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N 3, and β positive or negative corresponds to the defocusing or focusing NLS. Two≤ important invariants of GPE are the normalization of the wave function
N(ψ)= ψ(x, t) 2dx =1, t 0, (2) Ω | | ≥ and the energy R 1 β E(ψ(x, t)) = ψ 2 + V (x) ψ 2 + ψ 4 dx = E(ψ(x, 0)). (3) 2|∇ | | | 2 | | ZΩ To find stationary solution of (1), we substitute the formula ψ(x, t)= e−iλtφ(x) into (1) and (2) and obtain the time-independent Schr¨odinger equation with Dirichlet boundary condition and the normalized condition 1 λφ(x)= ∆φ(x)+ V (x)φ(x)+ βφ3(x), x Ω, (4) −2 ∈ φ(x)=0, x ∂Ω, (5) ∈ φ(x) 2dx =1, (6) | | ZΩ where λ is the chemical potential of the condensate and φ(x) is a real function independent of t [3]. (4)-(6) is a nonlinear eigenvalue problem. The eigenfunc- tion corresponding to the minimum energy is called ground state and other eigenfunctions corresponding to larger energy are called excited states in the literature. There have been many theoretical studies as well as numerical studies for the time-independent Schr¨odinger equation. Bao and Cai [4] pointed out when β > 0, the positive ground state is unique, and if V (x) is radially symmetric in 2D, the positive ground state must be radially symmetric. Bao and Tang [1] proposed methods by directly minimizing the energy functional via finite element approximation to obtain the ground state and by continuation method to obtain excited states. Edwards and Burnett [5] presented a Runge-Kutta type method and employed it to solve the spherically symmetric time-independent GPE. Adhikari [6] used this approach to get the ground state solution of GPE in 2D with radial symmetry. Chang and Chien [7] and Chang, Chien and Jeng [8] investigated stationary state solutions of (4) using numerical continuation method, where λ was treated as a continuation parameter. The solution curves branching from the first few bifurcation points of (4) were numerically traced using continuation method under the normalization condition (6). Since nonlinearity rather than discretization method is our main concern and finite difference discretization will lead to a simpler nonlinear structure, finite difference discretization is adopted in this paper. The finite difference dis- cretization of (4)-(6) is the following eigenvector-dependent nonlinear eigenvalue problem, Dϕ + βϕ3 = λϕ, (7) hϕTϕ 1=0, − 1 where D = 2 D1 +V , D1 is the coefficient matrix corresponding to ∆, V is the diagonal matrix corresponding to the potential V (x), λ and ϕ are the− unknowns, and h is a constant related to mesh size. ϕ3 represents the vector with elements being the corresponding elements of ϕ to the power 3. This convention will be used throughout this paper. HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 3
With respect to the theoretical aspects of eigenvector-dependent nonlinear eigenvalue problem, [9] and [10] studied the following general nonlinear eigen- value problem Ax + F (x)= λx, (8) where A is an n n irreducible Stieltjes matrix, i.e, an irreducible symmetric pos- × T itive definite matrix with off-diagonal entries nonpositive, F (x) = (f1(x1),...,fn(xn)) T and x = (x , , xn) . The functions fi(xi) are assumed to have the property 1 ··· that fi(xi) > 0, when xi > 0, i =1,...,n. It is shown that under certain condi- tions on F (x), there exists a positive eigenvector x(λ) if and only if λ>µ, where µ is the smallest eigenvalue of A, and for every λ>µ, the positive eigenvector is unique. Moreover, such a solution is a monotone increasing function of λ. The most popular numerical method to the eigenvector-dependent nonlinear eigen- value problems is the self-consistent field (SCF) iteration, which is suitable for computing the ground state; for instance, see [11, 12] and the references therein. In [13], inverse iteration method was applied to solve eigenvector-dependent non- linear eigenvalue problems. Most of the above papers concentrate on the ground state and the first excited state. As far as we know, there are only a few numer- ical works on other excited states, such as [1, 14, 15, 16]. The main purpose of this paper is to design algorithms for computing excited states of high energy. Homotopy method is one of the effective methods for solving eigenvalue problems. A great advantage of the homotopy method is that it is to a large degree parallel, in the sense that each eigenpath is traced independently of the others. There are several works on homotopy methods for linear eigenvalue problems. Remarkable numerical results have been obtained by using homotopy algorithm on eigenvalue problems of tridiagonal symmetric matrices [17, 18]. Solving eigenvalue problems of real nonsymmetric matrices with real homotopy was developed in [19, 20]. The homotopy method is also used to solve the generalized eigenvalue problem [21]. For eigenvalue-dependent nonlinear eigen- problems such as λ-matrix problems, a homotopy was given by Chu, Li and Sauer [22]. The major part of this paper is the construction of homotopy for comput- ing many eigenpairs of the eigenvector-dependent nonlinear eigen-problem. Key issues encountered in constructing the homotopy are the selection of the homo- topy parameter and that of an appropriate initial eigenvalue problem so that the homotopy paths determined by the homotopy equation are regular and the numerical work in following these paths is at reasonable cost. The parameter β in the original problem seems to be a natural choice for the homotopy pa- rameter. However, it seems difficult to prove that 0 is a regular value for such homotopy. In fact, 0 is probably not a regular value of the natural homotopy with parameter β. Instead, an artificial parameter t is chosen as the homotopy parameter to connect a constructed initial eigenvalue problem and the target one. As for the selection of an initial eigenvalue problem, random matrix with certain sparse structure is designed, which guarantees that 0 is a regular value of the homopoty with probability one and which renders the initial problem and the target problem possess similar structures. The rest of this paper is organized as follows. In Section 2, the time- independent GPE Dirichlet problem (4)-(6) is discretized by finite difference method and existence of certain types of solution of the discretized problems is derived. In Section 3, homotopies for (7) are constructed with Ω R and ⊂ HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 4
Ω R2 respectively, and regularity and boundedness of the homotopy paths are proved.⊂ In Section 4, numerical results are presented to verify the theoretical results derived for Ω R and Ω R2 respectively. Conclusions are drawn in the last section. ⊂ ⊂
2 Discretizations of the nonlinear eigenvalue prob- lem 2.1 Finite difference discretizations For one dimensional problem (4)-(6) with Ω = [a,b] R, the grid points are N+ b⊂−a xj = a + jh, j =0,...,n +1, where n and h = n+1 is the mesh size. The finite difference discretization of the differential∈ equation and a simple quadra- ture of the normalization condition lead to the following system of algebraic equations, Dϕ + βϕ3 λϕ =0, (9) 1 1 ϕT−ϕ =0, 2 h − T where ϕ = (ϕ1, , ϕn) , ϕj are the approximations of φ(xj ), vj = V (xj ), ··· 1 j =1,...,n, and D = 2 D1 + V with
2 1 − v .. 1 1 1 2 . . D1 = − , V = . . (10) h2 . . . .. .. 1 vn 1− 2 − The discretization of the normalized condition is rewritten so that the Jacobian matrix of the nonlinear mapping with respect to (ϕ, λ) is symmetric, as will be seen below. For two dimensional problem (4)-(6) with Ω = [a,b] [c, d], the domain is b−×a divided into a (m+1) (n+1) mesh with step size h1 = m+1 in x-direction, h2 = d−c × n+1 in y-direction. The grid points (xi,yj) are xi = a + ih1, i =0,...,m + 1, and yj = c + jh2, j =0,...,n + 1. Using central difference, we get
Dϕ + βϕ3 λϕ =0, − (11) 1 1 ϕTϕ =0, 2 h1h2 − T where ϕ = (ϕ , , ϕ n, , ϕm , , ϕmn) , ϕij are the approximations of 11 ··· 1 ··· 1 ··· φ(xi,yj), vij = V (xi,yj), i =1,...,m, j =1,...,n and D is a block tridiagonal matrix
D11 D12 V . 1 1 D D .. V2 D = 21 22 + . , 2 ...... D − m 1,m V D − D m m,m 1 mm HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 5
where 2 2 1 h2 + h2 h2 1 2 − 2 . 1 2 2 .. h2 h2 + h2 n×n Dii = − 2 1 2 R , (12) .. .. 1 ∈ . . 2 h2 1 2− 2 h2 h2 + h2 − 2 1 2
1 2 h1 − 1 h2 − 1 n×n Di− ,i = R ,Di− ,i = Di,i− , (13) 1 .. ∈ 1 1 . 1 h2 − 1
vi1 vi2 Vi = . , i =1, . . . , m. (14) .. v in Remark 2.1 For both one and two dimensional cases, D is an irreducible sym- metric diagonal dominant matrix and the diagonal entries of D are all positive. Therefore D is positive definite.
