Explicit formula of radiation fields of free waves with applications on channel of energy

In this work we give a few explicit formulas regarding the radiation fields of linear free waves. We then apply these formulas on the channel of energy theory. We characterize all the radial weakly non-radiative solutions in all dimensions and give a few new exterior energy estimates.


Background and topics
The semi-linear wave equation especially the energy critical case p = 1 + 4/(d − 2), has been extensively studied by many mathematicians in the past few decades.Please see, for example, Kapitanski [18] and Lindblad-Sogge [26] for local existence and well-posedness; Ginibre-Soffer-Velo [15], Grillakis [16,17], Kenig-Merle [22], Nakanishi [27,28] and Shatah-Struwe [29,30] for global existence, regularity, scattering and blow-up.Since the semi-linear wave equation can be viewed as a small perturbation of the homogenous linear wave equation in many situations, especially when we consider the asymptotic behaviours of solutions as spatial variables or time tends to infinity, it is important to first understand the asymptotic behaviours of solutions to the homogenous linear wave equation, i.e. free waves.This work is concerned with two important tools to understand the asymptotic behaviours of free waves: radiation field and channel of energy.We first introduce a few necessary notations.Throughout this work we consider the homogenous linear wave equation with initial data in the energy space In this work we also use the notation S L (u 0 , u 1 ) to represent the free wave u defined above.If it is necessary to mention the velocity u t , we use the notation It is well known that the linear wave propagation preserves the Ḣ1 × L 2 norm, i.e. the energy conservation law holds.(∇ x,t u = (∇u, u t )) Now we make a brief review of radiation field and channel of energy method.
Radiation field The asymptotic behaviour of free waves at the energy level can be characterized by the following theorem.
Theorem 1.1 (Radiation filed).Assume that d ≥ 3 and let u be a solution to the free wave equation ∂ 2 t u − ∆u = 0 with initial data (u 0 , u 1 ) ∈ Ḣ1 × L 2 (R d ).Then and there exist two functions G ± ∈ L 2 (R × S d−1 ) so that In addition, the maps (u 0 , u 1 ) → √ 2G ± are a bijective isometries form Ḣ1 × L 2 (R d ) to L 2 (R × S d−1 ).This has been known for more than 50 years, at least in the 3-dimensional case.Please see Friedlander [11,13], for example.The version of radiation field theorem given above and a proof for all dimensions d ≥ 3 can be found in Duyckaerts et al. [7].In addition, there is also a 2-dimensional version of radiation field theorem.The statement in dimension 2 can be given in almost the same way as in the higher dimensional case, except that the limit lim t→±∞ R 2 |u(x, t)| 2  |x| 2 dx = 0 no longer holds.A proof by Radon transform for all dimensions d ≥ 2 can be found in Katayama [19], where the statement of the theorem is slightly different.Throughout this work we call the function G ± radiation profiles and use the notations T ± for the linear map (u 0 , u 1 ) → G ± .

