Introduction to Uncertainty Quantification in Computational Science Handout #3

Transcription

1 Introduction to Uncertainty Quantification in Computational Science Handout #3 Gianluca Iaccarino Department of Mechanical Engineering Stanford University June 29 - July 1, 2009 Scuola di Dottorato di Ricerca in Ingegneria Industriale Dottorato di Ricerca in Ingegneria Aerospaziale, Navale e della Qualita Universita di Napoli Federico II Based on the Lecture Notes for ME470 prepared with Dr. A. Doostan

2 Uncertainty Propagation Using Stochastic Collocation

3 Uncertainty Quantification Computational Framework Consider a generic computational model (y R d with d large) How do we handle the uncertainties? 1. Data assimilation: characterize uncertainties in the inputs 2. Uncertainty propagation: perform simulations accounting for the identified uncertainties 3. Certification: establish acceptance criteria for predictions

4 Sampling Methods Sample the random input parameter vector according to its probability distributions Perform a sequence of independent simulations Compute statistics of the quantity of interest

5 Sampling Methods Sample the random input parameter vector according to its probability distributions Perform a sequence of independent simulations Compute statistics of the quantity of interest Now we will introduce an alternative way of computing the output statistics without random sampling!

6 Numerical Integration The basic idea is to use advanced numerical integration techniques Recall numerical quadrature: Problem: Compute the integral of f (x) over the interval [a : b].

7 Numerical integration Express integrals as a finite, weighted sum b a f (ξ)dξ N w i f (ξ i ) i=1 Examples: midpoint, trapezoidal, Simpson rules Remark: all use equispaced abscissas ξ i [a : b] (Newton-Cotes)

8 Numerical integration Express integrals as a finite, weighted sum b a f (ξ)dξ N w i f (ξ i ) i=1 Examples: midpoint, trapezoidal, Simpson rules Remark: all use equispaced abscissas ξ i [a : b] (Newton-Cotes) We can do better: Gauss quadrature rules. Observation: We can always fit an N-1 degree polynomial to a set N points (N = 2 line, N 3 parabola, etc.).

9 Numerical integration Express integrals as a finite, weighted sum b a f (ξ)dξ N w i f (ξ i ) i=1 Examples: midpoint, trapezoidal, Simpson rules Remark: all use equispaced abscissas ξ i [a : b] (Newton-Cotes) We can do better: Gauss quadrature rules. Observation: We can always fit an N-1 degree polynomial to a set N points (N = 2 line, N 3 parabola, etc.). By carefully choosing the abscissas and weights (ξ i, w i ), we can exactly evaluate the integral if f (ξ) is (2N 1) degree polynomial.

10 Numerical integration Express integrals as a finite, weighted sum b a f (ξ)dξ N w i f (ξ i ) i=1 Examples: midpoint, trapezoidal, Simpson rules Remark: all use equispaced abscissas ξ i [a : b] (Newton-Cotes) We can do better: Gauss quadrature rules. Observation: We can always fit an N-1 degree polynomial to a set N points (N = 2 line, N 3 parabola, etc.). By carefully choosing the abscissas and weights (ξ i, w i ), we can exactly evaluate the integral if f (ξ) is (2N 1) degree polynomial. What are the abscissas ξ i and the weights w i?

11 Numerical quadrature b a f (ξ)dξ N w i f (ξ i ) i=1

12 Numerical quadrature b a f (ξ)dξ N w i f (ξ i ) i=1 The abscissas are the roots of orthogonal polynomials (Legendre) 1 1 p j (x)p i (x)dx = C i δ ij Abscissas: impose p N (ξ) = 0 ξ 1,..., ξ N

13 Numerical quadrature b a f (ξ)dξ N w i f (ξ i ) i=1 The abscissas are the roots of orthogonal polynomials (Legendre) 1 1 p j (x)p i (x)dx = C i δ ij Abscissas: impose p N (ξ) = 0 ξ 1,..., ξ N The weights are the integrals of the Lagrange interpolating polynomials passing through the abscissas w i = 1 1 L i,n (ξ)dξ with L i,n (ξ) = N k=1 k j ξ ξ k ξ i ξ k

14 Legendre-Gauss quadrature Legendre Polynomials (n+1)p n+1 (ξ) = (2n+1)ξP n (ξ) np n 1 (ξ) 1 1 P j (x)p i (x)dx = 2 2n + 1 δ ij Three-term recurrence Orthogonality

