Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 3, 525–537

Some Surprising Results on a One-Dimensional Elliptic Boundary Value Blow-Up Problem

Y. J. Cheng

Abstract. In this paper we consider the one-dimensional elliptic boundary blow-up problem

\begin{matrix}Δ_{p}u = f(u) (a < t < b) \\ u(a) = u(b) = +∞ \end{matrix} \}

where Δ_{p}u = (|u′(t)|^{p−2}u′(t))′ is the usual p-Laplace operator. We show that the structure of the solutions can be very rich even for a simple function f which gives a leading that a simliar results might hold also in higher dimensional spaces

Keywords: Boundary blow-up, multiplicity, concave and convex nonlinearity AMS subject classification: 34 B, 35 J

1. Introduction and formulation of main results

In the last few years there is a great of interests in the investigation of boundary blowup solutions for elliptic equations [2, 3, 7], which comes originally from differential geometry [6] and electrohydrodynamics [5]. Very recently some existence results of two (one positive and one sign-changing) solutions have been established in [1, 8]. The purpose of the present paper is to show through one-dimensional examples that the structure of the solutions can be very rich even for a simple right-hand side. More precisely, we consider the problem

\begin{matrix}Δ_{p}u = λf(u) (a < t < b) \\ u(a) = u(b) = +∞ \end{matrix} \}  (1)

where

Δ_{p}u = (|u′(t)|^{p−2}u′(t))′

is the p-Laplace operator as usual, λ > 0 is a parameter, and f is a given continuous function. By a solution u = u(t) of problem (1) we mean that u satisfies the equation in (1), i.e. (|u′(t)|^{p−2}u′(t))′ = λf(u(t)) for all t ∈ (a, b), and lim_{t→a+} u(t) = lim_{t→b−} u(t) = +∞. By a sign-changing solution u(t) of (1) we mean that there exist t_{1}, t_{2} ∈ (a, b) such that u(t_{1}) > 0 and u(t_{2}) < 0.

The results for problem (1) in this paper are summarized in the followings three theorems.

Yuanji Cheng: Univ. of Malmö, School of Techn. and Management., 205 06 Malmö, Sweden ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

Theorem 1. For given positive constants q, ε, s and α, r such that q > p−1 > r > s let

f(u) = u^{q} + εu^{s} (u ≥ 0)  and  f(u) = α|u|^{r} (u ≤ 0).

Then the following statements hold:

(i) There exists a constant λ_{1} > 0 such that problem (1) has at least one signchanging solution if λ > λ_{1}, and no such kind of solutions can exist if λ < λ_{1}.

(ii) There exist a constant ε_{0} > 0 such that the solution is unique if ε ≥ ε_{0}.

(iii) If ε ∈ (0, ε_{0}), then there are constants λ_{2} < λ_{3} ≤ λ_{4} such that problem (1) has at least two sign-changing solutions if λ ∈ (λ_{1}, λ_{4}), has at least three sign-changing solutions if λ ∈ (λ_{2}, λ_{3})), and has a unique sign-changing solution if λ > λ_{4}.

Figure 1

Theorem 2. For given positive constants q, ε, s and α, r, δ, τ such that q, τ > p−1 > r > s let

f(u) = u^{q} + εu^{s} (u ≥ 0)  and  f(u) = α|u|^{r} + δ|u|^{τ} (u ≤ 0).

Then:

(i) There are constants Λ > 0 and Λ_{−} > 0 such that problem (1) has at least one sign-changing solution if λ ≤ Λ, has no sign-changing solutions if λ > Λ, and the solution is unique if λ < Λ_{−}.

Figure 2

(ii) For small ε and δ satisfying ε << δ there are λ_{1} < λ_{2} such that if λ ∈ (λ_{1}, λ_{2}), then problem (1) has at least four sign-changing solutions.

Next we consider for simplicity the semilinear (p = 2) problem

\begin{matrix}u′′ = λf(u) (a < t < b) \\ u(a) = u(b) = +∞. \end{matrix} \}  (2)

For this simple problem we have the following, a somehow surprising result.

