PC proxy: a method for dynamical tracer reconstruction

Abstract

A detailed development of the principal component proxy method of dynamical tracer reconstruction is presented, including error analysis. The method works by correlating the largest principal components of a matrix representation of the transport dynamics with a set of sparse measurements. The Lyapunov spectrum was measured and used to quantify the lifetime of each principal component. The method was tested on the 500 K isentropic surface with stratospheric ozone concentration measurements from the Polar Aerosol and Ozone Measurement III satellite instrument during October and November 1998 and compared with the older proxy tracer method which works by correlating measurements with a single other tracer or proxy. Using a 60 day integration time and five (5) principal components, cross validation of globally reconstructed ozone and comparison with ozone sondes returned root-mean-square errors of 0.16 and 0.36 ppmv, respectively. This compares favourably with the classic proxy tracer method in which a passive tracer equivalent latitude field was used for the proxy and which returned RMS errors of 0.22 and 0.59 ppmv for cross-validation and sonde validation respectively. The method seems especially effective for shorter lived tracers and was far more accurate than the classic method at predicting ozone concentration in the Southern hemisphere over the same time period. It is also more effective when reconstruction is performed over the entire Earth rather than a single hemisphere allowing for seamless reconstruction of global fields.

Appendices

Appendix 1: Model properties

1.1 Appendix 1.1: Mass conservation

Suppose that:

$$\begin{aligned} \sum _i q_i = const. \end{aligned}$$

(41)

This will be true for non-divergent flows on equal area grids. Then:

$$\begin{aligned} \sum _i \sum _j r_{ij} q_j= & {} \sum _j q_j \end{aligned}$$

(42)

$$\begin{aligned} \sum _j q_j \left( \sum _i r_{ij} - 1 \right)= &\, {} 0 \end{aligned}$$

(43)

Therefore:

$$\begin{aligned} \sum _i r_{ij} = 1 \end{aligned}$$

(44)

If (41) is true, then:

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d} t}\sum _i q_i = 0 \end{aligned}$$

(45)

is also be true. Continuing:

$$\begin{aligned} \sum _i \frac{\mathrm {d} q_i}{\mathrm {d} t}= & {} \,0 \end{aligned}$$

(46)

$$\begin{aligned} \sum _i \sum _j a_{ij} q_j= & {} \,0 \end{aligned}$$

(47)

$$\begin{aligned} \sum _j q_j \sum _i a_{ij}= & {} \,0 \end{aligned}$$

(48)

which shows the second part of (26) and (27):

$$\begin{aligned} \sum _i a_{ij} = 0. \end{aligned}$$

(49)

1.2 Appendix 1.2: Diffusion and the Lyapunov spectrum

A discrete tracer mapping will always require some amount of diffusion. This means that the tracer configuration will tend towards a uniform distribution over time, that is, it will “flatten out.” We can show that, given the constraint in (41), a tracer field with all the same values has the smallest magnitude. Suppose there are only two elements in the tracer vector, $\mathbf {q}=\lbrace q,~q \rbrace$. The magnitude of the vector is:

$$\begin{aligned} |\mathbf {q}|=\sqrt{q^2+q^2}=\sqrt{2} q \end{aligned}$$

(50)

Now we introduce a separation between the elements, $2{\varDelta }q$, that nonetheless keeps the sum of the elements constant:

$$\begin{aligned} |q+{\varDelta }q,~q-{\varDelta }q|= & {} \sqrt{(q+{\varDelta }q)^2+(q-{\varDelta }q)^2} \end{aligned}$$

(51)

$$\begin{aligned}= & {} \sqrt{2}\sqrt{q^2+({\varDelta }q)^2} \ge \sqrt{2} q \end{aligned}$$

(52)

This will generalize to higher-dimensional vectors. In general, we can say that:

$$\begin{aligned} \mathbf {q} R^T R \mathbf {q} \le | \mathbf {q} |^2 \end{aligned}$$

(53)

Implying that for the eigenvalue problem,

$$\begin{aligned} R^T R \mathbf {v}= & {} s^2 \mathbf {v} \nonumber \\ s^2\le & {} 1 \end{aligned}$$

(54)

Therefore the Lyapunov exponents are all either zero or negative. Note however that this does not constitute a proof; the actual proof is more involved.