15 Legendre-Gauss quadrature Both the abscissas and the weights are tabulated and can be computed in several ways For example: function I = gauss(f,n) beta =.5/sqrt(1-(2*(1:n))^(-2)); T = diag(beta,1) + diag(beta,-1); [V,D] = eig(t); x = diag(d); [x,i] = sort(x); w = 2*V(1,i)^2; I = w*feval(f,x); % (n+1)-pt Gauss quadrature % 3-term recurrence coeffs % Jacobi matrix % eigenvalue decomposition % nodes (= Legendre points) % weights % the integral The command gauss(cos,6) yields which is correct to double precision [Trefethen, 2008]

16 Quadrature for Uncertainty Analysis Stochastic Collocation What does quadrature have to do with uncertainty?

17 Quadrature for Uncertainty Analysis Stochastic Collocation What does quadrature have to do with uncertainty? Assume y is a uniform random variable describing the uncertainty in the input and Q is he quantity of interest

18 Quadrature for Uncertainty Analysis Stochastic Collocation What does quadrature have to do with uncertainty? Assume y is a uniform random variable describing the uncertainty in the input and Q is he quantity of interest The mean of Q is Q = Q(y)f y dy = Q(y)dy if y = U(-1,1) Similarly for the variance, etc.

19 Quadrature for Uncertainty Analysis Stochastic Collocation What does quadrature have to do with uncertainty? Assume y is a uniform random variable describing the uncertainty in the input and Q is he quantity of interest The mean of Q is Q = Q(y)f y dy = Q(y)dy if y = U(-1,1) Similarly for the variance, etc. Moments of a quantity of interest are integrals in the probability space defined by the uncertain variables!

20 Stochastic Collocation For an input variables that is U( 1, 1) Generate N values of the parameters y k k = 1,..., N: abscissas (the zeros of the Legendre polynomial P N )

21 Stochastic Collocation For an input variables that is U( 1, 1) Generate N values of the parameters y k k = 1,..., N: abscissas (the zeros of the Legendre polynomial P N ) Perform N simulations according to the selected abscissas and obtain Q(y k )

22 Stochastic Collocation For an input variables that is U( 1, 1) Generate N values of the parameters y k k = 1,..., N: abscissas (the zeros of the Legendre polynomial P N ) Perform N simulations according to the selected abscissas and obtain Q(y k ) Compute statistics as weighted sums (the weights are integrals of Lagrange polynomials through the abscissas) Q = 1 1 Q(y)dy = N Q(y k )w k k=1

23 Stochastic Collocation For an input variables that is U( 1, 1) Generate N values of the parameters y k k = 1,..., N: abscissas (the zeros of the Legendre polynomial P N ) Perform N simulations according to the selected abscissas and obtain Q(y k ) Compute statistics as weighted sums (the weights are integrals of Lagrange polynomials through the abscissas) Q = 1 1 Q(y)dy = N Q(y k )w k k=1 No randomness is considered! but convergence is exponential (cfr. MC)

24 Beyond Uniform rvs Gaussian rvs What do we do if the input variables are not distributed as uniform r.v.?

25 Beyond Uniform rvs Gaussian rvs What do we do if the input variables are not distributed as uniform r.v.? As you probably know, numerical quadrature is more than just Legendre-Gauss quadrature!

26 Beyond Uniform rvs Gaussian rvs What do we do if the input variables are not distributed as uniform r.v.? As you probably know, numerical quadrature is more than just Legendre-Gauss quadrature! Consider y distributed as a N(0, 1), we can build orthogonal polynomials w.r.t. to a Gaussian measure as: p j (x)p i (x)e x 2 dx = C i δ ij Hermite polynomials for normal r.v. play the same role as Legendre polynomials for uniform r.v.s!

27 Hermite-Gauss quadrature Hermite Polynomials H n+1 (ξ) = 2ξH n (ξ) 2nH n 1 (ξ) H j (x)h i (x)e x 2 dx = 2 i i! πδ ij Three-term recurrence Orthogonality

28 Stochastic Collocation Summary of the quadrature rules We can use Legendre or Hermite polynomials, can we do even more?

