Wave interactions J. Vanneste School of Mathematics University of Edinburgh, UK www.maths.ed.ac.uk/ vanneste Wave interactions p.1/50
Linear waves in conservative systems Small amplitude motion: linearize equations of motion and introduce solutions proportional to exp[i(k x ωt)] for wavevector k and frequency ω, related by a dispersion relation, ω = ω(k). Example: Rossby waves, large-scale atmospheric and oceanic waves, described by 2D fluid on a β-plane: ζ t + βψ x + ψ x ζ y ψ y ζ x = 0, with streamfunction ψ and vorticity ζ = 2 ψ. Wave interactions p.2/50
Linear waves: Rossby waves With ζ = A exp[i(k x ωt)] + c.c., the dispersion relation is found as ω = Note ω/k < 0: eastward propagation. βk k 2 + l 2 Wave interactions p.3/50
Linear waves: Rossby waves General solution: superposition ζ = a A a exp[i(k a x ω a t)], with a = or, and A a = A a. The nonlinear equation of motion conserves In terms of A a, these are energy E = 1 2 pseudomomentum P = 1 2β ψ 2 dx ζ 2 dx E = 1 2 E a A a 2, P = 1 2 a P a A a 2, a Wave interactions p.4/50
Wave interactions p.5/50 Linear waves: general formulation Consider the system governed by u t = Lu + N(u), where u n, L matrix operator, and N(u) contains nonlinear terms. Assume two conserved quantities: pseudoenergy and pseudomomentum E = E (2) + h.o.t. and P = P (2) + h.o.t., quadratic at leading order, with E (2) = 1 2 u Eu dx and P (2) = 1 2 u Pu dx. (L, E and P are related: L = JE for J skew-adjoint, and P is then such that u x = JPu.)
Linear waves: general formulation Introducing wave solutions u = Aû exp[i(k x ωt)] + c.c. leads to the eigenvalue problem iωû = ˆLû. This gives a dispersion relation ω = ω p (k) with n branches p = 1, 2,, n, and polarization relations û = û p. The general solution is again obtained by superposition u = a A a û a exp[i(k a x ω a t)], with a denoting wavenumber k a, and branch p a. If (ω a, û a ) is a solution for the wavevector k a, (ω a, û a ) = ( ω a, û a) is a solution for the wavevector k a. Take A a = A a. Wave interactions p.6/50
Linear waves: general formulation The conservation laws imply orthogonality relations: for each k, is constant only if E (2) = (2π)d 2 n A pa q u peu q exp[i(ω p ω q )t], p,q=1 (2π) d u peu q = E p δ p,q. with E p pseudoenergy of branch p: orthogonality in the sense of pseudoenergy. (Similarly, orthogonality in the sense of pseudomomentum.) Pseudoenergy and pseudomomentum are related: E a ω = P a k a = wave action Wave interactions p.7/50
Linear waves: general formulation Exercise: derive dispersion relation, polarization relation, and show orthogonality for internal gravity waves. With u = (ζ, b), they are governed by a system of the general form with L = 0 x N 2 x 2 0, E = 2 0 0 N 2, P = 0 N 2 N 2 0, (Solution: ω p = ( 1) p Nk/(k 2 + l 2 ) 1/2, etc.) Wave interactions p.8/50
Interaction equations Weak nonlinearity leads to modulation of wave amplitudes A a. Rossby waves: introduce ζ = a A a (t) exp[i(k a x ω a t)] into the nonlinear equation of motion lead to Ȧ a = 1 2 bc I bc a A ba c exp(2iω abc t)δ a+ b + c, where 2Ω abc = ω a + ω b + ω c, The strength of the interaction depends on the interaction coefficient I bc a = (k b l c k c l b ) [ (k 2 b + l 2 b) 1 (k 2 c + l 2 c) 1] (satisfying I bc a = I cb a ). Wave interactions p.9/50
