STABILITY OF GASEOUS STARS IN SPHERICALLY SYMMETRIC MOTIONS∗
SONG-SUN LIN†
Abstract. We study the linearized stability of stationary solutions of gaseous stars which are
in spherically symmetric and isentropic motion. If viscosity is ignored, we have following three types of problems: (EC), Euler equation with a solid core; (EP), Euler–Poisson equation without a solid core; (EPC), Euler–Poisson equation with a solid core. In Lagrangian formulation, we prove that any solution of (EC) is neutrally stable. Any solution of (EP) and (EPC) is also neutrally stable
when the adiabatic index γ ∈ (43, 2) and unstable for (EP) when γ∈ (1,43). Moreover, for (EPC)
and γ∈ (1, 2), any solution with small total mass is also neutrally stable. When viscosity is present
(ν > 0), the velocity disturbance on the outer surface of gas is important. For ν > 0, we prove that the neutrally stable solution (when ν = 0) is now stable with respect to positive-type disturbances,
which include Dirichlet and Neumann boundary conditions. The solution can be unstable with
respect to disturbances of some other types. The problems were studied through spectral analysis of the linearized operators with singularities at the endpoints of intervals.
Key words. stability, isentropic gas, self-gravitating, solid core, limit-point singularity AMS subject classifications. 35J65, 35P30, 85A15, 85A20
PII. S0036141095292883
1. Introduction. In this paper, we shall study the stability problem of gaseous
stars which are in spherically symmetric and isentropic motion. The problem orig-inated in Newtonian (nonrelativistic) astrophysical theory. A model equation for describing such motion is shown below:
∂ρ ∂t + v ∂ρ ∂r + ρ ∂v ∂r + 2 rρv = 0, (1.1) ρ ∂v ∂t + v ∂v ∂r +∂p ∂r =− ρ r2 M0+ 4πδ Z r R0 ρ(t, s)s2ds + ν ∂2v ∂r2 + 2 r ∂v ∂r − 2 r2v , (1.2) p = Aργ, (1.3)
where t≥ 0 and 0 ≤ R0 < r <∞; see, e.g., [6, 7, 8, 9, 10, 11, 12, 13, 14, 20]. Here
the unknown variable ρ is the density of the gas and v is the outward velocity. p is the pressure, A is a positive constant which is related to entropy, and γ∈ (1,2) is the adiabatic exponent.
The explanation of the physical parameters δ, M0, R0, and ν is as follows:
δ is the effect of self-gravitating of gas, the mutual graviational attraction among gas molecules, and is assumed to be either 0 or 1. If δ = 0, we ignore the effect of self-gravitating. This may happen when the total amount of gas is relatively small. If δ = 1, we then consider the self-gravitating of gas to be important.
∗Received by the editors October 6, 1995; accepted for publication January 31, 1996. This
research was partially supported by the National Science Council of the Republic of China. http://www.siam.org/journals/sima/28-3/29288.html
†Department of Applied Mathematics, National Chiao Tung University, Hsin-chu 30050, Taiwan,
Republic of China ([email protected]). 539
M0 is the total mass of the solid core surrounded by the gas. If M0 = 0, then
we assume that R0 = 0. This is the case when there is no solid core and also no
vacuum in the central part of the gaseous body. If M0> 0, we assume that there is
a stationary, spherical solid core surrounded by the gas. In this case, we normalize the radius of the solid core with R0 = 1. We also assume that the gas is in contact
the surface of the solid core, i.e., no vacuum exists between the core and the gas. A nonslip condition is now imposed at the interface, i.e.,
v(t, 1) = 0 for t≥ 0. (1.4)
We note that astrophysicists consider the solid core to be made of condensed gases in which there may be complicated activity that influences the surrounding gas. How-ever, for mathematical simplicity, we will consider these condensed gases to be a solid core and ignore their influence on the surface gas.
ν is viscosity coefficient. We are mainly concered with inviscid flow, i.e., ν = 0. After presenting a detailed study of inviscid flow, we will discuss the effect of viscosity on the stability of stationary solutions.
If viscosity is ignored, then according to the different combinations of parameters δ, M0, and R0, we have following three types of problems:
(EC): Euler equation with solid core (δ = 0, M0> 0, R0= 1, ν = 0);
(EP): Euler–Poisson equation without solid core (δ = 1, M0= 0, R0= 0, ν = 0);
(EPC): Euler–Poisson equation with solid core (δ = 1, M0> 0, R0= 1, ν = 0).
If viscosity is present, i.e., ν > 0, then the Euler equation will be replaced by a Navier– Stokes equation and we have problems (NSC), (NSP), and (NSPC), respectively.
The stationary solution (ρ(r), 0) of (1.1)–(1.3) satisfies dp dr =− ρ r2 M0+ 4πδ Z r R0 ρ(s)s2 . (1.5)
If we introduce the variable u(r) and the parameter µ > 0 in ρ = Cγuq and µ = dγM0, where q = 1 γ− 1, Cγ = Aγ 4π(γ− 1) 1 2−γ , and dγ= (4π)γ−1γ− 1 Aγ 1 2−γ , then (1.5) and (1.4) can be studied by considering the following initial-value problems: for (EC), u00+2 ru 0= 0, r > 1, u(1, α, µ) = α and u0(1, α, µ) =−µ ) ; (1.6) for (EP), u00+2 ru 0+ uq= 0, r > 0, u(0, α) = α and u0(0, α) = 0 ) ; (1.7)
and for (EPC),
u00+2 ru 0+ uq = 0, r > 1, u(1, α, µ) = α and u0(1, α, µ) =−µ ) . (1.8)
Here α > 0 is taken as a shooting parameter.
The total mass of the stationary solution u is given by ˜ M (u) = 4πCγ Z R R0 uq(r)r2dr, (1.9)
where R∈ (R0,∞] is the first zero of u, i.e.,
u(R) = 0 and u(r) > 0 in (R0, R).
From a physical point of view, we are only interested in a stationary solution with finite total mass.
The solution of (1.6) with finite total mass can be written explicitly as
u = µ 1 r− 1 R (1.10) for some R∈ (1, ∞].
The solution of (1.7) has been studied extensively by Lane et al.; see, e.g., [1]. Their solutions include the ball type (R <∞), the ground-state type (R = +∞), and the singularity type, i.e., limr→0+u(r) =∞.
Equation (1.8) has recently been studied in [5] and may have multiple solutions for certain µ and ˜M when q > 3.
The multiplicity results of these problems will be given in section 2.
In this paper, we mainly study the stability of stationary solutions obtained from (1.6), (1.7), and (1.8) since only the local existence and not the global-existance of the initial-value problem in (1.1)–(1.3) is known (see, e.g., [6, 7, 8, 9, 10, 11, 12, 13, 14]). We therefore need only study the linearized stability of these stationary solutions.
The linearized stability problem of the stationary solution ρ(r) will be studied in Lagrangian formulation. Indeed, equations (1.1)–(1.3) can be written in Lagrangian coordinates as ρt+ 4πρ(r2v)x= 0, (1.11) vt+ 4πr2px+ 1 r2(M0+ x) = 16π 2ν(r2ρv x)x− 2νv(r2ρ)−1, (1.12) r = R0+ 3 4π Z x 0 1 ρ(t, y)dy 1 3 and x = 4π Z r R0 ρ(s, t)s2ds,
where t ≥ 0 and x ∈ (0, ˜M ). We assume that the perturbation of (ρ(x), 0) is in a radial direction only and write
ρ(t, x) = ρ(x){1 + εeλtΦ(x)} and v(t, x) = εeλtΨ(x) (1.13)
in (1.11) and (1.12), where|ε| is small. Let φ(x) =
Z x
0
Φ(y) ρ(y)dy.
Then the linear equations for Φ and Ψ can be simplified as follows: (ρpφx)x− 1 πγr3pxφ = λ2 γ(4πr2)2φ − λν 4πγr2 16π2 r4ρ 1 4πr2φ x x − 2 r2ρ 1 4πr2φ (1.14)
with boundary condition
φ(0) = 0, (1.15) where r = 1 + 3 4π Z x 0 1 ρ(y)dy .
