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METHODS OF EFFICIENT MODELLING AND FORECASTING DIFFERENT SCALE ATMOSPHERIC PROCESSES
V. A. Prusov Professor of the Meteorology and Climatology Department of the Geography Faculty. Department Geography www.geo.univ.kiev.ua E-mail: vitaliy@softick.com . : + (380) - (44) 521-32-86 : + (380) - (44) 266-52-18 . : +(380) ­ (44) 544-42-51 . : 8-050-699-65-50


A COMPLEX MODEL OF ATMOSPHERIC STATE
Fundamental equations of atmosphere circulation are based on the universal physics laws: · of conservation of mass D + ( V ) = 0 Dt · conservation of momentum DV -1 · + 2 â V = - p - g + ( ) Dt · conservation of energy DT Dp rad c p - T = k T - F + QH Dt Dt · conservation of scalar entities = (, k , q , q L, qw ) D = (k ) + Qq Dt · and state equation p = RT

(

)


PROBLEM DEFINITION
Therefore problem of atmosphere circulation involve systems of convection-diffusion equations as a main constituent. It is the following vector form:
+ v1 + v2 + t x1 x 2 µ1 + x1 x1 v3 x 3 + x2 =F+
µ2 + x 2 x3 µ3 x 3

· with initial condition ( X ,0 ) = ( X ) , 0 X L · and boundary conditions X =0 = (t ) , X =L = (t ),

t >0


I. PROBLEM-SOLVING NUMERICAL PROCEDURE OF MACROSCALE FORECAST ON FORECAST ON BASIS OF UPSTREAM FINITE-DIFFERENCE SCHEME
· Recently in weather forecast problems for numerical integration of hydro-dynamical heat/mass transmission equations more often are are applied methods of a finite element and spectral methods. Yet we will consider one more finite-difference method what is explained by following reasons: basic concepts, underlying theory and main features for numerical applications (such as approximation, convergence and stability) are well understood and developed for finite-difference methods; these methods are treated universally in many applications areas; ll they allows decomposition of a complex multidimensional problem by reducing a numerical solution uniformly through spatial splitting into temporal sequence of one-dimensional problems; of the last feature is quite appropriate for parallelizing algorithms and their efficient parallel implementation in multiprocessor computer software.

· · · ·


SOLVING THE THREE-DIMENSIONAL EQUATION BY OPERATOR SPLITTING
The technique is based on directional operator splitting, which results in one-dimensional advection-diffusion equations for t (( k - 1), k ]

( 1) k

(t )

t

( 2) k

= -v


1

( 1) k 1

x

( 1) x k µ + 1 x x1 1

+F , x1

k (( k - 1) ) = sp (( k - 1) ),
(2) k

( 1)

(t )

t

( 2) (2) k xk µ = -v 2 + Fx , + 2 2 x2 x2 x2



(( k - 1) ) = sp (( k - 1) ),



(3) k

(t )

t

= -v


3

(3) k 3

x

(3) x k µ +F , + 3 x3 x 3 x 3



(3) k

(( k - 1) ) = sp (( k - 1) ),

1 sp ( k ) = 3

i =1



3

k ( (k - 1) )

(i )


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
· Consider the one-dimensional advection-diffusion equation
+v = t x x µ + F x
µ0

0xl

t >0

· with initial condition
( x ,0 ) = ( x )
0xl

· and boundary conditions
( 0, t ) = (t ) ( l ,t ) = (t )
( x ) (t )
(t )

t >0

· where

v (x,t )

µ( x , t )

are known functions, while the function ( x , t ) is unknown.


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
· Integrating equation at x

n +1 j

j

from t

n

to

t

n +1

yields

= j -

n

t

n +1


t

n

v x - x

µ - F dt x j

· Approximating the integral on the right-hand side by the mean-value theorem, we obtain

n
n +1 j

= - v - µ - F x x x
n j

t = j

· where

t <
n +1


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
t = · For the approximation of the derivatives ( x ) j and [(µ x ) x ] tj = we will use the following difference relations:

x

t = j

1 = h h j -1 + h j


j -1

j +1

-
j

j

h

+h

j -
j

j -1

h

j -1



t =

- 3
t =

-

h

j

3 h -1 j 6 x

x

µ

x

t = j

1 = µ h j -1 + h j

(

j +1



j

)



j +1

-
j

j

h

-
- hj - h 3
j -1

- µj +µ

(

j -1

)

j - h
j -1

j -1



t =

3 x

3

t =


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM The unilateral difference expressions ( j +1 - j ) h j and ( j - j -1 ) h j -1 in derivatives of order 1 and 2 will be taken at different time levels (n and n + 1). at different levels For construction of approximations only by two points it is natural for physical reasons to have it is reasons on the (n + 1)-th layer a point x j as central, and to select the second one from that side from where is transferred by advection to the central point. In this manner we gain the following form:


