Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 3, 517–523

On Uniqueness Conditions for Decreasing Solutions of Semilinear Elliptic Equations

Tadie

Dedicated to my uncle Toam Chatue J.B. (†14/08/1997)

Abstract. For f ∈ C([0, ∞)) ∩ C^{1}((0, ∞)) and b > 0, existence and uniqueness of radial solutions u = u(r) of the problem △u + f(u_{+}) = 0 in ℝ^{n} (n > 2), u(0) = b and u′(0) = 0 are well known. The uniqueness for the above problem with boundary conditions u(R) = 0 and u′(0) = 0 is less known beside the cases where lim_{r→∞} u(r) = 0. It is our goal to give some sufficient conditions for the uniqueness of the solutions of the problem D_{α}u + f(u_{+}) = 0 (r > 0), u(ρ) = 0 and u′(0) = 0 based only on the evolution of the functions f(t) and \frac{d}{dt} \frac{f(t)}{t}.

Keywords: Semilinear elliptic equations, comparison results for nonlinear differential equations

AMS subject classification: 35 J 65, 34 B 15

1. Introduction

Let the function f ∈ C([0, ∞)) ∩ C^{1}((0, ∞)) be such that it remains positive, or it has a finite number of positive zeros and changes its sign accross any of them. For a = n − 1 > 1 (n ∈ ℕ) and any u_{0} > 0, the problem

\begin{matrix}D_{a}u := u′′ + \frac{a}{r}u′ & = −f(u_{+}) \\ u(0) & = u_{0} \\ u′(0) & = 0 \end{matrix} \}  (E)

is known to have a unique solution u ∈ C^{2}([0, ∞)) which is positive in some interval [0, ρ) [3, 6]. For ρ > 0, finite or not, we will investigate some uniqueness conditions for the associated problem

\begin{matrix}D_{a}u + f(u_{+}) & = 0 (r > 0) \\ u(ρ) & = 0 \\ u′(0) & = 0. \end{matrix} \}  (BV)_{ρ}

For ease writing, the following notations will be used:

1) u_{+}(r) = max{0, u(r)} and F (t) = ∫_{0}^{t} f(s) ds.

Tadie: Matematisk Institut, Universitetsparken 5, 2100 Copenhagen, Denmark

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

2) f^{+} = {t > 0| f(t) > 0} and F^{+} = {t > 0| F (t) > 0}.

3) ψ(t) = \frac{d}{dt} \frac{f(t)}{t} and λ denotes any zero of ψ.

4) Ψ(t) = \frac{f(t)}{t}, ψ^{−} = \{t > 0| ψ(t) < 0\} and ψ^{+} = \{t > 0| ψ(t) > 0\}.

5) For connected components of ψ, define ψ_{i}^{±} = {(t_{i}, t_{i+1}) ⊂ ψ^{±}| 0 < t_{i} < t_{i+1} and ψ(t_{i}) = ψ(t_{i+1}) = 0} and similarly f_{i}^{±}.

6) For s ≠ t, {s, t} will denote the elements lying between s and t.

7) f will be said to satisfy the condition |f′; ψ^{+}| or |f′; ψ^{−}| if there exists α′ and C > 0 or B > 0 such that |f′| > C and ψ^{+} > B or ψ^{−} < −B, respectively, in (0, α′]. |f′; ψ|_{I} will have the same definition where I ⊂ ℝ_{+} replaces (0, α′].

2. Main results

Let ρ > 0 be finite or not.

Theorem 1. If f(t) is decreasing in t > 0, then problem (BV)_{ρ} has at most one solution.

Theorem 2. If \frac{d}{dt} \frac{f(t)}{t} > 0 or \frac{d}{dt} \frac{f(t)}{t} < 0 in t > 0, then problem (BV)_{ρ} has at most one solution.

Theorem 3. Assume that \frac{d}{dt} \frac{f(t)}{t} > 0 in (0, λ) and \frac{d}{dt} \frac{f(t)}{t} < 0 in (λ, ∞). If f′ ≤ 0 in some interval (0, α′] or condition |f′; ψ^{+}| holds, then problem (BV)_{ρ} has at most one solution.

