Numerical discretization of coupling conditions of hyperbolic - - PowerPoint PPT Presentation

Numerical discretization of coupling conditions

• f hyperbolic conservation laws by high-order

schemes

Axel-Stefan Häck LuF Mathematik - Kontinuierliche Optimierung IGPM RWTH Aachen May 25th, 2014 Joint work with

• Prof. Michael Herty (RWTH Aachen)

and

• Prof. Mapundi Banda (Stellenbosch University)

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Table of contents

1

Model

2

Numerical Scheme

3

Coupling Condition in Scheme

4

Algorithm

5

Linear Case and Numerical Results

6

Left to do and Outlook

2 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Consider a model of (spatial) 1D flow on graphs.

3 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Consider a model of (spatial) 1D flow on graphs. A singe vertex with n adjacent arcs (which we extend to infinity).

3 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Consider a model of (spatial) 1D flow on graphs. A singe vertex with n adjacent arcs (which we extend to infinity). All arcs are parameterized by [0, ∞), such that the junction is located at x = 0 for all arcs.

3 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Consider a model of (spatial) 1D flow on graphs. A singe vertex with n adjacent arcs (which we extend to infinity). All arcs are parameterized by [0, ∞), such that the junction is located at x = 0 for all arcs. Has broad applications e.g. gas, traffic, and blood flow.

3 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Consider a model of (spatial) 1D flow on graphs. A singe vertex with n adjacent arcs (which we extend to infinity). All arcs are parameterized by [0, ∞), such that the junction is located at x = 0 for all arcs. Has broad applications e.g. gas, traffic, and blood flow. We assume the flux f(·) ∈ C4(R2, R2) and the uj(t, x) : R+

0 × R+

to be the conserved states on the arcs j = 1, ..., n.

3 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Coupled PDEs

Setting: Coupled PDEs ∂tuj + ∂xf(uj) = 0, t ≥ 0, x ≥ 0, uj(0, x) = uj,o(x), x ≥ 0, Ψ(u1(t, 0+), . . . , un(t, 0+)) = 0, t ≥ 0, where Ψ : R2n → Rn is the (possibly nonlinear) coupling condition (CC). With Ψ fulfilling the transversality condition we have existence and uniqueness see [Colombo,Herty,Sachers2008].

4 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Transversality condition (TC)

Let Ψ fulfill the transversality condition det [D1Ψ(ˆ u)r2(ˆ u1), . . . , DnΨ(ˆ u)r2(ˆ un)] = 0, where DjΨ(ˆ u) =

∂ ∂uj Ψ(ˆ

u),

5 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Transversality condition (TC)

Let Ψ fulfill the transversality condition det [D1Ψ(ˆ u)r2(ˆ u1), . . . , DnΨ(ˆ u)r2(ˆ un)] = 0, where DjΨ(ˆ u) =

∂ ∂uj Ψ(ˆ

u), ˆ u ∈ R2n is a steady state solution to PDE-Setting (and Ψ(ˆ u) = 0),

5 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Transversality condition (TC)

Let Ψ fulfill the transversality condition det [D1Ψ(ˆ u)r2(ˆ u1), . . . , DnΨ(ˆ u)r2(ˆ un)] = 0, where DjΨ(ˆ u) =

∂ ∂uj Ψ(ˆ

u), ˆ u ∈ R2n is a steady state solution to PDE-Setting (and Ψ(ˆ u) = 0), Df(ˆ uj) has strictly negative λ1(ˆ uj) and strictly positive eigenvalue λ2(ˆ uj) with linearly independent (right) eigenvectors r1 and r2

5 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Transversality condition (TC)

Let Ψ fulfill the transversality condition det [D1Ψ(ˆ u)r2(ˆ u1), . . . , DnΨ(ˆ u)r2(ˆ un)] = 0, where DjΨ(ˆ u) =

∂ ∂uj Ψ(ˆ

u), ˆ u ∈ R2n is a steady state solution to PDE-Setting (and Ψ(ˆ u) = 0), Df(ˆ uj) has strictly negative λ1(ˆ uj) and strictly positive eigenvalue λ2(ˆ uj) with linearly independent (right) eigenvectors r1 and r2 Corresponding characteristic fields to be either genuine nonlinear or linearly degenerate.

5 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Relevance of Transversality and Lax curves to CC

By upper assumptions to the eigenvalues of the system on each arc: Ψ is implicitly Rn → Rn (Of each arc exactly one Lax curve ’enters’ the junction!).

6 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Relevance of Transversality and Lax curves to CC

By upper assumptions to the eigenvalues of the system on each arc: Ψ is implicitly Rn → Rn (Of each arc exactly one Lax curve ’enters’ the junction!). This and TC ⇒ Unique solution to CC.

