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An Application of Nash-Moser Theorem to Smooth Solutions of One-Dimensional Compressible Euler Equation with Gravity

We study one-dimensional motions of polytropic gas governed by the compressible Euler equations. The problem on the half space under a constant gravity gives an equilibrium which has free boundary touching the vacuum and the linearized approximation at this equilibrium gives time periodic solutions. But it is not easy to justify the existence of long-time true solutions for which this time periodic solution is the first approximation. The situation is in contrast to the problem of free motions without gravity. The reason is that the usual iteration method for quasilinear hyperbolic problem cannot be used because of the loss of regularities which causes from the touch with the vacuum. Interestingly, the equation can be transformed to a nonlinear wave equation on a higher dimensional space, for which the space dimension, being larger than 4, is related to the adiabatic exponent of the original one-dimensional problem. We try to find a family of solutions expanded by a small parameter. Applying the Nash-Moser theory, we justify this expansion.The application of the Nash-Moser theory is necessary for the sake of conquest of the trouble with loss of regularities, and the justification of the applicability requires a very delicate analysis of the problem.


Introduction
The aim of this paper is to study one-dimensional motions of polytropic gas governed by the compressible Euler equations for t, x ≥ 0 subject to the boundary condition Here ρ, u, P and g > 0 are density, velocity, pressure and gravitational acceleration constant respectively. Equations (1)∼(3) describe the atmosphere on the flat earth {x ≤ 0} moving in one direction under the constant gravitational force downward. In this work we assume that P = P (ρ) = Aρ γ for some constants A, γ such that 0 < A, 1 < γ ≤ 2. Then equilibria of (1) and (2) are of the form where A 1 = ((γ − 1)g/γA) 1/(γ−1) and x + is an arbitrary positive value, which represents the stratospheric depth.
Without loss of generality, we may assume x + = 1, A 1 = 1 and A = 1/γ. It can be seen easily by scale transformations of the variables. Since the interface with the vacuum would vary with the time, it is convenient to transform the equations (1) and (2) into the Lagrangian form. More precisely, we introduce the variable m = x 0 ρdx as the independent variable instead of x, then equations (1) and (2) can be transformed into the following second order equation: x tt + P m = −g, where P = γ −1 (x m ) −γ . Let us fix an equilibrium where Then we consider small perturbations of the equilibrium in (6) by putting x(t, m) =x(m) + y. Under such assumption, the equation (5) is reduced to where Takex(m) as the independent variable instead of m, and write it as x. Then the equation (7) is reduced to for 0 < x < 1 and the boundary condition is Hereρ Equation (8) is an apparently quasilinear hyperbolic equation. But it has a singularity at x = 1. Due to the singularity, the investigation for the existence of time periodic solutions becomes a difficult and challenging problem. To the best of our knowledge, the existence problem of time periodic solutions is still open.
For the sake of comparison, let us recall the results of [5], which considered the following simplified quasilinear wave equation y tt − (G(y x )) x = 0 for 0 < x < 1, y(t, 0) = y(t, 1) = 0. (10) This problem is derived from the Euler equations ρ t + (ρu) x = 0, (ρu) t + (ρu 2 + P ) x = 0, and the boundary condition ρu| x=0 = ρu| x=L = 0, for which the equilibria are positive constant densities. Hence there are no troubles caused by contact with vacuum. For any fixed arbitrarily long time, [5] shows that there are smooth small amplitude solutions of the problem (10) for which the periodic solutions of the linearized equation are the first-order approximation. This result was established by the usual iteration method for quasi-linear wave equations.
Therefore, similarly, we want to find smooth solutions for which a time periodic solution of the linearized equation around an equilibrium is the first approximation even for the present problem (8) (9). However, contrary to the case without gravity, the usual iteration method for quasilinear hyperbolic problem cannot be applied directly to the present problem because of the loss of regularities which causes from the touch with the vacuum. In this work we shall apply Nash-Moser theorem to establish long time existence of smooth solutions near time-periodic solution of the linearized equation.
More precisely speaking, we introduce the variable and small parameter ε, and we shall construct approximate solutions of the form where y k (t, z) are entire functions of t and z, while y 1 (t, z) is a non-trivial time periodic solution of the linearized equation. Then our aim is to find a true smooth solution y(t, z) of (8)(9) on 0 ≤ t ≤ T and 0 ≤ z ≤ 1, for arbitrarily fixed T , such that Of course for large T we should restrict ε sufficiently small. Then is a solution in the Lagrangian variable and the corresponding density distribution ρ = ρ(t, x), where x denotes the original Euler coordinate, satisfies and is the position of the free boundary. Since y(t, z) is smooth on 0 ≤ z ≤ 1, we have and This condition is that of "physical vacuum boundary" so called by the most recent works [7](2009) and [1](2011). This concept can be traced back to [8](1996), [9](2000), and [13](2006). Hence we can say that our purpose is to find long-time smooth solutions with "physical vacuum boundary". But [7] and [1] are interested in short-time solutions to the initial value problem for the case without external force. So the motivation, methods and results are different from those of this work. Now we have introduced the variable Moreover it is convenient to introduce the parameter Then the assumption 1 < γ ≤ 2 is equivalent to that 4 ≤ N < ∞. Hence, we assume N ≥ 4 in the following of this wok. Moreover, the equation (8) turns out to be where and [v] q denotes a convergent power series of the form j≥q a j v j . If we introduce the variable r by ∂ ∂r is the radial part of the Laplacian operator on the N -dimensional Euclidean space R N , provided that N is an integer. But we shall not assume that N is an integer in this work.
Here we would like to spend few words to explain why the usual iteration does not work although the equation (13) is apparently quasi-linear. For the sake of simplicity, let us assume N is an integer. Then a smooth function y of z can be regarded as a smooth function of r = x = ( j (x j ) 2 ) 1/2 , where x ∈ R N . Since y is smooth and spherically symmetric, we can assume that ∂y/∂r = 0 at r = 0 and as r → 0. In other words, v = −∂y/∂z is not of the first order, but of the second order, which is of the same order as the principal part △y = z ∂ 2 y ∂z 2 + N 2 ∂y ∂z . So, the loss of regularities cannot be recovered by one step of solving a (linear) wave equation. This is the reason why we try an application of the Nash-Moser theory. Note that this trouble comes from z = 0, that is, from the touch with vacuum at the free boundary.