Theorem 4.

(i) Assume that ψ^{−} = (0, λ), ψ^{+} = (λ, ∞), f is strictly monotone in (0, λ + k) for some k > 0 and condition |f′; ψ^{+}| holds. Then problem (BV)_{ρ} has at most one solution.

(ii) Assume that ψ has a finite number of zeros, f is strictly monotone in any \overline{ψ_{i}^{−}} and condition |f′; ψ^{+}| or |f′; ψ^{−}| holds. Then problem (BV)_{ρ} has at most one solution.

3. Preliminaries

Let u be a solution of problem (E), positive on I = (r_{0}, r_{2}). After multiplying the equation in problem (E) by u′, integration on I leads to the identity

\frac{u′(r_{2})^{2}}{2} + F (u(r_{2})) + a ∫_{r_{0}}^{r_{2}} \frac{u′(s)^{2}}{s} ds = F (u(r_{0})) + \frac{u′(r_{0})^{2}}{2} .  (1)

Lemma 1. Let u be a solution of problem (E), non-constant in some interval (R, R + τ) with R > 0 and τ > 0.

(i) If

u′(R) = 0 ⟹ u(r) ≠ u(R) for all r ∈ (R, R + τ),  (2)

then the solutions of problem (BV)_{ρ} are strictly decreasing in (0, ρ).

(ii) If u is a solution of problem (BV)_{ρ}, then F (u(0)) > 0 and u(0) ∈ f^{+} ∩ F^{+}.

Proof. (i) It suffices to notice that for r_{0} = R and r_{2} > R in (1), one cannot have u(r) = u(R) for r > R. Let u be a solution of problem (BV)_{ρ}. Then from (2), u has to be decreasing in some interval (0, r), u′ < 0 and decreasing as long as f(u) > 0 in (0, r). From the equation, if u′(r_{1}) = 0 for some r_{1} > 0 (r_{1} being the first such r), u′′(r_{1}) = −f(u(r_{1})). Identity (1) and (2) imply that u(r_{1}) cannot be a local minimun and obviously nor a local maximun. This reaches a contradiction as f has only simple zeros. Thus u′(r_{1}) = 0 cannot hold.

Statement (ii) is a direct consequence of identity (1)

Lemma 2. Let u and v be two distinct solutions of problem (E) which are positive in the interval I = (R, ρ). Then

\{r^{a}v(r)^{2} \frac{u}{v}′\}_{R}^{ρ} = ∫_{R}^{ρ} s^{a}uv\{Ψ(v) − Ψ(u)\}ds.  (3)

Consequently, if u and v are two distinct solutions of problem (E) strictly positive in I = (R, r), with u > v in I and (u′v − uv′)(R) = 0, then:

(i) u(I) ∪ v(I) ⊂ ψ^{+} ⟹ \frac{u}{v} is strickly decreasing on I.
(ii) u(I) ∪ v(I) ⊂ ψ^{−} ⟹ \frac{u}{v} is strickly increasing on I.

Note that the condition (u′v − uv′)(R) = 0 can be replaced by (u′v − uv′)(R) ≤ 0 for the case (i) and by (u′v − uv′)(R) ≥ 0 for the case (ii).

Proof. It is enough to notice that the function W = u′v − uv′ = v^{2}(\frac{u}{v})′ satisfies

(r^{a}W)′ = r^{a}uv\{Ψ(v) − Ψ(u)\} = r^{a}uv \{\frac{f(v)}{v} − \frac{f(u)}{u}\}

in (R, ρ). For statement (i), it is enough to notice that Ψ(v) − Ψ(u) < 0 on I by (3) whence W < 0 on I. Statement (ii) follows from a similar argument

Lemma 3. Let u and v be two distinct solutions of problem (E) which are nonnegative in I = (r_{1}, r_{2}).

(i) If (u′v − uv′)(r_{1}) = (u′v − uv′)(r_{2}) = 0 and u′v − uv′ ≠ 0 in I, then either ψ has a zero in {u(r), v(r)} for r ∈ I or u(r) = v(r) has a solution in I.

