hyunjoong kim 1 & chee han tan 1 January 4, 2018 contents list of figures

Transcription

1 T H E W E N T Z E L- K R A M E R S - B R I L L O U I N W K B ) M E T H O D hyunjoong kim 1 & chee han tan 1 January 4, 2018 contents 1 Introductory Example 3 2 Turning Points Transition layer Matching The opposite case: q t < Wave Propagation and Energy Methods Connection to energy methods Higher-Dimensional Waves - Ray Methods WKB Expansion Surfaces and wave fronts Solution of the Eikonal equation Solution of the transport equation Ray equation Choice of λ Summary for λ = 1/µ Breakdown of the WKB solution Exercises 20 list of figures Figure 1 An example of turning points: quantum tunneling Figure 2 Plot of Airy functions Figure 3 Instructive case of multi-dimensional wave equation Figure 4 Schematic figure of wave fronts in R Figure 5 Rays representing waves propagating inside the earth from a source on the surface of the earth Department of Mathematics, University of Utah, Salt Lake City, UT 1

2 list of figures 2 abstract These notes are largely based on Math 6730: Asymptotic and Perturbation Methods course, taught by Paul Bressloff in Fall 2017, at the University of Utah. Additional examples or remarks or results from other sources are added as we see fit, mainly to facilitate our understanding. These notes are by no means accurate or applicable, and any mistakes here are of course our own. Please report any typographical errors or mathematical fallacy to us by hkim@math.utah.edu or tan@math.utah.edu.

3 introductory example 3 introduction The WKB method, named after Wentzel, Kramers and Brillouin, is a method for finding approximate solutions to linear differential equations with spatially varying coefficients. The origin of WKB theory dates back to 1920s where it was developed by Wentzel, Kramers and Brillouin to study time-independent Schrodinger equation. This often arises from the following problem: d 2 y qɛx)y = 0, dx2 with the slowly varying potential energy. To handle this perturbation problem, the WKB method introduces an ansatz of the expansion term as a product of slowly varying and exponenetially rapidly varying terms. 1 introductory example Consider the differential equation ɛ 2 y qx)y = 0, x 0, 1], 1.1) where q is a smooth function. For constant q R, the general solution of 1.1) is yx) = a 0 e x q/ɛ + b 0 e x q/ɛ, 1.2) and the solution eigther blows up q > 0) or oscillates q < 0) rapidly on a scale of Oɛ). The hypothesis of the WKB method is that the exponential solutions 1.5) can be generalised to obtain an approximate solution of the full problem 1.1). We expect that the solutions of 1.1) to have rapid oscillations on a scale of Oɛ) with slowly-varying amplitude and phase. Then it is natural to start with the following general WKB ansatz: yx) e θx)/ɛα y 0 x) + ɛ α y 1 x) +... ], 1.3) for some α > 0. Here, we assume that the solution varies exponentially with respect to the fast variation. From 1.3) we obtain y { ɛ α θ x y 0 + y 0 + θ xy } e θ/ɛα, y {ɛ ) } 2α θxy ɛ α θ xx y 0 + 2θ x y 0 + θ2 xy e θ/ɛα. 1.4a) 1.4b) Substituting both 1.3) and 1.4) into 1.1) and cancelling the exponential term yields θ ɛ 2 2 x y 0 ɛ 2α + 1 ) ] ɛ α θ xx y 0 + 2θ x y 0 + θ2 xy qx) y 0 + ɛ α y ] = 0, 1.5) Note that Such cancellation is possible due to the linearity of the equation. Balancing leading-order terms in 1.5) we see that α = 1. The O1) equation is the well-known Eikonal equation: θ 2 x = qx), 1.6) and its solutions in one-dimensional) are θx) = ± x qs) ds. 1.7)

4 introductory example 4 To determine y 0 x), we need to solve the Oɛ) equation which is the transport equation: θ xx y 0 + 2θ x y 0 + θ2 xy 1 = qx)y 1, 1.8) The y 1 terms cancel out due to the Eikonal equation 1.6), so 1.8) reduces to θ xx y 0 + 2θ x y 0 = ) The equation 1.9) can be easily solved since it is separable y 0 y 0 = θ xx 2θ x, and it follows its general solution y 0 x) = C θx = ±Cqx) 1/4, 1.10) where C is an arbitrary nonzero constant and the last line follows from 1.7). Hence, a first-term asymptotic approximation of the general solution of 1.1) is yx) qx) a 1/4 0 exp 1 x ) 1 x )] qs) ds + b 0 exp qs) ds, ɛ ɛ 1.11) where a 0, b 0 are arbitrary constants, possibly complex. It is evident that 1.11) is valid if qx) = 0 on 0, 1]. The x-values where qx) = 0 are called turning points and this non-trivial issue will be addressed in the Section 2. Example 1. Choose qx) = e 2x. Then the WKB approximation 1.11) is yx) e x/2 a 0 e iex /ɛ + b 0 e iex /ɛ ] = e x/2 α 0 cosλe x ) + β 0 sinλe x )], where λ = 1/ɛ. With boundary conditions y0) = a, y1) = b, we obtain b e sin yx) e x/2 λe x 1)) a sin λe x ) e)). sin λe 1)) The exact solution of 1.1) with the given qx) can be solved as follows. Making a change of variable x = e x /ɛ = λe x, we obtain x = lnɛ) + ln x) = dx d x = 1 x. Setting Y x) = yx) and using Chain Rule gives dy d x = dy dx dx d x = 1 x Y, d 2 Y d x 2 = y x 2 + y x 2 = 1 x Y + y x 2. Consequently, the equation of Y x) is the zeroth-order Bessel s differential equation x 2 d2 Y dy + x d x 2 d x + x2 Y = 0, and the solution of this is Y x) = c 0 J 0 x) + d 0 Y 0 x) = c 0 J 0 λe x ) + d 0 Y 0 λe x ) = yx), where J 0 ) and Y 0 ) are the zeroth-order Bessel functions of the first and second kinds respectively. Finally, solving for c 0 and d 0 using the boundary conditions yields c 0 = 1 D by 0λ) ay 0 λe)], d 0 = 1 D aj 0λe) bj 0 λ)], where D = J 0 λe)y 0 λ) Y 0 λe)j 0 λ). One can plot the exact solution and the WKB approximation and see that their difference is almost zero, see Fig 4.1 and 4.2 in 2].

