The Averaged Gradient Technique inSensitivity Analysis

Jan Chleboun

Department of Mathematics, Faculty of Civil Engineering, Prague

Dolní Maxov, June 1–6, 2008

1 / 20

Outline

Averaged (a.k.a. Recovered) GradientIdeaProperties

Sensitivity AnalysisMotivationThe Core of the Sensitivity FormulaeObservations

Comments on Matlab and PDE Toolbox

2 / 20

Averaged (a.k.a. Recovered) Gradient

Idea (in 2D): Let (uax , ua

y ) be an approximation of ∇u. Byprocessing (ua

x , uay ), we wish to obtain (ua

x , uay ), a better (more

accurate) approximation of ∇u.

Application: In FEM, ‖∇u − (uhx , uhy )‖L2 ≤ Ch, where uh is aFE approximation to a BVP solution u. It can be‖∇u − (uhx , uhy )‖L2 ≤ Ch2.

Various methods. We will use the method proposed in

I. Hlavácek, M. Krížek, V. Pištora: How to recover the gradientof linear elements on nonuniform triangulations, Appl. Math. 41(1996), 241-267

3 / 20

Recovered Gradient: Technique

Z

A2

A1

B2

B1

Z

A2

A1

B2

B1

ai = (Ai − Z )i , bi = (Bi − Z )i , i = 1, 2(. . . d)

The weighted averaged gradient of function v at node Z

(Ghv(Z ))i = αiv(Ai) − (αi + βi)v(Z ) + βiv(Bi )

where

αi =bi

ai(bi − ai), βi =

ai

bi(ai − bi).

4 / 20

Properties IAssumptions:Ω ⊂ R

2 (R3), a bounded domain with a polygonal Lipschitzboundary.F , a strongly regular family of triangulations Thh→0, i.e.,

∃κ > 0 ∀Th ∈ F ∀K ∈ Th κh ≤ %K ,

where %K is the radius of a ball BK such that BK ⊂ K .

Theorem: Let q ∈ (1,∞) and v ∈ W 3q (Ω). Then exists a

constant C > 0 such that

‖∇v − Ghv‖0,q ≤ Ch2|v |3,q

for any Th ∈ F .

5 / 20

Properties IIAn elliptic BVP

− div(λ∇u) = f in Ω,

u = 0 on ∂Ω,

where λ is a matrix comprising sufficiently smooth functions(λ ∈ [W 2

2+ε(Ω)]2×2).

It is assumed that u ∈ W 3q (Ω), q > 2.

Let us consider uh, the finite element approximation to u (p.-w.linear basis functions).

Is Ghuh close to ∇u?

6 / 20

Properties IIAssumptions: The family F = Thh→0 is generated by smoothdistortions of uniform triangulations of square grids.

(Then the linear interpolation of uis sufficiently close to u.)

Ω0, a domain such that Ω0 ⊂ Ω

Theorem: ‖∇u − Ghuh‖0,2,Ω0≤ C(u)h2.

7 / 20

Sensitivity Analysis: MotivationThe weak formulation of the (simplified) BVP

∫Ω

λ(x)∇u · ∇v dx =

∫Ω

fv dx ∀v ∈ H10 (Ω), u ∈ H1

0 (Ω).

A unique solution u(λ; ·).

Criterion-functional: Ψ(λ) = Φ(u(λ))

Example (Parameter Identification)A function w is given. The goal is to find λ that minimizes

Ψ(λ) =

∫Ω

(w(x) − u(λ; x))2 dx .

Analogous settings appear in optimal design problems, theworst scenario method, or PDE/ODE-constrained optimization.

8 / 20

Sensitivity AnalysisIn the search for

arg minλ∈Uad

Ψ(λ),

where Uad is an admissible set of parameters, the derivative Ψ′

of Ψ w.r.t. λ is beneficial.

Remark: In discretized problems (FEM and a discretization ofUad), Ψ(λ) is a function of several variables λ1, . . . , λn and Ψ′

becomes ∇Ψ.

Let us focus on a particular approach to the Ψ′ calculation: theformula for Ψ′ is derived from the "continuous" weak problem,and then approximated by the solutions of discretized problems.

9 / 20

In our problem, letλ(x1, x2) = λ1 + λ2x1 + λ3x2 + λ4x2

1 + λ5x1x2 + +λ6x22 .

Then, for instance,

∂Ψ

∂λ4(λ) = −

∫Ω

x21∇u(x) · ∇z(x) dx ,

where z ∈ H10 (Ω) solves the adjoint equation

∫Ω

λ(x)∇z · ∇v dx =

∫Ω

2(u − w) dx ∀v ∈ H10 (Ω)

(the right-hand side equals the derivative of Φ(u) w.r.t. u).

10 / 20

Generally (not only in our example!), Ψ′ is evaluated throughthe integration of an expression containing ∇u and ∇z. Sincethe accuracy of Ψ′ depends on the accuracy of theapproximations of ∇u and ∇z, the idea of recovering ∇uh and∇zh, the FEM-computed gradients, emerged.