Transforming (1.14) into r-coordinates and writting φ(x) = ψ(r), we obtain Lψ≡ (r−2pψ0)0−4 γr −3p0ψ = λ2 γ r −2ρψ−λν γ (r −2ψ0)0 (1.16)
with ψ(R0) = 0, where p is the pressure in r-coordinates. Since ρ(R) = 0, if ν = 0,
then (1.16) is singular at r = R. We can prove that the singularity at R is a limit-point type and so L is self-adjoint. Therefore, λ2 is real for any eigenvalue λ when
ν = 0. Now ρ is called neutrally stable if λ2< 0 for any eigenvalue λ and unstable if
λ2
1 > 0 for some eigenvalue λ1. Hence if ν = 0, then neutrally stable is the best we
can hope for. Indeed, when ν = 0, we have our stability results for ball-type solutions as follows.
Theorem 1.1. Assume that ν = 0 and ball-type solutions have been considered. Then
(I) any solution of (EC) is neutrally stable;
(II) any solution of (EP) is neutrally stable if q ∈ (1, 3) and unstable if q > 3; and
(III) for (EPC), we have the following: (i) any solution is neutrally stable if q∈ (1, 3],
(ii) for any q > 1, u(·, α, µ) is neutrally stable if α ∈ (0, µ], and (iii) if|R − 1| is sufficiently small, then it is neutrally stable.
Some stability results concerning ground-state-type and singularity-type solutions are also presented in section 4.
When viscosity is present and λ /∈ [−γνp(R0), 0], then (1.16) is regular at R. In this
case, the viscosity term plays the dominant role in studying the eigenvalue problems. Now ρ is called stable if Reλ < 0 for any eigenvalue λ and unstable if Reλ1> 0 for some
eigenvalue λ1. Note that (1.16) is genuinely quadratic in λ (linear in λ2 when ν = 0)
and λ is complex in general. Hence when ν > 0, we may have better than the neutral stability that we have when ν = 0. Since the outer surface of gas is a free surface, the velocity disturbance Ψ on it will play an important role. For example, we have stability results for (EC), (EP), and (EPC) as follows.
Theorem 1.2. Let u be a neutrally stable, ball-type stationary solution of (EC), (EP), or (EPC) when ν = 0. Then for any ν > 0, u is stable with respect to Ψ = ψ1+ iψ2 if ψj( ˜M )ψ0j( ˜M )≤ 0 for both j = 1 and 2 on the gas surface. On the other
hand, there is a positive constant κ∗ depending on u such that u is unstable with respect to Ψ = ψ1+ iψ2 if ψ10( ˜M )/ψ1( ˜M )≥ κ∗ and some ψ2.
The precise definition of stability with respect to the boundary disturbance Ψ is given in section 5.
The paper is organized as follows. In section 2, we recall some useful multiplic-ity results for stationary solutions with finite total masses. Their stabilities will be investigated in subsequent sections. In section 3, we study the linearized operators L and prove that they have limit-point-type singularities at their endpoints. We also provide a useful comparison lemma to test for stability. In section 4, we prove various stability results, which include Theorem 1.1. The solutions for other types of stability problems are also studied. In section 5, we study the effect of viscosity on stability problems and prove some results, including Theorem 1.2. In Appendix A, we study the asymptotic behavior of solutions of (1.16) at R when ν = 0, which is very useful for studying ball-type solutions. In Appendix B, we recall Friedrichs’ criteria for the spectrum discreteness of differential operators that have singular endpoints. These criteria are very useful in studying ground-state-type and singularity-type solutions.
2. Stationary solutions. In this section, we recall some multiplicity results
for stationary solutions without interior vacuums and with finite total masses. Let R≤ ∞ be the first zero of solution u and ˜M (u) be the total mass given in (1.9). For notational simplicity, we omit the constant 4πCγ in (1.9) and then define
M (u) = Z R
R0
u(r)qr2dr, (2.1)
where R0= 0 for (EP) and R0= 1 for (EC) and (EPC).
Since the total mass of a gas remains constant while it is in motion and it may tend to a stationary state as time goes by, it is useful to know the numbers of stationary solutions for the same total mass. Hence we try to answer the following questions.
Questions. Given M > 0, how many solutions u are there for (EP) with M (u) = M ? Given µ > 0 and M > 0, how many solutions u are there for (EC) or (EPC) with M (u) = M ?
Complete answers of (EC) and (EP) can be provided; see, e.g., [1]. However, (EPC) has only recently been studied and the result is complete for 1 < q ≤ 3 but partial when q > 3; see [5].
First, for (EC), the solution of (1.6) is given by u(r, α, µ) = α− µ + µ1
r. (2.2)
If α∈ (0, µ), then u(R(α, µ), α, µ) = 0 with R(α, µ) = 1−α µ −1 . In this case, we may write u(·, α, µ) = uR,µ with
uR,µ(r) = µ 1 r− 1 R .
It is clear that M (uR,µ) is strictly increasing in R and tends to
Mq∗= +∞ if 1 < q≤ 3, 1 q− 3 · µ q if q > 3.
If α = µ, then R(µ, µ) = +∞ and
u(r, µ, µ) =µ r with
M (u(·, µ, µ)) = Mq∗.
If α > µ, then M (u(·, α, µ)) = ∞, which is not of physical interest. Hence we have the following unique result for (EC).
Proposition 2.1. For any q > 1, µ > 0, and M ∈ (0, Mq∗), there is a unique solution uR,µ for (EC) such that M (uR,µ) = M .
Next, for (EP), we consider the initial-value problem u00+2 ru 0+ uq = 0, r > 0, (2.3) u0(0, α) = 0 and u(0, α) = α > 0. (2.4)
It is known that solutions of (2.3) have similar properties. Indeed, if u(r) is a solution of (2.3), then for any β > 0,
uβ(r) = βσu(βr) (2.5)
is also a solution, where σ = 2
q−1. The total mass of uβ is
M (uβ) = β 3−q q−1M (u).
(2.6)
The property (2.5) is related to the following classical Lane–Emden–Fowler trans-formations:
Let
r = e−τ and z(τ ) = rσu(r). (2.7)
(2.3) can then be transformed into the autonomous equation z00+ (2σ− 1)z0+ σ(σ− 1)z + zq = 0 (2.8)
or, equivalently, the dynamic system
z0= y,
y0=−{2σ − 1)y + σ(σ − 1)z + zq}. (2.9)
If q∈ (1, 3], then 0 = (0, 0) is the only equilibrium for (2.9) on the right half-plane R2
+={(z, y) : z ≥ 0}. If q > 3, then there is another equilibrium S = (zσ, 0), where
zσ={σ(1 − σ)} 2
σ.
(2.10)
0 is always a saddle point with the unstable manifold Γ, which is leaving in the direction (1, 1− σ)t, and the stable manifold ˜Γ, which is arriving for the direction (1,−σ), where (a, b)t is the transpose of vector (a, b) in R2. Let
q+= 1 + 2
σ+ and σ
+=√2−1
2.
It is than easy to verify that q+∈ (3, 5).
We now list some useful properties of the equilibrium S and system (2.9) on the phase plane R2
+.
Proposition 2.2. (I)
(i) If q∈ (3, q+), then S is a stable improper node. (ii) If q = q+, then S is a stable proper node. (iii) If q∈ (q+, 5), then S is stable spiral. (iv) If q = 5, then S is a center.
(v) If q > 5, then S is an unstable spiral. (II)
(i) For q∈ (3, 5), the unstable manifold Γ of 0 is a heteroclinic orbit connecting 0 and S. There is no nontrivial periodic orbit on R2
+.
(ii) For q = 5, Γ = ˜Γ, i.e., Γ is a homoclinic orbit of 0. The inside of Γ is covered by a family of concentric periodic orbits centered around 0.
(iii) For q > 5, the stable manifold ˜Γ of 0 is a heteroclinic orbit connecting 0 and S.
The proofs are elementary and omitted; see [1] for details.
Every trajectory in the phase plane of (2.9) represents a family of self-similar solutions in (2.5). After carefully investigating the trajectories in the phase plane, we have exactly four types of solutions for (EP) with finite total mass for (EP):
(i) B-type solutions: ball-type solutions that lie on ˜Γ and appear when q ∈ (1, 5);
(ii) G-type solutions: ground-state solutions that also lie on ˜Γ and only appear when q≥ 5; they also have fast decay rates as r → +∞, i.e.,
lim
r→+∞ru(r)∈ (0, ∞); (2.11)
(iii) SB-type solutions: ball-type solutions with a singularity at r = 0 that appear when q ∈ (3, 5) and are trajectories between Γ and ˜Γ that have a weak singularity, i.e., u satisfies
lim r→0+r
σu(r)∈ (0, ∞); (2.12)
(iv) SG-type solutions: ground-state solutions with a singularity at r = 0 that lie on Γ and satisfy (2.12); they also appear when q∈ (3, 5);
Note that if the singularity at r = 0 is strong, i.e., lim
r→0+ru(r) > 0, (2.13)
then u has an infinite total mass: for example, the SB-type solution when q∈ (1, 3). If the ground-state solution has a slow-decay rate at∞, i.e.,
lim r→∞r
σu(r) > 0, (2.14)
then u also has an infinite total mass, which includes the following cases: (i) qσr−σ for q > 3; this corresponds to the equilibrium S = (zσ, 0); (ii) when q = 5, all trajectories lie on homoclinic orbit Γ;
(iii) when q > 5, all trajectories spiral out from S;
With this preparation complete, we can now state our unique results for (EP). Proposition 2.3. For (EP), we have the following:
(i) If q ∈ (1, 3) and any M > 0, there is a unique—B-type—solution u such that M (u) = M .
