Perfect observations in a reduced-order analysis

Perfect observations in a reduced-order 3D-Var/OI analysis

Introduction: 3D-Var/OI analysis

The linarised 3D-Var expression is an OI-type analysis. The analyis,

x^{a}

, is a correction of the background field,

x^{b}

x^{a} = x^{b} + δ x

where all vectors have length

I

. The number of observations is

M

M < I

The analysis increment,

δ x

, is a linear function of the innovation,

d

, a vector of length

M

δ x = K d

K

is the gain matrix, of order

(I, M)

d = y^{o} - H x^{b}

y^{o}

is the observation vector (of length

M

y^{b} \equiv H x^{b}

is the background estimate of the observations, obtained by applying the observation operator

H

to the background,

x^{b}

The gain matrix has two equivalent expressions:

K = B H^{T} {(H B H^{T} + R)}^{- 1}

K = {(B^{- 1} + H^{T} R^{- 1} H)}^{- 1} H^{T} R^{- 1}

B

(I, I)

background error covariance matrix,

R

(M, M)

observation error covariance matrix, and

H

(M, I)

, is the Jacobian matrix (evaluated in the background

x^{b}

) of the function

H

(which may be non-linear).

That the two expressions are equivalent is easily seen. In fact the relation:

{(B^{- 1} + H^{T} R^{- 1} H)}^{- 1} H^{T} R^{- 1} = B H^{T} {(H B H^{T} + R)}^{- 1}

is equivalent to:

H^{T} R^{- 1} (H B H^{T} + R) = (B^{- 1} + H^{T} R^{- 1} H) B H^{T}

which is an identity.

Now, in order to assume that observations are perfect, only the first of the two expressions can be used. By setting

R = 0

K^{p e r f e c t o b s} = B H^{T} {(H B H^{T})}^{- 1}

H B H^{T}

(M, M)

M < I

, then it can be invertible (though this is not guaranteed: for example it is not invertible when

H

has two equal rows).

In the perspective of the “perfect observation” assumption, in order to look more deeply into the two expressions, it is useful to define a couple of scalars:

σ_{o}^{2} \equiv \frac{1}{M} T r R

σ_{b}^{2} \equiv \frac{1}{I} T r B

T r

indicates the trace of the matrix. Then, define the following “tilde” matrices so that:

σ_{o}^{2} \tilde{R} = R

σ_{b}^{2} \tilde{B} = B

The two gain matrix expressions become:

K = \tilde{B} H^{T} {[H \tilde{B} H^{T} + \frac{σ_{o}^{2}}{σ_{b}^{2}} \tilde{R}]}^{- 1}

K = {[\frac{σ_{o}^{2}}{σ_{b}^{2}} {\tilde{B}}^{- 1} + H^{T} {\tilde{R}}^{- 1} H]}^{- 1} H^{T} \tilde{R}

Consider now the observations to be perfect with respect to the background field. This is obtained by assuming that:

\frac{σ_{o}^{2}}{σ_{b}^{2}} ≪ 1

σ_{o}^{2} / σ_{b}^{2}

, the first expression readily becomes, as above:

K_{p e r f e c t o b s} = \tilde{B} H^{T} {(H \tilde{B} H^{T})}^{- 1} = B H^{T} {(H B H^{T})}^{- 1}

In the second expression, though:

\frac{σ_{o}^{2}}{σ_{b}^{2}} {\tilde{B}}^{- 1} + H^{T} {\tilde{R}}^{- 1} H \approx H^{T} {\tilde{R}}^{- 1} H

{\tilde{R}}^{- 1}

(M, M)

, but the matrix

H^{T} {\tilde{R}}^{- 1} H

(I, I)

. So it must be rank-deficient, i. e. non-invertible, because

M < I

. The second expression cannot be used for perfect observations.

Reduced-order

In the general expression, the analysis is obtained as a linear combination of

M

vectors of length

I

M

(I, M)

B H^{T}

A reduced-order analysis is obtained as a combination of

N

N < M

, collected in the

N

(I, N)

E

. This is done when the matrix

B

can be approximated as:

B \approx E Γ E^{T}

Where the column of

E

are supposed to have been normalised, so that the magnitude of the background error is carried by the

(N, N)

Γ

, the “background error covariance matrix” in the subspace spanned by the columns of

E

. Again, two equivalent expressions are obtained for the gain matrix:

K_{E} = E Γ {(H E)}^{T} {[H E Γ {(H E)}^{T} + R]}^{- 1}

K_{E} = E {[Γ^{- 1} + {(H E)}^{T} R^{- 1} H E]}^{- 1} {(H E)}^{T} R^{- 1}

Note that the matrix

E

N

-dimensional basis) appears in both expressions on the left. That the two expressions are the same can be seen in a way similar to what was shown above. In fact, the relation:

{[Γ^{- 1} + {(H E)}^{T} R^{- 1} H E]}^{- 1} {(H E)}^{T} R^{- 1} = Γ {(H E)}^{T} {[H E Γ {(H E)}^{T} + R]}^{- 1}

is equivalent to:

{(H E)}^{T} R^{- 1} [H E Γ {(H E)}^{T} + R] = [Γ^{- 1} + {(H E)}^{T} R^{- 1} H E] Γ {(H E)}^{T}

which again is an identity.

