5 Linear Algebra and Inverse Problem 5.1 Introduction Direct problem ( Forward problem) is to find field quantities satisfying Governing equations, Bound- ary conditions, Initial conditions. The direct problem can be formulated as a well-posed problem. Inverse problem ( Backward problem) is to find some parameters related to Governing equation, Boundary conditions, Initial conditions, Geometry conditions from observation data. In general, inverse problems become nonlinear and ill-posed. So inverse problems are solved after linearization and regularization. Examples of inverse problems ² Example 1 — influence line fj (j=1,...,n) ui (i=1,...,m) Let Aij, yi and xj be influence functions, measured data and parameters to be determined, respec- tively. Then we have Xn ui = Aijfj (i = 1; : : : ; m) or fug = [A] ffg : (211) j=1 ² Example 2 — Green’s function in solid mechanics T t(z) u(x) Using the Green’s function G(»; ³), the displacement u due to the surface traction t is expressed by Z u(») = G(»; ³)t(³)dS: (212) S Due to the Saint-Venant’s principle, the distributed forces t(³), applied at far field from the obser- vation point », can be approximated by the resultant force T without any change of u. It means that it is difficult to find a unique solution t, i.e., the inverse problem is an ill-posed problem. ² Example 3 — geological prosecting s s q 1 t 0Dt 1 The problem is to determine the location, shape and constitution of subterranean bodies from measurements at the earth’s surface. 30 As shown in the above figure, consider a one dimensional problem that the mass density, x(t), distributed along the t-axis for 0 · t · 1 is determined from the vertical component of force, y(s), on the surface line s. The vertical force ∆y(s) due to a small mass element x(t)∆t is written as x(t)∆t x(t)∆t ∆y(s) = g cos θ = g : (213) (s ¡ t)2 + 1 ((s ¡ t)2 + 1)3=2 where g is the gravity constant. It then follows that Z 1 x(t)dt y(s) = g (214) 2 3=2 0 ((s ¡ t) + 1) ² Example 4 — simplified tomography y I x detector R (,)x y x emitter I 0 Consider the simplified tomography in which the radiation absorption coefficient f(x; y) of the object is determined from the measured intensity Ix of the radiation, as shown in the above figure. Then the radiation intensity Ix satisfies the following differential equation. dI x = ¡fI (215) dy x The solution Ix is obtained in the form of the function p(x) Z y(x) p(x) ´ ln(I0=Ix) = f(x; y)dy (216) ¡y(x) Suppose that the object hasp a circular region and the absorption coefficient f(x; y) has the form of f(x; y) = f(r), where r = (x2 + y2). Then we have Z R 2r p(x) = p f(r)dr (217) 2 2 x r ¡ x ² Example 5 — structural dynamics k m c As shown in the above figure, we consider here structural dynamics with single degree of freedom governed by the equation of motion as follows. mx¨ + cx˙ + kx = 0 orx ¨ + ax˙ + bx = 0 (218) subjected to the initial conditions x(0) = x0; x˙(0) =x ˙ 0: (219) The inverse problem is to find the constant values a and b from the time history of the mass x(t). Integrating eq.(218) twice using the initial conditions leads to Z Z t t x(t) ¡ x0 ¡ x˙ 0t + a (x(s) ¡ x0)ds + b x(s)(t ¡ s)ds = 0; t > 0: (220) 0 0 31 Suppose that displacements x1; x2; : : : ; xn at the time steps of t = h; 2h; : : : ; nh are measured. If the integrals in eq.(220) are evaluate by the trapezoidal rule, eq.(220) can be discretized as follows: Xk Ek(a; b) = xk ¡ x0 ¡ x˙ 0kh + a( xjh ¡ xkh=2 ¡ x0kh) j=1 kX¡1 2 2 +b( xj(k ¡ j)h + x0kh =2) k = 1; 2; : : : ; n (221) j=1 Since eq. (221) is an approximated equation to eq.(220), EPk(a; b) is generally nonzero. For n > 2, n 2 two parameters, a and b, are determined so that E(a; b) = k=1(Ek(a; b)) becomes minimum. To this end, least squares method is generally used to find optimal values for a and b. Problem 5.1 1. Solve eq.(218) with the initial conditions x(0) = 1,x ˙(0) = ¡1 with m = 1, c = 2, k = 5, and calculate the displacements x(tk) at the time steps of t = tk (k = 1; : : : ; n), where tn · 1. 2. Then generate noisy data by adding the noise ²k to the calculated displacement x(tk), namely, xk = x(tk) + ²k, where ²k is a uniformly distributed random number in [¡²; ²]. 