USA-Chile Workshop on Nonlinear Analysis, Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 359–367. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

Existence of non-negative solutions for a Dirichlet problem ^{∗}

Cecilia S. Yarur

Abstract The aim of this paper is the study of existence of non-negative solutions of fundamental type for some systems without sign restrictions on the non linearity.

1 Introduction

We study the existence of non-negative non-trivial solutions to the boundaryvalue problem

Δu = a_{2}v^{p_{2}} − a_{1}v^{p_{1}} in B′
Δv = b_{2}u^{q_{2}} − b_{1}u^{q_{1}} in B′  (1.1)
u = v = 0 on∂B ,

where a_{i}, b_{i} are non-negative constants, p_{i} > 0, q_{i} > 0 for i = 1, 2, B is the unit ball centered at zero in ℝ^{N}, N ≥ 3, and B′ = B \ {0}.

The above problem involves many problems of a quite different nature depending on the values of a_{i}, b_{i}. For instance, if a_{2} = b_{2} = 0 the solutions u, v are sub-harmonic functions, while if a_{1} = b_{1} = 0 the solutions are super-harmonic.

For a better understanding of this system, we recall that P.L. Lions [12], Ni and Sacks [15], and Ni and Serrin [16], studied conditions for existence or non existence of non-negative solutions u to

−Δu = u^{q} in B′, u = 0 on ∂B.  (1.2)

The range of existence of solutions to (1.2) is q < (N + 2)/(N − 2). On the other hand, the problem

Δu = u^{q} in B′, u = 0 on ∂B,

has a non-negative non-trivial solution if and only if q < N/(N − 2), see [5] for the non-existence and [18] for existence and related problems.

∗Mathematics Subject Classifications: 35C20, 35D10.

Key words: Dirichlet Problem, Non-negative solutions.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University.

Published January 8, 2001.

Partially supported by Fondecyt grant 1990877 and DICYT

We state next some known results concerning particular cases of problem (1.1). Assume first that a_{1} = b_{1} = 0. Thus, we are concerned with

−Δu = a_{2}v^{p_{2}} in B′
−Δv = b_{2}u^{q_{2}} in B′  (1.3)
u = v = 0 on ∂B.

The following result is well known.

Theorem 1.1 Assume that a_{1} = b_{1} = 0, p_{2}q_{2} > 1 and a_{2} > 0, b_{2} > 0. Then, there exists a classical solution to (1.3) if and only if

\frac{N}{p_{2} + 1} + \frac{N}{q_{2} + 1} > N − 2.

Troy [17] proved radial symmetry of positive classical solutions to problem (1.3). The existence of positive classical solutions of (1.3) was studied by Hulshof and van der Vorst [11] and de Figueiredo and Felmer [9]. The behavior of solutions was studied by Bidaut-Véron in [2]. The existence of some singular solutions, that is solutions with either lim sup_{x→0} u(x) = +∞ or lim sup_{x→0} v(x) = +∞, is given by García-Huidobro, Manásevich, Mitidieri and Yarur, see [10]. Using Pohozaev-Pucci-Serrin type identity, Mitidieri [13, 14] and van der Vorst [19] proved non existence of classical solutions of (1.3). Non-existence of radially symmetric singular positive solutions was given by Garcia-Huidobro, Manásevich, Mitidieri and Yarur in [10].

We note that since u and v are super-harmonic functions, and due to a result of Brezis and P.L.Lions [1], u^{q_{2}} ∈ L^{1}(B) , v^{p_{2}} ∈ L^{1}(B) and there exist c ≥ 0 and d ≥ 0 such that

−Δu = a_{2}v^{p_{2}} + cδ_{0} in D′(B)
−Δv = b_{2}u^{q_{2}} + dδ_{0} in D′(B)
u = v = 0 on ∂B.

If (c, d) ≠ (0, 0) we call this singularity of fundamental type.