Preparatory analysis of linear problems
First let us consider the linearized problem of (13): In [4] we showed that (17) admits a time periodic solution where θ is a constant, λ n is the eigenvalues of the operator −△ with the Dirichlet boundary condition, and is an entire function such that Φ N 2 −1 (λ n ) = 0. In fact, where j N 2 −1,n is the n-th positive zero of the Bessel function J N 2 −1 , and More precisely speaking, we consider the Hilbert space X which consists of functions of 0 ≤ z ≤ 1 endowed with the inner product (y 1 |y 2 ) X : The self-adjoint operator T = −△ with boundary condition is defined on D(T ) = {y ∈ X | ∃ η n ∈ C ∞ 0 (0, 1) such that η n → y in X, Q[η n − η m ] → 0 as m, n → ∞, and − △y ∈ X in distribution sense}. and "−△y = f ∈ X in distribution sense" means that for any η ∈ C ∞ 0 (0, 1) there holds (y| − △η) X = (f |η) X .
Here λ ≥ 0 and f are given.

Proposition 1
The inverse T −1 is a compact operator.
Proof. If f ∈ X, the solution of the problem (19) with λ = 0 is given by the formula we see that for some constant C > 0. Moreover, we have Therefore, Ascoli-Arzelà's theorem implies that a sequence y n converges on each compact subset of (0, 1] when f is confined in a bounded set of X. On the other hand, since , we see that y n converges in X, too. Therefore, T is a self-adjoint operator whose inverse is compact and the following assertion holds. See, eg., [2]. Proposition 2 If λ ≥ 0, then the range R(−λ + T ) is closed and Thus, if λ = λ n is an eigenvalue with an eigenfunction φ n , then the problem of (19) admits a solution y in X if and only if (f |φ n ) X = 0.
Proposition 3 If f (z) is an entire function, then there is an entire function y(z) which solves the equations of (19).
Proof. Let Let R be an arbitrarily large positive number. Since f is an entire function, there is a constant M such that |c k | ≤ M/R k for all k. We seek a solution y(z) of (19) in the form Substituting (21) into (19) and comparing the coefficients, we have the formula .
Taking a 0 arbitrarily, we claim that there is a constantM > 0 such that |a k | ≤ M /R k for all k. Suppose k > max{R, λ + 1} and |a k | ≤ M ′ /R k , then Hence the claim follows and the radius of convergence of a k z k is larger than R.
Proposition 4 Suppose λ > 0 and f is an entire function, then any solution of (19) in X is an entire function.
Proof. The homogeneous equation −λy − △y = 0 admits a pair of linearly independent solutions y 1 (z) = Φ N 2 −1 (λz) and y 2 (z) such that In fact, if we take the change of variables λz = r 2 4 and y = r −ν w, then the equation −λy − △y = 0 turns out to be the following Bessel equation: If ν is not an integer then J ν and J −ν are linearly independent solutions. On the other hand, if ν( = 0) is an integer, then J ν and the Bessel function of the second kind Y ν of the form k are linearly independent solutions. Here Ψ(x) := DΓ(x)/Γ(x). See [12]. Since N ≥ 4, we see y 2 does not belong to X. On the other hand, there is an entire function y = ψ 0 (z) which satisfies (19) due to Proposition 3. Of course ψ 0 ∈ X. Thus any solution y(z) of (19) can be written as in which, if y(z) ∈ X, then C 2 = 0, and therefore, y(z) is an entire function, too.

