Analysis Methods in Atmospheric and Oceanic Science

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1 Analysis Methods in Atmospheric and Oceanic Science AOSC 65 Partial Differential Equations Week 1, Day 3 1 Nov 014 1

2 Partial Differential Equation An equation that defines the relationship between an unknown function (or functions) of several independent variables and their partial derivatives with respect to those variables

3 3 Partial Differential Equations u (,) xt = K (,) xt (1)

4 4 Partial Differential Equations u (,) xt = K (,) xt (1) What is u?

5 5 Partial Differential Equations What is u? u (,) xt = K (,) xt Heat: measure of total energy that depends on mass Temperature: measure of molecular motion that is independent of mass (1) See

6 6 Partial Differential Equations What is u? What is K? u (,) xt = K (,) xt Heat: measure of total energy that depends on mass Temperature: measure of molecular motion that is independent of mass (1) See

7 7 Partial Differential Equations u (,) xt = K (,) xt (1) Alternate forms of this equation: cρ ( xyzt,,, ) = kxyz (,, ) ( xyzt,,, ) + kxyz (,, ) ( xyzt,,, ) y y + kxyz (,, ) ( xyzt,,, ) () z z

8 8 Partial Differential Equations u (,) xt = K (,) xt (1) Alternate forms of this equation: cρ ( xyzt,,, ) = kxyz (,, ) ( xyzt,,, ) + kxyz (,, ) ( xyzt,,, ) y y + kxyz (,, ) ( xyzt,,, ) () z z c is specific heat ρ is density and k is thermal conductivity where

9 9 Partial Differential Equations u (,) xt = K (,) xt (1) Alternate forms of this equation: cρ ( xyzt,,, ) = kxyz (,, ) ( xyzt,,, ) + kxyz (,, ) ( xyzt,,, ) y y + kxyz (,, ) ( xyzt,,, ) () z z Equation can be re-written (and solved in a straight forward manner) if c, ρ, and k are constant wrt spatial dimensions: cρ u u u ( xyzt,,, ) = ( xyzt,,, ) + ( xyzt,,, ) + ( xyzt,,, ) k y z (3)

10 10 Partial Differential Equations Equation can be re-written (and solved in a straight forward manner) if c, ρ, and k are constant wrt spatial dimensions: c u u u u ρ ( xyzt,,, ) = ( xyzt,,, ) + ( xyzt,,, ) + ( xyzt,,, ) k y z (3) How might a physicist write this same equation?

11 Partial Differential Equations Equation can be re-written (and solved in a straight forward manner) if c, ρ, and k are constant wrt spatial dimensions: c u u u u ρ ( xyzt,,, ) = ( xyzt,,, ) + ( xyzt,,, ) + ( xyzt,,, ) k y z (3) How might a physicist write this same equation? T ρ Cp = Q + ( k T) where T is temperature, T is the gradient of temperature is the divergence operator ρ is the density C is the heat capacity and Q k (4) 11 p is the heat added to the system is the thermal conductivity

12 1 Partial Differential Equations u (,) xt = K (,) xt Another application of this same equation in AOSC that does not involve Temperature or Heat: (1) Carbon Carbon (z, t) = K ( zt, ) (5) z where Carbon is the Dissolved Inorganic Carbon content of the ocean z is the depth of the ocean K is the eddy diffusion coefficient

13 13 Partial Differential Equations A classic example of PDEs from oceanography: For the barotropic, tropical ocean, the linear inviscid governing equations (Gill, 198) are: βyv v + βyu v + y v y = g = g = 1 H + βy η η y η v + y (1(6a) a), (1(6b) b), (6c) (1c ), + βv = 0 (6d) (1d )

14 14 Partial Differential Equations A classic example of PDEs from atmospheric sciences: For the barotropic atmosphere the governing equations (Anderson and McCreary, 1985) are: U βyv V + βyu U V + y P = ru P = rv y = r ( (7a) a), ((7b) b), P ((7c) c), c

15 15 Partial Differential Equations Coupled ocean/atmosphere system: = + = + + =, 1,, 0 t H y v x u y g yu t v H x g yv t u x η η β ρ τ η β + = + = + =,,, c Q c P r y V x U rv y P yu t V ru x P yv t U β β The ocean The atmosphere

16 16 Partial Differential Equations One more example of a classic PDE from AOSC: θ Potential Vorticity: PV = g ( ζ θ + f ) p (8a) where: PV is potential vorticity f is Coriolis Parameter, Ω sin(latitude), where Ω is angular speed of Earth s rotation (s 1 ) g is gravitational acceleration (m s ) p is pressure (mbar, or force/area kg m s / m ) θ is potential temperature (K) ζ θ is relative isentropic vorticity (s 1 ) ζ θ v = evaluated on an isentropic surface y (8b) Units: K m s 1 kg 1. The quantity 10 6 K m s 1 kg 1 is often called MKS PV units

