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球面上のベクトル場の推定とその応用

中野 慎也

 

モデリング研究系 准教授

2020年10月27日 統計数理研究所 オープンハウス

Localized basis functions for spherical vector field

 A divergence-free vector field can be represented by a stream function as follows:

 A curl-free vector field can be represented by a potential function as follows:

 According to the Helmholtz theorem, an arbitrary vector field on a sphere can be written as a sum of a divergence-free field and a curl-free field. We can thus represent any vector field in the

following form:

( ) .

df df

= − ×∇Ψ r

V r e

( ) .

cf = −∇Ψ cf

V r

( ) = df ( ) + cf ( ) = − ×∇Ψ − ∇Ψ r df cf .

V r V r V r e

Ψ df

Ψ cf

 We expand the stream function and the potential function by using localized basis functions and :

 Defining the following vector-valued basis functions:

an arbitrary spherical field can be represented using these basis functions :

Ψ df Ψ cf

( ) ( , ), ( ) ( , ).

df df df cf cf cf

i i i i

i i

w ψ w ψ

Ψ r = ∑ r r Ψ r = ∑ r r

( ) ( )

( ) ( , ) ( , )

( , ) ( , ).

df df cf cf

i r i i i

i i

df df cf cf

i i i i

i i

w w

w w

ψ ψ

= − ×∇ + −∇

= +

∑ ∑

∑ ∑

V r e r r r r

v r r v r r

ψ df ψ cf

( , ) ( , ), ( , ) ( , ),

df df cf cf

i = − ×∇ r ψ i i = −∇ ψ i

v r r e r r v r r r r

Modeling of divergence-free vector field

 We employ spherical Gaussian functions for obtaining a set of basis functions

 We then obtain

2 , 2

( , ) ( )exp 1 , ( , ) ( sin )exp 1 .

df i cf i

i

= η

i

×   η   R    

i

=

θ i

η ∆ θ   η   R    

r r r r

v r r r r v r r e

( )

( , ) ( , ) exp

2

1 exp cos 1 .

df cf i

i i

R

I

ψ = ψ =  η  =   η θ ′ −  

 

 

r r r r r r

 When we use the expansion of our basis functions

the node points r i can be placed arbitrarily.

 We placed 2500 node points randomly and uniformly distributed in the region above 40 degree in latitude, which approximates the following Monte Carlo convolution:

/2 2 2

0 0

( ) ( ) ( , ) ( ) ( , ) sin .

S w d π π w r R θ θ φ d d

= ∫ = ∫ ∫

V r r v r r r v r r

( ) i ( , ), i

i

w

= ∑

V r v r r

 Now we consider a divergence-free vector field (no source, no sink) and expand the field by using the divergence-free basis functions:

( ) i df df ( , ). i

i

w

= ∑

V r v r r

>> Next page

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Application to ionospheric physics Design of covariance matrices

 Now we apply this method for estimating the ionospheric plasma velocity distribution which can be assumed to be divergence-free.

 We fit the model to the data of SuperDARN, which is a radar network observing the ionospheric plasma velocity.

 The gaps of the data coverage of SuperDARN are filled with the empirical model (Weimber 2001).

 We assume the weight w can be decomposed into the model- based value ζ and the residual β :

and the residual β is estimated with the Kalman filter.

, w = + ζ β

 We assume the temporal evolution of the weights β

 The residual component can then be estimated with the following Kalman filter algorithm:

| 1 1| 1

| 1 2 1| 1

,

.

k k k k

k k k k

α α

− − −

− − −

=

= +

β β

P

P Q

Prediction:

Filtering:

1 1 1 1

| | 1 | 1

1 1 1

| | 1

( )

. (

( ,)

T T

k k k k k k k k k k k k k k

k k k k T k k k

− − − −

− −

− − −

+ +

+

= −

=

β β P H R H y H β

R H

H R

P P H )

1 1

( k | k ) ( k , ) . p β β =  α β Q

Results >> See our paper for detail ( https://doi.org/10.1186/s40623-020-01168-4 ).

 In order to ensure spatial smoothness, the covariance matrix Q k is set as follows:

where is the following correlation function:

where

1 1 1

2

1

( , ) ( , )

,

( , ) ( , )

Q Q n

Q

Q n Q n n

C C

C C

σ

 

 

=  

 

 

r r r r

r r r r

Q

  

 , )

( i

Q j

C r r

( , i j ) ( , exp i j ) 2 i 1 , C Q

ρ   κ   R    

= r r

r r r r

2

(sin sin 40 )(sin sin 40 )

( , .

(1 sin 40

) )

i j λ i λ j

ρ

° °

− °

= − −

r r

 The matrix R k is set as follows:

where b i and g i denote the beam number and the range gate of the i-th element of the observation y and

1 1

1

1 1

1 1 1

1 1 1 1

2

1 1 1 1

( , ) ( , )

,

( , ) ( , )

b b R b b R

k R

b b R b b R

C g g C g g

C g g C g g

δ δ

σ

δ δ

 

 

=  

 

 

R

  

( ) 2

, x

( ) e p i 2 j .

R g i j g g

C g  − 

=  − 

 

 

参照

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