(ii) If u(r_{1}) = v(r_{1}) and u(r_{2}) = v(r_{2}), then ψ has a zero in {u(r), v(r)} for r < r_{2}.

Proof. (i) From identity (3),

\frac{f(v)}{v} − \frac{f(u)}{u}

changes the sign at some R ∈ I and either u(R) = v(R) or there exist R_{1}, R_{2} ∈ I such that

\frac{f(u(R_{1}))}{u(R_{1})} = \frac{f(v(R_{2})}{v(R_{2})}.

The later case implies that ψ has a zero in {u, v} for r ∈ {R_{1}, R_{2}} by the mean value theorem.

(ii) Without loss of generality, suppose that u > v in I. For z(r) = u(r) − v(r), there exists R_{1} ∈ I such that

z′(R_{1}) = u′(R_{1}) − v′(R_{1}) = 0.

As u > v and u′ < 0 in I, we have

(u′v − uv′)(R_{1}) = u′(R_{1})(v − u)(R_{1}) > 0

whence u(r_{2}) = v(r_{2}) holds only if there exists R_{2} ∈ (R_{1}, r_{2}) such that (u′v−uv′)(R_{2}) = 0. The conclusion follows from statement (i) as (u′v − uv′)(0) = 0

If u and v are two distinct solutions of problem (E) and s > 0 is such that

U(r) = u(r) + s, V (r) = v(r) + s, Z(r) = u(r) − s, Y (r) = v(r) − s

are positive, then

X′′ + \frac{a}{r}X′ = −f(X − s)  for X = U, V  (4)_{a}
Φ′′ + \frac{a}{r}Φ′ = −f(Φ + s)  for Φ = Z, Y.  (4)_{b}

The next lemma is easy to verify.

Lemma 4. For 0 < s < t, define f_{±s}(t) = \frac{f(t±s)}{t}. Then

\begin{matrix}\frac{∂}{∂t}f_{s}(t) & = \frac{(t + s)^{2}Ψ′(t + s) − sf′(t + s)}{t^{2}} \\ \frac{∂}{∂t}f_{−s}(t) & = \frac{(t − s)^{2}Ψ′(t − s) + sf′(t − s)}{t^{2}}. \end{matrix} \}  (5)

Consequently, for I_{s}(t) = [t, t + s] and I_{−s}(t) = [t − s, t],

I_{τ} (t) ⊂ ψ^{+} ∩ \{f′ ≤ 0\} ⟹ \frac{∂f_{s}(t)}{∂t} > 0  (5)_{a}
I_{−τ} (t) ⊂ ψ^{+} ∩ \{f′ ≥ 0\} ⟹ \frac{∂f_{−s}(t)}{∂t} > 0  (5)_{b}
I_{τ} (t) ⊂ ψ^{−} ∩ \{f′ ≥ 0\} ⟹ \frac{∂f_{s}(t)}{∂t} < 0  (5)_{c}
I_{−τ} (t) ⊂ ψ^{−} ∩ \{f′ ≤ 0\} ⟹ \frac{∂f_{−s}(t)}{∂t} < 0  (5)_{d}

for 0 < s < τ.

Lemma 5. Let u and v be two distinct solutions of problem (E) with meas{r > 0| u(r) = v(r) > 0} = 0.

(i) As long as u and v remain in the same connected component of ψ, the problem u(r) = v(r) > 0 has at most one solution.

(ii) Suppose that ψ^{+} = (λ, A) and ψ^{−} = (0, λ). For u and v two solutions of problem (E) with u(A) > v(A) > λ and (u′v − uv′)(A) ≤ 0, if u(r_{1}) = v(r_{1}) ≤ λ, then u(r) = v(r) > 0 does not hold for r > r_{1}. If in addition f′ ≥ 0 in some interval [0, α′], then u(r) = v(r) ≥ 0 cannot hold for r > r_{1}. Consequently, if ψ has a finite number of components, then u(r) = v(r) ≥ 0 has a finite number of solutions.

Proof. (i) The claim follows from the fact that remaining in the same component ψ, if u − v has two distinct zeros, then \frac{u}{v} is strictly monotone between them with the same value 1 in both ends. That cannot hold.