6 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Relevance of Transversality and Lax curves to CC

By upper assumptions to the eigenvalues of the system on each arc: Ψ is implicitly Rn → Rn (Of each arc exactly one Lax curve ’enters’ the junction!). This and TC ⇒ Unique solution to CC. As an example: Subsonic states of isothermal gas dynamics fulfill those assumptions.

6 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Discretization and CFL

Equidistant spatial grid xi+1 − xi = ∆x.

7 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Discretization and CFL

Equidistant spatial grid xi+1 − xi = ∆x. Choose tm+1 − tm = ∆t such that CFL condition λmax∆t ≤ ∆x holds (as usual).

7 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Discretization and CFL

Equidistant spatial grid xi+1 − xi = ∆x. Choose tm+1 − tm = ∆t such that CFL condition λmax∆t ≤ ∆x holds (as usual). Center of the first cell of each arc is located at x0 = ∆x

2

(⇒ boundary of each arc is located at 0 = x0 − ∆x

2 ) and let t0 = 0.

7 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Finite Volume

Finite Volume Method: Discretization for each ui (seperately); cell average Um

i,j of uj in cell i at time tm.

8 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Finite Volume

Finite Volume Method: Discretization for each ui (seperately); cell average Um

i,j of uj in cell i at time tm.

Evolution over ∆t: Um+1

j,i

= Um

j,i − 1

∆x

• (Fj)i+ 1

2 − (Fj)i− 1 2

• .

8 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Finite Volume

Finite Volume Method: Discretization for each ui (seperately); cell average Um

i,j of uj in cell i at time tm.

Evolution over ∆t: Um+1

j,i

= Um

j,i − 1

∆x

• (Fj)i+ 1

2 − (Fj)i− 1 2

• .

(Fj)i− 1

2 is flux at (left) cell boundary to cell i. 8 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Finite Volume + Coupling

By the coupling condition we get boundary conditions for the PDEs at x = 0! (⇒ (Fj)0− 1

2 ) 9 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Finite Volume + Coupling

By the coupling condition we get boundary conditions for the PDEs at x = 0! (⇒ (Fj)0− 1

2 )

The cell average at the first cell i = 0 at time tm is given by the states Um

j,0, j = 1, ..., n. We assume them to be sufficiently close

to ˆ uj such that the TC holds.

9 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Let s → Lκ(uo, s) the κ-th Lax curve through the state uo for κ = 1, 2.

10 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Let s → Lκ(uo, s) the κ-th Lax curve through the state uo for κ = 1, 2. For (s∗

1, . . . , s∗ n) we solve (e.g. using Newton′s method)

Ψ

• L2(Um

1,0, s1), . . . , L2(Um n,0, sn)

• = 0

first-order accurate, which is unique thanks to TC.

10 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Boundary flux

We can now calculate the boundary value Um+1

j,0

at time tm+1 by standard finite volume for i = 0, using Um

j,−1 := L2(Um j,0, s∗ j ).

11 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Boundary flux

We can now calculate the boundary value Um+1

j,0

at time tm+1 by standard finite volume for i = 0, using Um

j,−1 := L2(Um j,0, s∗ j ).

Hence, we gain a first-order approximation to the coupling condition as well as to the solution uj.

11 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 1: p.w. Linear Reconstruction

Given the cell averages Um

j,i we reconstruct a p.w. linear function

uj(x, tm) an each cell.

12 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 1: p.w. Linear Reconstruction

Given the cell averages Um

j,i we reconstruct a p.w. linear function

uj(x, tm) an each cell. This is standard as in the MUSCL scheme, using slope limiters (for the scheme to be TVD).

12 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 1: p.w. Linear Reconstruction

Given the cell averages Um

j,i we reconstruct a p.w. linear function

uj(x, tm) an each cell. This is standard as in the MUSCL scheme, using slope limiters (for the scheme to be TVD). In first cell the slope reconstruction will be a forward finite difference, only to calculate σ1.

12 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 1: p.w. Linear Reconstruction

Given the cell averages Um

j,i we reconstruct a p.w. linear function

uj(x, tm) an each cell. This is standard as in the MUSCL scheme, using slope limiters (for the scheme to be TVD). In first cell the slope reconstruction will be a forward finite difference, only to calculate σ1. We obtain a p.w. linear uj(x, tm) = σj,i(x − xi) + Um

j,i,

xi− 1

2 ≤ x ≤ xi+ 1 2 , i = 1, . . . ,

σj,i is the vector of slopes.

