Asymptotics of a Free Boundary Problem^{∗}

F. V. Atkinson,^{†} H G. Kaper, Man Kam Kwong

Mathematics and Computer Science Division Argonne National Laboratory Argonne, IL 60439

May 18, 1995

Abstract

As was shown by Kaper and Kwong [Differential and Integral Equations 3, 353–362], there exists a unique R > 0, such that the differential equation

u′′ + \frac{2ν + 1}{r}u′ + u − u^{q} = 0, r > 0,

(0 ≤ q < 1, ν ≥ 0) admits a classical solution u, which is positive and monotone on (0, R) and which satisfies the boundary conditions

u′(0) = 0, u(R) = u′(R) = 0.

In this article it is shown that u(0) is bounded, but R grows beyond bounds as q → 1.

AMS(MOS) Subject Classification. Primary 34B15. Secondary 35J25, 35J65.

Key Words and Phrases. Reaction-diffusion equation, free boundary problem, radial solution, Emden-Fowler equation, positive solution, asymptotic behavior.

1 The Problem

In [1], the reaction-diffusion equation Δu + u^{1/2} − 1 = 0 was proposed as a simple model for Tokamak equilibria with magnetic islands. The equation motivated a study of free boundary problems for reaction-diffusion equations in R^{N} (N = 2, 3, …) of the general form Δu + u^{p} − u^{q} = 0 , where 0 ≤ q < p ≤ 1. In [2], we showed that there is a unique R (R > 0) and a unique positive-valued function u on (0, R), such that u is the radial solution of the differential equation which satisfies the boundary conditions u(R) = 0, u′(R) = 0. (A radial solution depends only on the radial variable r = |x|.) The solution (R, u) of the free boundary problem depends on the values of the exponents p and q.

∗This work was supported by the Mathematical, Information, and Computational Sciences Division subprogram of the Office of Computational and Technology Research, U. S. Department of Energy, under Contract W-31-109-Eng-38.

†Permanent address: Department of Mathematics, University of Toronto, Toronto, M5S 1A1, Ontario, Canada.

In this article we analyze the special case p = 1 in more detail and focus on the behavior of the solution as q → 1. That is, we are interested in the behavior as q → 1 (q < 1) of the pair (R, u), R a real number (R > 0), u a positive-valued function on (0, R), which satisfies the boundary value problem

u′′ + \frac{2ν + 1}{r}u′ + u − u^{q} = 0, 0 < r < R,  (1.1)
u′(0) = 0, u(R) = u′(R) = 0.  (1.2)

We consider ν as a real number, not necessarily half-integer (ν ≥ 0). The existence and uniqueness of such a solution follow from [2]. The function u is monotone on (0, R).

2 The Result

We prove the following result.

Theorem 1 For each q ∈ [0, 1), there is a unique R > 0 such that (1.1), (1.2) admits a (classical) solution u that is positive everywhere on (0, R). The function u is monotonically decreasing on (0, R); u(0) is bounded, but R grows beyond bounds as q → 1.

In the special case ν = \frac{1}{2} (N = 3) we have a lower bound on R,

R > \sqrt{\frac{2}{1 − q}} , 0 ≤ q < 1.  (2.1)

However, as we do not have a comparable upper bound, we cannot conclude that R = O((1 − q)^{−1/2}) as q → 1.

3 The Proof

Using a shooting argument, we replace the boundary value problem (1.1), (1.2) by the initial value problem

u′′ + \frac{2ν + 1}{r}u′ + u − u^{q} = 0, r > 0,  (3.1)
u(0) = γ, u′(0) = 0.  (3.2)

The results of [2] imply that, for any q ∈ [0, 1), there is a unique γ > 1 such that the solution of (3.1), (3.2) decreases from γ to meet the r-axis with zero slope at some value R > 0. Denoting this solution by u(·, γ), we have

u(R, γ) = 0, u′(R, γ) = 0.  (3.3)

The lower bound on γ can be sharpened to (2/(1 + q))^{1/(1−q)}, but 1 suffices for our purpose. The proof consists of a detailed investigation of the behavior of u(·, γ).

3.1 Down to 1

We begin by showing that u(r, γ) decreases monotonically from the value γ at r = 0 to the value 1 at some finite point r_{0}.