Formal solution expanded as power series of parameters
Now we construct formal power series solution of (13). Let us fix a non-trivial solution is the normalized eigenfunction in the Hilbert space X. According to the result of [4], we know that (φ n ) n=1,2,··· forms a complete orthonormal system in X.
Note that φ n (z) is an entire function of z. Our purpose is to find a formal solution of (13) of the form where ε stands for a small parameter. Substituting (23) into the equation (13) and comparing the coefficients, we get the following sequence of linear equations where Starting from the fixed y 1 , we can solve the equations (24) with the boundary condition y k (1) = 0 successively.

Solution for k = 2
The equation of (24) for k = 2 is in the form Since y 1 is an entire function, we can write the right-hand side of (25) by the form where Θ := λ n0 t + θ 0 , f 0 and f 1 are entire functions of z. Here we have used Then solutions of the problem can be represented by On the other hand, we consider the problem We need to consider the following two cases: Case-1: 4λ n0 is not an eigenvalue; Case-2: there is an eigenvalue λ q = 4λ n0 .
First, we consider the Case-1. Then (28) has a solution of the form According to Proposition 3, the first equation of (29) has a solution W 0 (z), which is an entire function of z such that W 0 (0) = 1. Then, for any constant C, Next, we consider Case-2, i.e. λ q = 4λ n0 for some integer q. We guess that this case could not happen actually for N ≥ 4. More generally, we have Conjecture Let ν ≥ 1 and θ be a positive zero of the Bessel function J ν . Then J ν (Lθ) = 0 for any integer L ≥ 2.
Note that the conclusion is not the case if ν = 1/2, for which J 1/2 (r) = 2 πr sin r. However we have not yet verified this conjecture. Therefore we should consider Case-2. By Proposition 2, there is a solution W 1 (z) of which is entire and satisfies the boundary condition. Then it is easy to see that Summing up, we have a solution y 2 of the form for Case-1, or for Case-2, where y 20 , y 21 and y 22 are entire functions of z.
Suppose Case-1. Then the 2nd order approximate solution is a time-periodic solution with period Ω = 2π/ λ n0 .
According to [4] , we know that any non-trivial true time-periodic solution of (10), if exists, with period T should satisfy as an effect of nonlinearity. This is true for the approximate solution y (2) , since Using integration by parts, the form of y 20 (0) can be checked by a tedious but direct computation.
We do not know whether Case-2 actually happens and the resonance could occur or not. This is an interesting open problem.

Solution for k = 3
For the sake of simplicity, we assume Case-1 for k = 2. Fixing y 2 of the form (30), we see that the right-hand side of the equation (24) is of the form where g 1 and g 3 are entire functions of z. Here we have used The equation has a solution of the form where W 1 (z) is an entire function which solves the following equation Then we can claim that w 1 (t, z) satisfies the boundary condition for any W 1 (z) with arbitrary W 1 (0). In fact, by Proposition 2, there is a solution W in X satisfying the equation and the boundary condition. Then U := W 1 − W belongs to X and satisfies the homogeneous equation As in the proof of Proposition 4, there is a constant C such that U = Cφ n0 . Hence, W 1 = W + U = W + Cφ n0 satisfies the boundary condition.
On the other hand, we consider the problem Similar to the discussion of k = 2, we need to consider the following two cases: Case-1.1: 9λ n0 is not an eigenvalue; Case-1.2: there is an eigenvalue λ q = 9λ n0 .
Let us consider the Case-1.1 for simplicity. The problem (34) has a solution of the form According to Proposition 3, the first equation of (35) has a solutionW 3 (z), which is an entire function of z such thatW 3 (0) = 1. Then, there exists a constant C = 0 such that is a solution of (35), too. Since 9λ n0 is not an eigenvalue, we have Φ N 2 −1 (9λ n0 ) = 0. Therefore, we can choose C so that satisfies the boundary value condition. Thus, we have a solution y 3 (t, z) of (24) of the form where y 31 and y 33 are entire functions of z and C 3 is a constant.
Now we can ask whether the approximate solution y 3 given by (36) is timeperiodic or not, or in other words, whether the resonance occurs actually or not. It depends on whether vanishes or not. We guess that C 3 = 0, but we have not yet verified it.