17 17 u (,) xt = K (,) xt Typically in atmospheric and oceanic sciences, PDEs are solved numerically rather than analytically: although in many cases analytic manipulation is key to understanding the solutions (e.g., wave propagation) or even being able to proceed with a numerical solution Three aspects of Numerical Solutions to PDEs that must be considered: 1. Initial Conditions and Boundary Conditions. Numerical Stability 3. Numerical Diffusion Note: material presented in this part of the lecture is based on Chapter 1 (Numerical Solutions to PDEs) of Numerical Analysis, 8 th Edition, by Richard L. Burden and J. Douglass Faires (1)

18 18 u (,) xt = K (,) xt Typically in atmospheric and oceanic sciences, PDEs are solved numerically rather than analytically: although in many cases analytic manipulation is key to understanding the solutions (e.g., wave propagation) or even being able to proceed with a numerical solution Three aspects of Numerical Solutions to PDEs that must be considered: 1. Initial Conditions and Boundary Conditions. Numerical Stability 3. Numerical Diffusion We ll consider all three aspects today: we re ambitious (1)

19 19 u (,) xt = K (,) xt 1. Initial Conditions (IC) and Boundary Conditions (BC) For the heat diffusion equation, must typically specify distribution of temperature (or heat) at all locations for t = 0 ux (,0) = f( x) Also, must specify either value of temperature (or heat) at boundary: 0 (1) (IC) u(0, t) = u and u( L, t) = u where the rod has length L (BCa) L or flux of temperature (or heat) at boundary: (0, t) = 0 and ( Lt, ) = 0 if the rod is insulated on both ends (BCb)

20 0 u (,) xt = K (,) xt 1. Initial Conditions (IC) and Boundary Conditions (BC) For the heat diffusion equation, must typically specify distribution of temperature (or heat) at all locations for t = 0 ux (,0) = f( x) Also, must specify either value of temperature (or heat) at boundary: 0 (1) (IC) u(0, t) = u and u( L, t) = u where the rod has length L (BCa) L or flux of temperature (or heat) at boundary: (0, t) = 0 and ( Lt, ) = 0 if the rod is insulated on both ends (BCb) Let s examine how our solution depends on IC and BC!

21 1 u (,) xt = K (,) xt (1) 1. Initial Conditions (IC) and Boundary Conditions (BC) For the heat diffusion equation, must typically specify distribution of temperature (or heat) at the all locations for t = 0 ux (,0) = f ( x ) (IC) Also, must specify either value of temperature (or heat) at boundary: u (0, t ) = u and u ( L, t ) = u where the rod has length L (BCa) 0 L or flux of temperature (or heat) at boundary: u u (0, t ) = 0 and ( Lt, ) = 0 if the rod is insulated (BCb) x x on both ends Let s examine how our solution depends on IC and BC! Can show that for BCa Can show that for BCb ul u lim uxt (, ) = u0 + x t L lim uxt (, ) = Constant t 0

22 1. Initial Conditions (IC) and Boundary Conditions (BC) For the heat diffusion equation, must typically specify distribution of temperature (or heat) at the all locations for t = 0 ux (,0) = f ( x ) (IC) Also, must specify either value of temperature (or heat) at boundary: u (0, t ) = u and u ( L, t ) = u where the rod has length L (BCa) 0 u (,) xt = K (,) xt (1) L or flux of temperature (or heat) at boundary: u u (0, t ) = 0 and ( Lt, ) = 0 if the rod is insulated (BCb) x x on both ends Let s examine how our solution depends on IC and BC! Can show that for BCa ul u0 lim uxt (, ) = u0 + x t L Can show that for BCb lim uxt (, ) = Constant What happened to the IC?!? t

23 3 u (,) xt = K (,) xt (1) u(0, t) = ult (, ) = 0, t> 0 and ux (,0) = f( x), 0 x L (IC) & (BCa). Numerical Stability Select a time-step size k ux ( i, tj) ux ( i, tj 1) k u ( xi, tj) + + k

24 4 u (,) xt = K (,) xt (1) u(0, t) = ult (, ) = 0, t> 0 and ux (,0) = f( x), 0 x L (IC) & (BCa). Numerical Stability Select a time-step size k ux ( i, tj) ux ( i, tj 1) k u ( xi, tj) + + k Select a spatial-step size h 4 u ux ( i + ht, j) ux ( i, tj) + ux ( i ht, j) h u ( x, ) i tj + h 1 4

25 5 u (,) xt = K (,) xt (1) u(0, t) = ult (, ) = 0, t> 0 and ux (,0) = f( x), 0 x L (IC) & (BCa). Numerical Stability Select a time-step size k ux ( i, tj) ux ( i, tj 1) k u ( xi, tj) + + k Select a spatial-step size h Then: 4 u ux ( i + ht, j) ux ( i, tj) + ux ( i ht, j) h u ( x, ) i tj + h 1 4 u u u u + u K k h i, j i, j 1 i + 1, j i, j i 1, j = 0

