ω 2 f 2 (ωk + ifl) and ˆ v = ω 2 2 u ʹ = Re[ u ˆ e i(kx+ly ωt ) ], v ʹ = Re[ˆ v e i(kx +ly ωt ) ], and h ʹ = Re[ h ˆ e i(kx+ly ωt ) ].

Transcription

1

2 28) Show that for the gravity wave modes supported in the shallow water system linearized about a basic state at rest, the complex amplitudes of the zonal and meridional winds are given by: u = g h ω 2 2 (ωk + ifl) and v = ω 2 2 (ωl ifk), respectively First, start with the linearized equations: u t v = g h x, v t + f u = g h y, h t + H( u x + v y ) = 0, where u = Re[ u e i(kx+ly ωt ) ], v = Re[ v e i(kx +ly ωt ) ], and h = Re[ h e i(kx+ly ωt ) ] After taking the derivatives, the linearized equations can be written as listed below u iωe i(kx +ly ωt ) v e i(kx +ly ωt ) = i(kx +ly ωt ) ike v iωe i(kx +ly ωt ) + f u e i(kx +ly ωt ) = i(kx +ly ωt ) ile h iωe i(kx +ly ωt ) + He i(kx+ly ωt ) ( u ik + v il) = 0 Simplifying these equations we get: A u iω v = ik B v iω + f u = il C h iω + H( u ik + v il) = 0 In order to find u let s take iωa B, which yields u i 2 ω 2 iωfv + iωfv 2 u = i 2 kω + fil => u (ω 2 2 ) = (ωk + ifl) => u = (ωk + ifl) (ω 2 2 ) To find v, we can take fa + iωb, which yields u fiω 2 v + i 2 ω 2 v + iωf u = fik ωi 2 l => v (ω 2 2 ) = (ωl ifk) => v = (ωl ifk) (ω 2 2 )

3 29) Using u = u = v = g h ω 2 2 (ωk + ifl) and v = ω 2 2 (ωl ifk), show that ω 2 2 [ωk cos(kx + ly ωt) lsin(kx + ly ωt)], and ω 2 2 [ωlcos(kx + ly ωt) + kf sin(kx + ly ωt)] Solve for u : u = Re[ u e i(kx+ly ωt ) ] = Re[ u (cos(kx + ly ωt) + isin(kx + ly ωt))] Substitute u into the equation to get: u = Re[ w 2 2 ((ωk + ifl)(cos(kx + ly ωt) + isin(kx + ly ωt)))] Expand to get Re[ w 2 (ωk cos(kx + ly ωt) + iωk sin(kx + ly ωt) + iflcos(kx + ly ωt) + 2 i2 flsin(kx + ly ωt))] Taking the real part of the above expression yields: u = w 2 (ωk cos(kx + ly ωt) lsin(kx + ly ωt)) 2 Solve for v : v = Re[ v e i(kx +ly ωt ) ] = Re[ v (cos(kx + ly ωt) + isin(kx + ly ωt))] Substitute v into the equation and expand to get: Re[ w 2 (ωlcos(kx + ly ωt) + iωlsin(kx + ly ωt) ifk cos(kx + ly ωt) 2 i2 fk sin(kx + ly ωt))] Finally, taking the real part yields: v = w 2 (ωlcos(kx + ly ωt) + fk sin(kx + ly ωt)) 2

4 31) Linearize the 2D, non- divergent, barotropic, vorticity equation on a β - plane and find the dispersion relationship for ψ = Uy and ψ = Re[ ψ e i(kx +ly wt ) ] First, linearize the vorticity equation: ζ ζ = u t x v ζ y βv Set ζ = ζ + ζ, u = u + u, and v = v + v Set the basic state as ζ t The linearized equation becomes: ζ = u x v ζ βv = 0 y t (ζ + ζ ) = (u + u ) x (ζ + ζ ) (v + v ) y (ζ + ζ ) β(v + v ) Expanding this, we now have: ζ t + ζ = (u ζ t x + u ζ x + u ζ x + u ζ ζ ) (v x y + v ζ y + v ζ y + v ζ ) βv β y v Now let s simplify Start by removing the basic state since it has been set to zero ζ t = (u ζ x + u ζ x + u ζ x ) (v ζ y + v ζ y + v ζ y ) βv Now get rid of the non- linear terms to get the final equation ζ t = u ζ x u ζ x v ζ y v ζ y βv In order to find the dispersion relationship, let us re- write this equation in terms of ψ Recall: ζ = 2 ψ, u = ψ ψ, and v = Substituting these in we get, y x t ( 2 ψ ) = ψ y x ( 2 ψ ) + ψ y x ( 2 ψ ) ψ x y ( 2 ψ ) ψ x y ( 2 ψ ) β ψ x However, we know that ψ = Uy, so ψ x = 2 ψ = 0 and ψ = U, giving us: y t ( 2 ψ ) = U x ( 2 ψ ) β ψ x

5 Now use ψ = Re[ ψ e i(kx +ly wt ) ] and take the appropriate derivatives to get: iω(i 2 k 2 + i 2 l 2 ) ψ e i(kx +ly ωt ) = Uik(i 2 k 2 + i 2 l 2 ) ψ e i(kx +ly ωt ) βik ψ e i(kx+ly ωt ) This simplifies to ω(k 2 + l 2 ) = Uk(k 2 + l 2 ) βk To get the dispersion relationship, solve for ω ω = Uk(k 2 + l 2 ) βk (k 2 + l 2 ), which can finally be simplified to ω = Uk βk (k 2 + l 2 ) The phase speed of these waves, c = ω k = U therefore, these waves are dispersive β, depends on the wave number; (k 2 + l 2 ) 32) (Holton 71) Show that F(x) = Re[Cexp(imx)] can also be written as F(x) = C cos(m(x x o )), where x o = m 1 sin 1 ( C i C ) First, we know that F(x) = Re[Cexp(imx)] can be expanded to F(x) = Re[Ccos(mx) + ic sin(mx)], where C = C r + ic i => F(x) = Re[C r cos(mx) + ic i cos(mx) + ic r sin(mx) C i sin(mx)] Taking the real part of the equation we get, F(x) = C r cos(mx) C i sin(mx) Now we need to find C r and C i From x o = m 1 sin 1 ( C i C ) we get C i = C sin(mx o ) Define C r = C cos(mx o ) and plug them into F(x) The resulting equation is: F(x) = Re[C cos(mx)cos(mx o ) C sin(mx)sin(mx o )]

6 Using the trig identity, cos(a)cos(b) sin(a)sin(b) = cos(a b), we get the final equation: F(x) = C cos(m(x x o )) 33) (Holton 72) Show that ψ = Acos(kx vt kx o )exp(αt) can be written as ψ = Re[Be ik(x ct ) ] Find B i,b r,c i,c r in terms of A, α, v, and x o By defining C = c r + ic i we can re- write ψ as, ψ = Re[Be ik(x c r t ) ]e kc i t Using the rule from problem 71 we can further write, ψ = Re[Be ik(x c r t ) ]e kc i t = Re[ B cos(k(x c r t) kx o )]e kc i t Now, let v = kc r, and α = kc i => ψ = Re[ B cos(kx vt kx o )]exp(αt) Similarly to 71, define x o = k 1 sin 1 ( B i B ) Giving us B r = Acos(kx o ) and B i = Asin(kx o ) Now we have, ψ = Acos(kx vt kx o )exp(αt)

7

8