On the radially symmetric solutions of a BVP for a class of nonlinear elliptic partial differential equations

JENŐ HEGEDŰS

(JATE, Bolyai Institute, Szeged, Hungary) hegedusj@math.u-szeged.hu

Abstract. Uniqueness and comparison theorems are proved for the BVP of the form

Δu(x) + g(x, u(x), |∇u(x)|) = 0, x ∈ B, u|_{Γ} = a ∈ ℝ (Γ := ∂B),

where B is the unit ball in ℝ^{n} centered at the origin (n ≥ 2). We investigate radially symmetric solutions, their dependence on the parameter a ∈ ℝ, and their concavity.

AMS subject classifications: 35B30, 35J60, 35J65.

Key words: radially symmetric solution, nonlinear elliptic partial differential equations, uniqueness, monotone dependence on the boundary value, concavity.

Introduction Radially symmetric solutions to Dirichlet problem for the nonlinearly perturbed

Laplace operator are investigated by many authors, see e.g. [1]-[4].

In [1] it is proven for a wide class of perturbations that the smooth positive solutions of the homogeneous Dirichlet problem in a ball are necessarily radially symmetric. The perturbation of the Laplacian in the paper [2], is f(u) with a locally Lipschitz function f; a BVP with a condition at infinity is considered, reduced to an ODE problem and sufficient conditions are given that guarantee the solvability of the original problem. In the papers [3], [4] nonlinear ODE-BVP-s (partly related to perturbed Laplacian) are considered on the intervals (a, b) and (0, 1) respectively; the term y′′ is perturbed with the sum

g(x, y′) + f(x, y), \frac{n − 1}{x}y′ + f(x, y)

respectively (where g is locally Lipschitz), and sufficient conditions (certain additional restrictions on f and g) of the existence and uniqueness of the positive solution y are presented. These cases do not cover the general case of perturbations y′′ of the form

\frac{n − 1}{x}y′ + f(x, y, y′) x ∈ (0, R), 0 < R < ∞,

i.e. the case of the perturbations Δu with f(|x|, u, ±|∇u|).

We remark that fundamental results for the investigations in [1] were already given in [5]. The contribution of the author of [5] to the theory of radially symmetric solutions of nonlinear elliptic PDE-s (mainly on the whole space ℝ^{n} and more generally for the m-Laplacian) can be found in [6].

This paper is in final form and no version of it will be submitted for publication elsewhere.

Recently G. Bognár [7] considered the following BVP in the unit ball B := {x ∈ ℝ^{n}|ρ ≡ |x| < 1} (Γ := ∂B) :

(A_{1}) Δu(x) + exp(λu(x) + κ|∇u(x)|) = 0 x ∈ B; κ, λ ≤ 0 are constants,

(A_{2}) u ∈ C^{2}(B) ∩ C(\overline{B}), u(x) = v(|x|) ≡ v(ρ),

(A_{3}) u|_{Γ} = a = const. a ≥ 0.

Existence and uniqueness results were established by the author and it was shown that the solution u depends monotonically on the parameter a.

The purpose of the present paper is to prove uniqueness, monotonicity, and concavity results for the solutions of the more general BVP: Problem 1:

(1.1)  Δu(x) + f(|x|, u(x), |∇u(x)|) = 0 x ∈ B,

(1.2)  u ∈ C^{2}(B) ∩ C(\overline{B}), ∃v : [0, 1] → ℝ : v(ρ) ≡ v(|x|) = u(x) ∀x ∈ \overline{B},

(1.3)  u|_{Γ} = a ∈ ℝ.

Here f ∈ C(G_{a}; (0, ∞)), a ∈ ℝ is arbitrarily fixed; G_{a} := [0, 1] × [a, ∞) × [0, ∞), B is the unit ball centered at the origin, and ρ := |x| x ∈ \overline{B}.

The method used here is, partly, a modification of that of [7]. Some results are proved without using radial symmetricity. These proofs are based upon the techniques communicated in [9].

To prepare our general results we formulate some of them in simplified versions: Theorem A. If f ∈ C(G_{a}; (0, ∞)) and f(ρ, t, β) is strongly decreasing in t ∈ [a, ∞),

then Problem 1 may have no more than one solution.

