Numerical Schemes for the Hamilton-Jacobi Equation ...jwcalder/NumSchemeTalk.pdf · Multi-query image retrieval Problem:Find images in a dataset S that are similar to multiple query

Numerical Schemes for the Hamilton-Jacobi EquationContinuum Limit of Non-dominated Sorting

Jeff Calder

Department of MathematicsUniversity of California, Berkeley

Applied Mathematics Seminar

McGill University

September 9, 2015

Calder (UC Berkeley) Numerical schemes September 9, 2015 1 / 63

Outline

1 IntroductionMotivating example: Image retrievalNon-dominated sorting

2 Continuum limit for nondominated sortingHamilton-Jacobi equation for layersPDE-based ranking

3 Numerical schemesAn O(h1/n) schemeTwo (formally) O(h) schemesRegularityConvergence rates

4 Experimental results

5 References


Motivating example: Google Goggles

Query image

Retrieved images


Multi-query image retrieval

Problem: Find images in a dataset S thatare similar to multiple query images.

Pareto method: Solve the multi-criteriaoptimization problem

argminI∈S

(dist(I ,Q1), . . . , dist(I ,Qd)).

Query 1 Query 2

Pareto points:



Problem: Find images in a dataset S thatare similar to multiple query images.

Pareto method: Solve the multi-criteriaoptimization problem

argminI∈S

(dist(I ,Q1), . . . , dist(I ,Qd)).

Query 1 Query 2

Pareto points:


Multi-objective optimizationHow do we solve the multi-objective optimization problem

argminI∈S

(f1(I ), . . . , fd(I ))?

Basic approach:

1 Choose some weights αi ∈ [0, 1] with∑d

i=1 αi = 1 and define

fα(I ) = α1f1(I ) + α2f2(I ) + · · ·+ αd fd(I ).

2 Solve the scalarized optimization problem

argminI∈S

fα(I ).

Problems:

1 Difficult to choose weights

2 Ignores relevant solutions



argminI∈S

(f1(I ), . . . , fd(I ))?

Basic approach:



fα(I ) = α1f1(I ) + α2f2(I ) + · · ·+ αd fd(I ).


argminI∈S

fα(I ).

Problems:





argminI∈S

(f1(I ), . . . , fd(I ))?

Basic approach:



fα(I ) = α1f1(I ) + α2f2(I ) + · · ·+ αd fd(I ).


argminI∈S

fα(I ).

Problems:





argminI∈S

(f1(I ), . . . , fd(I ))?

Basic approach:



fα(I ) = α1f1(I ) + α2f2(I ) + · · ·+ αd fd(I ).


argminI∈S

fα(I ).

Problems:




Basic approach


Basic approach


Basic approach


Basic approach


Non-dominated solutions















First Pareto front:

Query 1

Query 2

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

Hsiao, K.-J., Calder, J., and Hero III, A. O. (2015). Pareto-depth for multiple-queryimage retrieval. IEEE Transactions on Image Processing, 24(2):583–594.


Non-dominated sorting

Let X1, . . . ,XN be points in Rn and set S = X1, . . . ,XN.

Define the partial order

x 5 y ⇐⇒ xi ≤ yi for all i ∈ 1, . . . ,n.

DefinitionNon-dominated sorting is the process of arranging S into layers F1,F2,F3, . . . , definedby

F1 = Minimal elements of S ,

Fk = Minimal elements of S \⋃

j≤k−1

Fj .


Non-dominated sorting

Let X1, . . . ,XN be points in Rn and set S = X1, . . . ,XN.

Define the partial order

x 5 y ⇐⇒ xi ≤ yi for all i ∈ 1, . . . ,n.

DefinitionNon-dominated sorting is the process of arranging S into layers F1,F2,F3, . . . , definedby

F1 = Minimal elements of S ,

Fk = Minimal elements of S \⋃

j≤k−1

Fj .


Applications

Multi-objective optimization

Genetic algorithms [Deb et al., 2002]

Gene selection and ranking [Hero, 2003]

Database systems [Papadias et al., 2005]

Anomaly detection [Hsiao et al., 2012, Hsiao et al., 2015b]

Image retrieval [Hsiao et al., 2015a]

Combinatorics and probability

Longest chain in Euclidean space [Hammersley, 1972]

Patience sorting [Aldous and Diaconis, 1999]

Young Tableaux [Viennot, 1984]

Graph theory [Lou and Sarrafzadeh, 1993]

Polynuclear growth (crystals) [Prahofer and Spohn, 2000]

Other applications

Molecular biology [Pevzner, 2000]

Integrated circuit design [Adhar, 2007]


Applications













Other applications




Applications













Other applications




Demo: 50 Random samples

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Demo: Uniform distribution

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Demo: Gaussian distribution

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Demo: Uniform distribution on [0, 1]2 \ [0, 0.5]2

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N = 105 points



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Question

Can we characterize the asymptotic shapes of the Pareto fronts?


Outline





5 References


A PDE continuum limit for non-dominated sorting

Let X1, . . . ,XN be i.i.d. random variables in [0,∞)n with continuous density f .

Let UN : Rn → N0 be the function that ‘counts’ the layers F1,F2, . . .


Continuum limit

Theorem (Calder, Esedoglu, Hero, 2014)

With probability one

N−1d UN −→ u locally uniformly on [0,∞)n ,

where u ∈ C 0, 1n ([0,∞)n) is the unique nondecreasing viscosity solution of

(P1)

ux1 · · · uxn = cn f in Rn

+ := (0,∞)n

u = 0 on ∂Rn+.

Calder, J., Esedoglu, S., and Hero, A. O. (2014). A Hamilton-Jacobi equation for thecontinuum limit of non-dominated sorting. SIAM Journal on Mathematical Analysis,46(1):603–638.