5 introductory example 5 To measure the error of the WKB approximation 1.11), we look at the Oɛ 2 ) equation which has the form θ xx y 1 + 2θ x y 1 + θ2 xy 2 + y 0 = qx)y ) The y 2 terms vanish due to the Eikonal equation 1.6), so 1.12) reduces to θ xx y 1 + 2θ x y 1 + y 0 = ) Because the first two terms of 1.13) are similar to the transport equation 1.9), we make an ansatz y 1 x) = y 0 x)wx) and so 1.13) reduces to 2θ x y 0 w + y 0 = ) Suppose qx) > 0 so that θ x is a real-valued function. Substituting 1.10) into 1.14) gives 2Cθ x w θx Rearranging it in terms of w, w = 1 4 ) = d2 C θx dx 2 = d ) Cθxx dx 2θx 3/2. d dx θxx θ 3/2 x ) ) 1, θx and performing integration by parts with respect to x yields wx) = 1 4 x θxx θ 3/2 x ) ) 1 ds = d + 1 θx 4 θxx θ 2 x ) x θ 2 ) xx ds. where d is an arbitrary constant. On the other hand, θ x is a complex-valued function provided that qx) < 0. Substituting θ x = ±i q into wx) gives the result. θ xx = ± i ) qx = iq x 2 q 2 q θ xx θ 2 x θ 2 xx = = iq x 2q q q2 x 4 q) = ± iq x 2 q) 3/2 = q2 x 4q θ 3 x = ±i) 3 q ) 3 = i q) 3/2 θ 3 x θ 2 xx θ 3 x = q 2 x 4iq q) 3/2 = iq2 x 4 q) 5/2. Finally, for small ɛ the WKB ansatz 1.3) is well-ordered provided ɛy 1 x) y 0 x), or ɛwx) 1. In terms of the function qx) and its first derivatives, for x x 0, x 1 ] we will have an accurate approximation if ɛ d + 1 q x x1 )] 32 q 3/2 4 + q x x 0 q dx 1, where := over the interval x 0, x 1 ]. We stress that this condition holds if the interval x 0, x 1 ] does not contain a turning point. Remark 1. The constants a 0, b 0 in 1.11) and d in wx) are determined from boundary conditions. However, it is very possible that these constants depend on ɛ. It is therefore necessary to make sure this dependence does not interfere with the ordering assumed in the WKB ansatz 1.3).

6 turning points 6 Figure 1: An example of turning points: quantum tunneling. Depending on effective potential energy, the solutions have different behavior and need to be matched Taken from Wikipedia Commons). 2 turning points This section is devoted to the analysis of the effect provided that qx) = 0 for some x. One example of it is the tunneling effect in quantum physics. The quantum probability in the well potential, the quantum probability differs with the potential well, see Fig. 1. Then we need to match to different solutions with transition layer. Assume qx) is smooth and has a simple zero at x t, that is qx t ) = 0 and q x t ) = 0. For concreteness, we take q x t ) > 0 and so we expect solutions of 1.1) to be oscillatory for x < x t and exponential for x > x t. We can apply the WKB method on the regions {x < x t } and {x > x t }. More precisely, from 1.11) we have { y L x, x t ) if x < x t, y 2.1) y R x, x t ) if x > x t, where y L x, x t ) = y R x, x t ) = 1 qx) 1/4 a L exp 1 xt ) 1 xt )] qs) ds + b L exp qs) ds, ɛ x ɛ x 2.2a) 1 qx) 1/4 a R exp 1 x ) 1 x )] qs) ds + b R exp qs) ds. ɛ x t ɛ x t 2.2b) An important realization is that these coefficients a L, b L, a R, b R are not all independent. In addition to the two boundary conditions at x = 0 and x = 1, we also have matching conditions in a transition layer centered at x = x t. 2.1 Transition layer Following the boundary layer analysis, we introduce the stretched coordinate x = x x t ɛ β, or equivalently x = x t + ɛ β x. We can reduce 1.1) by expanding the function qx) around the turning point x t by Taylor series: qx) = qx t + ɛ β x) q x t )ɛ β x, since we assume x t is a simple zero. Denote the inner solution by Y x). Transforming 1.1) using d dx = 1 d ɛ β d x,

