The dynamics of a simple troposphere-stratosphere model

The dynamics of a simple troposphere-stratosphere model by Jessica Duin under supervision of Prof. Dr. A. Doelman, UvA Dr. W.T.M. Verkley, KNMI August 31, 25 Universiteit van Amsterdam Korteweg-de vries Instituut voor Wiskunde Plantage Muidergracht 24 118 TV Amsterdam

Contents Preface 5 Introduction 7 1 Primitive equations 9 2 The equations for an isentropic layer 11 3 The equations for an isothermal layer 15 4 The lower atmosphere 19 5 Linearisation 23 5.1 First case: f =..................................... 23 5.2 Second case: f.................................... 28 6 Initial value problem 35 6.1 One layer model..................................... 35 6.2 Two layer model..................................... 41 7 Circularly symmetric initial values 55 7.1 One layer model..................................... 55 7.2 Two layer model..................................... 61 Conclusions 75 A Derivation of the primitive equations 77 Bibliography 81 3

Preface This thesis is the result of a six month internship at the KNMI, the Royal Dutch Meteorological Institute, where I worked at the Climate Variability department. I wish to thank my supervisors Arjen Doelman from the University of Amsterdam for his suggestions from a mathematical point of view, and Wim Verkley from the KNMI for making this internship possible and for guiding me from a meteorological point of view. 5

Introduction The atmosphere can be divided into several layers, distinguished by different temperature profiles. The first layer is the troposphere, from the earth s surface to a height of 8 km at the poles, and 15 km at the equator. This layer is characterised by a linear decrease of temperature with height 6.5 K km 1. All the weather takes place in this layer. The second layer is the stratosphere, starting from the top of the troposphere, called the tropopause, to a height of 5 km. In the lower stratosphere the temperature is constant and then it increases with height. Above the stratosphere, we find the mesosphere, which reaches to a height of 85 km, and is characterised by a decrease of temperature. And on top of this layer,we find the thermosphere, with an increase of temperature with height. This structure of the atmosphere can also be seen in the figure below. Figure 1: The atmosphere divided into several layers, distinguished by different temperature profiles.holton [3] In this thesis I discuss a simple model of the lower atmosphere, containing the troposphere and the stratosphere. The question is whether this simple model has realistic dynamical properties. To find an answer to this question, we first consider the physical conservation laws that describe all the motions in the atmosphere. These so-called primitive equations are discussed in chapter 1. In chapter 2, the equations of motion are derived for the troposphere. For the troposphere we assume that the potential temperature is constant. The potential temperature is the temperature that a 7

parcel of dry air at pressure p and temperature T would have if it were expanded or compressed to a reference pressure p r, without energy transport to or from the parcel. This assumption implies the linear decrease of temperature with height and it simplifies the conservation laws of chapter 1. We see in figure 1 that the temperature is constant in the lower part of the stratosphere, and that the temperature increases with height, higher in the stratosphere. For the stratosphere we make the assumption that the temperature is constant within the layer, since this will be a good approximation. This also gives a set of simplified equations of motion. These equations are derived in chapter 3. In the next chapter, chapter 4, the two layers are added together into a simple model of the atmosphere. The two layers are coupled by the pressure at the tropopause. All the motions in the two layers are governed by the equations derived in chapters 2 and 3. We make a few assumptions for some of the variables at the boundaries and look at vertical profiles of temperature, pressure, density and potential temperature. One way to look at this model is to linearise the equations of motion and only look at small perturbation in the variables. Here we look at plane wave solutions with different frequencies. This is done in chapter 5. We consider two cases, the first case is for f, the Coriolis parameter, equal to zero. In the second case we take f constant and unequal to zero. For the second case, we can look if dynamics are realistic. In chapter 6 we discuss another way to look at the dynamics of the model. Here we look at an initial value problem. We look only at waves travelling in the zonal direction, first for the troposphere only, then for both layers. In chapter 7 we also consider waves travelling in the meridional direction. Again we look at initial value problems first for the troposphere only, and then for both layers. 8

Chapter 1 Primitive equations The motions in the atmosphere are described by equations that express the conservation of thermodynamic energy, mass and momentum. The thermodynamic state of the atmosphere at any point is determined by the values of pressure p, temperature T and density ρ at that point. For dry air we can express the equation of state as : ρ = p RT, 1.1 with R = 287.5 JK 1 kg 1, the gas constant of dry air. In the absence of atmospheric motions, the gravity force must be exactly balanced by the vertical component of the pressure gradient force. Thus, p = ρg, 1.2 z where g is the acceleration due to gravity. This equation is better known as the hydrostatic approximation, and remains valid if the atmospheric motions have horizontal scales that are large compared to the depth of the atmosphere. Now we can discuss the basic conservation laws for motions in the atmosphere. The derivation of the conservations laws can be found in Appendix A, and are based on Holton [3]. The first conservation law is Newton s second law of motion, the conservation of momentum. This equations states that the rate of change of momentum i.e., the acceleration is equal to the sum of all forces acting. The horizontal momentum equation for motions relative to the earth is given by: D z v Dt w v z = fk v 1 ρ zp F. 1.3 The first term on the left is the acceleration due to horizontal advection, where v is the horizontal velocity and is given by the vector u, v, T where u and v are the horizontal velocities. The second term is the acceleration due to vertical advection, where w is the vertical velocity. The first force ont the right-hand side is an apparent force, the Coriolis force, and is a result of the rotation of the earth. The Coriolis force deflects all motions in the Northern Hemisphere to the right and in the Southern Hemisphere to the left. Here f is the Coriolis parameter, and k is the unit vector perpendicular to the earth s surface. The second force is the pressure-gradient force and the last force is friction. The vertical momentum equation is reduced to the equation of hydrostatic equilibrium 1.2, see also Appendix A. The second conservation law is the conservation of thermodynamic energy. This law states that in order to change the temperature, we have to add or subtract heat. The equation for this conservation law is: D z θ Dt w θ z = θ Q, 1.4 c p T 9

1 Primitive equations in which Q represents the rate of heating/cooling, c p is the specific heat of dry air at constant pressure and θ, the potential temperature, is given by: κ pr θ = T. 1.5 p The potential temperature is the temperature that a parcel of dry air at pressure p and temperature T would have if it were expanded or compressed adiabatically to a reference pressure p r. Adiabatically means that there is no energy transport to or from the parcel. In the definition of potential temperature, κ = R/c p, with c p the specific heat of dry air at constant pressure, c p = 15 JK 1 kg 1 and p r is a reference pressure of 1 hpa. The third conservation law is the continuity equation: D z ρ Dt w ρ z = ρ z v ρ w z where ρ is the density and w is the vertical velocity. It states that the mass of the parcel of air is conserved during the motion. 1.6