Theorem B. If f ∈ C(G_{a}; (0, ∞)) and f(ρ, t_{1}, t_{2}) is nonincreasing both in t_{1} ∈ [a, ∞), and t_{2} ∈ [0, ∞), then for the (radially symmetric) solutions u_{1} and u_{2} of Problem 1 with the property:

u_{1}|_{Γ} ≡ a_{1} > u_{2}|_{Γ} ≡ a_{2} ≥ a

inequalities

v_{1}(ρ) ≡ u_{1}(|x|) ≥ v_{2}(ρ) ≡ u_{2}(|x|) x ∈ \overline{B}, v′_{1}(ρ) ≥ v′_{2}(ρ) ρ ∈ [0, 1)

hold.

Finally a concavity result:

Theorem C. Let f ∈ C(G_{a}; (0, ∞)), and let f(ρ, t_{1}, t_{2}) be nonincreasing both in t_{1}, ∈ [a, ∞), and t_{2} ∈ [0, ∞). Then there exists a constant K(a) such, that 0 < f ≤ K(a) < ∞, and any of the assumptions (C_{1}), (C_{2})

(C_{1})  f(t, a + \frac{K(a)}{n} \frac{1 − t^{2}}{2} , \frac{K(a)}{n}t) ≥ K(a)(1 − \frac{1}{n}) t ∈ [0, 1],

(C_{2})  f(t, a + \frac{K(a)}{2n}, \frac{K(a)}{n}) ≥ K(a)(1 − \frac{1}{n}) t ∈ [0, 1]

guarantees the concavity of the solution of Problem 1. For the case

f(ρ, u, |∇u|) ≡ exp (λu + κ|∇u|) λ, κ ≤ 0

considered in [7], the assumption (C_{1}) turnes into (C_{3}):

(C_{3})  exp \{λ[a + \frac{e^{λa}}{n} \frac{1 − t^{2}}{2} ] + κ\frac{e^{λa}}{n} t\} ≥ e^{λa}(1 − \frac{1}{n}) t ∈ [0, 1].

One of the simplest sufficient conditions for the concavity of u for this special case is

(C_{4})  κ ≤ λ(≤ 0), −1 ≤ κe^{λa}.

1. Uniqueness results.

We shall prove (under the corresponding conditions) two theorems on the uniqueness of solution of Problem 1. The first one will be a consequence of a classical, simple uniqueness theorem related to the problem more general than Problem 1.

Theorem 1. Let w(t) := f(α, t, β) t ∈ [a, ∞) for every α ∈ [0, 1], β ∈ [0, ∞) fixed be strongly decreasing in t on the interval [a, ∞); then Problem 1 has no more than one solution.

Instead of a direct proof consider Problem 2 (mentioned above) in an arbitrary bounded domain Ω of ℝ^{n} with the boundary Γ := ∂Ω :

Problem 2.

(4)  u ∈ C^{2}(Ω) ∩ C(\overline{Ω})

(5)  (Δu)(x) + g(x, u(x), u_{x_{1}}(x), …, u_{x_{n}}(x)) = 0 x ∈ Ω,

(6)  u|_{Γ} = φ ∈ C(Γ),

where g ∈ C(\overline{Ω} × ℝ^{n+1}).

Theorem 2. If w(t) := g(\underline{α}, t, \underline{β}) t ∈ ℝ is strongly decreasing in t for any

\underline{α} ∈ Ω, \underline{β} ∈ ℝ^{n} fixed, then Problem 2 has no more than one solution.

This theorem is very close to the Theorem 9.3 (p.208) of the book [8].

Proof. Suppose that there exist two different solutions of Problem 2: u_{1} and u_{2}. Define u := u_{1} − u_{2} and suppose that there exists a point y ∈ Ω such, that u(y) ≠ 0. Without loss of the generality it may be supposed that u(y) < 0. Letting

m := min_{x∈\overline{B}} u(x)

we see, that m < 0, and there exists a point x_{0} ∈ Ω of global minimum of the function u(x) x ∈ \overline{Ω} i.e. ∃ x_{0} ∈ Ω such that

0 > m = u(x_{0}) ≤ u(x) ∀x ∈ \overline{Ω}.

Consequently we have

(7)  (Δu)(x_{0}) ≥ 0, u_{x_{i}}(x_{0}) = 0 i = \overline{1, n}.

On the other hand we know, that

(8)  (Δu_{1})(x_{0}) + g(x_{0}, u_{1}(x_{0}), (grad u_{1})(x_{0})) = 0,

(9)  (Δu_{2})(x_{0}) + g(x_{0}, u_{2}(x_{0}), (grad u_{2})(x_{0})) = 0,

therefore subtracting (9) from (8) we have

(10) (Δu)(x_{0}) = g(x_{0}, u_{2}(x_{0}), (grad u_{2})(x_{0})) − g(x_{0}, u_{1}(x_{0}), (grad u_{1})(x_{0})).

