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Bifurcation for a free boundary problem modeling a protocell
Nonlinear Analysis 70 (2009) 2779–2795
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Bifurcation for a free boundary problem modeling a protocell Hua Zhang a,b , Changzheng Qu a,b , Bei Hu c,∗ a Center for Nonlinear Studies, Northwest University, Xi’an, 710069, PR China b Department of Mathematics, Northwest University, Xi’an, 710069, PR China c Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556, USA
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Article history: Received 25 February 2008 Accepted 7 April 2008 Keywords: Free boundary problem Stationary solution Bifurcation Symmetry breaking Protocell
a b s t r a c t In this paper we consider a free boundary problem for a physico-chemical model of a protocell. This model of a self-maintaining unity or a protocell is based on the reaction and diffusion process, and a mechanism of self-control of the boundary. For any positive radius R, there exists a radially symmetric solution with radius r = R. In the more realistic threespace-dimensional case, we give a proof that there exist symmetry-breaking bifurcation branches of solutions with free boundary r = R + Yn,0 (θ) + O(2 ) (n ≥ 2, even) for small ||, where Yn,0 is the spherical harmonic of mode (n, 0). © 2008 Elsevier Ltd. All rights reserved.
1. Introduction The protocell is a mathematical model of a self-maintaining unity based on the dynamics of simple reaction–diffusion process and a self-controlled dynamics of the surface. It has been developed and studied in many papers (see [15,16,9–11] and the references therein). In this paper, we shall establish the existence of symmetry-breaking bifurcations for the free boundary of this model in the three-space-dimensional case. The protocell can be visualized as having a porous structure maintained by building materials with concentration C . The structure is sustained only as long as C exceeds a critical concentration C ∗ . Metabolism is maintained by nutrient material with concentration σ which is distributed in the entire space with σ = τ at ∞ (τ > 0). C and σ satisfy a coupled system of reaction–diffusion equation:
∂C − ∆C = σ, −∆σ = −σ in the cell, ∂t −∆σ = 0 outside the cell, c
where c is a positive constant. The constant c is the quotient of the time scale of diffusion to the time scale of cell doubling. On the boundary of the protocell C = C ∗ . The various tasks that the cell continuously performs take their toll on the cell: they cause it to shrink. This is modeled by disintegration at the boundary at a rate β, β > 0. On the other hand the flux of building material at the boundary causes the cell to grow. The total result of these two effects is Vn = −
∂C − β on the boundary of the cell, ∂n
where n is the exterior normal, and Vn is the velocity of the boundary points in the direction n.
∗ Corresponding author. E-mail address: [email protected] (B. Hu). 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.04.003
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In the case of a spherical cell (in 3-dimensions), setting q t r=
x21 + x22 + x23 ,
u = C − C∗ ,
˜t =
c
and denoting the boundary of the cell by r = s(t), we have the following system for u = u(r, t), σ = σ(r, t) and r = s(t): ! ∂2 2 ∂ + σ = σχr
σ → τ as |x| → ∞, and ut −
2 ∂ ∂2 + ∂r2 r ∂r
! u=
σ if r < s(t), t > 0,
u=0
on r = s(t), t > 0,
u = u0 (r)
(1.2)
for t = 0,
and finally, the free boundary condition s0 (t) = −cur (s(t), t) − cβ.
(1.3)
It was established in [15] that there exist radially symmetric stationary solutions (R, σs , us ) to (1.1)–(1.3), and they are given explicitly by R − tanh R τ 1 − for r ≥ R, r σs (r) = (1.4) 1 sinh r τ for r < R, cosh R r τ sinh R sinh r us (r) = − for r < R, (1.5) cosh R R r where R satisfies R2 τ = . β R − tanh R
(1.6)
The asymptotic behavior for radially symmetric solutions was studied in [9]. It is very natural to ask whether there exist non-radially symmetric stationary solutions and what are their asymptotic behaviors. These are indeed very challenging questions with only few results in the literature. Using a very lengthy analytical expansion of the free boundary and the solution, the existence of non-radially symmetric stationary solutions was established in [10,11]: A sequence of symmetrybreaking bifurcation branches of stationary solutions of (1.1)–(1.3) was found in the two-space-dimensional case. Of course, the more realistic model is in three space dimension. However, due to the complexity of the analytic expansion used in [11], it is very difficult to extend this method into the three-space-dimensional case. The purpose of the present paper is as follows: first, we want to extend the bifurcation results of [10,11] into the more realistic three-space-dimensional case; second, we shall use a much more simplified and systematic method to establish the existence of symmetry-breaking bifurcation branches of stationary solutions by applying the Crandall–Rabinowitz bifurcation theorem [2]. In order to do so, we shall derive many delicate PDE estimates for our system. Note that for our system (1.1), the solution σ is only C 1+1 in the spatial variable, i.e., σrr is bounded, but has a jump across the free boundary {r = s(t)}. Therefore, a perturbation of this system involves a diffraction boundary condition (cf., (3.6)). Since we also need to perturb the free boundary {r = s(t)}, the extension of PDE estimates to our system is not a trivial matter. The application of the Crandall–Rabinowitz bifurcation theorem framework to free boundary problems started in [1,4,5]. This method was subsequently used successfully on other free boundary models in [12,13,3]. For systematic introduction of biological cancer models and bifurcation analysis we refer the readers to [6–8]. 2. Spherical harmonics and modified Bessel function In this paper we shall need Spherical harmonics and some inequalities for the modified Bessel function Im (ξ), where ξ is a complex variable and m is any nonnegative real number. For convenience, we collect all these required identities and inequalities in this section and refer the readers to [12] for the proof. The Spherical harmonics. The Spherical harmonics Yl,m (θ, φ) is defined by s (2l + 1)(l − m)! m eimφ m Yl,m (θ, φ) = (−1) Pl (cos θ) √ (m = −l, . . . , l), 2(l + m)! 2π
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where Plm (x) = 21l l! (1 − x2 )m/2 ddxl+m (x2 − 1)l . The family of functions Yl,m forms a complete orthonormal basis for L2 (Σ ), where Σ is the unit sphere, and l+m
∆ω Yl,m = −l(l + 1)Yl,m ,
where ∆ω =
1 ∂ sin θ ∂θ
(sin θ ∂θ∂ ) +
1 ∂2 sin2 θ ∂φ2
is the Laplace operator on the unit sphere Σ .
Bessel functions. For the modified Bessel functions Im (ξ), which is defined by m X 2k ∞ ξ 1 ξ Im (ξ) =
2
it is well known that s 2 I1/2 (ξ) =
πξ s
I3/2 (ξ) =
2
s
00 Im (ξ) + 0 Im (ξ) −
1
ξ
2
−
ξ
ξ
+ cosh ξ ,
+ 1 sinh ξ −
2
m2
I0 (ξ) − 1 +
ξ
!
ξ2
(2.2)
3
ξ
cosh ξ ,
(2.3)
Im (ξ) = 0,
(2.4)
m ≥ 0,
(2.5)
Im (ξ) = Im+1 (ξ),
Im (ξ)In (ξ) =
,
(2.1)
sinh ξ 3
πξ
m
2
sinh ξ,
πξ
I5/2 (ξ) =
k!Γ (m + k + 1)
k =0
∞ X
Γ (m + n + 2k + 1)(ξ/2)m+n+2k
k =0
k!Γ (m + k + 1)Γ (n + k + 1)Γ (m + n + k + 1)
.
(2.6)
Throughout this paper, we shall be working with the functions Pn (ξ) =
In+3/2 (ξ) , ξIn+1/2 (ξ)
n = 0, 1, 2, . . . .
(2.7)
It satisfies (cf. [12, page 297–299]) P0 (ξ) = P1 (ξ) =
1
ξ
1
coth ξ −
ξ2
1
ξ coth ξ − 1
,
(2.8) 3
−
ξ2
,
(2.9)
ξ2 + O(|ξ|4 ), |ξ| → 0, 2n + 3 (2n + 3)2 (2n + 5) 1 n+1 π + O(|ξ|−3 ), |arg(ξ)| < − δ, |ξ| → +∞, Pn (ξ) = − ξ ξ2 2 1
Pn (ξ) =
Pn (ξ) =
1
ξ2 Pn+1 (ξ) + (2n + 3) 1
Pn (0) =
d dξ
−
2n + 3
Pn (ξ) =
1
ξ
,
(2.11) (2.12)
,
−
(2.10)
(2.13)
2n + 3
ξ
Pn (ξ) − ξPn2 (ξ),
(2.14)
and for any real r > 0, n ≥ 0, Pn (r)
d dr 1 r
> Pn+1 (r),
Pn (r)
−
(2.15)
< 0,
n+1 r2
< Pn (r) <
(2.16) 1 r
−
n+1 r2
+
n2 + n + 1
2r 3
.
(2.17)
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3. The bifurcation problem Consider a family of domains with boundaries ∂Ω : r = R + R˜ (θ, φ), where R˜ (θ, φ) = S(θ, φ). Let (σ, u) be the solution of ∆σ = σχ{x∈Ω } (x)
in R3 ,
(3.1)
σ → τ as |x| → ∞, and
−∆u = σ in Ω , u=0
(3.2)
on ∂Ω .
We define F to be F (R˜ , R) = −
∂u R − tanh R τ. − ∂n ∂Ω R2
Then, recalling (1.6), we have F (0, R) ≡ 0
for any 0 < R < ∞.
It is also clear that (σ, u, R + R˜ ) is a stationary solution if and only if F (R˜ , R) = 0.
(3.3)
Using a scaling (u/τ, σ/τ) if necessary we can assume without loss of generality that τ = 1. We shall do so throughout the rest of this paper. The function S(θ, φ) may be viewed as a function defined on the unit sphere Σ = {x; |x| = 1}. We shall later assume that S(θ, φ) is in C m+α , that is, S is C m+α as a function defined on Σ ; note that this does not mean that S(θ, φ) is in C m+α in the variable (θ, φ). We shall see in Sections 3–5 that F maps the space C m+α (Σ ) × R into the space C m+α−1 (Σ ). In this section we shall formally compute the Fréchet derivatives of F (cf., (3.5)). The formal derivation will be rigorously justified in Section 4. In order to compute the Fréchet derivatives of F , we need the expansion of (σ, u) of order . We formally write,
σ = σs + σ1 + O(2 ), u = us + u1 + O(2 ),
(3.4)
∂Ω : r = R + S(θ, φ), where (σs , us ) is the stationary solution defined in (1.4) and (1.5). Using the relation ∂(us + u1 ) ∂u = + O(2 ) ∂n ∂Ω ∂n R + S ∂us ∂2 us ∂u1 = + O(2 ), + 2 S + ∂r ∂r ∂r r =R
we find that F (R˜ , R) = −
∂u1 ∂us ∂2 us + 2 S + ∂r ∂r ∂r
− r =R
R − tanh R R2
− O(2 ).
The Fréchet derivative is therefore given by ∂2 us ∂u1 [FR˜ (0, R)] · S = − S + . ∂r2 ∂r
(3.5)
r=R
We start now by computing explicitly (and formally) σ1 and u1 . 3.1. Computation of σ1 We shall use the notation σ + to denote the limit of σ from the exterior of Ω and σ − to denote the limit of σ from the interior of Ω . Clearly
σ+ |∂Ω = σ− |∂Ω ,
σr+ |∂Ω = σr− |∂Ω .
Using Taylor’s expansion at r = R from both sides, we obtain + − + + − −
σs + Sσsr + σ1 |r=R = σs + Sσsr + σ1 |r=R + O(2 ),
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and + − σsr+ + Sσsrr + σ1+r |r=R = σsr− + Sσsrr + σ1−r |r=R + O(2 ).
Dropping O(2 ) terms, we obtain + − +
σ1 − σ1 |r=R = −[σsr (R) − σsr− (R)]S = 0,
(3.6)
+ − σ1+r − σ1−r |r=R = −[σsrr (R) − σsrr (R)]S := h(R)S,
R + − where h(R) := −[σsrr (R) − σsrr (recalling that τ = 1). Substituting the first equation in (3.4) and (3.6) into (3.1) (R)] = tanh R and dropping the O() terms, we find that σ1 satisfies
−∆σ1 + σ1 = 0 in BR , ∆σ1 = 0
outside BR ,
(3.7)
σ1 − σ1 |r=R = 0, +
−
σ1+r − σ1−r |r=R = h(R)S.
If S(θ, φ) = Yn,m (θ, φ), we look for a solution of the form
σ1 = σ˜ 1 (r)S(θ, φ).
(3.8)
Then 2
n(n + 1)
r
2
r2 n(n + 1)
r
r2
σ˜ 100 + σ˜ 10 − σ˜ 100 + σ˜ 10 −
σ˜ 1 = σ˜ 1 , σ˜ 1 = 0,
r r
< R,
> R.