Chapter 2 The equations for an isentropic layer In this chapter we derive the equations of motion for the troposphere. We assume that the potential temperature θ is constant, since this implies a linear decrease of temperature with height. A layer with constant potential temperature is called an isentropic layer. The derivation of the equations in this chapter is based on the paper of Verkley [5]. First we obtain the temperature T, pressure p and density ρ as functions of height z. Consider the definition of potential temperature, 1.5. In an isentropic layer, we can write the temperature as a function of pressure: κ T p θ =. 2.1 p r If we differentiate this equation with respect to height, z, we get κ T p 1 p = κ z θ p z. 2.2 The term on the right-hand side of 2.2 is equal to 1/H, with H = c p θ/g constant, since 1 p p z = 1 κ pr κh p from the ideal gas law 1.1 and hydrostatic approximation 1.2. So p r T z θ = 1 H, 2.3 and we see that the temperature decreases linearly with height in an isentropic layer, which agrees well with observations. In fact, T z = g c p, which we better know as the dry adiabatic lapse rate. The solution of this differential equation 2.3 can be written alternatively as: T θ T θ = T l θ z l z H, 2.4a = T u θ z u z H, 2.4b where T l and T u are the temperatures at the lower and upper boundaries, respectively, and z l and z u are the heights of the lower and upper boundaries, respectively. From now on we denote 11

12 The equations for an isentropic layer values at upper and lower boundary with subscript u and l, respectively. Also from the definition of potential temperature, 1.5, we find for the pressure: If we define the density ρ r of air at the reference pressure p r as we can write for the density 1/κ p T =. 2.5 p r θ ρ r = p r Rθ ρ = p θ ρ r p r T,, using the ideal gas law 1.1. We can now use the definition of the potential temperature, 1.5, to write 1/κ 1 ρ T =. 2.6 ρ r θ If we substitute one of the solutions for the temperature, 2.4, in equations 2.5 and 2.6 we can write the pressure and the density as functions of height. When we know the temperature at one of the boundaries, the equations 1.1, 1.2 and 1.5 are satisfied in an isentropic layer. Now we can derive the conservation laws for a isentropic layer. First consider the equation of conservation of horizontal momentum 1.3. We assume that there is no friction, so F =. For the pressure-gradient term we find: 1 ρ zp = p r ρ r 1 κ z with M the Montgomery potential defined as T = z c p T = z M, θ M gz c p T. Both terms in the Montgomery potential are linear functions of height. From the dry adiabatic lapse rate we get that M z =. Thus in an isentropic layer M = M u = M l. Since the pressure-gradient term and the Coriolis force term are independent of height, the horizontal velocity must be independent of height if this is the case initially, thus v z =. The horizontal momentum equation 1.3 now becomes D z v Dt = fk v z M, since we also assume that the friction term F is zero. Now consider the continuity equation 1.6. We can rewrite this equation in the following form w z 1 ρ ρ z w = z v 1 D z ρ ρ Dt. 2.7

13 For the second terms on the left- and right-hand sides we can find the following expressions: 1 ρ θ 1 κ T =, ρ z T κ z θ 1 D z ρ θ 1 κ Dz T =, ρ Dt T κ Dt θ where D z T = D z zl Dt θ Dt H T l. θ A solution of the homogeneous part of equation 2.7 is 1 1/κ w T H = A, θ where A only depends on the horizontal coordinates. A particular solution of 2.7 is w T H = κ z v D z zl θ Dt H T l. θ So the general solution of 2.7 is 1 1/κ w T T H = A κ z v D z zl θ θ Dt H T l. 2.8 θ If we require that the vertical velocity at the lower boundary, w l, is equal to the material vertical velocity of the lower boundary: w l H = D z zl, 2.9 Dt H we find the following expression for A: A = 1/κ 1 [ Tl Dz θ Dt Tl κ z v θ ] Tl, θ which can be simplified as [ Dz A = κ Dt pl p r z v pl p r ]. Now use that from 2.4a and 2.1 it follows that z z l H = pl so that we can write for the vertical velocity: w H = D z zl zl z κ z v Dt H H κ 1 { p Dz pl κ Dt p r p r p r κ κ p, p r D z Dt z v Tl pl p r κ z v θ }. When we write the sum of the third and fourth terms on the right-hand side as: D z Dt Tl κ z v θ Tl = κ θ pl p r κ 1 { Dz Dt pl p r z v Tl θ pl p r },

14 The equations for an isentropic layer the equation for the vertical velocity becomes: w H = D z zl zl z κ z v Dt H H { κ 1 } κ 1 {Dz pl p κ p r p r Dt pl p r z v pl p r }. 2.1 We also require that the vertical velocity at the upper boundary w u is equal to D z z u /Dt, so w u H = D z Dt So the vertical velocity at the upper boundary equals: D z zu = D z zl zl z u κ z v Dt H Dt H H { κ 1 } κ 1 {Dz pl pu κ Dt p r p r zu. 2.11 H pl p r z v In the same way as above we get the following expression for the vertical velocity: w H = D z zu zu z κ z v Dt H H κ { pu p r κ 1 p p r κ 1 } {Dz Dt pu p r z v and with 2.9 we find for the vertical velocity at the lower boundary: D z zl = D z zu zu z l κ z v Dt H Dt H H { pu κ 1 } κ 1 {Dz pl pu κ z v Dt p r p r p r pu p r pl p r }. 2.12 }. 2.13 pu p r }. 2.14 If we substitute the expression for the vertical velocity at the lower boundary 2.14 into 2.12 we find that D z Dt p l p u z vp l p u =. This equation states that the mass in conserved in an isentropic layer following the horizontal motion of the column. The third conservation law, is the conservation of thermodynamic energy: D z θ Dt w θ z = θ Q. 2.15 c p T We assumed that the potential temperature is constant, so the left-hand side of 2.15 is equal to zero, which implies that Q is equal to zero. This means that we do not have to add or subtract heat, to keep the potential temperature constant.

Chapter 3 The equations for an isothermal layer In figure?? we see that the temperature is constant in the lower part of the stratosphere, and that the temperature increases with height in the upper part of the stratosphere. To simplify the equations of motion we make the assumption that the temperature is constant in the stratosphere. Such a layer is called an isothermal layer. First we obtain the pressure p, density ρ and potential temperature θ as functions of height. From the ideal gas law 1.1 and hydrostatic approximation 1.2 we find p z = ρg = g p. 3.1 RT At the lower boundary, the pressure is equal to p l, so the solution of 3.1 is equal to p = p l exp z l zg. 3.2 RT We see that the pressure decreases exponentially with height. We can derive a similar equation for the rate of change of density with height: so we can write the density as a function of height as ρ z = 1 p RT z = g ρ, 3.3 RT ρ = ρ l exp z l zg, 3.4 RT with ρ l = p l /RT, the density at the lower boundary. The density also decreases exponentially with height. For the rate of change of potential temperature we find the following equation κ θ z = T κ pr 1 p p p z = g θ, 3.5 c p T which has a solution with θ = θ l exp z z lg, 3.6 c p T θ l = T pr The potential temperature increases exponentially with height. The pressure, density and potential temperature are all exponential functions of height, and satisfy the ideal gas law, hydrostatic approximation and the definition of potential temperature in an isothermal layer. 15 p l κ.