29 Stochastic Collocation Summary of the quadrature rules We can use Legendre or Hermite polynomials, can we do even more? Distribution pdf Polynomials Weights Support Uniform 1/2 Legendre 1 [ 1 : 1] Gaussian (1/ 2π)e ( x2 /2) Hermite e ( x2 /2) [ : ] Exponential e x Laguerre e x [0 : ] (1 x) Beta (1+x) β B(α,β) Jacobi (1 x) α (1 + x) β [ 1 : 1] Table: Some polynomials in the Askey family

30 Stochastic Collocation Summary of the quadrature rules We can use Legendre or Hermite polynomials, can we do even more? Distribution pdf Polynomials Weights Support Uniform 1/2 Legendre 1 [ 1 : 1] Gaussian (1/ 2π)e ( x2 /2) Hermite e ( x2 /2) [ : ] Exponential e x Laguerre e x [0 : ] (1 x) Beta (1+x) β B(α,β) Jacobi (1 x) α (1 + x) β [ 1 : 1] Table: Some polynomials in the Askey family What if the random variables are not distributed according to any of the above?

31 Stochastic Collocation Summary of the quadrature rules We can use Legendre or Hermite polynomials, can we do even more? Distribution pdf Polynomials Weights Support Uniform 1/2 Legendre 1 [ 1 : 1] Gaussian (1/ 2π)e ( x2 /2) Hermite e ( x2 /2) [ : ] Exponential e x Laguerre e x [0 : ] (1 x) Beta (1+x) β B(α,β) Jacobi (1 x) α (1 + x) β [ 1 : 1] Table: Some polynomials in the Askey family What if the random variables are not distributed according to any of the above? 1. Szego (1939). Orthogonal Polynomials - American Mathematical Society. 2. Schoutens (2000). Stochastic Processes and Orthogonal Polynomial - Springer. 3. Gram-Schmidt Procedure

32 Nested rules The choice of abscissas The Gauss quadrature rules introduce different abscissas for each order N considered. They are not nested, no reuse of computed solutions for lower-order quadrature

33 Nested rules The choice of abscissas The Gauss quadrature rules introduce different abscissas for each order N considered. They are not nested, no reuse of computed solutions for lower-order quadrature Two extensions are possible Gauss-Kronrod rules Clenshaw-Curtis rules: express the integrand using Chebyshev polynomials (lower polynomial exactness) Figure: 32 abscissas in [ 1 : 1]

34 Clenshaw-Curtis vs. Gauss Figure: Black: Gauss, White: CC [Trefethen, 2008]

35 Why Nested rules? Assume you have a budget of N = 9 computations. With 9 Gauss abscissas (Legendre), we can obtain an estimate of the statistics of the solution which would be exact if the solution is a polynomial of degree 2N = 18

36 Why Nested rules? Assume you have a budget of N = 9 computations. With 9 Gauss abscissas (Legendre), we can obtain an estimate of the statistics of the solution which would be exact if the solution is a polynomial of degree 2N = 18 With Clenshaw-Curtis we can obtain again an estimate (only exact for polynomials of degree N = 9). On the other hand, with the same computations (N = 9 abscissas) we can also estiamte the solution statistics corresponding to N = 5 and N = 3 error estimate.

37 Multi-dimensional rules Tensor Product The extension of the previous 1D rules (Gauss or CC) is straightforward The abscissas are tensor products of the quadrature points in 1D The weights are the products of the 1D weights

38 Multi-dimensional rules Tensor Product The extension of the previous 1D rules (Gauss or CC) is straightforward The abscissas are tensor products of the quadrature points in 1D The weights are the products of the 1D weights The number of function evaluations increases as N d curse of dimensionality

39 Multi-dimensional rules Sparse Grids - Smolyak Grids Smolyak idea is to sparsify the construction of quadrature grids Figure: Sequence of grids used in 2D by a nested rule

40 Multi-dimensional rules Sparse Grids - Smolyak Grids The nominal accuracy can be preserved with much less points Figure: From Tensor grid to Sparse grid in 2D

41 Multi-dimensional rules Sparse Grids - Smolyak Grids The nominal accuracy can be preserved with much less points Figure: From Tensor grid to Sparse grid in 2D The trick is to use a linear combination of tensor products to build the actual sparse grid

42 Sparse Grids Stochastic Collocation Table: Abscissas for N = 5 in each dimension d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 Tensor Gauss Smolyak Gauss Smolyak CC

43 Stochastic collocation Consider the integral (with a i and b i U(0,1)) [Webster et al. 2006] exp [ ] N ai 2 (y i b i ) 2 d y i=1