Interaction equations Interactions are limited to wave triads satisfying the interaction condition k a k b k a + k b + k c = 0 k c This appears through the factor δ ka +k b +k c,0 δ la +l b +l c,0 δ a+ b + c = δ(k a + k b + k c )δ(l a + l b + l c ) periodic b.c. unbounded domains General formulation: interaction equations are derived using orthogonality relations, leading to I bc a = (2π) d û ae [ ˆN (2) (ûb, û c ) + ˆN (2) (û c, û b )] /Ea. Wave interactions p.10/50
Interaction equations The conservation laws imply some properties of the interaction coefficients. Writing E = E (2) +E (3) + = 1 2 E a A a 2 + 1 6 a S abc A aa ba c exp(2iω abc t)+, abc the cubic terms in E (2) + E (3) + = 0 are equal only if E a I bc a + E b I ca b + E c I ab c = 2iS abc Ω abc. Similarly, P a I bc a + P b I ca b + P c I ab c = 2iT abc Ω abc. These are useful when: (i) E and P are exactly quadratic (cf. Rossby and internal gravity waves), or (ii) the triad is resonant, i.e. ω a + ω b + ω c = 0. Wave interactions p.11/50
Interaction equations Assuming (i) or (ii), E a Ia bc = E bib ca = E cic ab s b s c s c s a s a s b and P a Ia bc = P bib ca = P cic ab, c b c c c c c a c a c b where s a = 1/c a = k a /ω a is the slowness of wave a. Assuming (ii), E a I bc a ω a = E bib ca ω b = E cic ab ω c and P a I bc a k a = P bib ca k b = P cic ab k c This can be verified explicitly, e.g., for Rossby waves. Wave interactions p.12/50
Interaction equations Remark: non-conservative systems are also governed by interaction equations of the same form, but with coefficients Ia bc, Ib ca and Ic ab that are independent. To derive these equations, use the orthogonality (2π) d u + p u q = E p δ p,q where u + is the left eigenvector of L. Exercise: derive the interaction coefficients for internal gravity waves, for which with 2 ψ = ζ. N(u) = ψ xζ y ψ y ζ x ψ x b y ψ y b x, Wave interactions p.13/50
Triad interactions The interaction equations Ȧ a = 1 2 bc I bc a A ba c exp(2iω abc t)δ a+ b + c, are an exact reformulation of the equations of motion. They can be approximated if we assume weak nonlinearity, i.e., A a = O(ɛ), ɛ 1. With this assumption, we can truncate the system, with the simplest truncation a wave triad obeying Ȧ a = Ia bc A ba c exp(2iω abc t), A b = I ca A ca a exp(2iω abc t), b A c = I ca A aa b exp(2iω abc t). c Wave interactions p.14/50
Triad interactions Over t = O(ɛ), the amplitudes change by O(1) only in the triad is near resonant: ω a + ω b + ω c = 2Ω abc = O(ɛ). If not, the transformation A a = B a iibc a 2Ω abc B b B c exp(2iω abc t), pushes nonlinear terms to O(ɛ 3 ). Hence A a = A a (0) + O(ɛ). Need to consider the interaction and resonance conditions vk a + k b + k c = 0 and ω a + ω b + ω c = 0 d + 1 algebraic equations for the 3d unknowns Wave interactions p.15/50
Triad interactions Graphical solution: Rossby waves k ω a ω c l =3 l =1 ω b k k a k b l =2 k c ω ω Wave interactions p.16/50
Solution of the triad equation Universal form: use scaled variables A a = ɛ Ib ca I ab c 1/2 α a, A b = ɛ Ic ab I bc a 1/2 α b, A c = ɛ I bc a Ib ca 1/2 α c and rescale t by ɛ to find α a = σ a α bα c exp(2iωt) α b = σ b α cα a exp(2iωt) α c = σ c α aα b exp(2iωt), with Ω = Ω abc /ɛ = O(1), σ a = sign I bc a, etc. Only two cases to consider: (i) σ a = σ b = σ c = 1 or (ii) σ a = σ b = σ c = 1. Wave interactions p.17/50