(ii) Let u ≥ v ≥ λ in some subset of ψ^{+}. Suppose that u(R) = v(R) ≤ λ and 0 < u < v in some interval I = (R, r). Then

(u′v − uv′)(R) = u(R)(u − v)′(R) < 0

as u′ < v′ at R. Therefore \frac{v}{u} is increasing in some r > R with the value 1 at R. We have v > u as long as u > 0. If u(ρ) = v(ρ) = 0 and f′ ≥ 0 in some interval [0, α′], then with Z(r) = v(r) + s and Y (r) = u(r) + s for some small s > 0 and X = Y or X = Z we have D_{a}X + f(X − s) = 0 in some interval J = (R, ρ) and Y (ρ) = Z(ρ) = s. From (4)_{a} and (5)_{b}, (\frac{Z}{Y})′ > 0 in J conflicting with the fact that (\frac{Z}{Y})(R) > 1 and \frac{Z}{Y} (ρ) = 1. Now statement (ii) follows from the fact that no component of ψ^{+} neither any of ψ^{−} can have more than two solutions of the problem u(r) = v(r) ≥ 0

Lemma 6. Let u and v be two distinct solutions of problem (E), A = \overline{u(I) ∪ v(I)} for I = (r_{0}, r_{1}) and some J = [t_{0}, t_{1}] ⊂ A with t_{0} > inf A.

(i) Suppose that A ⊂ ψ^{+} and

(α) u > v and u′v − uv′ ≤ 0 at r_{0}

(β) f′ ≤ 0 in J or condition |f′; ψ^{+}|_{J} holds.

Then u(r) = v(r) > inf A has a solution r_{1} in I with u′(r_{1}) ≠ v′(r_{1}). If in addition t_{0} = 0, then u ≠ v for r > r_{1} as long as u, v ≥ 0 in A.

(ii) Suppose that A ⊂ ψ^{−} and

(α) u > v and u′v − uv′ ≥ 0 at r_{0}

(β) f′ ≤ 0 in J or condition |f′; ψ^{−}|_{J} holds.

Then u > v ≥ 0 in I.

Proof. (i) From identity (3), \frac{u}{v} is decreasing in ψ^{+} as long as u > v > 0 there. Assume that u > v ≥ λ := inf ψ^{+}. Let s > 0 and t > 0 be such that t + s ∈ J and let v(R′) = s < u(R′) for some R′. The functions Y = v − s and Z = u − s satisfy Y (R′) = 0 and Z(R′) > 0; for X = Y and X = Z we have D_{a}X = −f(X + s) in (r_{0}, R′). From (5) and (5)_{a}, if f′ ≤ 0 in J, then (\frac{∂}{∂t})f_{s}(t) > 0. Applying Lemma 2 to Y and Z we find that \frac{Z}{Y} is decreasing in (r_{0}, R′) which conflicts with their values at R′. The assumption cannot hold. So there is an R′′ ∈ I such that u(R′′) = v(R′′). As

(u′v − uv′)(R′′) = u(R′′)(u − v)′(R′′) < 0,

we have u′(R′′) < v′(R′′).

The second part of statement (i) follows the same process as for s ∈ (0, t^{2} \frac{B}{4C}) and s ∈ (0, \frac{t}{2}),

\frac{∂}{∂t}f_{s}(t) > t^{−2} \{t^{2} \frac{B}{4} − sC\} > 0.

Let u(r_{1}) = v(r_{1}) and u > v in (r_{1}, ρ). If u = v = 0 at ρ, for W = U with U = u+s and W = V with V = v + s we have D_{a}W = −f(W − s) in (r_{1} + s, ρ) and \frac{U}{V} (r_{1}) = \frac{U}{V} (ρ). As condition |f′; ψ^{+}|_{J} holds, \frac{U}{V} is monotone in (r_{1}, ρ) and this cannot hold from their values at the both ends.