44 Stochastic collocation Consider the integral (with a i and b i U(0,1)) [Webster et al. 2006] exp [ ] N ai 2 (y i b i ) 2 d y i=1

45 Sparse Grids Summary A fundamental advance in the development of stochastic collocation approaches in multid Becoming more and more popular

46 Sparse Grids Summary A fundamental advance in the development of stochastic collocation approaches in multid Becoming more and more popular Not perfect Not straightforward to construct (implementation errors) Can lead to negative weights Does not solve the curse of dimensionality, although it is better than tensor grids

47 Uncertainty Propagation Using Response Surfaces

48 Response surface Stochastic collocation methods allows to compute moments of the solution using fewer function evaluations than MC. In same cases we still want to build the PDF, or evaluate a reliability constraint. One possibility is to build an interpolant of the collocation points, for example a Lagrange interpolating polynomial, and then to sample using MC The interpolant is called a response surface, as it represent the approximate dependency of the solution on the input paramters

49 Stochastic Collocation Example Figure: Unstructured grid used in the present computations showing the strong clustering at the cylinder surface to capture the boundary layer. The dark region is the actual computational domain: periodic conditions are used on the upper and lower boundary.

50 Stochastic Collocation Example Figure: Temperature distribution for the baseline case.

51 Stochastic Collocation Example Figure: Streamlines for the baseline case.

52 Stochastic Collocation: Example (a) GL rule on sparse grid. (b) GL rule on tensor grid (c) CC rule on sparse grid (d) CC rule on tensor grid Figure: Expectation of x 1 -velocity along the domain centerline (x 2 = 0).

53 Stochastic Collocation: Example (a) GL rule on sparse grid. (b) GL rule on tensor grid (c) CC rule on sparse grid (d) CC rule on tensor grid Figure: Expectation of the temperature on the cylinder surface.

54 Stochastic Collocation: Example (a) GL rule on sparse-grid. (b) GL rule on tensor grid (c) CC rule on sparse-grid (d) CC rule on tensor grid Figure: Variance of the temperature on the cylinder surface.

55 Stochastic Collocation: Example (a) GL rule on sparse-grid. (b) GL rule on tensor grid (c) CC rule on sparse-grid (d) CC rule on tensor grid Figure: Probability density function of the stochastic temperature at the stagnation point.

56 Stochastic Collocation: Example (a) GL rule on sparse-grid. (b) GL rule on tensor grid (c) CC rule on sparse-grid (d) CC rule on tensor grid Figure: Probability density function of the x 1 -velocity in the wake at the point (2,0).

57 Stochastic Galerkin Approach for Hyperbolic Systems

58 Polynomial Chaos Stochastic Galerkin Approach The unknown is expressed as a spectral expansion of "one" random variable ξ Ω (assumed to be a Gaussian) u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 The ψ i (ξ) are Hermite polynomials and form a complete set of orthogonal basis functions Orthogonal Polynomials ψ n ψ m = Ω ψ n(ξ)ψ m (ξ)w(ξ)dξ = h n δ n,m E[ψ 0 ] = Ω ψ 0(ξ)w(ξ)dξ = 1 E[ψ k ] = Ω ψ k(ξ)w(ξ)dξ = 0, k > 0 where w(ξ) is the pdf of ξ and h n are non-zero constants

59 Polynomial Chaos Stochastic Galerkin Approach Given u(x, t, ξ) = u i (x, t)ψ i (ξ) we can compute the moments Expectation of u u 0 E[u] = Ω Variance of u Ω i=0 uw(ξ)dξ = ψ 0 (ξ)w(ξ)dξ + Var[u] = E[u 2 ] (E[u]) 2 = Ω u i i=1 i=0 u 2 i ( ) u i ψ i w(ξ)dξ = Ω Ω i=0 ψ i (ξ)w(ξ)dξ = u 0 = E[u] ψi 2 w(ξ)dξ u0 2 = ui 2 ψi 2. i=1

60 Polynomial Chaos 1D Linear Convection Equations Consider the 1D linear convection equations u t + cu x = 0 0 x 1 The exact solution is u(x, t) = u initial (x ct)

61 Polynomial Chaos 1D Linear Convection Equations Consider the 1D linear convection equations u t + cu x = 0 0 x 1 The exact solution is u(x, t) = u initial (x ct) Assume the uncertainty is characterized by one parameter; let it be a Guassian random variable ξ Ω

62 Polynomial Chaos 1D Linear Convection Equations Consider the 1D linear convection equations u t + cu x = 0 0 x 1 The exact solution is u(x, t) = u initial (x ct) Assume the uncertainty is characterized by one parameter; let it be a Guassian random variable ξ Ω Consider a (truncated) spectral expansion of the solution in the random space u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 P u i (x, t)ψ i (ξ) i=0 where ψ i (ξ) are (1D) Hermite polynomials (ψ 0 = 1; ψ 1 = ξ; ψ 2 = ξ 2 1; ψ 3 = ξ 3 3ξ; etc.)