Let us consider now a_{1} = b_{2} = 0, in (1.1). Hence, we are looking for the solutions of:

−Δu = a_{2}v^{p_{2}} in B′
Δv = b_{1}u^{q_{1}} in B′  (1.4)
u = v = 0 on ∂B.

Since v is sub-harmonic, there exists no non-negative classical solutions to (1.4). The following result is given in [6] for p_{2}q_{1} > 1 and in [7] for p_{2}q_{1} < 1.

Theorem 1.2 Assume a_{1} = b_{2} = 0, p_{2} > 0, q_{1} > 0, and p_{2}q_{1} ≠ 1. Then there exists a non-trivial non-negative solution to (1.4) if and only if

\frac{N}{p_{2} + 1} + \frac{N − 2}{q_{1} + 1} > N − 2, and p_{2} < N/(N − 2).

The above result is based on the results given in [3].

We say that (u, v) has a strong singularity at 0 if either

lim sup_{x→0} |x|^{N−2}u(x) = +∞ or lim sup_{x→0} |x|^{N−2}v(x) = +∞ .

It can be proved that there exists a region in the plane p_{2} − q_{1} where there exist both strong and fundamental non-negative singular solutions, see [7]. This region is given by

\frac{N − 2}{p_{2} + 1} + \frac{N}{q_{1} + 1} > N − 2, and p_{2} < N/(N − 2) < q_{1} .

Assume now that a_{2} = b_{2} = 0, and thus the problem (1.1) is

Δu = a_{1}v^{p_{1}} in B′
Δv = b_{1}u^{q_{1}} in B′  (1.5)
u = v = 0 on ∂B,

Since u and v are sub-harmonic we have non existence of non-negative solutions with either u or v bounded.

In [4] and [20] it was proved non existence of positive solutions if either

\frac{N}{p_{1} + 1} + \frac{N − 2}{q_{1} + 1} ≤ N − 2, or \frac{N − 2}{p_{1} + 1} + \frac{N}{q_{1} + 1} ≤ N − 2.

If a_{1} = 0 ( similarly for b_{1} = 0) we have

−Δu = a_{2}v^{p_{2}} in B′
−Δv = b_{2}u^{q_{2}} − b_{1}u^{q_{1}} in B′  (1.6)
u = v = 0 on ∂B  (1.7)

The following result was proved in [8].

Theorem 1.3 Let p_{2} > 0, q_{1} > 0 and q_{2} > 0. Let a_{1} = 0, a_{2} ≥ 0, b_{1} ≥ 0 and b_{2} ≥ 0. Assume that for i = 1, 2 we have

p_{2} < \frac{N}{N − 2}, \frac{N}{p_{2} + 1} + \frac{N − 2}{q_{i} + 1} > N − 2.  (1.8)

Assume that one of the following holds:

(i) p_{2}q_{i} > 1 for all i = 1, 2.

(ii) p_{2}q_{i} < 1, for all i = 1, 2.

(iii) If p_{2}q_{i} = 1 for some i = 1, 2 then a_{2}^{p_{2}}b_{i} is sufficiently small.

(iv) p_{2}q_{i} < 1 < p_{2}q_{j}, for some i, j = 1, 2, i ≠ j, and a_{2}^{p_{2}}b_{i} is sufficiently small.

Then, there exist d_{∗} ≥ 0, d^{∗} > 0 with d_{∗} < d^{∗} such that for any d ∈ (d_{∗}, d^{∗}), there exists (u, v) a non-negative solution to (1.6) satisfying

lim_{x→0} |x|^{N−2}(u(x), v(x)) = (0, d).

Moreover, if p_{2}q_{i} ≥ 1, i = 1, 2 then d_{∗} = 0, and if p_{2}q_{i} ≤ 1, i = 1, 2 then d^{∗} = ∞.

For the general case we have the following previous result, see [8].