Here arguments (grad u_{2})(x_{0}), (gradu_{1})(x_{0}) are common in virtue of equalities u_{x_{i}}(x_{0}) = 0 i = \overline{1, n} (see (7)), therefore using the relations

0 > u(x_{0}) = m ≡ u_{1}(x_{0}) − u_{2}(x_{0})

and their consequence u_{2}(x_{0}) > u_{1}(x_{0}); from the monotonicity condition on w(t) we get

f(x_{0}, u_{2}(x_{0}), (grad u_{2})(x_{0})) − f(x_{0}, u_{1}(x_{0}), (grad u_{1})(x_{0})) < 0.

So, in (10) we have

(11)  (Δu) (x_{0}) < 0,

that contradicts inequality of (7). Theorem 2 is proved.

Remark 1. For the proof of Theorem 1 it is enough to apply Theorem 2 for the case Ω := B with the nonlinear part g (appearing in Problem 2) defined by the formula

g(x, u, u_{x_{1}}, …, u_{x_{n}}) := f (|x|, u, (∑_{i=1}^{n} u_{x_{i}}^{2})^{1/2}) x ∈ \overline{B},

where f is the nonlinearity appearing in Problem 1.

Next we explain another result on the uniqueness of the solution to the Problem 1 without assumption on strong decrease of f(α, t, β) in t. However we need that f(α, t_{1}, t_{2}) is nonincreasing both in t_{1} and t_{2}. Here in the proof we will use the radial symmetricity of the solutions.

Theorem 3. Let function f appearing in differential equation of Problem 1 satisfy conditions:

(i) w(t) := f(α, t, β) is nonincreasing in t ∈ [a, ∞) for every fixed α, β(α ∈ [0, 1], β ∈ [0, ∞)), and

(ii) w̃(t) := f(α, β, t) is nonincreasing in t ∈ [0, ∞) for every fixed α, β (α ∈ [0, 1], β ∈ [a, ∞)).

Then Problem 1 has no more, than one solution.

Proof. Suppose, that there exist two different solutions: u_{1}, u_{2}(u_{1}(x) = v_{1}(|x|), u_{2}(x) = v_{2}(|x|)x ∈ \overline{B}) of Problem 1 with the same boundary value a ∈ ℝ. We introduce the notation

v(ρ) := v_{1}(ρ) − v_{2}(ρ) ρ ∈ [0, 1].

From the assumption, thatf > 0 and u_{1}, u_{2} are solutions of Problem 1 (especially they are superharmonic and radially symmetric in B) easily follows that

(12)  \begin{matrix}v ∈ C^{2}([0, 1)) ∩ C([0, 1]), v(1) = 0, v′(0) = 0, \\ Δu_{i}(x) + f(|x|, u_{i}(x), |∇u_{i}(x)|) = \\ = v′′_{i}(ρ) + \frac{n − 1}{ρ}v′_{i}(ρ) + f(ρ, v_{i}(ρ), −v′_{i}(ρ)) = 0 x ∈ B, ρ ∈ (0, 1) i = 1, 2, \end{matrix}

and the multiplied by ρ^{n−1} version of the last equality of(12) holds:

(13)  (ρ^{n−1}v′_{i}(ρ))′ + ρ^{n−1}f(ρ, v_{i}(ρ), −v′_{i}(ρ)) = 0 ρ ∈ [0, 1), i = 1, 2.

It can be supposed – without loss of the generality – that there exists a point a_{1} ∈ [0, 1] such, that v(a_{1}) > 0. Using the continuity of v on [0, 1] it is trivial, that the interval (0, 1) also contains a point a_{1} such, that v(a_{1}) > 0. Let us fix such a point a_{1} for the sequel. Our aim is to construct an interval [α, β] ⊆ [0, 1] such that

v(ρ) > 0 ρ ∈ [α, β], v′(ρ) < 0 ρ ∈ (α, β], v′(α) = 0.

Let be

b := sup{ρ ∈ [0, 1)| v(ρ) > 0}

It is clear, that

b ∈ (0, 1], v(b) = 0,

and that

a_{1} ∈ (0, b).

Further let

d := inf{ρ ∈ (a_{1}, b]|v(ρ) = 0}.

It is clear, that

v(d) = 0.

Then let be

(14)  c := 0 if v(ρ) > 0 ρ ∈ [0, a_{1}],

and

(15)  c := sup{ρ ∈ [0, a_{1})|v(ρ) = 0} otherwise.