26 u (,) xt = K (,) xt (1) u(0, t) = ult (, ) = 0, t> 0 and ux (,0) = f( x), 0 x L (IC) & (BCa). Numerical Stability The relation: u u u u + u K k h i, j i, j 1 i + 1, j i, j i 1, j combined with the IC & BC imply: = 0 (1+ λ) λ 0 0 u u 1, j 1, j 1 λ (1 + λ) λ u, j u, j 1 λ (1 + λ) λ = λ λ 0 λ (1 + λ) um 1, j um 1, j 1 where λ = Kk ( / h) 6

27 u (,) xt = K (,) xt (1) u(0, t) = ult (, ) = 0, t> 0 and ux (,0) = f( x), 0 x L (IC) & (BCa). Numerical Stability The relation: u u u u + u K k h i, j i, j 1 i + 1, j i, j i 1, j combined with the IC & BC imply: 7 = 0 (1+ λ) λ 0 0 u u 1, j 1, j 1 λ (1 + λ) λ u, j u, j 1 λ (1 + λ) λ = λ λ 0 λ (1 + λ) um 1, j um 1, j 1 Another matrix equation, finally! where λ = Kk ( / h)

28 . Numerical Stability has everything to do with the properties of the matrix! It can be shown that, due to the properties of the eigenvalues of this matrix, it s inverse must exist: therefore, can always find values of u at time j based on knowledge of values at time j 1 (1+ λ) λ 0 0 u u 1, j 1, j 1 λ (1 + λ) λ u, j u, j 1 λ (1 + λ) λ = λ λ 0 λ (1 + λ) um 1, j um 1, j 1 where λ = Kk ( / h) 8

29 . Numerical Stability has everything to do with the properties of the matrix! It can be shown that, due to the properties of the eigenvalues of this matrix, it s inverse must exist: therefore, can always find values of u at time j based on knowledge of values at time j 1 This solution backward difference or implicit soln is said to be unconditionally stable Interestingly, had we defined: (1+ λ) λ 0 0 u1, j u1, j 1 we λ would (1 have λ) arrived λ at a different matrix that can only be inverted + when k, h, and K satisfy a specific relation. u This, j solution, u, j 1called the λ (1 + λ) λ = forward difference or explicit soln is said to be conditionally stable λ λ um 1, j um 1, j 1 0 λ (1 + λ) where λ = Kk ( / h) 9

30 30. Numerical Stability has everything to do with the properties of the matrix! It can be shown that, due to the properties of the eigenvalues of this matrix, it s inverse must exist: therefore, can always find values of u at time j based on knowledge of values at time j 1 This solution backward difference or implicit soln is said to be unconditionally stable Interestingly, had we defined: u ux ( i, tj+ 1) ux ( i, tj) k u ( xi, tj) + k we would have arrived at a different matrix that can only be inverted when k, h, and K satisfy a specific relation. This solution, called the forward difference or explicit soln is said to be conditionally stable

31 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Note: material presented in this part of the lecture is based on an excellent presentation by Anne Douglass (NASA GSFC) that was given at the University of Toronto, available on the web at: 31

32 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Physical solution 3

33 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Physical solution Upwind material mixes uniformly in neighboring grid box. Mathematical solution 33

34 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Physical solution Critical aspect: what happens at next time step! Upwind material mixes uniformly in neighboring grid box. Mathematical solution 34

35 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Physical solution Next time step: some of the material may appear here, which disagrees with physical solution (i.e., non-physical!) Upwind material mixes uniformly in neighboring grid box. Mathematical solution 35

36 3. Numerical Diffusion Consider the continuity equation: Chemical = v Chemical + Production Loss Let s examine the advective term & consider transport of a square wave to the right with velocity v: v t = 0 Move the material 0.1 grid box to the right: Upwind material mixes uniformly in neighboring grid box. NUMERICAL DIFFUSION Physical solution Next time step: some of the material will appear here, which disagrees with physical solution (i.e., is non-physical) Mathematical solution 36

37 37 Prather, M. J., Numerical advection by conservation of second-order moments, JGR, 91, , 1986 developed an algorithm that almost exactly accounts for the advective transport of a constituent in grid box of appropriate size of 3D climate model Advantages: Non-diffusive, mass-conserving, numerically stable slopes slopes first order moments (overshoot and undershoot)! 1 start with a step function upwind (very diffusive) second order moments with limiter (no undershoot) f 0 x second order moments (small overshoot and undershoot) Disadvantage: Computational requirements (need nine 3D arrays for each constituent) Take home message: if conducting 3D calculations of the distribution of a constituent that involves the solution of a PDE, will often have to consider (and account for) effects of numerical diffusion (there are numerous schemes to consider)

38 Prather, M. J., Numerical advection by conservation of second-order moments, JGR, 91, , 1986 developed an algorithm that almost exactly accounts for the advective transport of a constituent in grid box of appropriate size of 3D climate model Advantages: Non-diffusive, mass-conserving, numerically stable slopes slopes first order moments (overshoot and undershoot)! 1 start with a step function upwind (very diffusive) second order moments with limiter (no undershoot) f 0 x second order moments (small overshoot and undershoot) Disadvantage: Computational requirements (need nine 3D arrays for each constituent) Next Mon: review last two assignments, discuss project logistics, & quiz for those who have opted out of HW#1 Rest of Class: focus on projects 38