In this work, we present an algorithm to the recovery of the conductance of a n –dimensional grid. The algorithm is based in the solution of some overdetermined partial boundary value problems defined on the grid; that is, boundary value problem where the boundary conditions are set only in a part of the boundary (partial), and moreover in a fix subset of the boundary we prescribe both the value of the function and of its normal derivative (overdetermined). Our goal is to recover the conductance of a n –dimensional grid network with boundary using only boundary measurements and global equilibrium conditions. This problem is known as inverse boundary value problem . In general, inverse problems are exponentially ill–posed, since they are highly sensitive to changes in the boundary data. However, in this work we deal with a situation where the recovery of the conductance is feasible: grid networks. The recovery of the conductances of a grid network is performed here using its Schr ¨odinger matrix and boundary value problems associated to it. Moreover, we use the Dirichlet–to–Robin matrix, also known as response matrix of the network, which contains the boundary information. It is a certain Schur complement of the Schr ¨odinger matrix. The Schur complement plays an important role in matrix analysis, statistics, numerical analysis, and many other areas of mathematics and its applications.Postprint (author's final draft

Arauz Lombardía, Cristina

Carmona Mejías, Ángeles

Encinas Bachiller, Andrés Marcos

In this work, we present an algorithm to the recovery of the conductance of a n –dimensional grid. The algorithm is based in the solution of some overdetermined partial boundary value problems defined on the grid; that is, boundary value problem where the boundary conditions are set only in a part of the boundary (partial), and moreover in a fix subset of the boundary we prescribe both the value of the function and of its normal derivative (overdetermined). Our goal is to recover the conductance of a n –dimensional grid network with boundary using only boundary measurements and global equilibrium conditions. This problem is known as inverse boundary value problem . In general, inverse problems are exponentially ill–posed, since they are highly sensitive to changes in the boundary data. However, in this work we deal with a situation where the recovery of the conductance is feasible: grid networks. The recovery of the conductances of a grid network is performed here using its Schr ¨odinger matrix and boundary value problems associated to it. Moreover, we use the Dirichlet–to–Robin matrix, also known as response matrix of the network, which contains the boundary information. It is a certain Schur complement of the Schr ¨odinger matrix. The Schur complement plays an important role in matrix analysis, statistics, numerical analysis, and many other areas of mathematics and its applications

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Discrete inverse problem on grids