7 turning points 7 gives the inner equation ) ɛ 2 2β Y ɛ β xq t +... Y = 0, 2.3) where q t := q x t ). Balancing leading-order terms in 2.3) means we require 2 2β = β = β = 2 3. Since it is not clear what the asymptotic sequence should be, we take the asymptotic expansion to be The Oɛ 2/3 ) equation is Y ɛ γ Y 0 x) ) Y 0 xq ty 0 = 0, < x <. 2.5) Performing a coordinate transformation s = q t )1/3 x, 2.5) becomes Airy s equation: d 2 Y 0 ds 2 sy 0 = 0, < s <, 2.6) and this can be solved either using power series expansion or Laplace transform. The general solution of 2.6) is Y 0 s) = aais) + bbis), 2.7) where Ai ) and Bi ) are Airy functions of the first and the second kinds respectively. It is well-known that Aix) = 1 ) ) 1 k + 1 2π ) k 3 2/3 π k! Γ sin k + 1) 3 1/3 x 3 3 k=0 = Ai0) ) 6 x Ai 0) x + 1 ) 12 x Bix) = e iπ/6 Ai xe 2πi/3) + e iπ/6 Ai xe 2πi/3) = Bi0) ) 6 x Bi 0) x + 1 ) 12 x4 +..., where Γ ) is the gamma function. Setting ξ = 2 3 x 3/2, we also have that Aix) Bix) 1 π x 1/4 cos ξ π ) ξ sin ξ π ) ] 4 if x, 1 2 e ξ 1 5 π x 1/4 72 ξ] if x +, 1 π x 1/4 cos ξ + π ) a) 72ξ sin ξ + π ) ] 4 if x, 1 π x 1/4 e ξ ξ] if x b) 2.2 Matching From 2.7), the general solution of 1.1) in the transition layer is q ) ] 1/3 q ) ] Y 0 x) = aai t x 1/3 + bbi t x. 2.9) We now have 6 undetermined constants from 2.2) and 2.9), but these are all connected since the inner solution 2.9) must match the outer solutions 2.2) and this

8 turning points 8 Figure 2: Plot of Airy functions, Aix) in red and Bix) in blue. Wikipedia Commons. This figure comes from will results in two arbitrary constants in the general solution. Since the inner solution is unbounded, we introduce an intermediate variable x η = x x t ɛ η, 0 < η < 2 3, where the interval for η comes from the requirement that the scaling for the intermediate variable must lie between the outer scale, O1) and the inner scale, Oɛ 2/3 ) Matching for x > x t We first change the stretched variable x to the intermediate variable x η : x = x x t ɛ β = x x t ɛ η ɛ β η = ɛη β x η = ɛ η 2/3 x η. Note that x η > 0 since x > x t. From 2.4) and 2.9), the inner solution Y x) now becomes ) Y ɛ γ Y 0 ɛ η 2/3 x η +... = ɛ γ aai q t ) 1/3 ɛ η 2/3 x η ) + bbi = ɛ γ aair) + bbir)] +... ɛ γ a 2 exp πr1/4 23 r3/2 ) + q t ) 1/3 ɛ η 2/3 x η )] +... )] b 2 πr 1/4 exp 3 r3/2, 2.10) where r = q x t ) 1/3 ɛ η 2/3 x η > 0 and the last line follows from 2.8). On the other hand, since x xt +ɛ η x η qs) ds s x t )q t ds x t and = = 2 3 x t q t ] 2 x t +ɛ η x η 3 s x t) 3/2 x t q t ɛ η ) 3/2 x η = 2 3 ɛr3/2, qx) 1/4 qx t ) + x x t )q t] 1/4 = ɛ η x η q t] 1/4 = ɛ 1/6 q t) 1/6 r 1/4,

9 turning points 9 the right outer solution becomes ɛ 1/6 y R q t )1/6 r 1/4 a R exp 23 ) )] 2 r3/2 + b R exp 3 r3/ ) Consequently, matching 2.10) the right outer solution y R with 2.11) the inner solution Y yields the following: γ = 1 6, a R = a 2 ) q 1/6 π t, br = b ) q 1/6 π t. 2.12) Matching for x < x t Because x < x t, we have x η < 0 which introduces complex numbers into the outer solution y L. Using the asymptotic properties of Airy functions as r see 2.8)), the inner solution becomes Y ɛ γ aair) + bbir)] +... ɛ γ a 2 π r 1/4 cos 3 r 3/2 π 4 ) + b 2 π r 1/4 cos 3 r 3/2 + π )] 4 Using the identity cos θ = e iθ + e iθ )/2, a more useful form of the inner expansion Y as r is Y ɛ γ 2 π r 1/4 ae iπ/4 + be iπ/4) e iζ + ae iπ/4 + be iπ/4) e iζ], where ζ = 2 3 r 3/2. On the other hand, since xt x x xt qs) ds s x t )q x t +ɛ η t ds, x η and this follows that xt qs) ds 2 q 3 t ɛ η ) 3/2 2 x η = 3 iɛ r 3/2. Furthermore, qx) 1/4 ɛ η x η q t] 1/4 = ɛ 1/6 q t) 1/6 r 1/4 1) 1/4, the left outer solution becomes y L ɛ 1/6 e iπ/4 q t )1/6 r 1/4 = ɛ 1/6 q t) 1/6 r 1/4 e iπ/4, 2.13) a L e iζ + b L e iζ]. 2.14) Consequently, matching 2.14) the left outer solution y L with 2.13) the inner solution Y yields the following: a L = q t )1/6 2 π ia + b), b L = q t )1/6 2 π a + ib) = iā L. 2.15) From 2.12), it follows that or in matrix form, ] al = a L = ia R + b R 2, b L = a R + i 2 b R, 2.16) b L i 1/2 1 i/2 ] ar b R ]. 2.17)