63 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition Assume The exact solution is u initial (x, t = 0, ξ) = g(ξ)cos(x) u(x, t, ξ) = g(ξ)cos(x ct)

64 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition Assume The exact solution is u initial (x, t = 0, ξ) = g(ξ)cos(x) u(x, t, ξ) = g(ξ)cos(x ct) Plug in the truncated expansion is the original PDE: ( P P ) u i t ψ u i i(ξ) + c x ψ i(ξ) = 0 i=0 i=0

65 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition Assume The exact solution is u initial (x, t = 0, ξ) = g(ξ)cos(x) u(x, t, ξ) = g(ξ)cos(x ct) Plug in the truncated expansion is the original PDE: ( P P ) u i t ψ u i i(ξ) + c x ψ i(ξ) = 0 i=0 i=0 Multiply by ψ k (ξ) and integrate over the probability space Ω (Galerkin Projection) P u i P t ψ u i iψ k + c x ψ iψ k = 0 for k = 0, 1,..., P. i=0 i=0

66 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition We have P u i P t ψ iψ k + c i=0 i=0 u i x ψ iψ k = 0 for k = 0, 1,..., P.

67 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition We have P u i P t ψ iψ k + c i=0 i=0 u i x ψ iψ k = 0 for k = 0, 1,..., P. The orthogonality property ψ i ψ k = δ ik h k implies u 0 t u P t + c u 0 x = 0 + c u P x = 0 We obtain a system of P + 1 uncoupled & deterministic eqns.

68 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the initial condition Initial conditions for the u 0... u P equations are obtained by projection of the initial condition u initial (x, t = 0, ξ), ψ k = u k (x, t = 0) = = g(ξ), ψ k cos(x) k = 0,..., P The procedure is simply an approximation of g(ξ) on the polynomial basis ψ(ξ)

69 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the transport velocity Assume The exact solution is c = h(ξ) u(x, t, ξ) = cos(x h(ξ)t)

70 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the transport velocity Assume The exact solution is c = h(ξ) u(x, t, ξ) = cos(x h(ξ)t) Plug in the truncated expansion is the original PDE: ( P P ) u i t ψ u i i(ξ) + h(ξ) x ψ i(ξ) = 0 i=0 i=0

71 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the transport velocity Assume The exact solution is c = h(ξ) u(x, t, ξ) = cos(x h(ξ)t) Plug in the truncated expansion is the original PDE: ( P P ) u i t ψ u i i(ξ) + h(ξ) x ψ i(ξ) = 0 i=0 i=0 Multiply by ψ k (ξ) and integrate over the probability space Ω (Galerkin Projection) P u i P t ψ u i iψ k + x h(ξ)ψ iψ k = 0 for k = 0, 1,..., P. i=0 i=0

72 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the transport velocity If we assume P h h(ξ) = h j ψ j (ξ) j=0 The system of equations becomes P i=0 u i t ψ iψ k + P i=0 u i x P h h j ψ j ψ i ψ k = 0 for k = 0, 1,..., P. j=0 We obtain a system of P + 1 coupled & deterministic eqns.

73 Polynomial Chaos 1D Linear Convection Equations - uncertainty in the transport velocity If we assume P h h(ξ) = h j ψ j (ξ) j=0 The system of equations becomes P i=0 u i t ψ iψ k + P i=0 u i x P h h j ψ j ψ i ψ k = 0 for k = 0, 1,..., P. j=0 We obtain a system of P + 1 coupled & deterministic eqns. This is a much tougher non-linear problem and leads to the long-time integration issue [Gottlieb & Xiu, 2008]

74 Polynomial Chaos 1D Burgers Equations Consider the 1D Burgers equations u t + uu x = 0 0 x 1 Assume the uncertainty is characterized by one parameter; let it be a Guassian random variable ξ Ω Consider a (truncated) spectral expansion of the solution in the random space u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 P u i (x, t)ψ i (ξ) i=0 where ψ i (ξ) are (1D) Hermite polynomials (ψ 0 = 1; ψ 1 = ξ; ψ 2 = ξ 2 1; ψ 3 = ξ 3 3ξ; etc.)