Theorem 3. For given positive constants α, q and β such that q > 3 and β < 1 let

f(u) = αu^{q} (u ≥ 0)  and  f(u) = (1 + β sin |u|)|u| (u ≤ 0)

(λ(2) = (\frac{π}{b−a})^{2}). Then the following results hold:

(i) There are λ_{1}, λ_{2} > 0 such that problem (2) has at least one sign-changing solution if λ ≥ λ_{1} and no sign-changing solutions if λ < λ_{1}. The sign-changing solution is unique, when λ > λ_{2}.

(ii) For any integer n ≥ 1 there exists δ > 0 such that problem (2) has at least n distinct sign-changing solutions when λ ∈ (λ(2) − δ, λ(2) + δ).

(iii) For λ = λ(2) problem (2) has infinitely many sign-changing solutions.

Figure 3

Remark. Problem (2) has a unique positive solution for all λ > 0. The problems treated in Theorems 1 and 2 have also a unique positive solution if λ < λ_{0} where λ_{0} is defined by

∫_{0}^{∞} \frac{du}{\sqrt[p]{\frac{1}{q+1} u^{1+q} + \frac{ε}{s+1} u^{1+s}}} = \frac{b − a}{2} ^{p} \sqrt{p′λ_{0}}

and has no positive solution if λ ≥ λ_{0}.

2. Some basic analysis

It is easy to see that the equation in (1) has an first integral

\frac{1}{p′} |u′(t)|^{p} = λF (u) + C  with F (u) = ∫_{0}^{u} f(x) dx  (3)

where \frac{1}{p} + \frac{1}{p′} = 1. Let t_{0} ∈ (a, b) be a minimum point of u(t), which exists by the boundary condition. Then u′(t_{0}) = 0 and C = −λF (u_{0}), u_{0} = min u(t), and

|u′(t)|^{p} = λ p′(F (u) − F (u_{0})) (a < t < b).

If f = f(u) is non-negative, then we see that u = u(t) is convex and the minimum point t = t_{0} is unique. Consequetly, u′(t) ≤ 0 for t ∈ (a, t_{0}) and u′(t) ≥ 0 for t ∈ (t_{0}, b). Moreover,

u′(t) = ^{p} \sqrt{λ p′(F (u) − F (u_{0})}) sign (t − t_{0}) (a < t < b).

Direct integration yields

(b − t_{0})\sqrt[p]{p′λ} = ∫_{u_{0}}^{+∞} \frac{du}{\sqrt[p]{F (u) − F (u_{0})}} = (t_{0} − a) ^{p} \sqrt{p′λ}

which implies t_{0} = \frac{a+b}{2} and thus u = u(t) must be symmetric.

To establish the existence and the structure of solutions of problem (1) it suffices to study the nonlinear integral equation

∫_{u_{0}}^{+∞} \frac{du}{\sqrt[p]{F (u) − F (u_{0})}} = \frac{b − a}{2} ^{p} \sqrt{p′λ}.  (4)

Obviously, a necessary condition for the existence of solutions for problem (1) is

∫ ^{+∞} \frac{du}{\sqrt[p]{F (u)}} < +∞  (5)

and so throughout this paper we shall assume that this condition holds.

Rewriting the integral in (4) gives that it is equivalent to

∫_{0}^{+∞} \frac{du}{\sqrt[p]{F (u + u_{0}) − F (u_{0})}} = \frac{b − a}{2} ^{p} \sqrt{p′λ}.  (6)

It follows from here that problem (1) has at most one (positive) solution under the condition that f(u) is non-decreasing on ℝ (or on ℝ_{+}). On the other hand, if f(u) > 0 for u > 0, then it has at least one positive solution for all λ < λ_{0}, where λ_{0} ∈ (0, +∞] is defined by

∫_{0}^{+∞} \frac{du}{\sqrt[p]{F (u)}} = \frac{b − a}{2} ^{p} \sqrt{p′λ_{0}}.  (7)

By summarying, let λ_{0} be as above. Then we have the following

Theorem 4. If f(u) > 0 (u > 0), then problem (1) has at least one positive solution for all λ < λ_{0}. Moreover, if f is also non-decreasing on ℝ_{+}, then the solution is unique.

