Numerical discretization of tangent vectors of hyperbolic - - PowerPoint PPT Presentation

Numerical discretization of tangent vectors of hyperbolic conservation laws.

Michael Herty IGPM, RWTH Aachen www.sites.google.com/michaelherty joint work with Benedetto Piccoli MNCFF 2014, Bejing, 22.5.2014

Contents

◮ Originally on multi-phase fluid flow on networks (extension of

coupling conditions - NumHyp 2013 )

◮ Recent results on tangent vectors in collaboration with B.

Piccoli (Rutgers)

◮ Tangent vectors are used for optimization problems where the

dynamics is governed by 1–d hyperbolic systems

◮ Tangent vectors are a first–order sensitivity calculus for

solutions

∂ ∂u0 u(t, x) ◮ Examples are control of compressor stations in gas networks,

control of gates in open canals, parameter identification problems subject to transport problems, . . .

This presentation

◮ References and theoretical problem ◮ Example of tangent vectors for Burgers ◮ Numerical challenges ◮ Approximation of evolution of tangent vectors ◮ Numerical examples

References

Concept of tangent vectors present one possibility to compute and analyze sensitivity of systems on 1-d hyperbolic balance laws

◮ Existing calculus for spatially one–dimensional systems of

conservation laws

◮ Introduced by Bressan/Marson (1995), extended by Bressan

and co–workers (1997,2007), Bianchini (2000), Piccoli and co–workers (2000–)

◮ Lipschitz continuous dependence for 1–d system

Crasta/Bressan/Piccoli (2001)

◮ Related for scalar convex case: Ulbrich (2001), Giles (1996),

Ulbrich/Giles (2010), Zuazua/Castro et al (2008–).

◮ Application of sensitivity equations for general inverse

problems, optimal control problems, and control / controllability questions, . . .

Theoretical challenge

∂tu + ∂xf (u) = 0, u(t = 0, x) = u0(x)

◮ Solution operator for nonlinear conservation laws

u(t, ·) = Stu0 generically non–differentiable on L1

◮ No classical calculus for first–order variations of Stu0 with

respect to u0

◮ No characterizing conditions for parameter identification

problems or optimal control problems min

u0

• J(u(T, x))dx subject to ∂tu+∂xf (u) = 0, u(t = 0, x) = u0(x)

Example for Burger’s equation

∂tu + ∂x u2 2 = 0, u(t = 0, x) = uǫ

0 = ǫ · x · χ[0.1] ◮ uǫ 0 variations of initial data and interest in the behavior of the

solution Stuǫ

0 with respect to h at e.g. ǫ

Stuǫ+h ≈ Stuǫ

0 + hv + O(h) ◮ Exact solution uǫ(t, x) = (1+ǫ)x 1+ǫt χ[0,√1+ǫt] is Lipschitz in L1

wrt ǫ

◮ Shock position is 1 + ǫt ◮ t = 0: a first–order approximation in L1 exists and is

v(t, x) = lim

h→0 uǫ+h (x)−uǫ

0(x)

ǫ

= ǫ x · χ[0,1]

◮ But for any t > 0 the jump depends on ǫ + h and the previous

limit does not define any function in L1

Idea of tangent vectors for Burger’s equation

lim

h→0

uǫ+h(t, x) − uǫ(t, x) h dx = O(1)+1 h √

1+(ǫ+h)t √1+ǫt

(ǫ + h)x 1 + (ǫ + h)t dx = O( 1 √ h ) Use weak formulation to compute the limit h → 0 for arbitrary value of ǫ lim

h→0

uǫ+h(t, x) − uǫ(t, x) h φ(x)dx = lim

h→0

• x

(1 + (ǫ + h)t)(1 + ǫt)χ[0,√1+ǫt](x)φ(x)dx+ lim

h→0

t 2

• 1 + (ǫ + h)t

φ(y) (ǫ + h)y 1 + (ǫ + h)t , y ∈ [ √ 1 + ǫt,

• 1 + (ǫ + h)t]