16 The equations for an isothermal layer Now we can derive the conservation laws for an isothermal layer. Look at the equation of the conservation of the horizontal momentum, 1.3. We assume that there is no friction, so F =. For the pressure-gradient term we find: 1 ρ zp = 1 exp z z lg z p l exp z l zg ρ l RT RT = 1 g z p l p l ρ l RT zz l = p l ρ l zp l p l g RT zz l = RT z ln p l g RT zz l = RT z ln p l z gz l = z RT ln p l gz l = z M where we write RT ln p l gz l = M. We see that the pressure-gradient term is independent of the height z. Since the Coriolis parameter f is also independent of height we find from the conservation of horizontal momentum 1.3 that the horizontal velocity is also independent of z if this is the case initially, thus v z =. For the equation of conservation of horizontal momentum we now get D z v Dt = fk v z M = fk v z RT ln p l gz l. Let us now look at the continuity equation 1.6. We recall that it can be written as equation 2.7 w 1 z ρ w = z v 1 D z ρ ρ z ρ Dt. 3.7 The homogeneous solution of this first-order linear non-homogeneous differential equation is: since w = A exp z z lg, RT 1 ρ ρ z = g RT and A only depends on the horizontal coordinates. With the ideal gas law, the second term on the right-hand side of equation 3.7 becomes 1 D z ρ ρ Dt = RT D z p p Dt RT = 1 D z p p Dt = 1 p l D z p l Dt = D z Dt ln p l g D z z l RT Dt g D z z l RT Dt

17 and is independent of the height z. Since v is also independent of z, the right-hand side of 3.7 is independent of z, so there is a solution w of 3.7 which is independent of z, w/ z =. A particular solution is w = RT g A general solution of 3.7 is therefore w = A exp z z lg RT Dz Dt ln p l z v D zz l Dt. RT g Dz Dt ln p l z v D zz l Dt. If we require that the vertical velocity at the lower boundary is equal to the material vertical velocity, i.e. we find for A: Since A = RT g we can write for the vertical velocity w = D zz l Dt RT g w l = D zz l Dt Dz Dt ln p l z v. exp z z lg RT = p l p 3.8 Dz Dt ln p l z v 1 p l. 3.9 p We also require that the vertical velocity at the upper boundary is equal to the material vertical velocity, so If we substitute 3.1 into 3.9 we get D z z u Dt = D zz l Dt RT g In the same way we can also get the following solution of 3.7 w = D zz u RT Dz Dt g Dt ln p u z v w u = D zz u Dt. 3.1 Dz Dt ln p l z v 1 p l. 3.11 p u 1 p u p, 3.12 which, requiring that the vertical velocity at the lower boundary satisfies 3.8, gives D z z l = D zz u RT Dz Dt Dt g Dt ln p u z v 1 p u. 3.13 p l Now substitute the equation for D z z u /Dt, 3.11, in 3.13 and we get D z Dt p l p u z vp l p u = which implies that mass is conserved in an isothermal layer following the horizontal motion of the column. This result is the same as for an isentropic layer, which we derived in the previous chapter. The equation of thermodynamic energy is D z θ Dt w θ z = θ Q. 3.14 c p T

18 The equations for an isothermal layer In an isothermal layer we cannot assume that Q =. We have to add or subtract energy to the system in order to hold the temperature constant. If we rewrite the first term of this equation we find: D z θ Dt = D κ z Dt T pr exp z z lg = gθ D z z l κθ D z p l c p T c p T Dt p l Dt and for the second term we can write p l w θ z = w gθ c p T. So the equation for conservation of thermodynamic energy becomes: gθ D z z l κθ D z p l w gθ c p T Dt p l Dt c p T = θ c p T Q. From this equation and the equation for the vertical velocity, 3.9, we can derive the following expression for Q: Q = g D z Dt z l RT D z p l 3.9 = g D z Dt z l RT = RT [ 1 p p l D z p l Dt Dt p l wg D z Dt p l g D z Dt z l RT p l z v p l p z v RT z v RT p ] l. p l D z p Dt p l p l p RT z v To hold the temperature constant in an isothermal layer, we have to add or subtract energy, and the rate of cooling/heating Q is equal to [ RT 1 D z p l z v p ] l p Dt p z v. That concludes our discussion of the conservation laws for the stratosphere.

Chapter 4 The lower atmosphere In the previous chapters we derived equations that describe all the motions in the troposphere and stratosphere. Together they form a simple model of the lower atmosphere. The two layers are coupled by the pressure at the tropopause. See figure 4.1. The equations for conservation of horizontal momentum and mass for the troposphere are: D z Dt v1 = fk v 1 z M 1, D z Dt p 2 p 1 z v 1 p 2 p 1 =, where the superscript 1 indicates that they represent the values in the troposphere. The pressures p 2 and p 1 are the pressure at the surface and the tropopause, respectively. For z 2 we can take z B, the earth s orography. M 1 is the Montgomery potential in the troposphere and is equal to κ M 1 = c p θ 1 p2 gz 2, p r the Montgomery potential evaluated at the surface, since M 1 is independent of height. Montgomery potential is also equal to This κ M 1 = c p θ 1 p1 gz 1, p r T Z_,P_ Z_1,P_1 θ 1 Z_2,P_2 Figure 4.1: A layer with constant temperature and a layer with constant potential temperature, with the pressure and height of the boundaries and interfaces denoted by z i, and p i, i =, 1, 2 19

2 The lower atmosphere the Montgomery potential evaluated at the tropopause. Since these two expressions for the Montgomery potential are equal, we can derive a relation between the heights z 1 and z 2 : κ κ gz 1 = gz 2 c p θ 1 p2 p1. 4.1 p r p r For the isothermal layer we have the following equations: D z D z Dt v = fk v z M, Dt p 1 p z v p 1 p =, M = RT p1 ln gz 1, with p the pressure at the upper boundary and M and v are the Montgomery potential and horizontal velocity in the stratosphere. In figure 1 we can see that at the top of the stratosphere, the pressure is almost zero, so we will take p equal to zero. The temperature in the stratosphere is denoted by T. If we use the expression for z 1, 4.1, in the equation for the Montgomery potential M we can write κ M = c p θ 1 p2 gz 2 RT ln p r p r p1 p r κ c p θ 1 p1. p r The equations that describe all the motion in the lower atmosphere are: where D z Dt v1 = fk v 1 z M 1, D z Dt p 2 p 1 z v 1 p 2 p 1 =, D z Dt p 1 p z v p 1 p =, κ M 1 = c p θ 1 p2 gz 2 p r κ M = c p θ 1 p2 gz 2 RT ln p r D z Dt v = fk v z M, p1 p r κ c p θ 1 p1. p r We now have four equations that describe all the motions in the lower atmosphere, with four unknowns, v, v 1, p 1, p 2, and boundary conditions z 2 = z B and p =, so we have a complete dynamical system. In the previous chapters, we have also derived expressions for the pressure, density, potential temperature and temperature as functions of height so we can make a vertical profile of the lower atmosphere using these equations. To calculate the average height of the interface we can use 4.1, and assume that p 1 = 2 hpa, p 2 = 1 hpa and θ 1 = 3 K. The bars mean average values. We then find an average height of z 1 = 11.3 km. Before we make vertical profiles we assume that there is a temperature jump of 5 Kelvin at the interface. This makes the temperature in the stratosphere, T, equal to 194 K. The temperature jump follows from an article of Birner et al. [2], where the temperature at the tropopause in discussed. In figure 4.2 we see the vertical profiles of the pressure, temperature, density and potential temperature. We see that the pressure is continuous, while the density and potential temperature are discontinuous at the interface.

21 4 4 35 35 3 3 25 25 Height km 2 Height km 2 15 15 1 1 5 5 1 2 3 4 5 6 7 8 9 1 18 2 22 24 26 28 3 Pressure hpa Temperature K a b 4 4 35 35 3 3 25 25 Height km 2 Height km 2 15 15 1 1 5 5.2.4.6.8.1.12 2 4 6 8 1 12 14 Density kg m-3 Potential Temperature K c d Figure 4.2: The vertical profile of a the pressure, b the temperature, c the density and d the potential temperature. The green part represents the troposphere and the red part, the stratosphere.