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Discrete Inverse Problems on GridsCristina Araúz, Ángeles Carmona and Andrés M. EncinasDepartament de Matemàtica Aplicada IIIUniversitat Politècnica de Catalunya. Spain1 Recovering the conductances on a 3–dimensional gridA three dimensional grid is the discretization of any cuboid in R3. Let us take three integers `i ∈ N,i = 1, 2, 3. We define the three dimensional grid with boundary as the network Γ = (V, c) with vertex setV = {xijk : i = 0, . . . , `1 + 1, j = 0, . . . , `2 + 1, k = 0, . . . , `3 + 1}and conductivity function c given byc(xijk, xpqr) > 0 whenp = i± 1, q = j and r = k,p = i, q = j ± 1 and r = k,p = i, q = j and r = k ± 1for all i = 1, . . . , `1, j = 1, . . . , `2 and k = 1, . . . , `3, c(xij0, xij1) > 0, c(xij`3 , xij`3+1) > 0, c(x0jk, x1jk) > 0,c(x`1jk, x`1+1jk) > 0, c(xi0k, xi1k) > 0 and c(xi`2k, xi`2+1k) > 0 for all i = 1, . . . , `1, j = 1, . . . , `2 andk = 1, . . . , `3 and c(x, y) = 0 otherwise. We say that i, j and k are the first, the second and the thirdcomponent of the vertex xijk ∈ V , respectively.We define the following sets of verticesAj = {xijk : i = 1, . . . , `1, k = 1, . . . , `3} , j = 0, . . . , `2 + 1,Rj ={xijk : i = 0, `1 + 1, k = 1, . . . , `3} ∪ {xijk : i = 1, . . . , `1, k = 0, `3 + 1}, j = 1, . . . , `2.If we consider the set F =`2⋃j=1Aj , then its boundary is given by δ(F ) = A ∪R ∪B, whereA = A0, R =`2⋃i=1Ri and B = A`2+1.See Figure 1 for an illustration of a three dimensional grid with `1 = `3 = 2 and `2 = 3, and theassociated boundary sets A, B and R1, . . . , R`2 .Given an index i ∈ {1, . . . , a}, we consider the partial layers of verticesDi ={xijk ∈ F¯ : k = a+ 1− i, . . . , `, j = 1, . . . , p},Da+1 ={xa+1 jk ∈ F¯ : k = 1, . . . , `, j = 1, . . . , p}and D0 ={x0jk ∈ F¯ : k = a+ 1, . . . , `, j = 1, . . . , p}. Inparticular, Da+1, D0 ⊂ R.The recovery of conductances on a 3 dimensional grid is an iterative process, for we are not able to giveexplicit formulae for all the conductances at the same time but we can give a recovery algorithm instead.Hence, we describe the algorithm in steps, each of them requiring the information obtained in the last one.12To start with, let Nq be an irreducible and symmetric M–matrix of order n = 2(`1`2 + `1`3 + `2`3)satisfying that is a response matrix. Let λ ≥ 0 be the lowest eigenvalue of Nq and σ ∈ Ω(δ(F )) the eigenvectorassociated with λ. In addition, we choose ω ∈ Ω(F¯ ) such that ω = kσ on δ(F ) with 0 < k < 1.Step 0In this step we do not recover any conductance. However, we set the necessary tools to obtain them in futuresteps. Having fixed two indices r ∈ {1, . . . , p} and s ∈ {1, . . . , `}, we consider the overdetermined partialDirichlet–Neumann boundary value problem that consists in finding urs ∈ C(F¯ ) such thatLq(urs) = 0 on F, urs = εxa+1 rs on A ∪R and∂urs∂nF= 0 on A. (1)There exists a large set of vertices of the 3 dimensional grid Γ where urs = 0. We denote this set byZ(urs) ={x ∈ F¯ : urs(x) = 0}= supp(urs)c ⊆ F¯ .Clearly, A ⊆ Z(urs). The size of Z(urs), however, is much bigger than the size of A.Proposition 1.1. Fixed j = 1, . . . , `2 and xijk ∈ Rj, it is verified that for any x ∈ As, 0 ≤ s ≤ j, and forany x ∈ Aj+r \Nr(xij+rk), r = 1, . . . , `2 − j + 1P˜q(x, xijk) = 0.Moreover, fixed xi0k ∈ A, it is verified that for any s and x ∈ As \Ns−1(xisk), s = 1, . . . , `2 + 1P˜q(x, xi0k) = 0.In particular, P˜q(As;Rj) = 0 for any j = 1, . . . , `2 and 0 ≤ s ≤ j and P˜q(A;A) = P˜q(A1;A) = I.Proposition 1.2. The following equality is satisfied:Z(urs) ={xijk ∈ F¯ : i = 1, . . . , a, k = 1, . . . , `, j = 1, . . . , r −max{0, i− a + k − s}, r + max{0, i− a + k − s}, . . . , p}.Step 1Let us fix the indices r ∈ {1, . . . , p} and s ∈ {1, . . . , `} for this step and let us consider the unique solutionurs ∈ C(F¯ ) of problem (2). We already know that urs = 0 on A ∪ (R r {xa+1 rs}) and u(xa+1 rs) = 