Example. Consider the problem

\begin{matrix}Δ_{p}u = λ(u^{q} + εu^{s}) (a < t < b) \\ u(a) = u(b) = +∞ \end{matrix} \}  (8)

where 0 < r < q, ε > 0 and q > p − 1. Then condition (5) is satisfied, and further λ_{0} = +∞ if s ≥ p − 1 and λ_{0} < +∞ if s ∈ (0, p − 1). Hence problem (8) has a unique positive solution for all λ > 0 if q > r ≥ p − 1, and for all λ < λ_{0} if 0 < s < p − 1, where λ_{0} satisfies

∫_{0}^{+∞} \frac{du}{\sqrt[p]{\frac{1}{q+1} u^{1+q} + \frac{ε}{s+1} u^{1+s}}} = \frac{b − a}{2} ^{p} \sqrt{p′λ_{0}}.

3. Proofs

To investigate solutions of problem (1) which change its sign, we define

f_{+}(u) = f(u) (u ≥ 0)  and  f_{−}(u) = f(−u) (u ≤ 0).

Then we have

F (u) = \{ \begin{matrix}F_{+}(u) = ∫_{0}^{u} f_{+}(x) dx & for u ≥ 0 \\ −F_{−}(−u) = − ∫_{0}^{−u} f_{−}(x) dx & for u < 0. \end{matrix}

Now equation (6) becomes

∫_{0}^{+∞} \frac{du}{\sqrt[p]{F_{+}(u) + F_{−}(v_{0})}} + ∫_{0}^{v_{0}} \frac{du}{\sqrt[p]{F_{−}(v_{0}) − F_{−}(u)}} = \frac{b − a}{2} ^{p} \sqrt{p′λ}  (9)

where v_{0} = −u_{0} > 0. First observe that the first integral in (9) is strictly decreasing in v_{0} if f_{−}(u) is non-negative. For the second integral in (9), using

∫_{0}^{v_{0}} \frac{du}{\sqrt[p]{F_{−}(v_{0}) − F_{−}(u)}} = ∫_{0}^{1} \frac{v_{0}ds}{\sqrt[p]{F_{−}(v_{0}) − F_{−}(sv_{0})}}

and

F_{−}(v_{0}) − F_{−}(sv_{0}) = ∫_{0}^{1} f_{−}((s + t − st)v_{0})(1 − s)v_{0} dt,

we see that it is decreasing or increasing in v_{0} if f_{−}(u)u^{1−p} is increasing or decreasing, respectively. Therefore, if f_{−}(u) is non-negative and f_{−}(u)u^{1−p} is increasing, then (9) has at most one solution, which gives the uniqueness of sign-changing solutions of problem (1). On the other hand, if f_{−}(u)u^{1−p} is decreasing, then the first and the second integrals in (9) will compete to each other and thus the existence of a multiple solution is possible. In particular, if

∫_{0} \frac{du}{\sqrt[p]{F_{+}(u)}} = +∞  and  lim_{u→+∞} f_{−}(u)u^{1−p} = 0,

then the left side in (9) goes to infinity, when v_{0} goes either to zero or to infinity. Hence there is λ_{−} > 0 such that problem (1) has at least two sign-changing solutions for λ > λ_{−}, and no sign-changing solution if λ < λ_{−}. A typical example is

f_{+}(u) = ∑ α_{i}u^{q_{i}}  and  f_{−}(u) = ∑ β_{j}u^{r_{j}}

where α_{i}, β_{j} > 0 and q_{i} > p − 1 > r_{j}. When f_{+} and f_{−} are simply given by αu^{q} and βu^{r} (q > p − 1 > r), respectively, then we have a complete charaterization of signchanging solutions, namely, two solutions if λ > λ_{−}, a unique solution if λ = λ_{−}, and no solutions if λ < λ_{−}. In other situations the multiplicity of sign-changing solutions can be very complicated and three representive examples are as in Theorems 1 - 3.

