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Consider the Fredholm integral operator a)- Convert this integral operator to a

ID: 3141466 • Letter: C

Question

Consider the Fredholm integral operator

a)- Convert this integral operator to a Sturm-Liouville operator.

b)- Find the eigenvalues and associated eigenfunctions for the Sturm-Liouville problem. If there is no convenient formula for the eigenvalues, Indicate how many positive and negative ones there are and indicate the values graphically and indicate approximate values, if possible

c)_Use the eigenvalues and eigenfunctions from part (b) to solve the integral equation

If there are no convenient formulas for the eigenvalues, use the symbols ?n.

(u):/mar(z, y)u(y)dy K (u) = | mar(x, y)u(y)dy mal(T, y)u(Y )(ly

Explanation / Answer

Ans-

We consider a linear operator (Ku)(x) := Z b a k(x, z)u(z)dz = f(x), a x b, (1) where K : L 2 [a, b] L 2 [a, b] is a linear compact operator. We assume that k(x, z) is a smooth function on [a, b] × [a, b]. Since K is compact, the problem of solving equation (1) is ill-posed. Some applications of the Fredholm integral equations of the first kind can be found in [3], [5], [6]. There are many methods for solving equation (1): variational regularization, quasisolution, iterative regularization, the Dynamical Systems Method (DSM). A detailed description of these methods can be found in [4], [5], [6]. In this paper we propose an iterative scheme for solving equation (1) based on the DSM. We refer the reader to [5] and [6] for a detailed discussion of the DSM. When we are trying to solve (1) numerically, we need to carry out all the computations with finite-dimensional approximation Km of the operator K, limm kKm Kk = 0. One approximates a solution to (1) by a linear combination of basis functions vm(x) := Pm i=1 (m) j j (x), where (m) j are constants, and i(x) are orthonormal basis functions in L 2 [0, 1]. Here the constants (m) j can be obtained by solving the ill-conditioned linear 2 algebraic system: Xm j=1 (Km)ijj = gi , i = 1, 2, . . . , m, (2) where (Km)ij := R b a R b a k(x, s)j (s)dsi(x)dx, 1 i, j m, and gi := R b a f(x)i(x)dx. In applications, the exact data f may not be available, but noisy data f, kf fk , are available. Therefore, one needs a regularization method to solve stably equation (2) with the noisy data g i := R b a f(x)i(x)dx in place of gi . In the variational regularization (VR) method for a fixed regularization parameter a > 0 one obtains the coeffi- cients (m) j by solving the linear algebraic system: a(m) i + Xm j=1 (K mKm)ij (m) j = g i , i = 1, 2, . . . , m, (3) where (K mKm)ij := Z b a Z b a k(s, x)i(x) Z b a k(s, z)j (z)dzdsdx, kf fk , and k(s, x) is the complex conjugate of k(s, x). In the VR method one has to choose the regularization parameter a. In [4] the Newton’s method is used to obtain the parameter a which solves the following nonlinear equation: F(a) := kKmm g k 2 = (C) 2 , C 1, (4) where m = (aI + K mKm) 1K mg , and K m is the adjoint of the operator Km. In [2] the following iterative scheme for obtaining the coefficients (m) j is studied: n,m = q n1,m + (1 q)T 1 an,mK mg , d 0 = 0, an := 0q n , (5) where 0 > 0, q (0, 1), Ta,m := T (m) + aI, T(m) := K mKm, a > 0, (6) and I is the identity operator. Iterative scheme (5) is derived from a DSM solution of equation (1) obtained in [5, p.44]. In iterative scheme (5) adaptive regularization parameters an are used. A discrepancy-type principle for DSM is used to define the stopping rule for the iteration processes. 3 The value of the parameter m in (4) and (5) is fixed at each iteration, and is usually large. The method for choosing the parameter m has not been discussed in [2]. In this paper we choose the parameter m as a function of the regularization parameter an, and approximate the operator T := KK (respectively K ) by a finite-rank operator T (m) (respectively K m): lim m kT (m) Tk = 0. (7) Condition (7) can be satisfied by approximating the kernel g(x, z) of T, g(x, z) := Z b a k(s, x)k(s, z)ds, (8) with the degenerate kernel gm(x, z) := Xm i=1 wik(si , x)k(si , z), (9) where {si} m i=1 are the collocation points, and wi ,1 i m, are the quadrature weights. Quadrature formulas (9) can be found in [1]. Let K m be a finite-dimensional approximation of K such that lim m kK K mk = 0. (10) One may choose K m = PmK , where Pm is a sequence of orthogonal projection operators on L 2 [a, b] such that Pmx x as m , x L 2 [a, b]. We propose the following iterative scheme: u n,mn = qu n1,mn1 + (1 q)T 1 an,mnK mn f, u 0,m0 = 0, (11) where an := 0q n , 0 > 0, q (0, 1), kf fk , Ta,m is defined in (6) with T (m) satisfying condition (7), K m is chosen so that condition(10) holds, and mn in (11) is a parameter which measures the accuracy of the finite-dimensional approximations T (mn) and K mn at the nth iteration. We propose a rule for choosing the parameters mn so that mn depend on the parameters an. This rule yields a non-decreasing sequence mn. Since mn is a non-decreasing sequence, we may start to compute T 1 an,mnK mn f using a small size linear algebraic system Tan,mn g = K mn f,

(Sturm-Liouville Boundary Value Problem (SL-BVP)) With the notation L[y] d dx · p(x) dy dx¸ + q(x) y, (6.1) consider the Sturm-Liouville equation L[y] + r(x) y = 0, (6.2) where p > 0, r 0, and p, q, r are continuous functions on interval [a, b]; along with the boundary conditions a1y(a) + a2p(a)y 0 (a) = 0, b1y(b) + b2p(b)y 0 (b) = 0, (6.3) where a 2 1 + a 2 2 6= 0 and b 2 1 + b 2 2 6= 0. The problem of finding a complex number µ if any, such that the BVP (6.2)-(6.3) with = µ, has a non-trivial solution is called a Sturm-Liouville Eigen Value Problem (SL-EVP). Such a value µ is called an eigenvalue and the corresponding non-trivial solutions y(.; µ) are called eigenfunctions. Further, (i) An SL-EVP is called a regular SL-EVP if p > 0 and r > 0 on [a, b]. (ii) An SL-EVP is called a singular SL-EVP if (i) p > 0 on (a, b) and p(a) = 0 = p(b), and (ii) r 0 on [a, b]. (iii) If p(a) = p(b), p > 0 and r > 0 on [a, b], p, q, r are continuous functions on [a, b], then solving Sturm-Liouville equation (6.2) coupled with boundary conditions y(a) = y(b), y0 (a) = y 0 (b), (6.4)

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