Chapter 9. Resonant Interaction and Quasilinear Systems This chapter describes two extensions. First, we describe the resonant interaction of wave trains with distinct phases. This is multiphase nonlinear geometric optics. Second, the semilinear analysis is extended to the quasilinear case with the goal of discussing com- pressible inviscid fluid dynamics. §9.1. Introduction to resonance. Even at the level of formal asymptotic expansions, resonance poses a challenge. It was Majda and Rosales [1986] who got it right. The approach presented in this section is that of Joly, M´etivier, and Rauch in the Duke Math. J, 1994. The essence of the phenomenon is illustrated by the following simple example. Example 1. Consider the oscillatory semilinear initial value problem ix/ǫ (∂t + ∂x)u1 =0 u1 t=0 = a1(x) e ∂tu2 = u1u3 u2 t=0 =0 (9.1) ix/ǫ (∂t − ∂x)u3 =0 u3 t=0 = a3(x) e ∞ with initial amplitudes aj ∈ C0 (R). The exact solution is given by t i(x−t)/ǫ i(x+t)/ǫ u1(t,x) = a1(x − t) e , u3(t,x) = a3(x + t) e , u2 = u1(t,x) u3(t,x) dt. Z0 The key observation is that the phases, (x ± t)/ǫ that appear in the integrand for u2 add to 2ix/ǫ which is independent of t. The formula for u2 is, t i2x/ǫ u2 = e a1(x − t) a3(x + t) dt. (9.2) Z0 The oscillatory wave trains, u1 and u3 with phases φ1 := (x − t)/ǫ, φ2 := (x + t)/ǫ, respectively, interact to generate a third wave u2 with phase φ3 := 2x/ǫ. The phases satisfy the resonance relation φ1 + φ2 = φ3 . The amplitude of the new wave is of the same order, ǫ0, as the waves from which it is formed. 1 The linear operator 1 0 0 10 0 L(∂t,∂x) := 0 1 0 ∂t + 00 0 ∂x, 0 0 1 0 0 −1 in the background has principal symbol, τ + ξ 0 0 L(iτ, iξ) := i 0 τ 0 . 0 0 τ − ξ The characteristic variety of L has equation 0 = det L(τ,ξ) = τ(τ + ξ)(τ − ξ). The three phases satisfy the eikonal equation φt(φt + φx)(φt − φx)=0 . For the solutions with ∇t,xφ 6= 0, this is equivalent to exactly one of the equations φt =0, φt + φx =0, or φt − φx =0, at all points (assuming the domain of definition is connected). Variants of this example illustrate two properties of resonance. Example 2. iψ(x) Suppose that the initial condition for u3 is changed to u3 t=0 = a3(x)e with dψ(x)/dx nowhere equal to 1. Then the integral defining u is an oscillatory integral 2 in time with phase (x − t + ψ(x + t))/ǫ. The time derivative of the phase is O(1/ǫ) so the method of nonstationary phase shows that u2 = O(ǫ). The resonant interaction is destroyed. ′ Exercise. More generally, if {x : ψ (x)=1} has measure zero, then u2 = o(1) as ǫ → 0. Again the offspring wave is smaller than the parents. For those who know about Young measures, it is interesting to note that the Young mea- sures of the intitial data are independent of the function ψ so long as ψ′ 6= 0. Thus data with the same Young measures yield solutions with different Young measures. Introduce the symmetric form φj = 0 for resonance relations. If ψk satisfy nk ψk = 0, then the phases φk := nkψk satisfyP the symmetric form. The symmetric form isP often easier to manipulate. Example. We find all triples of resonant linear eikonal phases with pairwise independent differentials for L = ∂t + diag(λ1, λ2, λ3)∂x with λj distinct real numbers. Seek such φj satisfying the resonance relation φj = 0. The independent differentials together with the eikonal relation force (up to permutation),P 3 φj(t,x) = αj(x − λj t), (α1,α2,α3) ∈ R \ 0. 2 The α are determined up to scalar mulitplication by the resonance relation which is equiv- alent to the pair of equations, αj = 0, and, αj λj = 0 . Xj Xj Exercise. Show that if f,g,h ∈ C∞(R) each has nonvanishing derivative at the origin, and at least one of them has nonvanishing second derivate, then the three phases f(t), g(t − x), and h(t + x) cannot be resonant on a neighborhood of the origin. Discussion. This shows that linear phases are the only possibilities for resonant triples with pairwise independent differentials. The example and exercises show that the phenomenon of resonance is both rare and sen- sitive when viewed from the perspective of perturbing the phases. §9.2. The three wave interaction PDE. Consider the system (∂t + ∂x)u1 = c1 u3 u2 , ∂tu2 = c2 u1u3 , (9.3) (∂t − ∂x)u3 = c3 u1 u2 . with real cj ∈ R \ 0. This equation and its relatives maximizes the intermode interaction. 