10 turning points Conclusion Because we assume qt) < 0 for x < x t, this introduces complex numbers on y L : xt x qx) 1/4 = e iπ/4 qx) 1/4 xt qs) ds = i qs) ds In conclusion, we have { y L x, x t ) if x < x t, yx) = y R x, x t ) if x > x t, x where and with y L x, x t ) = = y R x, x t ) = θx) = xt x 1 qx) 1/4 ia R + b R 2 1 qx) 1/4 2a R cos ) e iθx)/ɛ e iπ/4 + a R + ib R 2 1 ɛ θx) π 4 1 qx) 1/4 a R e κx)/ɛ + b R e κx)/ɛ], x qs) ds, κx) = qs) ds. x t ) 1 + b R cos ɛ θx) + π 4 ) e iθx)/ɛ e iπ/4 ] )] Example 2. Consider qx) = x2 x), where 1 < x < 1. The simple turning point is at x t = 0, with q 0) = 2 > 0. One can compute and show that: θx) = x) xx 2) 1 ] 1 2 ln x + xx 2), x < 0 κx) = 1 2 x 1) x2 x) 1 2 arcsinx 1) + π 4, x > The opposite case: q t < 0 In the same fashion, one can obtain by change of variables with a new variable z = x t x see details in Section in 2]) al Consequently, b L ] = y L x) = y R x) = i/2 1 1/2 i ] ar b R ]. 1 qx) 1/4 a L e θx)/ɛ + b L e θx)/ɛ], 1 qx) 1/4 1 2b L cos ɛ κx) π ) 1 + a L cos 4 ɛ κx) + π )]. 4

11 wave propagation and energy methods 11 3 wave propagation and energy methods In this section, we study how to obtain an asymptotic approximation of a travellingwave solution of the following PDE which models the string displacement u xx = µ 2 x)u tt + αx)u t + βx)u, 0 < x <, t > 0, 3.1a) u0, t) = cosωt), 3.1b) The terms αx)u t and βu correspond to damping and elastic support respectively. From the initial condition, we see that the string is periodically forced at the left end and so the solution will develop into a wave that propagates to the right. Observe that there is no obvious small parameter ɛ, but we will extract one from the following observation. In the special case where α = β = 0 and µ equals some constant, 3.1) reduces to the classical wave equation and we obtain the rightmoving plane waves ux, t) = e iwt kx), where the wavenumber k satisfies k = ±ωµ. For higher temporal frequencies ω 1, these waves have short wavelength, that is λ = 2π/k 1. Motivated by this, we choose ɛ = 1/ω and construct an asymptotic approximation of the travelling-wave solution of 3.1) in the case of a high frequency. The WKB ansatz is assumed to be ux, t) exp i wt w γ θx) u 0 x) + 1 }{{} w fast oscillation γ u 1x) }{{} slowly-varying Substituting 3.2) into 3.1) we obtain 3.2) ω 2γ θx 2 u0 + w γ u ) + iw γ θ x x u ) + d dx iωγ θ x u ) = µ 2 ω 2 u 0 + ω γ u ) iωα u ) + β u ). Balancing the first terms on each side of this equation gives γ = 1. The Oω 2 ) = O1/ɛ 2 ) equation is the Eikonal equation: and its solutions are θ 2 x = µ 2 x), 3.3) θx) = ± x 0 µs) ds. 3.4) We choose the positive solution as we are considering the right-moving waves. The Oω) = O1/ɛ) equation is the transport equation: θ 2 xu 1 + iθ x x u 0 + i θ x x u 0 + θ xx u 0 ) = µ 2 u 1 iαu ) The u 1 terms cancel out due to the Eikonal equation 3.3), so 3.5) reduces to θ xx u 0 + 2θ x x u 0 = αu 0,. 3.6) With θ x = µx), we can rearrange 3.6) and obtain a first order ODE in u 0 : ) µx + α x u 0 + u 0 = 0, 3.7) 2µ