Lemma 1 There exists a point r_{0} < j_{ν,1}/(1 − q)^{1/2} such that u(·, γ) is monotonically decreasing on (0, r_{0}), with u(r_{0}, γ) = 1 and u′(r_{0}, γ) < 0. Here, j_{ν,1} is the first positive zero of J_{ν}—the Bessel function of the first kind of order ν.

Proof. As long as u > 1, we have u − u^{q} > (1 − q)u, so u(·, γ) oscillates faster than the solution v of the equation

v′′ + \frac{2ν + 1}{r}v′ + (1 − q)v = 0.  (3.4)

In particular, u(·, γ) reaches the value 1 before v does. Now, v(r) is a constant multiple of r^{−ν}J_{ν}(r(1 − q)^{1/2}), where J_{ν} is the Bessel function of the first kind of order ν—see, for example, [3]. Hence, v(r) = 1 for some value r < j_{ν,1}/(1 − q)^{1/2}, where j_{ν,1} is the first positive zero of J_{ν}. We conclude that there must be a point r_{0} < j_{ν,1}/(1 − q)^{1/2}, such that γ > u(r, γ) > 1 for 0 < r < r_{0} and u(r_{0}, γ) = 1.

Since u′(0, γ) = 0 and u′′(r, γ) < 0 near 0, it must be the case that u′(r, γ) < 0 near 0.

Suppose u(·, γ) were not monotone on (0, r_{0}). Then there exists a value r_{1} ∈ (0, r_{0}) where u(r, γ) has a local minimum, with u(r_{1}, γ) > 1. Because u(r, γ) reaches the value 1 at r_{0}, there must then exist a value r_{2} ∈ (r_{1}, r_{0}) such that u(r_{2}, γ) = u(r_{1}, γ) and u′(r_{2}, γ) ≤ 0. Multiplying the differential equation (3.1) by u′ and integrating over (r_{1}, r_{2}), we find that

\frac{1}{2}(u′(r_{2}, γ))^{2} = −(2ν + 1) ∫_{r_{1}}^{r_{2}} \frac{(u′(r, γ))^{2}}{r} dr.  (3.5)

But here we have a contradiction, as the two sides of this identity have opposite signs. It must therefore be the case that u(·, γ) is monotone on (0, r_{0}).

The monotonicity of u(·, γ) on (0, r_{0}) implies that u′(r_{0}, γ) ≤ 0. If u′(r_{0}, γ) = 0, then it follows from the Lipschitz continuity of the function u − u^{q} for u > 0 and the consequential uniqueness of the solution of the initial value problem for (3.1) in the direction of decreasing r starting at r = r_{0} that u(r, γ) = 1 for all r ∈ (0, r_{0}). But then we have a contradiction, as u(0, γ) = γ > 1. We conclude that u′(r_{0}, γ) < 0.

3.2 Beyond r_{0}

From Lemma 1 we know that u(r, γ) decreases monotonically until it reaches the value 1 with a negative slope at r = r_{0}. Beyond r_{0}, u(r, γ) decreases further until either it reaches the value 0 with a negative or zero slope, or it bottoms out at some finite value of r with a minimum value between 0 and 1.

Let r_{1} be the point where u(r, γ) ceases to be positive,

r_{1} = sup {r > r_{0} : u(s, γ) > 0, 0 < s < r}.  (3.6)

If r_{1} is finite and u(r_{1}, γ) = 0 , we do not consider u(·, γ) beyond r_{1}. In this case, we can use the same argument as in the proof of Lemma 1 to show that u(·, γ) is monotonically decreasing on the entire interval (0, r_{1}). In particular, if γ is such that not only u(r_{1}, γ) = 0, but also u′(r_{1}, γ) = 0, then u(·, γ) defines the (unique) solution u of the free boundary problem (3.1), (3.3), where R = r_{1}.

If r_{1} = ∞, then u(r, γ) has a positive minimum at some finite value of r, after which it oscillates with decreasing amplitude around the constant value 1.

Lemma 2 For 0 < r < r_{1}, we have 0 < u(r, γ) < γ.