2.2 Existence of certain types of solution In this subsection, we will study the existence of solution for the discretized nonlinear eigenvalue problem. From [4], we know for (4)-(6), when β > 0, the positive ground state is unique, and if V (x) is radially symmetric in 2D, the positive ground state must be radially symmetric. We will prove the existence of positive solution and the existence of antisymmetric solution for discretized nonlinear eigenvalue problem. For convenient reading, two underlying theorems from [9] are quoted as underlying lemmas. Lemma 2.2 ([9]) Let A be an irreducible Stieltjes matrix and µ be the smallest positive eigenvalue of A. Let λ>µ and let
f1(x1) . F (x)= . , (15) f (x ) n n 1 where for i = 1,...,n, fi(x) : [0, ) [0, ) are C functions satisfying the conditions: ∞ → ∞ f (t) f (t) lim i =0, lim i = . (16) t→0 t t→∞ t ∞ Then Ax + F (x)= λx has a positive solution. If, in addition, for i =1,...,n, f (s) f (t) i < i (17) s t whenever 0 Lemma 2.3 ([9]) Let the conditions (16) and (17) of Lemma 2.2 be satisfied and let x(λ) denote the unique positive eigenvector corresponding to λ (µ, ). Then: ∈ ∞ (i) x(λ ) < x(λ ), if µ < λ < λ < ; 1 2 1 2 ∞ (ii) x(λ) is continuous on (µ, ); ∞ (iii) lim xi(λ)= , i =1, ,n; λ→∞ ∞ ··· (iv) lim xi(λ)=0, i =1, ,n. λ→µ+ ··· Remark 2.4 Lemma 2.3 indicates for any given normalization r > 0, there exist a unique λ>µ and unique positive x(λ) such that x(λ) = r. || || Theorem 2.5 If β > 0, there exist unique positive eigenvectors for problem (9) and (11) respectively. Proof. From Remark 2.1, we know D in both (9) and (11) is an irreducible Stieltjes matrix. In addition it can be verified that βϕ3 in both (9) and (11) satisfies the conditions of F (x) in Lemma 2.2. From Remark 2.4, the claim is proved. Theorem 2.6 Let β > 0 and Ω = [ a,a] for problem (9). − (i) When n is odd and the grid points xi, i =1,...,n, satisfy x1 = xn, x2 = xn−1, , x n−1 = x n+3 , x n+1 =0, − − ··· 2 − 2 2 T there exists a unique solution ϕ = (ϕ , ϕ , , ϕn) with ϕ = ϕn, 1 2 ··· 1 − ϕ2 = ϕn−1, , ϕ n−1 = ϕ n+3 , ϕ n+1 =0, ϕj > 0, j =1,...,n. − ··· 2 − 2 2 (ii) When n is even and the grid points xi, i =1,...,n, satisfy x1 = xn, x2 = xn−1, , x n = x n +1, − − ··· 2 − 2 T there exists a unique solution ϕ = (ϕ , ϕ , , ϕn) with ϕ = ϕn, 1 2 ··· 1 − ϕ2 = ϕn−1, , ϕ n = ϕ n +1, ϕj > 0, j =1,...,n. − ··· 2 − 2 Proof. (i). When n is odd, consider the following equations D ϕ + βϕ3 = λϕ, 2 (18) ϕTϕ 1 =0, − 2h where 1 + 1 x2 1 h2 2 1 − 2h2 1 1 1 2 .. h2 h2 + x2 . n−1 × n−1 2 2 R 2 2 D2 = − . . . (19) .. .. 1 ∈ 2h2 1 1 − 1 2 2h2 h2 + 2 x n−1 − 2 HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 7 3 Note that D2 is an irreducible Stieltjes matrix and βϕ satisfies the condi- tions of F (x) in Lemma 2.2. Therefore there exists a unique positive solu- T tion ϕ = (ϕ1, ϕ2,...,ϕ n−1 ) for (18). Due to the relations x1 = xn, x2 = 2 − xn−1, , x n−1 = x n+3 , x n+1 = 0, set ϕn = ϕ1, ϕn−1 = ϕ2, , 2 2 2 − ··· − T− − ··· ϕ n+3 = ϕ n−1 , ϕ n+1 = 0. Then ϕ = (ϕ1, ϕ2,...,ϕn) is a solution of (9). 2 − 2 2 (ii). When n is even, consider the following equations D ϕ + βϕ3 = λϕ, 2 (20) ϕTϕ 1 =0, − 2h where 1 + 1 x2 1 h2 2 1 − 2h2 1 1 1 2 .. 2 2 + x2 . n × n 2h h 2 R 2 2 D2 = − . . . (21) .. .. 1 ∈ 2h2 1 3 − 1 2 2 2 + x n − 2h 2h 2 2 T Similarly there exists a unique positive solution ϕ = (ϕ , ϕ ,...,ϕ n ) for (20). 1 2 2 Due to the relations x = x , x = x − , , x n = x n , set ϕ = ϕ , 1 n 2 n 1 2 2 +1 n 1 − − ··· − T − ϕ − = ϕ , , ϕ n = ϕ n . Then ϕ = (ϕ , ϕ ,...,ϕ ) is a solution of n 1 2 2 +1 2 1 2 n (9). − ··· − 3 Homotopy methods In this section, in order to compute many eigenpairs, we construct homotopy equations for 1D discretized problem (9) and 2D discretized problem (11) re- spectively. We shall prove the regularity and boundedness of the homotopy paths. The regularity of homotopy paths can be usually obtained by random perturbations of appropriate parameters, so the most important feature of our construction is the choice of appropriate parameters. In addition, if the initial matrix can be chosen as close to the matrix D as possible, then most of the homotopy paths are close to straight lines and will be easy to follow [19]. 