Idea of tangent vectors (cont’d)

lim

h→0

• x

(1 + (ǫ + h)t)(1 + ǫt)χ[0,√1+ǫt](x)φ(x)dx+ lim

h→0

t 2

• 1 + (ǫ + h)t

φ(y) (ǫ + h)y 1 + (ǫ + h)t , y ∈ [ √ 1 + ǫt,

• 1 + (ǫ + h)t]

A suitable differential of uǫ(t, x) at ǫ may therefore consist of two components: an L1 part and a measure located at the jump of uǫ Stuǫ+h = uǫ+h(t, ·) = Stuǫ

0 + h

• v(t, ·) + δsǫ(t)(·)[uǫ]ξ(t)

Idea of tangent vectors for Burger’s equation (2/2)

lim

h→0

• x

(1 + (ǫ + h)t)(1 + ǫt)χ[0,√1+ǫt](x)φ(x)dx+ lim

h→0

t 2

• 1 + (ǫ + h)t

φ(y) (ǫ + h)y 1 + (ǫ + h)t , y ∈ [ √ 1 + ǫt,

• 1 + (ǫ + h)t]

◮ v(t, x) is the L1−part consists of the a.e. pointwise limit

v(t, x) = lim

h→0

uǫ+h(t, x) − uǫ(t, x) h = x (1 + ǫt)2

◮ ξ(t) is a real number and is the variation in the shift of the

shock position sǫ ξ(t) = lim

h→0

sǫ+h(t) − sǫ(t) h = t 2√1 + ǫt ,

◮ uǫ+h(t, ·) = uǫ(t, ·) + h

• v(t, ·) + δsǫ(t)(·)[uǫ]ξ(t)
• ◮ (v, ξ) is called generalized tangent vector

Generalized tangent vectors (v, ξ)

◮ v(t, x) is the L1−part consists of the a.e. pointwise limit

v(t, x) = lim

h→0 uǫ+h(t,x)−uǫ(t,x) h ◮ ξ(t) is a real number and is the variation in the shift of the

shock position sǫ ξ(t) = lim

h→0 sǫ+h(t)−sǫ(t) h ◮ uǫ+h(t, ·) = uǫ(t, ·) + h

• v(t, ·) + δsǫ(t)(·)[uǫ]ξ(t)
• Question

◮ Which variations ǫ → uǫ 0 lead to well–defined tangent vectors?

Different settings possible.

Here, we follow B/M with uǫ

0 piecewise Lipschitz with a finite number of isolated discontinuities

◮ A tangent vector a t = 0 may lead to a tangent vector at time

t by solving suitable evolution equations (first–order variations

• f hyperbolic system).

For hyperbolic balance laws and suitable variations this is proven up to shock interaction (B/M) and extended in case of conservation laws (C/B/P).

Suitable variations of initial data leading to well–defined tangent vectors (v, ξ)

A suitable variation uh of u0(x) is within an L1 equivalence class to uh(x) = u0(x)+hv(x)+

• ξi<0

[u0]χ[xi+hξ,xi](x)−

• ξi>0

[u0]χ[xi,xi+hξ](x) for some v ∈ L1 and ξ ∈ RN where N is the number of isolated discontinuities located at xi in u0(x).

Theoretical results (Theorem 2.2 (B/M)

◮ For previous variations the evolution equation for the tangent

vector is well–defined up to any point in time when two shocks coincide

◮ Result holds for system of balance laws

ut + f (u)x = h(t, x, u), A(u) = Df (u).

Numerical implementation of tangent vectors

◮ Requires knowledge on exact shock positions for evaluation of

the evolution of ξ(t)

◮ Requires solution of compatibility condition between waves of

different families

◮ Evolution of the L1 part v influences evolution of shift ξ(t) ◮ Non–conservative system for v

Approximation to the problem for numerical implementation

◮ Presentation on the simplest case of a 1–d scalar conservation

law y(1)

t

+ f (y(1))x = 0, y(1)(t = 0, x) = u0(x)

◮ Replace equation by Jin–Xin relaxation approximation for

0 < µ << 1 yt + 1 a2

• yx = 1

µ

• f (y(1)) − y(2)
• y(1)(0, x) = u0(x), y(2)(0, x) = f (u0(x))

◮ Relaxation is a viscous approximation

y(1)

t

+ f (y(1))x = µ

• a2 − f ′(y(1))2

y(1)

x

• x .