Theorem 1.4 Let p_{1} > 0, p_{2} > 0, q_{1} > 0 and q_{2} > 0. Let a_{i}, b_{i} i = 1, 2 be non-negative constants. Assume that

p_{i} < \frac{N}{N − 2}, q_{i} < \frac{N}{N − 2}, i = 1, 2  (1.9)

Assume that one of the following holds:

(i) p_{i}q_{j} > 1, for all i, j = 1, 2.

(ii) p_{i}q_{j} < 1, for all i, j = 1, 2.

(iii) If p_{i}q_{j} = 1 for some i = 1, 2 and some j = 1, 2 then a_{i}^{p_{i}}b_{j} is sufficiently small.

(iv) p_{i}q_{j} < 1 < p_{k}q_{l}, for some i, j, k, l = 1, 2 and a_{i}^{p_{i}}b_{j} is sufficiently small.

Then, there exist c > 0, d > 0 and (u, v) a non-negative solution to (1.1) such that

lim_{x→0} |x|^{N−2}(u(x), v(x)) = (c, d).

Here we prove the following general existence result of non negative nontrivial solutions to (1.1). Set

Γ(p, q) := \frac{N − 2}{p + 1} + \frac{N}{q + 1} − (N − 2).  (1.10)

Theorem 1.5 Let p_{i}, q_{i}, i = 1, 2, positive numbers. Then, there exists a nonnegative nontrivial solution (u, v) of (1.1) if one of the following holds:

(i) a_{1} > 0, b_{1} > 0, and p_{2} < N/(N − 2), q_{2} < N/(N − 2)

min{Γ(p_{1}, q_{1}), Γ(q_{1}, p_{1}), Γ(p_{2}, q_{1}), Γ(q_{2}, p_{1})} > 0,

with small coefficient a_{j} ( respectively b_{j} ) for some j = 1, 2 if p_{j} ≤ 1 ( respectively q_{j} ≤ 1) and 1 ≤ max_{i=1,2}{p_{i}, q_{i}}.

(ii) a_{1} = 0, b_{1} > 0, p_{2} < N/(N − 2) and

min{Γ(q_{1}, p_{2}), Γ(q_{2}, p_{2})} > 0,

with small coefficient a_{2} ( respectively b_{j}) if p_{2} ≤ 1 (respectively q_{j} ≤ 1) for some j = 1, 2, and 1 ≤ max_{i=1,2}{p_{2}, q_{i}}.

(iii) a_{1} = 0 = b_{1}, and

max{Γ(p_{2}, q_{2}), Γ(q_{2}, p_{2})} > 0,

with small coefficient a_{2} ( respectively b_{2}) if p_{2} ≤ 1 (respectively q_{2} ≤ 1) and 1 ≤ max{p_{2}, q_{2}}.

2 Proof of Theorem 1.5

We note that for p and q non-negative numbers the condition Γ(p, q) > 0 is equivalent to

p(2 − (N − 2)q) + N > 0.

Moreover, if pq > 1,

Γ(p, q) = (ζ − (N − 2))(pq − 1), ζ = \frac{2(p + 1)}{pq − 1}

Γ(q, p) = (ξ − (N − 2))(pq − 1), ξ = \frac{2(q + 1)}{pq − 1}.

Recall that u(x) = C_{1}|x|^{−ζ}, v(x) = C_{2}|x|^{−ξ} for some positive constants C_{1} and C_{2} is a non-negative solution of

−Δu = v^{p}, −Δv = u^{q}

if Γ(p, q) < 0 and Γ(q, p) < 0. This particular solution also plays a fundamental role for example for the system

−Δu = v^{p}, Δv = u^{q},

where this solution exists if Γ(p, q) < 0 and Γ(q, p) > 0.

Proof of Theorem 1.5. Set

f_{i}(t) = a_{i}t^{p_{i}}, g_{i}(t) = b_{i}t^{q_{i}}, i = 1, 2.