These ODEs can be solved explicitly: 1
σ˜ 1 = C1 r− 2 In+ 1 (r), 2
r
< R,
and σ˜ 1 = C2 r−n−1 ,
r
> R.
(3.9)
Using the last two relations in (3.7), we derive 1
C1 R− 2 In+1/2 (R) = C2 R−n−1 , C2 (−n − 1)R−n−2 − C1 [nR−3/2 In+1/2 (R) + R−1/2 In+3/2 (R)] = h(R),
so that (recalling the definition of Pn in (2.7)) C1 =
−h(R) , [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R)
C2 =
−h(R)Rn . (2n + 1)R−2 + Pn (R)
(3.10)
By (3.8), σ1 is given explicitly by
σ1 (r, θ, φ) =
−h(R)r−1/2 In+1/2 (r) Yn,m (θ, φ), [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R)
σ1 (r, θ, φ) =
−h(R)Rn r−n−1 Yn,m (θ, φ), (2n + 1)R−2 + Pn (R)
r
r
< R,
> R.
(3.11) (3.12)
3.2. Computation of u1 From (3.2) and the second equation in (3.4), we obtain
− ∆u1 = σ1 ,
in BR ,
(3.13)
and, recalling that u|r=R+S = 0 and us (R) = 0,
u1 |∂BR = u1 |∂Ω + O(2 ) = (u − us )|r=R+S + O(2 ) = −us (R + S) + O(2 ) = −[us (R) + usr (R)S] + O(2 ) = −λS + O(2 ),
(3.14)
where λ := usr (R) = −(tanh R)P0 (R). Dropping higher-order terms we find that u1 |r=R = −λS.
(3.15)
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If S = Yn,m (θ, φ), then u1 = u˜ 1 (r)S(θ, φ) satisfies 2
n(n + 1)
r
r2
(u˜ 1 + σ˜ 1 )00 + (u˜ 1 + σ˜ 1 )0 −
(u˜ 1 + σ˜ 1 ) = 0,
r
< R,
so that
˜ 1 = C3 r n , u˜ 1 + σ
r
< R.
Using conditions (3.9), (3.10) and (3.15), we obtain C3 =
−λ Rn
−
h(R)R−n−1
(2n + 1)R−2 + Pn (R)
.
Thus u1 is given by u1 (r, θ, φ) =
−λ
−
Rn
h(R)R−n−1
!
(2n + 1)R−2 + Pn (R)
rn Yn,m (θ, φ) − σ1 .
(3.16)
4. Rigorous justification Note that u is defined only on Ω , while us is defined on BR . The formulas that define σs and us actually define them in the whole space. In order to derive necessary estimates for σ1 and u1 , we transform all these functions to the same domain Ω and shall do so by the Hanzawa transformation, which is a diffeomorphism defined by
(r, θ, φ) = H (r0 , θ0 , φ0 ) ≡ (r0 + χ(R − r0 )S(θ0 , φ0 ), θ0 , φ0 ), where 0, χ(z) = 1,
χ∈C , ∞
if |z| ≥ 3δ0 /4, if |z| < δ0 /4,
dk χ k ≤ C /δk0 . dz
Observe that H maps BR onto Ω while keeping the ball {r < R − (3δ0 /4)} and the domain {r > R + (3δ0 /4)} fixed. The inverse Hanzawa transformation H−1 maps Ω onto BR . We set
σˆ 1 (r, θ, φ; ) = σ1 (H−1 (r, θ, φ)) in Ω , uˆ 1 (r, θ, φ; ) = u1 (H−1 (r, θ, φ)) in Ω . Note that an inverse Hanzawa transformation, if applied to σs or us , would destroy the boundary conditions on ∂Ω . Therefore, instead of using (1.4) and (1.5), we shall use the following functions R − tanh R 1 − in Ω , r σˆ s (r, θ, φ; ) = (4.1) sinh r 1 in R3 \ Ω , cosh R r 1 sinh R sinh r us (r) = − for 0 < r < ∞. (4.2) cosh R R r Remark 1. Note that with this definition, σˆ is discontinuous on ∂Ω . Lemma 4.1. There holds
kσ − σˆ s − σˆ 1 kC2+α (Ω ) + kσ − σˆ s − σˆ 1 kC2+α (R3 \Ω ) ≤ C 2 ,
(4.3)
where the constant C depends only on kSkC2+α (Σ ) . Proof. We have
−∆(σˆ s + σˆ 1 ) + (σˆ s + σˆ 1 ) = fˆ in Ω ,
(4.4)
−∆(σˆ s + σˆ 1 ) = fˆ in R3 \ Ω ,
(4.5)
where fˆ is compactly supported, 1 fˆ is a linear combination of at most second-order derivatives of σ1 and S. Thus
kfˆkCα (Ω ) ≤ C ||,
(4.6)
kfˆkCα (R3 \Ω ) ≤ C ||,
(4.7)
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where the constant C depends on kSkC2+α (Σ ) . Now we work on the boundary conditions. Since σˆ 1 is continuous,
(σˆ s + σˆ 1 )+ − (σˆ s + σˆ 1 )− = σˆ s+ − σˆ s− ≡ g1 on ∂Ω .
(4.8)
A direct computation shows that g1 = g11 (S),
where g11 (η) =
1
sinh(R + η)
cosh R
R+η
−1+
R − tanh R
R+η
.
(4.9)
0 0 00 Since g11 (0) = g11 (0) = 0, we have |g11 (η)| ≤ C |η|2 and |g11 (η)| ≤ C |η|, |g11 (η)| ≤ C . Using these estimates we can easily derive
kg1 kC2 (∂Ω ) ≤ C 2 ,
(4.10)
where the constant C depends on kSkC2 (Σ ) . Note that D g1 = (S)(DS) + g11 (S)D S, here we use D to denote all possible second derivatives and (DS)2 to denote all possible products of two first-order derivatives of S. Clearly, 2
2 00 g11
2
0
2
2
00 (S)(DS)2 ]Cα (∂Ω ) ≤ C (kSkC2+α (Σ ) ) 2 , [2 g11 0 and, using the standard interpolation on [g11 (S)]Cα (∂Ω ) , we obtain 0 0 0 [g11 (S)D2 S]Cα (∂Ω ) ≤ kg11 (S)kL∞ (∂Ω ) [D2 S]Cα (Σ ) + [g11 (S)]Cα (∂Ω ) kD2 SkL∞ (Σ ) 0 0 0 ≤ kg11 (S)kL∞ (∂Ω ) [D2 S]Cα (Σ ) + (kg11 (S)kL∞ (∂Ω ) + kg11 (S)kC1 (∂Ω ) )kD2 SkL∞ (Σ )
≤ C (kSkC2+α (Σ ) ) 2 . Combining all these estimates, we find that
kg1 kC2+α (∂Ω ) ≤ C 2 ,
(4.11)
where the constant C depends on kSkC2+α (Σ ) . We now write
ˆ s + σˆ 1 )− (σˆ s + σˆ 1 )+ on ∂Ω , r − (σ r = g2
(4.12)
where 0 g2 = g21 (S), where g21 (η) = g11 (η) − h(R)η.
(4.13)
By the definition of h(R) (cf. (3.6)) we clearly have h(R) = g11 (0), so that again we have g21 (0) = g21 (0) = 0, |g21 (η)| ≤ C |η|2 0 00 and |g21 (η)| ≤ C |η|, |g21 (η)| ≤ C . Following a similar argument as above, we conclude that 00
0
kg2 kC1+α (∂Ω ) ≤ C 2 ,
(4.14)
where the constant C depends on kSkC2+α (Σ ) . The function
ψ = σ − σˆ s − σˆ 1 clearly satisfies
−∆ψ + ψ = −fˆ in Ω , −∆ψ = −fˆ in R3 \ Ω ,
(4.15)
ψ+ − ψ− = −g1 on ∂Ω , − ψ+ on ∂Ω , r − ψr = −g2
where fˆ is compactly supported. Note that if S ∈ C 2+α (Σ ), then ∂Ω is in C 2+α uniformly in . Applying the Schauder estimates for the diffraction problem (4.15), we obtain, for small ||,
kψkC2+α (Ω ) + kψkC2+α (R3 \Ω
)
≤ C {||kfˆkCα (R3 \Ω ) + ||kfˆkCα (Ω ) + kg1 kC2+α (∂Ω ) + kg2 kC1+α (∂Ω ) }
≤ C 2 . This completes the proof.
Next we establish Lemma 4.2. There holds
ku − us − uˆ 1 kC2+α (Ω ) ≤ C 2 , where the constant C depends on kSkC2+α (Σ ) .
(4.16)
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Proof. The proof is very similar. We make use of (3.13), (3.14) and (4.3) and the elliptic Schauder estimates. Since this problem involves only Ω (does not involve R3 \ Ω ), the proof is actually much easier, and we omit the details. With the estimates in above two lemmas, it is clear that the computation of Frechét derivative in (3.5) is rigorous in an appropriate space, say, F : C 2+α → C 1+α . However, in order to apply the bifurcation theorem, further restrictions are necessary, and we shall discuss this in the next section. 5. Bifurcation We shall use the following Crandall–Rabinowitz theorem (see [2], Theorem 1.7). Theorem 5.1. Let X , Y be real Banach spaces and F (x, µ) a C p map, p ≥ 3, of a neighborhood (0, µ0 ) in X × R into Y . Suppose (i) (ii) (iii) (iv)
F (0, µ) = 0 for all µ in a neighborhood of µ0 , ker Fx (0, µ0 ) is one-dimensional space, spanned by x0 , Im Fx (0, µ0 ) = Y1 has codimension 1, Fµx (0, µ0 )x0 6∈ Y1 .
Then (0, µ0 ) is a bifurcation point of the equation F (x, µ) = 0 in the following sense: In a neighborhood of (0, µ0 ) the set of solutions of F (x, µ) = 0 consists of two C p−2 smooth curves Γ1 and Γ2 which intersect only at the point (0, µ0 ); Γ1 is the curve (0, µ) and Γ2 can be parameterized as follows: Γ2 : (x(), µ()), ||
small, (x(0), µ(0)) = (0, µ0 ), x0 (0) = x0 . In order to apply this theorem, we first compute explicitly expressions on the right-hand side of (3.5) when S = Yn,m . Recall that
λ = usr (R) = −(tanh R)P0 (R),
tanh R
h(R) =
R
.
(5.1)
[1 − 2P0 (R)].
(5.2)
A direct computation shows that usrr (R) = − tanh R
R2 − 2R coth R + 2 R3
=−
tanh R R
Differentiating (3.11) and (3.16) in r, using also (2.5) and (2.7) and letting r = R, we find that
σ1r (R, θ, φ) = =
−h(R)[nR−3/2 In+1/2 (R) + R−1/2 In+3/2 (R)] Yn,m (θ, φ) [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R) −h(R)[nR−2 + Pn (R)] Yn,m (θ, φ), (2n + 1)R−2 + Pn (R)
(5.3)
and u1r (R, θ, φ) =
=
−h(R)R−n−1 + nRn−1 Yn,m (θ, φ) − σ1r Rn (2n + 1)R−2 + Pn (R)
−λ
−
!
nλ R
+
h(R)R2 Pn (R)
!
(2n + 1) + R2 Pn (R)
Yn,m (θ, φ).
(5.4)
Substituting (5.1)–(5.4) into (3.5), using also (2.12) twice, we get
[FR˜ (0, R)] Yn,m = =
tanh R R
tanh R R
[1 − (n + 2)P0 − R2 Pn (R)Pn−1 (R)]Yn,m [(2n + 1)Pn−1 − (n + 2)P0 ]Yn,m .
(5.5)
It follows that
[FR˜ (0, R)]Yn,m = 0
(5.6)
if and only if
(2n + 1)Pn−1 (R) = (n + 2)P0 (R).
(5.7)
In the next section we shall prove For each n ≥ 2, there exists a unique solution R = Rn that solves
(2n + 1)Pn−1 (R) = (n + 2)P0 (R),
and
Rm
> Rl for all m > l ≥ 2,
(5.8)
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and d
tanh R
dR
R
[(2n + 1)Pn−1 (R) − (n + 2)P0 (R)]
6= 0 for n ≥ 2.
(5.9)
R =R n
We shall use R as the bifurcation parameter, i.e., in Theorem 5.1 we take µ = R. As in [4,12,13] we introduce the Banach spaces X m+α = {R˜ ∈ C m+α (Σ ), R˜
is π - periodic in θ, 2π - periodic in φ},
X1m+α = closure of the linear space spanned by{Yj,0 (θ), j = 0, 1, 2, . . .} in X m+α , X2m+α = closure of the linear space spanned by{Yj,0 (θ), j = 0, 2, 4, . . .} in X m+α .