Proof of Theorem 1. To this end we study the function in the left-hand side of (9). In this case we have

F_{+}(u) = \frac{1}{q + 1} u^{q+1} + \frac{ε}{1 + s} u^{1+s}  and  F_{−}(u) = \frac{α}{r + 1} u^{r+1}

and

∫_{0}^{1} \frac{v_{0} ds}{\sqrt[p]{F_{−}(v_{0}) − F_{−}(sv_{0})}} = c_{2} v_{0}^{1− \frac{1+r}{p}} with c_{2} = ∫_{0}^{1} \frac{ds}{\sqrt[p]{α(1 − s^{r+1})/(1 + r)}}

and the function in (9) is

∫_{0}^{∞} \frac{du}{\sqrt[p]{\frac{1}{q+1} u^{q+1} + \frac{ε}{1+s} u^{1+s} + \frac{α}{1+r} v_{0}^{r+1}}} + c_{2} v_{0}^{1− \frac{r+1}{p}} =: F_{ε}(v_{0}).  (10)

The first conclusion of Theorem 1 follows from that F_{ε}(v_{0}) > 0 is continuous on [0, +∞) and goes to infinity, as v_{0} → ∞. To get a complete picture for the existence of signchanging solutions we let

k_{1} = (1 + r)(\frac{1}{s + 1} − \frac{1}{p}),  k_{2} = \frac{(1 + r)(q − s)}{1 + s},  k_{3} = 1 − \frac{r + 1}{p}

and

g(v_{0}) = ∫_{0}^{+∞}(\frac{1}{q + 1} u^{q+1} + \frac{ε}{1 + s} u^{1+s} + \frac{α}{1 + r} v_{0}^{r+1})^{− \frac{1}{p}} du.

Then F_{ε}(v_{0}) = g(v_{0}) + c_{2}v_{0}^{k_{3}}. We first study the property of g(v_{0}) in a neighbourhood of the origin.

By the change of variable u = x v0^{\frac{1+r}{1+s}} we get for v_{0} > 0

\begin{aligned}g(v_{0}) & = ∫_{0}^{+∞} \frac{v_{0}^{k_{1}}dx}{\sqrt[p]{\frac{1}{q+1} x^{q+1}v_{0}^{k_{2}} + \frac{ε}{1+s} x^{1+s} + \frac{α}{1+r}}} \\ g(0) & = ∫_{0}^{+∞} \frac{v_{0}^{k_{1}}dx}{\sqrt[p]{\frac{1}{q+1} x^{q+1}v_{0}^{k_{2}} + \frac{ε}{1+s} x^{1+s}}} . \end{aligned}

Consequently,

\frac{g(v_{0}) − g(0)}{v_{0}^{k_{1}}} = − ∫_{0}^{+∞}G(v_{0}, x) dx

where

G(v_{0}, x) = \frac{1}{\sqrt[p]{\frac{1}{q+1} x^{q+1}v_{0}^{k_{2}} + \frac{ε}{1+s} x^{1+s}}} − \frac{1}{\sqrt[p]{\frac{1}{q+1} x^{q+1}v_{0}^{k_{2}} + \frac{ε}{1+s} x^{1+s} + \frac{α}{1+r}}} .

Writing the difference as an integral we see

G(v_{0}, x) = \frac{α}{p(r + 1)} ∫_{0}^{1} (\frac{1}{q + 1} x^{q+1}v_{0}^{k_{2}} + \frac{ε}{1 + s} x^{1+s} + \frac{α}{1 + r} θ)^{− \frac{p+1}{p}} dθ

which implies that G(v_{0}, x) is decreasing in v_{0}. Since

G(0, x) = \frac{α}{p(r + 1)} ∫_{0}^{1} (\frac{ε}{1 + s} x^{1+s} + \frac{α}{1 + r} θ)^{− \frac{p+1}{p}} dθ

is integrable over (0, +∞), due to s ∈ (0, p−1), we deduce by the dominate convergence theorem that

lim_{v_{0}→0} \frac{g(v_{0}) − g(0)}{v_{0}^{k_{1}}} = − ∫_{0}^{+∞}G(0, x) dx < 0.