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
1. for v > 0
x
t = j n +1 j n +1 j -1

1 h h j -1 + h j


j -1

n j +1

-
j

n j

h

+h


j

-
j -1

h
2

+

+
1 µ h j -1 + h j

t x
-
j

t =

j

x µ x



t = j

(

j +1



j

)



n j +1

n j

h

-
t =

- µj +µ

(

j -1

)



n +1 j

-
j -1

n +1 j -1

h

2 + tx

j


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
2. for v < 0
x
t = j

1 h h j -1 + h j


j -1

n +1 j +1

- h
j

n +1 j

+h

j -
j

n

n j -1

h

j -1

+
t =

+
n +1 j +1

t x
n +1 j

2

j

x µ x

t = j

1 µ h j -1 + h j

(

j +1



j

)



- h
j

-
t =

- µj +µ

(

j -1

)

j - h

n

n j -1

j -1

2 + tx

j


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
Difference scheme for the one-dimensional advection-diffusion th ti problem in the following form: · for v > 0

n +1 j

-

n j



1 h + h j -1 + h j

n vj j -1



n j +1

-
j

n j

h

+ hjv
n +1 j

n +1 j



n +1 j

-
j -1

n +1 j -1

h

- -Fn = 0 j

1 µ - h j -1 + h j

(

n j +1



n j

)



n j +1

-
j

n j

h



(



n +1 j -1

)



n +1 j

-
j -1

n +1 j -1

h

j = 1, 2,..., J - 1
j = x
0

( j)
n

n = 0,1, ...N
n

0 = t

n

()

J = t

()
n

j = 0,1,..., J

n = 0,1, ...N


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
· for v < 0

n +1 j

-

n j



1 h + h j -1 + h j

n +1 vj j -1



n +1 j +1

- h
j

n +1 j

+ hjv

n j

j - h
n

n

n j -1

j -1

-

1 µ - h j -1 + h j

(

n +1 j +1



n +1 j

)



n +1 j +1

-
j

n +1 j

h

- µ +µ

(

n j

n j -1

)

j - h

n j -1

j -1

- F jn = 0

j = J - 1, J - 2, ..., 2,1
=x
0 j

n = 0,1, ...N j = 0,1,..., J
n

( j)
n

0 = t

n

()

J = t

n

()

n = 0,1, ...N


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
Templates of difference networks: a) of the scheme v > 0 ; b) of the scheme v < 0


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
n +1

n +1 j



n +1 j +1



n +1 j +2

n



n +1 j -1



n j



n j +1



n j +2

j -1

j

j +1

j+2

j +3

·

Algorithm of solution of problem with the scheme

n +1 j

where
pj = h
j -1

= p j
j -1

n +1 j -1

- qj
n +1 j

n j +1

n n + 1 + q j j + F j 1 + p

(

)

(

j

)

h + hj hj

v

n +1 j

+

µ


j -1

n +1 j -1

h



n n h µ j +1 + µ j j -1 n qj = vj - h j -1 + h j h j hj


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
v <0 v >0

n +1

n
j -2

j -1

j

j +1

Geometric illustration of the flow velocity reversal from v < 0 to v > 0


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
v >0

n +1 j

= p j
j -1

n +1 j -1

- qj
n +1 j

n j +1

n n + 1 + q j j + F j 1 + p

(

)

(

j

)

pj =

h

j -1

h + hj hj

v

n +1 j

+

µ


j -1

n +1 j -1

h



n n h j -1 n µ j +1 + µ j qj = h vj - h j -1 + h j j hj

v <0

+h
n +1 j -1

= 1 - s
j -2 j -1

j -1



n j -1

+s
n +1 j -1

n j -2 j -1

-r

n +1 j j -1

+ f 1 - r
n j

j -1


n j -2

r

j -1

=

h

j -2

j -1

h h

v

n +1 j -1

-

µ

n +1 j


j -1

h



s

j -1

=

h

j -2

+h

j -1

h h

j -1 j -2

v

n j -1

+

µ

n j -1


j -2

h




A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
v >0
q
n +1 j -1

1 + s = 1 + s j

G - r j j -1 j 1 - r + s r j j
j

j

v <0
q
n +1 j

=



s j G j + 1 - 1 + s j 1 - r

r
j

j -1



j j

+ sjr

where
G j = 1 - s
j -1

q

n j -1

+s

j-

1q

n j -2

n j = 1 + d j q j - d j q

n j +1


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
Numerical viscosity of difference scheme is viscosity is
v vh j + 2µ = v - 2 h +h j -1 j 2q = -µ 2 x
eff