Calder, J. (2015). A direct verification argument for the Hamilton-Jacobi equationcontinuum limit of nondominated sorting. arXiv preprint:1508.01565


Demo: f = 1− χ[0,0.5]2

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Fast approximate sorting

Algorithm (PDE-based Ranking)1 Select k points from X1, . . . ,XN at random. Call them Y1, . . . ,Yk .

2 Select a grid spacing h for solving the PDE (P1) and estimate f with a histogramaligned to the grid [0, 1]nh , i.e.,

f (x) =1

khn·#Yi : x 5 Yi 5 x + h(1, . . . , 1)

for x ∈ [0, 1]nh .

3 Compute the numerical solution Uh on [0, 1]nh .

4 Evaluate Uh(Xi) for i = 1, . . . ,N via interpolation.

Notes:

Total complexity is O(k + h−n + N ).

If we fix k , h and n, independent of N , then Steps 1-3 have O(1) complexity.

Calder, J., Esedoglu, S., and Hero, A. O. (2015). A PDE-based approach tonondominated sorting. SIAM Journal on Numerical Analysis, 53(1):82–104.


Outline





5 References


Numerics

How do we numerically solve

(P1)

ux1 · · · uxn = f in Rn

+

u = 0 on ∂Rn+

efficiently and accurately (in dimensions n = 2, 3, 4)?


Recall: Viscosity solution

Consider the Hamilton-Jacobi equation

H (x , u,Du) = 0 on O ⊆ Rn , (1)

where H : O × R× Rn → R is continuous.

A continuous function u : O → R is a viscosity solution of (1) if

1 Subsolution: For every x ∈ O and ϕ ∈ C∞(Rn) such that u − ϕ has a localmaximum at x with respect to O

H (x , u(x),Dϕ(x)) ≤ 0.

2 Supersolution: For every x ∈ O and ϕ ∈ C∞(Rn) such that u − ϕ has a localminimum at x with respect to O

H (x , u(x),Dϕ(x)) ≥ 0.


Bounded domain

We first pose the PDE on a bounded domain

(P1)

ux1 · · · uxn = f in (0, 1]n

u = 0 on Γ

where Γ = [0, 1]n \ (0, 1]n .

Minor technicality: Viscosity solutions of (P1) do not exist due to a well-known issuewith viscosity solutions on boundaries of domains.

We actually need to slightly modify the PDE:

(P1’)

(ux1)+ · · · (uxn )+ = f in (0, 1]n

u = 0 on Γ,

where t+ = max(t , 0).


Bounded domain


(P1)

ux1 · · · uxn = f in (0, 1]n

u = 0 on Γ

where Γ = [0, 1]n \ (0, 1]n .



(P1’)

(ux1)+ · · · (uxn )+ = f in (0, 1]n

u = 0 on Γ,



Bounded domain


(P1)

ux1 · · · uxn = f in (0, 1]n

u = 0 on Γ

where Γ = [0, 1]n \ (0, 1]n .



(P1’)

(ux1)+ · · · (uxn )+ = f in (0, 1]n

u = 0 on Γ,



Upwind finite difference scheme for (P1)

The upwind scheme corresponds to using backward differences:

(S1)

D−1 uh(x) · · ·D−n uh(x) = f (x) for x ∈ (0, 1]nh

uh(x) = 0 for x ∈ Γh ,

where Ωh := Ω ∩ hZn and

D±i uh(x) = ±uh(x ± hei)− uh(x)

h.

The scheme (S) can be solved in a single pass (similar to fast sweeping or marching),and in dimension n = 2 we have the closed form expression

uh(x) =uh(x − he1) + uh(x − he2)

2+

1

2

√(uh(x − he1)− uh(x − he2))2 + 4h2f (x)2.

However, accuracy is poor O(h1n ).



Upwind finite difference scheme for (P1)

The upwind scheme corresponds to using backward differences:

(S1)

D−1 uh(x) · · ·D−n uh(x) = f (x) for x ∈ (0, 1]nh

uh(x) = 0 for x ∈ Γh ,

where Ωh := Ω ∩ hZn and

D±i uh(x) = ±uh(x ± hei)− uh(x)

h.

The scheme (S) can be solved in a single pass (similar to fast sweeping or marching),and in dimension n = 2 we have the closed form expression

uh(x) =uh(x − he1) + uh(x − he2)

2+

1

2

√(uh(x − he1)− uh(x − he2))2 + 4h2f (x)2.

However, accuracy is poor O(h1n ).



Accuracy of (S1)For f ≡ 1, u(x) = n(x1 · · · xn)

1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .

On the other hand, if xi = h then

D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .

Therefore, for any x ∈ (0, 1]nh such that xi = h for some i we have

D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).

By comparison, uh(x) ≤ ϕ(x) whenever xi = h for some i . For x = (h, 1, . . . , 1)

uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .

Thereforeu(x)− uh(x) ≥

(n − n1− 1

n

)h

1n .



1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .


D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .


D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).


uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .


(n − n1− 1

n

)h

1n .



1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .


D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .


D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).


uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .


(n − n1− 1

n

)h

1n .



1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .


D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .


D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).

By comparison, uh(x) ≤ ϕ(x) whenever xi = h for some i .

For x = (h, 1, . . . , 1)

uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .


(n − n1− 1

n

)h

1n .



1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .


D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .


D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).


uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .


(n − n1− 1

n

)h

1n .



1n . Set ϕ(x) = Cn(x1 · · · xn)

1n and by concavity

D−i ϕ(x) ≥ ϕxi (x) = C (x1 · · · xn)1n x−1

i .


D−i ϕ(x) =ϕ(x)

h= Cn(x1 · · · xn)

1n x−1

i .


D−1 ϕ(x) · · ·D−n ϕ(x) ≥ nC n := 1

(C :=

1

n1n

).


uh(x) ≤ ϕ(x) = n1− 1n h

1n and u(x) = nh

1n .