Solutions for k > 3
Similar to the computations for the conclusions of previous subsections, we can determine solutions y k of (24) for k > 3 successively in the form where V k,L,M and W k,L,M, are entire functions of z. In order to prove it, we need the following lemma.
Lemma 1 If f (z) is an entire function of z, then the problem admits a solution of the form respectively. Here Y m (z) are entire functions of z.
Proof. The proof can be done easily when √ λ m = L √ λ n for all m, n ∈ Z + . Here we only consider the case that there is a q ∈ Z + such that λ q = L √ λ n . This happens at least if L = 1. In this case a solution ψ(z) of which is an entire function of z satisfies the boundary condition. Then the problem admits a solution of the form where In fact, A(t) is a solution of the equation Then, using the mathematical induction with respect to M , the assertion of the Lemma follows .

Existence of smooth solutions
In this section, we will prove the existence of smooth solutions of (13), using the Nash-Moser theorem. In Section 3, we constructed the approximate solutions Fixing an arbitrarily large T and an integer K, we want to find a solution y(t, z) of the original problem of the form on the time interval 0 ≤ t ≤ T . First, we derive the problem of w. By (24), it follows that y (K) satisfies Let us denote v (K) := −∂y (K) /∂z and P := −∂w/∂z. Then w satisfies where Let us denote Then equation (41) can be written as Note that a(t, z, P, ε) and b(t, z, P, ε) are analytic functions of such that b(t, z, 0, ε) = 0, and c(t, z, ε) is an analytic function of Our goal is to seek a smooth solution w(t, z) of the equation (47) such that for sufficiently small ε. For completeness, we recall the Nash-Moser theorem as follows.
Nash-Moser Theorem(see [3], p.171, III.1.1.1) Let E 0 and E be tame spaces and P : U(⊂ E 0 ) → E a smooth tame map. Suppose that the equation for the derivative DP(w)h = g has a unique solution h = V P(w, g) for all w in U and all g, and that the family of inverses V P : U × E → E 0 is smooth tame map. Then P is locally invertible, and each local inverse P −1 is a smooth tame map.
Now we define the spaces E 0 , E and nonlinear mapping P : Note that P(0) = 0 and take a neighborhood U of 0 such that |ε| K ∂w/∂z L ∞ ≤ δ 0 for w ∈ U. Then the equation (47) can be written by P(w) = εc(t, z, ε).
By definition of P, it is easy to see the Fréchet derivative DP of the mapping P at a fixed w ∈ U ⊂ E 0 is of the form Suppose the following statements hold: (S1) P is a smooth tame map, E being endowed with a suitable system of graded norms (for definition, see [3]) ; (S2) for any w ∈ U ⊂ E 0 , g ∈ E, there is a unique solution h := V P(w, g) of the equation and the mapping V P : U × E → E 0 is a smooth tame map Then it follows from Nash-Moser Theorem that P is invertible in a neighborhood U of 0 in E 0 . Thus the inverse image w = P −1 (εc) is a solution of our problem (47) or (48), where ε is sufficiently small. More precisely, we have the following results.
But the term a 2 · ∂h ∂z in DP(w)h could cause trouble, since this term can have same order as the principal part △h = z ∂ 2 h ∂z 2 + N 2 ∂h ∂z . When we try to get the energy estimate, keeping in mind that (−△h|h) X = √ z∂h/∂z 2 X , we could not estimate ∂h/∂z X by √ z∂h/∂z X because of the singularity at z = 0. However we have fortunately the following observation: Proposition 5 For any fixed w in the neighborhood U ⊂ E 0 there is a smooth functionâ 2 (t, z, ε) of 0 ≤ t ≤ T, 0 ≤ z ≤ 1, |ε| ≤ ε 0 (U) such that a 2 (t, z, ε) = zâ 2 (t, z, ε).
Proof. By (50), we can write It follows from (42), (43) and (44) that Hence, we have Therefore, a 2 (t, z, ε) = zâ 2 (t, z, ε) by puttinĝ Thanks to Proposition 5, we can obtain the following energy inequality. then Proof. Let us consider the energy We claim that there is a constant A such that By Proposition 5, the coefficient a 2 (t, z, ε) is of the form a 2 (t, z, ε) = zâ 2 (t, z, ε), whereâ 2 is a smooth function of 0 ≤ t ≤ T, 0 ≤ z ≤ 1 and |ε| ≤ ε 0 . Then the equation (53) turns out to be Multiplying equation (55) by h t and integrating it by dν = z N 2 −1 dz from z = 0 to z = 1, we obtain that Using an integration by parts under the boundary condition, we have Since 1 + εa 1 ≥ 1/2, we have 1 0 z(h z ) 2 dν ≤ 2E. Therefore, it follows from (55), (56) and (57) that Hence, the Gronwall's argument implies Since E(0) = 0 from the initial condition for h ∈ E 0 , we get the required inequality (54).