Channel of energy
The second tool is the channel of energy method, which plays an important role in the study of wave equation in the past decade.This method is first introduced in 3-dimensional case by Duyckaerts-Kenig-Merle [3] and then in 5-dimensional case by Kenig-Lawrie-Schlag [20].This method was used in the proof of solition resolution conjecture of energy critical wave equation with radial data in all odd dimensions d ≥ 3 by Duyckaerts-Kenig-Merle [5,8].It can also be used to show the non-existence of minimal blow-up solutions in a compactness-rigidity argument in the energy super or sub-critical case.Please see, for example, Duyckaerts-Kenig-Merle [6] and Shen [31].Roughly speaking, the channel of energy method discusses the amount of energy located in an exterior region as time tends to infinity: Here the constant R ≥ 0. Since the energy travels at a finite speed, the energy in the exterior region {x : |x| > |t| + R} decays as |t| increases.Thus the limits above are always well-defined.We can also give the exact value of the limit in term of the radiation field: We first introduce a few already known results.We start with the odd dimensions.
Proposition 1.2 (see Duyckaerts-Kenig-Merle [4]).Assume that d ≥ 3 is an odd integer.All solutions to As a result, we have Corollary 1.3.Assume that d ≥ 3 is odd.Then u ≡ 0 is the only free wave u satisfying In the contrast, if R > 0, the subspace of Ḣ1 × L 2 (R d ) defined by does contain initial data (u 0 , u Ḣ1 ×L 2 (r≥R:r d−1 dr) . ( Here Π ⊥ P rad (R) is the orthogonal projection from Ḣ1 × L 2 (r ≥ R : r d−1 dr) onto the complement of the finite-dimensional subspace P rad (R).
The case of even dimensions is much more complicated and subtle.Côte-Kenig-Schlag [1] shows that in general the inequality does not hold for any positive constant C in even dimensions.But a similar inequality holds in the radial case for either initial data (u 0 , 0), if d = 0 mod 4, or (0, u 1 ), if d = 2 mod 4.More precisely we have lim In addition, Duychaerts-Kenig-Merle [9] shows that the only non-radiative solution is still zero solution in even dimensions d ≥ 4, i.e.Corollary 1.3 still holds for even dimensions d ≥ 4, even in the non-radial case.Much less is known about the exterior energy estimate in the region {x : |x| > R + |t|} with R > 0. Dyuchaerts at el. [2] proves the exterior energy estimate in dimension 4 and 6 if the initial data are radial: Here Π ⊥ R is the orthogonal projection from Ḣ1 ({x ∈ R 4 : |x| > R}) onto the complement space of span{|x| −2 }.While π ⊥ R is the orthogonal projection from L 2 ({x ∈ R 6 : |x| > R}) onto the complement space of span{|x| −4 }.

Main idea
According to (2) we may obtain exterior energy estimates conveniently from the radiation profiles G ± .Please note that G − and G + are not independent to each other.In fact the map T + • T −1 − : G − → G + is a bijective isometry.If we could find explicit expressions of the maps then we would be able to (a) Understand how the asymptotic behaviour in one time direction determines the behaviour in the other time direction.This is known in the odd dimensional case, as shown (although not stated explicitly) in the proof of Proposition 1.2 by Duyckaerts-Kenig-Merle [4].In this work we will try to figure out the even dimensional case.
(b) Characterize (weakly) non-radiative solutions, especially in the radial case.We first determine all the radiation profiles G − so that then we may obtain all the non-radiative solutions (as well as their initial data) by applying the formula of T −1 − .In particular we prove that radial non-radiative solutions in the even dimension can be characterized in the same way as in the odd dimensions.
(c) Prove exterior energy estimates.We generalize the radial exterior energy estimates in 4 and 6 dimension to all even dimensions; we also prove a non-radial exterior energy estimate in the odd dimensions.In both applications (b) and (c) we follow the same roadmap: exterior energy ↔ radiation profile ↔ initial data.

