In this work we show an error estimate for a first order Gaussian beam at a fold caustic, approximating time-harmonic waves governed by the Helmholtz equation. For the caustic that we study the exact solution can be constructed using Airy functions and there are explicit formulae for the Gaussian beam parameters. Via precise comparisons we show that the pointwise error on the caustic is of the order where k is the wave number in Helmholtz.
Gaussian beam superpositions is a high frequency asymptotic approximation for solutions of wave equations [8]. It is used in numerical methods to simulate waves in the high frequency regime. Unlike standard geometrical optics, the Gaussian beam approximation does not break down at caustics, which is one of its main advantages.
In this paper we consider error estimates for the approximation in terms of the wave number . Error estimates for Gaussian beams are known in a number of settings. See for instance [4,5] and the references therein. The main result is that, in and Sobolev norms, the relative error of first order beams decays as , independently of dimension and regardless of the presence of caustics. This has been shown for general strictly hyperbolic partial differential equations and the Schrödinger equation [4,5] as well as the Helmholtz equation [2]. The better rate is typically observed in numerical computations and has been shown in for the Schrödinger equation [9], and also in for the Schrödinger and the acoustic wave equation on sets strictly away from caustics [4]. Similar estimates have also been derived for higher order beams. For p-th order beams the rates are and respectively. There are, however, no precise, pointwise, error estimates for the solution at a caustic. In particular, for first order beams it has not been shown that this error vanishes as , although there is ample numerical evidence to this effect; see for instance [3].
The purpose of this paper is to show such an error estimate for a typical fold caustic in two dimensions. More precisely, we consider the Helmholtz equation
We assume there is an incident wave from making an angle with the x-axis. Moreover, at it has the amplitude envelope , so that
This wave will generate a fold caustic at the line where
Figure 1 shows a representative solution. In Section 3 we make this physical situation precise and derive an exact solution using Airy functions on . We subsequently study the solution in the region and compare it at to an approximation using Gaussian beams, denoted by . (Note that in this region; we do not make comparisons elsewhere, as the equation then no longer models the physical situation.) The main result is the following theorem.
The fold caustic: example of solution with ray tracing picture for .
Suppose A is a Schwartz class function,, and that. Let u be the exact solution of (
1
) as defined in Section
3
anda first order Gaussian beam approximation detailed in Section
4
. Then there is a constant C independent of k such that
This result hence confirms that first order Gaussian beams do converge pointwise at the caustic. Moreover, since the solution itself grows as at this caustic [6], the relative error is , the same as away from the caustic. We conjecture that this will be the case also for more general caustics.
The paper is organized as follows. In Section 2 notations are established and some preliminary results are discussed. In Section 3 the exact solution is defined and a formula for it is derived. In Section 4, the corresponding Gaussian beam approximation is introduced. Sections 5, 6 and 7 contain estimates of the Gaussian beam parameters, the phase, various oscillatory integrals, as well as the exact solution and the Gaussian beam approximation. In Section 8 the proof of the main result in Theorem 1.1 is carried out. Finally, in Section 9 some properties of Airy functions are presented.
Preliminaries
In the analysis we use a k-scaled Fourier transform and indicate it with a hat mark on the function,
The corresponding scaled inversion formula reads
We also have
We will frequently make use of a smooth, even, cut-off function which we denote . It is defined as
This is used to divide integrals into subdomains and to regularize the Fourier transform of functions in , the space of tempered distributions. For example, if f is in , but not in , the definition (2) must be interpreted in distributional sense. We then let and consider instead the Fourier transform of the compactly supported function , which is well-defined by (2) for all . The following Lemma shows that the limit as gives us the Fourier transform in .
Letand setfor any fixed real numbersand. Then, withas defined above,in. Moreover, ifthen, again in,
The short proof is found in the Appendix. In particular, if is defined pointwise, the Lemma shows that
We also introduce some notation that will prove useful later on in the paper. We let
so that is the unit vector pointing in the propagation direction of the incident wave. Following Theorem 1.1 we will assume, throughout the paper, that . This translates to bounds on and of the form
for some and . Moreover, we let
be the distance to the caustic. Finally, we introduce the polynomial q, which is related to the geometrical spreading of the rays,
It will be used frequently in the analysis.