3.1 One dimensional case For (9), a homotopy H : Rn R [0, 1] Rn R is defined as × × → × (1 t)A(K)ϕ + Dϕ + tβϕ3 λϕ H(ϕ, λ, t)= =0, (22) − 1 1 ϕTϕ − 2 h − T where A(K) = diag(a1, ,an) is a random diagonal matrix with K = (a1, ,an) Rn. At t = 0, H(ϕ, λ, 0)··· corresponds to the linear eigenvalue problem ··· ∈ A(K)ϕ + Dϕ λϕ H(ϕ, λ, 0) = =0, (23) 1 1 ϕTϕ− 2 h − while at t = 1, H(ϕ, λ, 1) = 0 corresponds to the problem (9). Assuming that (i) the eigenpairs of H(ϕ, λ, 0) = 0 are (ϕ , λi), i = 1,...,n, λ1 λn, (i) ≤ ··· ≤ we shall use these n points (ϕ , λi, 0) as our initial points when tracing the homotopy curves leading to the desired solutions of H(ϕ, λ, 1) = 0. HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 8 The choice of the initial matrix A(K)+ D provides some advantages. First, it makes sure that t [0, 1), what we need to solve is a sparse nonlinear eigenvalue problem with∀ ∈ constraint. Second, since K Rn, t [0, 1) and ϕ Rn, (1 t)A(K)+ D + tβdiag(ϕ2) is a real symmetric∀ ∈ matrix,∀ ∈ the solution curves∈ starting− from the initial points at t = 0 are real. Finally, since A(K)+ D is a tridiagonal matrix with all the subdiagonal and supdiagonal entries nonzero, all the eigenvalues of A(K)+ D are simple and the Jacobian matrix of H at (ϕ , λ , 0) is nonsingular for (ϕ , λ ) such that H(ϕ , λ , 0) = 0. Thus locally 0 0 ∀ 0 0 0 0 a unique curve around (ϕ0, λ0, 0) is guaranteed. The effectiveness of the homotopy is based on the following Parametrized Sard’s Theorem. Theorem 3.1 (Parametrized Sard’s Theorem) Let f : M P Rm Rq Rn be a Ck mapping with k > max(0,m n), where M and P×are⊂ open× sets in→Rm and Rq respectively. If y is a regular− value of f, then y is also a regular value of f( ,p) for almost all p P . · ∈ In the rest of this paper, we will denote the i-th row of a matrix M as M(i, :) and the j-th column of M as M(:, j). If I1 is a row index set, M(I1, :) will be the submatrix formed by the I1 rows of M. M(i : end, :) will be the submatrix formed by the rows from the i-th row to the last. Similarly M(:,J1) will be the submatrix formed by the J1 columns of M. If I1 and I2 are two index sets, [M(I1, :); M(I2, :)] denotes the submatrix formed by the I1 rows and the I2 rows of M. In Theorem 3.2, we prove regularity and boundedness of the homotopy paths determined by the homotopy equation (22). Theorem 3.2 For the homotopy H : Rn R [0, 1) Rn R in (22), for almost all K Rn, × × → × ∈ (i) 0 is a regular value of H and therefore the homotopy curves corresponding to different initial points do not intersect each other for t [0, 1); ∈ (ii) Every homotopy path (ϕ(s), λ(s),t(s)) H−1(0) is bounded. ⊂ Proof. (i) Define a mapping H˜ : Rn R [0, 1] Rn Rn R related to H as follows × × × → × (1 t)A(K)ϕ + Dϕ + tβϕ3 λϕ H˜ (ϕ,λ,t,K)= =0, (24) − 1 1 ϕTϕ − 2 h − ˜ ˜ ∂H˜ such that H(ϕ, λ, t)= H(ϕ,λ,t,K). The Jacobian matrix of H, ∂(ϕ,λ,t,K) , is (1 t)A(K)+ D +3tβdiag(ϕ2) λI ϕ βϕ3 A(K)ϕ (1 t)diag(ϕ) . − ϕT − −00− − 0 − Divide 1, ,n into two parts C0 and C∗, where C0 denotes the indices i { ··· } ∗ 0 such that ϕi = 0 and C denotes the indices i such that ϕi = 0. Since i C , both the columns 3tβdiag(ϕ2)(:,i) and tβdiag(ϕ2)(:,i) are equal6 to zero,∀ it∈ holds that, (1 t)A(K)+ D +3tβdiag(ϕ2) λI (:, C0) − − = (1 t)A(K)+ D + tβdiag(ϕ2) λI (:, C0). (25) − − HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 9 For the columns in C∗, the diagonal matrix diag(ϕ) is nonzero. By elementary ∂H˜ column transformations, the Jacobian matrix ∂(ϕ,λ,t,K) is transformed to the following, (1 t)A(K)+ D + tβdiag(ϕ2) λI ϕ βϕ3 A(K)ϕ (1 t)diag(ϕ) . − ϕT − −00− − 0 − Define (1 t)A(K)+ D + tβdiag(ϕ2) λI ϕ F = . (26) 1 − ϕT − −0 − For K Rn, (ϕ, λ, t) Rn R [0, 1) satisfying (24), since the subdiagonals and∀ supdiagonals∈ ∀ of the∈ matrix× (1× t)A(K)+D +tβdiag(ϕ2) are nonzero, λ is a − 2 simple eigenvalue of (1 t)A(K)+ D + tβdiag(ϕ ). Therefore F1 is nonsingular ∂H˜ − ˜ and ∂(ϕ,λ,t,K) is row full rank. As a result, 0 is a regular value of the mapping H. From Theorem 3.1, for almost all K Rn, 0 is a regular value of the restricted mapping H˜ ( , , ,K), i.e., H. ∈ (ii) From· (22),· · for fixed K Rn, (ϕ, λ, t) Rn R [0, 1) satisfying H(ϕ, λ, t)=0, ϕ = 1/√h and ∈λ = h ∀(1 t)ϕT∈A(K)ϕ×+ ϕ×TDϕ + tβϕTϕ3 . Then k k − λ = h (1 t)ϕTA(K)ϕ + ϕTDϕ + tβϕTϕ3 | | | − | tβ (1 t)ρ(A(K)) + ρ(D)+ ≤ − h β ρ(A(K)) + ρ(D)+ , ≤ h where ρ(M) denotes the spectral radius of M. 3.2 Two dimensional case 3.2.1 The homotopy with random tridiagonal matrix For two dimensional case, the homotopy H : Rmn R [0, 1] Rmn R is constructed as follows × × → × (1 t)A(K)ϕ + Dϕ + tβϕ3 λϕ H(ϕ, λ, t)= − − =0, (27) 1 1 ϕTϕ 2 h1h2 − ! where A(K) Rmn×mn is a random block diagonal matrix with tridiagonal blocks, namely,∈ (i) (i) a11 a12 (i) (i) (i) A1 a12 a22 a23 A2 (i) (i) .. A(K)= . with Ai = a23 a33 . , i =1, . . . , m, .. . . (i) Am .. .. an−1,n (i) (i) a − ann n 1,n HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 10 T (i) (i) (i) (i) (i) (i) (i) and K = (K1,K2,...,Km) with Ki = a11 ,a12 ,a22 ,a23 ,...,an−1,n−1,an−1,n,ann . H(ϕ, λ, 0) corresponds to the linear eigenvalue problem A(K)ϕ + Dϕ λϕ H(ϕ, λ, 0) = − =0, 1 1 ϕTϕ 2 h1h2 − ! while H(ϕ, λ, 1) corresponds to the problem (11). In order to show the effectiveness of this homotopy H by the Parametrized Sard’s Theorem, define a mapping H˜ : Rmn R [0, 1] R(2nm−m) Rmn R related to H as follows × × × → × (1 t)A(K)ϕ + Dϕ + tβϕ3 λϕ H˜ (ϕ,λ,t,K)= − − =0, (28) 1 1 ϕTϕ 2 h1h2 − ! ˜ ˜ ∂H˜ such that H(ϕ, λ, t)= H(ϕ,λ,t,K). The Jacobian matrix of H, ∂(ϕ,λ,t,K) , is 2 3 (1 t)A(K)+ D +3tβdiag(ϕ ) λI ϕ βϕ A(K)ϕ (1 t)B , − ϕT − −0− 0− 0 − (29) ∂(A(K)ϕ) R(mn)×(2nm−m) T where B = ∂K . Denote ϕi = (ϕi1, ϕi2,...,ϕin) . It can be verified that ∈ B1 B2 B = . (30) .. B m with ϕi1 ϕi2 ϕi1 ϕi2 ϕi3 ∂(Aiϕi) ϕi2 ϕi3 ϕi4 Bi = = . . ∂Ki .. ϕ − ϕ − ϕ i,n 2 i,n 1 in ϕ − ϕ i,n 1 in (31) For example, when m =2,n = 3, we have ϕ11 ϕ12 00000000 0 ϕ11 ϕ12 ϕ13 000000 0 0 0 ϕ ϕ 0 0 0 0 0 B = 12 13 . 