(ii) If q = 3, only a special ˆM of a stationary solution u—a B-type solution— admits. (All similar solutions of u also have the same total mass ˆM .)
(iii) If q∈ (3, 5) and any M > 0, there are unique B-type, SB-type, and SG-type solutions with the same total mass M .
(iv) If q = 5 and any M > 0, there is a unique—G-type—solution u such that M (u) = M .
(v) If q > 5, there is no stationary solution with finite total mass.
Proof. The proofs are based on the phase-plane analysis in (2.9) and the use of (2.6), and they are elementary. Thus the details are omitted.
As for (EPC), there are two types of solutions with finite total mass: (i) BC-type solutions: ball-type solutions with solid cores;
(ii) GC-type solutions: ground-state solutions with solid cores that satisfy (2.11). We recall some results from [5].
Proposition 2.4. For (EPC), we have the following:
(i) When q∈ (1, 3], for any µ > 0 and M > 0, there is a unique—BC-type— solution u that satisfies M (u) = M .
(ii) When q > 3, for any µ > 0, the solution set is the disjoint union of N many connected components Ck ={u(·, α, µ) : α ∈ (˜αk, ˆαk)}, k = 1, 2, . . . , N, where
N = N (µ, q) is a positive integer or infinity.
At Ck with k ≥ 2, M((u(·, α, µ)) tends to infinity at at least one end. At
C1, ˜α1= 0 and ˆα1> µ.
For detailed statements of Proposition 2.4(ii), see Theorems 3.5, 3.7, 3.9, and 3.13 in [5].
Remark 2.5. When there is a vaccum in the central part of the gaseous body that is also stationary, then u satisfies
u00(r) +2 ru 0(r) + uq(r) = 0, R 1< r < R2, (2.15) u(R1) = 0 = u(R2), (2.16)
where 0 < R1< R2≤ ∞. For any q > 1 and 0 < R1< R2 <∞, Ni and Nussbaum
[17] proved that there is a unique positive solution of (2.15) and (2.16). In contrast to Proposition 2.3(v), for any q > 1, the solution u of (2.15) and (2.16) with R2<∞
has a finite total mass. We can then ask the following questions: Given q > 1 and M > 0, how many solutions u are there for (2.15) and (2.16) with M (u) = M ? What is the stability of these annular-type solutions? These problems will be studied later.
3. Linearizations. In this section, we will use a Lagrangian formulation to
study the stability of the stationary solutions obtained in last section. Since we want to know the stability result when the outer surface of the gas is also perturbed, it is convenient to work in Lagrangian coordinates. We study only the inviscid flow in this section and defer study of the the viscous flow to section 5.
For notational simplicity, we replace r with r in (1.16) with ν = 0. We then obtain
Lψ = −`W ψ in (R0, R),
(3.1)
where Lψ ≡ (r−2pψ0)0− 4 γr −3p0ψ , W (r)≡ ρ(r) γr2, and ` =−λ 2 . ψ also satisfies the boundary condition
ψ (R0) = 0.
(3.2)
In terms of u, (3.1) can also be written as L0ψ ≡ ψ 00+ (1 + q)u 0 u − 2 r ψ 0−4q r u0 uψ =−`(γAC γ−1 γ )−1 ψ u. (3.3)
Since u(R) = 0,L is singular at R. Furthermore, L is also singular at r = 0 for (EP). When R <∞, we first study the asymptotic behavior of solution ψ of (3.1) at R. Indeed, we have the following result. (The proof is given in Appendix A.)
Lemma 3.1. Let R <∞. If ` is real and ψ is a (real) solution of (3.1) in (R0, R), then either ψ is bounded at r = R or ψ (r) = (R− r)−qψ ˆ(r) for r close to R, with
ˆ
ψ (R)6= 0, and ˆψ is continuous at R. Furthermore, in the former case, ψ is C2 at R,
and in the latter case,
ψ 0(r) = q(R− r)−q−1ψ ˆ(R) + o((R− r)−q−1) (3.4)
as r→ R−.
Similarly, if R0 = 0, then either ψ (0) 6= 0 or |ψ (r)| ≤ Cr3 for r close to 0 and
some C > 0.
To study the singularity type at R, it is convenient to remove the weight function W from right-hand side of (3.1). Indeed, if R0= 1, let r0= 1, and if R0= 0, choose
any r0∈ (0, R) and fix it. Then define
s = s(r) = Z r r0 W (τ )dτ = 1 γ Z r r0 τ−2ρ(τ )dτ (3.5) and S0= Z R0 r0 W (τ )dτ and S = Z R r0 W (τ )dτ.
It is clear that S0= 0 when R0= 1 and S0=−∞ when R0= 0. Furthermore, W > 0
in (R0, R) implies that the inverse function of s(r) exists. We may denote it by
r = r(s) for s∈ (S0, S). Let
χ(s) = ψ (r(s)). Then (3.1) is transformed into
˜ Lχ = −`χ in (S0, S), (3.6) where ˜ Lχ = 1 γ d ds r−4pρdχ ds − 4p0 rρχ.
Let L2(S
0, S) be the complex-valued L2-space on (S0, S) with the standard inner
product (χ, ˜χ)≡ Z S S0 χ ˜χds. (3.7) It is clear that (χ, ˜χ) = Z R R0 ψ ˜ψW (r)dr≡ (ψ, ˜ψ)w. (3.8)
Here ( , )w defines an inner product in space L2w(R0, R) by (3.8).
Now we can prove ˜L has limit-point-type singularity at S.
Lemma 3.2. If S < ∞, then ˜L is limit-point type at S. Furthermore, for (EP), ˜
L is also limit-point type at −∞.
Proof. From [2], it is known that ˜L is the limit-point-type singularity at S if we can find a solution pair{`, χ} for (3.6) in a neighborhood of S such that χ is not L2. This can be done as follows:
Since p = ACγγuq+1 (3.9) and p0= A(q + 1)Cγγuqu0, we have p0(r) rρ(r) = A(q + 1)C γ−1 γ u0(r) r . (3.10) Hence (3.10) implies lim r→R p0(r) rρ(r)= A(q + 1)C γ−1 γ u0(R) R . (3.11)
Fix ˆS∈ (S0, S). For any real `, let χ be the real solution of the following
initial-value problem: ˜ Lχ = −`χ in ( ˆS, S), (3.12) χ( ˆS) = 0 and χ0( ˆS) = 1. (3.13)
Denote ˆR = r( ˆS). Now (3.11) implies that there exists `0< 0 such that
`0rρ(r)− 4p0(r)≤ 0
(3.14)
in [ ˆR, R]. We claim that χ /∈ L2( ˆS, S) if `≤ ` 0.
Indeed, if χ∈ L2( ˆS, S), then Lemma 3.1 and (3.8) imply that χ is bounded at S.
From (3.12) and (3.13), we obtain 1 γ Z S ˆ S r−4pρ dχ ds 2 ds = ` Z S ˆ S χ2ds− 4 Z S ˆ S p0 rρχ 2ds. (3.15)
Now the left-hand side of (3.15) is positive and the right-hand side of (3.15) is non-positive when `≤ `0, a contradication. This implies that χ /∈ L2( ˆS, S) for ` ≤ `0.
Therefore, ˜L is limit-point type at S. For (EP), (3.10) and (1.7) imply that
lim r→0+ p0(r) rρ(r) =− A 3(q + 1)C γ−1 γ u q(0). (3.16)
Now using (3.11) and (3.16), we can choose `0 < 0 such that (3.14) holds in (0, ˆR).