1.Moreover, the values of urs on B are given by the matricial equationursB = −Nq(A;B)−1 · Nq(A; xa+1 rs).Notice that this means that all the values of urs on B are known, for the Dirichlet–to–Robin map isknown. In consequence, urs is known on all the boundary δ(F ). Let us define the vector uB = ursB for thesake of the simplicity of the notation. In Figure 1(b) we show all the information obtained at the end of thisstep.Step 2In this step we recover the conductances of all the boundary spikes by means of the boundary spike formula.However, we first need to determine some values of the modified Green matrix of a 3 dimensional grid.Lemma 1.3. For all j = 1, . . . , p and k = 1, . . . , `, the value G˜q(xajk, xajk) is always zero.3Now we are ready to determine the conductances c(x0j , x1j) for all j = 1, . . . , `, using the boundaryspike formula.Corollary 1.4. The conductances of the edges joining the vertices of Da+1 with the vertices of Da are givenbyc(xa+1 jk, xajk) =ω(xa+1 jk)ω(xajk)(Nq(xa+1 jk; xa+1 jk)− Nq(xa+1 jk;B) · Nq(A;B)−1 · Nq(A; xa+1 jk)− λ)for all j = 1, . . . , p and k = 1, . . . , `.Step 3Again, let us fix the indices r ∈ {1, . . . , p} and s ∈ {1, . . . , `} in this step and let us consider the uniquesolution urs ∈ C(F¯ ) of problem (2). Then, we know all the values of urs on Da, as the following result shows.Lemma 1.5. The values of urs on Da are given byurs(xajk) =1c(xa+1 jk, xajk)(λu(xa+1 jk)− Nq(xa+1 jk; xa+1 rs)− Nq(xa+1 jk;B) · uB)+ω(xajk)ω(xa+1 jk)urs(xa+1 jk)for all j = 1, . . . , p and k = 1, . . . , `.Step 4Here we find the conductances of all the edges with both ends in Da and such that the third compenent oftheir ends is different. However, we state a more general result.Proposition 1.6. Let i ∈ {0, . . . , a − 1}. For every r = 1, . . . , p and s = 1, . . . , `, let us suppose that weknow the values of urs on Di+2 and Di+1. Also, we suppose that the conductances of all the edges joiningvertices from Di+2 and Di+1 are known. Now we fix the indices r = 1, . . . , p and s = 1, . . . , `. Then, theconductances c(xi+1 r s+a−i−1, xi+1 r s+a−i) are also known. They are given byc(xi+1 r s+a−i−1, xi+1 r s+a−i) = −urs(xi+2 r s+a−i−1)urs(xi+1 r s+a−i)c(xi+1 r s+a−i−1, xi+2 r s+a−i−1).When i = a−1, Proposition 1.6 shows that c(xars, xar s+1) is known for all r = 1, . . . , p and s = 1, . . . , `.See Figure 1(e) in order to see all the known information at the end of this step.Step 5In this step we give the conductances of all the edges with both ends in Da that are still unknown. Further-more, we state a more general result.Proposition 1.7. Let i ∈ {0, . . . , a − 1}. For every r = 1, . . . , p and s = 1, . . . , `, let us suppose that weknow the values of urs on Di+2 and Di+1. Also, let us assume that we know the conductances of all theedges joining vertices from Di+2 and Di+1, and the ones of the edges with both ends in Di+1 and such thatthe ends have different third component. Now we fix the indices r = 1, . . . , p and s = 1, . . . , `. Then, theconductances c(xi+1 r+1 s+a−i, xi+1 r s+a−i) are also known. They are given byc(xi+1 r+1 s+a−i, xi+1 r s+a−i) = −urs(xi+1 r+1 s+a−i+1)urs(xi+1 r s+a−i)c(xi+1 r+1 s+a−i, xi+1 r+1 s+a−i+1).When i = a − 1, Proposition 1.7 shows that c(xa r+1 s+1, xar s+1) is known for all r = 1, . . . , p ands = 1, . . . , `. See Figure 1(f) in order to see all the information gathered at the end of this step.4Step 6Let us define the linear operator ℘ : C(F¯ ) −→ C(F ) given by the values℘(v)(xijk) = c(xijk, xij k−1)v(xij k−1) + c(xijk, xij k+1)v(xij k+1) + c(xijk, xi j−1 k)v(xi j−1 k)+ c(xijk, xi j+1 k)v(xi j+1 k) + c(xijk, xi+1 jk)v(xi+1 jk)for all v ∈ C(F¯ ) and xijk ∈ F . This operator will be useful in this and also in the following steps.In this step we give the conductances of all the edges joining the vertices from Da and Da−1. SeeFigure 1(g) in order to see all the information obtained