Then F (R˜ , µ) is a map from X m+2+α to X m+1+α . The operator A(µ) = [FR˜ (0, µ)] maps X12+α to X11+α , and, by (5.8), {Y1,0 } if µ 6= R2 , R3 , . . . , ker A(µ) = span{Yn,0 , Y1,0 } if µ = Rn . If we work with the space X12+α , then dim(ker A(µ)) = 2 when µ = Rn . In order to satisfy the requirement (ii) of Crandall–Rabinowitz theorem we shall henceforth work with the space X22+α . In the space X22+α , dim(ker A(µ)) = 1 if µ = Rn . If n is even and ≥ 2, then clearly the direct sum Im A(µ) ⊕ {Yn,0 } is the whole space X21+α and therefore codim(A(µ)) = 1 for µ = Rn . Differentiating (5.5) in R, we obtain d tanh R [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] Yn,m . [FR˜ µ (0, R)]Yn,m = dR R In particular, using (5.9) we find that d tanh R Yn,0 [FR˜ µ (0, Rn )]Yn,0 = [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] dR R R=Rn
6∈ Im A(Rn ). Thus, we have established the following result: Theorem 5.2. The points (0, Rn ) (n even and ≥ 2) are bifurcation points for the problem (3.1)–(3.3), and the corresponding free boundaries are of the form r = R + Yn,0 (θ) + o().
Note that this theorem gives non-radially symmetric stationary solutions to the original free boundary problem (1.1)–(1.3). 6. Monotonicity of Rn It remains to establish (5.8) and (5.9). We shall employ the ideas and methods from [10]. In [10], they deal with integer modified Bessel functions. Here, we deal with fraction modified Bessel functions. Theorem 6.1. (i) For any n ≥ 2, there exists a unique positive solution R = Rn of the equation Pn−1 (R)
=
P0 (R)
n+2
2n + 1
.
(ii) If 2 ≤ l < m, then Rl
< Rm .
6.1. Proof of (i) Consider the function fn−1 =
Pn−1 P0
=
In+1/2 I1/2 . In−1/2 I3/2
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From (2.10) and (2.11), we derive
fn−1 =
1 2n+1
(2n+1)2 (2n+3)
1 3 1
fn−1 =
R2
−
R
1 R
−
d
45
+ O (R 4 )
−
n R2
+ O ( R −3 )
−
1
O ( R −3 )
+
R2
n+2 2n+1
so that, if n ≥ 2, fn−1 −
dR
R2
fn−1 (R)
> 0,
+ O (R 4 )
=
→
3 2n + 1
R − n + O ( R −1 ) R − 1 + O(R−1 )
,
R → 0,
→ 1,
(6.1)
R → ∞,
(6.2)
is negative for small R and positive for large R. Hence, if
for all R > 0,
(6.3)
then, for any n ≥ 2, there exists a unique positive solution R = Rn of the equation Introduce the functions
Pn−1 (R) P0 (R)
=
n+2 . 2n+1
Sn = log In (R).
If we can prove Sn0 +1/2 − Sn0 −1/2
> S30 /2 − S10 /2 ,
(6.4)
then (6.3) follows. In order to prove (6.4), we need the fact Lemma 6.2. For any n ≥ 1, Sn0 +1/2 − S10 /2
<
n R
for all R > 0.
(6.5)
Proof. Using (2.5), we find that (6.5) is equivalent to In0 +1/2 In+1/2
−
I10 /2 I1/2
=
In+3/2 In+1/2
+
n + 1/2 R
−
I3/2 I1/2
+
1/2 R
!
n
< , R
which is the same as In+3/2 In+1/2
<
I3/2 I1/2
,
(6.6)
Using the product formula (2.6) it was proved in [14] that each term in the power series of In+3/2 In−1/2 is smaller than each term in the power series of (In+1/2 )2 , so that
In+3/2 In+1/2
<
In+1/2
In−1/2
and (6.6) then follows by iterating this inequality.
Introduce the function
φ(R) = (Sn+1/2 − Sn−1/2 ) − (S3/2 − S1/2 ), then the assertion (6.3) is equivalent to the inequality
φ0 (R) > 0,
if R > 0.
From (2.4), we have Sn00+1/2 + (Sn0 +1/2 )2 +
1 R
Sn0 +1/2 = 1 +
(n + 1/2)2 R2
.
Similarly, Sn00−1/2 + (Sn0 −1/2 )2 +
1 R
Sn0 −1/2 = 1 +
(n − 1/2)2 R2
,
so that 1
(n + 1/2)2 − (n − 1/2)2
R
R2
(Sn+1/2 − Sn−1/2 )00 + (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) + (Sn0 +1/2 − Sn0 −1/2 ) =
.
(6.7)
Writing (6.7) for n = 1 and subtracting it from (6.7), we obtain 1
2n − 2
R
R2
φ00 + (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) + φ0 =
.
(6.8)
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For small R, (6.1) implies In+1/2 I1/2
fn−1 =
In−1/2 I3/2
=
3
2n + 1
+
1 15
−
3
(2n + 1)2 (2n + 3)
R2 + O(R4 ),
since n ≥ 2, fn0 −1 > 0 for R small. If (6.3) does not hold for all R > 0, then there is a smallest R = R0 such that φ0 (R0 ) = 0, and, clearly, φ00 (R0 ) ≤ 0, so that, by (6.8) 2n − 2 R2
≤ (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) at R = R0 .
(6.9)
We write φ0 (R0 ) = 0 in the form Sn0 +1/2 − Sn0 −1/2 = S30 /2 − S10 /2
at R = R0 .
Then the right-hand side of (6.9) is equal to
(Sn0 +1/2 + Sn0 −1/2 )(Sn0 +1/2 − Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) = [(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )](S30 /2 − S10 /2 ) = [(Sn0 −1/2 + S30 /2 − S10 /2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )](S30 /2 − S10 /2 ) = 2(Sn0 −1/2 − S10 /2 )(S30 /2 − S10 /2 ). Recalling (6.5), the right-hand side of the above equality is <2 ·
n−1 R0
·
1 R0
, which contradicts to (6.9).
6.2. Proof of (ii) For fn−1 =
Pn−1 P0
Pn−1 (R) P0 (R) Pn−1 (R) P0 (R) Pn−1 (R) P0 (R)
=
< >
we just established fn0 −1 > 0, so that n+2
2n + 1
,
if R = Rn ,
(6.10)
,
if R < Rn ,
(6.11)
,
if R > Rn .
(6.12)
n+2
2n + 1 n+2
2n + 1
If for any n ≥ 2, we can establish Pn (R)
<
P0 (R)
n+3
2n + 3
,
R = Rn ,
(6.13)
then by combining this with (6.12) we conclude that Rn+1 > Rn , and this would complete the proof of Theorem (ii). Recalling (6.10), the inequality (6.13) can be written as follows: W (Rn )
< Pn−1 (Rn ),
(6.14)
where W (R) =
(2n + 3)(n + 2) Pn (R). (n + 3)(2n + 1)
(6.15)
It now suffices to prove that (6.14) holds. For n ≥ 5, this inequality is actually a consequence of the next lemma. Lemma 6.3. There holds W (R) < Pn−1 (R) for 0 < R ≤ Rn , n ≥ 5. Proof. By (2.14), W satisfies the differential equation W0 =
1 (2n + 3)(n + 2) R (n + 3)(2n + 1)
−
2n + 3 R
W−R
(n + 3)(2n + 1) 2 W , (n + 2)(2n + 3)
(6.16)
and by (2.10) and (6.15), W =
=
(2n + 3)(n + 2) 1 + O(R2 ) (n + 3)(2n + 1) 2n + 3 n+2
(n + 3)(2n + 1)
+ O(R2 ),
R → 0.
(6.17)
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It follows that W (R) < Pn−1 (R) for R near 0. If we can show that, for 0 < R ≤ Rn , 1 (2n + 3)(n + 2) R (n + 3)(2n + 1)
2n + 3
−
R
W−R
(n + 3)(2n + 1) 2 1 2n + 1 W < − W − RW 2 , (n + 2)(2n + 3) R R
(6.18)
then by comparison, we deduce that W (R) < Pn−1 (R) for all 0 < R ≤ Rn . Indeed, otherwise, there is a smallest R satisfies, 0 < R ≤ Rn , W (R) = Pn−1 (R), (W − Pn−1 )0 ≥ 0 at R, however, from (6.18) we have W 0 (R) < Pn0 −1 (R), which is a contradiction. To establish the inequality (6.18) and therefore the proof of (ii) of Theorem 6.1 (monotonicity of Rn ), we note that (6.18) is equivalent to 3
1
R (n + 3)(2n + 1)
3R
+
(n + 2)(2n + 3)
W2
<
2W R
,
0 < R ≤ Rn .
(6.19)
To complete the proof of Lemma 6.3 (and hence the proof of Theorem 6.1(ii)), it suffices to establish (6.19). The rest of this section is just devoted to this task. We shall first prove a sequence of auxiliary lemmas. Lemma 6.4. There holds, for all 0 < R < ∞, 3(n + 2) P0 (R) < W (R). (n + 3)(2n + 1)
(6.20)
˜ (R) = λP0 , λ > 0, to the function W . By (6.16), this means that λ has to be such Proof. We shall construct a subsolution W that
λP00 +
2n + 3 R
(n + 3)(2n + 1) 2 2 1 (2n + 3)(n + 2) Rλ P0 < , (n + 2)(2n + 3) R (n + 3)(2n + 1)
λP0 +
in view of (2.14) for n = 0, the above inequality is equivalent to 2n + 3 3λ 1 (2n + 3)(n + 2) 2 (n + 3)(2n + 1) 2
λ
If λ ≤
(n + 2)(2n + 3)
3(n+2) (n+3)(2n+1) ,
P0
2nλ R
<
− λ RP0 +
R
λ−
P0
R
<
R
(n + 3)(2n + 1)
−λ .
then the first term in the above inequality is negative, and this inequality is a consequence of
(2n + 3)(n + 2) −λ , R (n + 3)(2n + 1)
1
i.e. 2P0 <
(2n + 3)(n + 2) n(n + 3)(2n + 1)λ
−
1 n
.
This inequality is valid, since, by (2.16) and (2.13) and the fact that λ ≤ 2P0 (R) < 2P0 (0) =
2 3
≤
(2n + 3)(n + 2) n(n + 3)(2n + 1)λ
This completes the proof of Lemma 6.4.
−
1 n
3(n+2) (n+3)(2n+1) ,
.
Let G(R) =
I3/2 (R)
I1/2 (R)
.
(6.21)
Lemma 6.5. The function G(R) satisfies, for all 0 < R < ∞, 0 < G(R) < 1,
G0 (R)
> 0,
G00 (R)
< 0.
Proof. From (2.8) we find that G(R) = coth R − 1R , so that G0 (R) = G(R) = RP0 (R) ∼ G(R) = RP0 (R) ∼
1 3 1 R
R,
R → 0,
R = 1,
R → ∞.
sinh2 R−R2 R2 sinh2 R
> 0. It is easily seen that
H. Zhang et al. / Nonlinear Analysis 70 (2009) 2779–2795
It follows that 0 < G(R) < 1. Differentiating G0 (R) = 2(R3 cosh R − sinh R) 3
G00 (R) =
R3 sinh R
−
1 sinh2 R
once more we obtain
2 cosh R
=
3
1 R2
3
R3 sinh R
2 cosh R
:=
3
R3 sinh R
2791
R3 − (cosh R)−1 sinh R 3
g(R).
(6.22)
A direct computation shows that g(0) = g0 (0) = g00 (0) = 0 and 4
g000 (R) =
2
2
sinh R · {−14 cosh R + 6 sinh R} 4
cosh R
<0
so that g(R) < 0 for all 0 < R < ∞. This establishes that G00 (R) < 0. Vn−1 (R) , R
We write Pn−1 (R) =
In+1/2 (R)
Vn−1 (R) =
In−1/2 (R)
where
.
(6.23)
By (2.14), we have Vn0 −1 = 1 −
2n R
Vn−1 − Vn2−1 .
(6.24)
Introduce the positive solution V n−1 of 1=
2n R
2
V n−1 + V n−1 ,
(6.25)
i.e. V n−1 (R) = −
n R
+
" n 2 R
#1 2
+1
.
(6.26)
Note that, by (6.25) V n−1
R
<
2n
.
(6.27)
Differentiating (6.25), we obtain 0
2V n−1 V n−1 +
2n R
0
V n−1 =
2n R2
V n−1 ,
(6.28)
and hence upon using (6.27), we obtain 2n V R2 n−1
0
V n−1 =
2V n−1 +
2n
<
R
2n R R2 2n 2n R
=
1 2n
.