Proof. The lemma is true for 0 < r ≤ r_{0} (cf. Lemma 1). Beyond r_{0}, we use a simple energy argument. The energy E of any solution u of (3.1), defined by the expression

E(r) = \frac{1}{2}(u′(r))^{2} + \frac{1}{2}(u(r))^{2} − \frac{1}{q + 1}(u(r))^{q+1},  (3.7)

is a monotonically decreasing function of its argument, as E′(r) = −((2ν +1)/r)(u′(r))^{2} ≤ 0 for all r ≥ 0.

Suppose the lemma were false for r_{0} < r < r_{1}. Then u(r_{2}, γ) = γ for some r_{2} ∈ (r_{0}, r_{1}), where E(r_{2}) ≥ γ^{2}/2 − γ^{q+1}/(q + 1) = E(0), and we have a contradiction.

Let w be defined in terms of u(·, γ) by the expression

w(r) = \frac{ru(r, γ)}{γ}.  (3.8)

This function is a solution of the initial value problem

w′′ + \frac{2ν − 1}{r}w′ + (1 − \frac{1}{(u(r, γ))^{1−q}} − \frac{2ν − 1}{r^{2}}) w = 0, r > 0,  (3.9)
w(0) = 0, w′(0) = 1.  (3.10)

It vanishes when u vanishes, while its derivative vanishes when both u and u′ vanish. Furthermore,

0 < w(r) < r, 0 < r < r_{1}.  (3.11)

The following lemma gives a lower bound for r_{1}.

Lemma 3 We have r_{1} > j_{ν,1}/δ, where

δ = \sqrt{1 − γ^{−(1−q)}}.  (3.12)

Proof. Because u(r, γ) < γ, w oscillates less than the solution v of the initial value problem

v′′ + \frac{2ν − 1}{r}w′ + (δ^{2} − \frac{2ν − 1}{r^{2}}) v = 0, r > 0; v(0) = 0, v′(0) = 1,  (3.13)

at least as long as v is positive. Since v(r) = 2^{ν}Γ(ν + 1)δ^{−ν}r^{1−ν}J_{ν}(δr), the first zero of v occurs at j_{ν,1}/δ. It must therefore be the case that r_{1} > j_{ν,1}/δ.

3.3 Bounds on (0, j_{ν,1}/δ)

We rewrite the equation (3.9) in the form

w′′ + \frac{2ν − 1}{r}w′ + (δ^{2} − \frac{2ν − 1}{r^{2}}) w = f(w),  (3.14)

where

f(w) = \frac{1}{γ^{1−q}} \{(\frac{r}{w} )^{1−q} − 1\} w.  (3.15)

Using the method of variation of parameters, we obtain the integral equation for w,

w(r) = rg(δr) + \frac{π}{2} ∫_{0}^{r} r^{1−ν}s^{ν} \{J_{ν}(δs)Y_{ν}(δr) − Y_{ν}(δs)J_{ν}(δr)\} f(w(s)) ds,  (3.16)

where

g(ρ) = 2^{ν}Γ(ν + 1)ρ^{−ν}J_{ν}(ρ).  (3.17)

J_{ν} and Y_{ν} are the Bessel functions of the first and second kind, respectively, of order ν. The expression (3.16) holds for all r ∈ [0, r_{1}) or, if r_{1} is finite, for all r ∈ [0, r_{1}]. We now restrict r to the interval [0, j_{ν,1}/δ].

Lemma 4 For 0 < r < j_{ν,1}/δ, we have

0 < rg(δr) < w(r) < r (g(δr) + \frac{ϕ(δr)}{log γ}) ,  (3.18)

where g is defined in (3.17) and

ϕ(ρ) = \frac{ρ^{2}(g(ρ))^{−1} log(g(ρ))^{−1}}{4(ν + 1)} .  (3.19)

Proof. Take any r ∈ (0, j_{ν,1}/δ). It follows from the Kneser-Sommerfeld expansion [3, Section 15.42] that