Main results
Now we give the statement of our results.The details and proof can be found in subsequent sections.
Theorem 1.5.Let u be a finite-energy free wave with an even spatial dimension d ≥ 2 and G + , G − be the radiation profiles associated with u.Then we may give an explicit formula of the operator Here H is the Hilbert transform in the first variable, i.e.
Remark 1.6.A similar but simpler argument shows that if d is odd, then This can also be verified by assuming that the initial data is smooth and compactly-supported, and considering the expression of G − and G + in terms of (u 0 , u 1 ) if d is odd.Please refer to Duyckaerts-Kenig-Merle [4].Since we have H 2 = −1.We may write the odd and even dimensions in a universal formula Remark 1.7.It has been proved in Section 3.2 of Duychaerts-Kenig-Merle [9] (in the language of Hankel and Laplace transforms) that the zero function is the only It immediately follows that Let Ω be a region in S d−1 .If a finite-energy solution u to homogenous linear wave equation satisfies This is an angle-localized version of Corollary 1.3.
Applications on channel of energy By following the idea described above, we obtain the following results about the channel of energy.
Proposition 1.9 (Radial weakly non-radiative solutions).Let d ≥ 2 be an integer and R > 0 be a constant.If initial data (u 0 , u 1 ) ∈ Ḣ1 × L 2 are radial, then the corresponding solution to the homogeneous linear wave equation u is R-weakly non-radiative, i.e.
Here the notation ⌊q⌋ is the integer part of q.In particular, all radial R-weakly non-radiative solution in dimension 2 are supported in {(x, t) , thus our result here is the same as the already known result in odd dimension, as given in Theorem 1.4.
Proposition 1.11 (Radial exterior estimates in even dimensions).Let d = 4k with k ∈ N and R > 0. If initial data u 0 ∈ Ḣ1 (R d ) are radial, then the corresponding solution u to the homogenous linear wave equation with initial data (u 0 , 0) satisfies Here radial, then the corresponding solution u to the homogenous linear wave equation with initial data (0, u 1 ) satisfies Here gradually vanishes as R → 0 + .Therefore if we make R → 0 + in Proposition 1.11, we immediately obtain (6) and (7).Proposition 1.13 (Non-radial exterior energy estimates).Let d ≥ 3 be an odd integer and R > 0 be a constant.Then the following inequality holds for all (u 0 , u 1 ) .
Here Π ⊥ P (R) is the orthogonal projection from Ḣ1 × L 2 (R d ) onto the complement of the closed linear space

Structure of this work
This work is organized as follows.In section 2 we deduce an explicit formula of T −1 − in all dimensions.Then in Section 3 we prove the explicit formula of given in Theorem 1.5.The rest of the paper is devoted to the applications in channel of energy.We characterize radial weakly non-radiative solutions in Section 4, prove radial exterior energy estimate for all even dimensions in Section 5 and finally give a short proof of non-radial exterior energy estimate in odd-dimensional space in Section 6.The appendix is concerned with Hilbert transform of a family of special functions, since the Hilbert transform is involved in the even dimensions.
Notations In this work we use the notation C(d) for a nonzero constant determined solely by the dimension d.It may represent different constants in different places.This avoid the trouble of keeping track of the constants when unnecessary.

From Radiation Profile to Solution
Now we assume that G − (r, θ) is smooth and compactly supported and give an explicit formula of the operator T −1 − .We consider the odd dimensions first.

Odd dimensions
Here the notation − represents the partial derivative Remark 2.2.This formula in 3-dimensional case was previously known.Please refer to Friedlander [12], for example.
Here µ = (d − 1)/2 and χ : R → [0, 1] is a smooth center cut-off function satisfying It is clear that the data (v 0,t , v 1,t ) are smooth and compactly-supported in {x : Thus by radiation field we have Since the linear propagation operator S L (t) preserves the Ḣ1 × L 2 norm, we have Next we use the explicit expression of linear propagation operator (see, for instance, Evans [10]) and write v = S L (v 0 , v 1 ) in terms of (v 0 , v 1 ) when the initial are sufficiently smooth. Here ) are all constants.We may differentiate and obtain Now we plug in (v 0 , v 1 ) = (v 0,t , v 1,t ) with large time t.We observe that and Please note that the implicit constants in (13), O(t −µ−1 ) and O(1/t) above may depend on x but remain to be uniformly bounded if x is contained in a compact subset of R d .Next we observe the facts and further simplify the formula Finally we make t → +∞, utilize (12) and obtain We plug in the value of c d and finish the proof.
3. An explicit formula of the free wave u = S L T −1 − G − can be given by This can be verified by a straight-forward calculation.One may check • The function u above is a smooth solution to the homogenous linear wave equation; • The initial data of u are exactly those given in Lemma 2.1.
We may differentiate and obtain