Thus we obtain that near the origin F_{ε} is increasing if k_{1} > 1 − \frac{r+1}{p}, which is equivalent to r > s, and is decreasing when r < s. Similarly, we get that

lim_{v_{0}→∞} g′(v_{0}) v_{0}^{1+(r+1)( \frac{1}{p} − \frac{1}{q+1} )} = −\frac{α}{p} ∫_{0}^{1} (\frac{1}{q + 1}x^{q+1} + \frac{α}{1 + r} )^{− \frac{p+1}{p}} dx

which implies that F_{ε} is increasing for large v_{0} > 0, because q > p − 1. Since \frac{dF_{ε}}{dv_{0}} is

increasing in ε and tends to (1 − \frac{r+1}{p}) c_{2} v_{0}^{− \frac{r+1}{p}} uniformly in v_{0} as ε → ∞, we deduce in the case r > s that there is an ε_{0} ≥ 0 such that min_{v_{0}} \frac{dF_{ε}}{dv_{0}} > 0 for ε > ε_{0}. Moreover, when ε = 0,

F_{0}(v_{0}) = c_{1}v_{0}^{1− \frac{q+1}{p}} + c_{2}v_{0}^{1− \frac{r+1}{p}}

for 1 − \frac{q+1}{p} < 0 < 1 − \frac{r+1}{p} and thus \frac{dF_{0}}{dv_{0}} < 0 for small v_{0} > 0. Further, F_{ε}(v_{0}) → F_{0}(v_{0}) on compacta in C^{1}(0, +∞) as ε → 0 and we assert that ε_{0} > 0 and min_{v_{0}} \frac{dF_{ε}}{dv_{0}} < 0 for ε ∈ (0, ε_{0}). Whence, we obtain that the sign-changing solution is unique when ε ≥ ε_{0}. If ε ∈ (0, ε_{0}), using the fact that F_{ε} is increasing for both small and large v_{0}, we deduce that there are 0 < v_{1} < v_{2} ≤ v_{3} < +∞ such that F_{ε} is increasing on (0, v_{1}) and (v_{3}, +∞) and is decreasing on (v_{1}, v_{2}). Thus let

λ_{1} = min F_{ε}(v_{0}), λ_{2} = F_{ε}(v_{1}), λ_{3} = F_{ε}(v_{2}), λ_{4} = max_{F′_{ε}(v_{0})=0} F_{ε}(v_{0}).

Then problem (1) has no sign-changing solutions if λ < λ_{1} and has at least two signchanging solutions λ ∈ (λ_{1}, λ_{4}). In particular, it has at least three sign-changing solutions if λ ∈ (λ_{2}, λ_{3}) and has a unique sign-changing solution as λ > λ_{4}. The proof is complete

Proof of Theorem 2. The idea is the same as in the proof of Theorem 1, so we will be sketch in many places. In view of

F_{+}(u) = \frac{1}{q + 1} u^{q+1} + \frac{ε}{1 + s} u^{1+s}  and  F_{−}(u) = \frac{α}{r + 1} u^{r+1} + \frac{δ}{τ + 1} u^{τ+1}

it follows that equation (9) takes the form

\begin{aligned}∫_{0}^{+∞} & \frac{du}{\sqrt[p]{\frac{1}{q+1} u^{q+1} + \frac{ε}{1+s} u^{1+s} + F_{−}(v_{0})}} \\ & + ∫_{0}^{1} \frac{v_{0}dx}{\sqrt[p]{r(x) v_{0}^{s+1} + τ(x) v_{0}^{τ+1}}} = \frac{b − a}{2} ^{p} \sqrt{αp′λ} \end{aligned}  (11)

where

r(x) = \frac{α}{r + 1} (1 − x^{r+1})  and  τ(x) = \frac{δ}{τ + 1} (1 − x^{τ+1}).