Lemma 6.6. There holds Vn−1 (R) ≥
2n
12
2n + 1
V n−1
12
2n 2n + 1
R ,
(6.29)
for all R > 0. Proof. The function H (R ) =
λV n−1 (λR),
λ > 0, ξ = λR
satisfies H0 +
2n R
H + H2 − 1 =
0
λ2 V n−1 +
0
2n R
= λ2 V n−1 +
< λ2
1 2n
2
λV n−1 + λ2 V n−1 − 1
2n
ξ
2 V n−1 + V n−1 − 1 + λ2 − 1
+ λ2 − 1 =
2n + 1 2n
λ2 − 1 ≤ 0 ,
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provided λ2 < H(R) ∼
2n , 2n+1
λ2
R
2n
so that H is a subsolution of (6.24). Since also, for R near 0
<
R
2n + 1
∼ Vn−1 (R).
We conclude that by comparison, that H(R) < Vn−1 (R) for all R > 0 and (6.29) follows by letting λ → 0
2n 2n+1
1/2
.
00
Lemma 6.7. The function V n−1 (R) satisfies V n−1 (R) > 0, V n−1 (R) < 0. Proof. The first inequality follows from (6.28). To prove the second inequality, we set V = V n−1 and differentiate (6.28) to obtain 2VV 00 + 2(V 0 )2 +
2n R
V 00 −
2n R2
V0 = −
4n R3
V+
2n R2
V0.
Since V > 0, we have 4n 4n sgnV 00 = sgn −2(V 0 )2 + 2 V 0 − 3 V R R 4n 0 V ≤ sgn 2 (V − ) R R 2 0 = sgn (−2VV ) by (6.28) R
< 0. This completes the proof of Lemma 6.7.
We introduce the function Φ (z, R) =
=
1
3
R (n + 3)(2n + 1)
3R
+
1
3
R
(n + 3)(2n + 1)
(n + 2)(2n + 3) +
z2 P02 −
3z2 G2
(n + 2)(2n + 3)
−
2 R
2 R
zP0
! zG
.
Lemma 6.8. For n ≥ 5, there holds Φ (z, R) < 0,
0 < R ≤ Rn , z ∈
3(n + 2)
,
n+2
(n + 3)(2n + 1) 2n + 1
.
(6.30)
Proof. Set z1 =
3(n + 2) , (n + 3)(2n + 1)
z2 =
n+2
2n + 1
.
Since the function z → Φ (z, R) is a parabola, it is sufficient to prove (6.30) just at the extreme points z1 and z2 , Φ (z1 , R) < 0 reduces to 1 +
9(n + 2)G2
<
2(n + 2)G
(n + 3)(2n + 1)(2n + 3) R 3 3(n + 2)G2 2(n + 2)G Φ (z2 , R) < 0 reduces to + < . n+3 (2n + 1)(2n + 3) R
,
Since 0 < G < 1, it suffices to prove 9(n + 2) 2(n + 2)G < , (n + 3)(2n + 1)(2n + 3) R 3 3(n + 2) 2(n + 2)G + < . n+3 (2n + 1)(2n + 3) R
1+
Noting that (6.32) is a consequence of (6.31), it remains to prove that 9(n + 2) M(R) := 2(n + 2)G(R) − 1 + R > 0. (n + 3)(2n + 1)(2n + 3)
(6.31) (6.32)
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Observe that for R near 0, 2(n + 2)R
M (R ) ∼
3
9(n + 2)
− 1+
(n + 3)(2n + 1)(2n + 3)
8n4 + 44n3 + 74n2 + 18n − 45
=
3(n + 3)(2n + 1)(2n + 3)
R
R,
so that M(0) = 0, M0 (0) > 0, and M00 (R) < 0 (since G00 < 0). Hence in order to prove that M(R) > 0 for all 0 < R ≤ Rn , it suffices to show that M(Rn ) > 0, i.e., 9(n + 2) Rn > 0. (6.33) 2(n + 2)G(Rn ) − 1 + (n + 3)(2n + 1)(2n + 3) Set Qn−1 (R) =
2n + 1
Kn−1 (R) =
2n + 1
Ln−1 (R) =
n+2
12
2n
2n + 1
V n−1
2n
12
2n + 1
Vn−1 (R), n+2 (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
2(n + 2)(n + 3)(2n + 1)(2n + 3)
R ,
R.
By (6.10), (6.21) and (6.23), we have Kn−1 (Rn ) = G(Rn ).
(6.34)
It is also clear that (6.33) is equivalent to Ln−1 (Rn )
< G(Rn ).
(6.35)
To finish the proof of this lemma, it suffices to establish (6.35). Since M(0) = 0 and M0 (0) > 0, we obtain G(0) = Ln−1 (0) = 0,
G0 (0)
> L0n−1 (0),
whereas, for R large, G(R) < 1 < Ln−1 (R). Since G is concave (G00 < 0), there exists a unique point R˜ n such that Ln−1 (R)
< G(R) if R < R˜ n ,
Ln−1 (R)
> G(R) if R > R˜ n .
(6.36)
To prove (6.35), let R∗n be the point where Ln−1 (R∗n ) = 1,
i.e., R∗n =
2(n + 2)(n + 3)(2n + 1)(2n + 3) . (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
(6.37)
Since G(R) < 1, (6.36) implies that R˜ n
< R∗n .
Suppose Qn−1 (R∗n )
> 1 = Ln−1 (R∗n ),
(6.38)
then from the concavity of Qn−1 (i.e., Qn00−1 < 0 by Lemma 6.7) and the fact that Qn−1 (0) = Ln−1 (0) = 0, it follows that Qn−1 (R)
> Ln−1 (R),
if R < R∗n .
(6.39)
By monotonicity of Qn−1 (i.e., Qn−1 > 0 by Lemma 6.7) 0
Qn−1 (R)
> Qn−1 (R∗n ) > 1,
if R > R∗n ,
whereas by (6.39) and (6.36), Qn−1 (R)
> Ln−1 (R) > G(R),
if R˜ n < R < R∗n ,
thus, altogether, Qn−1 (R) > G(R) if R˜ n < R < ∞. Since Kn−1 (R) > Qn−1 (R) for all R > 0 (by Lemma 6.6), we conclude that Kn−1 (R)
> G(R),
if R˜ n < R < ∞.
Hence, by (6.34), Rn < R˜ n and, recalling (6.36), the assertion (6.35) follows, and this completes the proof of (6.33).
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It remains to prove that (6.38) holds, that is, V n−1 (σn−1 )
> An−1 ,
(6.40)
where
σn−1 =
12
2n
2n + 1
R∗n ,
An−1 =
n+2
2n + 1
2n + 1 2n
12
.
(6.41)
Since, by (6.25), 2n
2
V n−1 (σn−1 ) +
σn−1
V n−1 (σn−1 ) = 1.
If A2n−1 +
2n
σn−1
An−1
< 1,
(6.42)
then (6.40) holds. Substituting R∗n from (6.37) into the formula for σn , we obtain
σn−1 =
2n 2n + 1
12
2(n + 2)(n + 3)(2n + 1)(2n + 3) , (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
(6.43)
Substituting σn−1 from (6.43) and An from (6.41) into the left-hand side of (6.42), we find that (6.42) is equivalent to
−
2n4 + 3n3 − 35n2 − 81n − 36 2n(2n + 1)(n + 3)(2n + 3)
< 0,
which is valid if n ≥ 5. This completes the proof of Lemma 6.8.
Proof of (6.19) for the case n ≥ 5. From (6.17) we conclude that (6.19) is valid for small R. Let (0, R# ) be the largest interval with R# ≤ Rn on which (6.19) is valid. We want to claim that R# = Rn . If this is not the case, then R# < Rn , and, following the proof of Lemma 6.3 that we have W (R )
< Pn−1 (R) for 0 < R ≤ R# .
By (6.11) and (6.20), 3(n + 2) n+2 P0 (R) < W (R) < Pn−1 (R) < P0 (R) for 0 < R ≤ R# . (n + 3)(2n + 1) 2n + 1 It follows from Lemma 6.8 that (6.19) is also valid for 0 < R ≤ R# (with strict inequality at R = R# ). Thus the interval (0, R# ) can be further extended, and this contradicts the maximality of this interval. This establishes the estimate (6.19) for the case n ≥ 5. We have proved that Rn+1
> Rn for n ≥ 5.
The proof for n ≤ 4 can be obtained by explicit calculation. Indeed, solving Pn−1 (R) = 2nn++21 P0 (R) gives R = Rn , with R2 = 4.63512, R3 = 5.99911, R4 = 7.34806 and R5 = 8.68091. Clearly they are also monotonically increasing. To complete the bifurcation proof we shall now establish (5.9): Lemma 6.9. There holds d tanh R [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] >0 dR R R =R n
for n ≥ 2.
Proof. Since (2n + 1)Pn−1 (Rn ) − (n + 2)P0 (Rn ) = 0, we derive d tanh R d tanh R Pn−1 (R) [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] = P0 (R) (2n + 1) − (n + 2) dR R dR R P0 (R) R =R n R =R n tanh Rn d Pn−1 (R) = P0 (Rn )(2n + 1) . Rn dR P0 (R) R=Rn Now the lemma follows from (6.3).
(6.44)
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Acknowledgements The second author was partially supported by National Science Foundation of China (Grant No. 10671156) and the Program for New Century Excellent Talents in University (NCET-04-0968) References [1] A. Borisovich, A. Friedman, Symmetry-breaking bifurcations for free boundary problems, Indiana Univ. Math. J. 54 (3) (2005) 927–947. [2] M.G. Crandall, L.H. Rabinowitz, Bifurcation from simple eigenvalues, J. Funct. Anal. 8 (2) (1971) 321–340. [3] S. Cui, J. Escher, Bifurcation analysis of an elliptic free boundary problem modelling the growth of vascular tumors, SIAM J. Math. Anal. 39 (1) (2007) 210–235. [4] M. Fontelos, A. Friedman, Symmetry-breaking bifurcations of free boundary problems in three dimensions, Asymptotic Anal. 35 (3–4) (2003) 187–206. [5] M. Fontelos, A. Friedman, Symmetry-breaking bifurcations of charged drops, Arch. Ration. Mech. Anal. 172 (2) (2004) 267–294. [6] A. Friedman, A hierarchy of cancer models and their mathematical challenges, in: Mathematical Models in Cancer (Nashville, TN, 2002), Discrete Contin. Dyn. Syst. Ser. B 4 (1) (2004) 147–159. [7] A. Friedman, Symmetry-breaking bifurcations for free boundary problems, in: Nonlinear Partial Differential Equations and Related Analysis, in: Contemp. Math., vol. 371, Amer. Math. Soc, Providence, RI, 2005, pp. 153–161. [8] A. Friedman, Cancer models and their mathematical analysis, in: Tutorials in Mathematical Biosciences. III, in: Lecture Notes in Math., vol. 1872, Springer, Berlin, 2006, pp. 223–246. [9] A. Friedman, B. Hu, A Stefan problem for a protocell model, SIAM J. Math. Anal. 30 (4) (1999) 912–926. [10] A. Friedman, B. Hu, J.J.L. Velazquez, On the zeros of quotients of bessel functions, Chinese Ann. Math. 21B (3) (2000) 1–12. [11] A. Friedman, B. Hu, J.J.L. Velazquez, A Stefan problem for a protocell model with symmetry-breaking bifurcations of analytic solutions, Interfaces Free Bound. 3 (2) (2001) 143–199. [12] A. Friedman, B. Hu, Bifurcation from stability to instability for a free boundary problem arising in a tumor model, Arch. Ration. Mech. Anal. 180 (2) (2006) 293–330. [13] A. Friedman, B. Hu, Bifurcation for a free boundary problem modelling tumor growth by Stokes equation, SIAM J. Math. Anal. 39 (1) (2007) 174–194. [14] A. Friedman, F. Reitich, Symmetry-breaking bifurcation of analytic solutions to free boundary problems: An application to a model of tumor growth, Trans. Amer. Math. Soc. 353 (4) (2000) 1587–1634. [15] H. Schwegler, K. Tarumi, B. Gerstmann, Physico-chemical model of protocell, J. Math. Biol. 22 (3) (1985) 335–348. [16] K. Tarumi, H. Schwegler, A nonlinear treatment of the protocell by boundary layer approximation, Bull. Math. Biol. 49 (3) (1987) 307–320.