Chapter 5 Linearisation We now have a two layer model represented by four equations. We wish to know if this model has realistic dynamical properties. In this chapter we look at the dynamics of this model by linearising the governing equations and by considering small perturbations from a mean state, for which we take the state of rest. First we assume that f, the Coriolis parameter, is equal to zero, which simplifies our model. In the second paragraph we take f constant and unequal to zero. 5.1 First case: f = We assume that f =. Define η = p/p r. We write the variables M, η and v as a sum of a constant part and a small perturbation: M = M M, η = η η, v = v v, where the bar denotes a constant part, and the prime denotes a perturbation. The horizontal velocity v has constant part equal to zero and consists only of a small perturbation. We therefore write v = v. First we linearise the equations for the troposphere. If we write η 2 = η 2 η 2 and assume that z 2 is small, z 2 = z 2, the Montgomery potential in the troposphere can be written as: with M 1 = c p θ 1 η2 κ gz 2 = c p θ 1 η 2 κ κ 1 Rθ 1 η 2 η 2 gz 2 = M 1 M 1, M 1 = c p θ 1 η κ 2, and M 1 κ 1 = Rθ 1 η 2 η 2 gz 2. Now look at the horizontal momentum equation. We can split the horizontal momentum equation into two parts, a zonal and meridional part: and t u1 u 1 x u1 v 1 = y u1 x M 1, t v1 u 1 x v1 v 1 = y v1 y M 1. If we take, for simplicity, z 2 =, the linearised Montgomery potential is equal to Aη 2 with κ 1 A = Rθ1 η 2. When we linearise the horizontal momentum equations, we get t u1 = x M 1 23 5.1

24 Linearisation and t v1 = y M 1. 5.2 If we take the divergence, by taking the / x of 5.1 and the / y of 5.2 and adding them, we find: u 1 t x v1 = = 2 Aη y t 2, D1 where we write D 1 = u1 x v1 y, where D 1 is the divergence. If we take the rotation, i.e. take the / y of 5.1 and add this to the / x of 5.2, we get: t ζ1 =, where ζ 1 = v1 x u1 y is the vorticity which we can take equal to zero, since the vorticity is constant in time. For the equation of conservation of mass we get t η 2 η 1 D 1 η 2 η 1 =. For the troposphere we thus have the following set of linearised equations: t D1 = 2 Aη 2, t η 2 η 1 D 1 η 2 η 1 =. 5.3 We also have to linearise the governing equations for the stratosphere. We assume that the pressure at the upper boundary is constant and equal to zero, so take η = η =. When we linearise the Montgomery potential for an isothermal layer we find: M = RT η 1 κ 1 Rθ 1 η 2 η κ 1 2 Rθ 1 η 1 η 1 η 1 = Aη 2 Bη 1. κ 1 with A = Rθ1 η 2 and B = RT κ 1 / η 1 Rθ1 η 1. When we now linearise the zonal and meridional part of the horizontal momentum equation and take the divergence and rotation, we get for the isothermal layer: t D = 2 Aη 2 Bη 1, t η 1 D η 1 =. 5.4 In this layer, the vorticity is also equal to zero. These four equations 5.3 and 5.4 have four unknowns D, D 1, η 1 and η 2.

First case: f = 25 Suppose equations 5.3 and 5.4 have solutions D 1 = D 1 exp ikx ly ωt D = D exp ikx ly ωt η 1 = E 1 exp ikx ly ωt η 2 = E 2 exp ikx ly ωt, where k and l are the zonal and meridional wavenumbers, respectively, and ω is the angular frequency. These solutions can be seen as Fourier modes. We can substitute these solutions in the linearised equations, 5.3 and 5.4. We see they are indeed solutions if iωd k 2 l 2 BE 1 k 2 l 2 AE 2 = iωd 1 k 2 l 2 AE 2 = η 1 D iωe 1 = η 2 η 1 D 1 iωe 1 iωe 2 =. We can add the third equation to the last and calculate the determinant of the matrix. For the determinant we find iω k 2 l 2 B k 2 l 2 A iω k 2 l 2 A η 1 iω = ω 4 k 2 l 2 B η 1 A η 2 ω 2 k 2 l 2 2 AB η 1 η 2 η 1. η 1 η 2 η 1 iω For the values of ω for which the determinant is zero, the equations are dependent. So if we know one of the amplitudes D, D 1, E 1 or E 2 we know them all. If the determinant is unequal to zero, we will find that the solutions are zero, and we do not want that. The determinant is zero if Introduce ω = ± ω 1 = ω 2 = k 2 l 2 2 k 2 l 2 2 k 2 l 2 2 k ω 3 = 2 l 2 2 k ω 4 = 2 l 2 2 A η 2 B η 1 ± A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2. A η 2 B η 1 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2 A η 2 B η 1 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2 A η 2 B η 1 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2 A η 2 B η 1 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2. If ω is positive negative, the waves travel in the opposite direction of the wavevector k, l. Suppose we know the amplitude of the perturbation in the pressure at the surface, E 2, then we can write the other amplitudes in terms of E 2. We can get the following expressions for the amplitudes D = iω η 1 i η 2 η 1 k 2 l 2 A ω η 1 ω 2 E 2 D 1 = ik2 l 2 A E 2 ω ω 2 η 2 η 1 k 2 l 2 A E 1 = E 2.

26 Linearisation η 1 T T ω 1 ω 2.1 155.4..26756 Nan.1 16.4 5..26977.3199.1 165.4 1..26124.4587.2 189.4..26756 Nan.2 194.4 5..261196.359.2 199.4 1..261647.42436.3 212.7..26756 Nan.3 217.7 5..261415.2894.3 222.7 1..26287.39629 Table 5.1: The values for ω 1 and ω 2 for different values of η 1 and T. For T = ω 2 is undefined. If we choose η 1 = 2/1 =.2 and η 2 = 1 and T = 194 K, θ 1 = 3 K we find for A and B, A = 86115 m 2 s 2 and B = 6539.46 m 2 s 2. The typical wavelength of a gravity wave is λ = 1 5 m, so the corresponding wavenumbers are k = l = 2π/λ = 6.2832 1 5 m 1. This gives: ω = ±.2611 rad s 1 D = ±.26436iE 2 D 1 = ±.2636iE 2 E 1 =.2245E 2. These waves have a period of P = 2π/ω = 24.59 seconds, which is about 4 minutes. For the other angular frequencies we find: ω = ±.36 rad s 1 D =.933348iE 1 D 1 = ±.236924iE 2 E 1 = 65.451E 2 and the period is 34.8 minutes. We see that for large ω, ω = ±.2611 rad s 1, the amplitudes D and D 1 have the same sign and magnitude. For these ω s the amplitudes E 1 and E 2 also have the same sign, but the amplitude of the pressure at the surface E 2 is much larger than the pressure at the interface E 1. This implies that for ω = ±.2611 rad s 1 the waves are barotropic. For the smaller ω, ω =.36 rad s 1, the signs of the amplitudes at the interface and surface are opposite and the magnitudes of the amplitudes at the surface are much smaller than at the interface. These waves are called baroclinic. Here we assumed that the pressure at the tropopause is 2 hpa, and that the temperature makes a jump of 5 K. How do the frequencies change, when we change the pressure at the interface and the temperature jump? In the table 5.1, we find the frequencies for some different values of η 1 and the temperature jump. We let η 1 vary between.1 and.3 and the temperature jump T between and 1 K. For T =, we do not have an expression for ω 2. This is because ω 2 is imaginary. If ω is imaginary and we substitute this into the solution, iω will be real, and we get exponential decay or growth. We only want real values for ω, that is why ω 2 is undefined if T =. We see that when the tropopause is lower, which means that η 1 becomes larger, the temperature at the tropopause increases. In figure 5.1, ω 1 and ω 2 are plotted for different values of η 1 and T. In figure 5.1a and 5.1c we took η 1 constant and only varied the temperature jump. We see that both ω s increase when the temperature jump increases. In 5.1b and 5.1d we only varied the height of the interface, η 1, and took the temperature jump constant. For the barotropic waves, waves with frequency ω 1, we

