Use Riemann Solver for Fine Scale Model

Use Riemann Solver for Fine Scale Model

Hann-Ming Henry Juang

RSM workshop 2011

Contents• Introducing Riemann Solver

– Simple 1D for easy illustration

• Possible 2D– In shallow water equation– In nonhydrostatic system

• How about 3D?– With splitting, 2D then 1D in vertical

• Discussion– advantage– Future work

€

∂u

∂t+ u

∂u

∂x+ g

∂h

∂x= 0

On dimensional shallow water equation

can be written as

€

∂∂t

h

u

⎛

⎝ ⎜

⎞

⎠ ⎟+

u h

g u

⎛

⎝ ⎜

⎞

⎠ ⎟∂

∂x

h

u

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

And we can have a matrix L and it inverse matrix L-1 for the above matrix be diagonal matrix with eigenvector as

€

∂h

∂t+

∂(uh)

∂x= 0

€

L−1 u h

g u

⎛

⎝ ⎜

⎞

⎠ ⎟L =

C1 0

0 C2

⎛

⎝ ⎜

⎞

⎠ ⎟

€

LL−1 =1

&

&

then the shallow water equation can be written as

€

∂∂t

R1

R2

⎛

⎝ ⎜

⎞

⎠ ⎟+

C1 0

0 C2

⎛

⎝ ⎜

⎞

⎠ ⎟∂

∂x

R1

R2

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

so

where

€

R1 = gh + u /2

R2 = gh − u /2

C1 = u + gh

C2 = u − gh

€

L−1 ∂

∂t

h

u

⎛

⎝ ⎜

⎞

⎠ ⎟+ L−1 u h

g u

⎛

⎝ ⎜

⎞

⎠ ⎟LL−1 ∂

∂x

h

u

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

L−1 u h

g u

⎛

⎝ ⎜

⎞

⎠ ⎟L =

C1 0

0 C2

⎛

⎝ ⎜

⎞

⎠ ⎟ & LL−1 =

1 0

0 1

⎛

⎝ ⎜

⎞

⎠ ⎟

we can find L to be

€

∂Ri

∂t+ Ci

∂Ri

∂x= 0

∂R2

∂t+ C2

∂R2

∂x= 0

or

So the procedure to solve the 1D SWE by

1) Obtain R and C from u and h at any model grid as

€

R1(t) = gh(t) + u(t) /2

R2(t) = gh(t) − u(t) /2

€

C1(t) = u(t) + gh(t)

C2(t) = u(t) − gh(t)&

2) Use advection eq to solve next time step of R by

€

∂Ri

∂t+ Ci

∂Ri

∂x= 0

3) Obtain next time step u and h and C by

€

u(t + Δt) = R1(t + Δt) − R2(t + Δt)

h(t + Δt) =1

g

R1(t + Δt) + R2(t + Δt)

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

4) Back to 2) for the next time

to get

€

R1(t + Δt) and R2(t + Δt)

&

€

C1(t + Δt) = u(t + Δt) + gh(t + Δt)

C2(t + Δt) = u(t + Δt) − gh(t + Δt)

Nonhydrostatic system

€

∂u

∂t= −u

∂u

∂x− RT

∂Q

∂x∂Q

∂t= −u

∂Q

∂x− γ

∂u

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟

For Riemann solver, we let the above be

€

∂∂t

Q

u

⎛

⎝ ⎜

⎞

⎠ ⎟+

u γ

RT u

⎛

⎝ ⎜

⎞

⎠ ⎟∂

∂x

Q

u

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

∂R1

∂t= −c1

∂R1

∂x∂R2

∂t= −c2

∂R2

∂x

where

€

R1 = RT /γ Q + u

R2 = RT /γ Q − u

c1 = u + γRT

c2 = u − γRT

or

€

dR1

dt= 0

dR2

dt= 0

The initial acoustic spread

After 800s with different CFL

2D nonhydrostatic tests in x-z with isotherm

€

∂u

∂t= −u

∂u

∂x− w

∂u

∂z− RT

∂ ′ Q

∂x∂w

∂t= −u

∂w

∂x− w

∂w

∂z− RT

∂ ′ Q

∂z

∂ ′ Q

∂t= −u

∂ ′ Q

∂x− w

∂ ′ Q

∂z− γ

∂u

∂x+

∂w

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟+

gw

RT

where

€

∂Q

∂z= −

g

RT

Q = Q + ′ Q

2D tests in x-z (non-forcing) – Q’(t=30s)

2D tests in x-z (non-forcing) – Q’(t=60s)

€

∂u

∂t+ u

1

acosφ

∂u

∂λ+ v

1

a

∂u

∂φ+

g

acosφ

∂H

∂λ− f +

u tanφ

a

⎛

⎝ ⎜

⎞

⎠ ⎟v = 0

∂v

∂t+ u

1

acosφ

∂v

∂λ+ v

1

a

∂v

∂φ+

g

a

∂H

∂φ+ f +

u tanφ

a

⎛

⎝ ⎜

⎞

⎠ ⎟u = 0

∂h

∂t+ u

1

acosφ

∂h

∂λ+ v

1

a

∂h

∂φ+

h

acosφ

∂u

∂λ+

∂v cosφ

∂φ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

u = acosφdλ

dt

where

€

H = h + hs

SWE on spherical coordinates can be written as

€

v = adφ

dt

€

1

acosφ

∂

∂λ=

∂

∂ ′ λ

1

a

∂

∂φ=

∂

∂ ′ φ

let

rewrite the previous SWE into

€

∂∂t

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

u h 0

g u 0

0 0 u

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

∂

∂ ′ λ

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

v 0 h

0 v 0

g 0 v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

∂

∂ ′ φ

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

−tanφ

ahv

− fv −tanφ

auv + g

∂hs

∂ ′ λ

fu +tanφ

auu + g

∂hs

∂ ′ φ

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

= 0

Then dimensional split into

€

∂∂t

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

u h 0

g u 0

0 0 u

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

∂

∂ ′ λ

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

1

2

0

− fv −tanφ

auv + g

∂hs

∂ ′ λ

fu +tanφ

auu + g

∂hs

∂ ′ φ

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

= 0

€

∂∂t

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

v 0 h

0 v 0

g 0 v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

∂

∂ ′ φ

h

u

v

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

1

2

−2tanφ

ahv

− fv −tanφ

auv + g

∂hs

∂ ′ λ

fu +tanφ

auu + g

∂hs

∂ ′ φ

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

= 0

Discussion• 2D system should be evaluated carefully.• 3D system can apply splitting method, do 2D

first then 1D vertical, so it is no problem if 2D isno probelm.

• Since it becomes an advection equation, semi-Lagrangian with splitting can be used to solve it with stable integration.

• The solution becomes non isotropic while CFL is larger then 1 or 2, more to examine and resolve with additional method.