Let ψ(r) = χ(s(r)); then Lemma 3.1 implies either ψ(0)6= 0 (3.17)
or
|ψ(r)| ≤ Cr3 and |ψ0(r)| ≤ Cr2
(3.18)
for some C > 0. Now we can rule out the possibility of (3.18) when `≤ `0. Indeed,
if (3.18) holds, then 0 < 1 γ Z Sˆ −∞ r−4pρ dχ ds 2 ds = Z Sˆ −∞ `−4p 0 rρ χ2ds < 0, a contradication.
Hence we must have (3.17) when `≤ `0, i.e., χ /∈ L2(−∞, ˆS). Therefore, ˜L is a
limit-point-type at−∞. The proof is complete.
An immediate consequence of Lemma 3.2 is that ˜L is self-adjoint. Indeed, we have the following result (for the proof, see [2]).
Corollary 3.3. For (EC) and (EPC), if R < ∞, let D1 be the set of all functions χ such that
(i) χ is differentiable and χ0 is absolutely continuous on [0, ˆS] for any ˆS < S, (ii) χ and ˜Lχ ∈ L2(0, S), and
(iii) χ(0) = 0.
Then ˜L is self-adjoint, i.e.,
( ˜Lχ, ˆχ) = (χ, ˜Lˆχ) (3.19)
for all χ and ˆχ inD1.
Similarly, for (EP), letD0 be the set of all functions χ such that
(i)0 χ is differentiable and χ0 is absolutely continuous over (−∞, ˆS] for any ˆS ∈ (−∞, S) and
(ii)0 χ and ˜Lχ ∈ L2(−∞, S).
Then ˜L is self-adjoint.
Furthermore, using Friedrichs’ criteria, we can prove that ˜L has only a discrete spectrum.
Theorem 3.4. Let u be a stationary solution of (EC), (EP), or (EPC) with R < ∞. The spectra of ˜L consist of sequences of strictly increasing eigenvalues {`j}j=1 with associated eigenfunctions {χj}∞j=1in D1 (or D0).
Proof. We first claim that no continuous spectrum comes out of S. Indeed, using (3.5), it can be verified that
s(r) = S− c1(R− r)q+1+ o((R− r)q+1)
(3.20)
for r close to R, where c1> 0 depends on R, u0(R), and γ. Let a(s) = 1 γr −4p(r)ρ(r), b(s) = 4p0 rρ, and c(s) = 1. (3.20) then implies a(s) = c2(S− s)2−ε+ o((S− s)2−ε), (3.21)
where ε = q+11 and c2> 0. Let
h(s) = Z s 0 1 a(τ )dτ. Then (3.21) implies h(s) = c3(S− s)ε−1+ o((S− s)ε−1) (3.22)
for s close to S, where c3> 0.
Hence (3.21) and (3.22) imply that
4ah2= c4(S− s)ε+ o((S− s)ε),
(3.23)
where c4> 0. (3.11) now implies that b(s) is bounded at S. Therefore, (3.23) implies
that Z(s) = 1 c b + 1 4ah2 → +∞ (3.24)
as s→ S. By Proposition B.3 in Appendix B, no continuous spectrum comes out of S, and ˜L is totally descrete in R1. In particular, for (EC) and (EPC), the spectrum
of ˜L is a sequence of eigenvalues {`j}∞j=1such that lim
j→∞`j= +∞. (3.25)
For (EP), we also need to prove that (3.24) holds when s→ −∞. From (3.5), we have s =−c0r−1+ o(r−1)
(3.26)
for r→ 0+, where c0> 0. Therefore,
a(s) = c5s4+ o(s4) (3.27) as s→ −∞, where c5> 0. Let h(s) = Z 0 s 1 a(τ )dτ. Then (3.27) implies h(s) = c6(−s)−3+ o(s−3) (3.28)
as s→ −∞, where c6> 0. Hence (3.27) and (3.28) imply
4ah2= c7s−2+ o(s−2)
as s→ −∞. (3.24) then follows from last equation and (3.16). Hence no continuous spectrum comes out from−∞ for (EP). The proof is complete.
From Lemma 3.4, ˜L has only the real eigenvalue `. Therefore, λ is either real or purely imaginary for any eigenvalue λ.
From these observations, we then introduce the following notion of stability. Definition 3.5. Let u be a ball-type stationary solution of (EC), (EP), or (EPC), and let {`j}∞j=1 be the associated eigenvalues of ˜L given in Theorem 3.4. u is then
called neutrally stable if `1 > 0 (i.e., λ1 = ±i
√
`1 is purely imaginary), is called
unstable if `1 < 0 (i.e., λ1 = ±
p
|`1| is real), and is called marginally stable if
`1= 0.
A similar definition can also be given for ground-state- and singularity-type solu-tions.
Remark 3.6. From Lemma 3.1 and Theorem 3.4, if ψ is an eigenfunction, then ψ(R) is bounded. Furthermore, for (EP), ψ(r) = O(r3) as r → 0+. Moreover, the least eigenvalue `1 can be obtained by a variational method; see, e.g., [3]. Indeed, for
(EC) or (EPC), we have `1= inf Q(ψ) I(ψ) : ψ(1) = 0 and ψ∈ C 1[1, R] , where Q(ψ) = Z R 1 r−2p(r)ψ02(r) + 4 γr −3p0(r)ψ2(r) dr and I(ψ) = 1 γ Z R 1 r−2ρ(r)ψ2(r)dr.
A similar formulation also holds for (EP) with ψ(r) = O(r3) as r→ 0+.
The following comparison lemma is very useful for testing the stability of station-ary solutions.
Lemma 3.7. Let u be a BC-type stationary solution for (EC) or (EPC). Then the following hold:
(i) If there exists a ˜ψ∈ C2([1, R]) with ˜ψ(1) = 0, ˜ψ > 0 in (1, R], that satisfies L0ψ˜≤ 0 (but not ≡) in (1, R),
then u is neutrally stable.
(ii) If there exists a ψ ∈ C2([1, R]) with ψ(1) = 0, ψ > 0 in (1, R], that satisfies
L0ψ≥ 0 (but not ≡) in (1, R),
then u is unstable.
A similar result also holds for a B-type stationary solution u for (EP) provided the comparison function ˜ψ (or ψ) satisfies ˜ψ (or ψ) ∈ L2
w(0, R) and L ˜ψ (or Lψ) ∈
L2
w(0, R).
Proof. Let ψ1 > 0 in (R0, R) be the associated eigenfunction with respect to `1
in (3.20). If there is a ˜ψ that satisfies all conditions in (i), then it is easy to see that L ˜ψ≤ 0 in (1, R), which implies that
0 = Z R 1 ( ˜ψLψ1− ψ1L ˜ψ)dr >−`1 Z R 1 W ˜ψψ1dr.
Therefore, `1> 0. This proves (i). (ii) and the cases for (EP) can be proved
analo-gously. The proof is complete.
4. Stability results. In this section, we shall use the methods developed in the
preceding section to study the stability of various stationary solutions. We begin with ball-type solutions, proceed to ground-state solutions, and finally conclude with singular solutions.
4.1. Ball-type solutions. We first introduce an auxiliary operator ˜L, defined as ˜ Lψ≡ ψ00−(q + 3) r ψ 0+4q r2ψ. (4.1) ˜
L is closed related to L0, as can be seen from the following:
L0ψ = ˜Lψ + (1 + q)ψ0−4q r ψ u0 u + 1 r . (4.2)
The following results for operator ˜L are very useful in constructing the comparison functions ˜ψ and ψ according to Lemma 3.7.
Lemma 4.1. For any q > 1, we have ˜L(r4) = 0 and ˜L(rq) = 0. Moreover, if q = 4, we also have ˜L(r4log r) = 0.
Furthermore, if we let (i) ˜ψ = r4− rq if q∈ (1, 4), (ii) ˜ψ = r4log r if q = 4, and
(iii) ˜ψ = rq− r4 if q∈ (4, ∞), then we have (a) ˜L ˜ψ = 0 for r > 1, (b) ˜ψ(1) = 0, and (c) the following: (1 + q) ˜ψ0−4q r ˜ ψ > 0 for r > 1. (4.3)
Proof. The computations are straightforward, so we verify only the last inequality and omit the others. Indeed, for q6= 4,
(1 + q)(rq− r4)0−4q r (r q− r4) = q(q− 3)rq−1− 4r3, and for q = 4, (1 + q)(r4log r)0−4q r (r 4log r) = 4r3log r + 5r3.
The result follows.