Likely as in the proof of Theorem 1 we define

L(v_{0}) = ℓ_{1}(v_{0}) + ℓ_{2}(v_{0})

where

\begin{aligned}ℓ_{1}(v_{0}) & = ∫_{0}^{+∞} \frac{du}{\sqrt[p]{\frac{1}{q+1} u^{q+1} + \frac{ε}{1+s} u^{1+s} + F_{−}(v_{0})}} \\ ℓ_{2}(v_{0}) & = ∫_{0}^{1} \frac{v_{0}dx}{\sqrt[p]{r(x) v_{0}^{r+1} + τ(x) v_{0}^{τ+1}}} . \end{aligned}

The key points to prove that the function L(v_{0}) have a graph as shown in Figure 2 are

L(v_{0}) = F_{0}(v_{0}) = c_{1} v0^{\frac{p−q−1}{p}} + c_{2} v0^{\frac{p−r−1}{p}} when ε = δ = 0

which has the property F′_{0}(v_{0})(v_{0} − v) > 0 for some v > 0 and L converges to F_{0} on compacta of (0, +∞) in C^{1} as ε, δ → 0 and L → 0 as v_{0} → +∞.

To complete the proof we need to show two more things: Monotonicity of L near the origin and L(0) ≤ L(v_{2}) where v_{2} is the second local maximum of L. But

L(0) = c_{1}ε^{\frac{q−p−1}{p(q−s)}}  and  L(δ^{\frac{1}{r−τ}}) ≥ ℓ_{2}(δ^{\frac{1}{r−τ}}) = c_{2}δ^{\frac{r+1−p}{p(τ−r)}}

for some constants c_{1} > 0 and c_{2} > 0 from which it follows that if ε and δ satisfy c_{1}ε^{\frac{q−p−1}{p(q−s)}} ≤ c_{2}δ^{\frac{r+1−p}{p(τ−r)}} , then we are done.

To get the monotonicity of L in the nearby of the origin, we exploit the technique in the proof of Theorem 1 and can show that

lim_{v_{0}→0} \frac{ℓ_{1}(v_{0}) − ℓ_{1}(0)}{v_{0}^{k_{1}}} = −\frac{α}{p(r + 1)} ∫_{0}^{+∞}dx ∫_{0}^{1} (\frac{ε}{1 + s} x^{s+1} + \frac{α}{r + 1} θ)^{− \frac{p+1}{p}} dθ < 0

where we have used that r > τ and F_{−}(v_{0}) = \frac{α}{r+1} v_{0}^{r+1}(1 + o(1)) as v_{0} → 0,

lim_{v_{0}→0} \frac{ℓ_{2}(v_{0}) − ℓ_{2}(0)}{v_{0}^{k_{3}}} = ∫_{0}^{1} \frac{dx}{\sqrt[p]{r(x)}} > 0  and  lim_{v_{0}→0} \frac{L(v_{0}) − L(0)}{v_{0}^{k_{1}}} < 0.

Finally, the uniqueness follows from the results in [4] that the function ℓ_{2} has the following property: there is v > 0 such that ℓ′_{2}(v_{0}) > 0 for v_{0} ∈ (0, v) and ℓ′_{2}(v_{0}) < 0 for v_{0} ∈ (v, +∞). Then L′(v_{0}) < 0 for v_{0} ∈ [v, +∞) since clearly ℓ′_{1}(v_{0}) < 0 for v_{0} > 0. The proof is complete

Proof of Theorem 3. By virtue of p = 2 and

F_{+}(u) = \frac{α}{q + 1} u^{q+1}  and  F_{−}(u) = \frac{1}{2} u^{2} + β(sin u − u cos u)

we have that

\begin{aligned}∫_{0}^{+∞}\frac{du}{\sqrt{F_{+}(u) + F_{−}(v_{0})}} = c(q)(\frac{1}{2}v_{0}^{2} + β(sin v_{0} − v_{0} cos v_{0}))^{\frac{1}{q+1} − \frac{1}{2}} & =: g_{1}(v_{0}) \\ ∫_{0}^{v_{0}} \frac{du}{\sqrt{F_{−}(v_{0}) − F_{−}(u)}} = ∫_{0}^{1} \frac{v_{0} dx}{\sqrt{h(x, v_{0})}} & =: g_{2}(v_{0}) \end{aligned}

534 Y. J. Cheng

where

h(x, v_{0}) = e(x) v_{0}^{2} + β ε(x, v_{0})
ε(x, v_{0}) = sin v_{0} − sin(xv_{0}) − v_{0}( cos v_{0} − x cos(xv_{0}))
c(q) = ∫_{0}^{+∞} (1 + \frac{α}{q + 1} u^{q+1})^{− \frac{1}{2}} du
e(x) = \frac{1 − x^{2}}{2}

and the function in the left-hand side of (9) takes the form

g_{1}(v_{0}) + g_{2}(v_{0}) := H(v_{0}).