(n − n1− 1

n

)h

1n .


Towards a new schemeThis suggests that we should try to remove the gradient singularity. Consider

v :=un

nn.

Then

vxi =un−1

nn−1uxi = v

n−1n uxi .

Thereforevx1 · · · vxn = vn−1 ux1 · · · uxn︸︷︷︸

f

= vn−1f .

We find that v is a viscosity solution of

(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ

Since f ≥ 0, (P2) has a zeroth order term of the wrong sign for a comparison principle.The method of vanishing viscosity takes the form

vεx1 · · · vεxn − ε∆vε = (vε)n−1f .



v :=un

nn.

Then

vxi =un−1

nn−1uxi = v

n−1n uxi .


f

= vn−1f .


(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ





v :=un

nn.

Then

vxi =un−1

nn−1uxi = v

n−1n uxi .


f

= vn−1f .


(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ





v :=un

nn.

Then

vxi =un−1

nn−1uxi = v

n−1n uxi .


f

= vn−1f .


(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ




Nonuniqueness for (P2)

(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ

When f ≡ 1, v(x) = u(x)n/nn = x1 · · · xn , but

vy(x) := (x1 − y1)+ · · · (xn − yn)+

is also a viscosity solution of (P2) for any y ∈ (0, 1)n .

LemmaAssume f is continuous on [0, 1]n . Then v := un/nn is the unique maximal viscositysolution of (P2).


Nonuniqueness for (P2)

(P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ

When f ≡ 1, v(x) = u(x)n/nn = x1 · · · xn , but

vy(x) := (x1 − y1)+ · · · (xn − yn)+

is also a viscosity solution of (P2) for any y ∈ (0, 1)n .

LemmaAssume f is continuous on [0, 1]n . Then v := un/nn is the unique maximal viscositysolution of (P2).


Numerical scheme for (P2)

An upwind scheme for (P2) uses backward difference quotients

(S2)

D−1 vh(x) · · ·D−n vh(x) = vn−1

h (x)f (x) for x ∈ (0, 1]nh

vh(x) = 0 for x ∈ Γh ,

We define vh by taking the largest solution of (S2) at each x ∈ (0, 1]nh .

The scheme can be solved efficiently in a single pass, and in dimension n = 2 thescheme can be solved in closed form

vh(x) =A + h2f (x)

2+

1

2

√B2 + 2h2f (x)A + h4f (x)2,

where

A = vh(x − he1) + vh(x − he2) and B = vh(x − he1)− vh(x − he2).

Notice when f ≡ 1, vh(x) = x1 · · · xn is the exact solution of (P2).




(S2)

D−1 vh(x) · · ·D−n vh(x) = vn−1

h (x)f (x) for x ∈ (0, 1]nh




vh(x) =A + h2f (x)

2+

1

2

√B2 + 2h2f (x)A + h4f (x)2,

where






(S2)

D−1 vh(x) · · ·D−n vh(x) = vn−1

h (x)f (x) for x ∈ (0, 1]nh




vh(x) =A + h2f (x)

2+

1

2

√B2 + 2h2f (x)A + h4f (x)2,

where




Rate of convergence for (S2)

(S2)

D−1 vh(x) · · ·D−n vh(x) = vn−1

h (x)f (x) for x ∈ (0, 1]nh


The scheme (S2) has formal accuracy of O(h) and we have

Theorem

Suppose f ∈ C 0,1([0, 1]n) and f > 0. Then

|nnvh − un | ≤ C√h in [0, 1]nh , (2)

and

|nv1nh − u| ≤ C δ1−n

√h in [δ, 1]nh , (3)

where δ > 0 and C = C(n, [f ]1;[0,1]n ,min[0,1]n f

).

Calder, J. (2015). Numerical schemes and rates of convergence for the Hamilton-Jacobiequation continuum limit of nondominated sorting. arXiv preprint:1508.01557.


One-sided rate for (S1)Recall uh satisfies

D−1 uh(x) · · ·D−n uh(x) = f (x) for x ∈ (0, 1]nh .

Let ψ(x) = unh /n

n . By convexity and monotoncity

D−i ψ(x) =uh(x)n − uh(x − hei)

n

nnh≤ uh(x)n−1

nn−1D−i uh(x).

ThereforeD−1 ψ(x) · · ·D−n ψ(x) ≤ ψ(x)n−1f (x),

so ψ is a subsolution of (S2). By comparison we have unh /n

n = ψ ≤ vh .

Corollary


unh − un ≤ C

√h.




Let ψ(x) = unh /n



n

nnh≤ uh(x)n−1

nn−1D−i uh(x).



n = ψ ≤ vh .

Corollary


unh − un ≤ C

√h.




Let ψ(x) = unh /n



n

nnh≤ uh(x)n−1

nn−1D−i uh(x).



n = ψ ≤ vh .

Corollary


unh − un ≤ C

√h.




Let ψ(x) = unh /n



n

nnh≤ uh(x)n−1

nn−1D−i uh(x).



n = ψ ≤ vh .

Corollary


unh − un ≤ C

√h.


Regularity

To prove the O(√h) convergence rate, we need a Lipschitz regularity result for (P2) or

(S2). Textbook regularity results are based on (p = Du)

(Coercivity) lim|p|→∞

H (x , p) =∞ uniformly in x ,

or(Zeroth order term) u + H (x ,Du) = 0.

Refer to [Bardi and Dolcetta, 1997]

The Hamiltonian for (P2) is (z = u, p = Du)

H (x , z , p) = p1 · · · pn − zn−1f (x)

which satisfies neither.


Regularity

To prove the O(√h) convergence rate, we need a Lipschitz regularity result for (P2) or

(S2). Textbook regularity results are based on (p = Du)

(Coercivity) lim|p|→∞

H (x , p) =∞ uniformly in x ,

or(Zeroth order term) u + H (x ,Du) = 0.