First case: f = 27 see that the frequency increases very little when η 1 becomes larger, and also increases when T becomes larger. For the baroclinic waves, ω 2, we see that the frequency increases for increasing T, and decreases when η 1 becomes larger..1851.26145.185.2614.26135.1849.2613 Omega rad/s.1848.1847.1846 Omega rad/s.26125.2612.26115.2611.1845.2615.1844.261.1843 2 4 6 8 1.2695.1.12.14.16.18.2.22.24.26.28.3 Temperature jump K eta_1 a b.35.32.3.315.25.31.35 Omega rad/s.2.15 Omega rad/s.3.295.1.29.5.285 2 4 6 8 1.28.1.12.14.16.18.2.22.24.26.28.3 Temperature jump K eta_1 c d Figure 5.1: In a ω 1 for different values of T, η 1 constant, b ω 1 for different values of η 1 T constant and in c ω 2 for different values of T and η 1 constant and in d ω 2 for different values of η 1 T constant

28 Linearisation 5.2 Second case: f We now assume that the Coriolis parameter f is constant and unequal to zero. Here we also assume that the pressure at the upper boundary is constant and equal to zero, so η = η =. The linearised equations for conservation of horizontal momentum and mass, for the troposphere and stratosphere, now read: t u fv = x M t v fu = y M D z Dt η 1 z v η 1 = t u1 fv 1 = x M 1 t v1 fu 1 = y M 1 D z Dt η 2 η 1 z v 1 η 2 η 1 =, with M = Aη 2 Bη 1 and M 1 = Aη 2. Again we can take the divergence and rotation of the horizontal momentum equations to get: t D t D1 fζ = 2 Bη 1 Aη 2 fζ 1 = 2 Aη 2 t η 1 D η 1 = t η 2 η 1 D 1 η 2 η 1 = t ζ fd = t ζ1 fd 1 =. 5.5 We see that for f the vorticity is unequal to zero. Again we can look for plane wave solutions of the form: D = D exp ikx ly ωt D 1 = D 1 exp ikx ly ωt η 1 = E 1 exp ikx ly ωt η 2 = E 2 exp ikx ly ωt ζ = F exp ikx ly ωt ζ 1 = F 1 exp ikx ly ωt. 5.6 We substitute these solutions into the set of linearised equations 5.5. This leads to a homogeneous set of six equations. If we add the third equation to the fourth, we get: iωd k 2 l 2 BE 1 k 2 l 2 AE 2 ff = iωd 1 k 2 l 2 AE 2 ff 1 = η 1 D iωe 1 = η 1 D η 2 η 1 D 1 iωe 2 = fd iωf = fd 1 iωf 1 =.

Second case: f 29 We can calculate the determinant of the coefficient matrix to determine the angular frequencies ω for which the set of equations is dependent. The determinant is zero for and ω = ω = ± 1 k 2 l 2 A η 2 B η 1 2f 2 ± k 2 l 2 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 1 2. 5.7 2 The waves with frequencies ω = are steady waves. These waves are the most interesting ones. The other waves, with frequencies unequal to zero, are gravity waves. If we take f =, we get the same values of ω as in the previous paragraph. First look at the wave with ω. Suppose ω and we know for instance E 2, then the expressions for the other amplitudes are: D = iω η 1 ω 2 f 2 Ak 2 l 2 η 2 η 1 ω 2 f 2 E 2 D 1 = iω Ak2 l 2 ω 2 f 2 E 2 E 1 = ω2 f 2 Ak 2 l 2 η 2 η 1 ω 2 f 2 E 2 F = f η 1 ω 2 f 2 Ak 2 l 2 η 2 η 1 ω 2 f 2 E 2 F 1 = Ak2 l 2 f ω 2 f 2 E 2. For f the amplitudes converge to the amplitudes in section 5.1. We can substitute the values A = 86115 m 2 s 2, B = 6539.46 m 2 s 2, η 1 =.2, η 2 = 1 and f = 1 4 s 1 in the equation for ω, 5.7, and we get ω 1 = 1 8 86379.8k 2 l 2 ω 2 = 1 8 143.11k 2 l 2 ω 3 = 1 8 86379.8k 2 l 2 ω 4 = 1 8 143.11k 2 l 2. A typical wavelength for gravity waves is 1 km. This gives frequencies ω 1,3 = ±.2611 rad s 1 and ω 2,4 = ±.36 rad s 1. The periods are 4 minutes and 34.8 minutes, respectively. The corresponding amplitudes are: D = ±.26436iE 2 D 1 = ±.2636iE 2 E 1 =.22452E 2 F =.11226E 2 F 1 =.996935E 2 for ω = ±.2611 rad s 1. These are the barotropic waves. And D =.933914iE 2 D 1 = ±.23768iE 2 E 1 = 65.451E 2 F =.325225E 2 F 1 =.825563E 2

3 Linearisation η 1 T T ω 1 ω 2.1 155.4..26758.998.1 16.4 5..26978.31925.1 165.4 1..26125.4598.2 189.4..26758.999.2 194.4 5..261198.375.2 199.4 1..261649.42448.3 212.7..26758.1.3 217.7 5..261417.28112.3 222.7 1..26289.39641 Table 5.2: The values for ω 1 and ω 2 for different values of η 1 and T. for ω = ±.36 rad s 1. These waves are called baroclinic. The difference between the barotropic and baroclinic waves can be seen in figure 5.2. For the perturbation in the surface pressure we take a sine wave with amplitude equal to one. In the left column we see the barotropic waves, for ω =.2611 rad s 1, and in the right column we see the baroclinic waves, for ω =.36 rad s 1. We see that the divergence is out of phase with the pressure and vorticity. This can also be seen from the factor i in the relation between the amplitudes. Here we can also look at the influence of the parameters η 1 and T on the ω s. In table 5.2 we see the values for the ω s for different values of η 1 and T. We see that ω 1 is increasing if η 1 is increasing and when T is increasing; ω 2 is increasing when T is increasing, and ω 2 is decreasing when η 1 is increasing. This can also be seen in figure 5.3. In 5.3a and 5.3c η 1 is kept constant and in 5.3b and 5.3d, T is kept constant. We now look at the steady waves. First look at the relations between the amplitudes. For ω = we find D = D 1 =, so there is no divergence. For the other amplitudes we have the following relations: k 2 l 2 BE 1 k 2 l 2 AE 2 ff = These two equations induce geostrophy: k 2 l 2 AE 2 ff 1 =. 5.8 fv i = x M i fu i = y M i, 5.9 for i =, 1. If we substitute the plane wave solutions 5.6 in equations 5.9, we get equations 5.8. Geostrophy means that the pressure-gradient term and the Coriolis force are in balance. When we let f go to zero, the right-hand sides of equations 5.9 must go to zero also. This means that the amplitude of the pressure perturbation must go to zero. So if f, the steady wave disappears, and we have the same situation as in the previous paragraph. Now we wish to take a closer look at the dynamics of the steady waves. We wish to know if this steady wave has realistic pressure perturbations for realistic values of the vorticity. A typical value for the vorticity is 1.2 1 4 s 1 and a typical wavelength is λ = 1 km. For these values we find for E 2 the value -.35, which means a perturbation in the surface pressure of -3.5 hpa. Similarly we find for the perturbation in the pressure at the tropopause -46.5 hpa. What does this mean in terms of height? In the troposphere the height is given by z z l H = pl p r κ κ p p r