Even dimensions
The formula of T −1 − in even dimensions are a little more complicated.
Lemma 2.4.Assume that d ≥ 2 is even and Proof.Without loss of generality let us assume supp It is sufficient to show that given any R 2 > 0, the formula above holds for almost everywhere x ∈ B(0, R 2 ).Let us use the notations (u 0 , u 1 ) = T −1 − (G − ) and u = S L (u 0 , u 1 ).We consider the approximated data and Here χ is the center cut-off function as given in the previous subsection.A basic calculation shows lim Let us first recall the explicit formula of v = S L (v 0 , v 1 ) in the even dimensional case: Here B d is the unit ball in R d and We differentiate and obtain We plug in (v 0 , v 1 ) = (v 0,t , v 1,t ) and observe This gives the approximation Here r = |x + ty|, θ = x+ty |x+ty| .Furthermore, we observe (k = d/2, d/2 − 1) and write As a result, we may restrict the domain of integral to Because in the region we have We can simplify the formula Next we utilize the change of variables and the approximations Finally we recall ( 16), make t → +∞ and conclude This finishes the proof.
Remark 2.5.If d ≥ 4, the convergence (16) implies that (w 0,t , w 1,t ) converges to (u 0 , u 1 ) in L 2d d−2 × L 2 by Sobolev embedding.We may combine this convergence with the local uniform convergence given above to verify the identities above.This argument breaks down in dimension 2. We given another argument below in dimension 2. Given any test function ϕ ∈ C ∞ 0 (R 2 ), integration by parts gives an identity w 0,t (x)∇ϕ(x)dx = − ∇w 0,t (x)ϕ(x)dx.
We recall the local uniform convergence of w 0,t given above and the L 2 convergence of ∇w 0,t → ∇u 0 , then obtain This finishes the proof.Finally the author would like to mention that we have Thus Proof.A basic calculation shows that u(x, t) solves the free wave equation with initial data given in Lemma 2.4.

Universal formula
Now let us give a universal formula of T −1 − for all dimensions.We first define two convolution operators (1/ √ πx is understood as zero if x < 0) Their Fourier symbols are As a result, we may understand Here µ = d−1 2 .This formula holds for both odd and even dimensions.

Between Radiation Profiles
In this we give an explicit expression of the operator T + • T −1 − in the even dimension case, without the radial assumption.Theorem 3.1.Assume that d ≥ 2 is an even integer.The operator T + • T −1 − can be explicitly given by the formula Here H is the Hilbert transform in the first variable, i.e.
) to itself.We only need to prove this formula for smooth and compactly supported data G − .Without loss of generality let us assume supp , then we may apply Corollary 2.6 and obtain 1 be a large constant, we may split the integral above into two parts We may find an upper bound of J 2 .In this region we have Thus we may integrate by parts and obtain dωdρ Thus when t is sufficiently large In the integral region of J 1 , we have the approximation ω = −θ + O(t −1/2 ).Thus we have we utilize the change of variables (please refer to figure 1 for a geometrical meaning) and obtain We observe that the integrand is independent of ϕ and integrate by parts Figure 1: Change of variables We next change the variables τ = ρ ′ − ρ, η = ρ ′ + ρ, and write The integrals above can be split into two parts: and In summary we have Now we may combine J 1 and J 2 This finishes the proof.

Radial Weakly Non-radiative Solutions
In this section we prove Proposition 1.9.First of all, we briefly show that any initial data in P rad (R) leads to a R-weakly non-radiative solution.By linearly we only need to consider the case solves the linear wave equation with initial data (|x| 2k1−d , 0) in the region R d \ {0}.By finite speed of propagation, we have A simple calculation shows that this is indeed a non-radiative solution.The case (u 0 , u 1 ) = (0, r 2k2−d ) can be dealt with in the same manner by considering the solution Thus it is sufficient to show initial data of any non-radiative solution are contained in the space P rad (R).We first consider the odd dimensions.