Next, it is easy to verify the following lemma, so we omit the proof. Lemma 4.2. If u > 0 in (1, R) and satisfies the equation
u00+2 ru 0+ f (u) = 0 for r > 1, then d dr u0 u + 1 r =− 1 r2u2{(ru0+ u) 2+ r2uf (u)}.
In particular, if u(·, α, µ) is a solution of (1.6) or (1.8), then α ≤ µ implies u0 u + 1 r < 0 in (1, R). (4.4)
We can now establish the stability results for (EC) and (EPC) when α≤ µ.
Theorem 4.3.
(i) For any q > 1, µ > 0, and R > 1, the solution uR,µ of (EC) is neutrally
stable.
(ii) For any q > 1, let u(·, α, µ) be the solution of (EPC). Then u(·, α, µ) is neutrally stable if α≤ µ.
Proof. It is not difficult to verify R(α, µ) <∞ when α ≤ µ in (ii). Let ˜ψ be given as in Lemma 4.1. Then for both (i) and (ii), Lemmas 4.1 and 4.2 imply L ˜ψ < 0 in (1, R).
Thus Lemma 3.7 implies that uR,µ and u(·, α, µ) with 0 < α ≤ µ are neutrally stable. The proof is complete.
We can also establish other stability results for (EPC) by choosing appropriate comparison functions and applying Lemma 3.7. For example, we can prove the fol-lowing theorem.
Theorem 4.4. For (EPC), we have the following:
(i) If q∈ (1, 3], then all BC-type solutions are neutrally stable.
(ii) For any q > 1, there is Rq > 1 such that u is neutrally stable whenever the
first zero R of u is less than Rq.
Proof. (i) It is known that R(α, µ) <∞ for any α > 0 and µ > 0 when q ∈ (1, 3]; see, e.g., [18]. Let ˜ψ = r3− 1. Then ˜ψ(1) = 0, ˜ψ > 0 in (0,∞), and
L0ψ =˜ u0 u(3− q)r 2+4q r ,
which is negative in (0,∞) if q ∈ (1, 3]. Thus by Lemma 3.7(i), u is neutrally stable. (ii) Let ˜ψ = log r. Then ˜ψ(1) = 0, ˜ψ > 0 in (1,∞), and
L0ψ =˜ −3r−2+
1 r
u0
u{(1 + q) − 4q log r}.
Therefore, L0ψ < 0 in (1, R) if R˜ ≤ Rq ≡ exp(1+q4q ). The result also follows from Lemma 3.7(i). The proof is complete.
Remark 4.5. By picking a comparison function ˜ψ different from log r in Theorem 4.4(ii), we can also obtain another ˜Rq, which ensures that u is neutrally stable when
R≤ ˜Rq.
By choosing an appropriate comparison function, we obtain the following stability results for (EP).
Theorem 4.6. For (EP), we have the following:
(i) If q∈ (1, 3), then any B-type solution is neutrally stable. (ii) If q = 3, then any B-type solution is marginally stable. (iii) If q∈ (3, 5), then any B-type solution is unstable.
Proof. Let ˜ψ = r3. Then ˜ψ(0) = ˜ψ0(0) = 0 and ˜ψ > 0 in (0,∞). Furthermore,
we have
L0ψ = (3˜ − q)r2
u0 u.
Hence the result follows by Lemma 3.7. The proof is complete.
Proof of Theorem 1.1. Combining the results from Theorems 4.3, 4.4, and 4.6, we obtain Theorem 1.1.
4.2. Ground-state solutions. From section 2, we know that if a
ground-state-type solution u has a finite total mass, then it is necessary that u have a fast decay rate, i.e.,
lim
r→∞ru(r) = m∈ (0, ∞). (4.5)
In this section, we will prove that the linearized operatorL associated with u has a continuous spectrum (0,∞). Therefore, u cannot be neutrally stable. In fact, it is either marginally stable or unstable.
Lemma 4.7. If u is a G- or GC-type solution and satisfies (4.5), then the lin-earized operator L of u is discrete below 0 and has a continuous spectrum (0, ∞).
Proof. In Lemma 3.4, we have shown that no continuous spectrum comes from r = 0 for L in (EP). Therefore, we need only study L as r → ∞. We may assume that m = 1 in (4.5). We then have
p(r) = ˜Ar−1−q+ o(r−1−q) and
p0(r) =−(1 + q) ˜Ar−q+ o(r−q)
as r→ ∞, where ˜A = ACγγ. As before, we have the following asymptotic expansions for the coefficients ofL as r → ∞:
a(r) = r−2p(r) = ˜Ar−3−q+ o(r−3−q), b(r) = 4 γr −3p0(r) =−4q ˜Ar−5−q+ o(r−5−q), and c(r) = 1 γr −2ρ(r) = ˆAr−2−q+ o(r−2−q)
as r→ ∞, where ˆA > 0 is a constant. Therefore, for large fixed ˆr, we have h(r) = Z r ˆ r dτ a(τ )={ ˜A(4 + q)} −1r4+q+ o(r4+q) as r→ ∞. We claim that Z(r) = 1 c(r) b(r) + 1 4a(r)h2(r) → 0 as r→ ∞. (4.6)
Indeed, it is clear that
4a(r)h2(r) = 4 ˜A−1(4 + q)−2rq+5+ o(r5+q) as r→ ∞. Therefore, we have b(r) + 1 4a(r)h2(r)= ˜A (4 + q)2 4 − 4q r−5−q+ o(r−5−q) = ˜ A 4(q− 4) 2r−5−q+ o(r−5−q).
Hence
Z(r) = A∗r−3(q− 4)2+ o(r−3)
for some A∗> 0. (4.6) follows. Now by Proposition B.3 (II)–(III) in Appendix B, the linearized operator L of u has a continuous spectrum (0, ∞) and is descrete below 0. The proof is complete.
An immediate consequence of Lemma 4.7 is the following theorem for ground-state-type stationary solutions.
Theorem 4.8. Any ground-state-type solution of (EC), (EP), or (EPC) is either marginally stable or unstable.
4.3. Singular solutions. In this section, we will continuously apply Friedrichs’
criteria to study the stability of singularity-type solutions. We know that if q∈ (3, 5) and u is a singular solution of (EP) with finite total mass, then u has a weak singularity at r = 0, i.e.,
lim r→o+r
σu(r) = m∈ (0, ∞). (4.7)
As in section 4.2, we are interested in the limit of Z(r) as r→ 0+. (4.7) now implies
the following expansions:
a(r) = r−2p(r) = ˜Ar−σ(q+1)−2+ o(r−σ(q+1)−2), b(r) = 4 γr −3p0(r) =−4σq ˜Ar−σ(q+1)−4+ o(r−σ(q+1)−4), and c(r) = ˆAr−2−σq+ o(r−2−σq) as r→ 0+ for some positive constants ˜A and ˆA.
Therefore, h(r) = Z r 0 ds a(s)= ˜A −1{3 + σ(q + 1)}−1rσ(q+1)+3, with h(0) = 0. It is straightfoward to compute Z(r) = A∗r−σ−2 1 4[3 + σ(q + 1)] 2 − 4σq + o(r−σ−2) = A ∗ 4 r −σ−2{4σ2+ 4σ− 7} + o(r−σ−2)
for some positive constant A∗.
Hence we obtain the following lemma.
Lemma 4.9. Let u be a singular solution of (EP) satisfying (4.7). Then we have the following:
(i) if q∈ (3, q+), then lim
r→0+Z(r) = +∞; (ii) if q = q+, then lim
r→0+Z(r) = 0; (iii) if q∈ (q+, 5), then lim
r→0+Z(r) =−∞;
and Ω0= Z rˆ 0 c(r) −a(r)Z(r) 1 2 dr <∞.
Therefore, by applying Theorem 4.6(iii), Lemma 4.9, and Proposition B.3, we obtain the following theorem for singularity-type solutions.
Theorem 4.10. For problem (EP), we have the following:
(i) If q∈ (3, q+), then any SB-type solution is unstable and has no continuous
spectrum. Any SG-type solution is also unstable but has a continuous spectrum (0,∞). (ii) If q = q+, then any SB-type and SG-type solution is unstable and has a
continuous spectrum (0,∞).
(iii) If q∈ (q+, 5), then any SB-type and SG-type solution is unstable, and there is a sequence of pure imaginary eigenvalues {λk} such that limk→∞λ2k=−∞.
Proof. For any q∈ (3, 5) and for an SB-type solution u, choose ˜ψ = r3. Then we
have
L ˜ψ = (3− q)r2u
0
u > 0.