In the case of (2.43)

v(c) = 0

holds automatically. Further, denoting by M the maximum of the function v on [c, d] (M > 0) let us introduce

e := sup{ρ ∈ [c, d]|v(ρ) = M}.

We remark that for the case of (14) e ∈ [c, d] ≡ [0, d] and

(16)  v′(e) ≡ v′(0) = 0

in virtue of (12) if e = c = 0; and v′(e) = 0 if e ∈ (c, d) ≡ (0, d) using the fact that v(e) = M i.e. e is a point of interior global maximum of the function v on the interval [c, d]. In the case of (15)

(17)  e ∈ (c, d), v(e) = M, v′(e) = 0

hold automatically because e is an interior point of global maximum of v on [c, d].

The assumption v′(ρ) ≥ 0 ρ ∈ [e, d) leads to contradiction in both cases (14) and (15), because if d_{1} < d, and d_{1} → d then

v(e) + ∫_{e}^{d_{1}} v′(ρ) dρ → v(d) = 0

and

v(e) + ∫_{e}^{d_{1}} v′(ρ) dρ ≡ M + ∫_{e}^{d_{1}} v′(ρ) dρ ≥ M > 0 (∀d_{1} ∈ (e, d)).

Consequently there exists a point β ∈ (e, d) such, that v′(β) < 0. Fixing such a point, it is easy to show – using (16) and continuity of v′ on [0, 1) – that there exists an interval [α, β] ⊆ [c, d] such, that

(18)  v′(ρ) < 0 ρ ∈ (α, β], v′(α) = 0, v(ρ) > 0 ρ ∈ [α, β].

Namely, for the both of the cases (14) and (15) α may be choosen as

(19)  α := sup{ρ ∈ [e, β)|v′(ρ) = 0} ≡ sup M

because the set M is non empty (e ∈ M), and using the property v′ ∈ C[0, 1) (see (12))

(20)  α ∈ [e, β), v′(α) = 0.

The next step of the proof is the using of the validity of differential equation of Problem 1 for v_{1}, v_{2} on the interval I ≡ (α, β) choiced above:

(ρ^{n−1}v′_{1}(ρ))′ + ρ^{n−1}f(ρ, v_{1}(ρ), −v′_{1}(ρ)) = 0,
(ρ^{n−1}v′_{2}(ρ))′ + ρ^{n−1}f(ρ, v_{2}(ρ), −v′_{2}(ρ)) = 0,

from which after subtracting we get

(ρ^{n−1}v′(ρ))′ + ρ^{n−1}[f(ρ, v_{1}, −v′_{1}) − f(ρ, v_{2}, −v′_{2})] = 0

i.e.

(ρ^{n−1}v′(ρ))′ = ρ^{n−1}[f(ρ, v_{2}, −v′_{2}) − f(ρ, v_{1}, −v′_{1})].

Subtracting and adding in the brackets of the right hand side the term

f(ρ, v_{1}(ρ), −v′_{2}(ρ))

we get

(21)  (ρ^{n−1}v′(ρ))′ = δ_{1}(ρ) + δ_{2}(ρ) ≡ δ(ρ) ρ ∈ [α, β],

where

δ_{1}(ρ) := ρ^{n−1}[f(ρ, v_{2}(ρ), −v′_{2}(ρ)) − f(ρ, v_{1}(ρ), −v′_{2}(ρ)] ρ ∈ [α, β],
δ_{2}(ρ) := ρ^{n−1}[f(ρ, v_{1}(ρ), −v′_{2}(ρ)) − f(ρ, v_{1}(ρ), −v′_{1}(ρ)] ρ ∈ [α, β].

Of course δ_{i} ∈ C[α, β] i = 1, 2. Moreover, taking into account the choice of the interval [α, β] we have the relations

(22) v(ρ) ≡ v_{1}(ρ) − v_{2}(ρ) > 0 ρ ∈ [α, β], v′(ρ) ≡ v′_{1}(ρ) − v′_{2}(ρ) < 0 ρ ∈ (α, β].

They imply the inequalities

(23)  δ_{i}(ρ) ≥ 0 ρ ∈ [α, β]

in virtue of the monotonicity - assumptions (i), (ii) of the theorem. Summarising the precedings results, we get

(24)  δ ∈ C[α, β], δ(ρ) ≥ 0 ρ ∈ [α, β].