0 0 0 0 0 ϕ21 ϕ22 0 0 0 000000 ϕ21 ϕ22 ϕ23 0 00000000 ϕ22 ϕ23 Denote by Ga all the inner grid points and by Ra the ordering of the grid points in Ga, i.e., Ga = (i, j): i =1,...,m,j =1,...,n , (32) { } Ra = j + (i 1) n : (i, j) Ga , (33) { − ∗ ∈ } HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 11 and by G0 the grid points with function value being zero, 0 G = (i, j) Ga : ϕij =0 . (34) { ∈ } Note that Ga and Ra have a one to one correspondence. Define such correspon- dence as a mapping Γ : Ga Ra, → Γ(i, j)= j + (i 1) n. (35) − ∗ Denote by R0 the indices of rows in which B is zero, by R∗ the indices of rows, 0 ∗ in which B is not zero, and Si and Si with similar meanings for Bi, R0 = r : B(r, :) = 0 , R∗ = r : B(r, :) =0 , (36) { } { 6 } 0 ∗ S = r : Bi(r, :) = 0 , S = r : Bi(r, :) =0 . (37) i { } i { 6 } It can be verified that m m 0 0 ∗ ∗ R = Si , R = Si . (38) i=1 i=1 [ [ Lemma 3.3 Let B be the matrix defined in (30). Then the nonzero rows of B are linearly independent, i.e., the submatrix B(R∗, :) is row full rank. ∗ ∗ ∗ Proof. Note that B(R , :) = [B(S1 , :); ... ; B(Sm, :)]. Due to the block struc- ∗ ture of B, it suffices to prove that for any i, 1 i m, Bi(Si , :) is row full ∗ ≤ ≤ rank if Si is not empty. ∗ Now suppose that S is not empty, that is ϕi = 0. Let the nonzero com- i 6 ponents of ϕi be ϕi,i1 , ϕi,i2 ,...,ϕi,ir , with i1 < i2 < ... < ir. Denote by Z(ϕi,i1 ,...,ϕi,ik ) the rows of Bi containing ϕi,i1 , ϕi,i2 ,...,ϕi,ik , 1 k r. ∗ ≤ ≤ Then Bi(Si , :) = Z(ϕi,i1 ,...,ϕi,ir ). The claim will be proved by successively adding rows with nonzero component of ϕi to Z. (1) Prove Z(ϕi,i1 ) is row full rank. If i1 = 1, Z(ϕi,i1 ) is the first two rows of Bi and it is row full rank. If 1 ϕi,i1 0 0 ϕi,i ∗ 1 ∗ 0 0 ϕi,i1 where stands for an element which may be zero or nonzero. Therefore Z(ϕi,i ) ∗ 1 is row full rank. If i1 = n, Z(ϕi,i1 ) is the last two rows of Bi and is row full rank too. (2) Prove that when Z(ϕi,i1 ,...,ϕi,ik ) is row full rank, Z(ϕi,i1 ,...,ϕi,ik+1 ) is also row full rank, 1 k < r. If ik = n 1, then ik = n and Z(ϕi,i ,...,ϕi,i )= ≤ − +1 1 k+1 Z(ϕi,i ,...,ϕi,i ). Next suppose ik ϕi,i ϕi,i 0 00 ∗ k k+1 0 0 ϕi,i ϕi,i 0 k k+1 ∗ 00 0 0 ϕi,i k+1 ∗ HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 12 Thus Z(ϕi,i1 ,...,ϕi,ik+1 ) is row full rank. If ik+1 = ik+2 consists of Z(ϕi,i1 ,...,ϕi,ik ) and two new rows and similarly these two new rows are linearly independent with Z(ϕi,i ,...,ϕi,i ). If ik ik + 3 and ik Proof. Let f(λ)= det(A(K)+D λI). The polynomial f(λ) has no multiple roots if and only if its discriminant R−(K) is nonzero [22]. It is obvious that R(K) is not identically zero. Furthermore, since R(K) is a polynomial in the elements of vector K, it can vanish only on a hypersurface of real codimension 1. The hypersurface is U = K R(K)=0 . 1 { | } In the following Lemma 3.5, we prove that for any matrix F consisting of several submatrices by row, if every submatrix of F is row full rank and the index sets of nonzero columns do not intersect for any two submatrices, then F is row full rank. p×q s Lemma 3.5 Let F R be a matrix. Suppose 1,...,p = Ii, with ∈ { } i=1 Ii Ij = , if i = j. Denote ∅ 6 S T Ei = k 1,...,q : F (Ii, k) =0 . { ∈{ } 6 } Suppose that for any 1 i s, F (Ii, :) is row full rank and Ei Ej = , if i = j. Then F is row full≤ rank.≤ ∅ 6 T Proof. Since for any 1 i s, F (Ii, :) is row full rank, there exist a column ≤ ≤ index set Ji Ei such that F (Ii,Ji) is a nonsingular submatrix. Correspond- ⊂ ingly, the matrix [F (I1, :); F (I2, :); ... ; F (Is, :)] has a nonsingular submatrix as follows, F (I1,J1) F (I2,J2) . . .. F (I ,J ) s s As a result, the matrix [F (I1, :); F (I2, :); ... ; F (Is, :)] is row full rank and so is F . To prove that 0 is a regular value of H˜ (ϕ,λ,t,K): Rmn R [0, 1) R(2nm−m) (U U ) Rmn+1, we need to prove (ϕ,λ,t,K)× Rmn× R × \ 1 ∪ 2 → ∀ ∈ × × HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 13 (2nm−m) [0, 1) R (U1 U2) satisfying H˜ (ϕ,λ,t,K) = 0, the Jacobian matrix of H˜ (×ϕ,λ,t,K) is\ row full∪ rank. (ϕ,λ,t,K) satisfying H˜ (ϕ,λ,t,K)= 0, for B defined in (30), we have B(R0, :)∀ = 0. Correspondingly for the Jacobian matrix defined in (29), we have ((1 t)A + D +3tβdiag(ϕ)2 λI)(R0, :) = ((1 t)A + D λI)(R0, :). − − − − ˜ ∂H˜ Through row permutations, the Jacobian matrix of H(ϕ,λ,t,K), ∂(ϕ,λ,t,K) , can be rewritten as ((1 t)A + D +3tβdiag(ϕ)2 λI)(R∗, :) ϕ(R∗) (βϕ3 Aϕ)(R∗) (1 t)B(R∗, :) − ((1 t)A + D λI)(R−0, :) −ϕ(R0) (βϕ3 − Aϕ)(R0)− 0 . − ϕT− − 0− 0 0 ! − (ϕ,λ,t,K) satisfying H˜ (ϕ,λ,t,K) = 0, from Lemma 3.3, we know B(R∗, :) is row∀ full rank. Therefore, if we can prove ((1 t)A + D λI)(R0, :) (39) − ϕT− − ˜ is row full rank, then ∂H is row full rank. From H˜ = 0, we have ((1 ∂(ϕ,λ,t,K) − t)A + D λI)(R0, :)ϕ = 0, i.e., ϕ is orthogonal to the rows of ((1 t)A + D λI)(R0, :).