We will construct radially symmetric non-negative solutions to (1.1), by monotone iteration as follows. Let d > 0, (u_{1}, v_{1}) = (0, dm), where m(r) := |x|^{2−N} −1 and let (u_{n}, v_{n}) be given by (u_{n+1}, v_{n+1}) = T (u_{n}, v_{n}) where T = (T_{1}, T_{2}) is the operator given by

T_{1}(u, v)(r) = ∫_{r}^{1} s^{1−N} ∫_{s}^{1} t^{N−1}f_{1}(v(t))dtds
+ ∫_{r}^{1} s^{1−N} ∫_{0}^{s} t^{N−1}f_{2}(v(t))dtds,  (2.1)
T_{2}(u, v)(r) = dm(r) + ∫_{r}^{1} s^{1−N} ∫_{s}^{1} t^{N−1}g_{1}(u(t))dtds
+ ∫_{r}^{1} s^{1−N} ∫_{0}^{s} t^{N−1}g_{2}(u(t))dtds,

We are looking for α, δ and C such that

T_{1}(Cr^{α}, Cr^{δ}) ≤ Cr^{α}, T_{2}(Cr^{α}, Cr^{δ}) ≤ Cr^{δ}

and v_{1} = dm(r) ≤ Cr^{δ}. Hence, the sequence (u_{n}, v_{n}) satisfies

u_{n} ≤ Cr^{α}, v_{n} ≤ Cr^{δ} for all n ∈ ℕ,

and the convergence of (u_{n}, v_{n}) to a solution of (1.1) follows.

To find C, d, α and δ we use the following: Let κ be any number such that κ + N ≠ 0, and define

ϕ(κ) := min{2 − N, κ + 2}.

Then

m_{κ}(r) := ∫_{r}^{1} s^{1−N} ∫_{s}^{1} t^{N−1+κ}dtds ≤ Kr^{ϕ(κ)},  (2.2)

where K = K(N, κ).

Moreover, for any κ satisfying κ + N > 0, and κ + 2 ≠ 0, set

ψ(κ) := min{0, κ + 2}.

We have

h_{κ} := ∫_{r}^{1} s^{1−N} ∫_{0}^{s} t^{N−1+κ}dtds ≤ Kr^{ψ(κ)},  (2.3)

where K = K(N, κ). Hence,

T_{1}(Cr^{α}, Cr^{δ}) = a_{1}C^{p_{1}}m_{p_{1}δ} + a_{2}C^{p_{2}}h_{p_{2}δ}.

From (2.2) and (2.3) and if we choose p_{1}δ + N ≠ 0 and p_{2}δ + N > 0 we obtain

T_{1}(Cr^{α}, Cr^{δ}) ≤ K (a_{1}C^{p_{1}}r^{ϕ(p_{1}δ)} + a_{2}C^{p_{2}}r^{ψ(p_{2}δ)}) .  (2.4)

We note that ϕ(p_{1}δ) ≤ 2 − N < ψ(p_{2}δ), and thus

T_{1}(Cr^{α}, Cr^{δ}) ≤ K (a_{1}C^{p_{1}} + a_{2}C^{p_{2}}) r^{σ},  (2.5)

where

σ := \{ \begin{matrix}ϕ(p_{1}δ) & if a_{1} ≠ 0, \\ ψ(p_{2}δ) & if a_{1} = 0. \end{matrix}

Therefore, if α ≤ σ and K(a_{1}C^{p_{1}} +a_{2}C^{p_{2}}) ≤ C, we obtain T_{1}(Cr^{α}, Cr^{δ}) ≤ Cr^{α}. Arguing as above, we have