Expression of the exact Helmholtz equation solution
In this section we define an exact solution to the Helmholtz equation for the physical setup described in the introduction. Using a property of the Airy function we deduce a decomposition of the solution into forward and backward going waves.
We consider a solution u to (1) which is a tempered distribution on , i.e. . The solution then has a k-scaled Fourier transform in y which we denote . Upon Fourier transforming also (1) in y, we obtain an ODE for ,
The only tempered distribution solution to this ODE is given by
where Ai is the Airy function of the first kind, and is a function to be determined. This solution is thus a bounded solution. The Airy function in this expression contains waves going both forward and backward. In the sequel, we will choose the function as defined in (10). To arrive at this choice, we first note that when
it holds for all z that [7, Eq. (9.2.14)],
This follows since and solve the same ODE as given that . We then introduce the scaled variables
such that
To further understand this decomposition, we note that the asymptotics of the Airy function (in the angular sector ) is
Therefore, upon defining the phase we have, when ,
Thus, the phases of the two expressions, and their gradients, have opposite signs, meaning that the two terms in (9) represent a decomposition into forward and backward going waves in the region . More precisely, the solution is fully known when is given, and one assumes that this solution of for all x (including where the velocity grows) is the sum of an incoming and an outgoing wave of the form and , respectively. This leads us to define
which are three particular solutions of (8), since each term in (9) solve the ODE. These solutions are normalized such that they equal one for . The solutions and represent forward and backward going waves. Among the three solutions, only is bounded, since is bounded, while are not even in , since includes also the unbounded second kind Airy function , cf. (72).
We are looking for the solution
We do not know , just that the incoming part of represents the incident plane wave. We therefore write it as a sum of an incoming and an outgoing part
where we define as the k-scaled Fourier transform in y of the incoming wave with amplitude A and direction Θ (recall ),
We then want to find such that
To achieve this, it is necessary and sufficient that the values of the functions and their derivatives agree at , since both sides satisfy the same second order ODE. This gives us the linear relations
from which we can deduce
It follows that
The decomposition is not valid at the roots of . Another form is available, which is valid at all points. It is given in the following Lemma.
One can expressas follows
To simplify notation we write . Then, using (9),
Hence,
The first identity follows upon differentiating this expression with respect to x. The second identity is then given by another application of (9). □
It follows now that
Since we know we can thus express the full solution as
where we defined
In this expression, one notices that the denominator never vanish because all the roots of the Airy function are on the negative real axis.
Finally, since
we get
We write this as
where
Construction of the Gaussian beam approximation
In this section we derive expressions for a first order Gaussian beam approximation to the solution of (1). A Gaussian beam is a high frequency asymptotic solution to the Helmholtz equation. To model a general solution of (1), superpositions of Gaussian beams are used. We give the general form of a Gaussian beam and their superposition in below. The derivations of the expressions can be found in [2].
The Helmholtz equation with a general index of refraction reads
When is real, the equation models wave propagation, but it has a well-defined solution also when is imaginary. However, Gaussian beams can only be defined for real .
A first order Gaussian beam for (11) has the form
where is the amplitude, the reference phase, the phase gradient and the phase Hessian. Moreover, is the central ray, which agrees with the rays of geometrical optics. An example of a Gaussian beam is shown in Fig. 2. In (12) the parameter s depends on the point of evaluation . Normally one takes the s-value for the point on the central ray that is closest to , as indicated in Fig. 2. However, in the analysis below we make a simpler choice. By a result in [8] the Hessian M will always have a positive definite imaginary part. The solution will therefore be a “fattened” version of the central ray, with a Gaussian profile normal to the ray with a width determined by M.
A Gaussian beam starting at with direction . The central ray γ is indicated with a solid line.
The s-dependent parameters in the Gaussian beam are all given by ODEs [2], as follows
The initial data for γ and is given by the starting point and direction of the beam,
In order to form an admissible Gaussian beam, must always satisfy
The choice of and the precise form of the incoming wave finally determine the initial data for S and a. We come back to this issue below.