To prove (54) from (53), we first expand $\mathbf {q}$ in terms of the right singular vectors, $\lbrace \mathbf {v}_i \rbrace$:

$$\begin{aligned} \mathbf {q} = \sum _i c_i \mathbf {v}_i \end{aligned}$$

(55)

where $\lbrace c_i \rbrace$ are a set of coefficients. Substituting this into the left-hand-side of (53):

$$\begin{aligned} \mathbf {q} R^T R \mathbf {q}= & {} \sum _i c_i \mathbf {v}_i \sum _i c_i s_i^2 \mathbf {v}_i \end{aligned}$$

(56)

$$\begin{aligned}= & {} \sum _i \sum _j c_i c_j s_i^2 \mathbf {v}_i \mathbf {v}_j \end{aligned}$$

(57)

$$\begin{aligned}= & {} \sum _i \sum _j c_i c_j s_i^2 \delta _{ij} \end{aligned}$$

(58)

$$\begin{aligned}= & {} \sum _i c_i^2 s_i^2 \end{aligned}$$

(59)

where $\delta$ is the Kronecker delta. Similarly, we can show that:

$$\begin{aligned} \mathbf {q} \cdot \mathbf {q} = \sum _i c_i^2 \end{aligned}$$

(60)

If we assume that $s_i \le 1$ for every i, then:

$$\begin{aligned} \sum _i c_i^2 s_i^2 \le \sum _i c_i^2 \end{aligned}$$

(61)

since each term on the left side is less-than-or-equal-to the corresponding term on the right side. Note that in order for the inequality in (61) to be broken, at least one singular value must be greater-than one. Therefore (53) is true for every $\mathbf {q}$ if-and-only-if (54) is true for every s. In the language of set theory and first-order logic:

$$\begin{aligned} \forall \mathbf {q} \in \mathfrak {R}^n ~ (| R \mathbf {q} |^2 \le |\mathbf {q} |^2) \iff \forall s \in \mathfrak {R}| ~R^T R \mathbf {v} = s^2 \mathbf {v} ~ (s \le 1). \end{aligned}$$

(62)

Appendix 2: Deviation from equal area

Here we calculate the ratio between the largest and smallest grid boxes in the azimuthal equidistant coordinate system. First we show that there is no distortion at the pole:

$$\begin{aligned} \lim _{x\rightarrow 0, ~y\rightarrow 0} \left( \frac{\mathrm {d}s}{\mathrm {d} x} \right) ^2= & {} \lim _{x\rightarrow 0, ~y\rightarrow 0} \left( \frac{\mathrm {d}s}{\mathrm {d} y} \right) ^2 \end{aligned}$$

(63)

$$\begin{aligned}= & {} \lim _{x\rightarrow 0, ~y\rightarrow 0} \frac{1}{r^2} \left[ \frac{R_E^2}{r^2} \sin ^2 \left( \frac{r}{R_E} \right) y^2 + x^2 \right] \end{aligned}$$

(64)

$$\begin{aligned}= & {} \frac{1}{r^2} \left[ \frac{R_E^2}{r^2} \left( \frac{r}{R_E} \right) ^2 y^2 + x^2 \right] \end{aligned}$$

(65)

$$\begin{aligned}= & {} 1 \end{aligned}$$

(66)

hence the ratio between projected and unprojected areas is 1. Grid areas become progressively smaller the further from the pole you get. Since the projection is hemi-spherical, r takes on a maximum value at the equator:

$$\begin{aligned} r=\pi R_E/2 \end{aligned}$$

(67)

Hence the largest possible values for x and y are:

$$\begin{aligned} x = y = \frac{\pi R_E}{2 \sqrt{2}} \end{aligned}$$

(68)

which represents a point on the equator along a diagonal from the origin in the projected coordinate system. The metric coefficients can be calculated:

$$\begin{aligned} \left( \frac{\mathrm {d}s}{\mathrm {d} x} \right) ^2= & {} \left( \frac{\mathrm {d}s}{\mathrm {d} y} \right) ^2 \end{aligned}$$

(69)

$$\begin{aligned}= & {} \frac{4}{\pi ^2 R_e^2} \left[ \frac{4 R_E^2}{\pi ^2 R_E^2} \sin ^2 \left( \frac{\pi R_E}{2 R_E} \right) \frac{\pi ^2 R_E^2}{8} + \frac{\pi ^2 R_E^2}{8} \right] \end{aligned}$$

(70)

$$\begin{aligned}= & {} \frac{1}{\pi ^2 R_E^2} \left( 2 R_E^2 + \frac{\pi ^2 R_E^2}{2} \right) \end{aligned}$$

(71)

$$\begin{aligned}= & {} \frac{2}{\pi ^2} + \frac{1}{2} \end{aligned}$$

(72)

$$\begin{aligned}\approx &\, {} 0.703 \end{aligned}$$

(73)