Integrating equality (21) over the interval (α, β) we get after rearranging:

β^{n−1}v′(β) = α^{n−1}v′(α) + ∫_{α}^{β} δ(ρ) dρ,

from which using equality v′(α) = 0 we get

β^{n−1}v′(β) = ∫_{α}^{β} δ(ρ) dρ ≥ 0,

consequently v′(β) ≥ 0 that contradicts the choice of β as a point, such, that v′(β) < 0. Theorem is proved.

Now, let us formulate a weakly generalized Problem 1, namely Problem 3:

(25)  u ∈ C^{2}(B_{0}^{R}) ∩ C(\overline{B_{0}^{R}}),

(26)  Δu(x) + f(|x|, u(x), |∇u(x)|) = 0 x ∈ B_{0}^{R},

(27)  ∃v : [0, R] → ℝ, v(|x|) = u(x) x ∈ \overline{B}_{0}^{R} (|x| ∈ [0, R]),

(28)  u|_{Γ} = a ∈ ℝ,

where a ∈ ℝ is arbitrarily fixed; R ∈ (0, ∞),

B_{0}^{R} := {x ∈ ℝ^{n}| |x| < R}, Γ = ∂B_{0}^{R} ≡ {x ∈ ℝ^{n}| |x| = R},

and

(29)  f ∈ C(G_{a,R}; (0, ∞),

where

G_{a,R} := [0, R] × [a, ∞) × [0, ∞).

Theorem 4. Let function f satisfy the monotonicity conditions: (i) w(t) := f(α, t, β) is nonincreasing in t ∈ [a, ∞) for every fixed α, β (α ∈ [0, R], β ∈ [0, ∞)),

(ii) w̃(t) := f(α, β, t) is nonincreasing in t ∈ [0, ∞) for every fixed α, β (α ∈ [0, R], β ∈ [a, ∞)).

Then Problem 3 has no more than one solution.

Proof. The arguments used in the proof of Theorem 3 applied to [0, R] instead of [0, 1] show the validity of Theorem 4.

2. Comparison results

Theorem 5. Suppose that all of the assumptions included in the formulation of Problem 3 are fulfilled, moreover assumptions (i), (ii) of Theorem 4 hold. Consider the problems

(30)  u_{i} ∈ C^{2}(B_{0}^{R}) ∩ C(\overline{B_{0}^{R}}) i = 1, 2,

(31)  Δu_{i}(x) + f(|x|, u_{i}(x), |∇u_{i}(x)|) = 0 x ∈ B_{0}^{R}, i = 1, 2,

(32)  ∃v_{i} : [0, R] → ℝ, v_{i}(|x|) = u_{i}(x) x ∈ \overline{B}_{0}^{R} (|x| ∈ [0, R]), i = 1, 2,

(33)  u_{i}|_{Γ} = a_{i} ∈ ℝ i = 1, 2,

where

a_{1}, , a_{2} ∈ ℝ, a_{1} > a_{2} ≥ a.

If u_{i} ∼ v_{i} i = 1, 2 are solutions of problems (30) - (33), then

(34)  v_{1}(ρ) ≥ v_{2}(ρ) ρ ∈ [0, R],

(35)  (0 ≥)v′_{1}(ρ) ≥ v′_{2}(ρ) ρ ∈ [0, R], v′_{1}(0) = v′_{2}(0) = 0.

Proof. Let us begin with the proof of inequality (34). We introduce the notation

v(ρ) := v_{1}(ρ) − v_{2}(ρ) ρ ∈ [0, R].

The arguments used for the derivation of the relations (12), (13) applied to B_{0}^{R} instead of B_{0}^{1} give

(36)  v ∈ C^{2}([0, R)) ∩ C([0, R]), v(R) = a_{1} − a_{2} > 0, v′(0) = 0,

and

(37)  (ρ^{n−1}v′_{i}(ρ))′ + ρ^{n−1}f(ρ, v_{i}(ρ), −v′_{i}(ρ)) = 0 ρ ∈ [0, R); i = 1, 2.

If v(ρ) > 0 ρ ∈ [0, R] is also fulfilled, then v_{1}(ρ) > v_{2}(ρ) ρ ∈ [0, R] and (34) is proved. In the case, when there exists a point b_{1} ∈ [0, R) such that v(b_{1}) = 0 let

(38)  b := sup{ρ ∈ [0, R)|v(ρ) = 0}.

It is clear that

b ∈ [0, R), v(b) = 0.

If b = 0, then

(39)  v_{1}(ρ) > v_{2}(ρ) ρ ∈ (0, R], v_{1}(0) = v_{2}(0),

so (34) is fulfilled. If b > 0, then b ∈ (0, R) and Theorem 4 applied to the ball B_{0}^{b} gives

(40)  v_{1}(ρ) = v_{2}(ρ) ρ ∈ [0, b].