Second case: f 31 Barotropic Baroclinic 1 1.5.5 E_2 E_2 -.5 -.5-1 1 2 3 4 5 6-1 1 2 3 4 5 6 1 6.5 4 2 E_1 E_1-2 -.5-4 -6-1 1 2 3 4 5 6 1 2 3 4 5 6 1.4.5.2 D^ D^ -.2 -.5 -.4 1 2 3 4 5 6-1 1 2 3 4 5 6 1.4.2.5 D^1 D^1 -.2 -.5 -.4 1 2 3 4 5 6-1 1 2 3 4 5 6.1.4.5.2 F^ F^ -.2 -.5 -.4 -.1 1 2 3 4 5 6 1 2 3 4 5 6.1.4.5.2 F^1 F^1 -.2 -.5 -.4 -.1 1 2 3 4 5 6 1 2 3 4 5 6 Figure 5.2: The waves with different amplitudes, for a sine wave in the surface pressure. The left column is for barotropic waves and the right for baroclinic waves. And from top to bottom we see E 2,E 1,D,D 1,F and F 1.

32 Linearisation.1851.3697.185.3696.1849.3695.1848 Omega rad/s.1847 Omega rad/s.3694.3693.1846.1845.3692.1844.3691.1843 2 4 6 8 1.369.1.12.14.16.18.2.22.24.26.28.3 Temerature jump K eta_1 a b.35.46.3.45.25.44 Omega rad/s.2.15 Omega rad/s.43.42.1.41.5.4 2 4 6 8 1.39.1.12.14.16.18.2.22.24.26.28.3 Temerature jump K eta_1 c d Figure 5.3: In a and b ω 1 for different values of T and η 1, respectively, and in c and d ω 2 for different values of T and η 1.

Second case: f 33 2 2 15 Pressure hpa 4 6 Height km 1 8 5 1-1 -5 5 1-1 -5 5 1 x-axis a-axis km a b Figure 5.4: a The perturbation in the pressure with the pressure as vertical axis and b the perturbation with height as vertical axis. which follows from 2.4a and 2.1. For the earth s surface we have z l = km, p l = 1 hpa, so that the amplitude of the perturbation at the tropopause is equal to 1438.9 m. From 3.2 it follows that in the stratosphere the amplitude of the perturbation in height is given by z l z = RT g For p l we have 2 hpa, so we find an amplitude of 152.13 m. So a perturbation in the pressure at the tropopause of 46.5 hpa, gives a perturbation in height of approximately 1.5 km. If we assume that the perturbations are sines, the perturbations at the surface and interface are shown in figure 5.4. A small perturbation in the surface pressure results in a much larger perturbation in the pressure at the tropopause. These values for the amplitudes of the pressure perturbations are realistic, for this typical value for the vorticity. ln pl p.

Chapter 6 Initial value problem Another way to look at the dynamics of the steady waves is by solving an initial value problem. First we do this for the troposphere only, then for the two layer model of the lower atmosphere. 6.1 One layer model First we reduce our model to one layer, the troposphere. We assume that there are no perturbations in the pressure at the interface, so η 1 = η 1. Our set of linearised equations is now reduced to three equations for divergence, pressure and vorticity in the isentropic layer. The set of linearised equations for the layer is: t D1 fζ 1 = 2 M 1 t η 2 D 1 η 2 η 1 = t ζ1 fd 1 =, where M 1 = Aη 2, κ 1 with A = Rθ1 η 2. To make it more simple, we assume that l =, so that we are only looking at waves propagating in the zonal direction. We can write the solutions as D 1 = D 1 exp ikx ωt η 2 = E 2 exp ikx ωt ζ 1 = F 1 exp ikx ωt and find for the angular frequencies ω 1 = and ω 2,3 = ± Ak 2 η 2 η 1 f 2. For the relations between the amplitudes we find D 1 = and F 1 = Ak2 f E 2 for ω = and D 1 = F 1 = iω η 2 η 1 E 2 f η 2 η 1 E 2 and 35

36 Initial value problem for ω. We can write D 1, η 2 and ζ 1 as a superposition of waves with the different frequencies and we can sum over a finite number of wavenumbers m, so D 1 = A 1 m D 1 1 m A 2 m D 1 2 m A 3 m D 1 3 m η 2 = ζ 1 = m= m= m= A 1 m η 1 2,m A2 m η 2 2,m A3 m η 3 2,m A 1 m ζ 1 1 m A 2 m ζ 1 2 m A 3 m ζ 1 3 m. The superscripts i, i = 1, 2, 3, implies that D 1 i, η i 2 and ζ 1 i are evaluated with frequency ω i. Here we replaced the zonal wavenumber k by k = mπ/l, since we consider the finite domain [ L, L], and assume that the perturbations are periodic with period 2L. We consider a finite domain, since we cannot use an infinite domain for calculations. Suppose that at t = there is a small perturbation in the pressure at the surface and the divergence and vorticity are zero. The initial perturbation in the pressure can be written in the following form: η 2t = = c m expimπx/l. m= The initial conditions can also be written as a superposition of the waves with different frequencies at t =, so: m= A 1 m D 1 1 m t = A 2 m D 1 2 m t = A 3 m D 1 3 m t = m= m= A 1 m η 1 2,m t = A2 m η 2 2,m t = A3 m η 3 2,m t = A 1 m ζ 1 1 m t = A 2 m ζ 1 2 m t = A 3 m ζ 1 3 m t = For every wavenumber m it holds that = A 1 m D 11 m A 2 m D 12 m A 3 m D 13 m = A 1 m E 1 2,m A2 m E 2 2,m A3 m E 3 2,m c m = = =. A 1 m F 11 m A 2 m F 12 m A 3 m F 13 m =. m= c m expimπx/l We can solve this set of equations using the relations between the amplitudes and using that ω 2 = ω 3 and get for the A i m s: and A 1 f 2 L 2 c m m = f 2 L 2 Am 2 π 2 η 2 η 1 A 2 m = A 3 m = c m Am 2 π 2 η 2 η 1 2 f 2 L 2 Am 2 π 2 η 2 η 1. Now we can write D 1, η 2 and ζ 1 as a Fourier series in sines and cosines: D 1 = m=1 m=1 2A 2 m 2A 3 m ω 2 mπx η 2 η 1 sin L ω 3 mπx η 2 η 1 sin L ω 2t ω 3t,