− Therefore, if we can prove ((1 t)A + D λI)(R0, :)− is row full− ∂H˜ − − rank, ∂(ϕ,λ,t,K) is row full rank. Now the problem is turned into proving that ((1 t)A + D λI)(R0, :) is row full rank. −For easy exposition,− some concepts concerning the topology of the grid points with zero function value are introduced. In addition, (i, j) will be considered as grid point in the rest of this section, representing (xi,yj). Definition 3.6 In the grid Ga, a zero valued node (i, j) is a grid point with ϕi,j =0. Definition 3.7 Two zero valued nodes (i1, j1) and (ip, jp) are said to be zero valued connected, if there exist a sequence of zero valued nodes, (i1, j1), (i2, j2),..., (ip, jp), (40) such that for any two successive nodes (ik, jk) and (ik+1, jk+1) of the sequence, the distance of these two nodes is 1 in the sense that ik ik + jk jk =1. (41) | − +1| | − +1| Definition 3.8 A set S consisting of zero valued nodes is called a zero valued connected set if any two nodes of S are zero valued connected. Definition 3.9 A set S consisting of zero valued nodes is called a zero valued connected component if S is connected and S is the largest zero connected set containing S. From (27), the discretization of the differential equation in a stencil is written explicitly, (ij) (ij) (ij) (ij) (ij) α1 ϕij + α2 ϕi−1,j + α3 ϕi+1,j + α4 ϕi,j+1 + α5 ϕi,j−1 =0, (42) ϕ0j = ϕm+1,j =0, (43) ϕi0 = ϕi,n+1 =0, (44) 1 ϕTϕ =0, (45) − h1h2 HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 14 (ij) (i) 1 1 2 (ij) (ij) where α1 = (1 t)ajj + h2 + h2 + vij + tβϕij λ, α2 = α3 = (1 − 1 2 − − (i) 1 (ij) (ij) 1 t)ajj+1 2h2 , α4 = α5 = 2h2 , and i = 1,...,m, j = 1,...,n. When − 2 − 1 (2nm−m) K R (U1 U2), the sign relationships among the components of ϕ are∈ given in the\ following remark. S Remark 3.10 Let (i, j) be an inner zero valued node. Assume that all except two of its neighbouring points are known to be zero valued nodes. If one of the rest two points is a zero valued node, then so is the other; If one of the rest two points is not a zero valued node, then neither is the other. For any set G of zero valued nodes, denote by RG the set of indices of rows corresponding to G, in which the matrix B is zero, i.e., RG = r : B(r, :) = 0, r = j + (i 1) n, (i, j) G . (46) { − ∗ ∈ } Note that if a zero valued node (i, j) G is such that B(s, :) = 0 with s = j + (i 1) n, then the neighbouring inner∈ grid points in y-direction should be zero valued− ∗ nodes, that is, both (i, j 1) and (i, j + 1) are zero valued nodes if 1 CG = c Ra : M(RG,c) =0 . (47) { ∈ 6 } −1 If RG = , define CG = . For any s RG, let (i, j)=Γ (s). Then the Γ images of∅ (i, j) and its neighbouring∅ inner∈ grid points are possibly included in CG. Denote by gi the i-th column of grid points in x-direction, i.e., gi = (i, j):1 j n , (48) { ≤ ≤ } and by O1 (Om) the ordering of all the inner grid points except the first (last) column in y-direction, O = r = j + (i 1) n : (i, j) Ga g , (49) 1 { − ∗ ∈ \ 1} Om = r = j + (i 1) n : (i, j) Ga gm . (50) { − ∗ ∈ \ } Lemma 3.11 Both M(O1, :) and M(Om, :) are row full rank. Proof. It is obvious that M(O1, :) has the following form: ...... ∗ ...... ∗ . , ...... ... ∗ HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 15 where represents nonzero element. Therefore M(O , :) is row full rank. ∗ 1 M(Om, :) has the following form: ...... ∗ ∗ . ...... ...... ∗ Therefore M(Om, :) is row full rank. 0 0 0 Note that if G g1 = , then R O1, from Lemma 3.11, M(R , :) is row ∅ ⊂ 0 0 full rank. In the following, we consider the case G g1 = . The set G g1 can have its own zeroT valued connected components. 