Therefore, by modifying the proof of Lemma 3.5, we can prove that u is unstable. The remaining results follow from Lemma 4.9 and Proposition B.3. The details of the proof are omitted and the proof is complete.
5. Effects of viscosity. In this section, we shall study the effect of viscosity
on the stability problem of stationary solutions. From equation (1.2), it is clear that stationary solutions for inviscid flow are also solutions for viscous flow. As we have seen in the previous sections, the best possibilities for stationary solutions are neutrally stable in the inviscid case. It is known that neutral stability is very sensitive to disturbances. Therefore, we need to know what effect viscosity has on neutrally stable stationary solutions.
Since the gaseous mass is not confined from outside, its outer surface is a free surface maintained by the attraction of the core and its own gravitational forces. Presumably, the surface of the gas should be very sensitive to a direct disturbance of it. In this section, we show that this is the case, as mentioned in Theorem 1.2.
When viscosity its present, the linearized equation is Lψ = λ2W ψ− λν ˆLψ, (5.1) where ˆ Lψ ≡ 1 γ(r −2ψ0)0 (5.2) or, equivalently, r−2 p(r) + λν γ ψ0 0 −1 γ{4r −3p0(r) + λ2r−2}ψ = 0. (5.3)
When ν > 0, the eigenvlaue equation (5.1) is linear for ψ but quadratic for λ, which is different from ordinary eigenvalue problems. Indeed, if ν = 0 in (5.1), then (5.1) is linear for ` =−λ2. Since the coefficients of L, ˆL, and W are real, it is easy
to see that if{λ, ψ} is a solution of (5.1), then its conjugate {λ, ψ} is also a solution.
This property does not affect the stability, which depends on the sign of Reλ in the stationary solution.
In this section, we concentrate on the effects of viscosity and boundary distur-bances. Therefore, we restrict our study to ball-type solutions which are neutrally stable. The problems of unstable stationary solutions, ground-state solutions, and singularity-type solutions will be left for future study.
We first consider (EC) and (EPC) and then continue by studying (EP). Let
λ∗=γ νP (R0). (5.4)
When R0= 1, we will prove that (5.1) is regular on [1, R] when λ /∈ [−λ∗, 0]. Indeed,
for λ 6= −λ∗, let ψ(·, λ) = ψ(·, λ, ν) be the solution of (5.3) that satisfies the initial conditions ψ(1, 0) = 0 (5.5) and ψ0(1, λ) = 1. (5.6)
We can then prove the following result.
Lemma 5.1. Let u be a BC-type stationary solution of (EC) or (EPC). If λ /∈ [−λ∗, 0], then ψ(·, λ) is C2on [1, R] and is analytic in λ∈ C−[−λ∗, 0]. Furthermore,
if λ∈ (−λ∗, 0), then either ψ(·, λ) is bounded at r = ˆr or |ψ(r, λ)| grows like | log |r−ˆr|| as r → ˆr, where ˆr ∈ (1, R) satisfies p(ˆr) + λνγ = 0. If λ =−λ∗, then any nontrivial solution ψ of (5.3) is unbounded in a neighborhood of r = 1. The case in which λ = 0 was studied in Lemma 3.1.
Proof. Let λ = λ1+ iλ2and ψ = ψ1+ iψ2 in (5.3) and denote
a = r−2 p + λ1νγ , b = λ2 ν γr −2, c = 1 γ{4r −3p0+ r−2(λ2 1− λ 2 2)}, d = 2 γλ1λ2r −2.
Then it is clear that a2(ˆr) + b2 = 0 for some ˆr ∈ [1, R] if and only if λ
2 = 0 and
λ1∈ [−λ∗, 0]. In this case, p(ˆr) + λ1γν = 0.
Now (5.3) can be written as the following system of equations: (aψ10 − bψ20)0 = cψ1− dψ2,
(bψ01+ aψ20)0 = dψ1+ cψ2.
(5.7)
For λ /∈ [−λ∗, 0], denote ˜
ψ1= aψ1− bψ2 and ψ˜2= bψ1+ aψ2.
(5.8)
We then have
ψ1= (a ˜ψ1+ b ˜ψ2)(a2+ b2)−1,
ψ2= (−b ˜ψ1+ a ˜ψ2)(a2+ b2)−1.
(5.9)
By a straightforward but lengthy computation on (5.7), we obtain the following system of equations for ˜ψ1and ˜ψ2:
˜ ψ001 = ˜A ˜ψ01+ ˜B ˜ψ1+ ˜C ˜ψ20 + ˜D ˜ψ2, ˜ ψ002 = ˜A ˜ψ02+ ˜B ˜ψ2− ˜C ˜ψ10 − ˜D ˜ψ2, (5.10) where ˜ a = a(a2+ b2)−1, ˜b = b(a2+ b2)−1, (5.11) and ˜ A = a0a + b˜ 0˜b, ˜ B = (a00+ c)˜a + (b00+ d)˜b + a0˜a0+ b0˜b0, ˜ C = a0˜b− b0˜a, ˜ D = (a00+ c)˜b− (b00+ d)˜s + a0˜b0− b0˜a0. (5.12)
Since the coefficients of (5.10) are continuous on [1, R], then ˜ψ01 and ˜ψ02 are C2 on
[1, R] and analytic in λ∈ C − [−λ∗, 0]. Hence ψ1and ψ2have the same properties as
˜
ψ1and ˜ψ2. This proves the first part of the lemma.
To study λ∈ (−λ∗, 0), we write (5.3) as ψ00+ 1 r− ˆr+ g(r) ψ0+ 1 r− ˆrf (r)ψ = 0 for r < ˆr,
where g and f are analytic at ˆr. Hence ˆr is a regular singular point. Therefore, by a standard theorem (see, e.g., [2]), this implies that ψ either is bounded at ˆr or grows logarithmically at ˆr.
Finally, if λ =−λ∗, then p(1) = λ∗ νγ. Let s = r− 1; then (5.3) can be written as
ψ00+ 2 s + g ψ0+ c 2 s2 + c1 s + f ψ = 0 for s > 0, where g and f are continuous at s = 0 and c2> 0.
Let µ1= 1 2(−1 + √ 1− 4c2) and µ2= 1 2(−1 − √ 1− 4c2).
If µ16= µ2, then ψ behaves asymptotially like sµ1 or sµ2 as s→ 0+. If µ1= µ2=−12,
then|ψ(s)| behaves asymptotically like s−12 or s− 1
2| log s| as s → 0+. In any case, ψ is unbounded at s = 0. The case in which λ = 0 was studied in Lemma 3.1. The proof is complete.
Considering (1.13) and Lemma 5.1, we introduce the following notion.
Definition 5.2. For λ /∈ [−λ∗, 0], ψ(·, λ) is called a stable mode if Reλ < 0, an unstable mode if Reλ > 0, and a marginally stable mode if Reλ = 0.
In the following, we shall study the relationship between the sign of Reλ and ψ(R, λ), i.e., how the disturbance of the gas surface influences the stability of the stationary solution u.
Since ψ(·, λ) is C2 in [1, R] for λ /∈ [−λ∗, 0], ψ(R, λ) and ψ0(R, λ) satisfies
homo-geneous boundary conditions at R, i.e.,
ajψj0(R) + bjψj(R) = 0, (5.13)
where ψ1= Reψ and ψ2= Imψ , aj= aj(λ) and bj= bj(λ) are analytic in λ /∈ [−λ∗, 0] for j = 1, 2. When aj 6= 0, denote κj(λ) = bj(λ) aj(λ) . (5.14)
Then (5.13) can be written as
ψ 0j(R) ψ j(R)
=−κj. (5.15)
When aj = 0, i.e., ψ j satisfies the Dirichlet boundary condition ψ j(R) = 0, we adopt the convention κj = +∞.
We can now introduce the notion of the stability of stationary solutions with respect to the boundary conditions (5.5) and (5.15) (or (5.13)).
Definition 5.3. Let u be a BC-type stationary solution for (EC) or (EPC). Then u is called stable with respect to (5.5) and (5.15) if any eigenvalue λ of (5.1), (5.5), and (5.15) satisfies Reλ < 0. u is called unstable if there is an eigenvalue ˜λ of (5.1), (5.5), and (5.15) such that Re˜λ > 0. u is called marginally stable if any eigenvalue λ of (5.1), (5.5), and (5.15) satisfies Reλ≤ 0 and equality holds for some ˜
λ.