12 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 1: p.w. Linear Reconstruction

Given the cell averages Um

j,i we reconstruct a p.w. linear function

uj(x, tm) an each cell. This is standard as in the MUSCL scheme, using slope limiters (for the scheme to be TVD). In first cell the slope reconstruction will be a forward finite difference, only to calculate σ1. We obtain a p.w. linear uj(x, tm) = σj,i(x − xi) + Um

j,i,

xi− 1

2 ≤ x ≤ xi+ 1 2 , i = 1, . . . ,

σj,i is the vector of slopes. We can get distinct values at the interface which we will denote by the values at xi+ 1

2 ∓. 12 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 2: Solve CC

Reconstruct for each arc j a piecewise linear function vj(t) for tm ≤ t ≤ tm+1 such that Ψ(v1(tm), . . . , vn(tm)) =0 and d dt Ψ(v1(t), . . . , vn(t))|t=tm =0.

13 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 3: Numerical Flux

Using the Midpoint rule for the flux, Taylor expansion (on f ±) and the reconstruction we get:

14 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 3: Numerical Flux

Using the Midpoint rule for the flux, Taylor expansion (on f ±) and the reconstruction we get: Um

j,i− :=Um j,i + σj,i

∆x 2 , Um

j,i+ := Um j,i+1 − σj,i+1

∆x 2 , 1 ∆x (Fj)i+ 1

2 ≈ ∆t

∆x

• f +
• Um

j,i− − ∆t

2 Df(Um

j,i−)σj,i

• +

f −

• Um

j,i+ − ∆t

2 Df(Um

j,i+)σj,i+1

,

14 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 3: Numerical Flux

Using the Midpoint rule for the flux, Taylor expansion (on f ±) and the reconstruction we get: Um

j,i− :=Um j,i + σj,i

∆x 2 , Um

j,i+ := Um j,i+1 − σj,i+1

∆x 2 , 1 ∆x (Fj)i+ 1

2 ≈ ∆t

∆x

• f +
• Um

j,i− − ∆t

2 Df(Um

j,i−)σj,i

• +

f −

• Um

j,i+ − ∆t

2 Df(Um

j,i+)σj,i+1

, where the flux is splitted as f(u) = f +(u) + f −(u) := 1 2 (f(u) + au) + 1 2 (f(u) − au) , with a = λmax.

14 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 4 Flux at Boundary

We need to evaluate the flux at the boundary (i = 0). (Fj)− 1

2

remains of interest.

15 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 4 Flux at Boundary

We need to evaluate the flux at the boundary (i = 0). (Fj)− 1

2

remains of interest. By CC and Step 2 of the algorithm the characteristic speed of information is non-negative and by the midpoint rule we get: (Fj)− 1

2 =

tm+1

tm

f(uj(x− 1

2 , s)ds

≈ tm+1

tm

f(vj(s))ds = ∆tf(vj(tm+ 1

2 )) + O((∆t)3). 15 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 4 Flux at Boundary

We need to evaluate the flux at the boundary (i = 0). (Fj)− 1

2

remains of interest. By CC and Step 2 of the algorithm the characteristic speed of information is non-negative and by the midpoint rule we get: (Fj)− 1

2 =

tm+1

tm

f(uj(x− 1

2 , s)ds

≈ tm+1

tm

f(vj(s))ds = ∆tf(vj(tm+ 1

2 )) + O((∆t)3).

Therefore, we have as as boundary flow: 1 ∆x (Fj)− 1

2 ≈ ∆t

∆x f

• vj(tm) + ∆t

2 d dt vj(t)|t=tm

• 15 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Step 5

Evolve the dynamics according to FV boundary update for arcs i = 1, . . . , n to obtain the new cell averages at time tm+1 and proceed with STEP1.

16 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Calculate Step 2

To solve Step 2 we use the p.w. linear reconstruction uj(x− 1

2 , tm) = Um

j,0 − ∆x

2 σj,0.

17 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Calculate Step 2

To solve Step 2 we use the p.w. linear reconstruction uj(x− 1

2 , tm) = Um

j,0 − ∆x

2 σj,0. We determine the vector s = (s1, . . . , sn) ∈ Rn by solving the (possibly nonlinear) equation: Ψ

• L2(Um

1,0 − ∆x

2 σ1,0, s1), . . . , L2(Um

n,0 − ∆x

2 σn,0, sn)

• = 0.

17 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Calculate Step 2

To solve Step 2 we use the p.w. linear reconstruction uj(x− 1

2 , tm) = Um

j,0 − ∆x

2 σj,0. We determine the vector s = (s1, . . . , sn) ∈ Rn by solving the (possibly nonlinear) equation: Ψ

• L2(Um

1,0 − ∆x

2 σ1,0, s1), . . . , L2(Um

n,0 − ∆x

2 σn,0, sn)

• = 0.

And 0 = d dt Ψ(u1(t, 0+), . . . , un(t, 0+)) =

n

• k=1

Duk Ψ

• u1(t, 0+), . . . , un(t, 0+)
• ∂tuk(t, 0+),

since we want Ψ to hold for t > tm.