J_{ν}(δs)Y_{ν}(δr) − Y_{ν}(δs)J_{ν}(δr) = \frac{4δrJ_{ν}(δr)}{πJ_{ν}(δs)} ∑_{n=1}^{∞} \frac{(J_{ν}(j_{ν,n}s/r))^{2}}{(j_{ν,n}^{2} − (δr)^{2})j_{ν,n}J_{ν}^{′2}(j_{ν,n})},  (3.20)

for 0 ≤ s ≤ r. All the terms in the right member are positive, so the expression in the left member is positive. Furthermore, f(w(s)) is positive for 0 ≤ s ≤ r. Therefore, the integral in (3.16) is positive. Obviously, g(δr) is positive, so

w(r) > rg(δr) > 0, 0 < r < j_{ν,1}/δ.  (3.21)

It remains to establish the upper bound on w(r) in (3.18). From (3.21) and the fact that g is decreasing on (0, j_{ν,1}) we deduce that

\frac{s}{w(s)} < \frac{1}{g(δs)} ≤ \frac{1}{g(δr)}, 0 ≤ s ≤ r.  (3.22)

Therefore,

f(w(s)) < \frac{(g(δr))^{−(1−q)} − 1}{γ^{1−q}} w(s), 0 ≤ s ≤ r.  (3.23)

Furthermore, w(s) ≤ s, cf. (3.11), so

w(r) − rg(δr) = \frac{π}{2} ∫_{0}^{r} r^{1−ν}s^{ν} \{J_{ν}(δs)Y_{ν}(δr) − Y_{ν}(δs)J_{ν}(δr)\} f(w(s)) ds

≤ r \frac{(g(δr))^{−(1−q)} − 1}{γ^{1−q} − 1} [ρ^{−ν} \frac{π}{2} ∫_{0}^{ρ} \{J_{ν}(z)Y_{ν}(ρ) − Y_{ν}(z)J_{ν}(ρ)\} z^{ν+1} dz]_{ρ=δr} . (3.24)

The expression in square brackets can be evaluated by means of the recurrence formulae for Bessel functions [3, Section 3.2] and the resulting expression can be simplified further by means of the Wronskian [3, Section 3.63],

ρ^{−ν} \frac{π}{2} ∫_{0}^{ρ} \{J_{ν}(z)Y_{ν}(ρ) − Y_{ν}(z)J_{ν}(ρ)\} z^{ν+1} dz = 1 − g(ρ).  (3.25)

We estimate this expression by substituting the series expansion for the Bessel function J_{ν} and truncating after the first term,

1 − g(ρ) = 1 − 2^{ν}Γ(ν + 1)ρ^{−ν}J_{ν}(ρ) ≤ \frac{ρ^{2}}{4(ν + 1)}.  (3.26)

Thus,

[ρ^{−ν} \frac{π}{2} ∫_{0}^{ρ} \{J_{ν}(z)Y_{ν}(ρ) − Y_{ν}(z)J_{ν}(ρ)\} z^{ν+1} dz]_{ρ=δr} ≤ \frac{(δr)^{2}}{4(ν + 1)}.  (3.27)

To estimate the factor in front of the square brackets in (3.24), we observe that 0 < g(δr) < 1 on (0, j_{ν,1}/δ) and γ > 1. Furthermore, one readily verifies that

\frac{1 − x^{1−q}}{log x^{−1}} ≤ \frac{y^{1−q} − 1}{log y}

for any pair (x, y) with 0 < x ≤ 1 ≤ y. Therefore,

\frac{(g(δr))^{−(1−q)} − 1}{γ^{1−q} − 1} = (g(δr))^{−(1−q)} \frac{1 − (g(δr))^{1−q}}{γ^{1−q} − 1}

≤ \frac{(g(δr))^{−(1−q)} log(g(δr))^{−1}}{log γ} ≤ \frac{(g(δr))^{−1} log(g(δr))^{−1}}{log γ} .  (3.28)

Using (3.27) and (3.28) in (3.24), we obtain the estimate

w(r) − rg(δr) ≤ r \frac{ϕ(δr)}{log γ},  (3.29)

where ϕ is defined in (3.19). The upper bound for w(r) given in (3.18) follows.

In terms of u, we have the following bounds:

0 < γg(δr) < u(r, γ) < γ [g(δr) + \frac{ϕ(δr)}{log γ}] , 0 < r < \frac{j_{ν,1}}{δ} .  (3.30)

Because ϕ(ρ) increases beyond bounds as g(ρ) decreases to 0, the upper bound in (3.18) or (3.30) increases indefinitely as r approaches the right endpoint of the interval (0, j_{ν,1}/δ).