Refer to [Bardi and Dolcetta, 1997]

The Hamiltonian for (P2) is (z = u, p = Du)

H (x , z , p) = p1 · · · pn − zn−1f (x)

which satisfies neither.


Regularity for n = 2

Differentiate both sides of (P2) ∂∂x

(vxvy) = ∂∂x

(vf ) to find that

vxxvy + vxvxy = vx f + vfx .

Set ϕ = vx and rearrange

ϕxvy + (ϕy − f )ϕ = vfx ≤ ‖f ‖L∞ [f ]1;[0,1]n .

We can compare ϕ against a supersolution of the form

ϕ(x , y) = C (1 + y).

The bound v(x) ≤ ‖f ‖L∞xy yields boundary gradient estimates.

Lemma

Let f ∈ C 0,1([0, 1]2) be nonnegative. Then there exists C > 0 such that

[v ]1;[0,1]2 ≤ C‖f ‖C0,1([0,1]2). (4)




(vxvy) = ∂∂x

(vf ) to find that





ϕ(x , y) = C (1 + y).


Lemma


[v ]1;[0,1]2 ≤ C‖f ‖C0,1([0,1]2). (4)




(vxvy) = ∂∂x

(vf ) to find that





ϕ(x , y) = C (1 + y).


Lemma


[v ]1;[0,1]2 ≤ C‖f ‖C0,1([0,1]2). (4)




(vxvy) = ∂∂x

(vf ) to find that





ϕ(x , y) = C (1 + y).


Lemma


[v ]1;[0,1]2 ≤ C‖f ‖C0,1([0,1]2). (4)


Sketch of proofDifferentiate numerical scheme D−x (D−x vD−y v) = D−x (vf ) (x = (x , y))

D−x D−x v(x)D−y v(x− hex ) + D−x v(x)D−x D−y v(x) = v(x)D−x f (x) + f (x− hex )D−x v(x).

Set ϕ = D−x v to find

D−x ϕ(x)D−y v(x− hex ) +(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.

Set ϕ(x , y) = C (1 + y) and compute(D−y ϕ(x)− f (x− hex )

)ϕ(x) ≥ C

(C − ‖f ‖L∞

).

We can selectC = ‖f ‖L∞ +

√‖f ‖L∞ [f ]1

and we have (D−y ϕ(x)− f (x− hex )

)ϕ(x) > ‖f ‖L∞ [f ]1 .






)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) ≥ C

(C − ‖f ‖L∞

).


√‖f ‖L∞ [f ]1


)ϕ(x) > ‖f ‖L∞ [f ]1 .






)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) ≥ C

(C − ‖f ‖L∞

).


√‖f ‖L∞ [f ]1


)ϕ(x) > ‖f ‖L∞ [f ]1 .






)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) ≥ C

(C − ‖f ‖L∞

).


√‖f ‖L∞ [f ]1


)ϕ(x) > ‖f ‖L∞ [f ]1 .


Sketch of proofWe claim ϕ ≤ ϕ := C (1 + y).

Since v(x , y) ≤ ‖f ‖L∞xy , ϕ ≤ ϕ on Γ = [0, 1]2 \ (0, 1]2.Assume to the contrary that ϕ(x) > ϕ(x) for some x ∈ (0, 1]2. Then there exists xsuch that

ϕ(x) > ϕ(x), ϕ(x− hex ) ≤ ϕ(x− hex ), and ϕ(x− hey) ≤ ϕ(x− hey).

ThenD−x ϕ(x) ≥ D−x ϕ(x) = 0 and D−y ϕ(x) ≥ D−y ϕ(x) ≥ ‖f ‖L∞ .

Hence

(D−y ϕ(x)− f (x− hex ))ϕ(x) ≥ (D−y ϕ(x)− f (x− hex ))ϕ(x).

Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.

(D−y ϕ(x)− f (x− hex )

)ϕ(x) > ‖f ‖L∞ [f ]1,

we have a contradiction.Thus

D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .


Sketch of proofWe claim ϕ ≤ ϕ := C (1 + y). Since v(x , y) ≤ ‖f ‖L∞xy , ϕ ≤ ϕ on Γ = [0, 1]2 \ (0, 1]2.

Assume to the contrary that ϕ(x) > ϕ(x) for some x ∈ (0, 1]2. Then there exists xsuch that



Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,


D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .


Sketch of proofWe claim ϕ ≤ ϕ := C (1 + y). Since v(x , y) ≤ ‖f ‖L∞xy , ϕ ≤ ϕ on Γ = [0, 1]2 \ (0, 1]2.Assume to the contrary that ϕ(x) > ϕ(x) for some x ∈ (0, 1]2. Then there exists xsuch that



Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,


D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .





Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,


D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .





Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,


D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .





Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,

we have a contradiction.

Thus

D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .





Hence


Recalling

D−x ϕ(x)D−y v(x− hex )︸︷︷︸≥0

+(D−y ϕ(x)− f (x− hex )

)ϕ(x) = vD−x f ≤ ‖f ‖L∞ [f ]1.


)ϕ(x) > ‖f ‖L∞ [f ]1,


D−x v ≤ 2C = 2(‖f ‖L∞ +

√‖f ‖L∞ [f ]1

)≤ C‖f ‖C0,1 .




(vxvyvz ) = ∂∂x

(v2f ) to find that

vxxvyvz + vxvyxvz + vxvyvzx = 2vvx f + v2fx .


F (ϕ) := ϕxvyvz + (vzϕy + vyϕz − 2vf )ϕ = v2fx ≤ ‖f ‖2L∞ [f ]1.

Perhaps we should be looking for a supersolution of the form

ϕ(x , y , z ) = C (1 + y + z )?

Then F (ϕ) ≥ C (Cvz + Cvy − 2vf ). . .

At a maximum of ϕ, ϕx = ϕy = ϕz = 0 so

−2vf ϕ ≤ ‖f ‖2L∞ [f ]1.