◮ System is linear hyperbolic with stiff source term

Tangent vectors for relaxation system

yt + 1 a2

• yx = 1

µ

• f (y(1)) − y(2)
• ◮ L1−part of tangent vector does not depend on y(1) and is

linear system v(0, ·) = ¯ v(·), vt + 1 a2

• vx = 1

µ

• f ′(y(1))v(1) − v(2)
• ,

v = (v(1), v(2)) outside the discontinuities of y

◮ N discontinuities and lj jth eigenvector and ∆iv =

v(xi(t)+, t) − v(xi(t)−, t) jump across ithe discontinuities ξi(t) = ¯ ξi and lj · (∆iv + ∆iyx ξi) = 0 j = k.

Variation of general cost functional wrt initial data

◮ Given yd ∈ L1(R), T > 0 given and y = y(1) solution to

relaxation system J(y(·, ·), yd(·)) =

• χI(x) (y(T, x) − yd(x))2 dx

◮ Variation of J with respect to variations in u0 generating the

tangent vectors (v, ξ) and a fixed number N of discontinuities ∇u0J(y, yd) = 2

• χI(x)
• y(1)(T, x) − yd(x)
• v(1)(T, x)dx+

N(u0)

• i=1

y(1)(T, xi+) − yd(xi+)

• +
• y(1)(T, xi−) − yd(xi−)
• ∆iy(1)(T, ·)ξi(T)

Variation of cost and tangent vectors used for maximal descent

∇u0 J(y, yd ) = 2

• χI (x)
• y(1)(T, x) − yd (x)
• v(1)(T, x)dx+

N(u0)

• i=1

y(1)(T, xi +) − yd (xi +)

• +
• y(1)(T, xi −) − yd (xi −)
• ∆i y(1)(T, ·)ξi (T)

◮ Update of control u0 using gradient based information with

stepsize ρ > 0 ˜ u0(x) = u0(x)−  ρv(0, x) −

N(u0)

• i=1

χ[xi+ρξi(0),xi+1+ρξi+1(0)](x)u0(xi+)  

◮ Choice of v for maximal descent in J

v(1)(T, x) =

• y(1)(T, x) − yd(x)
• , v(2)(T, x) = 0

◮ Choice of ξ for maximal descent in J

ξi (T) =

• y(1)(T, xi +) − yd (xi +)
• +
• y(1)(T, xi −) − yd (xi −)
• ∆i y(1)(T)

Numerical implementation of tangent vectors

∇u0 J(y, yd ) = 2

• χI (x)
• y(1)(T, x) − yd (x)
• v(1)(T, x)dx+

N(u0)

• i=1

y(1)(T, xi +) − yd (xi +)

• +
• y(1)(T, xi −) − yd (xi −)
• ∆i y(1)(T, ·)ξi (T)

vt + 1 a2

• vx =

1 µ

• f ′(y(1))v(1) − v(2)
• , ξi (t) = ¯

ξi and lj · (∆i v + ∆i yx ξi ) = 0 j = k.

◮ Introduce a spatial grid (xi)n i=1 and intervals Ii = [xi− 1

2 , xi+ 1 2 ]

◮ Choose a control u0 a piecewise constant on Ii leading to

ξ ∈ R2n possible shock variations

◮ Diagonalise matrix

1 a2

• and derive tangent vectors for

system in diagonal form

◮ Splitting of transport and source term integration

Advantage of relaxation system over nonlinear system

vt + 1 a2

• vx =

1 µ

• f ′(y(1))v(1) − v(2)
• , ξi (t) = ¯

ξi and lj · (∆i v + ∆i yx ξi ) = 0 j = k.