75 Polynomial Chaos 1D Burgers Equations Plug in the governing equations ( P u i P t ψ P i(ξ) + u j ψ j (ξ) i=0 j=0 i=0 u i x ψ i(ξ) ) = 0

76 Polynomial Chaos 1D Burgers Equations Plug in the governing equations ( P u i P t ψ P i(ξ) + u j ψ j (ξ) i=0 j=0 i=0 u i x ψ i(ξ) ) = 0 Multiply by ψ k (ξ) and integrate over the probability space Ω (Galerkin Projection) P i=0 u i t ψ iψ k + P P i=0 j=0 u i u j x ψ iψ j ψ k = 0 for k = 0, 1,..., P. We obtain a system of P + 1 coupled & deterministic equations (independently of the type of uncertainty)

77 Polynomial Chaos 1D Burgers Equations PC expansion for the Burgers equations P i=0 u i t ψ iψ k + P P i=0 j=0 Double/Triple products are numbers ψ i ψ j ψ k = and s = (i + j + k)/2 u i u j x ψ iψ j ψ k = 0 for k = 0, 1,..., P. ψ i ψ j = δ ij i! { 0 if i + j + k is odd or max(i, j, k) > s i!j!k! (s i)!(s j)!(s k)! otherwise

78 Polynomial Chaos 1D Burgers Equations PC Expansion for the Burgers equations P=1 u 0 t u 1 t u 0 + u 0 x + u u 1 1 x + u 1 u 0 x + u 0 u 1 x = 0 = 0

79 Polynomial Chaos 1D Burgers Equations PC Expansion for the Burgers equations P=1 u 0 t u 1 t u 0 + u 0 x + u u 1 1 x + u 1 u 0 x + u 0 u 1 x = 0 = 0 PC Expansion for the Burgers equations P=2 u 1 t u 2 t u 0 t u 0 + u 0 x + u u 1 1 x + 2u u 2 2 x + u 1 u 0 x + (u 0 + 2u 2 ) u 1 + u 2 u 0 x + u 1 u 2 x x + 2u 1 u 1 x + (u 0 + 4u 2 ) u 2 x = 0 = 0 = 0

80 Simple example 1D Viscous Burgers Governing equation; note the modified convective flux: Exact solution 1 u (1 u) 2 x = u µ 2 u 2 u(x) = 1 2 [ ( )] x 1 + tanh 4µ Assume uncertainty in the viscosity - Gaussian r.v. with E[µ] = 0.25 and Var[µ] =

81 Monte Carlo Sampling 1D Viscous Burgers Expectation of the solution: Computed solution (32 points) Exact solution

82 Monte Carlo Sampling 1D Viscous Burgers Variance of the solution: Computed solution (32 points) Exact solution

83 Polynomial Chaos 1D Viscous Burgers Statistics of the solution: Expectation Variance

84 Polynomial Chaos 1D Viscous Burgers Polynomial chaos modes of the solution (P = 3) Coefficients 0 and 1 Coefficients 2 and 3 Mode 0 is the mean (as expected) Mode 1 is dominant with respect to the others (u 2 1 closely approximates the variance)

85 Polynomial Chaos Observations Let s consider the PC expansion with P = 1 u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 1 u i (x, t)ψ i (ξ) = u 0 + u 1 ξ i=0

86 Polynomial Chaos Observations Let s consider the PC expansion with P = 1 u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 1 u i (x, t)ψ i (ξ) = u 0 + u 1 ξ i=0 The output is a linear function of the input uncertainty!

87 Polynomial Chaos Observations Let s consider the PC expansion with P = 1 u(x, t, ξ) = u i (x, t)ψ i (ξ) i=0 1 u i (x, t)ψ i (ξ) = u 0 + u 1 ξ i=0 The output is a linear function of the input uncertainty! Remark: ξ = N (0, 1) = u = N (u 0, u 2 1 ) As a consequence f u = e (u u 0) 2 /2u 2 1 2π u1 independently of the governing PDE!