Solution of the triad equation Manley Rowe relations: ( σa α a 2 σ b α b 2) t = ( σ b α b 2 σ c α c 2) t = ( σ c α c 2 σ a α a 2) t = 0 In case (i), amplitudes are bounded: stable; in case (ii), amplitude can become unbounded: explosive instability. Recall that Ia bc E a /ω a = Ib cae b/ω b = Ic ab E c /ω c and ω a + ω b + ω c = 0: (ii) only possible if wave with the largest ω has E oppositely signed to the other two waves. Thus, sign-definite pseudoenergy implies stable interactions (i). Sign-indefinite pseudoenergy is possible for waves propagating on shear flows: purely nonlinear mechanism of instability. Wave interactions p.18/50
Solution of the triad equation Integrate triad equations: let [ Z(t) = σ a αa (t) 2 α a (0) 2] [ = σ b αb (t) 2 α b (0) 2] [ = σ c αa (t) 2 α a (0) 2] W = α a α b α c sin(arg α a + arg α b + arg α c 2Ωt) + ΩZ(t), and verify Ẇ = 0. Then, 1 2Ż2 + V (Z) = 0, V (Z) = 2 [ (W ΩZ) 2 (σ a Z + α a (0) 2 )(σ b Z + α b (0) 2 )(σ c Z + α c (0) 2 ) ] Dynamics of a particle with coordinate Z(t), with zero energy starting at Z(0) = 0 in potential V (Z). Wave interactions p.19/50
Solution of the triad equation Stable case for Ω = W = 0: V 0.1 α 0.8 a 0.05 0.4 b 0-0.4-0.2 0 0.2-0.05 0.4 0.6 z -0.4 0 0 c 5 10 15 20 25 30 t -0.1-0.8 Energy exchange between the three waves, complete between a and c, partial with b. Non-zero Ω or W inhibit energy exchanges Wave interactions p.20/50
Solution of the triad equation Explosive case for Ω = W = 0: -0.04-0.03-0.02-0.01 V 0 0-0.00002 0.01 0.02 z α 1.6 1.2-0.00004-0.00006 0.8-0.00008 0.4-0.0001 0 0 2 4 6 8 10 t Finite time blow-up of the three amplitudes: explosion Non-zero W : slows down blow up Non-zero Ω: slows down blow up or suppresses it for sufficiently small initial amplitudes. Wave interactions p.21/50
Wave instability Consider a stable fluid, with sign-definite pseudoenergy. Plane Rossby waves and internal gravity waves are exact nonlinear solutions of the equations of motion. Are they stable to small perturbations? This can be studied by linear stability analysis: write ζ = A exp[i(x vx ωt) + c.c. + ζ, and derive a linear evolution equation for ζ. This has periodic coefficients: need to use Floquet theory ζ = exp(µθ)p (θ) exp(il x), θ = x vx ωt, for P (θ) periodic. Instability if Rµ 0. Wave interactions p.22/50
Wave instability For small-amplitude waves, this reduces to a wave interaction problem wave a = primary waves whose stability is studied waves b and c = perturbation to wave a Assuming A b, A c A a 1, the interaction equations reduce to A b = I ca b A aa c exp(2iω abc t) Ȧ c = I ab c A aa b exp(2iω abc t) where A a is taken as a constant: pump-wave approximation. These have solutions A b = exp[(λ + iω)t]âb and A c = exp[(λ + iω)t]âc, for constant  b,  c and λ = ± [ Ib ca Ic ab A a 2 Ω 2 1/2 abc]. Wave interactions p.23/50
Wave instability Instability for Ω abc = 0, resonant triads, if This is the case if ω a > ω b, ω c. Ib ca Ic ab > 0. Hasselmann s criterion: a wave is unstable if it is the highest frequency member of a resonant triad. In practice, most small amplitude waves are unstable. Example: inertia-gravity waves Wave interactions p.24/50
Wave instability Wave interactions p.25/50
Wave instability Limiting cases: assume k b, k c k a, then k b k c. If b and c are on the same branch of the dispersion relation, resonance implies ω(k a ) = ω(k b ) ω( k b k a ) k a ω k (k b). In 1D, group velocity ω k (k b) of secondary waves matches phase velocity ω(k a )/k a of the primary wave. If b and c are on branches of the dispersion relation corresponding to oppositely signed ω, resonance implies ω b ω c ω a /2. Wave interactions p.26/50