One layer model 37 and η 2 = A ζ 1 = m=1 m=1 m=1 m=1 m=1 m=1 mπx 2A 1 m cos L mπx 2A 2 m cos L ω 2t mπx 2A 3 m cos L ω 3t, 2A 1 Am 2 π 2 mπx m fl 2 cos L 2A 2 m 2A 3 m f mπx η 2 η 1 cos L f mπx η 2 η 1 cos L ω 2t ω 3t, where we sum over a finite number of wavenumbers m, since we cannot take an infinite number, and this gives a good approximation. Taking a finite number of wavenumbers will result in wiggles in the figures. These wiggles will be smaller, if M is larger. We see that the solutions consist of waves with different frequencies, and, since ω 2 = ω 3 two waves are travelling in opposite direction. For η 2 and ζ 1 we also see the steady wave with frequency ω 1 =. Take as initial perturbation in the surface pressure η 2t = = { b cosπx/2a a x a otherwise, a cosine with an amplitude of b over an interval of 2a km. This gives c m = b sinaπ1/2a m/l sinaπ1/2a m/l 2L π1/2a m/l π1/a m/l and A = 2ab πl. In figures 6.1, 6.2 and 6.3, η 2, D 1 and ζ 1 are shown, respectively, for different values of t. Here we used L = 4 km, b =.1, a = 1 km and M = 1. The time is given in 1 seconds, so t = 1 means t = 1 s. In figure 6.1 we see the cosine at t =, which we took as initial function. We see that this perturbation splits into two waves travelling in opposite direction, when t becomes larger. The divergence for t = is zero, but when t >, there appears a perturbation around x = and splits into two waves travelling in opposite direction. For the vorticity, see figure 6.3, we see also two waves travelling in opposite direction, and in the last two figures we see the steady wave around x =. We cannot recognise the steady wave in the figures for the pressure perturbation, but this is because the amplitude of this wave is too small. We can take a closer look at the steady wave. In figure 6.4 we see the steady waves in the surface pressure and the vorticity. The wave in the pressure has a strange form, compared to the wave in the vorticity, but the pressure and vorticity satisfy geostrophy: fv 1 = x M 1 where ζ 1 = v 1 / x and M 1 = Aη 2. The vorticity of the steady wave is -1.2 1 6 s 1 and the pressure perturbation is η 2 = 1 1 4. This means that for a typical value for the vorticity of 1.2 1 4 s 1, the perturbation in the surface pressure would be η 2 = 1 1 2, which means a perturbation of 1 hpa, which is a realistic value.

38 Initial value problem t= t=.1.1.1.8.8.6.6 eta_2.4 eta_2.4.2.2-1 -5 5 1-1 -5 5 1 Distance km Distance km t=.2 t=.3.1.1.8.8.6.6 eta_2.4 eta_2.4.2.2-1 -5 5 1-1 -5 5 1 Distance km Distance km t=.4 t=.5.1.1.8.8.6.6 eta_2.4 eta_2.4.2.2-1 -5 5 1-1 -5 5 1 Distance km Distance km t=1 t=2.1.1.8.8.6.6 eta_2.4 eta_2.4.2.2-1 -5 5 1-1 -5 5 1 Distance km Distance km Figure 6.1: The surface pressure perturbation for t =, t =.1, t =.2, t =.3, t =.4, t =.5, t = 1 and t = 2. Recall that the time is in 1 s.

One layer model 39 t= t=.1.12.12 1e-4 1e-4 8e-5 8e-5 6e-5 6e-5 Divergence 1/s 4e-5 2e-5 Divergence 1/s 4e-5 2e-5-2e-5-2e-5-4e-5-4e-5-6e-5-1 -5 5 1-6e-5-1 -5 5 1 Distance km Distance km t=.2 t=.3.12.12 1e-4 1e-4 8e-5 8e-5 6e-5 6e-5 Divergence 1/s 4e-5 2e-5 Divergence 1/s 4e-5 2e-5-2e-5-2e-5-4e-5-4e-5-6e-5-1 -5 5 1-6e-5-1 -5 5 1 Distance km Distance km t=.4 t=.5.12.12 1e-4 1e-4 8e-5 8e-5 6e-5 6e-5 Divergence 1/s 4e-5 2e-5 Divergence 1/s 4e-5 2e-5-2e-5-2e-5-4e-5-4e-5-6e-5-1 -5 5 1-6e-5-1 -5 5 1 Distance km Distance km t=1 t=2.12.12 1e-4 1e-4 8e-5 8e-5 6e-5 6e-5 Divergence 1/s 4e-5 2e-5 Divergence 1/s 4e-5 2e-5-2e-5-2e-5-4e-5-4e-5-6e-5-1 -5 5 1-6e-5-1 -5 5 1 Distance km Distance km Figure 6.2: The divergence for t =, t =.1, t =.2, t =.3, t =.4, t =.5, t = 1 and t = 2. Recall that the time is in 1 s.

4 Initial value problem t= t=.1 5e-7 5e-7 Vorticity 1/s -5e-7 Vorticity 1/s -5e-7-1e-6-1e-6-1 -5 5 1-1 -5 5 1 Distance km Distance km t=.2 t=.3 5e-7 5e-7 Vorticity 1/s -5e-7 Vorticity 1/s -5e-7-1e-6-1e-6-1 -5 5 1-1 -5 5 1 Distance km Distance km t=.4 t=.5 5e-7 5e-7 Vorticity 1/s -5e-7 Vorticity 1/s -5e-7-1e-6-1e-6-1 -5 5 1-1 -5 5 1 Distance km Distance km t=1 t=2 5e-7 5e-7 Vorticity 1/s -5e-7 Vorticity 1/s -5e-7-1e-6-1e-6-1 -5 5 1-1 -5 5 1 Distance km Distance km Figure 6.3: The vorticity for t =, t =.1, t =.2, t =.3, t =.4, t =.5, t = 1 and t = 2. Recall that the time is in 1 s.

Two layer model 41.11 2e-7.1 9e-5-2e-7 8e-5-4e-7 eta_2 7e-5 Vorticity 1/s -6e-7 6e-5-8e-7 5e-5-1e-6 4e-5-1.2e-6 3e-5-1 -5 5 1-1.4e-6-1 -5 5 1 Distance km Distance km a b Figure 6.4: The steady waves in a the pressure and b the vorticity. 6.2 Two layer model We will now consider the two layer model. Again we are only looking at waves propagating in the zonal direction, so l =. The set of linearised equations is: t D t D1 fζ = 2 Aη 2 Bη 1 fζ 1 = 2 Aη 2 t η 1 D η 1 = t η 2 η 1 D 1 η 2 η 1 = t ζ fd = t ζ1 fd 1 =, κ 1 where A = Rθ1 η 2 and B = RT κ 1 / η 1 Rθ1 η 1. For the solutions we can write: D = D exp ikx ωt D 1 = D 1 exp ikx ωt η 1 = η 2 = E 1 exp ikx ωt E 2 exp ikx ωt ζ = F exp ikx ωt ζ 1 = F 1 exp ikx ωt with angular frequencies ω = 2x and ω = ± 1 2 k 2 A η 2 B η 1 2f 2 ± k 2 A 2 η 2 2 2AB η 1 η 2 B 2 4AB η 2 1.