To build more general solutions we use superpositions of Gaussian beams. We assume that the incoming wave is known along a curve Γ in , which we can parameterize with the parameter z, so that . For each point on Γ we launch one Gaussian beam in the direction of the wave at that point. The parameters of the beams then also depend on z and we write , , etc. This gives the beams , from which we finally construct the Gaussian beam superposition
See [2] for more details. In a numerical scheme the z-variable is discretized and for each discrete value, the ODEs (13) are solved with a numerical ODE method. The superposition (15) is subsequently computed using numerical quadrature.
Expressions for Gaussian beam parameters
In (1) we have the index of refraction . The ODEs (13) can be solved explicitly and we get analytic formulae for all parameters in the Gaussian beam. To show this we let and and also recall that , where . Since we let the beam start at , we set . With the particular choice of n it follows from (13) that and . Hence,
For the positions, we get and . By also using the initial data we obtain
The caustic is located at the point where the ray turns back, i.e. where , which gives and
Note that all the rays are confined to the region , where the index of refraction is real-valued. The fact that is complex-valued for therefore does not affect the Gaussian beams.
We also need to compute the coefficients corresponding to the phase S, the second derivative of the phase M and the amplitude a. We have
so the phase is a third order polynomial,
For M we have the Riccati equation
with the solution
The matrix must satisfy the conditions in (14). We pick
Note that P, Q are symmetric, P is the orthogonal projection on , and . Moreover, .
Next, one checks that
where
We note that q is related to the geometrical spreading of the beams. Further manipulations, using the facts that , and reveals that can be written simply as
We let be the elements of M and deduce that
Finally, for a,
We note that if and are eigenvalues of , then and
It follows that
The last thing needed to make the expression (12) for the Gaussian beam well defined, is to decide which s-value to use for a given , i.e. the function . As mentioned above, this is normally taken to be the s-value for the point on the central ray that is closest to . Here, however, to simplify, we just take the s-value for the point of the curve that has the same y-coordinate; see Fig. 3.
The simplified map .
With this leads to
Then and (12) becomes
with , , , , and given above.
The value of crosses the negative real axis when . To find a better branch cut for the square root in the expression (18) for we note first that the equation
has the unique solution . Therefore, the equation with only has a solution if . Hence, never crosses the line , which we therefore use as branch cut. This guarantees a smooth dependence of the Gaussian beam on s for all . It can be written as if is the usual square root with branch cut along the negative real axis.
Gaussian beam superposition
We will now prepare the superposition. The initial curve Γ is simply the y-axis so that . By assumption the incoming wave propagates in the same direction for all z. Moreover, the same initial data for M is used for all z. This means that x, ξ, η, M and the geometrical spreading parameter q are all independent of z. Only y, S and a depend on z. We obtain from (12),
To derive the initial data , for and we consider the trace of and on . To get explicit formulae we also let be the s-value for the point on the curve that has the same x-value, i.e. . That gives
and for ,
To match this with the incoming wave on , i.e. , we take
and
Thus the expressions for the Gaussian beam coefficients are
This gives us the simplified expression for ,
which, together with (19a)–(19g), define via
We will continue now to simplify (20) and compute the k-scaled Fourier transform of in y. Since and by (6) we have
where
Then we can write
Since the beam v is always integrable in z, so that in (21) is well-defined. However, there is no guarantee that is in ; in general it is not. When we compute its Fourier transform we therefore use Lemma 2.1 and (3). By Lemma 5.1 and 5.2 below, q is bounded away from zero and is strictly positive. Hence, . Therefore,
where we can choose , that is, and in Lemma 2.1. We then compute
Here we made the change of variables and . Moreover,
where we have introduced the Gaussian beam phase as,
Since by (17) and , this finally gives
where
and
Then
with
Properties of the amplitude and phases
In this section we collect a series of estimates that we will need for the final proof of the magnitude of the Gaussian beam error.
Geometrical spreading
Here we show some properties of the q-polynomial in (7) that relates to the geometrical spreading, repeated here for convenience,
We have
There are positive constantsand, independent ofand, such thatFurthermore, there are constantsindependent of θ andsuch that
We first show that q has no real root for the considered values of . Suppose therefore that q has a real root . Then the real and imaginary parts of reads
The only solutions to this system are , which are both ruled out by (5). Let , which is then continuous and non-zero for all θ. For large θ it is bounded from below and above since . In fact, there is a constant such that
uniformly in , because of the bound (5). We can then take and as
The stated bound then follows.