Clearly, H is well defined and continuous on (0, +∞). Since q > 3, we see that

g_{1}(v_{0}) → +∞ as v_{0} → 0+

and

\begin{matrix}g_{1}(v_{0}) & → 0 \\ g_{2}(v_{0}) & → ∫_{0}^{1} \frac{dx}{\sqrt{e(x)}} = \frac{π}{\sqrt{2}} \end{matrix} \} as v_{0} → +∞

and therefore H(0+) = +∞ and H(+∞) = \frac{π}{\sqrt{2}}.

To show the results (ii) and (iii) we need to prove that H oscillates around the line H = \frac{π}{\sqrt{2}}, and first show that g_{1} oscillates around H = \frac{π}{\sqrt{2}}.

Choosing v_{0} = nπ (n ∈ ℕ is even) we get

h(x, nπ) = (nπ)^{2}e(x) − β( sin(nπx) + nπ(1 − x cos(nπx)))
≤ (nπ)^{2}e(x) − β(nπ(1 − x) − sin (nπ(1 − x)))
≤ (nπ)^{2}e(x)

thereafter

g_{2}(nπ) > ∫_{0}^{1} \frac{nπdx}{\sqrt{e(x)(nπ)^{2}}} = \frac{π}{\sqrt{2}}.

Analogously, for odd integer n, g_{2}(nπ) < \frac{π}{\sqrt{2}}.

To show the function H is also oscillating, it suffices to show that, for odd n, H(nπ) < \frac{π}{\sqrt{2}} since g_{1} > 0 and g_{2}(nπ) > \frac{π}{\sqrt{2}} for even n. For this purpose, first we have

\frac{π}{\sqrt{2}} − g_{2}(nπ) = ∫_{0}^{1} (\frac{nπ}{\sqrt{e(x)(nπ)^{2}}} − \frac{nπ}{\sqrt{e(x)(nπ)^{2} + β ε(x, nπ)}}) dx

= \frac{1}{2} ∫_{0}^{1} dx ∫_{0}^{1} \frac{nπβ ε(x, nπ)}{(e(x)(nπ)^{2} + θβ ε(x, nπ))^{\frac{3}{2}}} dθ

≥ \frac{1}{2} ∫_{0}^{1} \frac{nπβ ε(x, nπ)}{(e(x)(nπ)^{2} + β ε(x, nπ))^{\frac{3}{2}}} dx,

Some Results on an Elliptic Blow-Up Problem 535

due to ε(x, nπ) > 0 for all x ∈ (0, 1). Using f_{−}(u) ≥ (1 − β)u, we deduce that

e(x)(nπ)^{2} + β ε(x, nπ) = nπ(1 − x) ∫_{0}^{1} f_{−}((x + t − tx)nπ) dt

≥ nπ(1 − x) ∫_{0}^{1}(1 − β)(x + t − tx)nπ dt

≥ \frac{1}{2}(1 − β)(1 − x)(nπ)^{2}

for all x ∈ (0, 1). Thereafter

\frac{π}{\sqrt{2}} − g_{2}(nπ) ≥ \frac{2^{\frac{3}{2}} β}{(1 − β)^{\frac{3}{2}}} (nπ)^{−2} ∫_{0}^{1} \frac{ε(x, nπ)}{(1 − x)^{\frac{3}{2}}} dx,

and by a change of variable z = \frac{1}{\sqrt{1−x}} (then x = 1 − \frac{1}{z^{2}} )