42 Initial value problem Suppose we know E 2, the other amplitudes are then given by: D = iω η 1 ω 2 f 2 Ak 2 η 2 η 1 ω 2 f 2 E 2 D 1 = iω Ak2 ω 2 f 2 E 2 E 1 = ω2 f 2 Ak 2 l 2 η 2 η 1 ω 2 f 2 E 2 F = f η 1 ω 2 f 2 Ak 2 η 2 η 1 ω 2 f 2 E 2 F 1 = Ak 2 f ω 2 f 2 E 2 for ω and k 2 BE 1 k 2 AE 2 ff = k 2 AE 2 ff 1 = for ω =. We can write D, D 1, η 1, η 2, ζ and ζ 1 as a superposition of waves with the different frequencies and we can sum over the wavenumbers m: D = D 1 = η 1 = η 2 = ζ = ζ 1 = m= = m= m= m= m= A 1 m D 1 m A 2 m D 2 m A 3 m D 3 m A 4 m D 4 m A 5 m D 5 m A 6 m D 6 m A 1 m D 1 1 m A 2 m D 1 2 m A 3 m D 1 3 m A 4 m D 1 4 m A 5 m D 1 5 m A 6 m D 1 6 m A 1 m η 1 1,m A2 m η 2 1,m A3 m η 3 1,m A4 m η 4 1,m A5 m η 5 1,m A6 m η 6 1,m A 1 m η 1 2,m A2 m η 2 2,m A3 m η 3 2,m A4 m η 4 2,m A5 m η 5 2,m A6 m η 6 2,m A 1 m ζ 1 m A 2 m ζ 2 m A 3 m ζ 3 m A 4 m ζ 4 m A 5 m ζ 5 m A 6 m ζ 6 m A 1 m ζ 1 1 m A 2 m ζ 1 2 m A 3 m ζ 1 3 m A 4 m ζ 1 4 m A 5 m ζ 1 5 m A 6 m ζ 1 6 m. Here we also used that k = mπ/l. Suppose there is a small perturbation in η 2 at t = and the initial values for D, D 1, η 1, ζ, ζ 1 are zero. The initial perturbation in the surface pressure can be written as a Fourier series: η 2t = = c m expimπx/l. m=

Two layer model 43 Again, we can write the initial conditions as a superposition of waves, with the six different frequencies, at t = : m= m= A 1 m D 1 m A 2 m D 2 m A 3 m D 3 m A 4 m D 4 m A 5 m D 5 m A 6 m D 6 m A 1 m D 1 1 m A 2 m D 1 2 m A 3 m D 1 3 m A 4 m D 1 4 m A 5 m D 1 5 m A 6 m D 1 6 m m= m= m= m= A 1 m η 1 1,m A2 m η 2 1,m A3 m η 3 1,m A4 m η 4 1,m A5 m η 5 1,m A6 m η 6 1,m A 1 m η 1 2,m A2 m η 2 2,m A3 m η 3 2,m A4 m η 4 2,m A5 m η 5 2,m A6 m η 6 2,m A 1 m ζ 1 m A 2 m ζ 2 m A 3 m ζ 3 m A 4 m ζ 4 m A 5 m ζ 5 m A 6 m ζ 6 m A 1 m ζ 1 1 m A 2 m ζ 1 2 m A 3 m ζ 1 3 m A 4 m ζ 1 4 m A 5 m ζ 1 5 m A 6 m ζ 1 6 m = = = = c m expimπx/l = =. m= For every wavenumber m we have A 1 m Dm 1 A 2 m Dm 2 A 3 m Dm 3 A 4 m Dm 4 A 5 m Dm 5 A 6 m Dm 6 = A 1 m Dm 11 A 2 m Dm 12 A 3 m Dm 13 A 4 m Dm 14 A 5 m Dm 15 A 6 m Dm 16 = A 1 m E 1 1,m A2 m E 2 1,m A3 m E 3 1,m A4 m E 4 1,m A5 m E 5 1,m A6 m E 6 1,m = A 1 m E 1 2,m A2 m E 2 2,m A3 m E 3 2,m A4 m E 4 2,m A5 m E 5 2,m A6 m E 6 2,m c m = A 1 m Fm 1 A 2 m Fm 2 A 3 m Fm 3 A 4 m Fm 4 A 5 m Fm 5 A 6 m Fm 6 = A 1 m Fm 11 A 2 m Fm 12 A 3 m Fm 13 A 4 m Fm 14 A 5 m Fm 15 A 6 m Fm 16 =. Using the relations between the amplitudes and taking L = 4 km, f = 1 4 s 1, η 1 =.2, η 2 = 1, A = 86115 m 2 s 2 and B = 6539.46 m 2 s 2, we can solve this set of equations using Mathematica and the A i m s will be: A 1 m = c m 2.91675.23534m 2 2.91675 15.72913m 2 m 4 A 2 m = c m 3.9876m 2 2.91675 15.72913m 2 m 4 A 3 m = A 5 m = c m.49845m 2.18768 m 2 A 4 m = A 6 m = c m.155m 2.18768 m 2 2.91675 15.72913m 2 m 4. Now we can write D, D 1, η 1, η 2, ζ and ζ 1 as a Fourier series in sines and cosines:

44 Initial value problem D = D 1 = η 1 = m=1 m=1 m=1 m=1 m=1 m=1 m=1 m=1 m=1 2A 3 ω 3 L 2 ω3 2 f 2 Am 2 π 2 η 2 η 1 m η 1 L 2 ω3 2 f 2 sinmπx/l ω 3 t 2A 4 ω 4 L 2 ω4 2 f 2 Am 2 π 2 η 2 η 1 m η 1 L 2 ω4 2 f 2 sinmπx/l ω 4 t 2A 5 ω 5 L 2 ω5 2 f 2 Am 2 π 2 η 2 η 1 m η 1 L 2 ω5 2 f 2 sinmπx/l ω 5 t 2A 6 ω 6 L 2 ω6 2 f 2 Am 2 π 2 η 2 η 1 m η 1 L 2 ω6 2 f 2 sinmπx/l ω 6 t 2A 3 m ω 3 Am 2 π 2 L 2 ω 2 3 f 2 sinmπx/l ω 3t 2A 4 m ω 4 Am 2 π 2 L 2 ω 2 4 f 2 sinmπx/l ω 4t 2A 5 m ω 5 Am 2 π 2 L 2 ω 2 5 f 2 sinmπx/l ω 5t 2A 6 m ω 6 Am 2 π 2 L 2 ω 2 6 f 2 sinmπx/l ω 6t 2A 2 m cosmπx/l m=1 m=1 m=1 m=1 2A 3 L 2 ω3 2 f 2 Am 2 π 2 η 2 η 1 m L 2 ω3 2 f 2 cosmπx/l ω 3 t 2A 4 L 2 ω4 2 f 2 Am 2 π 2 η 2 η 1 m L 2 ω4 2 f 2 cosmπx/l ω 4 t 2A 5 L 2 ω5 2 f 2 Am 2 π 2 η 2 η 1 m L 2 ω5 2 f 2 cosmπx/l ω 5 t 2A 6 L 2 ω6 2 f 2 Am 2 π 2 η 2 η 1 m L 2 ω6 2 f 2 cosmπx/l ω 6 t