In the following analysis we also need an estimate of the quantity r^{1−2ν}(r^{2ν−1}w)′(r). It is given by the expression

r^{1−2ν}(r^{2ν−1}w)′(r) = h(δr) + δ \frac{π}{2} ∫_{0}^{r} r^{1−ν}s^{ν} \{J_{ν}(δs)Y_{ν−1}(δr) − Y_{ν}(δs)J_{ν−1}(δr)\} f(w(s)) ds,

(3.31)

where

h(ρ) = 2^{ν}Γ(ν + 1)ρ^{1−ν}J_{ν−1}(ρ).  (3.32)

Like (3.16), (3.31) holds for all r ∈ [0, r_{1}) or, if r_{1} is finite, for all r ∈ [0, r_{1}]. The following lemma gives an estimate on (0, j_{ν,1}/δ).

Lemma 5 For 0 < r < j_{ν,1}/δ, we have

|r^{1−2ν}(r^{2ν−1}w)′(r) − h(δr)| < 2(ν + 1)\frac{ϕ(δr)}{log γ},  (3.33)

where h is defined in (3.32) and ϕ is defined in (3.19).

Proof. The proof is similar to, although slightly more involved than, the proof of Lemma 4. Instead of (3.16), we use (3.31). The analog of (3.24) is

δ \frac{π}{2} ∫_{0}^{r} r^{1−ν}s^{ν} \{J_{ν}(δs)Y_{ν−1}(δr) − Y_{ν}(δs)J_{ν−1}(δr)\} f(w(s)) ds

≤ \frac{(g(δr))^{−(1−q)} − 1}{γ^{1−q} − 1} [ρ^{1−ν} \frac{π}{2} ∫_{0}^{ρ} \{J_{ν}(z)Y_{ν−1}(ρ) − Y_{ν}(z)J_{ν−1}(ρ)\} z^{ν+1} dz]_{ρ=δr} .(3.34)

The expression in square brackets can again be evaluated; instead of (3.25) we have

ρ^{1−ν} \frac{π}{2} ∫_{0}^{ρ} \{J_{ν}(z)Y_{ν−1}(ρ) − Y_{ν}(z)J_{ν−1}(ρ)\} z^{ν+1} dz = 2(ν + 1) − h(ρ),  (3.35)

where

2(ν + 1) − h(ρ) = 2(ν + 1) − 2^{ν}Γ(ν + 1)ρ^{1−ν}J_{ν−1}(ρ) ≤ \frac{1}{2}ρ^{2}.  (3.36)

The lemma follows from (3.31), (3.34), (3.35), (3.36), and (3.27).

3.4 Estimates at r_{0}

We use the results of Lemmas 4 and 5 to estimate r_{0} and r^{1−2ν}(r^{2ν−1}w)′ at r_{0}.

Lemma 6 Let a ∈ (j_{ν−1,1}, j_{ν,1}) be fixed. Then there exists a constant γ_{1} > 1 that does not depend on q, such that

r_{0}^{1−2ν}(r^{2ν−1}w)′(r_{0}) < −\frac{1}{2}|h(a)|  (3.37)

and

\frac{a}{δ} < r_{0} < (1 + \frac{4ν}{|h(a)|})^{1/(2ν)} \frac{a}{δ},  (3.38)

for all γ ≥ γ_{1}.

Proof. With the choice of a indicated in the statement of the lemma, we have g(a) > 0 and h(a) < 0. These inequalities follow from the interlacing property of the zeros of Bessel functions,

0 < j_{ν,1} < j_{ν+1,1} < j_{ν,2} < j_{ν+1,2} < j_{ν,3} < …;

cf. [3, Section 15.22].

We begin by observing that w oscillates less than v, where v(r) = rg(δr), g defined by (3.17). Therefore r_{0}, which is defined by the identity w(r) = r/γ, is certainly beyond the point r_{2}, where g(δr_{2}) = 1/γ. Therefore, if

γ_{0} = 1/g(a),  (3.39)

then g(δr_{2}) ≤ g(a) for all γ ≥ γ_{0}. Now, g is monotonically decreasing between a and j_{ν−1,1}, so then we also have δr_{2} ≥ a for all γ ≥ γ_{0}. Since r_{0} > r_{2}, we have thus achieved that

a/δ < r_{0}.  (3.40)

for all γ ≥ γ_{0}.