(vxvyvz ) = ∂∂x

(v2f ) to find that





ϕ(x , y , z ) = C (1 + y + z )?



−2vf ϕ ≤ ‖f ‖2L∞ [f ]1.




(vxvyvz ) = ∂∂x

(v2f ) to find that





ϕ(x , y , z ) = C (1 + y + z )?



−2vf ϕ ≤ ‖f ‖2L∞ [f ]1.




(vxvyvz ) = ∂∂x

(v2f ) to find that





ϕ(x , y , z ) = C (1 + y + z )?



−2vf ϕ ≤ ‖f ‖2L∞ [f ]1.


Regularity for n ≥ 3?

Unfortunately this argument fails for n ≥ 3 due to the additional nonlinearity.

Furthermore, we cannot directly prove rates for (P2) due to the wrong sign on zerothorder term.

Back to the drawing board!


Regularity for n ≥ 3?

Unfortunately this argument fails for n ≥ 3 due to the additional nonlinearity.

Furthermore, we cannot directly prove rates for (P2) due to the wrong sign on zerothorder term.

Back to the drawing board!


Another PDERecall for f ≡ 1, u(x) = n(x1 · · · xn)

1n . For general f , we are tempted to make the

ansatzu(x) = n(x1 · · · xn)

1n w(x).

Proceeding formally

uxi = (x1 · · · xn)1nw

xi+ n(x1 · · · xn)

1n wxi =

1

xi(x1 · · · xn)

1n (w + nxiwxi ).

Therefore

f = ux1 · · · uxn =n∏

i=1

(w + nxiwxi ).

We can show that w is a bounded viscosity solution of

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

Notice:

Zeroth order term has correct sign! (since uxi ≥ 0)

v = un/nn = x1 · · · xnwn .





1n w(x).

Proceeding formally

uxi = (x1 · · · xn)1nw

xi+ n(x1 · · · xn)

1n wxi =

1

xi(x1 · · · xn)

1n (w + nxiwxi ).

Therefore

f = ux1 · · · uxn =n∏

i=1

(w + nxiwxi ).


(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

Notice:


v = un/nn = x1 · · · xnwn .





1n w(x).

Proceeding formally

uxi = (x1 · · · xn)1nw

xi+ n(x1 · · · xn)

1n wxi =

1

xi(x1 · · · xn)

1n (w + nxiwxi ).

Therefore

f = ux1 · · · uxn =n∏

i=1

(w + nxiwxi ).


(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

Notice:


v = un/nn = x1 · · · xnwn .


Regularity for (P3)Let us rewrite (P3) as

n∑j=1

log(w + nxjwxj ) = log(f ) in (0, 1]n .

Differentiate each side in xi :

n∑j=1

wxi + nδi,jwxj + nxjwxixj

w + nxjwxj

=fxif.

At a maximum of wxi we have wxixj = 0 hence

wxi

n∑j=1

1 + nδi,jw + nxjwxj

=fxif.

By the inequality of arithmetic and geometric means

n∑j=1


≥n∑

j=1

1

w + nxjwxj

≥ n

n∏j=1

(w + nxjwxj )−1n = nf −

1n .



n∑j=1



n∑j=1


w + nxjwxj

=fxif.


wxi

n∑j=1


=fxif.


n∑j=1


≥n∑

j=1

1

w + nxjwxj

≥ n

n∏j=1


1n .



n∑j=1



n∑j=1


w + nxjwxj

=fxif.


wxi

n∑j=1


=fxif.


n∑j=1


≥n∑

j=1

1

w + nxjwxj

≥ n

n∏j=1


1n .



n∑j=1



n∑j=1


w + nxjwxj

=fxif.


wxi

n∑j=1


=fxif.


n∑j=1


≥n∑

j=1

1

w + nxjwxj

≥ nn∏

j=1


1n .


Regularity for (P3)

Hence, at a positive maximum of wxi we have

wxi ≤1

nf

1n−1fxi = (f

1n )xi ≤ [f

1n ]1;[0,1]n .

Problem: We have no boundary condition for w , how do we know wxi attains amaximum or minimum?

Recall

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

When f ≡ 1, the solution of (P3) that we care about is w ≡ 1. For any C > 0

w(x) =n∏

i=1

(1 + Cx−1i )

1n

is an unbounded solution of (P3).


Regularity for (P3)


wxi ≤1

nf

1n−1fxi = (f

1n )xi ≤ [f

1n ]1;[0,1]n .


Recall

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .


w(x) =n∏

i=1

(1 + Cx−1i )

1n



Regularity for (P3)


wxi ≤1

nf

1n−1fxi = (f

1n )xi ≤ [f

1n ]1;[0,1]n .


Recall

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .


w(x) =n∏

i=1

(1 + Cx−1i )

1n



Extension property for (P3)

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

LemmaLet f ≥ 0 be continuous on [0, 1]n and let u be the solution of (P1). Then

w(x) =u(x)

n(x1 · · · xn)1n

can be extended to a continuous function w ∈ C ((−∞, 1]n) satisfying

w(x) = w(x+) for all x ∈ (−∞, 1]n ,

where x+ = (max(x1, 0), . . . ,max(xn , 0)). Furthermore, if we extend f by settingf (x) := f (x+) then w is a viscosity solution of

n∏i=1

(w + nxiwxi ) = f in (−∞, 1]n .


Regularity for (P3)

The extension property allows us to complete the regularity result.

Theorem

Let f be a nonnegative function for which f1n ∈ C 0,1([0, 1]n), and let w be the maximal

bounded viscosity solution of (P3). Then w ∈ C 0,1([0, 1]n) and

[w ]1;[0,1]n ≤√n [f

1n ]1;[0,1]n . (5)

Theorem is sharp: When f (x) = (x1 · · · xn)n−1, w(x) = 1n

(x1 · · · xn)1−1n 6∈ C 0,1([0, 1]n).