Interaction of wavepackets Instead of plane waves, we can consider wavepackets, localized over long distances. These are obtained by superposition of plane waves, with amplitudes that are narrowly peaked around a central wavevector, for instance A(k, t) exp( k k a 2 /δ 2 ), with δ 1. The wavepacket envelope is described by B a (x, t) = A(k a, t) exp [i(k a k a ) x (ω(k a ) ω a )t] dk a. This is localized over a distance x = O(δ 1 ). Take δ = O(ɛ). Wave interactions p.27/50
Interaction of wavepackets Consider the interaction of three wavepackets with central wavenumbers k a, k b and k c which are resonant. Derive evolution equations for the corresponding envelope amplitudes B a (x, t), B b (x, t) and B c (x, t). Start with t B a (x, t) = [Ȧa ia a (ω a ω a )] e i( a a) (ω a ω a )t dk a. Use ω a ω a c a (k a k a ), with c a = ω k (k a), and the interaction equations with I b c a I bc a to find: Wave interactions p.28/50
Interaction of wavepackets t B a + c a B a = I bc a B b B c exp(2iω abc t), t B b + c b B b = Ib ca Bc Ba exp(2iω abc t), t B c + c c B c = Ic ca BaB b exp(2iω abc t). Similar to amplitude equations for plane waves Can scale amplitudes to replace the interaction coefficients by their sign Partial rather than ordinary differential equations Additional terms: advection with the group velocity Wave interactions p.29/50
Interaction of wavepackets The interaction equations are completely integrable by inverse scattering: solution consists of solitons + waves solitons are exchanged between envelopes a and b and c wave part does not disperse interactions are significant only if the three wavepackets collide energy exchanges depend on signs of interaction coefficients and relative group velocities explosive interaction is possible for sufficiently large initial amplitudes Wave interactions p.30/50
Interaction of wavepackets From Kaup et al. 1979: stable interaction explosive interaction Wave interactions p.31/50
Quartet interactions Triad interactions do not control wave interactions when the nonlinearity is cubic at lowest order, or resonant triads are impossible In both cases, need to consider resonant quartets: four waves satisfying k a + k b + k c + k d = 0 and ω a + ω b + ω c + ω d = 0. Surface gravity waves: dispersion relation ω = ±(g k ) 1/2, does not admit resonant triads but resonant quartets are possible. Wave interactions p.32/50
Resonant quartets: surface waves For surface waves, can take k a + k b = k c k d = 1 0. Then the resonance relation k a 1/2 + k b 1/2 = k c 1/2 + k d 1/2 can be represented graphically in the (k a, l a )-plane on the level curves of k a 1/2 + k b 1/2, with k b = 1 k a and l b = l c. k a k b k d k c Wave interactions p.33/50
Quartet interactions: interaction equations Amplitude equations are derived as in the triad case: wave expansion and use of orthogonality. Non-resonant quadratic nonlinearities are eliminated by variable transformation. This leads to A a = 1 6 abcd Ia bcd A ba ca d exp(2iω abcd t)δ a+ b + c + d A single quartet of modes (a, b, c, d) cannot be isolated: modes (a, a, a, a), (a, a, b, b) and permutations are always resonant. (Wave-mean flow effect: can think of b b mean flow followed by mean flow a a.) Wave interactions p.34/50