With r_{3} = a/δ, it follows from (3.33) that

r_{3}^{1−2ν}(r^{2ν−1}w)′(r_{3}) < −|h(a)| + 2(ν + 1)\frac{ϕ(a)}{log γ} .  (3.41)

Here, h(a) and ϕ(a) do not depend on q or γ. Therefore, if we now define γ_{1},

γ_{1} = min\{γ_{0}, exp (2(ν + 1)\frac{ϕ(a)}{|h(a)|})\},  (3.42)

then γ_{1} is independent of q and

r_{3}^{1−2ν}(r^{2ν−1}w)′(r_{3}) < −\frac{1}{2}|h(a)|  (3.43)

for all γ ≥ γ_{1}. Writing the differential equation (3.9) in the form

(r^{1−2ν} (r^{2ν−1}w)′)′ = − (1 − u^{−(1−q)}) ,  (3.44)

we observe that the function r^{1−2ν}(r^{2ν−1}w)′ is decreasing as long as u(r, γ) > 1—that is, up to r_{0}. Therefore, the bound (3.43) extends to the entire interval [r_{3}, r_{0}], and we have

r^{1−2ν}(r^{2ν−1}w)′(r) < −\frac{1}{2}|h(a)|, r_{3} ≤ r ≤ r_{0}.  (3.45)

for all γ ≥ γ_{1}. In particular, the inequality holds at r_{0}, as asserted in (3.37).

Multiplying both sides of the inequality (3.45) by r^{2ν−1} and integrating over the interval (r_{3}, r_{0}), we find

(\frac{w(r_{3})}{r_{3}} + \frac{|h(a)|}{4ν}) r_{3}^{2ν} − \frac{|h(a)|}{4ν}r_{0}^{2ν} > \frac{w(r_{0})}{r_{0}}r_{0}^{2ν}.  (3.46)

Here, we estimate the expression in the right member from below by 0. In the left member, we estimate the ratio w(r_{3})/r_{3} from above by 1; cf. (3.11). Thus,

r_{0}^{2ν} < (1 + \frac{4ν}{|h(a)|}) r_{3}^{2ν}.  (3.47)

The inequalities (3.38) now follow from (3.40) and (3.47).

3.5 Down to 0

We are now in a position to prove that the continuation of u beyond r_{0} decreases to 0 for all sufficiently large γ, independently of q.

Lemma 7 There exists a constant γ_{2} that does not depend on q (γ_{2} ≥ γ_{1}, where γ_{1} is the constant introduced in Lemma 6), such that r_{1} < ∞ for all γ ≥ γ_{2}.

Proof. The proof is by contradiction, where we assume that, for some γ ≥ γ_{1}, the solution u(·, γ) of (3.1), (3.2) is positive for all r ≥ 0.

Consider the function w defined by (3.8). By assumption, w is positive for all r > 0. Because (r^{1−2ν}(r^{2ν−1}w)′)′ = (r/γ)(u^{q} − u) and u^{q} − u < 1 − q for u > 0, we have

(r^{1−2ν}(r^{2ν−1}w)′)′(r) < \frac{(1 − q)r}{γ} , r > 0.  (3.48)

Integrating (3.48) from r_{0} to any point r > r_{0}, and using the estimate (3.37) at r_{0}, we find

r^{1−2ν}(r^{2ν−1}w)′(r) < −\frac{1}{2}|h(a)| + \frac{(1 − q)r^{2}}{2γ} , r > r_{0},  (3.49)

for all γ ≥ γ_{1}. Because γδ^{2} = γ − γ^{q} > γ^{1−q} − 1 > (1 − q) log γ, it follows that

r^{1−2ν}(r^{2ν−1}w)′(r) < −\frac{1}{2}|h(a)| + \frac{r^{2}δ^{2}}{2 log γ} , r > r_{0},  (3.50)

for all γ ≥ γ_{1}.