The stability problem with respect to boundary condition (5.15) can also be studied by making the following observation:
Denote
C+={λ ∈ C : Reλ > 0}, C−={λ ∈ C : Reλ < 0}, and C0={λ ∈ C : Reλ = 0}.
For any stationary solution u and any (κ1, κ2)∈ R
2
≡ R2∪ {(k
1,∞) : κ1∈ R1} ∪ {(∞, k2) : κ2∈ R1} ∪ {(∞, ∞)},
denote by σ(κ1, κ2) the set of eigenvalues of (5.1), (5.5), and (5.15). Then define
Ks= Ks(u)≡ {(κ1, κ2) : σ(κ1, κ2)⊂ C−},
Ku= Ku(u)≡ {(κ1, κ2) : σ(κ1, κ2)∩ C+6= φ},
and
Km= Km(u) ={(κ1, κ2) : σ(κ1, κ2)∩ C+= φ and σ(κ1, κ2)∩ C06= φ}.
From Lemma 5.1, we know that any one of Ks, Ku, and Kmis nonempty. Hence the stability of u with respect to a given (κ1, κ2) is equivalent to deciding to which
set—Ks, Ku, or Km—(κ1, κ2) belongs. In general, for a given u, it is not easy to
completely identify Ks, Ku, and Km. However, we shall find some subsets of Ksand
Kuthat will give us sufficient conditions to determine whether u is stable or unstable with respect to given (κ1, κ2).
We first prove the following stability result.
Theorem 5.4. Let u be a neutrally stable BC-type stationary solution of (EC) or (EPC) when ν = 0. Then for any ν > 0, we have
{(κ1, κ2) : κ1≥ 0, κ2≥ 0} ⊂ Ks(u), (5.16) i.e., u is stable if κ1≥ 0 and κ2≥ 0 (5.17) or, equivalently, if ψ0j(R)ψj(R)≤ 0 (5.18) for j = 1, 2.
Proof. Since u is assumed to be neutrally stable when ν = 0, 0 is not an eigenvalue of (5.1), (5.5), and (5.15). If λ∈ (−λ∗, 0), then there is nothing to prove. Hence we consider the case where λ /∈ [−λ∗, 0) and is an eigenvalue with respect to (5.15) such that (κ1(λ), κ2(λ)) satisfies (5.17). We must prove that
Reλ < 0. (5.19)
Indeed, multiply (5.1) by ψ and then integrate from 1 to R; λ satisfies aλ2+ bλ + c = 0, (5.20) where a = 1 γ Z R 1 r−2ρ(r)(ψ21+ ψ22)dr > 0, (5.21) b =−ν Z R 1 ψ ˆLψdr, and c = − Z R 1 ψLψdr.
Since u is assumed to be neutrally stable when ν = 0, we have `1 > 0 in (3.1).
Moreover, ψ is C2 on [1, R]. Hence Remark 3.6 implies that c > 0. (5.22) Now let b = b1+ ib2, (5.23) where b1=−ν Z R 1 (ψ1Lψˆ 1+ ψ2Lψˆ 2) = ν 2 X j=1 Z R 1 r−2(ψj0)2− 2 X j=1 R−2ψ0j(R)ψj(R) and b2=−ν Z R 1 (ψ1Lψˆ 2− ψ2Lψˆ 1) =ν γR −2(ψ 2(R)ψ10(R)− ψ1(R)ψ20(R)).
Assumption (5.18) implies that
b1> 0.
(5.24)
Now we are going to show that the root λ of (5.20) satisfies (5.19) provided the coefficients satisfy (5.21)–(5.24). It is clear that the roots λ of (5.20) are given by
λ± = 1 2a{−(b1+ ib2)± (b 2− 4ac)1 2}. (5.25) Let X = b21− b22− 4ac and Y = b1b2. Then b2− 4ac = X + 2iY. Moreover, if x and y are real numbers such that (x + iy)2= X + iY, then x2=1 2{X + (X 2 + 4Y2)12}. To show (5.19), it suffices to prove that b1>|x|, i.e.,
b21> x2. (5.26)
By (5.21) and (5.22), we have
2b21− X = b21+ b22+ 4ac > 0. (5.27)
It is easy to check that
(2b21− X)2− (X2+ 4Y2) = 16ac. (5.28)
Hence (5.26) follows from (5.21), (5.22), (5.27), and (5.28). The proof is com-plete.
Next, we prove the following instability results.
Lemma 5.5. Let u be a neutrally stable BC-type stationary solution of (EC) or (EPC) when ν = 0. For any ν > 0, if λ is real and λ > 0, we have
ψ(R, λ) > 0 and ψ0(R, λ) > 0. (5.29) Furthermore, we have lim λ→0+ ψ0(R, λ) ψ(R, λ) = +∞ (5.30) and lim λ→∞ ψ0(R, λ) ψ(R, λ) = +∞. (5.31)
Proof. If λ > 0, then p(r) + λνγ > 0 in [1, R]. Integrating (5.3) from 1 to r and using (5.5) and (5.6), we obtain
(5.32) r−2 p(r) + λν γ ψ0(r) = p(1) + λν γ + 1 γ Z r 1 {4s−3p0(s) + λ2s−2}ψ(s, λ)ds.
If λ2is large enough that
λ2+ 4s−1p0(s)≥ 0 in [1, R], (5.33)
then (5.5), (5.6), and (5.32) imply
ψ(r, λ) > 0 and ψ0(r, λ) > 0 in [1, R]. (5.34)
In particular, (5.29) holds.
Now by applying Theorem 5.4, we claim that (5.29) also holds for any λ > 0. Otherwise, by the continuous dependence of ψ(R, λ) with respect to λ, we have ψ0(R, λ1) = 0 or ψ(R, λ1) = 0 for some λ1 > 0. Since (5.18) is satisfied by this λ1,
Theorem 5.4 implies λ1< 0, a contradiction. Hence (5.29) holds for any λ > 0.
To show (5.30), we note that u is neutrally stable and by Proposition A.1 in Appendix A, we have lim r→R−(R− r) qψ(r, 0) = c 0> 0 (5.35) and lim r→R−(R− r) q+1ψ0(r, 0) = c 1> 0. (5.36)
From (5.35), (5.36), and (5.29), it is not difficult to prove that (5.30) holds. The details of the proof are omitted.
Finally, it remains to prove (5.31). If λ > 0 and is large enough, then (5.3), (5.33), and (5.34) imply that
ψ00(r, λ) > 0 in [1, R]. (5.37)
Moreover, by (5.32), there is a positive constant c2that is independent on λ such
that for a large λ, we have
ψ0(R, λ)≥ λc2
Z R r
ψ(s, λ)ds (5.38)
for r∈ [12R, R]. Now for any s∈ [12R, R], write ψ(s, λ) = ψ(R, λ) + ψ0(R, λ)(s− R) +1
2ψ
00(˜r, λ)(s− R)2
(5.39)
for some ˜r∈ (s, R). Subsituting (5.39) into (5.38) and using (5.37), we obtain
ψ0(R, λ) 1 +1 2λc2(R− r) 2 ≥ λc2ψ(R, λ)(R− r). (5.40)
If we choose r such that (R−r)λ12 = 1, then (5.40) implies that for a large λ, we have ψ0(R, λ)≥ c3λ
1
2ψ(R, λ),
where the positive constant c3 is independent of λ. Hence (5.31) follows. The proof
is complete.
For any real λ /∈ [−λ∗, 0], ψ(r, λ) is a real function, i.e., ψ2= Imψ≡ 0. κ2(λ) is
then undetermined for all real λ /∈ [−λ∗, 0]. However, we can define κ2(λ) for these λ
by going through the following limiting process.
Let λ1 and λ2 be real numbers such that λ1 ∈ [−λ/ ∗, 0] and |λ2| 6= 0 and is
sufficiently small. We then have
ψ(r, λ1+ iλ2) = ψ(r, λ1) + iλ2
∂ψ
∂λ(r, λ1) + o(|λ2|
2)
(5.41)
as λ2→ 0. Therefore, for any real λ1∈ [−λ/ ∗, 0], we can define
κ2(λ1) = ∂2ψ ∂r∂λ(R, λ1)/ ∂ψ ∂λ(R, λ1). (5.42)
It is not difficult to prove that κ2(λ) is well defined and is continuous for λ∈ C −
[−λ∗, 0].
Now by applying Lemma 5.5, we have that following instability result.