Odd dimensions
Assume that u = S L (u 0 , u 1 ) is a radial R-weakly non-radiative solution.Let G − = T − (u 0 , u 1 ).By radial assumption G − is independent of the angle ω ∈ S d−1 .Let us first consider smooth functions G − .We may calculate (r > R, e 1 = (1, 0, is the area of the sphere S d−2 .We may integrate by parts and rescale Here A d,k 's are nonzero constants.Similarly we have This clearly shows that if u = S L (u 0 , u 1 ) is a radial R-weakly non-radiative solution, then (u 0 , u 1 ) ∈ P rad (R).

Even dimensions
The even dimensions involve Hilbert transform, thus are much more difficult to handle with.The general idea is the same.If the initial data (u 0 , u 1 ) are radial, then G ± (s) = T ± (u 0 , u 1 ) is independent to the angle.We also have G + (s) = (−1) d/2 HG − (−s).Thus S L (u 0 , u 1 ) is R-weakly non-radiative if and only if G − is contained in the space Now recall the operators Q, Q ′ and D defined in Subsection 2.3.We claim Lemma 4.2.
Proof.In order to avoid technical difficulties, we use an approximation technique.Given any G − ∈ P rad , we may utilize a local smoothing kernel to generate a sequence G k , so that Let us consider the properties of the function ) by smooth approximation.A simple calculation of Fourier symbols shows that Q ′−1 = −Q ′ D and HQ ′−1 = QD.A combination of these identities with the convolution expressions of Q and Q ′ immediately verifies Q ′−1 g ∈ P rad .
We also need to use the following explicit formula of T − for radial data Here P d is an even or odd polynomial of degree d/2 − 1 defined by Proof.If G ∈ C ∞ 0 (R), we use the polar coordinates and integrate by parts: This verifies the formula if G ∈ C ∞ 0 (R).In order to deal with profile G without compact support, we use standard smooth cut-off techniques.More precisely, we may choose First of all, if u 0 ∈ Ḣ1 rad (R d ), then free wave u = S L (u 0 , u 1 ) is radial and satisfies u(x, t) = u(x, −t); u t (x, t) = −u t (x, −t).
We define a linear bounded operator T from L 2 (R + ) to itself 1 We may further rewrite it as Here L is the Laplace transform LG(s) = ∞ 0 G(τ )e −sτ dτ, which is self-adjoint operator in L 2 (R + ) with an operator norm √ π.More details about the Laplace transform can be found in Lax [25].As a result, we have LG, LG LG 2 L 2 .
Thus the operator norm of T less or equal to 1/2.This means that the function satisfies the equation G − TG = g and G L 2 (R + ) ≤ 2 g L 2 (R + ) .Finally we naturally extend the domain of G to R by defining G(s) = 0 if s < 0. We have G(s) − HG(−s) 2 = g(s), s > 0; (−1/2)HG(−s), s < 0.
Therefore we may find an upper bound of the L 2 norm G(s) − HG(−s) Proof of Theorem 1.11 Let G − = T − (u 0 , 0) and g(s) be its cut-off version: Then radiation field implies that the free wave u = S L (u 0 , 0) satisfies Here agian σ 4k−1 is the area of the sphere S 4k−1 .According to Lemma 5.1 and Lemma 5.2, there exists a function ũ0 ∈ Ḣ1 rad (R 4k ), so that T − (ũ 0 , 0)(s) = g(s), s > 0; ũ0 As a result, we may apply Proposition 1.9 and conclude u 0 − ũ0 ∈ Q k (R).This means A combination of this inequality and identity (21) immediately verifies the conclusion of Proposition 1.11 in the negative time direction.The positive time direction follows the time symmetry.

Non-radial Exterior Energy Estimates
In this section we give a short proof of Proposition 1.13.We start by Lemma 6.1.Let d ≥ 3 be an odd integer.Then In particular, we have (see (4) for the definition of P (R)) Hilbert transform of the second term in the right hand side has been known to be a polynomial of degree n − 2. Thus we only need to consider the first term.We have This is a polynomial of degree n − 1 by induction hypothesis.A simple integration then finish the proof of case κ = n + 1.