As a corollary of Proposition 6, g = 0 implies h = 0 by the boundary condition and which implies that (51) has a unique solution. Moreover, this consideration of energy is sufficient to claim that the inverse V P(w, ·) of DP(w, ·) exists. This can be verified by the standard method on solving the initial boundary value problem to linear wave equations with smooth coefficients. See, e.g. Chapter 2 of [6]. In fact, for any fixed t 0 , if we consider the Hilbert space H = X 1 × X and the operator A(t 0 ), whose domain D(A(t 0 )) is then the problem where g(t) = (0, g(t, ·)) T , allows the application of Hille-Yosida theory. Note that −(1 + εa 1 )△h = f means that for any test function φ or φ ∈ X 1 . Here X 1 denotes the space of functions y(z) ∈ X such that √ zdy/dz ∈ X and X 2 denotes the space of functions y ∈ X 1 such that −△y ∈ X. For more details we refer the reader to [6].
Next, we show that the Fréchet space E is tame for some gradings of norms. For y ∈ E, n ∈ N, let us define Then we can claim that E turns out to be tame by this grading ( · (∞) n ) n (see [3], p.136, II.1.3.6 and p.137, II 1.3.7). In fact, even if N is not an integer, we can define the Fourier transformation F y(ζ) of a function y(z) for 0 ≤ z < ∞ by Here K(X) is an entire function of X ∈ C given by , J ν being the Bessel function. Then we have F (−△y)(ζ) = 4ζ · F y(ζ) and the inverse of the transformation F is F itself. See, e.g. [11]. Then it is easy to see E endowed with the grading ( y (∞) n ) n of the form (58) is a tame direct summand of the tame space and its inverse applied to the space C ∞ 0 ((−2T, 2T ) × [0, T + 1)), into which functions of E can be extended (see, e.g. [10], p.189, Theorem 3.13) and the spaceĊ for which functions of E are restrictions. For the details, see the proof of [3], p.137, II.1.3.6.Theorem.
On the other hand, let us define by the Sobolev imbedding theorem (see Appendix A), provided that 2σ > N/2. The derivatives with respect to t can be treated more simply. Then we see that the grading ( · n ) n is equivalent to the grading ( · (∞) n ) n . Hence E is tame with respect to ( · (2) n ) n . The grading ( · (2) n ) n is suitable for energy estimates. Note that E 0 is a closed subspace of E endowed with these gradings. Now we show the statement (S1) by verifying the nonlinear mapping P is tame for the grading ( · (∞) n ) n . To do so, we write P(w) = F (t, z, Dw, w tt , △w), where D = ∂/∂z, F is a smooth function of t, z, Dw, w tt , △w and linear in w tt , △w. According to [3] (see p.142, II.2.1.6 and p.145, II.2.2.6), it is sufficient to prove the linear differential operator w → Dw = ∂w/∂z is tame. But it is clear because of the following result.
Proposition 7 For any m ∈ N we have the formula As a corollary it holds that, for any m, k ∈ N, Proof. It is easy by integration by parts in induction on m starting from the formula In parallel with the results of [3] (see p.144, II.2.2.3.Corollary and p.145, II.2.2.5.Theorem), we should use the following two propositions. Proofs for these propositions are given in Appendix B.
Proposition 8 For any positive integer m, there is a constant C such that where · stands for · L ∞ . Proposition 9 Let F (z, y) be a smooth function of z and y and C 0 be a positive number. Then for any positive integer m, there is a constant C > 0 such that Now, we are going to deduce an energy estimate of the higher order derivatives of h ∈ E 0 satisfying (51) for g ∈ E, that is, whereĎ := z · ∂/∂z. We shall use the following lemma.
Proof. Putting Z(t) := e −λt Y (t), it is easy to see that Here λ := A 2 + B + A 2 4 is the positive root of the equation Using the condition Z(0) = 0, we have Hence, the estimate of this lemma follows.
Now we look at the equation satisfied by the higher order derivative of h. Let us denote H m := △ m h for m ∈ N. Differentiating the equation (59), we can show that H m satisfies the equation Here b and so on. Let us write (60) as Then we see that the energy Here Now we estimate the integral 1 0 HH t dν. Since H| t=0 = H t | t=0 = 0, then we see Thus we have Therefore, we get the integro-differential inequality Taking Y := · F (s) X ds.
Therefore, we get which implies the energy estimate Thus we have the following estimate: and the constant C depends only upon In order to verify that the mapping V P : more concretely. The recurrence formula for these coefficients are: We remark that these coefficients, depending on m, do not depend on higher derivatives of w. So, we can claim the following result.