12 wave propagation and energy methods 12 which can be solved using the method of integrating factor. The integrating factor is given by x ) ) µs s) + αs) 1 x ) αs) Ix) = exp ds = µx) exp 2µs) 2 0 µs) ds, 0 and so 3.7) can be written as d dx Ix)u 0) = 0, u 0 = a 0 Ix) = a 0 exp 1 x ) αs) µx) 2 0 µs) ds. 3.8) Finally, imposing the boundary condition at x = 0 we obtain a first-term asymptotic expansion of the travelling-wave solution of 3.1) µ0) ux, t) µx) exp 1 x ] αs) x ) 2 0 µs) ds cos ωt ω µs) ds. 3.9) 0 Observe that in 3.9) the amplitude and phase of the travelling wave depend on the spatial position x. Interestingly, 3.9) is independent of βx)! 3.1 Connection to energy methods Energy methods are extremely powerful in the study of wave-related problems. To determine the energy equation in this case, we multiply 3.1) by u t u t u xx = µ 2 x)u t u tt + αx)u 2 t + βx)uu t, and one can rewrite it as x u t u x ) 1 2 t ) u 2 x = 1 ) 2 µ2 x) t u 2 t + αx)u 2 t + βx) t u 2). Collecting time derivative and spatial derivative terms yields 1 ) t 2 µ2 x) u 2 t βx)u2 + 1 ) ] u 2 x x u t u x ) = αx)u 2 t, 2 }{{}}{{}}{{} :=Sx,t) :=Φx,t) :=Ex,t) where Ex, t) is the energy density, Sx, t) is the energy flux and Φx, t) is the dissipation function at position x and time t. Subsequently, we can achieve the energy equation t Ex, t) + x Sx, t) = Φx, t). 3.10) We are interested in the energy over some spatial interval of the form x 1 t), x 2 t)] with assuming x 1 t) < x 2 t). It follows from Leibniz s rule d x2 t) x2 t) Ex, t) dx = Ex 2 t), t) x 2 Ex dt 1 t), t) x 1 + t Ex, t) dx. x 1 t) x 1 t) Imposing the energy equation 3.10) gives d x2 t) x2 t) E dx = Ex 2 t), t) x 2 Ex dt 1 t), t) x 1 Sx 2 t), t) + Sx 1 t), t) Φx, t) dx. x 1 t) x 1 t) 3.11) The term Ex j t), t) x j is the change of energy due to the motion of the endpoint, Sx j t), t) is the flux of energy across the endpoint due to wave motion and the last integration term on the right-hand side is the energy loss over the interval due to dissipation.

13 wave propagation and energy methods 13 The WKB solution can be written in the more general form: ux, t) Ax) cos wt ϕx)], ϕx) = ωθx). 3.12) with slowly changing amplitude Ax) and rapidly changing phase ϕx). It follows that Ex, t) 1 ) 2 A2 µ 2 ω 2 + ϕ 2 x sin 2 ωt ϕx)], 3.13a) Sx, t) ωϕ x A 2 sin 2 ωt ϕx)], Φx, t) αω 2 A 2 sin 2 ωt ϕx)]. 3.13b) 3.13c) Suppose we choose x i t) satisfying x i = ω ϕ x x i ), 3.14) that is the reference frame moves as phase velocity. It is worth noting that the phase velocity of plane waves is v p = w x kx) = w k. A curve satisfying 3.14) in the x t plane are called phase lines. Now, imposing 3.14) to 3.13) gives Eẋ S 1 2 and it follows that Eẋ S 1 2 ωa 2 ] µ 2 ω 2 + ϕ 2 x sin 2 ωt ϕx)] ωϕ x A 2 sin 2 ωt ϕx)], ϕ x ωa 2 ] µ 2 ω 2 ϕ 2 x sin 2 ωt ϕx)] = 0, ϕ x since θx) = ϕx)/ω satisfies the Eikonal equation 3.3). Hence, if x 2 x 1 = O1/ω) then it follows from 3.11) that de tot 0, that is the total energy remains dt constant to the first term) between any two phase lines x 1 t), x 2 t) that are O1/ω) apart. Recall the energy equation that t E + x S = Φ. Averaging the energy equation over one period in time results in x 2π/ω 0 ) 2π/ω Sx, t) dt = Φx, t) dt, 0 where the average of t E over one period vanishes using 3.13) for E. Substituting 3.13) for S and Φ, we obtain and it follows that x ϕ x A 2) = αωa 2, θ xx A + 2θ x A x = αa. This implies that A = u 0 since the last equation is precisely the transport equation 3.6). Physically, this means that the transport equation corresponds to the balance of energy over one period in time.

14 higher-dimensional waves - ray methods 14 Figure 3: Instructive case of multi-dimensional wave equation. In R 2, the wave propagates from the circle with radius a. 4 higher-dimensional waves - ray methods The extension of the WKB method to higher dimensions is relatively straightforward, but the equations could be difficult to solve explicitly. Consider the multidimensional wave equation 2 u = µ 2 x) 2 t u, x R n, n = 2, ) We look for time-harmonic solutions ux, t) = e iωt Vx) and 4.1) reduces to the Helmholtz equation 2 V + ω 2 µ 2 x)v = ) It is more instructive to have some understanding of what properties the solution has and how the WKB approximation takes advantage of them. Suppose µ is some constant and we want to solve 4.2) in the region exterior to the circle x = a in R 2. Exploiting the geometry leads to the choice of polar coordinates x = ρ cosϕ), y = ρ sinϕ). We impose the Dirichlet boundary condition V = f ϕ) at ρ = a and the Sommerfeld radiation condition which ensures that waves only propagate outward from the circle: ρ ρ V iωµv ] = 0 for ρ. Using separation of variables, the general solution of 4.2) is given by Vρ, ϕ) = n= α n H 1) n ) ωµρ) e inϕ, 4.3) H n 1) ωµa) where H n 1) is the Hankel function of first kind and the α n are determined from the boundary condition at ρ = a. It is known that for large values of z H n 1) 2 z) πz exp i z nπ 2 π )). 4 Consequently, in the regime of higher frequency ω 1 4.3) reduces to Vρ, ϕ) f ϕ) a ρ eiωµρ a). 4.4)