6 ∅ T T Lemma 3.12 For m n 6, suppose that there is a zero valued connected component γ of G0 g≥ with≥s points, s 2. Then 1 ≥ (i) s (i) If s = n, we have ϕ11 = ϕ12 = ... = ϕ1n = 0. From the sign relationships among the components of ϕ as stated in Remark 3.10, the grid points in the column 2 are all zero valued nodes, namely, ϕ21 = ϕ22 = ... = ϕ2n = 0. By induction, all the grid points are zero valued nodes, namely, ϕ11 = ... = ϕ1n = ... = ϕmn = 0, a contradiction with the condition that ϕT ϕ = 0. Therefore s Figure 1: Illustration of a flag (1,n) (1,s + 1) (1,s) (1,s − 1) (1, 2) (2, 1) (s − 1, 1) (s, 1) (s + 1, 1) (m, 1) (1, 1) Figure 2: Illustration of a flag (1,n) (1, t + s) (1, t + s − 1) (1, t + s − 2) (1, t + s + 1) 2 (1, t + s ) 2 (1, t + s − 1) 2 (1, t + s − 2) 2 (1, t + 1) (1, t) (1, t − 1) ( s − 1, 1) ( s , 1) (2, 1) 2 2 m, (1, 1) ( 1) (iii) Assume that γ starts from the point (1, j1) and ends at the point (1, j2) with j1 2 and j2 n 1. From the sign relationships among the components≥ of ϕ, the≤ zero− valued nodes connecting γ on the column 2 of grid points are (2, j), 3 j s 2. The zero valued nodes connecting γ are illustrated in Fig 2.≤ The≤ zero− valued nodes connecting γ will end on column s/2 with two zero valued nodes and form a zero valued component G. (iv) Similar to the case (3), the zero valued nodes connecting γ are illustrated in Fig 3. However, since s is odd, the zero valued nodes connecting γ arrives at one single point on column (s + 1)/2. All these zero valued nodes connecting γ from column 1 to column (s + 1)/2 form a zero val- ued connected set G. It may be or may not be a zero valued connected component. −1 (v) From the figures for cases (2), (3), (4), it can be seen that Γ (RG) is HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 17 Figure 3: Illustration of a flag (1,n) (1, t + s) (1, t + s − 1) (1, t + s − 2) s+1 (1, t + ) 2 s−1 (1, t + ) 2 s−3 (1, t + ) 2 (1, t + 1) (1, t) (1, t − 1) s−1 s+1 (2, 1) ( , 1) ( , 1) m, (1, 10) 2 2 ( 1) the set of zero valued nodes by shrinking G one layer from its nonzero −1 −1 boundary. Γ (CG) is the set of zero valued nodes by extending Γ (RG) one layer towards the nonzero boundary, which is G itself. 0 Definition 3.13 Let γ be a zero valued connected component of G g1 with s points, s 2. The zero valued connected set G mentioned in (2), (3), (4) in ≥ T Lemma 3.12 is called a flag of γ, denoted as Fγ . 0 Lemma 3.14 Let γ be a zero valued connected component of G g1 of s points with s> 2 an odd number and Fγ be its flag. Let G be another set of zero valued T nodes. Let I1 = RFγ and I2 = RG be the row index sets as defined in (46) such that M(I2, :) is row full rank. Denote J = j Ra : M(I , j) =0 , J = j Ra : M(I , j) =0 , (51) 1 { ∈ 1 6 } 2 { ∈ 2 6 } J12 = J1 J2. (52) \ Suppose that J12 has only one element k and that both the column M(I1, k) and the column M(I2, k) have only one nonzero element, denoted as M(s1, k) = 0 and M(s , k) = 0 respectively. Let (i , j ) and (i , j ) be the grid point corre-6 2 6 1 1 2 2 sponding to s1 and s2 respectively. Suppose i2 = i1 +2 and j1 = j2. Then [M(I1, :); M(I2, :)] is row full rank. Proof. Note that since J12 is not empty, the zero valued connected com- ponent γ corresponds to the case (4) of Lemma 3.12. Geometrically, for that k, the two grid points (i , j ) and (i , j ) corresponding to M(s , k) = 0 and 1 1 2 2 1 6 M(s2, k) = 0 lie in the same row of grid points with distance 2 in the sense that i i +6 j j = 2, as illustrated locally in Figure 4. Note that by Lemma | 2 − 1| | 2 − 1| 3.11, M(I1, :) is row full rank. By row permutations and column permutations, [M(I1, :); M(I2, :)] is transformed to the following form, denoted as W , HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 18 Figure 4: Local illustration of connection with a flag (i1, j1) (i2, j2) s s 2 s 4 3 − − ∗∗∗ ∗ ∗∗∗ ∗ z ∗∗∗. }|. . { z }| ∗ . { z }| { z }| { ...... (s−1)2 ∗∗∗∗ ∗∗∗ ∗ ∗ ...... 4 . . . . . . ∗. ∗∗∗. . ...... ∗ ∗ ∗ ··· ∗∗∗∗ ∗ ∗ · · · ∗ .. . (s−1)2 (s+1)2 Note that the diagonal entries of the submatrix W (1 : 4 ,s + 1 : 4 ) are all nonzero. Therefore, by elementary matrix transformations on the first (s+1)2 4 columns, the above matrix W is transformed to the following, denoted as W , s s 2 s 4 3 − − ∗ ∗ z }| { z ∗}|. { z }| { z }| { .. ∗ (s−1)2 ∗ .. 4 . . ∗ .. . ∗ ∗ ∗ ∗ ×∗∗··· ∗∗∗∗∗··· ∗ ∗∗··· ∗··· ∗∗∗∗∗··· ∗ .. . (s−1)2 By elementary matrix transformations on the first 4 +1 rows, the above matrix W is further transformed to the following, denoted as W , HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 19 s s 2 s 4 3 − − ∗ ∗ z }| {z ∗}|. {z }| { z }| { .. ∗ (s−1)2 ∗ .. 4 . . ∗ .. . ∗ ∗ ∗ ∗ ×∗∗···∗∗∗ ∗··· ∗ .. . (s−1)2 (s−1)2 Note that in the row 4 +1, at least one element W ( 4 +1, 1), denoted − 2 as ′ ′, is not zero. Therefore, the lower submatrix W ( (s 1) + 1 : end, :) is still × 4 row full rank. Now Lemma 3.5 can be applied to conclude that W is row full rank. Therefore [M(I1, :); M(I2, :)] is row full rank. Theorem 3.15 For the homotopy H : Rmn R [0, 1) Rmn R in (27), n N+, m n 6, for almost all K R(2nm× −m×) (U →U ), × ∀ ∈ ≥ ≥ ∈ \ 1 ∪ 2 (i) 0 is a regular value of H defined in (27) and therefore the homotopy paths corresponding to different initial points do not intersect each other for t [0, 1); ∈ (ii) Every homotopy path (ϕ(s), λ(s),t(s)) H−1(0) is bounded. ⊂ Proof. (i). It suffices to prove that (ϕ,λ,t,K) Rmn R [0, 1) (2nm−m) ∀ ∈ × × 0 × R (U1 U2) satisfying H˜ (ϕ,λ,t,K) = 0, ((1 t)A + D λI)(R , :) is row full\ rank,∪ or in short hand notation M(R0, :) is row− full rank.