3. Write a program to estimate a and b by means of the least square method to minimize E(a; b). 4. Compare results calculated from data with various noise levels and discuss the sta- bility of the method with respect to ². 5.2 Linearized inverse problems As seen in the last section, many inverse problems are formulated in integral forms, which can be reduced to the following system of equations after discretization. n fbg = m [A] fxg: (222) Note that in general, [A] is a non-square matrix with the size m £ n. The system of equation (222) is classified into three categories depending on the numbers of observation points and measurement points, namely, m and n, as follows. (a) m > n Consider the case of m = 3 and n = 2. Then we have three equations for two unknowns x1 and x2; 8 9 2 3 ½ } < y1 = A11 A12 4 5 x1 y2 = A21 A22 : (223) : ; x2 y3 A31 A32 The above equation is an overdetermined set of linear equations. As illustrated in Fig. 8, there exists no exact solution, but the least squares solutions may be found in this case. X 2 X 2 X X 1 1 Figure 8: Figure 9: (b) m = n 32 (b)¡1 If A is regular, a unique solution is obtained as x = A¡1b (224) For m = n = 2, the behavior of equations is shown in Fig. 9. (b)¡2 If A is singular, there exists no solution as shown in Fig. 10. X 2 X 1 Figure 10: (b)¡3 When A is nearly singular, the solution is very sensitive to erros involved observed data. For example, we here consider the system of equations given as follows. ½ } · ¸ ½ } y A A x 1 = 11 12 1 (225) y2 A11 A12 + ² x2 Then the solution x is obtained by ½ } · ¸ ½ } x1 1 A12 + ² ¡A12 y1 = ¡ (226) x2 A11² A11 A11 y2 As seen in Fig. 11, the solution is very sensitive to the error ² involved in the component of X 2 X 2 X 1 X 1 Figure 11: Figure 12: the matrix. The stability of solution will be discussed later. (c) m < n In this case, the number of equations is less than the number of unknowns. As seen in Fig. 12 for m = 1 and n = 2, we cannot obtain unique solution. Problem 5.2.1 Discuss the existence, uniqueness and stability of solutions for the following system of equations. ½ } £ ¤ x © ª 1 1 1 = 2 (227) x · ¸ ½ 2 } ½ } 1 1 x 2 1 = (228) 1 1 x 3 8 2 9 · ¸ < x = ½ } 2 1 1 1 3 x = (229) 1 1 0 : 2 ; 2 x 2 3 3 8 9 2 1 ½ } < 4 = x 4 1 1 5 1 = 3 (230) x : ; 1 0 2 1 2 3 8 9 2 1 ½ } < 4 = x 4 1 1 5 1 = 3 (231) x : ; 1 0 2 2 33 5.3 Singular Value Decomposition (SVD) 5.3.1 SVD of a square regular matrix If A is a real symmetric m £ m matrix, then a useful decomposition is based on its eigenvalues and eigenvectors. The statement that ui is an eigenvector associated with eigenvalue ¸i can be written Aui = ui¸i: (232) If we now write the column vectors u1;:::; um next to each other to form the square matrix 0 1 . B . C B C U = @ u1 u2 ::: um A ; (233) . the relations (232) for i = 1; : : : ; m may be written as AU = UΛ: (234) where Λ is the diagonal matrix of eigenvalues. The eigenvectors have a convenient mathematical property of orthogonality, U T U = I, where I is the identity matrix, and span the entire space of A as a basis or minimum spanning set. The set of eigenvalues is called the spectrum of A. If two or more eigenvalues of A are identical, the spectrum of the matrix is called degenerate. The “spectrum” nomenclature is an exact analogy with the idea of the spectrum of light as depicted in a rainbow. The brightness of each color of the spectrum tell us “how much” light of that wavelength exists in the undispersed white light. For this reason, the procedure of SVD is often referred to as a spectral decomposition. Since U T U = UU T = I, multiplying eq.(234) by U T from the right hand side yields Xm T T T AUU = A = UΛU = ¸kukuk : (235) k=1 For a real symmetric matrix A, therefore, it is easy to compute any power of A An = (UΛU T )n = UΛnU T ; (236) because of U T U = UU T = I. Since Λ is diagonal, raising Λ to the nth power simply raises each of its diagonal elements to the nth power. The decomposition (235) means that the action of the real symmetric matrix A on an input vector x 2 Rm may be understood in terms of three steps: 1.