For the second statement, we observe that there exists a sequence of polynomials of degree n such that
given by the recursion
thanks to
Then, by the lower bound on and (5), there are constants and depending on n but independent of θ and such that
This shows the lemma. □
Phase
In this section we define the two phase functions and that turn up in our analysis and show a few properties of them. The first one, , is defined as
This has a close connection to the Airy function and we call it the Airy phase. Indeed, is the Fourier transform of in and therefore, using the regularization of Lemma 2.1,
The second phase is the Gaussian beam phase (22) derived in the previous section. It can be written as a sum of and a complex correction, by (16),
We start by looking at the part of the Gaussian beam phase.
Forit holds thatfor some constants,independent of θ and.
From (27) we have
showing the first limit in (28). Moreover,
which similarly implies the second limit in (28). For (29) we have by (16),
To show (31) with we observe that,
where we used Lemma 5.1 as well as the facts that and for .
For (30) we note that if and p are any polynomials of degrees and ℓ, then
where the degree of is . By induction,
and has degree . Since is the quotient of the first order polynomial , and the second order polynomial , its n-th derivative, is
and is of degree . Using Lemma 5.1 and (5) we then obtain the required estimate,
where is independent of . □
We are now ready to estimate the full phases and .
Let ϕ be eitheror. Suppose. Then there are constantsand, independent of η, θ,and δ such thatAdditionally,Forwe havewhen.
We first prove the statements for . Suppose . Then and by (5)
which gives (32) when . Similarly,
showing (33) for . The bounds for larger n follow easily from an explicit calculation, yielding for and for .
To prove the statements for we denote the correction term by . Lemma 5.2 gives
Consequently, there is a K such that for all , uniformly in and δ thanks to (5). We now take . Then for , we have
and (32) is proved.
For (33) we use (30) in Lemma 5.2. When we have, as above,
which shows the result for with . For we get
For ,
which shows the result for as and for .
Finally, statement (34) for is trivial, as , and for it follows from (29) in Lemma 5.2 and (5), since
This concludes the proof. □
In the final part of this section we consider a space of function that is used in Lemma B.1. For a fixed phase function ϕ and order p we first introduce the basis functions
when and let be the constant function equal to one. Second, we denote by the linear span of these functions over the complex numbers,
Functions in appear in Lemma B.1. Here we show that when ϕ is either or , these functions are bounded on subsets where the phase gradient grows at least quadratically.
Let, withand, for some,where c andare independent of δ and η. Then for each, where ϕ is eitheror, there is a uniform boundThe constant C depends on c andbut is independent of,, δ and η.
We get from Lemma 5.3,
while for we have
The constants and are independent of δ and η. Consider next and assume it has factors and, without loss of generality, that the first factors have . Then, for all ,
where is independent of δ, η, and , but depends on c and . Since is a linear combination of functions in the same bound holds true for u on K. □
Estimates of oscillatory integrals
We consider integrals of the type
where ϕ is either or and , whose norm is defined by
In general the integrand is then not in and the integral must be defined in a generalized sense as an oscillatory integral. In this section, however, we only estimate the integral over bounded intervals that are defined using a smooth cutoff function , which takes values in , is equal to one on and has . This leaves us with integrals over compact domains with smooth integrands. Our main tool for estimating them are the identities stated in Lemma B.1. They are used to rewrite the integral on the domain where . Lemma 5.3 in the previous section tells us when this is true. Lemma B.1 uses the space of functions defined in (36). Functions in are bounded on the domains we consider here, which is proved in Lemma 5.4. Together, Lemma B.1 and Lemma 5.4 constitute a precise version of the non-stationary phase lemma.
We will also make use of the simple inequalities
for all , and
We start with an estimate of the integral between R and where t is arbitrarily large. For this we consider a smooth cutoff around and and obtain bounds that are independent of t.