17 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Lemma: Second order CC

Lemma: Consider a single node with n connected arcs and let tm be some positive time. Let Ψ ∈ C2(R2n; Rn) and let ˆ uj := Um

j,0 − ∆x 2 be such that

condition TC holds true. Then, for vj(t) as in the the previous construction, the coupling condition is satisfied up to second order in time Ψ (v1(t), . . . , vn(t)) = O((t − tm)2).

18 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Linear Scenario

Consider two connected arcs j = 1, 2 with f1(u) = −c u, f2(u) = c u and c > 0.

19 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Linear Scenario

Consider two connected arcs j = 1, 2 with f1(u) = −c u, f2(u) = c u and c > 0. Assuming the coupling condition Ψ(u1(t, 0+), u2(t, 0+)) = u2(t, 0+) − u1(t, 0+) = 0,

19 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Linear Scenario

Consider two connected arcs j = 1, 2 with f1(u) = −c u, f2(u) = c u and c > 0. Assuming the coupling condition Ψ(u1(t, 0+), u2(t, 0+)) = u2(t, 0+) − u1(t, 0+) = 0, which is equivalent to the Cauchy Problem ∂ty(x, t) + c∂xy(x, t) = 0, y(x, 0) = yo(x) := u1,o(−x) x ≤ 0 u2,o(x) x > 0.

• 19 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Evaluate linear CC

We have that the admissible boundary states for arc j = 2 are given by L2(ˆ u2, s2) = s2 + ˆ u2, s2 ∈ R. In arc j = 1 we have L2(ˆ u1, s1) = ˆ u1. (s2 + ˆ u2 = v2 for the linear case!)

20 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Evaluate linear CC

We have that the admissible boundary states for arc j = 2 are given by L2(ˆ u2, s2) = s2 + ˆ u2, s2 ∈ R. In arc j = 1 we have L2(ˆ u1, s1) = ˆ u1. (s2 + ˆ u2 = v2 for the linear case!) Solve Ψ =0 and d dt Ψ =0

20 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Evaluate linear CC

We have that the admissible boundary states for arc j = 2 are given by L2(ˆ u2, s2) = s2 + ˆ u2, s2 ∈ R. In arc j = 1 we have L2(ˆ u1, s1) = ˆ u1. (s2 + ˆ u2 = v2 for the linear case!) Solve Ψ =0 and d dt Ψ =0 For t ∈ [tm, tm+1] we have by v2(t) = Um

1,0 − ∆x

2 σ1,0 + c σ1,0(t − tm).

20 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Boundary Flux

The numerical flux in cell i = 1 is then given by (F2)− 1

2 = ∆t c

• Um

1,0 − ∆x

2 σ1,0 + ∆t 2 c σ1,0

• .

21 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Boundary Flux

The numerical flux in cell i = 1 is then given by (F2)− 1

2 = ∆t c

• Um

1,0 − ∆x

2 σ1,0 + ∆t 2 c σ1,0

• .

From first order CC and the boundary flux we get Um+1

2,0

= Um

2,0 − ∆t

∆x c

• Um

2,0 − Um 1,0 + (σ2,0 + σ1,0)∆x − c∆t

2

• as the updated value of the first cell average in u2.

21 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Convergence 1

We implemented a cyclic linear transport u(t, x) on x ∈ [0, 1] where u(t, 0) = u(t, 1) for all t ≥ 0.

Abbildung : No Coupling

22 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Convergence 1

Same scenario as above but coupled by the linear CC as above.

Abbildung : Using Coupling Condition

23 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Convergence Cyclic vs. Coupled

Compare the two solutions:

Abbildung : Cyclic vs. Coupled

24 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Convergence Table

Nk MUSCL cyclic Rate MUSCL nodal Rate 25 −3.0768 / −3.1472 / 26 −4.6573 1.5805 −4.7055 1.5583 27 −6.4464 1.7891 −6.4689 1.7635 28 −8.3080 1.8616 −8.3198 1.8509 29 −10.1955 1.8876 −10.2018 1.8820 210 −12.1139 1.9184 −12.1171 1.9153 211 −14.0533 1.9394 −14.0549 1.9378 212 −16.0073 1.9540 −16.0081 1.9532 213 −17.9715 1.9642 −17.9719 1.9638 214 −19.9429 1.9714 −19.9431 1.9712

25 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Left to do and Outlook

Finish nonlinear system code (isothermal Euler for gas dynamics)

26 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Left to do and Outlook

Finish nonlinear system code (isothermal Euler for gas dynamics) Apply to other second order methods

26 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Left to do and Outlook

Finish nonlinear system code (isothermal Euler for gas dynamics) Apply to other second order methods Extend to order higher than two

26 / 27

Model Numerical Scheme Coupling Condition in Scheme Algorithm Linear Case and Numerical Results Left to do and Outlook

Thank you for your attention! And for the opportunity of this talk and visit.

27 / 27