Quartet interactions: interaction equations A a = ia a j=a,d Ȧ b = ia b j=a,d Ȧ c = ia c j=a,d A d = ia d j=a,d Ja A j j 2 + Ia bcd A ba ca d exp(2iω abcd t), J j b A j 2 + Ib cda A ca da a exp(2iω abcd t), Jc j A j 2 + Ic dab A da aa b exp(2iω abcd t), J j d A j 2 + Id abc A aa ba c exp(2iω abcd t), where ija b = I a ( a)b( b) /3, etc. Pseudoenergy conservation imposes that interactions coefficients purely imaginary Ja: j for a single wave a, nonlinear frequency ω a + J a a A a0 3. A a (t) = A a0 exp( ij a a A a0 3 t), Wave interactions p.35/50
Quartet interactions: interaction equations Conservation of pseudoenergy and 2 components of pseudomomentum impose E a I bcd a ω a = E bib cda ω b = E cic dab ω c = E did abc ω d The interaction coefficients can be reduced to signatures by scaling Interaction equations integrable in closed form Wavepackets can be considered Wave interactions p.36/50
Quartet interactions: wave instability Consider the stability of the primary wave a, to perturbations consisting of secondary waves c and d, with k c = k a + K and k d = k a K. Then, a, b = a, c and d form a resonant quartet. k a k a k d K k c Wave interactions p.37/50
Quartet interactions: wave instability The pump-wave approximation assumes that A a (t) = A a0 exp( ij a a A a0 3 t) remains unchanged. The amplitudes of c and d obey the linear equations Ȧ c = 2iJc a A c A 2 a0 + Ic aad A da 2 a0 exp[2i(ω + Ja a A 2 a0)t] A d = 2iJd a A d A 2 a0 + I aac A ca 2 a0 exp[2i(ω + Ja a A 2 a0)t]. Solutions can be sought in the form d A c = Âc exp[i(ω + J a a A 2 a0)t] exp(λt) A d =  d exp[i(ω + J a a A 2 a0)t] exp(λ t) for constant  c,  d and growth rate λ. Wave interactions p.38/50
Growth rate Quartet interactions: wave instability λ = i(ja c Ja) b ± [ Ic aad Id aac A 4 a0 [Ω + (Ja a Jc a Jd a )A 2 a0] 2] 1/2. Instability is possible provided that Ω/A 2 a0 + (Ja a Jc a Jd a ) is small enough. Side-band instability: limiting case where K k a, i.e. the perturbation represents a large-scale modulation of the amplitude of a (cf. NLS equation). In this limit, λ = [ 2ΩIA 2 a0 Ω 2] 1/2 with Ω = 1 2 K 2 ω k k (k a) K. Instability occurs if Ω(IA 2 a0 Ω/2) > 0: condition in terms of nonlinear frequency shift I and linear dispersion relation. Wave interactions p.39/50
Alternative derivations: Lagrangian systems Use the Lagrangian or Hamiltonian structure to derive interaction equations, with explicitly symmetric interaction coefficients. Lagrangian system: derived from variational principle δ dt dx L(u, u, u, ) = 0, Example: Klein Gordon equation ψ tt ψ xx + ψ + 4σψ 3 = 0 follows from δ [ψ 2 t ψ 2 x ψ 2 2σψ 4] dx = 0. To derive interaction equations, introduce ψ = a A a (t) exp[i(k a x ω a )t]. Wave interactions p.40/50
Alternative derivations: Lagrangian systems The variational principle becomes δ ( ) L (2) + L (3) + = 0, where the variations are taken with respect to the amplitudes A a. Find L (2) = a D(k a, ω a ) A a 2, The leading-order variation gives D(k a, ω a ) = 0, a form of the dispersion relation. At next order (with Ȧ a = O(ɛ)), integrate over t keeping Ω abc t fixed: [ ] L (3) 1 = i dt D ω (k a, ω a )ȦaA a 1 Γ 2 6 abca a A b A c exp( 2iΩ abc t) a abc Wave interactions p.41/50
Alternative derivations: Lagrangian systems Taking variations give D ω (k a, ω a )Ȧa = 1 2 Interaction equations, with Γ abc A ba c exp(2iω abc t)δ bc a+ b + c. I bc a = Γ abc D ω (k a, ω a ). Since mode pseudoenergy is E a = ω a D ω (k a, ω a ), the interaction coefficients explicitly satisfy E a I bc a ω a = E bib ca ω b = E cic ab. ω c Wave interactions p.42/50