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Bifurcation for a free boundary problem modeling a protocell Hua Zhang a,b , Changzheng Qu a,b , Bei Hu c,∗ a Center for Nonlinear Studies, Northwest University, Xi’an, 710069, PR China b Department of Mathematics, Northwest University, Xi’an, 710069, PR China c Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556, USA
article
info
Article history: Received 25 February 2008 Accepted 7 April 2008 Keywords: Free boundary problem Stationary solution Bifurcation Symmetry breaking Protocell
a b s t r a c t In this paper we consider a free boundary problem for a physico-chemical model of a protocell. This model of a self-maintaining unity or a protocell is based on the reaction and diffusion process, and a mechanism of self-control of the boundary. For any positive radius R, there exists a radially symmetric solution with radius r = R. In the more realistic threespace-dimensional case, we give a proof that there exist symmetry-breaking bifurcation branches of solutions with free boundary r = R + Yn,0 (θ) + O(2 ) (n ≥ 2, even) for small ||, where Yn,0 is the spherical harmonic of mode (n, 0). © 2008 Elsevier Ltd. All rights reserved.
1. Introduction The protocell is a mathematical model of a self-maintaining unity based on the dynamics of simple reaction–diffusion process and a self-controlled dynamics of the surface. It has been developed and studied in many papers (see [15,16,9–11] and the references therein). In this paper, we shall establish the existence of symmetry-breaking bifurcations for the free boundary of this model in the three-space-dimensional case. The protocell can be visualized as having a porous structure maintained by building materials with concentration C . The structure is sustained only as long as C exceeds a critical concentration C ∗ . Metabolism is maintained by nutrient material with concentration σ which is distributed in the entire space with σ = τ at ∞ (τ > 0). C and σ satisfy a coupled system of reaction–diffusion equation:
∂C − ∆C = σ, −∆σ = −σ in the cell, ∂t −∆σ = 0 outside the cell, c
where c is a positive constant. The constant c is the quotient of the time scale of diffusion to the time scale of cell doubling. On the boundary of the protocell C = C ∗ . The various tasks that the cell continuously performs take their toll on the cell: they cause it to shrink. This is modeled by disintegration at the boundary at a rate β, β > 0. On the other hand the flux of building material at the boundary causes the cell to grow. The total result of these two effects is Vn = −
∂C − β on the boundary of the cell, ∂n
where n is the exterior normal, and Vn is the velocity of the boundary points in the direction n.
∗ Corresponding author. E-mail address: [email protected] (B. Hu). 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.04.003
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In the case of a spherical cell (in 3-dimensions), setting q t r=
x21 + x22 + x23 ,
u = C − C∗ ,
˜t =
c
and denoting the boundary of the cell by r = s(t), we have the following system for u = u(r, t), σ = σ(r, t) and r = s(t): ! ∂2 2 ∂ + σ = σχr
σ → τ as |x| → ∞, and ut −
2 ∂ ∂2 + ∂r2 r ∂r
! u=
σ if r < s(t), t > 0,
u=0
on r = s(t), t > 0,
u = u0 (r)
(1.2)
for t = 0,
and finally, the free boundary condition s0 (t) = −cur (s(t), t) − cβ.
(1.3)
It was established in [15] that there exist radially symmetric stationary solutions (R, σs , us ) to (1.1)–(1.3), and they are given explicitly by R − tanh R τ 1 − for r ≥ R, r σs (r) = (1.4) 1 sinh r τ for r < R, cosh R r τ sinh R sinh r us (r) = − for r < R, (1.5) cosh R R r where R satisfies R2 τ = . β R − tanh R
(1.6)
The asymptotic behavior for radially symmetric solutions was studied in [9]. It is very natural to ask whether there exist non-radially symmetric stationary solutions and what are their asymptotic behaviors. These are indeed very challenging questions with only few results in the literature. Using a very lengthy analytical expansion of the free boundary and the solution, the existence of non-radially symmetric stationary solutions was established in [10,11]: A sequence of symmetrybreaking bifurcation branches of stationary solutions of (1.1)–(1.3) was found in the two-space-dimensional case. Of course, the more realistic model is in three space dimension. However, due to the complexity of the analytic expansion used in [11], it is very difficult to extend this method into the three-space-dimensional case. The purpose of the present paper is as follows: first, we want to extend the bifurcation results of [10,11] into the more realistic three-space-dimensional case; second, we shall use a much more simplified and systematic method to establish the existence of symmetry-breaking bifurcation branches of stationary solutions by applying the Crandall–Rabinowitz bifurcation theorem [2]. In order to do so, we shall derive many delicate PDE estimates for our system. Note that for our system (1.1), the solution σ is only C 1+1 in the spatial variable, i.e., σrr is bounded, but has a jump across the free boundary {r = s(t)}. Therefore, a perturbation of this system involves a diffraction boundary condition (cf., (3.6)). Since we also need to perturb the free boundary {r = s(t)}, the extension of PDE estimates to our system is not a trivial matter. The application of the Crandall–Rabinowitz bifurcation theorem framework to free boundary problems started in [1,4,5]. This method was subsequently used successfully on other free boundary models in [12,13,3]. For systematic introduction of biological cancer models and bifurcation analysis we refer the readers to [6–8]. 2. Spherical harmonics and modified Bessel function In this paper we shall need Spherical harmonics and some inequalities for the modified Bessel function Im (ξ), where ξ is a complex variable and m is any nonnegative real number. For convenience, we collect all these required identities and inequalities in this section and refer the readers to [12] for the proof. The Spherical harmonics. The Spherical harmonics Yl,m (θ, φ) is defined by s (2l + 1)(l − m)! m eimφ m Yl,m (θ, φ) = (−1) Pl (cos θ) √ (m = −l, . . . , l), 2(l + m)! 2π
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where Plm (x) = 21l l! (1 − x2 )m/2 ddxl+m (x2 − 1)l . The family of functions Yl,m forms a complete orthonormal basis for L2 (Σ ), where Σ is the unit sphere, and l+m
∆ω Yl,m = −l(l + 1)Yl,m ,
where ∆ω =
1 ∂ sin θ ∂θ
(sin θ ∂θ∂ ) +
1 ∂2 sin2 θ ∂φ2
is the Laplace operator on the unit sphere Σ .
Bessel functions. For the modified Bessel functions Im (ξ), which is defined by m X 2k ∞ ξ 1 ξ Im (ξ) =
2
it is well known that s 2 I1/2 (ξ) =
πξ s
I3/2 (ξ) =
2
s
00 Im (ξ) + 0 Im (ξ) −
1
ξ
2
−
ξ
ξ
+ cosh ξ ,
+ 1 sinh ξ −
2
m2
I0 (ξ) − 1 +
ξ
!
ξ2
(2.2)
3
ξ
cosh ξ ,
(2.3)
Im (ξ) = 0,
(2.4)
m ≥ 0,
(2.5)
Im (ξ) = Im+1 (ξ),
Im (ξ)In (ξ) =
,
(2.1)
sinh ξ 3
πξ
m
2
sinh ξ,
πξ
I5/2 (ξ) =
k!Γ (m + k + 1)
k =0
∞ X
Γ (m + n + 2k + 1)(ξ/2)m+n+2k
k =0
k!Γ (m + k + 1)Γ (n + k + 1)Γ (m + n + k + 1)
.
(2.6)
Throughout this paper, we shall be working with the functions Pn (ξ) =
In+3/2 (ξ) , ξIn+1/2 (ξ)
n = 0, 1, 2, . . . .
(2.7)
It satisfies (cf. [12, page 297–299]) P0 (ξ) = P1 (ξ) =
1
ξ
1
coth ξ −
ξ2
1
ξ coth ξ − 1
,
(2.8) 3
−
ξ2
,
(2.9)
ξ2 + O(|ξ|4 ), |ξ| → 0, 2n + 3 (2n + 3)2 (2n + 5) 1 n+1 π + O(|ξ|−3 ), |arg(ξ)| < − δ, |ξ| → +∞, Pn (ξ) = − ξ ξ2 2 1
Pn (ξ) =
Pn (ξ) =
1
ξ2 Pn+1 (ξ) + (2n + 3) 1
Pn (0) =
d dξ
−
2n + 3
Pn (ξ) =
1
ξ
,
(2.11) (2.12)
,
−
(2.10)
(2.13)
2n + 3
ξ
Pn (ξ) − ξPn2 (ξ),
(2.14)
and for any real r > 0, n ≥ 0, Pn (r)
d dr 1 r
> Pn+1 (r),
Pn (r)
−
(2.15)
< 0,
n+1 r2
< Pn (r) <
(2.16) 1 r
−
n+1 r2
+
n2 + n + 1
2r 3
.
(2.17)
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3. The bifurcation problem Consider a family of domains with boundaries ∂Ω : r = R + R˜ (θ, φ), where R˜ (θ, φ) = S(θ, φ). Let (σ, u) be the solution of ∆σ = σχ{x∈Ω } (x)
in R3 ,
(3.1)
σ → τ as |x| → ∞, and
−∆u = σ in Ω , u=0
(3.2)
on ∂Ω .
We define F to be F (R˜ , R) = −
∂u R − tanh R τ. − ∂n ∂Ω R2
Then, recalling (1.6), we have F (0, R) ≡ 0
for any 0 < R < ∞.
It is also clear that (σ, u, R + R˜ ) is a stationary solution if and only if F (R˜ , R) = 0.
(3.3)
Using a scaling (u/τ, σ/τ) if necessary we can assume without loss of generality that τ = 1. We shall do so throughout the rest of this paper. The function S(θ, φ) may be viewed as a function defined on the unit sphere Σ = {x; |x| = 1}. We shall later assume that S(θ, φ) is in C m+α , that is, S is C m+α as a function defined on Σ ; note that this does not mean that S(θ, φ) is in C m+α in the variable (θ, φ). We shall see in Sections 3–5 that F maps the space C m+α (Σ ) × R into the space C m+α−1 (Σ ). In this section we shall formally compute the Fréchet derivatives of F (cf., (3.5)). The formal derivation will be rigorously justified in Section 4. In order to compute the Fréchet derivatives of F , we need the expansion of (σ, u) of order . We formally write,
σ = σs + σ1 + O(2 ), u = us + u1 + O(2 ),
(3.4)
∂Ω : r = R + S(θ, φ), where (σs , us ) is the stationary solution defined in (1.4) and (1.5). Using the relation ∂(us + u1 ) ∂u = + O(2 ) ∂n ∂Ω ∂n R + S ∂us ∂2 us ∂u1 = + O(2 ), + 2 S + ∂r ∂r ∂r r =R
we find that F (R˜ , R) = −
∂u1 ∂us ∂2 us + 2 S + ∂r ∂r ∂r
− r =R
R − tanh R R2
− O(2 ).
The Fréchet derivative is therefore given by ∂2 us ∂u1 [FR˜ (0, R)] · S = − S + . ∂r2 ∂r
(3.5)
r=R
We start now by computing explicitly (and formally) σ1 and u1 . 3.1. Computation of σ1 We shall use the notation σ + to denote the limit of σ from the exterior of Ω and σ − to denote the limit of σ from the interior of Ω . Clearly
σ+ |∂Ω = σ− |∂Ω ,
σr+ |∂Ω = σr− |∂Ω .
Using Taylor’s expansion at r = R from both sides, we obtain + − + + − −
σs + Sσsr + σ1 |r=R = σs + Sσsr + σ1 |r=R + O(2 ),
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and + − σsr+ + Sσsrr + σ1+r |r=R = σsr− + Sσsrr + σ1−r |r=R + O(2 ).
Dropping O(2 ) terms, we obtain + − +
σ1 − σ1 |r=R = −[σsr (R) − σsr− (R)]S = 0,
(3.6)
+ − σ1+r − σ1−r |r=R = −[σsrr (R) − σsrr (R)]S := h(R)S,
R + − where h(R) := −[σsrr (R) − σsrr (recalling that τ = 1). Substituting the first equation in (3.4) and (3.6) into (3.1) (R)] = tanh R and dropping the O() terms, we find that σ1 satisfies
−∆σ1 + σ1 = 0 in BR , ∆σ1 = 0
outside BR ,
(3.7)
σ1 − σ1 |r=R = 0, +
−
σ1+r − σ1−r |r=R = h(R)S.
If S(θ, φ) = Yn,m (θ, φ), we look for a solution of the form
σ1 = σ˜ 1 (r)S(θ, φ).
(3.8)
Then 2
n(n + 1)
r
2
r2 n(n + 1)
r
r2
σ˜ 100 + σ˜ 10 − σ˜ 100 + σ˜ 10 −
σ˜ 1 = σ˜ 1 , σ˜ 1 = 0,
r r
< R,
> R.
These ODEs can be solved explicitly: 1
σ˜ 1 = C1 r− 2 In+ 1 (r), 2
r
< R,
and σ˜ 1 = C2 r−n−1 ,
r
> R.