Now, use again:

R = σ_{o}^{2} \tilde{R}

And use the scalar

γ^{2}

γ^{2} \equiv \frac{1}{N} T r (Γ)

γ^{2} \tilde{Γ} = Γ

The two expressions become:

K_{E} = E \tilde{Γ} {(H E)}^{T} {[H E \tilde{Γ} {(H E)}^{T} + \frac{σ_{o}^{2}}{γ^{2}} \tilde{R}]}^{- 1}

K_{E} = E {[\frac{σ_{o}^{2}}{γ^{2}} {\tilde{Γ}}^{- 1} + {(H E)}^{T} {\tilde{R}}^{- 1} H E]}^{- 1} {(H E)}^{T} {\tilde{R}}^{- 1}

In order to assume that observations are perfect with respect to the background field, assume:

\frac{σ_{o}^{2}}{γ^{2}} ≪ 1

N < M,

σ_{o}^{2} / γ^{2}

in the second expression is not a problem:

{K_{E}}^{p e r f e c t o b s} = E {[{(H E)}^{T} {\tilde{R}}^{- 1} H E]}^{- 1} {(H E)}^{T} {\tilde{R}}^{- 1} = E {[{(H E)}^{T} R^{- 1} H E]}^{- 1} {(H E)}^{T} R^{- 1}

In the first expression, though, the matrix

H E \tilde{Γ} {(H E)}^{T}

(M, M)

, then it is rank-deficient and not invertible, because

\tilde{Γ}

(N, N)

N < M

In conclusions, in the reduced-order analysis, it is the second expression that has to be used when the case of “perfect” observations is considered.

Illustrative case

N = 1

M > N = 1

the matrix $E$ has a single column, the vector $e$ (of length $I$ );

the matrix $Γ$ is reduced to a scalar, $γ^{2}$ ;

the matrix $H E = H e$ becomes a vector of length $M$ , with components $η_{m}$ . ${(H e)}^{T}$ is then a row vector;

Let $d_{m} = y_{m}^{o} - y_{m}^{b}$ be the components of the innovation vector $d$ (of length $M$ );

Assume $R$ , which has order $(M, M)$ , to be diagonal, with diagonal elements $σ_{m}^{2}$ .

The analysis increment, obtained applying the gain matrix to the innovation, is then a multiple of the vector

e

δ x = K_{E} d = e \frac{{(H e)}^{T} R^{- 1} d}{\frac{1}{γ^{2}} + {(H e)}^{T} R^{- 1} H e} = e \frac{\sum_{m = 1}^{M} \frac{η_{m} d_{m}}{σ_{m}^{2}}}{\frac{1}{γ^{2}} + \sum_{m = 1}^{M} \frac{η_{m}^{2}}{σ_{m}^{2}}}

Its expression for “perfect” observations is obtained by defining:

σ_{o}^{2} = \frac{1}{M} \sum_{m = 1}^{M} σ_{m}^{2}

R = σ_{o}^{2} \tilde{R}

\tilde{R}

also is diagonal, with elements

μ_{m}^{2} \equiv σ_{m}^{2} / σ_{o}^{2}

. The analysis increment is:

δ x = K_{E} d = e \frac{{(H e)}^{T} {\tilde{R}}^{- 1} d}{\frac{σ_{o}^{2}}{γ^{2}} + {(H e)}^{T} {\tilde{R}}^{- 1} H e} = e \frac{\sum_{m = 1}^{M} \frac{η_{m} d_{m}}{μ_{m}^{2}}}{\frac{σ_{o}^{2}}{γ^{2}} + \sum_{m = 1}^{M} \frac{η_{m}^{2}}{μ_{m}^{2}}}

σ_{o}^{2} / γ^{2}

to obtain the analysis increment in the perfect observations case:

δ x = K_{E} d = e \frac{{(H e)}^{T} {\tilde{R}}^{- 1} d}{{(H e)}^{T} {\tilde{R}}^{- 1} H e} = e \frac{\sum_{m = 1}^{M} \frac{η_{m} d_{m}}{μ_{m}^{2}}}{\sum_{m = 1}^{M} \frac{η_{m}^{2}}{μ_{m}^{2}}}

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Francesco Uboldi 2014,2015,2016,2017