2q x 2

If v = 0 or µ = 0 and Inasmuch as
µ
eff

h

j -1

= h j = const then

=0

= vh j + 2µ

(

) (h
h

j -1

+ hj -v 2 > 0
+h 1 2

)

under v > 0 and µ > 0

therefore

h j + 2µ v
j -1 j

scheme is stable


A FINITE-DIFFERENCE SCHEME FOR THE ADVECTION-DIFFUSION PROBLEM
The scheme stability condition is satisfied if ti fi
1 + 2q sin 2 2 + q 2 sin 2 j j = 2 2 1 + 2 p sin 2 + p 2 sin 2 j j
2

We have q j p j . steps h j and takes place. It condition of a

It means that irrespective of a ratio of grid It means ratio from this expression the inequality 1 follows from here the stability requirement difference grid steps h j and . diff id


A NUMERICAL EXPERIMENT
For experimental estimation of such important characteristics of difference schemes like: accuracy, stability and efficiency we consider a problem of propagation of some physical quantity q (x1 , x 2 , x 3 , t ) in viscous continuum µ = {µ1 , µ 2 , µ 3 }
µ1 = sin (x1
2

)

µ 2 = sin (x
2

2

)
)

µ3 = x
3

2a 3

that moves with a speed V = {v 1 , v 2 , v
v1 = 1 sin (2 x1 2

}
v3 = a x
2a - 1 3

)

v2 =

1 sin (2 x 2
2

2

A function

1- a x1 x 2 bx 3 (2 - b )t tg tg sin q (x1 , x 2 , x 3 , t ) = e 2 2 (a - 1) is a precise solution of the problem:


THE RESULTS OF A NUMERICAL EXPERIMENT
q q q q + v1 + v2 + v3 = t x1 x 2 x 3 q µ1 = x1 x1 0 x1 2 t >0 0
q x1 , x 2 , x 3 , t

x2 2
2 sin bx
2 2

+ x

2

q µ2 + x x 2
1- a 3

3

0 x3 2
1

q µ3 x 3

(

)
)

q ( x1 , x 2 , x 3 , t
q x1 , x 2 , x 3 , t

t =0

= tg x1 2 tg x

(

)(

2

)

(a - 1)-
1- a 3


1

(

)

x1 = 0

=0
=e

2 - b2 t

q ( x1 , x 2 , x 3 , t

q (x1 , x 2 , x 3 , t

q x1 , x 2 , x 3 , t

(

)

x1 = 2

tg x

2 sin

bx

(a - 1)-



)

x2 = 0

=0
=e
=e

2 - b2 t

q x1 , x 2 , x 3 , t

(

)

x2 = 2

)

x3 = 0

=0

2 tg x1 2 sin bx

1- a 3

(a - 1)-

1


1

2 - b2 t

x3 = 2

tg x

2 1

2 tg x

2 2

2 sin

b 4

( )1- a (a - 1)-




THE RESULTS OF A NUMERICAL EXPERIMENT
t 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.12307 0.12307 12 0.12307 0.12307 0.12307 0.12307 0.12307 0.12307 0.12307 0.12307 12 Maximum fractional error of the task solution
= 4.7124·10
-2

= 2.3562·10

-2

= 4.7124·10

-3

= 4.7124·10

-4

6.3636·10-2 6.3636·10-2 3636 10 6.3636·10-2 6.3636·10-2 6.3636·10-2 6.3636·10-2 6.3636·10-2 6.3636·10-2 6.3636·10-2 6.3636·10-2 3636 10

5.2971·10-3 5.2972·10-3 2972 10 5.2972·10-3 5.2972·10-3 5.2972·10-3 5.2972·10-3 5.2972·10-3 5.2972·10-3 5.2972·10-3 5.2972·10-3 2972 10

1.7102·10-4 1.7106·10-4 7106 10 1.7107·10-4 1.7108·10-4 1.7108·10-4 1.7108·10-4 1.7108·10-4 1.7108·10-4 1.7108·10-4 1.7108·10-4 7108 10


II. PROBLEM-SOLVING NUMERICAL PROCEDURE OF MESOSCALE FORECAST ON FORECAST ON BASIS OF MULTIPLE NODES INTERPOLATION TECHNIQUE
· Regional atmospheric processes are influenced by macroscale atmospheric circulation where modeling ti li meteorological values in restricted area is considered as a part some whole with time dependent, transitional boundary conditions. To achieve demanded level of accuracy of the solutions for a model in places of heavy gradients of related functions it is often needed to have variable grid step of numerical solution for restricted terrains. However common techniques of mathematical physics often can not satisfy these requirements because of low accuracy, slow divergence and suffering from stability problems, so some dedicated numerical modeling are needed to make computational methods efficient.


