Notes Regarding the Karhunen Loève Expansion
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1 Notes Regarding the Karhunen Loève Expansion 1 Properties of the Karhunen Loève Expansion As noted in Section 5.3 of [3], the Karhunen Loève expansion for a correlated second-order random field α(x, ω), with x, mean ᾱ(x) and covariance function C(x, y), is α(x, ω) = ᾱ(x) + λn φ n (x)q n (ω). (1) Here λ n and φ n are the eigenvalues and orthonormal eigenfunctions of C; that is, they solve the integral equation C(x, y)φ n (y)dy = λ n φ n (x) () for x. To motivate (1) and (), we start with Mercer s Theorem, which states that because C(x, y) is bounded, symmetric and positive definite, it has the spectral decomposition where C(x, y) = λ n φ n (x)φ n (y) C(x, y)φ n (y)dy = λ n φ n (x). The eigenfunctions are orthogonal and form a complete set so φ n (x)φ m (x) = δ mn. (3) To determine statistical properties of the component of α(x, ω) quantified by the covariance function C(x, y), we consider α(x, ω) = ᾱ(x) + β(x, ω) where β(x, ω) has zero mean and covariance function C(x, y). We employ the expansion so that β(x, ω)β(y, ω) = β(x, ω) = m=1 Since β(x, ω) has zero mean, it follows that C(x, y) = Q n (ω) λ n φ n (x) Q n (ω)q m (ω) λ n λ m φ n (x)φ m (y). E(β(x, ω)β(y, ω)) = β(x, ω)β(y, ω) = Q n (ω)q m (ω) λ n λ m φ n (x)φ m (y) m=1 1
2 where X denotes the expectation E(X). Since the eigenfunctions are orthonormal, it follows that λ k φ k (x) = C(x, y)φ k (y)dy = Q n (ω)q k (ω) λ n λ k φ n (x). We then multiply by φ l (x), integrate, and again invoke orthogonality to obtain Since it follows that λ k φ k (x)φ l (x)dx = Q n (ω)q k (ω) λ n λ k δ nl λ k δ kl = λ k λ l Q k (ω)q l (ω). Hence the random variables satisfy the relations k = l Q k (ω)q l (ω) = 1 k l Q k (ω)q l (ω) = 0, Q k (ω)q l (ω) = δ kl. E(Q n ) = 0, E(Q n Q m ) = δ mn. (4) Choices for the Correlation Function C(x, y) Two common choices for the correlation function are where L is the correlation length, and C(x, y) = e x y /L, = [ 1, 1], (5) C(x, y) = min(x, y), = [0, 1], (6) which arises when studying 1- Weiner processes or Brownian motion. One reason these are commonly considered is because both have analytic solutions. Consider first (6). From the relation 1 0 min(x, y)φ n (y)dy = λ n φ n (x), we obtain the boundary condition φ n (0) = 0. For fixed x, we note that x which, when differentiated to obtain 0 yφ n (y)dy + x 1 x 1 x φ n (y)dy = λ n φ n (x) φ n (y)dy = λ n φ n(x),
3 yields the second boundary condition φ n(1) = 0. We differentiate again to obtain the Sturm Louiville problem φ n (x) = λ n φ n(x) The solution 1 λ n = ( ) n + 1 π φ n (0) = φ n(1) = 0., φ n (x) = [( sin n + 1 ) ] πx = sin(x/ λ n ) (7) yields the eigenvalues and eigenfunctions for (6). It is illustrated in [4] that the eigenvalues and eigenfunctions of (5) are λ n = { L 1+L wn L 1+L vn, if n is even,, if n is odd, and the eigenfunctions are φ n (x) = sin(w nx) 1 sin(wn) wn cos(v nx) 1+ sin(vn) vn, if n is even,, if n is odd. Here w n and v n are the solutions of the transcendental equations { Lw + tan(w) = 0, for even n, 1 Lv tan(v) = 0, for odd n. You can get good initial guesses by plotting the functions and zooming to approximate the roots. You can obtain accurate values using the Matlab command fzero.m. For both correlation functions, it is important to note the decay in λ n, which allows one to truncate the expansion (1). You can find a more mathematical discussion of the Karhunen Loève expansion in [1], which is posted in the class references. 3 Approximation of the Karhunen Loève Expansion ue to the decay of the eigenvalues λ N, we can employ approximate relations α(x, ω) = ᾱ(x) + N λn φ n (x)q n (ω). (8) To approximate ᾱ(x) and φ n (x), one can employ the expansions ᾱ(x) ᾱ h (x) = α j Φ j (x) φ n (x) φ h n(x) = φ jn Φ j (x) (9) 3
4 where Φ i (x) are, for example, finite element basis functions. The Karhunen Loève expansion is then approximated by α N (x, ω) αh N (x, ω) = α j Φ j (x) + N λ h nq n (ω) φ jn Φ j (x). (10) The coefficients {α j } J are obtained from (8) by solving the linear system M α = f where α = [α 1,..., α J ] T and M ij = Φ i (x)φ j (x)dx, f i = The approximation of the eigenvalue problem C(x, y)φ n (y)dy = λ n φ n (x) yields C(x, y) φ jn Φ j (y)dy = λ h n φ jn φ jn Φ j (x) C(x, y)φ j (y)φ i (x)dxdy = λ h n ᾱ(x)φ i (x)dx. φ jn This requires the solution of the generalized matrix eigenvalue problem Kφ h = λ h nmφ h Φ j (x)φ i (x)dx. where and (φ h ) j = φ h j. K ij = C(x, y)φ j (y)φ i (x)dxdy 4 Random Field for the Heat Equation Consider the heat equation T t = x ( α N (x, ω) T ) + f(t, x), 1 < x < 1, t > 0 x T (t, 1, ω) = T l (ω), T (t, 1, ω) = T r (ω) t > 0 T (0, x, ω) = T 0 (ω) 1 < x < 1. (11) To establish the well-posedness of solutions, one typically requires that 0 < α min α N (x, ω) α max. Unfortunately, this cannot be ensured for Gaussian random fields. The lower bound can be enforced by employing the log-normal Karhunen Loève expansion α(x, ω) = α min + eᾱ(x)+ λnφ n(x)q n(ω). However, this does not provide an upper bound. To address this, one can instead prove well-posedness in a stochastic sense as detailed in []. 4
5 References [1] A. Alexanderian, A brief note on the Karhunen Loève expansion. [] J. Charrier, Strong and weak error estimates for elliptic partial differential equations with random coefficients, SIAM Journal of Numerical Analysis, 50(1), pp , 01. [3] R.C. Smith, Uncertainty Quantification: Theory, Implementation, and Applications, SIAM, Philadelphia, PA, 014. [4] H.L. Van Trees, etection, Estimation and Modulation Theory, Part 1, John Wiley & and Sons, New York,
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