∫_{0}^{1} \frac{ε(x, nπ)}{(1 − x)^{\frac{3}{2}}} dx = 2 ∫_{1}^{+∞}ε(1 − \frac{1}{z^{2}} , nπ)dz = 2 ∫_{1}^{+∞}( − sin (\frac{nπ}{z^{2}} ) + nπ(1 − (1 − \frac{1}{z^{2}} ) cos \frac{nπ}{z^{2}} ))dz = 2 ∫_{1}^{+∞}(nπ(1 − cos \frac{nπ}{z^{2}} ) + (− sin \frac{nπ}{z^{2}} + \frac{nπ}{z^{2}} cos \frac{nπ}{z^{2}} ))dz cr

where we have used n being an odd integer. Change variables once more, we obtain

∫_{1}^{+∞}(1 − cos \frac{nπ}{z^{2}} )dz = ∫_{\frac{1}{\sqrt{nπ}}}^{+∞}(1 − cos \frac{1}{w^{2}} )\sqrt{nπ} dw ≥ C\sqrt{nπ}

where C = ∫_{1}^{+∞}(1 − cos \frac{1}{z^{2}} ) dz. On the other hand,

∫_{1}^{+∞}( − sin \frac{nπ}{z^{2}} + \frac{nπ}{z^{2}} cos \frac{nπ}{z^{2}} )dz ≤ ∫_{1}^{+∞}\frac{2nπ}{z^{2}} dz = 2nπ.

Whence

∫_{0}^{1} \frac{ε(x, nπ)}{(1 − x)^{\frac{3}{2}}} dx ≥ 2(Cnπ\sqrt{nπ} − 2nπ)

and consequetly for large n the estimate

\frac{π}{\sqrt{2}} − g_{2}(nπ) ≥ \frac{2^{\frac{3}{2}} β}{(1 − β)^{\frac{3}{2}}} C(nπ)^{− \frac{1}{2}}

holds. Combing the above estimates, we obtain by noting g_{1}(nπ) ≤ C_{1}(nπ)^{\frac{2}{q+1} −1} that

\frac{π}{\sqrt{2}} − H(nπ) = \frac{π}{\sqrt{2}} − g_{2}(nπ) − g_{1}(nπ) ≥ C_{2}(nπ)^{− \frac{1}{2}} − C_{1}(nπ)^{\frac{2}{q+1} −1} > 0

since −\frac{1}{2} > \frac{2}{q+1} − 1 (this is the source for the condition on q).

Assertions (ii) and (iii) follow from the oscillatory property of H. Using

lim_{v_{0}→∞} H(v_{0}) = \frac{π}{\sqrt{2}},  lim_{v_{0}→0+} H(v_{0}) = +∞,  H(nπ) < \frac{π}{\sqrt{2}}

we deduce from the continuity of H that the minimum of H(v_{0}) on (0, +∞) achives and thus the equation H(v_{0}) = μ is solvable if and only if μ ≥ min H(v_{0}) which complete the proof of assertion (i)

4. Final remarks

In this note we have only carried out some basic calculations to exhibit the rich structure for boundary blow-up problems, even it is very elementary (just calculus), but it is certainly not easy to give a complete bifurcation picture for all involved parameters, for instance q, r, s, τ, ε, δ, α in Theorem 2. From our one-dimensional examples we can see that there is a big difference between Dirichlet boundary value problems and boundary bolw-up problems. If the boundary is Dirichlet, then there are infinitely many sign solutions in the superlinear case, but it can have only finite number of sign solutions for boundary bolw-up problems (note, the function in (2) is not superlinear at −∞). As our examples are of one-dimensional character, one may say that it would not be representive, so it will be interesting to study those equations in two-dimensional or higher dimensional domains.

References

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[3] Bandle, C. and M. Marcus: Asymptotic behaviour of solutions and their derivatives, for semilinear elliptic problems with boundary blow-up. Ann. Inst. Henri Poicaré 12 (1995), 155 – 171.

Some Results on an Elliptic Blow-Up Problem 537

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[7] Matero, J.: Nonlinear Elliptic Problems with Boundary Blow-Up. Diss. in Math. 6. Uppsala: Univ. 1997.

[8] Mckenna, P. J., Reichel, W. and W. Walter: Symmetry and multiplicity for nonlinear elliptic differential equations with boundary blow-up. Nonlin. Anal. Theory, Methods and Appl. 28 (1997), 1213 – 1225.

Received 26.10.1998; in revised form 21.04.1999