On the other hand v(R) > 0, and the definition of b implies the inequality

(41)  v_{1}(ρ) > v_{2}(ρ) ρ ∈ (b, R].

Relations (40) combined with (41) give

v_{1}(ρ) ≥ v_{2}(ρ) ρ ∈ [0, R].

Next we prove the inequality (35). Suppose the contrary. Then using also the first one of the relations in (36) there exists a point c_{1} ∈ (0, R) such that v′(c_{1}) < 0. Introduce the notation

(42)  c := sup{c_{1} ∈ (0, R]|v′(c_{1}) < 0}.

It is clear that c ∈ (0, R] and v′(c) ≤ 0. Then we consider the three possible cases

(A)  v(ρ) > 0 ρ ∈ [0, R],

(B)  v(0) = 0, v(ρ) > 0 ρ ∈ (0, R] (b = 0),

(C)  v(ρ) ≡ 0 ρ ∈ [0, b], v(ρ) > 0 ρ ∈ (b, R] (b ∈ (0, R)).

In the cases (A),(B) let us choose a point d ∈ (0, c) such, that v′(d) < 0. Then we define the set M:

M := {ρ ∈ [0, d)| v′(ρ) = 0}.

It is obvious, that M ≠ ∅ because v′(0) = 0 (see the last of the relations in (36)). Then let

e := sup M.

It is trivial that

e ∈ [0, d), v′(e) = 0, v′(ρ) < 0 ρ ∈ (e, d].

Summarising, in the cases (A), (B) we have

v(ρ) > 0 ρ ∈ (e, d], v(e) ≥ 0; v′(ρ) < 0 ρ ∈ (e, d], v′(e) = 0,

consequently, the same arguments as in the proof of Theorems 3, 4, applied to the interval (α, β) := (e, d) lead to the inequality v′(d) ≥ 0 that contradicts the choice of d for which v′(d) < 0.

For the case (C), first, remark that in virtue of the inequality (41)

(43)  v(ρ) > 0 ρ ∈ (b, R],

moreover

(44)  v(b) = 0, v(ρ) = 0 ρ ∈ [0, b) (v′_{1}(ρ) = v′_{2}(ρ) ρ ∈ [0, b))

according to the definition of b and to Theorem 4 on the uniqueness in the ball B_{0}^{b}. Now (44) - using the property v ∈ C^{2}[0, 1)- implies v′(b) = 0, consequently we have the same situation as in the case (B), but on the interval [b, R] instead of interval [0, R]. The theorem is proven.

Remark 2. In fact, we proved a stronger result, than inequality (34) : namely, may occour three and only the following three cases:

(A)  v_{1}(ρ) > v_{2}(ρ) ρ ∈ [0, R],

or

(B)  v_{1}(ρ) > v_{2}(ρ) ρ ∈ (0, R], v_{1}(0) = v_{2}(0),

or there exists a number b ∈ (0, R) such that

(C)  v_{1}(ρ) = v_{2}(ρ) ρ ∈ [0, b], v_{1}(ρ) > v_{2}(ρ) ρ ∈ (b, R].

On the other hand inequality (35)

(0 ≥) v′_{1}(ρ) ≥ v′_{2}(ρ) ρ ∈ [0, R] (v′_{1}(0) = v′_{2}(0) = 0)

– in general – cannot be replaced by another, stronger one under assumptions of Theorem 5 (see e.g. the case, when f does not depend on argument u).

Theorem 6. All of the statements of Theorem 5 remain - except for inequality (35) - if in conditions of Theorem 5 assumptions (i), (ii) of Theorem 4 are replaced by condition:

w(t) := f(|x|, t, |∇u|) ∼ f(α, t, β)

is strongly decreasing in t ∈ [a, ∞) for every fixed α, β (α ∈ [0, R], β ∈ [0, ∞)). This theorem is a corollary of a general comparison result, namely:

Theorem 7. Let u_{1}, u_{2} be solutions of Problem 2 satisfying conditions

u_{i}|_{Γ} = φ_{i} ∈ C(Γ) i = 1, 2; φ_{1} ≥ φ_{2},

and suppose that function

w(t) := f(\underline{α}, t, \underline{β}) t ∈ ℝ

is strongly decreasing in t ∈ ℝ for any \underline{α} ∈ Ω, \underline{β} ∈ ℝ^{n} fixed. Then

u_{1}(x) ≥ u_{2}(x) x ∈ Ω.

