Atmospheric moisture transport: stochastic dynamics of the - PowerPoint PPT Presentation

Atmospheric moisture transport: stochastic dynamics of the advection-condensation equation Yue-Kin Tsang School of Mathematics University of Edinburgh Jacques Vanneste

Moisture parameters specific humidity of an air parcel: q = mass of water vapor total air mass saturation specific humidity, q s ( T ) when q > q s , condensation occurs excessive moisture precipitates out, q → q s q s ( T ) decreases with temperature T T 2 < T 1 q s ( T 1 ) q s ( T 2 )

Moisture field of the atmosphere y = latitude, temperature decreases with y model : q s ( y ) = q 0 exp( − α y ) moist air parcels being advected around in the troposphere Figure 3. Schematic of the overturning circulation with emphasis on the mechanism controlling the hu- midity distribution in the subtropics. (Sherwood et al., Reviews of Geophysics, 2010)

Atmospheric moisture and climate Earth’s radiation budget: absorption of incoming shortwave radiation generates heat heat carried away by outgoing longwave radiation (OLR) water vapor is a greenhouse gas that traps OLR OLR ∼ − � log q � OLR ∼ − � log[ � q � + q ′ ] � ≈ − log � q � + 1 2 � q � 2 � q ′ 2 � how fluctuation q ′ is generated? what is the probability distribution of water vapor in the atmosphere?

Advection-condensation model ∂ q ∂ t + � � u · ∇ q = S − C , u = ( u , v ) S = moisture source (evaporation) C = condensation sink saturation profile: q s ( y ) = q 0 exp( − α y ) rapid condensation limit: C : q ( x , y , t ) = min [ q ( x , y , t ) , q s ( y ) ] Initial-value problem: S = 0 entire domain saturated at t = 0 : q ( x , y , 0) = q s ( y ) what is the PDF of q at location ( x , y ) and time t ? how fast does the total moisture content decay?

Coherent circulation + random transport Figure 3. Schematic of the overturning circulation with emphasis on the mechanism controlling the hu- midity distribution in the subtropics. coherent circulating component: u ( X , Y ) = − Ω ( R ) Y v ( X , Y ) = Ω ( R ) X √ X 2 + Y 2 where R = random component is δ -correlated in time (Brownain): U ∼ ˙ W ( t ) where W ( t ) is a Wiener process ( ˙ W ( t ) ∼ white noise)

A stochastic transport model √ d X ( t ) = u ( X , Y ) d t + 2 κ d W 1 ( t ) u = − Ω ( R ) Y √ d Y ( t ) = v ( X , Y ) d t + 2 κ d W 2 ( t ) v = Ω ( R ) X d Q ( t ) = − C ( Q , Y )d t 6 0.8 4 2 0.6 Ω 0 = 1 y 0 κ = 10 − 2 0.4 −2 0.2 −4 −6 0 −6 −4 −2 0 2 4 6 x

Evolution of the moisture field √ d X ( t ) = u ( X , Y ) d t + 2 κ d W 1 ( t ) u = − Ω ( R ) Y √ d Y ( t ) = v ( X , Y ) d t + 2 κ d W 2 ( t ) v = Ω ( R ) X d Q ( t ) = − C ( Q , Y )d t t = 10 t = 250 6 6 0.08 0.08 4 4 0.06 0.06 2 2 y y 0 0 0.04 0.04 −2 −2 0.02 0.02 −4 −4 −6 −6 0 −6 −4 −2 0 2 4 6 −6 −4 −2 0 2 4 6 x x

Maximum excursion of an air parcel maximum excursion , λ = max t ∈ [0 , t 1 ] y ( t ) q ( x , y , t 1 ) = q s ( λ ) statistics of q ⇐⇒ statistics of maximum excursion Pierrehumbert, Brogniez & Roca 2007 : obtain P ( q | y , t ) for an ensemble of particles execute independent random walks in a 1D domain

Theory: maximum excursion statistics The backward Fokker-Planck equation: ∂ P u · ∇ P + κ ∇ 2 P , P ≡ P ( x ′ , y ′ , t | x , y , 0) ∂ t = � Boundary conditions y = λ , y = −∞ and x = ±∞ : P ( x ′ , y ′ , t | x , y , 0) = 0 Initial conditions P ( x ′ , y ′ , 0 | x , y , 0) = δ ( x − x ′ ) δ ( y − y ′ ) For a parcel starts at ( x , y ) and time τ = 0, the probability that the maximum excursion Λ = max τ ∈ [0 , t ] y ( τ ) < λ : � λ −∞ d y ′ � ∞ −∞ d x ′ P ( x ′ , y ′ , t | x , y , 0) F ( x , y , t ; λ ) = Probability density function of Λ : P Λ ( λ, t | x , y ) = ∂ F ∂λ

Asymptotics: fast advection limit ∂ F u · ∇ F + κ ∇ 2 F , ∂ t = � F ( x , y , t ; λ ) F ( x , y = λ, t ; λ ) = 0 B.C.: � if y < λ 1 F ( x , y , t = 0; λ ) = I.C.: 0 otherwise Fast advection limit : ǫ = κ/ ( Ω 0 L 2 ) ≪ 1 Scaling : x → L x , t → ( L 2 /κ ) t , � u → ( Ω 0 L ) � u ∂ F ∂ t = ǫ − 1 � u · ∇ F + ∇ 2 F Expand : F = F 0 + ǫ F 1 , ǫ − 1 : � u ( r ) · ∇ F 0 = 0 ⇒ F 0 = F 0 ( r , t ; λ ) axisymmetric

PDF of specific humidity ∂ F 0 ǫ − 1 : u · ∇ F 1 + ∇ 2 F 0 , ∂ t = � F 0 ( r , t ; λ ) , F 1 ( r , θ, t ; λ ) Averaging over θ with � � � u · ∇ F 1 θ = 0, we get ∂ F 0 ∂ r ∂ F 0 ∂ t = 1 � � ∂ r ∂ r r Boundary conditions: F 0 ( r , t ; λ ) = 0 at r = λ P Λ ( λ, t | r ) ≈ ∂ F 0 ∂λ q ( r , t ) = q s ( λ ) = q 0 exp( − αλ ) � � � d λ � � � P Q ( q , t | r ) = P Λ ( λ, t | r ) � � � � d q q 0 � � λ = α − 1 ln q

Results: PDF of λ and q r = π /2 r = π /2 0.5 400 t = 10 0.4 t = 10 t = 10 2 300 t = 10 2 t = 10 3 t = 10 3 Λ ( λ, t | r ) Q ( q,t | r ) 0.3 t = 10 4 t = 10 4 200 0.2 P P 100 0.1 0 0 0 10 20 30 40 50 0.0 0.2 0.4 0.6 0.8 1.0 q / q s (r ) λ − r � � 1 P Q ( q , t | r ) = α q P Λ ( λ, t | r ) q 0 λ = α − 1 ln q

Results: decay of total moisture content � � Q ( t ) = 1 ¯ d q q P Q ( q , t | r ) d A A -1 10 simulation theory -2 10 asymptotics t 1/6 exp( − 0.59 t 1/3 ) -3 10 ( t ) _ Q -4 10 -5 10 -6 10 0 2000 4000 8000 10000 6000 time