88 Polynomial Chaos 1D Heat Conduction ( ) u(x, ξ) a(x, ξ) = f x D (0, 1) x x u(0, ξ) = u(1, ξ) = 0 where the diffusivity a(x, ξ) is a random field over D where: 0 < a min < a(x, ξ) < a max

89 Polynomial Chaos 1D Heat Conduction ( ) u(x, ξ) a(x, ξ) = f x D (0, 1) x x u(0, ξ) = u(1, ξ) = 0 where the diffusivity a(x, ξ) is a random field over D where: 0 < a min < a(x, ξ) < a max How to define the conductivity? We need data!

90 Polynomial Chaos 1D Heat Conduction We need to represent a random field to describe the conductivity a(x, ξ)

91 Polynomial Chaos 1D Heat Conduction We need to represent a random field to describe the conductivity a(x, ξ) One option is to use a polynomial-chaos type representation: P a a(x, ξ) = a k (x)ψ k k=0 For P a sufficiently high we can represent the random process Figure: Microstructures in a metal (From Professor H. Fujii, JWRI)

92 Polynomial Chaos 1D Heat Conduction ( ) u(x, ξ) a(x, ξ) = f x D (0, 1) x x u(0, ξ) = u(1, ξ) = 0 Plug in the expressions for the (unknown) solution u(x, ξ) = P k=0 u k(x)ψ k and the conductivity a(x, ξ) = P a k=0 a k(x)ψ k, and perform the Galerkin projection: * XP a a k (x)ψ k ( ) P P j=0 u! + j(x)ψ j ( ) f, ψ i = 0 i = 0, 1,, P x x k=0 (1)

93 Polynomial Chaos 1D Heat Conduction Since ψ i ψ j = δ ij ψ i ψ i and ψ 0 = 1 and ψ i = 0, i = 1,, P,!! PX P a X u j (x) a k (x)c ijk = f i = 0 x x j=0 PX x j=0 with C ijk = ψ i ψ j ψ k k=0 XP a k=0 u i (0) = u i (1) = 0 a k (x)c ijk! u j (x) x! i = 0, 1,, P = 0 i = 1,, P This is a system of P + 1 coupled elliptic PDEs

94 Polynomial Chaos 1D Heat Conduction The coupling is due to the randomness in the diffusion term and the fact that in general C ijk δ ij, k Notice the similarity of the terms corresponding to indices i, j and the deterministic version of the original PDE, i.e. when the diffusivity is just a function of space x. However, due to the coupling, one cannot use deterministic codes without changing them accordingly. This is why the Stochastic Galerkin scheme is called an intrusive technique. The space discretization of the coupled system can be done through a suitable scheme of one s choice, e.g. Finite Element, Finite Difference, etc. Due to orthogonality of basis some terms cancel out

95 Polynomial Chaos 1D Heat Conduction K 0,0 K 0,1 K 0,P K 1,0 K 1,1 K 1,P K P,0 K P,1 K P,P u 0 u 1. u P = f The left-hand-side (l.h.s) matrix is block symmetric The l.h.s matrix is symmetric if the space discretization is symmetric If the number of degrees-of-freedom for the space discretization of each u i (x) is N then: Each block K i,j has size N N Thus the l.h.s. matrix has size N(P + 1) N(P + 1)

96 Polynomial Chaos 1D Heat Conduction When a(x, y) is linear w.r.t. y the l.h.s matrix is sparse (C ijk = ψ i ψ j ψ k is sparse) d = 4, p = 1, nnz = 13 d = 4, p = 2, nnz = 55 d = 4, p = 3, nnz = 155 When a(x, y) is non-linear w.r.t. y and p a 2p the l.h.s matrix is full (easy to show)

97 Polynomial Chaos Navier-Stokes Equations Consider the NS equations for an incompressible fluid u i x i = 0 u i t + u j u i x j = 1 ρ p x i + µ 2 u i x j x j

98 Polynomial Chaos Navier-Stokes Equations Consider the NS equations for an incompressible fluid u i t u i x i = 0 u i + u j = 1 p + µ 2 u i x j ρ x i x j x j Assuming that the uncertainty is represented with one uncertain variable ξ, the usual polynomial chaos expansion reads u i (x, t, ξ) = P j=0 u (j) i (x, t)ψ j (ξ) p(x, t, ξ) = P p (j) (x, t)ψ j (ξ) j=0