INTRODUCTION
· For forecasting of values of meteorological quantities (components v1, v 2 , v 3 of velocity V , pressure p, temperature , specific humidity , specific liquid water content q , concentration of pollutants and others) of an atmosphere above the limited territory we will follow basics of the method of "unilateral influence" where results of analysis and forecast received within lts ed ithin macroscale (hemisphere or global) model are used as boundary conditions for a regional model. conditions model


PROBLEM STATEMENT AND A METHOD OF ITS NUMERICAL SOLUTION · Let a state of atmosphere in space r = (, , ) of macrospace area G(r ) G(r ) is defined by a vector of meteorological quantities (r , t ) of discrete values of discrete values of the analysis and forecast r ,t m +1 = m +1(r ) received on a basis of macrospace model at the model moment of time t = t m +1 (m = 0,1,..., M ) with a step m +1 m =t -t . · Then for definition of a state of atmosphere in the limited territory G at t t m , t m +1 we will solve a task of following kind in vector representation:


PROBLEM STATEMENT (continuation)
= D t

,

m m +1 t t , t

,

r G

m +1 m +1 r , t (r ) = ,

m = 0, 1,..., M

where

1 1 2 1 D = + + r cos r cos r r r
v1 v 2 - - - v3 +F r r cos r

3 r

-


APPROXIMATION OF DIFFERENTIAL OPERATORS BY GRID ONES
· Computation of grid values of partial derivative of the first 2 2 i = order i = ( )i and of the second order included in the differential operator D , will be performed in the on the basis of relations: · first order order
h 1 + i i +1 + 2 hi -1
3 = hi i +1 2

i

h i + i i -1 = hi -1
hi hi -1
2

- 1-

i

-

hi hi -1

2 i -1 -

-

hi h

i -1 1 24

2 hi 4 - h i -1 4


APPROXIMATION OF DIFFERENTIAL OPERATORS BY GRID ONES (continuation)
· second order
hi -1 hi hi -1 hi hi -1 1 - + 1 i +1 + hi

h h h h h 1 + i -1 i -1 3 + i -1 + 1 i + i -1 1 + i -1 - 1 i -1 = + hi hi hi hi hi

h 12 hi -1 1 + i -1 i + i -1 + = i +1 - 2 h hi hi i
2 2 hi hi -1 hi -1 1 - + h 360 i

h i -1 + 21 5 hi

2 hi -1 5 - h 5 i


APPROXIMATION OF DIFFERENTIAL OPERATORS BY GRID ONES (continuation)
· It is obvious, that the derived relations have the third order at · hi hi -1 and fourth order at hi = hi -1. These systems are the algebraic equations with three-diagonal matrixes, so solutions can be found with boundary conditions: · ·
h - 1 - 1 (2 - 1) + 1 + 2 = 2 2 , 6 h1 hN - 1 N - N -1 ( N - N -1 ) + N -1 + N = 2 - hN - 1 6

.

· The main advantage of the offered method of approximation of derivatives. As a solution of the system of algebraic equations in all points depends on values in other points, it ti ll th it depends on i globally instead of locally that means smooth filling up and approximation of differential operators by grid ones.


PROPOSED PROBLEM-SOLVING PROCEDURE



jkl

t

=

n jkl

f

n jkl



jkl

(t p )

t t n , t

n +1


jkl

=

p jkl

f

p jkl

=

p jkl

=

(t p )

p = n + 1, n , n - 1


PROBLEM-SOLVING PROCEDURE
· After computation of values of right parts m +1 m +1 m +1 m +1 , m = 1,2,..., M f t =f = D t = of the grid ( j , k , l ) , 1 j J , 1 k K , 1 search for a solution of the problem for with Hermite polynomial like above for number of
(t ) = + t -t m t -t + 4 4
m m

in all nodes l L , we will the help of points:

(

f
m +1

m

+
m m -1

- 2 +

)- (f

m +1

-f

m -1

)

+

t -t + 4 m t -t - 4 m t -t + 4

m

5 m +1 - m -1 - f m +1 + 8f m + f m -1 - 2 m +1 - 2m + m -1 - f m +1 - f m -1 +

3 m +1 - m -1 - f m +1 + 4f m + f m -1


PROBLEM DEFINITION
A function u = sin(t + ) exp(at ) is a precise solution of the problem: du - a u = exp(a t ) cos( t + ) dt

t [2, 3

]

u (t i ) = ui ,

i = 1, 2, 3 ,

where a = 0 .2 , = 1.4 are known constants


THE RESULTS OF A NUMERICAL EXPERIMENT

given values

,

development, ----- computational solution


PROGRAMMIC INTERFACE


PROGRAMMIC INTERFACE


PROGRAMMIC INTERFACE


PROGRAMMIC INTERFACE


PROGRAMMIC INTERFACE


PROGRAMMIC INTERFACE


EXAMPLE WEATHER FORECAST


VERTICAL PROFILE OF WEATHER QUANTITY


=0,000015