Since v = x1 · · · xnwn we have the following corollary.

Corollary

Let v be the maximal viscosity solution of (P2). Then v ∈ C 0,1([0, 1]n) and

[v ]1;[0,1]n ≤ C‖fn−1n ‖L∞([0,1]n )‖f

1n ‖C0,1([0,1]n ). (6)


Regularity for (P3)

The extension property allows us to complete the regularity result.

Theorem

Let f be a nonnegative function for which f1n ∈ C 0,1([0, 1]n), and let w be the maximal

bounded viscosity solution of (P3). Then w ∈ C 0,1([0, 1]n) and

[w ]1;[0,1]n ≤√n [f

1n ]1;[0,1]n . (5)

Theorem is sharp: When f (x) = (x1 · · · xn)n−1, w(x) = 1n

(x1 · · · xn)1−1n 6∈ C 0,1([0, 1]n).

Since v = x1 · · · xnwn we have the following corollary.

Corollary

Let v be the maximal viscosity solution of (P2). Then v ∈ C 0,1([0, 1]n) and

[v ]1;[0,1]n ≤ C‖fn−1n ‖L∞([0,1]n )‖f

1n ‖C0,1([0,1]n ). (6)


Numerical scheme for (P3)An upwind scheme for (P3) uses backward difference quotients

(S3)n∏

i=1

(wh(x) + nxiD

−i wh(x)

)= f (x) for x ∈ [0, 1]nh .

We define wh by taking the largest solution of (P2) at each x ∈ [0, 1]nh .Notice

wh(0) = f (0)1n .

When f ≡ 1, wh ≡ 1 which is the exact solution of (P3).


wh(x) = C +√

D2 + (2x1 + h)(2x2 + h)h2f (x),

whereC = x1(2x2 + h)wh(x − he1) + x2(2x1 + h)wh(x − he2),

andD = x1(2x2 + h)wh(x − he1)− x2(2x1 + h)wh(x − he2).



(S3)n∏

i=1

(wh(x) + nxiD

−i wh(x)

)= f (x) for x ∈ [0, 1]nh .


wh(0) = f (0)1n .



wh(x) = C +√

D2 + (2x1 + h)(2x2 + h)h2f (x),





(S3)n∏

i=1

(wh(x) + nxiD

−i wh(x)

)= f (x) for x ∈ [0, 1]nh .


wh(0) = f (0)1n .



wh(x) = C +√

D2 + (2x1 + h)(2x2 + h)h2f (x),




Convergence rate

Since (P3) has a zeroth order term of the correct sign, and w and wh are Lipschitzcontinuous on [0, 1]n we can prove

Theorem

Suppose that f ∈ C 0,1([0, 1]n) and f > 0. Then

|w(x)− wh(x)| ≤ C√h for all x ∈ [0, 1]nh , (7)

where C = C(n, [f ]1;[0,1]n ,min[0,1]n f

).


Directed Hamiltonians

All of the PDE in this talk have a common useful property.

H (x , u(x),Du(x)) = 0.

DefinitionWe say that H (x , z , p) is directed if

H (x , z , p) ≤ H (x , z , q) whenever pi ≤ qi for all i .

Recall:

(P1) H (x , z , p) = (p1)+ · · · (pn)+ − f (x)

(P2) H (x , z , p) = (p1)+ · · · (pn)+ − zn−1f (x)

(P3) H (x , z , p) =n∏

i=1

(z + nxipi)+ − f (x)


Directed Hamiltonians

All of the PDE in this talk have a common useful property.

H (x , u(x),Du(x)) = 0.

DefinitionWe say that H (x , z , p) is directed if

H (x , z , p) ≤ H (x , z , q) whenever pi ≤ qi for all i .

Recall:

(P1) H (x , z , p) = (p1)+ · · · (pn)+ − f (x)

(P2) H (x , z , p) = (p1)+ · · · (pn)+ − zn−1f (x)

(P3) H (x , z , p) =n∏

i=1

(z + nxipi)+ − f (x)


Monotone (upwind) scheme for directed H

A monotone scheme for a directed Hamiltonian H is always to use backward differences

H (x , u(x),D−u(x)) = 0,

whereD−u(x) = (D−1 u(x), . . . ,D−n u(x)).

Indeed, if u ≥ v and u(x) = v(x) then D−i u(x) ≤ D−i v(x) for all i and

H (x , u(x),D−u(x)) ≤ H (x , v(x),D−v(x)),

which is exactly monotonicity of the scheme.


Monotone (upwind) scheme for directed H

A monotone scheme for a directed Hamiltonian H is always to use backward differences

H (x , u(x),D−u(x)) = 0,

whereD−u(x) = (D−1 u(x), . . . ,D−n u(x)).

Indeed, if u ≥ v and u(x) = v(x) then D−i u(x) ≤ D−i v(x) for all i and

H (x , u(x),D−u(x)) ≤ H (x , v(x),D−v(x)),

which is exactly monotonicity of the scheme.


Convergence rate

(S)

uh + H (D−uh) = f for x ∈ (0, 1]nh

uh = g for x ∈ Γh ,

(H)

u + H (Du) = f in (0, 1]n

u = g on Γ,

Theorem ([Crandall and Lions, 1984, Souganidis, 1985])

Assume H ∈ C 0,1(Rn) is directed and f , g ∈ C 0,1([0, 1]n). Let uh be a solution of (S)and let u be a viscosity solution of (H). If u ∈ C 0,1([0, 1]n) and

suph>0

supx 6=y

|uh(x)− uh(y)||x − y | <∞,

then|uh − u| ≤ C

√h.


Sketch of proof that uh ≤ u + C√h

For α > 0 setΦ(x , y) = uh(x)− u(y)− α

2|x − y |2,

and let (xα, yα) ∈ (0, 1]nh × (0, 1]n such that Φ(xα, yα) = max[0,1]nh×[0,1]n Φ.

Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).

Then ϕ(xα) = uh(xα) and uh(x) ≤ ϕ(x) for all x ∈ [0, 1]nh . Since H is directed

f (xα) = uh(xα) + H (D−uh(xα)) ≥ uh(xα) + H (D−ϕ(xα)).

Since H ∈ C 0,1(Rn) and D−ϕ(xα) = pα − αh2

(1, . . . , 1) where pα = α(xα − yα)

uh(xα) + H (pα) ≤ f (xα) + Cαh.

Let ψ(y) = α2|xα − y |2. Since u + ψ has a minimum at yα and −Dψ(yα) = pα

u(yα) + H (pα) ≥ f (yα).

Hence: uh(xα)− u(yα) ≤ f (xα)− f (yα) + Cαh.




2|x − y |2,


Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).

Then ϕ(xα) = uh(xα) and uh(x) ≤ ϕ(x) for all x ∈ [0, 1]nh .

Since H is directed



(1, . . . , 1) where pα = α(xα − yα)



u(yα) + H (pα) ≥ f (yα).





2|x − y |2,


Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).




(1, . . . , 1) where pα = α(xα − yα)



u(yα) + H (pα) ≥ f (yα).





2|x − y |2,


Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).




(1, . . . , 1) where pα = α(xα − yα)



u(yα) + H (pα) ≥ f (yα).





2|x − y |2,


Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).




(1, . . . , 1) where pα = α(xα − yα)



u(yα) + H (pα) ≥ f (yα).





2|x − y |2,


Setϕ(x) =

α

2|x − yα|2 −

α

2|xα − yα|2 + uh(xα).




(1, . . . , 1) where pα = α(xα − yα)



u(yα) + H (pα) ≥ f (yα).




Since Φ(x , y) = uh(x)− u(y)− α2|x − y |2

max[0,1]n

h

(uh − u) ≤ Φ(xα, yα) ≤ uh(xα)− u(yα) ≤ f (xα)− f (yα) + Cαh.

Since Φ(xα, xα) ≤ Φ(xα, yα) and u ∈ C 0,1([0, 1]n)

uh(xα)− u(xα) ≤ uh(xα)− u(yα)− α

2|xα − yα|2

or α

2|xα − yα|2 ≤ u(xα)− u(yα) ≤ C |xα − yα|.

Hence |xα − yα| ≤ C/α. Therefore

max[0,1]n

h

(uh − u) ≤ C

(1

α+ αh

).

The best possible choice for α is α = 1/√h which yields

uh − u ≤ C√h.




max[0,1]n

h




2|xα − yα|2

or α


Hence |xα − yα| ≤ C/α.

Therefore

max[0,1]n

h

(uh − u) ≤ C

(1

α+ αh

).


uh − u ≤ C√h.




max[0,1]n

h




2|xα − yα|2

or α



max[0,1]n

h

(uh − u) ≤ C

(1

α+ αh

).


uh − u ≤ C√h.




max[0,1]n

h




2|xα − yα|2

or α



max[0,1]n

h

(uh − u) ≤ C

(1

α+ αh

).


uh − u ≤ C√h.


Convergence rate for (S3)

(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .

The proof that |w − wh | ≤ C√h is similar, with the extension property replacing the

boundary condition.

Zeroth order term: Notice we can write (P3) as H (x , u,Du) = f (x) where

H (x , z , p) =n∏

i=1

(z + nxipi)

Therefore

∂

∂zH (x , z , p) = H (x , z , p)

n∑i=1

1

z + nxipi

≥ nH (x , z , p)

(n∏

i=1

1

z + nxipi

) 1n

= nH (x , z , p)n−1n

= nf (x)n−1n .

Proof requires f ≥ m > 0.



(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .


boundary condition.Zeroth order term: Notice we can write (P3) as H (x , u,Du) = f (x) where

H (x , z , p) =n∏

i=1

(z + nxipi)

Therefore

∂

∂zH (x , z , p) = H (x , z , p)

n∑i=1

1

z + nxipi

≥ nH (x , z , p)

(n∏

i=1

1

z + nxipi

) 1n

= nH (x , z , p)n−1n

= nf (x)n−1n .

Proof requires f ≥ m > 0.



(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n .


boundary condition.Zeroth order term: Notice we can write (P3) as H (x , u,Du) = f (x) where

H (x , z , p) =n∏

i=1

(z + nxipi)

Therefore

∂

∂zH (x , z , p) = H (x , z , p)

n∑i=1

1

z + nxipi

≥ nH (x , z , p)

(n∏

i=1

1

z + nxipi

) 1n

= nH (x , z , p)n−1n

= nf (x)n−1n .

Proof requires f ≥ m > 0.Calder (UC Berkeley) Numerical schemes September 9, 2015 52 / 63


(P3)n∏

i=1

(w + nxiwxi ) = f in (0, 1]n (P2)

vx1 · · · vxn = vn−1f in (0, 1]n

v = 0 on Γ

Recall v = x1 · · · xnwn . This identity approximately holds for wh and vh :

Lemma

Assume f ∈ C 0,1([0, 1]n) and f > 0. Then

|x1 · · · xnwh(x)n − vh(x)| ≤ Ch for all x ∈ [0, 1]nh , (8)

whereC = C

(n, [f ]1;[0,1]n , min

[0,1]nf).

The rate of convergence |nnvh(x)− u(x)n | ≤ C√h follows from the lemma and the rate

|wh(x)− w(x)| ≤ C√h.