(ii) Identity (3) implies that (\frac{u}{v})′ > 0 as long as u > v in I whence they cannot intersect there nor intersect at some r_{1} with u(r_{1}) = v(r_{1}) > 0. Assume that u(r_{1}) = v(r_{1}) = 0. Let s > 0 and t > 0 such that t−s ∈ J. The functions U = u+s and V = v+s satisfy for some R_{1} > r_{1} and W = U or W = V the relation D_{a}W = −f(W − s) in (R_{1}, r_{1}) = K with W (r_{1}) = s. If f′ ≤ 0 in J, \frac{U}{V} is increasing in K with a value greater than 1 at R_{1}. This conflicts with their values at r_{1}.

The last part follows the same process as before. In fact, for s ∈ (0, \frac{t}{2}),

(t − s)^{2}ψ(t − s) + sf′(t − s) < −t^{2} \frac{B}{4} + s sup_{I} |f′|

and it suffices to take s ∈ (0, t^{2} \frac{B}{4C}) for (5) and (5)_{d} to apply

Lemma 7. Let A < λ < B, ψ_{0}^{−} = (A, λ), ψ_{0}^{+} = (λ, B), u and v two distinct solutions of problem (E) such that for some 0 ≤ r_{1} < r_{2}

(i) u(r_{1}), v(r_{1}) > λ and (u′v − uv′)(r_{1}) < 0
(ii) u(r_{2}) = v(r_{2}) < λ with u > v in (r_{1}, r_{2}).

Then if f′ is strictly monotone in ψ_{0}^{−}, we have u′(r_{2}) < v′(r_{2}).

Proof. Let v(r_{λ}) = λ. As u′v −uv′ = u′(v −u)+u(u−v)′, (u−v)′ < 0 and strictly decreasing in (r_{1}, r_{λ}) (see (3)). If (u − v)′(r_{2}) = 0, then by the mean value theorem, there is R ∈ (r_{λ}, r_{2}) such that (u − v)′′(R) = 0. In that case, from the equations of u and v,

a(u − v)′(R) = R{f(v(R)) − f(u(R))} < 0

and this cannot hold if f′ < 0 in ψ_{0}^{−} whence (u − v)′(r_{2}) < 0 in this case. If f′ > 0 in ψ_{0}^{−}, then (r^{a}(u − v)′)′ < 0 in (R, r_{2}) and (u − v)′(R) < 0 which leads to (u − v)′(R_{2}) ≤ (u − v)′(R) < 0

On Uniqueness Conditions for Decreasing Solutions 523

4. Proof of the theorems

The lemmae established in Section 3 enable us to prove now the theorems.

Proof of Theorem 1. Let u and v be two solutions of problem (E), with u > v in some interval [0, r), say. From the equations,

(u − v)′(r) = ∫_{0}^{r} (\frac{s}{r})^{a} \{f(v) − f(u)\} ds ≥ 0

whence u(r)−v(r) ≥ u(0)−v(0) > 0. Therefore they cannot intersect as long as v ≥ 0

Proof of Theorem 2. 1. In any of the cases, if u > v in [0, ρ) and u(ρ) = v(ρ) = 0, then the left-hand side of identity (3) is 0 while the right-hand side is non-zero as the integrand there does not change sign. So u(ρ) ≠ v(ρ) = 0.

2. If there is R ∈ (0, ρ) with u(R) = v(R) > 0 and R_{1} ∈ (R, ρ] with u(R_{1}) = v(R_{1}), then there is R_{2} ∈ (R, R_{1}) with (u′v − uv′)(R_{2}) = 0 and this cannot hold following similar an argument as in part 1

Proof of Theorem 3. Let u and v be two distinct solutions of problem (E). If there is r < ρ such that u(r) = v(r), then u(r) < λ. Lemma 6/(i) implies that r ≠ ρ

Proof of Theorem 4. 1. Let u and v be two distinct solutions of problem (E). The problem u(r) = v(r) > λ has at most one solution by Lemma 5/(i). Lemmae 6/(ii) and 7 imply that u(ρ) ≠ v(ρ).

2. Lemmae 6/(i) and 7 imply that u − v changes sign accross any r where u(r) = v(r) > 0. The ends of Theorems 2 and 3 complete the proof

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Received 15.05.1998