Proposition 11
The coefficients A m and B m do not depend on the derivative of w of higher order than 4.
Let us take the neighborhood U such that sup 0≤k≤2 △ k w L ∞ ≤ C 0 for w ∈ U.
Now we analyze the coefficients c Then the last term of equation (60) can be rewritten as where in the summation k, j run so that k + j = m under the conditions 1 ≤ k and j ≤ m − 1. We have the following recurrence formula: We start the above recurrence formula from is a linear combination of components of a. In fact we see Therefore, we see that and γ (m) 01 0 ≤ C a 1 for any m. Then, keeping in mind these, the recurrence formula show that γ (m) 1k 0 + γ (m) 0k 0 ≤ C a k for any m, since the second order differential operations occur only with descent of k, provided that k ≥ 2.
We shall use the following Proposition 12 If non-negative integers j, k, m satisfy j + k = m, then there is a constant C such that A proof of this proposition will be given in Appendix B. Now we assume that for 0 ≤ j ≤ m − 1 and assume g 0 + a 0 ≤ C 0 . Then Proposition 12 implies that, for j + k = m, Thus the energy estimate gives us since e εAmT and B m are independent of higher derivatives of w. By induction we can claim that the inequality holds for any m ∈ N. We note that Hence by Sobolev imbedding theorem we have the tame estimate n+σ ′ ), provided that σ = 4 + [N/4], σ ′ = 1 + [N/4], for h = V P(w)g. Now, for R > 1, we denote by X(0, R) the Hilbert space of functions y(z) of 0 ≤ z ≤ R endowed with the inner product Moreover, for m ∈ N, we denote by X 2m (0, R) the space of functions y(z) of 0 ≤ z ≤ R for which the derivatives (−△) j y ∈ X exist in the sense of distribution for 1 ≤ j ≤ m. And we use the norm Let us denote by Z 2m (0, R) the closure of C ∞ 0 (0, R) in the space X 2m (0, R). It is well-known in the theory of elliptic equations that there is a continuous linear extension Ψ : X 2m (0, 1) → Z 2m (0, 2) such that y X2m(0,1) ≤ Ψy X2m(0,2) ≤ C y X2m(0,1) .

B. Nirenberg-Moser type inequalities
Let us prove Proposition 8, Proposition 9 and Proposition 12 in the sequel.
Repeating this estimate, we get On the other hand, sinceḊ 2 = △ − N −1 2 D and D△ − △D = D 2 , we havė with some constants C kµ = C(k, µ, N ). Then it follows from (B.3) and Proposition 7 that Ḋ 2µ D j y ≤ C △ µ+j y . with some constants C kℓjm . The summation is taken for 0 ≤ j ≤ k ≤ m, 0 ≤ ℓ ≤ 2(m − k). By estimating each term of the right-hand side of (B.6), we can obtain the assertion of Proposition 8. In fact, we consider the term provided that ℓ ′ + ℓ + 2(j ′ + j) = 2m. By (B.2) and (B.4) we have for some positive constants C, C ′ and C ′′ . Here we have used the Nirenberg interpolation forḊ = ∂/∂r. The same estimate holds for Ḋ ℓ ′ D j ′ f . Therefore we have Proof of Proposition 9.