Moreover, if there exists a point y ∈ Γ such that φ_{1}(y) > φ_{2}(y), then may occour two, and only the following two cases:

(A)  u_{1}(x) > u_{2}(x) ∀x ∈ Ω,

or there exists a subset Ω_{1} ≠ ∅ of Ω such that

0 < μ(Ω_{1}) ≤ μ(Ω)

( μ is the n -dimensional Lebesgue measure) and

(B)  u_{1}(x) > u_{2}(x) ∀x ∈ Ω_{1}; u_{1}(x) = u_{2}(x) ∀x ∈ Ω\Ω_{1}.

Proof. Let u := u_{1} − u_{2}, and suppose that there exists a point y ∈ Ω such that u(y) < 0. Then there is a point x_{0} ∈ Ω with the property:

u(x_{0}) = min_{x∈\overline{Ω}} u(x) ≡ m < 0,

and all that remains is to repeat the proof of Theorem 2 for to get a contradiction. Theorem is proven.

3. Concavity results.

Here we will present certain results on the concavity of the function v : [0, 1] → ℝ, defined in the Introduction ((1.2)) by the relation v(|x|) = u(x) x ∈ \overline{B}, where the function u is supposed to be a solution of Problem 1.

Theorem 8. Let a ∈ ℝ in Problem 1 be fixed, and suppose that

(i)  w(t) := f(α, t, β)

is nonincreasing in t ∈ [a, ∞) for every α, β fixed (α ∈ [0, 1], β ∈ [0, ∞)),

(ii)  w̃(t) := f(α, β, t)

is nonincreasing in t ∈ [0, ∞) for every α, β fixed (α ∈ [0, 1], β ∈ [a, ∞)).

If, in addition,

(iii)  f(t, a + \frac{K_{a}}{n} \frac{1 − t^{2}}{2} , \frac{K_{a}}{n} t) ≥ K_{a}(1 − \frac{1}{n}) t ∈ [0, 1),

where

K_{a} := sup_{Ga} f(= max_{ρ∈[0,1]} f(ρ, a, 0))

then function v is concave (in non strong sense) on the interval [0, 1).

In other words - if γ = (γ_{1}, γ_{2}, γ_{3}) is a curve in ℝ^{3} :

γ : γ_{1} = t, γ_{2} = a + \frac{K_{a}(1 − t^{2})}{2n}, γ_{3} = \frac{K_{a}t}{n} t ∈ [0, 1),

then condition (iii) means that

(iv)  f|_{γ} ≥ K_{a}(1 − \frac{1}{n}).

Proof. Assumptions of the Theorem guarantee the uniqueness (see Theorem 3 in above) of the solution u ∼ v to the Problem 1. We know (see (12), (13)) that v has the following properties:

(45)  \begin{matrix}v ∈ C^{2}[0, 1) ∩ C[0, 1], v(1) = a, v′(0) = 0, \\ Δu(x) + f(|x|, u(x), |∇u(x)|) = v′′(ρ) + \frac{n − 1}{ρ}v′(ρ) + f(ρ, v(ρ), −v′(ρ)) = 0 \\ x ∈ B, ρ ∈ (0, 1), \end{matrix}

and

(46)  (ρ^{n−1}v′(ρ))′ + ρ^{n−1}f(ρ, v(ρ), −v′(ρ)) = 0 ρ ∈ [0, 1).

Integrating equality (46) over the inteval [δ, t](0 < δ < t < 1) we get

(47)  t^{n−1}v′(t) = δ^{n−1}v′(δ) − ∫_{δ}^{t} ρ^{n−1}f(ρ, v(ρ), −v′(ρ)) dρ

from which passing to the limit as δ → 0 + 0 we obtain

(48)  t^{n−1}v′(t) = − ∫_{0}^{t} ρ^{n−1}f(ρ, v(ρ), −v′(ρ)) dρ ∀t ∈ (0, 1).