15 higher-dimensional waves - ray methods 15 Thus we have a WKB-like solution for constant µ. Radial lines in this example correspond to rays and from 4.4) we see that along a ray so that ϕ is fixed), the solution has a highly oscillatory component that is multiplied by a slowly varying amplitude V 0 = f ϕ) a/ρ that decays as ρ increases. 4.1 WKB Expansion We first specify the domain and boundary conditions. Equation 4.2) is to be solved in a region exterior to a smooth surface S, where S encloses a bounded convex domain. This means that there is a well-defined unit outward normal at every point on the surface. We impose the Dirichlet boundary condition Vx 0 ) = f x 0 ), x 0 S, 4.5) and focus only on outward propagating waves. For higher frequency waves, we take a WKB ansatz of the form Vx) e V iωθx) 0 x) + 1 ] ω V 1x) ) Then we have V {iω θv 0 + i θv 1 + V } e iωθ, { 2 V ω 2 θ θv 0 + ω Substituting 4.7) into 4.2) gives 4.7a) ) θ θv 1 + 2i θ V θv ) ω 2 θ θv 0 + µ 2 V 0 + ] ω θ θv 1 + 2i θ V 0 + i 2 θv 0 + µ 2 V 1 + O1) = 0, and rearranging we find that θ θ µ 2) V ω θ θ µ 2) V 1 i 2 θv 0 2i θ V 0 ] + O 4.7b) ) 1 ω 2 = 0. The O1) equation is the Eikonal equation which is now nontrivial to solve: θ θ = µ ) After cancelling the V 1 term using the Eikonal equation 4.8), the O1/ω) equation is the transport equation: ) 2 θ V θ V 0 = ) Both ±θ are solutions to the Eikonal equation and we choose positive θ since this corresponds to the outward propagating waves. } e iωθ. 4.2 Surfaces and wave fronts The usual method method for solving the nonlinear Eikonal equation 4.8) is to introduce characteristic coordinates. More precisely, we use curves that are orthogonal to the level surfaces of θx) which are also known as wave fronts or phase fronts.

16 higher-dimensional waves - ray methods 16 Figure 4: Schematic figure of wave fronts in R 3 and path followed by one of the points in the wave front. This figure comes from Fig 4.11 in 2]. First, note that the WKB approximation of 4.1) has the form ux, t) e iωθx) ωt) V 0 x). We introduce the phase function Θx, t) = ωθx) ωt. Suppose we start at t = 0 with the surface S c = {θx) = c}, so that Θx, 0) = ωc. As t increases, the points where Θ = ωc change, and therefore points forming S c move and form a new surface S c+t = {θx) = c + t}. We still have Θx, t) = ωc, that is S c to S c+t has same phase. The path each point takes to get from S c to S c+t is obtained from the solution of the Eikonal equation and in the WKB method these paths are called rays. The evolution of the wave front generates a natural coordinate system s, α, β) where α, β comes from parameterising the wave front and s from parameterising the rays. Note that these coordinates are not unique as there are no unique parameterisation for the surfaces and rays. It turns out that determining these coordinates is crucial in the derivation of the WKB approximation. Example 3. Suppose we know a-priori that θx) = x x. In this case, the surface S c+t is described by the equation x 2 = c + t, which is just the sphere with radius c + t. The rays are now radial lines and so the points forming S c move along radial lines to form the surface S c+t. To this end, we use a modified version of spherical coordinates: with x, y, z) = ρs) sin α cos β, sin α sin β, cos α), 0 α < π, 0 β 2π, 0 s. The function ρs) is required to be smooth and strictly increasing. Examples are ρ = s, ρ = e s 1 or ρ = ln1 + s). An important property of the preceeding modified spherical coordinates is that s, α, β) forms an orthogonal coordinate system. That is, under the change of variables x = Xs, α, β), the vector s X tangent to the ray is orthogonal to the wave front S c+t. We now in the opposite case: we need to find θx) given conditions on the map Xs, α, β). Observe the degree of freedom on specifiy X!

17 higher-dimensional waves - ray methods Solution of the Eikonal equation In what follows, we will assume that s, α, β) forms an orthogonal coordinate system. This means that a ray s tangent vector s X points in the same direction as θ when x = Xs, α, β), or equivalently X s = λ θ, 4.10) where λ is a smooth positive function, to be specified later. Without loss of generality, we assume that the rays are parameterized so that s 0. One should not confuse s with the arclength parameterization. Along a ray, s θx) = s θx) = θ s X = λ θ θ. Therefore we can rewrite the Eikonal equation as s θ = λµ 2, 4.11) which can be integrated directly to yield s θs, α, β) = θ0, α, β) + λµ 2 dσ, 4.12) 0 assuming we can find such a coordinate system s, α, β). This amounts to solving 4.10) which is generally nonlinear and requires the assistance of numerical method. Nonetheless, we still have the freedom of choosing the function λ. 4.4 Solution of the transport equation It remains to find the first term V 0 of the WKB approximation 4.6). Using 4.10) we have del s V 0 = V 0 s X = λ V 0 θ. Consequently we can also rewrite the transport equation 4.9) as ) 2 s V 0 + λ 2 θ V 0 = ) Using the identity ) J s = J 2 θ, 4.14) λ where J = x, y, z) s, α, β) is the Jacobian of the transformation x = Xs, α, β), we can rewrite 4.13) as ) J 2J s V 0 + λ s V 0 = 0. λ This follows that ) 1 s λ JV2 0 = 0, and its general solution is V 0 x) = a 0 λx) Jx). 4.15)