◮ Due to the linearity of system the equations for L1 and real

part ξ decouple

◮ Nonlinear source term does not influence evolution of t → ξ(t) ◮ Variations of initial data on each grid cell Ii leading to shock

variations ξi at each grid cell and an L1 variation piecewise constant vi on each grid cell

◮ Evaluation of source term may lead to additional

discontinuities and therefore additional shock variations

◮ Apply a splitting technique and use piecewise constant control

in each component (v(1)

0 , v(2) 0 ) spanning two consecutive

intervals

Admissible controls and discretization of tangent vectors (v, ξ)

vt + 1 a2

• vx =

1 µ

• f ′(y(1))v(1) − v(2)
• ,

∂tη(1) + a∂x η(1) = 0, ∂tη(2) = 0, ∂tη(1) = S(η(1), η(2)), ∂tη(2) = S(η(1), η(2)), ∂tη(2) − a∂x η(2) = 0, ∂tη(1) = 0, (η(1) )2i = (η(1) )2i+1, (η(2) )2i−1 = (η(2) )2i , i = 0, . . . , n 2

◮ Diagonalise the system in variables η with source term S and

discretization according to the splitting above using a transformation T

◮ Choose control η0 in first component constant on even and in

the second on odd gridpoints

◮ Choose ∆x and ∆t according to the CFL condition with

constant one

◮ Exact evaluation of first–order Upwind scheme for the

transport part

Admissible controls and discretization of tangent vectors (v, ξ) (cont’d)

∂tη(1) + a∂x η(1) = 0, ∂tη(2) = 0, ∂tη(1) = S(η(1), η(2)), ∂tη(2) = S(η(1), η(2)), ∂tη(2) − a∂x η(2) = 0, ∂tη(1) = 0, x2i−1(tn) = x2i−1(0) + a tn, x2i (tn) = x2i (0) − a tn, i = 0, . . . , n,

◮ Splitting does not need introduce additional shocks since

exact Upwind leads to piecewise constant functions on even and odd grid cells

◮ Shock position at time tn is given xi(tn) = x0 i ± atn

depending if discontinuity at even or odd cell boundary

◮ Condition lj · (∆iv + ∆iyx ξi) = 0 j = k, automatically

satisfied since there is only one discontinuity at each cell boundary – at even grid points in the first and odd in the second component

Tests and algorithm for control problems with cost J

min

y0=(y(1)

0 ,y(2) 0 )

J sbj to relaxation system

◮ Unregularized cost J(y(·, ·), yd(·)) =

• χI(x) (y(T, x) − yd(x))2 dx

◮ Update of control u0 using gradient based information with stepsize

ρ > 0 ˜ u0(x) = u0(x)−  ρv(0, x) −

N(u0)

• i=1

χ[xi+ρξi(0),xi+1+ρξi+1(0)](x)u0(xi+)  

◮ Choice of v for maximal descent in J

v (1)(T, x) =

• y (1)(T, x) − yd(x)
• , v (2)(T, x) = 0

◮ Choice of ξ for maximal descent in J

ξi(T) =

• y (1)(T, xi+) − yd(xi+)
• +
• y (1)(T, xi−) − yd(xi−)
• ∆iy (1)(T

Algorithm for computing a gradient

• 1. Set terminal time T > 0, a2 ≥ max

y(1) (f ′(y(1))2 and choose an

equidistant spatial discretization with Nx gridpoints. Choose ∆t = ∆x

a and k = 0. Let ηk 0,i =

• (η(1)

0,i )k, (η(2) 0,i )k

for i = 0, . . . , Nx, be an arbitrary initial control such that ηk

0 is

admissible Let (yd)i be a discretization of the given function yd(·).

• 2. Solve forward relaxation equations with (y0)i := T ηk

0,i to

• btain ηNt

i

= T −1yNt

i

.

• 3. Set initial data ϕ0

i and shock variations ξi where xi(tNt) are

given by equation shock positions

• 4. Solve sensitivity equations backwards in time for initial data

˜ v0

i := T ϕ0 i to obtain ϕNt i

= T −1˜ vNt

i

.

• 5. Update the current iterate ηk

0,i by setting ηk+1 0,i

:= ˜ η0

i by

η0

i := ηk 0,i.

• 6. Provided that J(T η, yd) is sufficiently small we terminate.

Otherwise set k → k + 1 and continue with step (2).

Numerical results: Grid convergence for splitting

Numerical results: Convergence for a linear system without source term

Numerical results: Convergence table using (v, ξ) components

Numerical results: Convergence table using only v components

Numerical results: Relaxation for Burger’s equation

Numerical results: Convergence

Thank you for your attention.