Now we restrict r to a compact interval [r_{0}, r_{2}], where

r_{2} = b/δ,  (3.51)

and b > a is a suitably chosen constant. Defining the constant γ_{2} by

γ_{2} = min{γ_{1}, e^{2b^{2}/|h(a)|}},  (3.52)

we then have

\frac{r^{2}δ^{2}}{2 log γ} ≤ \frac{1}{4}|h(a)|, r_{0} ≤ r ≤ r_{2},  (3.53)

for all γ ≥ γ_{2}, so (3.50) reduces to

r^{1−2ν}(r^{2ν−1}w)′(r) < −\frac{1}{4}|h(a)|, r_{0} ≤ r ≤ r_{2},  (3.54)

for all γ ≥ γ_{2}. Hence,

w(r_{2}) < [(a/b)^{2ν} \frac{w(r_{0})}{r_{0}} − (1 − (a/b)^{2ν})\frac{|h(a)|}{8ν}] r_{2}.  (3.55)

Using (3.38) to estimate w(r_{0})/r_{0} and writing the inequality in terms of u, we thus find that

u(r_{2}, γ) < (a/b)^{2ν} (1 + \frac{4ν}{|h(a)|})^{1/(2ν)} − γ (1 − (a/b)^{2ν}) \frac{|h(a)|}{8ν},  (3.56)

for all γ ≥ γ_{2}.

But now we have a contradiction, as the expression in the right member of this inequality certainly becomes negative for sufficiently large values of γ. We conclude therefore that u(·, γ) reaches the value 0 at some finite point r_{1}, as claimed.

3.6 Completion of the Proof

According to Lemma 7, u(·, γ) ceases to be positive at a finite point r_{1} for all γ ≥ γ_{2}, where γ_{2} is a constant that does not depend on q. Obviously, r_{1} depends on the value of γ; in fact, it decreases as γ increases. Let

Γ = inf{γ > 1 : r_{1} < ∞}.  (3.57)

If γ = u(0) = Γ, then u(·, γ) reaches the r-axis with a horizontal slope, so u(·, Γ) defines the unique solution u of the free boundary problem (1.1), (1.2), where

R = r_{1}(Γ).  (3.58)

Obviously, Γ depends on q. However, it follows from Lemma 7 that 1 < Γ ≤ γ_{2}, so u(0) is bounded as q → 1 (q < 1).

It remains to investigate the behavior of R as q → 1 (q < 1). Because Γ is bounded, lim_{q→1} Γ^{1−q} = 1. Then it follows from (3.12) that lim_{q→1} δ = 0, and therefore, by Lemma 3, lim_{q→1} R = ∞. Thus, the proof of the theorem is complete.

3.7 Special Case: N = 3

In the special case N = 3 (ν = \frac{1}{2}), it is actually possible to find a lower bound for R that shows that R grows beyond bounds as q → 1.

A simple energy argument gives the inequality

0 = E(R) < E(0) = \frac{Γ^{2}}{2} − \frac{Γ^{1+q}}{1 + q};  (3.59)

cf. (3.7). Hence,

Γ^{1−q} > \frac{2}{1 + q} .  (3.60)

Next, we use an energy argument for (3.9). If ν = \frac{1}{2}, this equation reduces to

w′′ + w − Γ^{−(1−q)}r^{1−q}w^{q} = 0.  (3.61)

Hence,

(w^{′2} + w^{2} − \frac{2}{1 + q} Γ^{−(1−q)}r^{1−q}w^{1+q})′ = −2\frac{1 − q}{1 + q} Γ^{−(1−q)}r^{−q}w^{1+q}.  (3.62)

Upon integration over (0, R), the left member yields −1; in the right member we use the inequality w(r) < r to obtain the estimate

∫_{0}^{R} r^{−q}w^{1+q} dr < \frac{1}{2}R^{2}.  (3.63)

Thus, using (3.60), we find that

R > \sqrt{\frac{2}{1 − q}} .  (3.64)

References

[1] G. Miller, V. Faber, A. B. White, Jr., Finding plasma equilibria with magnetic islands, J. Computational Physics 79 (1988), 417–435.

[2] H. G. Kaper and M. K. Kwong, Free Boundary Problems for Emden–Fowler Equations, Differential and Integral Equations 3 (1990), 353–362.

[3] G. N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. Cambridge University Press (1966).