Using the notation

ν := −v′ (ν(t) := −v′(t) ∀t ∈ [0, 1])

the first and second of the relations of (45) give

v(t) − v(t_{1}) = ∫_{t}^{t_{1}} ν(s)ds 0 ≤ t < t_{1} < 1, v(t) − v(t_{1}) → v(t) − a as t_{1} → 1 − 0,

consequently there exists the improper integral

∫_{t}^{1} ν(s)ds := lim_{t_{1}→1−0} ∫_{t}^{t_{1}} ν(s)ds t ∈ [0, 1),

and

(49)  v(t) = a + ∫_{t}^{1} ν(s) ds ∀t ∈ [0, 1],

From (48),(49) we obtain that function ν satisfies equality

(50)  ν(t) = ∫_{0}^{t} (\frac{ρ}{t})^{n−1} f(ρ, a + ∫_{ρ}^{1} ν(s) ds, ν(ρ)) dρ t ∈ [0 + 0, 1)

which is understood at t = 0 + 0 in the limit sense. From the definition of K_{a} and equality (50) we get the inequality

(51)  (0 ≤)ν(t) ≤ \frac{K_{a}}{n} t ∀t ∈ [0, 1).

To prove the theorem we have to show that

(52)  ν′(t) ≥ 0  t ∈ [0, 1),

i.e.-using the last of the equalities in (45) for ρ ∈ [0 + 0, 1) combined with (50) - the inequality

(53)  \begin{matrix}ν′(t) ≡ f(t, a + ∫_{t}^{1} ν(s)ds, ν(t))− \\ − & \frac{n − 1}{t} ∫_{0}^{t} (\frac{ρ}{t})^{n−1} f(ρ, a + ∫_{ρ}^{1} ν(s)ds, ν(ρ)) dρ ≥ 0 ρ ∈ [0 + 0, 1). \end{matrix}

From (51) we obtain that

ν′(t) ≥ f(t, a + \frac{K_{a}}{n} \frac{1 − t^{2}}{2} , \frac{K_{a}}{n} t) − \frac{n − 1}{t}ν(t) ≥

(54)

≥ f(t, a + \frac{K_{a}}{n} \frac{1 − t^{2}}{2} , \frac{K_{a}}{n} t) − \frac{n − 1}{t} \frac{K_{a}}{n} t ≥ 0 t ∈ [0, 1)

in virtue of conditon (iii). Theorem is proven.

Some concrete sufficient conditions for the special case of Problem 1, when

(55)  f(ρ, u, |∇u|) = e^{λu+K|∇u|} λ, K ∈ ℝ; λ, K ≤ 0

are presented in the following

Theorem 9. Let a ∈ ℝ be arbitrarily fixed in Problem 1 with nonlinearity f of the form in (55). Then solution u ∼ v of Problem 1 exists ([7]), is unique, and any of the following conditions (i) - (vi) guarantees the nonstrong concavity of solution v on [0, 1); where we use the notation

d_{n} := ln [(1 − \frac{1}{n})^{n}] n ∈ ℕ, n is fixed n ≥ 2 (d_{n} < 0),

(i)  λ = K = 0,

(ii)  λ = 0, 0 > K ≥ d_{n},

(iii)  K = 0, 0 > \frac{λe^{λa}}{2} ≥ d_{n},

(iv)  K < λ < 0, Ke^{λa} ≥ d_{n},

(v)  K = λ < 0, λe^{λa} ≥ d_{n},

(vi)  λ < K < 0, \frac{e^{λa} · λ}{2} (1 + \frac{K^{2}}{λ^{2}} ) ≥ d_{n}.

Proof. It is enough to prove that inequality (iii) of Theorem 8 is fulfilled in every of the cases (i) - (vi) of the present Theorem. Using that

f(t_{1}, t_{2}, t_{3}) ∼ f(t_{2}, t_{3}) = e^{λt_{2}+Kt_{3}} t_{2} ∈ [a, ∞), t_{3} ∈ [0, ∞)

and relations

f(a, 0) = e^{λa} ≥ f(t_{2}, t_{3}) t_{2} ∈ [a, ∞), t_{3} ∈ [0, ∞)

we get that K_{a} = e^{λa}. Substituting this value into inequality (iii) of Theorem 8, the desirable inequality gains the form

e^{λ[a+ \frac{e^{λa}}{n} \frac{1−t^{2}}{2} ]+K \frac{e^{λa}}{n} t} ≥ e^{λa}(1 − \frac{1}{n}) t ∈ [0, 1)

i.e.

e^{e^{λa}[λ \frac{1−t^{2}}{2} +Kt] \frac{1}{n}} ≥ (1 − \frac{1}{n}) t ∈ [0, 1)

i.e.

e^{λa}[λ\frac{1 − t^{2}}{2} + Kt] ≡ g(t) ≥ ln[(1 − \frac{1}{n})^{n}] ≡ d_{n} t ∈ [0, 1).

It is easy to prove in every of the cases (i) - (vi) that

min_{t∈[0,1]} g(t) ≥ d_{n},

which completes the proof.

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