Eigenstructure of nonselfadjoint complex discrete vector Sturm-Liouville problems
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We present a study of complex discrete vector Sturm-Liouville problems, where coefficients of the difference equation are complex numbers and the strongly coupled boundary conditions are nonselfadjoint. Moreover, eigenstructure, orthogonality, and eigenfunctions expansion are studied. Finally, an example is given. 1. Introduction and motivation Consider the parabolic coupled partial differential system with coupled boundary value conditions ut (x,t) − Auxx (x,t) = 0,
0 < x < 1, t > 0,
(1.1)
A1 u(0,t) + B1 ux (0,t) = 0,
t > 0,
(1.2)
A2 u(1,t) + B2 ux (1,t) = 0,
t > 0,
(1.3)
u(x,0) = F(x),
0 ≤ x ≤ 1,
(1.4)
where u = (u1 ,u2 ,...,um )T , F(x) are vectors in Cm , and A,A1 ,A2 ,B1 ,B2 ∈ Cm×m . We divide the domain [0,1] × [0, ∞[ into equal rectangles of sides ∆x = h and ∆t = l, introduce coordinates of a typical mesh point p = (kh, jl) and represent u(kh, jl) = U(k, j). Approximating the partial derivatives appearing in (1.1) by the forward difference approximations U(k, j + 1) − U(k, j) , l U(k + 1, j) − U(k, j) Ux (k, j) ≈ , h U(k + 1, j) − 2U(k, j) + U(k − 1, j) Uxx (k, j) ≈ , h2 Ut (k, j) ≈
Copyright © 2005 Hindawi Publishing Corporation Advances in Difference Equations 2005:1 (2005) 15–29 DOI: 10.1155/ADE.2005.15
(1.5)
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Nonselfadjoint discrete vector Sturm-Liouville problems
(1.1) takes the form U(k, j + 1) − U(k, j) U(k + 1, j) − 2U(k, j) + U(k − 1, j) =A , l h2
(1.6)
where h = 1/N, 1 ≤ k ≤ N − 1, j ≥ 0. Let r = l/h2 and we can write the last equation in the form
rA U(k + 1, j) + U(k − 1, j) + (I − 2rA)U(k, j) − U(k, j + 1) = 0,
1 ≤ k ≤ N − 1, j ≥ 0, (1.7)
where I is the identity matrix in Cm×m . Boundary and initial conditions (1.2)–(1.4) take the form
A1 U(0, j) + NB1 U(1, j) − U(0, j) = 0,
j ≥ 0,
A2 U(N, j) + NB2 U(N, j) − U(N − 1, j) = 0, U(k,0) = F(kh),
j ≥ 0,
0 ≤ k ≤ N.
(1.8) (1.9) (1.10)
Once we discretized problem (1.1)–(1.4), we seek solutions of the boundary problem (1.7)–(1.9) of the form (separation of variables) U(k, j) = G( j)H(k),
G( j) ∈ Cm×m , H(k) ∈ Cm .
(1.11)
Substituting U(k, j) given by (1.11) in expression (1.7), one gets
rAG( j) H(k + 1) + H(k − 1) + (I − 2rA)G( j)H(k) − G( j + 1)H(k) = 0.
(1.12)
Let ρ be a real number and note that (1.12) is equivalent to
rAG( j) H(k + 1) + H(k − 1) + G( j)H(k) − 2rAG( j)H(k) + ρAG( j)H(k) − ρAG( j)H(k) −G( j + 1)H(k) = 0,
(1.13)
or
rAG( j) H(k + 1) + − 2 −
ρ H(k) + H(k − 1) + (I + ρA)G( j) − G( j + 1) H(k) = 0. r (1.14)
Note that (1.14) is satisfied if sequences {G( j)}, {H(k)} satisfy G( j + 1) − (I + ρA)G( j) = 0, j ≥ 0, ρ H(k + 1) + − 2 − H(k) + H(k − 1) = 0, 1 ≤ k ≤ N − 1. r
(1.15) (1.16)
The solution of (1.15) is given by G( j) = (I + ρA) j ,
j ≥ 0.
(1.17)
´ R. J. Villanueva and L. Jodar
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Now, we deal with boundary conditions (1.8)-(1.9). Using (1.11), we can transform them into
NB1 G( j)H(1) + A1 − NB1 G( j)H(0) = 0,
j ≥ 0,
A2 + NB2 G( j)H(N) − NB2 G( j)H(N − 1) = 0,
j ≥ 0.
(1.18)
By the Cayley-Hamilton theorem [7, page 206], if q is the
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