(3.9)
Using the last two relations in (3.7), we derive 1
C1 R− 2 In+1/2 (R) = C2 R−n−1 , C2 (−n − 1)R−n−2 − C1 [nR−3/2 In+1/2 (R) + R−1/2 In+3/2 (R)] = h(R),
so that (recalling the definition of Pn in (2.7)) C1 =
−h(R) , [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R)
C2 =
−h(R)Rn . (2n + 1)R−2 + Pn (R)
(3.10)
By (3.8), σ1 is given explicitly by
σ1 (r, θ, φ) =
−h(R)r−1/2 In+1/2 (r) Yn,m (θ, φ), [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R)
σ1 (r, θ, φ) =
−h(R)Rn r−n−1 Yn,m (θ, φ), (2n + 1)R−2 + Pn (R)
r
r
< R,
> R.
(3.11) (3.12)
3.2. Computation of u1 From (3.2) and the second equation in (3.4), we obtain
− ∆u1 = σ1 ,
in BR ,
(3.13)
and, recalling that u|r=R+S = 0 and us (R) = 0,
u1 |∂BR = u1 |∂Ω + O(2 ) = (u − us )|r=R+S + O(2 ) = −us (R + S) + O(2 ) = −[us (R) + usr (R)S] + O(2 ) = −λS + O(2 ),
(3.14)
where λ := usr (R) = −(tanh R)P0 (R). Dropping higher-order terms we find that u1 |r=R = −λS.
(3.15)
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If S = Yn,m (θ, φ), then u1 = u˜ 1 (r)S(θ, φ) satisfies 2
n(n + 1)
r
r2
(u˜ 1 + σ˜ 1 )00 + (u˜ 1 + σ˜ 1 )0 −
(u˜ 1 + σ˜ 1 ) = 0,
r
< R,
so that
˜ 1 = C3 r n , u˜ 1 + σ
r
< R.
Using conditions (3.9), (3.10) and (3.15), we obtain C3 =
−λ Rn
−
h(R)R−n−1
(2n + 1)R−2 + Pn (R)
.
Thus u1 is given by u1 (r, θ, φ) =
−λ
−
Rn
h(R)R−n−1
!
(2n + 1)R−2 + Pn (R)
rn Yn,m (θ, φ) − σ1 .
(3.16)
4. Rigorous justification Note that u is defined only on Ω , while us is defined on BR . The formulas that define σs and us actually define them in the whole space. In order to derive necessary estimates for σ1 and u1 , we transform all these functions to the same domain Ω and shall do so by the Hanzawa transformation, which is a diffeomorphism defined by
(r, θ, φ) = H (r0 , θ0 , φ0 ) ≡ (r0 + χ(R − r0 )S(θ0 , φ0 ), θ0 , φ0 ), where 0, χ(z) = 1,
χ∈C , ∞
if |z| ≥ 3δ0 /4, if |z| < δ0 /4,
dk χ k ≤ C /δk0 . dz
Observe that H maps BR onto Ω while keeping the ball {r < R − (3δ0 /4)} and the domain {r > R + (3δ0 /4)} fixed. The inverse Hanzawa transformation H−1 maps Ω onto BR . We set
σˆ 1 (r, θ, φ; ) = σ1 (H−1 (r, θ, φ)) in Ω , uˆ 1 (r, θ, φ; ) = u1 (H−1 (r, θ, φ)) in Ω . Note that an inverse Hanzawa transformation, if applied to σs or us , would destroy the boundary conditions on ∂Ω . Therefore, instead of using (1.4) and (1.5), we shall use the following functions R − tanh R 1 − in Ω , r σˆ s (r, θ, φ; ) = (4.1) sinh r 1 in R3 \ Ω , cosh R r 1 sinh R sinh r us (r) = − for 0 < r < ∞. (4.2) cosh R R r Remark 1. Note that with this definition, σˆ is discontinuous on ∂Ω . Lemma 4.1. There holds
kσ − σˆ s − σˆ 1 kC2+α (Ω ) + kσ − σˆ s − σˆ 1 kC2+α (R3 \Ω ) ≤ C 2 ,
(4.3)
where the constant C depends only on kSkC2+α (Σ ) . Proof. We have
−∆(σˆ s + σˆ 1 ) + (σˆ s + σˆ 1 ) = fˆ in Ω ,
(4.4)
−∆(σˆ s + σˆ 1 ) = fˆ in R3 \ Ω ,
(4.5)
where fˆ is compactly supported, 1 fˆ is a linear combination of at most second-order derivatives of σ1 and S. Thus
kfˆkCα (Ω ) ≤ C ||,
(4.6)
kfˆkCα (R3 \Ω ) ≤ C ||,
(4.7)
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where the constant C depends on kSkC2+α (Σ ) . Now we work on the boundary conditions. Since σˆ 1 is continuous,
(σˆ s + σˆ 1 )+ − (σˆ s + σˆ 1 )− = σˆ s+ − σˆ s− ≡ g1 on ∂Ω .
(4.8)
A direct computation shows that g1 = g11 (S),
where g11 (η) =
1
sinh(R + η)
cosh R
R+η
−1+
R − tanh R
R+η
.
(4.9)
0 0 00 Since g11 (0) = g11 (0) = 0, we have |g11 (η)| ≤ C |η|2 and |g11 (η)| ≤ C |η|, |g11 (η)| ≤ C . Using these estimates we can easily derive
kg1 kC2 (∂Ω ) ≤ C 2 ,
(4.10)
where the constant C depends on kSkC2 (Σ ) . Note that D g1 = (S)(DS) + g11 (S)D S, here we use D to denote all possible second derivatives and (DS)2 to denote all possible products of two first-order derivatives of S. Clearly, 2
2 00 g11
2
0
2
2
00 (S)(DS)2 ]Cα (∂Ω ) ≤ C (kSkC2+α (Σ ) ) 2 , [2 g11 0 and, using the standard interpolation on [g11 (S)]Cα (∂Ω ) , we obtain 0 0 0 [g11 (S)D2 S]Cα (∂Ω ) ≤ kg11 (S)kL∞ (∂Ω ) [D2 S]Cα (Σ ) + [g11 (S)]Cα (∂Ω ) kD2 SkL∞ (Σ ) 0 0 0 ≤ kg11 (S)kL∞ (∂Ω ) [D2 S]Cα (Σ ) + (kg11 (S)kL∞ (∂Ω ) + kg11 (S)kC1 (∂Ω ) )kD2 SkL∞ (Σ )
≤ C (kSkC2+α (Σ ) ) 2 . Combining all these estimates, we find that
kg1 kC2+α (∂Ω ) ≤ C 2 ,
(4.11)
where the constant C depends on kSkC2+α (Σ ) . We now write
ˆ s + σˆ 1 )− (σˆ s + σˆ 1 )+ on ∂Ω , r − (σ r = g2
(4.12)
where 0 g2 = g21 (S), where g21 (η) = g11 (η) − h(R)η.
(4.13)
By the definition of h(R) (cf. (3.6)) we clearly have h(R) = g11 (0), so that again we have g21 (0) = g21 (0) = 0, |g21 (η)| ≤ C |η|2 0 00 and |g21 (η)| ≤ C |η|, |g21 (η)| ≤ C . Following a similar argument as above, we conclude that 00
0
kg2 kC1+α (∂Ω ) ≤ C 2 ,
(4.14)
where the constant C depends on kSkC2+α (Σ ) . The function
ψ = σ − σˆ s − σˆ 1 clearly satisfies
−∆ψ + ψ = −fˆ in Ω , −∆ψ = −fˆ in R3 \ Ω ,
(4.15)
ψ+ − ψ− = −g1 on ∂Ω , − ψ+ on ∂Ω , r − ψr = −g2
where fˆ is compactly supported. Note that if S ∈ C 2+α (Σ ), then ∂Ω is in C 2+α uniformly in . Applying the Schauder estimates for the diffraction problem (4.15), we obtain, for small ||,
kψkC2+α (Ω ) + kψkC2+α (R3 \Ω
)
≤ C {||kfˆkCα (R3 \Ω ) + ||kfˆkCα (Ω ) + kg1 kC2+α (∂Ω ) + kg2 kC1+α (∂Ω ) }
≤ C 2 . This completes the proof.
Next we establish Lemma 4.2. There holds
ku − us − uˆ 1 kC2+α (Ω ) ≤ C 2 , where the constant C depends on kSkC2+α (Σ ) .
(4.16)
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Proof. The proof is very similar. We make use of (3.13), (3.14) and (4.3) and the elliptic Schauder estimates. Since this problem involves only Ω (does not involve R3 \ Ω ), the proof is actually much easier, and we omit the details. With the estimates in above two lemmas, it is clear that the computation of Frechét derivative in (3.5) is rigorous in an appropriate space, say, F : C 2+α → C 1+α . However, in order to apply the bifurcation theorem, further restrictions are necessary, and we shall discuss this in the next section. 5. Bifurcation We shall use the following Crandall–Rabinowitz theorem (see [2], Theorem 1.7). Theorem 5.1. Let X , Y be real Banach spaces and F (x, µ) a C p map, p ≥ 3, of a neighborhood (0, µ0 ) in X × R into Y . Suppose (i) (ii) (iii) (iv)
F (0, µ) = 0 for all µ in a neighborhood of µ0 , ker Fx (0, µ0 ) is one-dimensional space, spanned by x0 , Im Fx (0, µ0 ) = Y1 has codimension 1, Fµx (0, µ0 )x0 6∈ Y1 .
Then (0, µ0 ) is a bifurcation point of the equation F (x, µ) = 0 in the following sense: In a neighborhood of (0, µ0 ) the set of solutions of F (x, µ) = 0 consists of two C p−2 smooth curves Γ1 and Γ2 which intersect only at the point (0, µ0 ); Γ1 is the curve (0, µ) and Γ2 can be parameterized as follows: Γ2 : (x(), µ()), ||
small, (x(0), µ(0)) = (0, µ0 ), x0 (0) = x0 . In order to apply this theorem, we first compute explicitly expressions on the right-hand side of (3.5) when S = Yn,m . Recall that
λ = usr (R) = −(tanh R)P0 (R),
tanh R
h(R) =
R
.
(5.1)
[1 − 2P0 (R)].
(5.2)
A direct computation shows that usrr (R) = − tanh R
R2 − 2R coth R + 2 R3
=−
tanh R R
Differentiating (3.11) and (3.16) in r, using also (2.5) and (2.7) and letting r = R, we find that
σ1r (R, θ, φ) = =
−h(R)[nR−3/2 In+1/2 (R) + R−1/2 In+3/2 (R)] Yn,m (θ, φ) [(2n + 1)R−3/2 + R1/2 Pn (R)]In+1/2 (R) −h(R)[nR−2 + Pn (R)] Yn,m (θ, φ), (2n + 1)R−2 + Pn (R)
(5.3)
and u1r (R, θ, φ) =
=
−h(R)R−n−1 + nRn−1 Yn,m (θ, φ) − σ1r Rn (2n + 1)R−2 + Pn (R)
−λ
−
!
nλ R
+
h(R)R2 Pn (R)
!
(2n + 1) + R2 Pn (R)
Yn,m (θ, φ).
(5.4)
Substituting (5.1)–(5.4) into (3.5), using also (2.12) twice, we get
[FR˜ (0, R)] Yn,m = =
tanh R R
tanh R R
[1 − (n + 2)P0 − R2 Pn (R)Pn−1 (R)]Yn,m [(2n + 1)Pn−1 − (n + 2)P0 ]Yn,m .
(5.5)
It follows that
[FR˜ (0, R)]Yn,m = 0
(5.6)
if and only if
(2n + 1)Pn−1 (R) = (n + 2)P0 (R).
(5.7)
In the next section we shall prove For each n ≥ 2, there exists a unique solution R = Rn that solves
(2n + 1)Pn−1 (R) = (n + 2)P0 (R),
and
Rm
> Rl for all m > l ≥ 2,
(5.8)
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and d
tanh R
dR
R
[(2n + 1)Pn−1 (R) − (n + 2)P0 (R)]
6= 0 for n ≥ 2.
(5.9)
R =R n
We shall use R as the bifurcation parameter, i.e., in Theorem 5.1 we take µ = R. As in [4,12,13] we introduce the Banach spaces X m+α = {R˜ ∈ C m+α (Σ ), R˜
is π - periodic in θ, 2π - periodic in φ},
X1m+α = closure of the linear space spanned by{Yj,0 (θ), j = 0, 1, 2, . . .} in X m+α , X2m+α = closure of the linear space spanned by{Yj,0 (θ), j = 0, 2, 4, . . .} in X m+α .