− Note that 0 0 0 0 R corresponds to the zero valued nodes G . If G g1 = , then R O1. By 0 0 ∅ ⊂ Lemma 3.11 M(R , :) is row full rank. Next G g1 = is assumed. If the zero 0 T6 ∅ 0 valued connected components of the set G g1 are all single point sets, R is T0 a subset of O1, and again by Lemma 3.11, M(R , :) is row full rank. T 0 Assume that there are q zero valued connected components of the set G g1, each of which has more than one point. These connected components of the set 0 T G g1 are denoted as γi, with corresponding flags Gi = Fγi , i = 1, , q. Denote ··· T q G = Ga ( Gi), (53) \ i=1 [ R = R , C = C , Ri = RG , Ci = CG , 1 i q. (54) G G i i ∀ ≤ ≤ By Lemma 3.12, it can be verified that q 0 R = R Ri , (55) i=1 ! [ [ Ri R = , 1 i q, (56) ∅ ∀ ≤ ≤ Ri Rj = ,\ Ci Cj = , i = j. (57) ∅ ∅ ∀ 6 \ \ HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 20 Figure 5: Local illustration of a connection Gi G 0 Note that if a zero valued connected component γ of G g1 has only two nodes, then the flag of γ is the γ itself, and Rγ = . Assume that all the zero valued ∅ T connected components γi have more than two nodes. Since R O , Ri Om, 1 i q, (58) ⊂ 1 ⊂ ∀ ≤ ≤ by Lemma 3.11, all of M(R, :) and M(Ri, :), 1 i q, are row full rank. It is ≤ ≤ possible that for some γi, Ci C = . If so, γi is the case described in case (4) 6 ∅ in Lemma 3.12, Ci C has only one element and the corresponding grid points T are illustrated in Figure 5. Without loss of generality, suppose that all the γi T satisfying such property are the first p γi, i.e., Ci C = , 1 i p, (59) 6 ∅ ≤ ≤ Ci C\= , p +1 i q. (60) ∅ ≤ ≤ Denote \ q G = G Gi , (61) i=p+1 [ [ Z(Gk,...,G , G) = [M(Rk, :); ... ; M(R , :); M(RG, :)], 1 k p. (62) 1 1 ∀ ≤ ≤ 0 0 Then M(R , :) = Z(Gp,...,G1, G). The claim that M(R , :) is row full rank will be proved by recursion. Firstly, prove Z(G1, G) is row full rank. Take I1 = R1 and I2 = RG. The conditions of Lemma 3.14 are satisfied. Thus [M(R1, :); M(RG, :)] is row full rank. Secondly, prove Z(Gk+1,...,G1, G) is row full rank if Z(Gk,...,G1, G) is, 1 k 3.2.2 The homotopy with random pentadiagonal matrix A homotopy with random pentadiagonal matrix is also possible. Specifically, for H˜ defined in (28), replacing A(K) with A(K), where A(K) Rmn×mn is a ∈ HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 21 random pentadiagonal matrix with the same sparse structure as D, namely, A1 A1 A1 A2 A2 A(K)= .. , . Am−1 Am−1 Am where a(i) a(i) (i) 11 12 b1 a(i) a(i) a(i) (i) 12 22 23 b2 . (i) (i) (i) . b Ai = a23 a33 . , Ai = 3 , . .. .. (i) .. . . an−1,n (i) (i) (i) a a bn n−1,n nn T K = K1,K2,...,Km, K1,..., Km−1 , (i) (i) (i) (i) (i) (i) (i) Ki = a11 ,a12 ,a22 ,a23 ,...,an −1,n−1,an−1,n,ann , i =1, . . . , m, (i) (i) (i) Ki = b ,b ,...,b , i =1,...,m 1. 1 2 n − ˜ The Jacobian matrix of H˜ (ϕ, λ, t, K), ∂H , is : ∂(ϕ,λ,t,K) 2 3 (1 t)A(K)+ D +3tβdiag(ϕ ) λI ϕ βϕ A(K)ϕ (1 t)B , − ϕT − −0− 0− 0 − where B = ∂(A(K)ϕ) R(mn)×(3nm−m−n). Recall ϕ = (ϕ , ϕ ,...,ϕ )T. It ∂K i i1 i2 in can be verified that ∈ B1 B2 B2 B1 B3 . B = .. B B 2 4 . . . ...... B B − m m 1 B2 B1 B3 = B B2 B4 , (63) . . .. .. B − m 1 with ϕi1 ϕi2 Bi = . . (64) .. ϕ in Note that the left part B of B is nothing but the matrix in (30). HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 22 Lemma 3.16 The eigenvalues of A(K)+D are simple for K almost everywhere except on a subset of real codimension 1. Proof. Similar to the proof of Lemma 3.4. + (3nm−m−n) For our discussion, U 2 = (R ) is removed. To prove that (ϕ, λ, t, K) mn (3nm−m−n) ∀ ∈ R R [0, 1) R (U 1 U 2) satisfying H˜ (ϕ, λ, t, K) = 0, the Jacobian× matrix× of×H˜ (ϕ, λ, t, K) in\ (63)∪ is row full rank, it suffices to prove that the following submatrix of the Jacobian matrix in (63) (1 t)A(K)+ D +3tβdiag(ϕ2) λI ϕ βϕ3 A(K)ϕ (1 t)B − ϕT − −0− 0− 0 − (65) is row full rank. With the notations R0 and R∗ as in (36) and similar arguments as in Subsection 3.2.1, it can be proved that ((1 t)A + D λI))(R0, :) is row full rank. Therefore the matrix (65) is row full− rank. That− is 0 is a regular value of the homotopy H with random pentadiagonal matrix for almost all K R(3nm−m−n) (U U ) with m n 6. ∈ \ 1 ∪ 2 ≥ ≥ 4 Algorithm and numerical results 4.1 Algorithm Thanks to Theorems 3.2 and 3.15, since 0 is a regular value of the homotopies constructed, the homotopy paths determined by the homotopy equations (22) and (27) have no bifurcation points with probability one. Therefore the usual path following algorithm, i.e., the predictor-corrector method as in [23, 24], can be adapted to trace the homotopy paths of the homotopy equations (22) and (27). The adapted algorithm is stated in Algorithm 1. For notation convenience, (k) in Algorithm 1, we denote the iterate at step k as xk = (ϕ , λk), k =1,.... 