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Atmospheric Dynamics 2

by: Ida Auer

Atmospheric Dynamics 2 MET 4306

Ida Auer
Florida Tech
GPA 3.7

Steven Lazarus

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Steven Lazarus
Class Notes
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This 5 page Class Notes was uploaded by Ida Auer on Monday October 12, 2015. The Class Notes belongs to MET 4306 at Florida Institute of Technology taught by Steven Lazarus in Fall. Since its upload, it has received 72 views. For similar materials see /class/221688/met-4306-florida-institute-of-technology in Meteorology at Florida Institute of Technology.

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Date Created: 10/12/15
Geostrophic vorticity Holton discusses the concept of the invertibility of the above relationship What does the great and powerful Holton mean by this Rewriting the above relationship 2 2 Vqf0 EF gt V 7F0 P01ssonEqn g Laplace Eqn F0 The above equation is an elliptic equation recall from your Engineering Math class that there are three basic types of partial differential equations depending on the value of their coefficients the coefficients are notdo not necessarily constant but if they are it simplifies things greatly as it keeps the nature of the equation of one type instead of dependent on varying values of the coef ficients For a second order quasilinear PDE in two independent variables X y we have where if AC 732 gt 0 elliptic if AC 732 0 parabolic if AC 732 lt 0 hyperbolic Thus from above 2 2 Aa DCa D F 3x2 By2 l l SoAC1B0ACB2101gt0Elliptic What is the advantage of distinguishing these types of equations Solutions of each type vary For the class of Elliptic equations these are generally boundary value problems as opposed to initial value problems and can be solved using various numerical techniques One technique that is comonly used to solve an Elliptic equation is called the Relaxation Method see Haltiner and Williams Numerical Methods pg 158 Note that Elliptic boundary value problems have 3 principle types of boundary conditions 1 Dirichlet BC 1 is known everywhere on a boundary 2 Neuman BC EMan is known everywhere on a boundary 3 Mixed BC both ofthe above An practical example of the rst condition I observed temperatures t 0 grid poin I I use EmpleEng mt n and Laplace eqn to get faux temps inside of mtn sea level use underground temps to get est of sea level pressure in mtns We are going to look at the solution of the Poisson Equation with Dirichlet boundary conditions as a followup to Holton s comments on pg 154 Eq 615 Let s look at the simpler problem Laplace s equation ie F 0 zero forcing In this case the geopotential eld is completely speci ed by the boundary conditions For example we have Let s make the following substitution so that we only deal with deviations from the mean geopo tential 1306 y c132 D OCJ Thus our equation becomes 6 BZCD 7 2 2 i 0 Bx By Now let s suggest a function I that is a solution ie satis es the above equation BCID i By 75 D xy Ky where KEconstant This equation satis es the Laplace equation above correct The linear relation given above yield the following perturbation geopotential height eld y4 020 y3 qgt 15 y2 310 So we know what the true geopotential height eld looks like so now let s take a step back and assume that we know only values on the boundary of our domain Thus for a regularly spaced Cartesian grid drawn above we know the value of 13 at pts AH 325 e 7 1 2 N 1720g dy 20 3 4 131Jh err15 a b 310 Now we want to estimate the interior grid points from the exterior or BC s ie a boundary value problem We can derive using Taylor Series expansion the discretized form of the Laplace equa tion which is just the 4 surrounding points minus 4 times the center point see notes from class ie c13i717jlt13i7j d2 Where we have assumed that the grid stencil is uniform in X and y ie AX Ay d and we have dropped the primes but the geopotential above remains the perturbation geopotential 2 I Vq 141 1 D 71 4 1313139 F We can use the discretized apX version of the Poisson equation or if F0 the Laplace equation to solve for the interior grid points using only the boundary conditions given above We do this by rearranging the above equation and making a rst estimate the superscript m m m m I I w m m m 2 H17 cIgti7J1cDi7j7174ltDiyde R 1717 What is Rij Because we do NOT know the interior values of I that is what we are trying to get we made a first guess an educated hopefully Therefore the equation above will no longer equal zero there will be some residual or leftover After the first guess iteration n1 for the entire grid we make a second guess iteration for DU ieDiJ ml over the entire grid m m m m m l 131171 13471 1317111 1313171 74c1gti7j 7Fd2 0 where the second guess CIDLJ ml is exactly what is needed to force the residual to be tempo rarily zero this is what we want zero residual because the actual solution should have no RiJ If we subract the new equation above from the first guess equation we have the value of the new guess that we need to make the residual zero at the point in question ie R m 41 1317 Did 4 You can plug this relationship into above and see that we have our original Laplace equation V I 7F 0 So now we are ready to solve for our interior grid points using the above two equations we can march through the grid above for points l4 note for our purposes here F 0 Iteration 1 first guess new guess pointl 20002574gtlt0R391quot 45 q5 Iquot104541125 point2 02025074gtlt0R392quot45 c153 1 o454 1125 point3 15001074gtlt0R393quot25 13393quot10254625 ml point4 01501074gtlt0R1quot25 CD4 0254625 qgt 25 e 7 1 2 N7 D 7 20 g 1125 1125 d quot 20 7 3 4 Ixi lt1 715 h 625 625 y 715 a b But we know that these are not the correct values at these points Why is this Well everytime we set a residual at one point to zero it effects the residual at surrounding points ie they re only zero when we apply the grid stencil at that point and as soon as we move on to the next point and do the same thing there it changes the residual at the previous point which we had just set to zero So we step through the process again with the new grid estimates ie Second iteration pointl 2011256252574gtlt1125 12391quot 175 cin39lquot2 11251754 15625 point2 1125202562574gtlt1125 12392quot1 175 cin39zquot2 11251754 15625 point3 1562511251074gtlt 625 Rgquot1 175 clngquot2 6251754 10625 point4 6251511251074gtlt 625R2quot1 175 of 6251754 10625 Hey now we are closer So what about the 3rd iterationHave students nish this one off This is not the fastest method of solution but it is relatively straightforward You can expedite the convergence by applying sequential relaxation ie using the new values as they become available overrelax multiply the new guess by some factor ie overcorrect Also a better rstguess eld would help too Thus we now see what H olton means by invertibility One can back out the geopotential height eld iftl39Jor its normal derivative is known on the perimeter of the domain


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