Then F (R˜ , µ) is a map from X m+2+α to X m+1+α . The operator A(µ) = [FR˜ (0, µ)] maps X12+α to X11+α , and, by (5.8), {Y1,0 } if µ 6= R2 , R3 , . . . , ker A(µ) = span{Yn,0 , Y1,0 } if µ = Rn . If we work with the space X12+α , then dim(ker A(µ)) = 2 when µ = Rn . In order to satisfy the requirement (ii) of Crandall–Rabinowitz theorem we shall henceforth work with the space X22+α . In the space X22+α , dim(ker A(µ)) = 1 if µ = Rn . If n is even and ≥ 2, then clearly the direct sum Im A(µ) ⊕ {Yn,0 } is the whole space X21+α and therefore codim(A(µ)) = 1 for µ = Rn . Differentiating (5.5) in R, we obtain d tanh R [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] Yn,m . [FR˜ µ (0, R)]Yn,m = dR R In particular, using (5.9) we find that d tanh R Yn,0 [FR˜ µ (0, Rn )]Yn,0 = [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] dR R R=Rn
6∈ Im A(Rn ). Thus, we have established the following result: Theorem 5.2. The points (0, Rn ) (n even and ≥ 2) are bifurcation points for the problem (3.1)–(3.3), and the corresponding free boundaries are of the form r = R + Yn,0 (θ) + o().
Note that this theorem gives non-radially symmetric stationary solutions to the original free boundary problem (1.1)–(1.3). 6. Monotonicity of Rn It remains to establish (5.8) and (5.9). We shall employ the ideas and methods from [10]. In [10], they deal with integer modified Bessel functions. Here, we deal with fraction modified Bessel functions. Theorem 6.1. (i) For any n ≥ 2, there exists a unique positive solution R = Rn of the equation Pn−1 (R)
=
P0 (R)
n+2
2n + 1
.
(ii) If 2 ≤ l < m, then Rl
< Rm .
6.1. Proof of (i) Consider the function fn−1 =
Pn−1 P0
=
In+1/2 I1/2 . In−1/2 I3/2
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From (2.10) and (2.11), we derive
fn−1 =
1 2n+1
(2n+1)2 (2n+3)
1 3 1
fn−1 =
R2
−
R
1 R
−
d
45
+ O (R 4 )
−
n R2
+ O ( R −3 )
−
1
O ( R −3 )
+
R2
n+2 2n+1
so that, if n ≥ 2, fn−1 −
dR
R2
fn−1 (R)
> 0,
+ O (R 4 )
=
→
3 2n + 1
R − n + O ( R −1 ) R − 1 + O(R−1 )
,
R → 0,
→ 1,
(6.1)
R → ∞,
(6.2)
is negative for small R and positive for large R. Hence, if
for all R > 0,
(6.3)
then, for any n ≥ 2, there exists a unique positive solution R = Rn of the equation Introduce the functions
Pn−1 (R) P0 (R)
=
n+2 . 2n+1
Sn = log In (R).
If we can prove Sn0 +1/2 − Sn0 −1/2
> S30 /2 − S10 /2 ,
(6.4)
then (6.3) follows. In order to prove (6.4), we need the fact Lemma 6.2. For any n ≥ 1, Sn0 +1/2 − S10 /2
<
n R
for all R > 0.
(6.5)
Proof. Using (2.5), we find that (6.5) is equivalent to In0 +1/2 In+1/2
−
I10 /2 I1/2
=
In+3/2 In+1/2
+
n + 1/2 R
−
I3/2 I1/2
+
1/2 R
!
n
< , R
which is the same as In+3/2 In+1/2
<
I3/2 I1/2
,
(6.6)
Using the product formula (2.6) it was proved in [14] that each term in the power series of In+3/2 In−1/2 is smaller than each term in the power series of (In+1/2 )2 , so that
In+3/2 In+1/2
<
In+1/2
In−1/2
and (6.6) then follows by iterating this inequality.
Introduce the function
φ(R) = (Sn+1/2 − Sn−1/2 ) − (S3/2 − S1/2 ), then the assertion (6.3) is equivalent to the inequality
φ0 (R) > 0,
if R > 0.
From (2.4), we have Sn00+1/2 + (Sn0 +1/2 )2 +
1 R
Sn0 +1/2 = 1 +
(n + 1/2)2 R2
.
Similarly, Sn00−1/2 + (Sn0 −1/2 )2 +
1 R
Sn0 −1/2 = 1 +
(n − 1/2)2 R2
,
so that 1
(n + 1/2)2 − (n − 1/2)2
R
R2
(Sn+1/2 − Sn−1/2 )00 + (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) + (Sn0 +1/2 − Sn0 −1/2 ) =
.
(6.7)
Writing (6.7) for n = 1 and subtracting it from (6.7), we obtain 1
2n − 2
R
R2
φ00 + (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) + φ0 =
.
(6.8)
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For small R, (6.1) implies In+1/2 I1/2
fn−1 =
In−1/2 I3/2
=
3
2n + 1
+
1 15
−
3
(2n + 1)2 (2n + 3)
R2 + O(R4 ),
since n ≥ 2, fn0 −1 > 0 for R small. If (6.3) does not hold for all R > 0, then there is a smallest R = R0 such that φ0 (R0 ) = 0, and, clearly, φ00 (R0 ) ≤ 0, so that, by (6.8) 2n − 2 R2
≤ (Sn0 +1/2 − Sn0 −1/2 )(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) at R = R0 .
(6.9)
We write φ0 (R0 ) = 0 in the form Sn0 +1/2 − Sn0 −1/2 = S30 /2 − S10 /2
at R = R0 .
Then the right-hand side of (6.9) is equal to
(Sn0 +1/2 + Sn0 −1/2 )(Sn0 +1/2 − Sn0 −1/2 ) − (S30 /2 + S10 /2 )(S30 /2 − S10 /2 ) = [(Sn0 +1/2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )](S30 /2 − S10 /2 ) = [(Sn0 −1/2 + S30 /2 − S10 /2 + Sn0 −1/2 ) − (S30 /2 + S10 /2 )](S30 /2 − S10 /2 ) = 2(Sn0 −1/2 − S10 /2 )(S30 /2 − S10 /2 ). Recalling (6.5), the right-hand side of the above equality is <2 ·
n−1 R0
·
1 R0
, which contradicts to (6.9).
6.2. Proof of (ii) For fn−1 =
Pn−1 P0
Pn−1 (R) P0 (R) Pn−1 (R) P0 (R) Pn−1 (R) P0 (R)
=
< >
we just established fn0 −1 > 0, so that n+2
2n + 1
,
if R = Rn ,
(6.10)
,
if R < Rn ,
(6.11)
,
if R > Rn .
(6.12)
n+2
2n + 1 n+2
2n + 1
If for any n ≥ 2, we can establish Pn (R)
<
P0 (R)
n+3
2n + 3
,
R = Rn ,
(6.13)
then by combining this with (6.12) we conclude that Rn+1 > Rn , and this would complete the proof of Theorem (ii). Recalling (6.10), the inequality (6.13) can be written as follows: W (Rn )
< Pn−1 (Rn ),
(6.14)
where W (R) =
(2n + 3)(n + 2) Pn (R). (n + 3)(2n + 1)
(6.15)
It now suffices to prove that (6.14) holds. For n ≥ 5, this inequality is actually a consequence of the next lemma. Lemma 6.3. There holds W (R) < Pn−1 (R) for 0 < R ≤ Rn , n ≥ 5. Proof. By (2.14), W satisfies the differential equation W0 =
1 (2n + 3)(n + 2) R (n + 3)(2n + 1)
−
2n + 3 R
W−R
(n + 3)(2n + 1) 2 W , (n + 2)(2n + 3)
(6.16)
and by (2.10) and (6.15), W =
=
(2n + 3)(n + 2) 1 + O(R2 ) (n + 3)(2n + 1) 2n + 3 n+2
(n + 3)(2n + 1)
+ O(R2 ),
R → 0.
(6.17)
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It follows that W (R) < Pn−1 (R) for R near 0. If we can show that, for 0 < R ≤ Rn , 1 (2n + 3)(n + 2) R (n + 3)(2n + 1)
2n + 3
−
R
W−R
(n + 3)(2n + 1) 2 1 2n + 1 W < − W − RW 2 , (n + 2)(2n + 3) R R
(6.18)
then by comparison, we deduce that W (R) < Pn−1 (R) for all 0 < R ≤ Rn . Indeed, otherwise, there is a smallest R satisfies, 0 < R ≤ Rn , W (R) = Pn−1 (R), (W − Pn−1 )0 ≥ 0 at R, however, from (6.18) we have W 0 (R) < Pn0 −1 (R), which is a contradiction. To establish the inequality (6.18) and therefore the proof of (ii) of Theorem 6.1 (monotonicity of Rn ), we note that (6.18) is equivalent to 3
1
R (n + 3)(2n + 1)
3R
+
(n + 2)(2n + 3)
W2
<
2W R
,
0 < R ≤ Rn .
(6.19)
To complete the proof of Lemma 6.3 (and hence the proof of Theorem 6.1(ii)), it suffices to establish (6.19). The rest of this section is just devoted to this task. We shall first prove a sequence of auxiliary lemmas. Lemma 6.4. There holds, for all 0 < R < ∞, 3(n + 2) P0 (R) < W (R). (n + 3)(2n + 1)
(6.20)
˜ (R) = λP0 , λ > 0, to the function W . By (6.16), this means that λ has to be such Proof. We shall construct a subsolution W that
λP00 +
2n + 3 R
(n + 3)(2n + 1) 2 2 1 (2n + 3)(n + 2) Rλ P0 < , (n + 2)(2n + 3) R (n + 3)(2n + 1)
λP0 +
in view of (2.14) for n = 0, the above inequality is equivalent to 2n + 3 3λ 1 (2n + 3)(n + 2) 2 (n + 3)(2n + 1) 2
λ
If λ ≤
(n + 2)(2n + 3)
3(n+2) (n+3)(2n+1) ,
P0
2nλ R
<
− λ RP0 +
R
λ−
P0
R
<
R
(n + 3)(2n + 1)
−λ .
then the first term in the above inequality is negative, and this inequality is a consequence of
(2n + 3)(n + 2) −λ , R (n + 3)(2n + 1)
1
i.e. 2P0 <
(2n + 3)(n + 2) n(n + 3)(2n + 1)λ
−
1 n
.
This inequality is valid, since, by (2.16) and (2.13) and the fact that λ ≤ 2P0 (R) < 2P0 (0) =
2 3
≤
(2n + 3)(n + 2) n(n + 3)(2n + 1)λ
This completes the proof of Lemma 6.4.
−
1 n
3(n+2) (n+3)(2n+1) ,
.
Let G(R) =
I3/2 (R)
I1/2 (R)
.
(6.21)
Lemma 6.5. The function G(R) satisfies, for all 0 < R < ∞, 0 < G(R) < 1,
G0 (R)
> 0,
G00 (R)
< 0.
Proof. From (2.8) we find that G(R) = coth R − 1R , so that G0 (R) = G(R) = RP0 (R) ∼ G(R) = RP0 (R) ∼
1 3 1 R
R,
R → 0,
R = 1,
R → ∞.
sinh2 R−R2 R2 sinh2 R
> 0. It is easily seen that
H. Zhang et al. / Nonlinear Analysis 70 (2009) 2779–2795
It follows that 0 < G(R) < 1. Differentiating G0 (R) = 2(R3 cosh R − sinh R) 3
G00 (R) =
R3 sinh R
−
1 sinh2 R
once more we obtain
2 cosh R
=
3
1 R2
3
R3 sinh R
2 cosh R
:=
3
R3 sinh R
2791
R3 − (cosh R)−1 sinh R 3
g(R).
(6.22)
A direct computation shows that g(0) = g0 (0) = g00 (0) = 0 and 4
g000 (R) =
2
2
sinh R · {−14 cosh R + 6 sinh R} 4
cosh R
<0
so that g(R) < 0 for all 0 < R < ∞. This establishes that G00 (R) < 0. Vn−1 (R) , R
We write Pn−1 (R) =
In+1/2 (R)
Vn−1 (R) =
In−1/2 (R)
where
.
(6.23)
By (2.14), we have Vn0 −1 = 1 −
2n R
Vn−1 − Vn2−1 .
(6.24)
Introduce the positive solution V n−1 of 1=
2n R
2
V n−1 + V n−1 ,
(6.25)
i.e. V n−1 (R) = −
n R
+
" n 2 R
#1 2
+1
.
(6.26)
Note that, by (6.25) V n−1
R
<
2n
.
(6.27)
Differentiating (6.25), we obtain 0
2V n−1 V n−1 +
2n R
0
V n−1 =
2n R2
V n−1 ,
(6.28)
and hence upon using (6.27), we obtain 2n V R2 n−1
0
V n−1 =
2V n−1 +
2n
<
R
2n R R2 2n 2n R
=
1 2n
.