4.2 Numerical results The first numerical example is the 1D discretized problem (9) with β = 20, 1 2 V (x) = 2 x , Ω = [ 2, 2] and n = 999. Some eigenvectors, i.e., approximate eigenfunctions are plotted− in Figures 7 to 15. The eigenvector in Figure 7 is the unique positive solution of (9) as stated in Theorem 2.5, which corresponds to the unique positive ground state, as proved in [4] for the continuous nonlinear eigenvalue problem (4)-(6) when β > 0. The approximate eigenfunction in Fig- ure 8 is antisymmetric as described in Theorem 2.6 and is an approximate first excited state. Others are approximate excited states corresponding to higher energy. The order preserving property of the eigenvalue curves as stated in [17] was observed, that is, if λ(0) is the kth smallest eigenvalue of the initial problem, then λ(t) is the kth smallest eigenvalue of the intermediate problem for each t [0, 1). However, we are not able to prove such property for the eigenvector-dependent∈ nonlinear eigen-problem yet. HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 23 Algorithm 1: Predictor-Corrector method Initialization: Set (x0,t0) = (x0, 0), k = 0, the minimum step size dsmin, the initial step size ds. Compute the tangent vector (x ˙ 0, t˙0) such that t˙0 > 0 and record the orientation ori H H H x˙ + H t˙ =0, ori = sign x t . (66) x 0 t 0 x˙ t˙ 0 0 while tk < 1 do Predictor: (¯xk+1, t¯k+1) = (xk,tk)+ ds(¯xk, t¯k); if t¯k+1 > 1 then change ds such that t¯k+1 = 1; end Corrector: if t¯k+1 =1 then (v, τ) =(0, 1); else (v, τ) = (x ˙ k, t˙k); end Employ Newton method to solve the following nonlinear equations: H(x, t) T = 0. (67) v (x x¯k )+ τ(t t¯k ) − +1 − +1 Judgement: if the above iteration converges to (xk+1,tk+1) then compute the tangent vector (x ˙ k+1, t˙k+1) satisfying Hx Ht Hxx˙ k+1 + Htt˙k+1 =0, sign = ori (68) x˙ k t˙k +1 +1 else 1 ds = 2 ds; go to Predictor; end Compute angle θ of (¯xk, t¯k) and (¯xk+1, t¯k+1); if θ> 180 then 1 ds = 2 ds, go to Predictor; end Accept iterates: (xk,tk) = (xk+1,tk+1),(x ˙ k, t˙k) = (x ˙ k+1, t˙k+1); if θ< 60 then ds =2ds; end if ds < dsmin then stop the algorithm; end end HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 24 0.6 0.8 0.8 0.6 0.6 0.5 0.4 0.4 0.4 0.2 0.2 0.3 0 0 -0.2 -0.2 0.2 -0.4 -0.4 0.1 -0.6 -0.6 0 -0.8 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 7: λ =6.76 Figure 8: λ =8.39 Figure 9: λ = 10.35 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 -0.2 -0.2 -0.2 -0.4 -0.4 -0.4 -0.6 -0.6 -0.6 -0.8 -0.8 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 10: λ = 12.73 Figure 11: λ = 15.62 Figure 12: λ = 19.08 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 -0.2 -0.2 -0.2 -0.4 -0.4 -0.4 -0.6 -0.6 -0.6 -0.8 -0.8 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 13: λ = 23.13 Figure 14: λ = 27.79 Figure 15: λ = 33.05 The second numerical example is the 2D discretized problem (11) with 1 2 2 β = 20, V (x) = 2 (x1 + x2), Ω = [0, 1] [0, 1] and m = n = 29. Some ap- proximate eigenfunctions are collected in× Figures 16 to 27. The approximate eigenfunction in Figure 16 corresponds to the unique positive ground state. Oth- ers are approximate excited states corresponding to higher energy. The order preserving property of the eigenvalue curves is also observed for the 2D case. 1.5 2 2 1 1 1 0 0 0.5 -1 -1 0 -2 -2 1 1 1 1 1 1 0.8 0.8 0.8 0.5 0.6 0.5 0.6 0.5 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 0 0 0 Figure 16: λ = 43.36 Figure 17: λ = 60.89 Figure 18: λ = 65.31 3 3 2 2 2 1 1 1 0 0 0 -1 -1 -1 -2 -2 -3 -3 -2 1 1 1 1 1 1 0.8 0.8 0.8 0.5 0.6 0.5 0.6 0.5 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 0 0 0 Figure 25: λ = 129.67 Figure 26: λ = 133.81 Figure 27: λ = 138.68 HOMOTOPY METHODS FOR NONLINEAR EIGENVALUE PROBLEM 25 2 2 3 2 1 1 1 0 0 0 -1 -1 -1 -2 -2 -2 1 1 1 1 1 1 0.8 0.8 0.8 0.5 0.6 0.5 0.6 0.5 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 0 0 0 Figure 19: λ = 78.81 Figure 20: λ = 86.28 Figure 21: λ = 94.06 2 3 2 2 1 1 1 0 0 0 -1 -1 -1 -2 -2 -3 -2 1 1 1 1 1 1 0.8 0.8 0.8 0.5 0.6 0.5 0.6 0.5 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0 0 0 0 0 0 Figure 22: λ = 104.4 Figure 23: λ = 108.82 Figure 24: λ = 120.47 5 Conclusion Solutions to the discretized problem with the finite difference disretization for the GPE inherit certain properties of the solutions to the continuous problem, such as the existence and uniqueness of positive eigenvector (eigenfunction). 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