Hamiltonian formulation For many fluid systems, the Lagrangian formulation is not the most natural one: in usual coordinates, the systems are non-canonical Hamiltonian systems. Rossby waves: ζ t + βψ x + (ψ, ζ) can be written ζ t = J δe, J = (ζ, ). ζ The derivation of interaction equations is simple only for canonical systems: J is constant. Zakharov Piterbarg: introduce a near-identity transformation of ζ to make the system canonical. Let ζ be ζ = ζ + ζζ y /β + O(ζ 3 ). Wave interactions p.43/50
The equation of motion becomes Hamiltonian formulation ζ t + β x (ψ + ψ y ζ) = O(ɛ 3 ). This has the canonical Hamiltonian form H ζ t = β x δ ζ, with H = 1 ψ 2 dx = 1 2 2 ψζ dx. Introducing the expansion ζ = a C a (t) exp(ik a x) with C a = 1 (2π) 2 ζ(x, t) exp( ik x) dx, the equations of motion follow from Ċ a = iβk a (2π) 2 H C a. Wave interactions p.44/50
Hamiltonian formulation From ζ = a ψ = a + i 2β C a (t) exp(ik a x) i 2β 1 ka 2 + la 2 C a (t) exp(ik a x) bc (l b + l c )C b C c exp[i(k b + k c ) x], bc l b + l c (k b + k c ) 2 + (l b + l c ) 2 C bc c exp[i(k b + k c ) x], the Hamiltonian follows as ( H = (2π)2 2 a C a 2 k 2 a + l 2 a + iβ 3 ) CaC b Cc, abc abc = ( la k 2 a + l 2 a + l b k 2 b + l2 b + l c k 2 c + l 2 c ). Wave interactions p.45/50
The interaction equations are then Hamiltonian formulation Ċ a = iω a C a + k a abc Cb Cc. With C a = A a exp( iω a t), these are equivalent to those found in Lecture 1 near resonance since bc k a abc l ak b k 2 b + l2 b + l ak c k 2 c + l 2 c k al b k 2 b + l2 b k al c k 2 c + l 2 b = I bc a where I bc a is given in. Wave interactions p.46/50
Waves in shear flow When equations of motion depend on a coordinate, y, waves take the modal form u = Aû(y) exp[i(kx ωt). The dispersion and polarisation relations are found by solving a differential eigenvalue problem. Example: Rossby waves in a shear flow (U(y), 0) obey the linearized equation conserving ( t + U x )ζ + (β U ) x ψ = 0, 2 ψ = ζ, E (2) = 1 2 [ ψ 2 Uζ 2 /(β U ) ] dx. and P (2) = 1 2 [ζ 2 /(β U ) ] dx. Wave interactions p.47/50
Waves in shear flow The eigenvalue problem (U c)( ˆψ k 2 ˆψ) + (β U ) ˆψ = 0 with ˆψ = 0 for y = 0, L has two types of solutions: Rossby waves for c = c n, n = 1, 2,, singular modes for U min < c < U max, singular at y c : U(y c ) = c, where [ ] y + dψ c = λ c, ζ λ c δ(y y c ) + dy y c Superposition of singular modes is smooth and represents a sheared disturbance: ζ = Â(t; c)ζ(y; c)e ik(x ct) dc A(t; U(y))e ik(x U(y)t). Wave interactions p.48/50
Waves in shear flows Orthogonality relations remain valid, although in a generalized sense for singular modes. For fixed k, pseudomomentum orthogonality, ˆζa (y)ˆζ b (y) β U (y) dy = P aδ a,b for regular modes a, b = 1, 2,, and ˆζ(y; ca )ˆζ(y; c b ) β U dy = P (c a )δ(c a c b ) (y) for singular modes with U min < c a, c b < U max. Wave interactions p.49/50
Waves in shear flow Interaction equations can be derived formally but presence of terms secular in t ( y ζ t for sheared disturbances can be remedied by expanding ζ = ζ ζ y ζ/(β U ) cannot truncate the continuous spectrum of singular modes simple truncations break down for t = O(ɛ 1 ) Open problem: explosive interaction in shear flows (must involve singular modes). Wave interactions p.50/50