) |
With this notation, we present a more algorithmic way of reconstructing
\begin{equation} \mathsf{A}^+ = (\boldsymbol{a}^r_i)_{i\in \mathbb{I}^+}, \qquad \mathsf{A}^- = \left((\boldsymbol{a}^c_j)_{j\in \mathbb{I}^-}\right)^T. \end{equation} | (A.3) |
We see now that each row of
Proposition 5.
Proof. Since each column of
The adjacency matrix
Remark 5. Assume that
where
\bar A^+(\bar A^-)^T = (A^+P)(A^-P)^T = A^+PP^T(A^-)^T = A^+A^-, |
as
(A.4) |
Example 4 Consider the networks
\begin{equation} \mathsf{A} = \left(\begin{array}{ccccccc}0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 1&1&0&1&0&0&0\\ 0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0\\ 0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0\end{array}\right). \end{equation} | (A.5) |
Then,
\begin{equation} \mathsf{A}^+ = \left(\begin{array}{ccccccc} 1&1&0&1&0&0&0\\ 0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0\end{array}\right) \end{equation} | (A.6) |
and we see that there are two transient (internal) vertices
\begin{equation} \mathsf{A}^- = \left(\begin{array}{ccccccc} 0&0&1&0&0&0&0\\ 0&0&0&1&0&1&0\\ 0&0&0&0&0&0&0\end{array}\right). \end{equation} | (A.7) |
The last row corresponds to sinks and the zero columns inform us that arcs
If we want to reconstruct the original graph with one source and one sink, then
A^+ = \left(\begin{array}{ccccccc} 1&1&0&1&0&0&0\\ 0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&1&1\end{array}\right), \qquad A^- = \left(\begin{array}{ccccccc} 0&0&1&0&0&0&0\\ 0&0&0&1&0&1&0\\ 1&1&0&0&1&0&1\\ 0&0&0&0&0&0&0\end{array}\right) |
and
A^+(A^-)^T = \left(\begin{array}{cccc} 0&1&2&0\\ 1&0&1&0\\ 0&0&0&0\\ 0&1&1&0\end{array}\right), |
which describes the right multi digraph in Fig. 5. On the other hand, we can consider two sinks (maximum number, as there are two zero columns in
A^+ = \left(\begin{array}{ccccccc} 1&1&0&1&0&0&0\\ 0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0\end{array}\right), \qquad A^- = \left(\begin{array}{ccccccc} 0&0&1&0&0&0&0\\ 0&0&0&1&0&1&0\\ 1&0&0&0&0&0&1\\ 0&1&0&0&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\end{array}\right) |
and
A^+(A^-)^T = \left(\begin{array}{ccccccc} 0&1&1&1&0&0&0\\ 1&0&0&0&1&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0\end{array}\right) |
which describes the left multi digraph in Fig. 5.
It is easily seen that both digraphs have the same line digraph, shown on Fig. 6, whose adjacency matrix is
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