11 / 20

Generally (not only in our example!), Ψ′ is evaluated throughthe integration of an expression containing ∇u and ∇z. Sincethe accuracy of Ψ′ depends on the accuracy of theapproximations of ∇u and ∇z, the idea of recovering ∇uh and∇zh, the FEM-computed gradients, emerged.

The idea failed.

The gradient of Ψ computed by means of the recoveredgradient technique is not better than the gradient computeddirectly.

11 / 20

Why?

Reasons:

1. A bug in the code?

12 / 20

Why?

Reasons:

1. A bug in the code?

2. A "boundary layer", where the averaging is inaccurate.

12 / 20

Interpolation features of Gh, uniform mesh

0

0.5

1

0

0.5

10

0.01

0.02

0.03

0.04

0.05

Eucl. norm of the diff. in the gradients: pwc − exact (centroids)

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0

0.5

1

0

0.5

10

0.2

0.4

0.6

0.8

1

x 10−16

Eucl. norm of the diff. in the gradients: rpwc − exact (nodes)

0

1

2

3

4

5

6

7

8

x 10−17

Interpolated function: xy(1 − x)(1 − y)Euclid. norm of the difference between the calculated and theexact gradient: p.-w. constant gradient (left), recovered gradient(right).

13 / 20

Interpolation features of Gh, distorted mesh

0

0.5

1

0

0.2

0.4

0.6

0.8

10

0.02

0.04

0.06

Eucl. norm of the diff. in the gradients: pwc − exact (centroids)

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0

0.5

1

0

0.5

10

0.02

0.04

0.06

Eucl. norm of the diff. in the gradients: rpwc − exact (nodes)

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

Interpolated function: xy(1 − x)(1 − y)Euclid. norm of the difference between the calculated and theexact gradient: p.-w. constant gradient (left), recovered gradient(right).

14 / 20

FEM features of Gh, uniform mesh

0

0.5

1

0

0.2

0.4

0.6

0.8

10

0.02

0.04

0.06

Eucl. n. of the diff. in the gradients: pwc FEM − exact (centroids)

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0

0.5

1

0

0.5

10

1

2

3

4

x 10−3

Eucl. n. of the diff. in the gradients: rpwc FEM − exact (nodes)

0.5

1

1.5

2

2.5

3

3.5x 10

−3

FEM solution of the Poisson equation with the exact solutionxy(1 − x)(1 − y).Euclid. norm of the difference between the calculated and theexact gradient: p.-w. constant gradient (left), recovered gradient(right).

15 / 20

FEM features of Gh, distorted mesh

0

0.5

1

0

0.2

0.4

0.6

0.8

10

0.02

0.04

0.06

Eucl. n. of the diff. in the gradients: pwc FEM − exact (centroids)

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0

0.5

1

0

0.5

10

0.02

0.04

0.06

Eucl. n. of the diff. in the gradients: rpwc FEM − exact (nodes)

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

FEM solution of the Poisson equation with the exact solutionxy(1 − x)(1 − y).Euclid. norm of the difference between the calculated and theexact gradient: p.-w. constant gradient (left), recovered gradient(right).

16 / 20

Gradient of the adjoint state, uniform mesh

0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

10

0.5

1

Eucl. n. of the adj. gradient (FEM, centroids)

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0

0.5

1

0

0.2

0.4

0.6

0.8

10

0.5

1

Eucl. n. of the recov. adj. gradient (FEM, nodes)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

The Euclid. norm of the p.-w. constant (left) and recoveredgradient of the adjoint solution.

Observations: A boundary layer exists where (a) the recoveredgradient of ∇uh (and ∇zh) is inaccurate, and (b) the amount ofsensitivity information stored in ∇zh is substantial.Conclusion: The performance of ∇uh · ∇zh is poor even forrecovered gradients.

17 / 20

Work in progress

I Exact integration in the FEM code.I Focus not on ∇uh · ∇zh in the sensitivity formula, but on

special criterion-functionals as

Ψ(λ) =

∫Ω0

(w,x1 − u,x1)2 + (w,x2 − u,x2)

2 dx ,

where u solves the Poisson equation, w is a given function,and Ω0 ⊂ Ω.

18 / 20

Matlab and PDE ToolboxTwo errors discovered in R2007b and R2008a (R2007a too?,not in R2006a):

I The format of the boundary condition matrix "b" as used inthe past and as described in the PDE Toolbox help systemor handbook causes an error message and the breakdownof the code.Will be fixed in R2008b. DIY fixing is easy.

I The PDE Toolbox GUI (pdetool) does not solve BVPswith a Dirichlet homogeneous boundary condition. Itdelivers a solution that is nonzero along the boundary.Will be fixed?

19 / 20

Acknowledgement: I thank dr. Tomáš Vejchodský for his routinefor numerical integration.

Thank you for your attention

20 / 20