Let ϕ be eitherorand setwherewithas in Lemma
5.3
,and. Ifand, there is a constantindependent of k, R, δ, η and t such that
On this domain the results in Section 5 show that the phase gradient does not vanish, and . Since the integrand is smooth and compactly supported we can therefore use the non-stationary phase lemma to estimate the integral. For sufficiently regular r the repeated partial integrations in this lemma enables us to offset the growing factor and obtain a bound that is independent of t.
To be precise, let
which is supported in the compact set . Then by Lemma 5.3 we have on K, independent of δ, η and t, since . We apply Lemma B.1 with and let D be any bounded open set containing K. This gives
where the space is defined in (36). Since K satisfies all conditions in Lemma 5.4 and we obtain a uniform bound,
where depends on , but is independent of δ, η, R and t. This allows us to estimate as
where we also used the fact that and . Moreover, by (37) and (38),
The result in the lemma follows. □
Next we consider the main part of the integral for small η and δ with the Airy phase. The estimate involves the norm of r with an argument scaled by .
Letwhereandwith. Moreover, supposeThen there is a constant C, depending onand, but independent of k, R, η and δ such that
For the phase has a degenerate stationary point at the origin. We will therefore treat the integral in the vicinity of the origin separately. Away from the origin we have the same type of lower bound as in Lemma 6.1 and we can therefore once again use the non-stationary phase lemma to estimate the integral.
For the proof we use the rescaled variables , , . Since
we can rewrite the integral as
We then divide the integral into two pieces,
where is the part close to the origin containing the stationary point, and is the remaining part. Note that matches the general form of the integral in Lemma 6.1 if we take .
For we simply have
with C independent of k. If we have and the proof is complete. We assume henceforth that and let
the support of which lies in the compact set . Then for , by (40),
Thus, since has no stationary points on K we can use Lemma B.1 with and D an open bounded set containing K. This gives
where , with defined in (36). This expression is now estimated in the same way as (39) above. Since and K satsfies the assumptions of Lemma 5.4 we obtain a uniform bound for on K. Then
since and . Moreover, as in the proof of Lemma 6.1, by (37) and (38), since ,
Together the estimates of and then prove the lemma. We finally note that since the norm is bounded because of (38) and (40). □
Finally, we show that the derivatives of the Airy function are well approximated by an oscillatory integral with a monomial factor and the Airy phase.
Letwherewithas in Lemma
5.3
and. Ifand, there is a constant, independent of k, R, δ and η, such that
The Fourier transform of in is
Therefore, using Lemma 2.1, and noting that ,
After rescaling we get
It follows that if
then
Since is odd and ψ is even in θ we obtain
The result now follows from Lemma 6.1, with . □
Solution estimates
In Sections 3 and 4 it was shown that the partial Fourier transform in y for both the exact solution and the Gaussian beam approximation can be written on the form
where A is the amplitude function and for the two cases are given in (10) and (25). In this section we prove bounds of those in terms of k and η, which are valid for all , and . We start with the Gaussian beam superposition case and estimate as follows.
Fordefined in (
25
), there is a constant M such thatfor all,and.
From (24) in Section 4.2 we have
We note that for all n by Lemma 5.1 and that . We divide the integral into two parts. Let with as in Lemma 5.3 and define
For we have, again using Lemma 5.1, and that fact that by Lemma 5.3,
For we use Lemma 6.1 with , which says that for any and ,
where is independent of k, , δ, η and t. By (23) we have for all η and k. Then we get
and upon taking the Lemma follows with . □
Next, for the exact solution, we estimate .
Fordefined in (
10
) there is a constant M such thatfor all,and.
From (10) we have
Then, using (68) and (70) in Lemma 9.1,
The result follows as . (The estimate is sharp for .) □
Proof of the main result
In this section we prove the main result Theorem 1.1 estimating the error between the exact solution and the Gaussian beam solution. To estimate the difference between and the exact solution , it is enough to control the norm of the difference between their scaled Fourier transforms since
We will use this strategy. From (25) and (10) we get
We divide the expression into two parts, one for smaller than and one for larger than . Thus, for c to be determined below, we let
For the large values of we can immediately get a bound of by using the fact that has very rapid decay, being the Fourier transform of . This is used in the following lemma.