Theorem 5.6. Let u be a neutrally stable BC-type stationary solution of (EC) or (EPC) when ν = 0. Then for any ν > 0, there is a positive constant κ∗= κ∗(ν, u) such that for any κ1 <−κ∗, there is a nonempty open set U (κ1, ν, u) such that u is
unstable with respect to (5.5) and (5.15) for (κ1, κ2) with κ2∈ U(κ1, ν, u).
Proof. For any ν > 0, let κ∗(ν, u) = min ψ0(R, λ, ν) ψ(R, λ, ν) : λ∈ (0, ∞) .
By (5.30) and (5.31), we have κ∗(ν, u) > 0. If κ1 <−κ∗, then there is λ1 > 0 such
that ψ0(R, λ1) ψ(R, λ1) =−κ1. Let U (κ1, ν, u) ={κ2∈ (−∞, ∞] : (κ1, κ2)∈ Ku}. Then (5.42) implies that
κ2(λ1)∈ U(κ1, ν, u).
Thus U (κ1, ν, u) is nonempty. It is clear that U (κ1, ν, u) is open, and the result follows.
The proof is complete.
Proof of Theorem 1.2. Theorem 1.2 follows from Theorems 5.4 and 5.6.
We now come to (EP). In this case, (5.3) has a singularity at r = 0 even for λ /∈ [−λ∗, 0). Therefore, we need to modify our argument to obtain a result as in Lemma 5.1. Indeed, the initial conditions (5.5) and (5.6) will be replaced with
r−2ψ(r, λ) = 0 at r = 0 (5.43)
and
(r−2ψ(r, λ))0= 1 at r = 0. (5.44)
We then have the following result.
Lemma 5.7. Let u be a B-type solution of (EP). Then the solution ψ(r, λ) of (5.3), (5.43), and (5.44) exists in a neighborhood of r = 0 if λ 6= −λ∗ ≡ γνp(0). Furthermore, ψ(·, λ) has the same property as in Lemma 5.1.
Proof. Following the same argument as in the proof of Lemma 5.1, we have equation (5.9) for ˜ψ1 and ˜ψ2 in r > 0. For λ 6= −λ∗, after a careful computation,
(5.9) can be written as ˜ ψ001 = −2 r + g1 ˜ ψ01+ 2 r2+ f1 ˜ ψ1+ g2ψ˜20 + f2ψ˜2 (5.45) and ˜ ψ002 = −2 r + g1 ˜ ψ02+ 2 r2+ f1 ˜ ψ2− g2ψ˜10 − f2ψ˜1, (5.46)
where gj(r) and rfj(r) are continuous (in fact, C2) at r = 0 for j = 1, 2. Now r = 0 is a regular singular point in (5.45) and (5.46). By a standard argument, we can prove that there is bounded solution{ ˜ψ1, ˜ψ2} of (5.45) and (5.46). Moreover, they satisfy
˜
ψ1(r) = a0r + o(r),
˜
ψ2(r) = b0r + o(r)
(5.47)
as r → 0+. The details of the proof are omitted. Now the initial conditions (5.43)
and (5.44) imply that a0 and b0satisfy
a0= p(0) + λ1 ν γ and b0= λ2 ν γ. (5.48)
Subsituting (5.47) into (5.9), we obtain
ψ1(r) = r3+ o(r3),
ψ2(r) = O(r3)
(5.49) as r→ 0+.
The other properties of ψ(·, λ) can also be obtained as in proving Lemma 5.1; the details are omitted. The proof is complete.
By arguing as in Theorems 5.4 and 5.6, we can obtain the following stability result for problem (EP).
Theorem 5.8. Let u be a neutrally stable B-type stationary solution of (EP) when ν = 0. Then for any ν > 0, u is stable with respect to (5.43) and (5.15) if κ1 ≥ 0 and κ2 ≥ 0. On the other hand, there is a positive constant κ∗ = κ∗(ν, u)
such that for any κ1 <−κ∗, there is a nonempty open set U (κ1, ν, u) such that u is
unstable with respect to (5.43) and (5.15) for (κ1, κ2) with κ2∈ U(κ1, ν, u).
Proof. The proof is the same as was used for Theorems 5.4 and 5.6. Therefore, the details are omitted.
Remark 5.9. In their recent work on (EC), Makino et al. [15, 16, 19] showed that when γ > 43 and ν > 0, uR,µis nonlinearly asymptotically stable with respect to small perturbations. Their result is consistent with ours.
Appendix A. Asymptotic behavior at R. In this section, we shall study the
asymptotic behavior of a real solution ψ at R for (3.3) with real `. Let τ = R− r and ˜ψ(τ ) = ψ(r).
Then ˜ψ satisfies ˜
ψ00+{(1 + q)τ−1+ g(τ )} ˜ψ0+ τ−1f (τ ) ˜ψ = 0. For simplicity, we omit the∼’s and write the last equation as
ψ00+{(1 + q)τ−1+ g(τ )}ψ0+ τ−1f (τ )ψ = 0, τ > 0, (A.1)
where g and f are continuous at τ = 0.
Then we have the following result concerning the behavior of ψ at 0.
Proposition A.1. For any q > 1, let ψ be a solution of (A.1). Then either ψ is bounded at 0 or
ψ(τ ) = τ−qψ(τ )ˆ (A.2)
with ˆψ continuous at 0 and ˆψ(0)6= 0. Furthermore, in the former case, ψ is C2 at 0,
and in the latter case, we have
ψ0(τ ) =−qτ−q−1ψ(0) + o(τˆ −q−1) (A.3)
as τ→ 0+.
Proof. If g and f are analytic in a neighborhood of τ = 0, then the result is well known; see, e.g., [2]. For completeness, we provide a proof here that assumes only that g and f are continuous at τ = 0. Since the proof is elementary, some details are omitted. For τ > 0, let ψ(τ ) = τ−qω(τ ). (A.4) ω then satisfies ω00+{(1 − q)τ−1+ g}ω0+ (f− qg)τ−1ω = 0, τ > 0. (A.5)
Let G(0) = 0 and G0(τ ) = g(τ ), (A.5) can then be written as (τ1−qeGω0)0+ τ−qeG(f− qg)ω = 0. (A.6)
Fix τ1> 0 and let τ0∈ (0, τ1) be chosen later. After integrating (A.6) from τ to τ0,
we have ω0(τ ) = τq−1E(τ )C0+ τq−1E(τ ) Z τ0 τ F (s)s−qω(s)ds, (A.7) where
E(τ ) = exp(−G(τ)), F (τ ) = (f− qg) exp(G(τ)), and C0= τ
1−q
0 ω0(τ1) exp(G(τ1)).
We first claim that
ω and ω0 are bounded on (0, τ ]. (A.8)
Indeed, let
C1=|C0| max
τ∈[0,τ1]
|E(τ)| and C2= max
τ∈[0,τ1]
|E(τ)| · max τ∈[0,τ1]
|F (τ)|. Then from (A.7), we have
|ω0(τ )| ≤ C 1τq−1+ C2τq−1 Z τ0 τ s−q|ω(s)|ds, (A.9)
which implies that
|ω0(τ )| ≤ C 1τq−1+ C3 max s∈[τ,τ0] |ω(s)|, where C3= C2· (q − 1)−1.
Now for any τ ∈ (0, τ0), substituting (A.9) into
ω(τ ) = ω(τ0) + Z τ τ1 ω0(s)ds, we obtain |ω(τ)| ≤ |ω(τ0)| + C4+ C3τ0 max s∈[τ,τ0] |ω(s)|, (A.10) where C4= C1τ1q.
Now if we choose C3τ0< 1, (A.10) then implies
|ω(τ)| ≤ (1 − C3τ0)−1{|ω(τ0)| + C4}.
Hence ω is bounded on [0, τ1]. By (A.9), ω0 is bounded on [0, τ1], which also implies
that ω is continuous at 0. Now if ω(0) 6= 0, then (A.4) and (A.8) imply (A.3). If ω(0) = 0, we shall claim that
|ω(τ)| ≤ C5τq
(A.11)
for some C5> 0. Indeed, ω(0) = 0 and ω0 bounded on [0, τ1] implies that
|ω(τ)| ≤ C6τ.
(A.12)
Now substituting (A.12) into (A.9), we have |ω0(τ )| ≤ C
7τq−1+ C8τ
(A.13)
for some C7> 0 and C8> 0. Substituting (A.13) into
ω(τ ) = Z τ
0
ω0(s)ds,
we obtain a better estimate for ω than (A.12). After repeating the processes a finite number of times, (A.15) follows. The proof is complete.