18 higher-dimensional waves - ray methods 18 Imposing the boundary condition V 0 x 0 ) = f x 0 ), we obtain λx)jx V 0 x) = f x 0 ) 0 ) λx 0 )Jx), 4.16) and this is true provided θ0, α, β) = 0 in 4.12). Otherwise we will get an exponential term!) We now prove the identity 4.14) in 2D but this easily extends to 3D. Main idea comes from Exercise 4.43 in 2]. The transformation in 2D is x = Xs, α) and its Jacobian is J = x, y) s, α) = sx α y α x s y. Using the ray equation 4.10) s J = s λ x θ) α y + s x α λ y θ) α λ x θ) s y α x s λ y θ), and the chain rule yields Rearranging it and this follows that s J = α y s x x + s y y ]λ x θ) α x s x x + s y y ]λ y θ) s y α x x + α y y ]λ x θ) + s x α x x + α y y ]λ y θ). s J = J x λ x θ) + J y λ y θ), s J = J λ θ). 4.17) For a smooth function qx), one can achieve from the chain rule s qj) = q s xj + q s J. Imposing the ray equation 4.10) and 4.17) gives and it follows that s qj) = q λ θ) + q λ θ)] J, s qj) = J qλ θ). 4.18) Finally, setting q = 1/λ leads the equation to the identity 4.14). 4.5 Ray equation Set X = X 1, X 2, X 3 ), divide the ray equation 4.10) by λ and differentiating the resulting equation yields ] 1 X i = ) θx), s λ s s x i and applying the chain rule 1 s λ ] X i = s 3 x i s j=1 Imposing the ray equation again ] 1 X i = s λ s x i θ θx) x j x i ) λ θ),

19 higher-dimensional waves - ray methods 19 and it follows that ] 1 X i = 1 s λ s 2 λ θ θ). x i Invoking the Eikonal equation 4.8) gives ] 1 X i = 1 s λ s 2 λ µ 2. x i In vector form, this equals ) 1 s λ s X = λµ µ. 4.19) We require two boundary conditions as 4.19) is a second order equation in s. Each ray starts on the initial surface S. Given any point on x 0 S, its ray satisfies X s=0 = x ) The second boundary condition is typically X s = λ 0 µ 0 n 0, 4.21) s=0 where n 0 is the unit outward normal at x 0, λ 0 = λ0, α, β) and µ 0 = µ0, α, β). 4.6 Choice of λ Taking the dot product of the ray equation against s X, we have X s X s = λ2 θ θ = λ 2 µ 2. Let l be the arc length along a ray. Then l = s 0 s X ds = s 0 λµds. Hence, if we choose λ = 1/µ then s equals the arclength. Another common choice is λ = Summary for λ = 1/µ The ray equation 4.10) becomes µ s ) s X = µx), 4.22) with X s=0 = x 0 S and s X s=0 = n 0. Once this is solved, the phase function becomes ΘX) = and the amplitude is s 0 V 0 x) = f x 0 ) µx) dσ, 4.23) µx 0 )Jx 0 ) µx)jx). 4.24) Finally, the WKB approximation is µx ux, t) f x 0 ) 0 )Jx 0 ) s )] µx)jx) exp iω t + µxσ)) dσ, 4.25) 0 where s is the value for which the solution of satisfies Xs) = x.

20 exercises 20 Example 4. Suppose µ is some constant µ 0. The ray equation 4.7) becomes Also, 2 X s 2 = 0 = Xs) = x 0 + sn 0. s θ = µ 0 dσ = µs. 0 Thus, given a point x on the ray s = n 0 x x 0 ) the WKB approximation is Jx ux, t) f x 0 ) 0 ) Jx) exp i k x x 0) ωt)], where k = µ 0 ωn 0. In R 2, when the boundary surface is a circle of radius a, one finds that the Jacobian in polar coordinates is just ρ and aρ ux, t) f x 0 ) e iωµρ a) t). 4.8 Breakdown of the WKB solution Fails at turning points x where µx) = 0. But this can be handled analogously to 1D case boundary layer). A more likely complication arises when J = 0. Points where this occur are called caustics - arise when two or more rays intersect. Breakdown of the characteristic coordinates s, α, β). If a ray passes through a caustic, one picks up an additional factor in WKB solution of the form e imπ/2, where m depends on the rank of J at the caustic. A less obvious breakdown occurs when Xs) = x has no solution. This happens with shadow regions - ray splitting. 5 exercises Problem 1. Use the WKB method to find an approximation of the following problem on x 0, 1]: ɛy + 2y + 2y = 0, y0) = 0, y1) = 1. Proof. The WKB ansatz gives yx) e θx)/ɛα y 0 x) + ɛ α y 1 x) + ]. To apply it into the given equation, find y e θx)/ɛα ɛ α θ y 0 + y 0 + θ y 1 ) + ɛ α y 1 + ], and ] y e θx)/ɛα ɛ 2α θ ) 2 y 0 + ɛ α θ y 0 ) + θ y 0 + θ y 1 )] +. To balance the given equation, we require α = 1. Substituting them into the given equation and collecting Oɛ 1 ) gives the Eikonal equation θ ) 2 + 2θ = )