Supposefor eachand. Letbe given. If there existandsuch that,then, for each, there exists a constant, independent of k, but dependent on A, M, q, r, c, such that
Since for all there exists such that . Hence,
Then for , with ,
For the given p we now take . Then and
which is the desired estimate. □
In our case we let for fixed x. Then it follows from Lemma 7.1 and Lemma 7.2 that f satisfies (45), as for ,
where is in fact independent also of . Lemma 8.1 with , , , and now shows that
The choice of c will be motivated below in the next step.
To estimate we will use more precise estimates of for small and the following lemma.
Supposefor eachand. Letbe given. If there existandsuch that,then there exists a constant C, independent of k, but dependent on A, M, q, r, c, such that
As in the proof of Lemma 8.1, when we get
This proves the lemma. □
As for we apply this lemma with and to get the bound (47) we need to estimate the difference between and when . This estimate is the main part of the proof.
To examine more carefully we first recall the expressions:
where
and
We divide the difference into three parts
In three Lemmas below we show that when and , there is a constant M such that
It follows that
Then applying Lemma 8.2 with , , and shows that
when . Together with (43), (44) and (46) this proves Theorem 1.1.
Note that the estimates of and above are shown to be valid for all , while the estimate is considered, in this paper, only for . Furthermore, note that and exhibit the same loss of decay through the term . In this comes from the estimate (52) and has embedded in (54).
We now turn to proving the lemmas about .
Estimate of
There is a constant M independent of η and, such that
Since we have
Moreover, from (69) in Lemma 9.1 we get
Then, since
we obtain
It follows that where with being the constant in (69). □
Estimate of
There is a constant M dependent on x, but independent of η and, such that
We start by proving two estimates of . We use the inequalities which hold for . The definition (48) together with the assumption on η and the fact that , then gives
Moreover, since ,
Clearly, we also have , and therefore, in summary,
Next, we rewrite , adopting the definition
from Lemma 9.1. Then for ,
and
We get
We can then estimate as
We will now study the different parts of this expression separately.
Estimate of.
This is given directly by (67) in Lemma 9.1 with , as then . We get
where is the constant in (67).
Estimate of.
Using Taylor’s formula for around , we compute
Therefore,
and consequently, by (49),
This gives us
for .
Estimate of.
We divide this into three subcases. Suppose first that
Then by (49),
By (68) in Lemma 9.1 and (50),
where is the constant in (68). On the other hand, if
then by (49),
and we obtain the same estimate as above, via
Finally, when (caustic case) we can not get better than
In summary, we have with ,
We can now put the estimates together and apply them to in (51). We get
where
This proves the lemma. □
Estimate of at the caustic
This is the main estimate. Here we assume that .
Forthere is a constant M independent of η and, such that
We consider , which amounts to taking in (41). By the assumption on η and k it is bounded as
Moreover, since and as before, , we get
Hence, we need to estimate .
Let
As in the proof of Lemma 7.1 we then use the fact that where is defined and divided as
With as in Lemma 5.3 we choose here , independent of η, which implies that for all η which we consider,
Moreover, we take such that . To analyze we then first note that it can be written as
We next expand in terms of θ, first using the Taylor expansion of ,
This gives
Furthermore, let
and Taylor expand these functions as
Then
From this expansion of we now get a corresponding expansion of ,
where
and
with
We will next show that the last four terms in (57) are at most of size . To see this, we note that by Lemma 5.1 and Lemma 5.2, both r and belong to for all n, , and their -norms are bounded independent of . By (37) the same is true for , for all ℓ. Therefore, by (38),
showing that also for all n, p, ℓ. Since (40) is satisfied under the assumptions on η, δ and k, we can use Lemma 6.2 with to estimate
For we first observe that
By appealing to Lemma C.1 with we conclude that is bounded uniformly for . Consequently, we can use Lemma 6.2 with together with (37) and (38) to show that
We have thus proved that
From Lemma 6.3 we know that and we therefore introduce the approximation of obtained by replacing with the corresponding Airy function,
By (56) we can use Lemma 6.3 with large enough n to obtain
where C is uniform in ρ. Then
as uniformly in .