at the end of this step.Step 7In this step we are able to obtain the unknown values of urs on Da−1 for all r = 1, . . . , p and s = 1, . . . , `. Infact, let us state a more general result.Proposition 1.8. Let i ∈ {0, . . . , a − 1}. For every r = 1, . . . , p and s = 1, . . . , `, let us suppose that weknow the values of urs on Di+2 and Di+1. Also, let us suppose that we know the conductances of all theedges joining vertices from Di+2 and Di+1, from Di+1 and Di and the ones of the edges with both ends inDi+1. Now fix the indices r = 1, . . . , p and s = 1, . . . , `. Then, the values of urs on Di are also known. Theyare given byurs(xijk) =℘(urs)(xi+1 jk)c(xi+1 jk, xijk)− ℘(ω)(xi+1 jk)ω(xi+1 jk)c(xi+1 jk, xijk)urs(xi+1 jk)− ω(xijk)ω(xi+1 jk)urs(xi+1 jk)for all j = 1, . . . , ` and k = r + a− i, . . . , `.Proof. Fixed three indices i ∈ {0, . . . , a − 1}, r = 1, . . . , p and s = 1, . . . , `, let j ∈ {1, . . . , p} and k ∈{r + a− i, . . . , `}. Observe that ℘(ω)(xi+1 jk) and ℘(urs)(xi+1 jk) are already known. Then,0 = Lq(urs)(xi+1 jk) = urs(xi+1 jk)ω(xi+1 jk)℘(ω)(xi+1 jk)− ℘(urs)(xi+1 jk)− c(xi+1 jk, xijk)urs(xijk) + ω(xijk)ω(xi+1 jk)c(xi+1 jk, xijk)urs(xi+1 jk)and hence urs(xijk) is the unique unknown term of this equality.In particular, when i = a − 1, Proposition 1.8 shows that urs is known on Da−1 for all r = 1, . . . , pand s = 1, . . . , `. Observe that we already knew some of the values of urs on Da−1, which are those of thevertices in Z(urs). Figure 1(h) shows the information obtained until this step.Step 8 and beyondWe keep repeating the same process to obtain more conductances, that is, we keep applying Proposition 1.6from Step 4, then Proposition 1.7 from Step 5, ?? from Step 6 and then Proposition 1.8 from Step 7 foreach i = a− 2, . . . , 1. We stop when we have obtained all the conductances between and all the vertices in⋃ai=1Di, see Figure 1(j).The final step left is to rotate the grid (see Figure1(k)), that is, to consider the new boundary setsA ={xij `+1 ∈ F¯ : i = 1, . . . , a, j = 1, . . . , p}and B ={xij0 ∈ F¯ : i = 1, . . . , a, j = 1, . . . , p}instead of theprevious ones, and consider now the overdetermined partial Dirichlet–Neumann boundary value problemLq(urs) = 0 on F, urs = εx0rs on A ∪R and∂urs∂nF= 0 on A. (2)5for r ∈ {1, . . . , p} and s ∈ {1, . . . , `}. By proceeding analogously to the last steps, we obtain the lackingconductances of the 3 dimensional grid, see Figure 1(l).(a) (b) (c) (d)(e) (f) (g) (h)(i) (j) (k) (l)Figure 1: The bold items are the ones known at the end of each step for the case r = s = 1.References[1] G. Alessandrini. Stable determination of conductivity by boundary measurements. Appl. Anal., 27: 153–172,1998.[2] V. Anandam. Harmonic functions and potentials on finite or infinite networks. Lecture Notes of the UnioneMatematica Italiana, 12, Springer-Verlag, 2011.[3] E. Bendito, A. Carmona, A. M. Encinas: Potential Theory for Schrödinger operators on finite networks.Rev. Mat. Iberoamericana 21 (2005), 771–818.[4] E. Bendito, A. Carmona, A. M. Encinas, J.M. Gesto: Potential Theory for boundary value problems onfinite networks. Appl. Anal. Discrete Math. 1 (2007), 299–310.[5] A. Carmona, A. M. Encinas and M. Mitjana. On the Moore–Penrose inverse of singular, symmetric andperiodic Jacobi M-matrices. SIAM 2012, personal communication.[6] E. B. Curtis, D. Ingerman, J. A. Morrow: Circular planar graphs and resistor networks. Linear AlgebraAppl. 283 (1998), 115–150.[7] E. Curtis and J. Morrow. Inverse Problems for Electrical Networks. Series on Applied Mathematics, 13,2000.[8] J. Friedman and J. P. Tillich. Wave equations for graphs and the edge–based laplacian. Pacific Journal ofMathematics, 216 (2004), 229–266.[9] A. I. Nachman. Reconstructions from boundary measurements. Annals of Math., 128: 531–587, 1988.

Discrete inverse problem on grids

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