21 exercises 21 Balancing O1) terms yields and it follows that θ y 0 ) + θ y 0 + θ y 1 ) + 2y 0 + θ y 1 ) + 2y 0 = 0, θ + 1)y 0 + y 0 = ) Then one can achieve the first term of its WKB expansion yx) A 1 e x x 2x + A 2 e ɛ. 5.3) Imposing the left-hand boundary condition gives A 1 + A 2 = 0, and the right-hand boundary condition Then one can achieve A 1 e 1 + A 2 e 1 2 ɛ = 1. A 1 = 1 e 1 e 1 2/ɛ, A 2 = A 1. Substituting it into the general solution yields y W x) A 1 ɛ) e x e x1 2/ɛ)). 5.4) It is known that the match asymptotic expansion of given problem is y M x) e e x e 2x/ɛ). 5.5) One clear difference is that y W satisfies both boundary conditions but y M 1) = 1 e 1 2/ɛ, It implies that y M does not satisfies the right-hand boundary condition exactly but O1) order. Moreover, for sufficiently small ɛ > 0, A 1 ɛ) e, 1 2 ɛ 2 ɛ, and it implies that y W and y M are similar when ɛ is sufficiently small. Problem 2. Consider seismic waves propagating through the upper mantle of the earth from a source on the earth s surface see Fig. 5). We want to use a WKB approximation in R 3 to solve the equation 2 v + w 2 µ 2 r)v = 0, where µ has spherical symmetry. Take λ = 1/µ. 1. Use the ray equation to show the vector p = r µ s r), is independent of s. Hence, show that rµ sinχ) is constant along a ray, where χ is the angle between r and s r.

22 exercises 22 Figure 5: Rays representing waves propagating inside the earth from a source on the surface of the earth. Proof. It suffices to show that the partial derivative of p with respect to s is zero. Applying product rule gives s p = s r µ s r) + r s µ s r), and one can rewrite it by ray equation s p = µ s r s r) + r µr). Using the fact that µ = µ r/r gives s p = µ s r s r) + µ r) r r) = 0, r because the product of parallel vectors is zero. This implies that p is independent of s. Then it follows that p is also independent of s. One can obtain p = r µ s r sin χ = rµ s r sin χ. 5.6) Taking the dot product of the ray equation with respect to s r gives s r s r = λ 2 µ 2 = ) Combining 5.6) and 5.7) yields p = rµ sin χ, and it shows that rµ sin χ is constant along a ray. 2. Part 1 implies that each ray lies in a plane containing the origin of the sphere. Let ρ, ϕ be polar coordinates of this plane. It follows that for a polar curve ρ = ρϕ), the angle χ satisfies sinχ) = ρ ρ 2 + ϕ ρ) 2. Assuming ϕ = 0, show that ρ dr ϕ = ϕ 0 + ρ 0 r µ 2 r 2 κ, 2 where ρ 0, ϕ 0, κ are constants.

23 exercises 23 Proof. On the polar coordinate, it follows that r = ρcos ϕ, sin ϕ), s r ϕ r = ϕ ρcos ϕ, sin ϕ) + ρ sin ϕ, cos ϕ), and r = ρ, ϕ r = ρ 2 + ϕ ρ) 2. This follows that sin χ = r ϕr r ϕ r = ρ ρ 2 + ϕ ρ) ) From part a), rµ sin χ is constant along a ray and this gives that rµ sin χ = ρ 2 µ ρ 2 + ϕ ρ) 2 = κ, 5.9) where κ is an constant. Then one can solve it for ϕ ρ ϕ ρ = ρ κ ρ 2 µ 2 κ 2, and performing chain rule yields that ρ ϕρ) = Integrating it with respect to ρ gives that κ ρ ρ 2 µ 2 κ ) ρ dr ϕ = ϕ 0 + κ ρ 0 r r 2 µ 2 κ. 5.11) 2 3. Using the definition of arc length, show that for a polar curve ds = ρ 2 + ϕ ρ) 2 dϕ. Combining this result with part 2, show that the solution of the Eikonal equation is given by θ = 1 κ ϕ ϕ 0 µ 2 ρ 2 dϕ. Proof. Since the choice of λ, then s r ds = 1 ds. Moreover, ϕ r dϕ = ρ 2 + ϕ ρ) 2 dϕ. By the definition of arc length, one can obtain that dl = 1 ds = ρ 2 + ϕ ρ) 2 dϕ. 5.12) In case of choosing λ = 1, then we have s r ds = µ ds and it follows that dl = µds = ρ 2 + ϕ ρ) 2 dϕ. Recall the Eikonal equation s θ = λµ 2 = µ.

24 references 24 Performing integration with the change of variables in 5.12) yields θ = ϕ µ ϕ 0 ρ 2 + ϕ ρ) 2 dϕ, and one can obtain θ by imposing 5.9) θ = ϕ ϕ 0 µ ρ2 µ κ dϕ = 1 κ ϕ ϕ 0 µ 2 ρ 2 dϕ. 5.13) references 1] Bressloff P C 2014 Waves in Neural Media Springer 2] Holmes M H 2013 Introduction to Perturbation Methods Springer 2/e 3] Pikovsky A, Rosenblum M and Kurths J 2003 Synchronization: A Universal Concept in Nonlinear Sciences Cambridge University Press