The next step is to show that is close to . Upon using the identities for with given in Remark 9.1 we can simplify the expression for as follows
From (55), (7) and (16), we obtain
Hence, and
Since ρ is bounded by (53) and Ai is smooth around , this shows that
Note that the dependence on which appears here, also appears in the estimate of in Lemma 8.4.
It remains to estimate . By (56) we get from Lemma 6.1 with and , for all , that
where the constant is independent of t. In conclusion, using (58), (59), (60) and (61) we have shown that
Together with (54) this concludes the proof of Lemma 8.5.
Finally note that, away from the caustic point, i.e. , the method used here to estimate will not give sharp results; if the stationary phase method is applied directly to extra decay in k follows. □
Properties of the Airy function
Here we show some known properties of the Airy function and we derive a few consequences in two lemmas. A more complete source for information about Airy functions is [7], which we frequently cite below. We consider the Airy function of the first kind Ai and second kind Bi.
The Airy functions are linearly independent solutions of the Airy differential equation
Ai and only have zeros on the negative real line. The zeros do not coincide. is positive and decreasing for .
Bi and also only have zeros on the negative real line. The zeros do not coincide. is positive and increasing for .
Let
Then, for real ,
and it follows easily from [7, Section 9.7 (ii,iii)] that
We can now prove the following lemmas.
Letwith. There is a constant C such thatMoreover, for eachthere is a constantsuch thatMoreover, forandthere is a constant C such that
We define by the relation . It has an asymptotic expansion,
From the estimates on P and Q of [1, Appendix A, Lemma 7] one obtains the uniform estimate
valid for . Then (66) follows directly with ,
Now suppose . We first note that for all ,
Hence, for we have , and since only has zeros on the negative real line, there is a positive infimum,
Therefore, as above,
when , proving (67) with .
Next, to show (68) and (69) we note first that, for real ,
Then (64), (65) readily give
and
which extend to (68) and (69) as is bounded.
The lower bound (70) follows for from the previous estimates,
as then
By (P2) we also have for and some . Moreover, the identity [7, Eq. (9.2.11)]
and (P3) implies that for . This gives the bound (70) also for .
For (71) we consider and use (64), (65)
Hence,
Since the zeros of and Ai do not coincide (no double roots) we get
and the estimate (71) for follows. Moreover, by (P2), when ,
which gives (71) for . Here we also used the fact that by (64) and (65). □
For the Airy function we havewhereandare polynomials given by the recursionsThe degree of their sum satisfiesand, for,Furthermore,
Using the form of given and using (62) we note that
where we used the Airy differential equation . This gives the recursion (73). The statement about the degree is easily checked for . Suppose it holds up to a general . Then
That follows by induction. Since the polynomials all have positive coefficients, it also follows that . For a polynomial p, let denote its largest coefficient in magnitude. Then and . Consequently,
Therefore, since ,
Finally, we have for a polynomial p of degree d, and ,
The last statement of the lemma is known for . Suppose it holds for p and use the Airy differential equation . That gives , which shows the claim. □
The first few polynomials and in the theorem are given by
Footnotes
Proof of Lemma 2.1
We show this for so that . The case with general k follows from a simple rescaling. Let be a test function and the duality pairing between and . Then since and, by dominated convergence,
where we used the facts that , for all t and pointwise. This is true for all and therefore in , proving the first statement.
That follows from the first statement since when and . The last part of the second statement is true, since and therefore the first part gives
This shows the lemma.
Proof of the non-stationary phase identities
Below is a proof of identities used in the non-stationary phase lemma. The identities show how the rewritten integral depends on the derivatives of the phase function. In order to do that we use the spaces of functions defined in (35) and (36).
Supposeis a bounded open set andis compact. Letand. Ifon K and, then there exist functionssuch that
Define the differential operator,
Then, since and on K integration by parts gives
Since and u and ϕ are sufficiently regular, this can be repeated n times, giving
We thus need to show that there exist such that
When this simply says that . Suppose the claim holds for n and consider
For the first term we have
For the third term
For the second term, consider one basis function ,
for some M. Then
Hence, , as it is a linear combination of derivatives of functions in . This shows that is of the correct form and the lemma is proved. □
Boundedness of Z
Here we consider the scaled remainder term in the Taylor expansion of ,
We have the following lemma.
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