Outline





5 References


Test cases

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(a) u1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(b) u2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(c) u3

We consider 3 test cases:

u1(x) = n maxi∈1,...,n

(xi −

1

2

)+

∏j 6=i

xj

1n

u2(x) =1

k + 1(x1 · · · xn)

1n

(n∑

j=1

sin(kxj )2 + nk

)(k = 20)

u3(x) = n(x1 · · · xn)1n

(C maxx1, . . . , xn+

n∑j=1

xj

)(C = 10)


n=2 (S1) (S2) (S3)Mesh size h `∞ Error Order `∞ Error Order `∞ Error Order2.5× 10−2 7.1× 10−2 2.1× 10−2 6.7× 10−2

6.3× 10−3 3.4× 10−2 0.54 5.7× 10−3 0.93 3.3× 10−2 0.511.6× 10−3 1.6× 10−2 0.51 1.5× 10−3 0.97 1.6× 10−2 0.503.9× 10−4 8.2× 10−3 0.50 3.8× 10−4 0.98 8.2× 10−3 0.509.8× 10−5 4.1× 10−3 0.50 9.7× 10−5 0.99 4.1× 10−3 0.502.4× 10−5 2.0× 10−3 0.50 2.4× 10−5 1.00 2.0× 10−3 0.50

n=3 (S1) (S2) (S3)Mesh size h `∞ Error Order `∞ Error Order `∞ Error Order5× 10−2 3.1× 10−1 9.1× 10−2 2.1× 10−1

2.5× 10−2 2.3× 10−1 0.41 5.3× 10−2 0.79 1.7× 10−1 0.341.3× 10−2 1.8× 10−1 0.38 3.0× 10−2 0.80 1.3× 10−1 0.346.3× 10−3 1.4× 10−1 0.36 1.7× 10−2 0.82 1.1× 10−1 0.333.1× 10−3 1.1× 10−1 0.35 9.5× 10−3 0.84 8.5× 10−2 0.331.6× 10−3 8.5× 10−2 0.34 5.3× 10−3 0.85 6.7× 10−2 0.33


1.3× 10−1 7.9× 10−1 0.46 2.4× 10−1 0.68 4.1× 10−1 0.266.3× 10−2 6.0× 10−1 0.40 1.5× 10−1 0.64 3.5× 10−1 0.253.1× 10−2 4.7× 10−1 0.36 9.5× 10−2 0.70 2.9× 10−1 0.251.6× 10−2 3.7× 10−1 0.32 5.7× 10−2 0.72 2.5× 10−1 0.257.8× 10−3 3.1× 10−1 0.29 3.4× 10−2 0.75 2.1× 10−1 0.25



6.3× 10−3 4.6× 10−2 0.53 6.1× 10−3 0.99 5.9× 10−3 1.011.6× 10−3 2.3× 10−2 0.50 1.6× 10−3 0.97 1.4× 10−3 1.023.9× 10−4 1.1× 10−2 0.50 4.1× 10−4 0.98 3.5× 10−4 1.029.8× 10−5 5.6× 10−3 0.50 1.0× 10−4 0.99 8.8× 10−5 1.002.4× 10−5 2.8× 10−3 0.50 2.6× 10−5 1.00 2.2× 10−5 0.99


2.5× 10−2 2.8× 10−1 0.39 4.8× 10−2 0.46 4.0× 10−2 0.481.3× 10−2 2.2× 10−1 0.36 2.4× 10−2 1.02 2.0× 10−2 1.016.3× 10−3 1.7× 10−1 0.35 1.2× 10−2 0.94 1.0× 10−2 0.963.1× 10−3 1.3× 10−1 0.34 6.2× 10−3 0.98 5.3× 10−3 0.971.6× 10−3 1.1× 10−1 0.34 3.2× 10−3 0.96 2.7× 10−3 0.96


1.3× 10−1 1.2× 10−0 0.42 3.9× 10−1 3.8× 10−1

6.3× 10−2 6.9× 10−1 0.79 1.4× 10−1 1.49 1.1× 10−1 1.813.1× 10−2 5.3× 10−1 0.37 7.2× 10−2 0.93 5.8× 10−2 0.881.6× 10−2 4.3× 10−1 0.30 3.7× 10−2 0.94 3.0× 10−2 0.967.8× 10−3 3.4× 10−1 0.28 1.9× 10−2 0.98 1.5× 10−2 0.96



6.3× 10−3 4.2× 10−2 0.49 1.9× 10−2 1.00 8.0× 10−3 0.981.6× 10−3 2.1× 10−2 0.50 4.7× 10−3 1.00 2.0× 10−3 1.003.9× 10−4 1.1× 10−2 0.50 1.2× 10−3 1.00 5.0× 10−4 1.009.8× 10−5 5.3× 10−3 0.50 2.9× 10−4 1.00 1.3× 10−4 1.002.4× 10−5 2.7× 10−3 0.50 7.4× 10−5 1.00 3.1× 10−5 1.00


2.5× 10−2 2.5× 10−1 0.31 1.3× 10−1 0.96 6.8× 10−2 0.951.3× 10−2 2.0× 10−1 0.32 6.7× 10−2 0.99 3.5× 10−2 0.976.3× 10−3 1.6× 10−1 0.33 3.3× 10−2 0.99 1.8× 10−2 0.983.1× 10−3 1.2× 10−1 0.33 1.7× 10−2 1.00 8.8× 10−3 0.991.6× 10−3 9.9× 10−2 0.33 8.4× 10−3 1.00 4.4× 10−3 1.00


1.3× 10−1 6.6× 10−1 0.37 7.7× 10−1 0.84 3.8× 10−1 0.736.3× 10−2 5.5× 10−1 0.26 4.1× 10−1 0.89 2.1× 10−1 0.863.1× 10−2 4.6× 10−1 0.25 2.2× 10−1 0.94 1.1× 10−1 0.911.6× 10−2 3.9× 10−1 0.25 1.1× 10−1 0.98 5.6× 10−2 0.967.8× 10−3 3.2× 10−1 0.25 5.5× 10−2 0.99 2.8× 10−2 0.99


Outline





5 References


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