2 The absence of a term in u1 in the first equation has a consequence that harmonics are not generated in the first mode. Harmonic generation is absent in the other equations too. Multiplying the first equation by a1 u1, the second by a2 u2, and the third by a3 u3, shows that if a1, a2, and a3 are real numbers so that ajcj = 0 then for solutions one has the differential conservation laws P ∂ ∂ a u2 + a u2 + a u2 + a u2 − a u2 = 2 a c u u u = 0 . ∂t 1 1 2 2 3 3 ∂x 1 1 3 3 j j 1 2 3 Xj Integrating dx yields the integral conservation laws, d 2 2 2 a1 u + a2 u + a3 u dx = 0 . dt Z 1 2 3 This is a two dimensional space of conservation laws parameterized by the a. We are interested in complex solutions. For complex solutions, conservation laws involving 2 2 2 |uj| are more interesting than those involving uj as they yield L bounds. The complex analogue of (9.3) with such conservation laws is, ∗ (∂t + ∂x)u1 = c1 u2 u3 ∂tu2 = c2 u1u3 (9.4) ∗ (∂t − ∂x)u3 = c3 u1 u2 . 3 with cj ∈ R \ 0. th ∗ Mutiplying the j equation by ajuj and taking the real part shows that if aj ∈ R satisfy aj cj = 0, then solutions satisfy P 2 2 2 ∗ ∗ ∂t aj|uj| + ∂x a1 |u1| − a3 |u3| = 2 aj cj Re u1 u2 u3 = 0 . Xj Xj If the cj do not all have the same sign, then there are conservation laws of this type with 2 all the aj > 0. This yields an L bound on the solution. On the other hand, if the cj are all positive then initial data with uj(0,x) real and positive yield real solutions such that for all j, uj is nondecreasing along j characteristics. For sufficiently positive data there is finite time blow up. Proposition. Suppose that cj ≥ c > 0 and the real valued initial data satisfy ∀j, ∀|x| ≤ R, uj(0,x) ≥ A > 0. ∞ i. If u(t,x) ∈ C ([0,t∗[×R) is a solution, then uj(t,x) ≥ y(t) for t∗ >t ≥ 0 and |x| ≤ R−t, where y = A/(1 − cAt) is the solution of y′ = cy2, y(0) = A. −1 ii. If T∗ := (cA) is the blow up time for y, and R > T∗, then u blows up on or before time T∗ in the sense that one must have t∗ ≤ T∗. Proof. The second assertion follows from the first. Since the speed of propagation is no larger than 1, the values of u in |x| ≤ R − t are unaffected by the values of the Cauchy data for |x| > R. Therefore, it suffices to prove that uj ≥ y(t) when the data satisfy uj (0,x) ≥ A for all x. Define m(t) := min uj(t,x) . x∈R ,j Since the uj are nondecreasing on j-characteritics, it follows that m(t) is nondecreasing. And, m(0) ≥ A > 0. In addition one has the lower bound obtained by integration along j characteristics, t 2 uj(t,x) ≥ m(0) + c m(t) dt. Z0 Taking the infinum on x yields t t m(t) ≥ m(0) + c m(t)2 d ≥ A + c m(t)2 dt. Z0 Z0 The function y is characterized as the solution of t y(t) = A + c y(t)2 dt. Z0 4 For ǫ > 0 small, let yǫ be the solution of (yǫ)′ = c(yǫ)2 with yǫ(0) = A − ǫ so t yǫ(t) = A − ǫ + c yǫ(t)2 dt. Z0 ǫ It follows that m(t) > y (t) for all 0 ≤ t<t∗. For, if this were not so there would be a ǫ smallest t ∈]0,t[∗ where m(t) = y (t). Then t t yǫ(t) = m(t) ≥ A + c m(t)2 dt > A − ǫ + c (yǫ(t))2 dt = yǫ(t) . Z0 Z0 This contradiction establishes m>yǫ. Passing to the limit ǫ → 0 proves m ≥ y which is the desired conclusion. Only the signs of the cj play a roll in the qualitative behavior of the equation (9.4). Proposition. There is exactly one positive diagonal linear transformation u := (d1 v1 , d2 v2 , d3 v3) , dj > 0 , which transform the the system to the analogous system with interaction coefficients c1, c2, c3 replaced by c c c 1 , 2 , 3 . |c1| |c2| |c3| Proof. The change of variables yields an anlogous system for v with the interaction coefficients replaced by d2 d3 d3 d1 d1 d2 c1 , c2 , c3 . n d1 d2 d3 o Need dj so that d2 d3 c1 d3 d1 c2 d1 d2 c3 c1 = , c2 = , c3 = .