99 Polynomial Chaos Navier-Stokes Equations The PC expansion for the velocity plugged in the continuity ( u i / x i = 0) gives u (k) i x i = 0 k = 0,..., P

100 Polynomial Chaos Navier-Stokes Equations The PC expansion for the velocity plugged in the continuity ( u i / x i = 0) gives u (k) i x i = 0 k = 0,..., P The momentum equation instead becomes (for each k component) u (k) i t + P P m=0 n=0 u (m) u (n) i ψ m ψ n ψ k j = 1 p (k) x j ψ k ψ k ρ x i + µ 2 u (k) i x j x j

101 Polynomial Chaos Navier-Stokes Equations The PC expansion for the velocity plugged in the continuity ( u i / x i = 0) gives u (k) i x i = 0 k = 0,..., P The momentum equation instead becomes (for each k component) u (k) i t + P P m=0 n=0 u (m) u (n) i ψ m ψ n ψ k j = 1 p (k) x j ψ k ψ k ρ x i + µ 2u (k) i x j x j We obtain P + 1 equations for the velocity-mode vectors and P + 1 constraints.

102 Polynomial Chaos Navier-Stokes Equations The PC expansion for the velocity plugged in the continuity ( u i / x i = 0) gives u (k) i x i = 0 k = 0,..., P The momentum equation instead becomes (for each k component) u (k) i t + P P m=0 n=0 u (m) u (n) i ψ m ψ n ψ k j = 1 p (k) x j ψ k ψ k ρ x i + µ 2u (k) i x j x j We obtain P + 1 equations for the velocity-mode vectors and P + 1 constraints. Not dissimilar from deterministic system Can be solved by projection and results in a coupled system of 3 (P + 1) momentum-like equations with P + 1 constraints.

103 Summary of Probabilistic Uncertainty Propagation Methods

104 Probabilistic Approaches to UQ Define/characterize the input uncertainty in terms of a set of d random variables Propagation methodologies Sampling-based techniques: MC, LHS Quadrature-based methods: SC Galerkin-based approaches: PC Compute statistics of the output of interest

105 Probabilistic Approaches to UQ Computing statistics Sampling methods compute statistics and pdf by interrogating the parameter space SC and PC are radically different, because the moments are computed explicitly SC: from quadrature rules PC: by construction from the assumed solution representation

106 Probabilistic Approaches to UQ Computing statistics Sampling methods compute statistics and pdf by interrogating the parameter space SC and PC are radically different, because the moments are computed explicitly SC: from quadrature rules PC: by construction from the assumed solution representation How do we compute the pdf in SC and PC? We interpret both SC and PC as approximate (surrogate) representations of the solution that depend continuously on the input random variables

107 Probabilistic Approaches to UQ Computing pdfs We interpret both SC and PC as approximate (surrogate) representations of the solution that depend continuously on the input random variables

108 Probabilistic Approaches to UQ Computing pdfs We interpret both SC and PC as approximate (surrogate) representations of the solution that depend continuously on the input random variables P PC: it is obvious that u(x, t, ξ) u P G = u i (x, t)ψ i (ξ) we can sample (MC) ug P (x, t, ξ) directly i=0

109 Probabilistic Approaches to UQ Computing pdfs We interpret both SC and PC as approximate (surrogate) representations of the solution that depend continuously on the input random variables P PC: it is obvious that u(x, t, ξ) u P G = u i (x, t)ψ i (ξ) we can sample (MC) ug P (x, t, ξ) directly SC: Gaussian quadrature rules are weighted sums where the weights are integrals of Lagrange interpolating polynomials, therefore we can interpret SC as M u(x, t, ξ) uc M = u(x, t, ξ i )L i,p (ξ) i=0 we can sample (MC) uc M (x, t, ξ) directly i=0

110 Probabilistic Approaches to UQ Computing pdfs We interpret both SC and PC as approximate (surrogate) representations of the solution that depend continuously on the input random variables P PC: it is obvious that u(x, t, ξ) u P G = u i (x, t)ψ i (ξ) we can sample (MC) ug P (x, t, ξ) directly SC: Gaussian quadrature rules are weighted sums where the weights are integrals of Lagrange interpolating polynomials, therefore we can interpret SC as M u(x, t, ξ) uc M = u(x, t, ξ i )L i,p (ξ) i=0 we can sample (MC) uc M (x, t, ξ) directly Note: sampling u, u G and u C is NOT in general equivalent! i=0