Lemma 6.6. There holds Vn−1 (R) ≥
2n
12
2n + 1
V n−1
12
2n 2n + 1
R ,
(6.29)
for all R > 0. Proof. The function H (R ) =
λV n−1 (λR),
λ > 0, ξ = λR
satisfies H0 +
2n R
H + H2 − 1 =
0
λ2 V n−1 +
0
2n R
= λ2 V n−1 +
< λ2
1 2n
2
λV n−1 + λ2 V n−1 − 1
2n
ξ
2 V n−1 + V n−1 − 1 + λ2 − 1
+ λ2 − 1 =
2n + 1 2n
λ2 − 1 ≤ 0 ,
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provided λ2 < H(R) ∼
2n , 2n+1
λ2
R
2n
so that H is a subsolution of (6.24). Since also, for R near 0
<
R
2n + 1
∼ Vn−1 (R).
We conclude that by comparison, that H(R) < Vn−1 (R) for all R > 0 and (6.29) follows by letting λ → 0
2n 2n+1
1/2
.
00
Lemma 6.7. The function V n−1 (R) satisfies V n−1 (R) > 0, V n−1 (R) < 0. Proof. The first inequality follows from (6.28). To prove the second inequality, we set V = V n−1 and differentiate (6.28) to obtain 2VV 00 + 2(V 0 )2 +
2n R
V 00 −
2n R2
V0 = −
4n R3
V+
2n R2
V0.
Since V > 0, we have 4n 4n sgnV 00 = sgn −2(V 0 )2 + 2 V 0 − 3 V R R 4n 0 V ≤ sgn 2 (V − ) R R 2 0 = sgn (−2VV ) by (6.28) R
< 0. This completes the proof of Lemma 6.7.
We introduce the function Φ (z, R) =
=
1
3
R (n + 3)(2n + 1)
3R
+
1
3
R
(n + 3)(2n + 1)
(n + 2)(2n + 3) +
z2 P02 −
3z2 G2
(n + 2)(2n + 3)
−
2 R
2 R
zP0
! zG
.
Lemma 6.8. For n ≥ 5, there holds Φ (z, R) < 0,
0 < R ≤ Rn , z ∈
3(n + 2)
,
n+2
(n + 3)(2n + 1) 2n + 1
.
(6.30)
Proof. Set z1 =
3(n + 2) , (n + 3)(2n + 1)
z2 =
n+2
2n + 1
.
Since the function z → Φ (z, R) is a parabola, it is sufficient to prove (6.30) just at the extreme points z1 and z2 , Φ (z1 , R) < 0 reduces to 1 +
9(n + 2)G2
<
2(n + 2)G
(n + 3)(2n + 1)(2n + 3) R 3 3(n + 2)G2 2(n + 2)G Φ (z2 , R) < 0 reduces to + < . n+3 (2n + 1)(2n + 3) R
,
Since 0 < G < 1, it suffices to prove 9(n + 2) 2(n + 2)G < , (n + 3)(2n + 1)(2n + 3) R 3 3(n + 2) 2(n + 2)G + < . n+3 (2n + 1)(2n + 3) R
1+
Noting that (6.32) is a consequence of (6.31), it remains to prove that 9(n + 2) M(R) := 2(n + 2)G(R) − 1 + R > 0. (n + 3)(2n + 1)(2n + 3)
(6.31) (6.32)
H. Zhang et al. / Nonlinear Analysis 70 (2009) 2779–2795
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Observe that for R near 0, 2(n + 2)R
M (R ) ∼
3
9(n + 2)
− 1+
(n + 3)(2n + 1)(2n + 3)
8n4 + 44n3 + 74n2 + 18n − 45
=
3(n + 3)(2n + 1)(2n + 3)
R
R,
so that M(0) = 0, M0 (0) > 0, and M00 (R) < 0 (since G00 < 0). Hence in order to prove that M(R) > 0 for all 0 < R ≤ Rn , it suffices to show that M(Rn ) > 0, i.e., 9(n + 2) Rn > 0. (6.33) 2(n + 2)G(Rn ) − 1 + (n + 3)(2n + 1)(2n + 3) Set Qn−1 (R) =
2n + 1
Kn−1 (R) =
2n + 1
Ln−1 (R) =
n+2
12
2n
2n + 1
V n−1
2n
12
2n + 1
Vn−1 (R), n+2 (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
2(n + 2)(n + 3)(2n + 1)(2n + 3)
R ,
R.
By (6.10), (6.21) and (6.23), we have Kn−1 (Rn ) = G(Rn ).
(6.34)
It is also clear that (6.33) is equivalent to Ln−1 (Rn )
< G(Rn ).
(6.35)
To finish the proof of this lemma, it suffices to establish (6.35). Since M(0) = 0 and M0 (0) > 0, we obtain G(0) = Ln−1 (0) = 0,
G0 (0)
> L0n−1 (0),
whereas, for R large, G(R) < 1 < Ln−1 (R). Since G is concave (G00 < 0), there exists a unique point R˜ n such that Ln−1 (R)
< G(R) if R < R˜ n ,
Ln−1 (R)
> G(R) if R > R˜ n .
(6.36)
To prove (6.35), let R∗n be the point where Ln−1 (R∗n ) = 1,
i.e., R∗n =
2(n + 2)(n + 3)(2n + 1)(2n + 3) . (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
(6.37)
Since G(R) < 1, (6.36) implies that R˜ n
< R∗n .
Suppose Qn−1 (R∗n )
> 1 = Ln−1 (R∗n ),
(6.38)
then from the concavity of Qn−1 (i.e., Qn00−1 < 0 by Lemma 6.7) and the fact that Qn−1 (0) = Ln−1 (0) = 0, it follows that Qn−1 (R)
> Ln−1 (R),
if R < R∗n .
(6.39)
By monotonicity of Qn−1 (i.e., Qn−1 > 0 by Lemma 6.7) 0
Qn−1 (R)
> Qn−1 (R∗n ) > 1,
if R > R∗n ,
whereas by (6.39) and (6.36), Qn−1 (R)
> Ln−1 (R) > G(R),
if R˜ n < R < R∗n ,
thus, altogether, Qn−1 (R) > G(R) if R˜ n < R < ∞. Since Kn−1 (R) > Qn−1 (R) for all R > 0 (by Lemma 6.6), we conclude that Kn−1 (R)
> G(R),
if R˜ n < R < ∞.
Hence, by (6.34), Rn < R˜ n and, recalling (6.36), the assertion (6.35) follows, and this completes the proof of (6.33).
H. Zhang et al. / Nonlinear Analysis 70 (2009) 2779–2795
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It remains to prove that (6.38) holds, that is, V n−1 (σn−1 )
> An−1 ,
(6.40)
where
σn−1 =
12
2n
2n + 1
R∗n ,
An−1 =
n+2
2n + 1
2n + 1 2n
12
.
(6.41)
Since, by (6.25), 2n
2
V n−1 (σn−1 ) +
σn−1
V n−1 (σn−1 ) = 1.
If A2n−1 +
2n
σn−1
An−1
< 1,
(6.42)
then (6.40) holds. Substituting R∗n from (6.37) into the formula for σn , we obtain
σn−1 =
2n 2n + 1
12
2(n + 2)(n + 3)(2n + 1)(2n + 3) , (n + 3)(2n + 1)(2n + 3) + 9(n + 2)
(6.43)
Substituting σn−1 from (6.43) and An from (6.41) into the left-hand side of (6.42), we find that (6.42) is equivalent to
−
2n4 + 3n3 − 35n2 − 81n − 36 2n(2n + 1)(n + 3)(2n + 3)
< 0,
which is valid if n ≥ 5. This completes the proof of Lemma 6.8.
Proof of (6.19) for the case n ≥ 5. From (6.17) we conclude that (6.19) is valid for small R. Let (0, R# ) be the largest interval with R# ≤ Rn on which (6.19) is valid. We want to claim that R# = Rn . If this is not the case, then R# < Rn , and, following the proof of Lemma 6.3 that we have W (R )
< Pn−1 (R) for 0 < R ≤ R# .
By (6.11) and (6.20), 3(n + 2) n+2 P0 (R) < W (R) < Pn−1 (R) < P0 (R) for 0 < R ≤ R# . (n + 3)(2n + 1) 2n + 1 It follows from Lemma 6.8 that (6.19) is also valid for 0 < R ≤ R# (with strict inequality at R = R# ). Thus the interval (0, R# ) can be further extended, and this contradicts the maximality of this interval. This establishes the estimate (6.19) for the case n ≥ 5. We have proved that Rn+1
> Rn for n ≥ 5.
The proof for n ≤ 4 can be obtained by explicit calculation. Indeed, solving Pn−1 (R) = 2nn++21 P0 (R) gives R = Rn , with R2 = 4.63512, R3 = 5.99911, R4 = 7.34806 and R5 = 8.68091. Clearly they are also monotonically increasing. To complete the bifurcation proof we shall now establish (5.9): Lemma 6.9. There holds d tanh R [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] >0 dR R R =R n
for n ≥ 2.
Proof. Since (2n + 1)Pn−1 (Rn ) − (n + 2)P0 (Rn ) = 0, we derive d tanh R d tanh R Pn−1 (R) [(2n + 1)Pn−1 (R) − (n + 2)P0 (R)] = P0 (R) (2n + 1) − (n + 2) dR R dR R P0 (R) R =R n R =R n tanh Rn d Pn−1 (R) = P0 (Rn )(2n + 1) . Rn dR P0 (R) R=Rn Now the lemma follows from (6.3).
(6.44)
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Acknowledgements The second author was partially supported by National Science Foundation of China (Grant No. 10671156) and the Program for New Century Excellent Talents in University (NCET-04-0968) References [1] A. Borisovich, A. Friedman, Symmetry-breaking bifurcations for free boundary problems, Indiana Univ. Math. J. 54 (3) (2005) 927–947. [2] M.G. Crandall, L.H. Rabinowitz, Bifurcation from simple eigenvalues, J. Funct. Anal. 8 (2) (1971) 321–340. [3] S. Cui, J. Escher, Bifurcation analysis of an elliptic free boundary problem modelling the growth of vascular tumors, SIAM J. Math. Anal. 39 (1) (2007) 210–235. [4] M. Fontelos, A. Friedman, Symmetry-breaking bifurcations of free boundary problems in three dimensions, Asymptotic Anal. 35 (3–4) (2003) 187–206. [5] M. Fontelos, A. Friedman, Symmetry-breaking bifurcations of charged drops, Arch. Ration. Mech. Anal. 172 (2) (2004) 267–294. [6] A. Friedman, A hierarchy of cancer models and their mathematical challenges, in: Mathematical Models in Cancer (Nashville, TN, 2002), Discrete Contin. Dyn. Syst. Ser. B 4 (1) (2004) 147–159. [7] A. Friedman, Symmetry-breaking bifurcations for free boundary problems, in: Nonlinear Partial Differential Equations and Related Analysis, in: Contemp. Math., vol. 371, Amer. Math. Soc, Providence, RI, 2005, pp. 153–161. [8] A. Friedman, Cancer models and their mathematical analysis, in: Tutorials in Mathematical Biosciences. III, in: Lecture Notes in Math., vol. 1872, Springer, Berlin, 2006, pp. 223–246. [9] A. Friedman, B. Hu, A Stefan problem for a protocell model, SIAM J. Math. Anal. 30 (4) (1999) 912–926. [10] A. Friedman, B. Hu, J.J.L. Velazquez, On the zeros of quotients of bessel functions, Chinese Ann. Math. 21B (3) (2000) 1–12. [11] A. Friedman, B. Hu, J.J.L. Velazquez, A Stefan problem for a protocell model with symmetry-breaking bifurcations of analytic solutions, Interfaces Free Bound. 3 (2) (2001) 143–199. [12] A. Friedman, B. Hu, Bifurcation from stability to instability for a free boundary problem arising in a tumor model, Arch. Ration. Mech. Anal. 180 (2) (2006) 293–330. [13] A. Friedman, B. Hu, Bifurcation for a free boundary problem modelling tumor growth by Stokes equation, SIAM J. Math. Anal. 39 (1) (2007) 174–194. [14] A. Friedman, F. Reitich, Symmetry-breaking bifurcation of analytic solutions to free boundary problems: An application to a model of tumor growth, Trans. Amer. Math. Soc. 353 (4) (2000) 1587–1634. [15] H. Schwegler, K. Tarumi, B. Gerstmann, Physico-chemical model of protocell, J. Math. Biol. 22 (3) (1985) 335–348. [16] K. Tarumi, H. Schwegler, A nonlinear treatment of the protocell by boundary layer approximation, Bull. Math. Biol. 49 (3) (1987) 307–320.