T_{2}(Cr^{α}, Cr^{δ}) ≤ dr^{2−N} + K (b_{1}C^{q_{1}}r^{ϕ(q_{1}α)} + b_{2}C^{q_{2}}r^{ψ(q_{2}α)}) ,  (2.6)

with q_{1}α + N ≠ 0, q_{2}α + 2 ≠ 0, and q_{2}α + N > 0. Therefore,

T_{2}(Cr^{α}, Cr^{δ}) ≤ (d + K (b_{1}C^{q_{1}} + b_{2}C^{q_{2}})) r^{ρ},  (2.7)

where

ρ := \{ \begin{matrix}ϕ(q_{1}α) & if b_{1} ≠ 0, \\ 2 − N & if b_{1} = 0. \end{matrix}

Hence,

T_{2}(Cr^{α}, Cr^{δ}) ≤ Cr^{δ},

if δ ≤ ρ and d + K(b_{1}C^{q_{1}} + b_{2}C^{q_{2}}) ≤ C.

Next we prove the existence of α, δ, C and d under the hypothesis of the theorem. The existence of C and d, is classical. We can choose d = C/2 and for i = 1, 2

Ka_{i}C^{p_{i}} ≤ C/2, Kb_{i}C^{q_{i}} ≤ C/4.

Therefore, if either for all i, p_{i} < 1 and q_{i} < 1, or p_{i} > 1 and q_{i} > 1, the existence of C follows. By the contrary if max{p_{i}, q_{i}, i = 1, 2} ≥ 1 and min{p_{i}, q_{i}, i = 1, 2} ≤ 1, we obtain existence with a restriction on the coefficients.

We summarize the conditions that α and δ must satisfy as follows:

α ≤ \{ \begin{matrix}min\{2 − N, p_{1}δ + 2\} & if a_{1} ≠ 0 \\ min\{0, p_{2}δ + 2\} & if a_{1} = 0 \end{matrix}

δ ≤ \{ \begin{matrix}min\{2 − N, q_{1}α + 2\} & if b_{1} ≠ 0 \\ 2 − N & if b_{1} = 0. \end{matrix}

Moreover, we need that

p_{2}δ + N > 0, q_{2}α + N > 0,  (2.8)

We also used that p_{1}δ + N ≠ 0, q_{1}α + N ≠ 0, p_{2}δ + 2 ≠ 0, q_{2}α + 2 ≠ 0. These last conditions are not relevant since we can take α and δ smaller and hence these new α and δ satisfy the conditions.

Case (i). Assume first that a_{1} > 0 and b_{1} > 0. If p_{1} < N/(N − 2), and since Γ(p_{1}, q_{1}) > 0, and Γ(p_{2}, q_{1}) > 0, we can take

α = 2 − N, δ = min{2 − N, 2 − (q_{1} − ε)(N − 2)},

where ε > 0 is such that

Γ(p_{1}, q_{1} − ε) > 0, and Γ(p_{2}, q_{1} − ε) > 0.

Now, since q_{2} < N/(N − 2), we have that q_{2}α + N > 0. From p_{2} < N/(N − 2) and Γ(p_{2}, q_{1} − ε) > 0, we also have p_{2}δ + N > 0. It remains to prove that α = 2 − N ≤ p_{1}δ + 2, which follows easily from Γ(p_{1}, q_{1} − ε) > 0.

If p_{1} ≥ N/(N − 2), from Γ(p_{1}, q_{1}) > 0 we deduce that q_{1} < N/(N − 2). Thus, we may proceed as before but now with

δ = 2 − N, α = p_{1}(2 − N) + 2.

Case (ii). Assume that a_{1} = 0 and b_{1} > 0. Let us choose

δ = 2 − N, α = min{0, 2 − (p_{2} − ε)(N − 2)},

where ε > 0 is such that Γ(q_{1}, p_{2} − ε) > 0 and Γ(q_{2}, p_{2} − ε) > 0. Then the conclusion follows as in the above case.

Case (iii). Assume that b_{1} = 0 and a_{1} = 0. Assume that p_{2} ≤ q_{2}. Since Γ(q_{2}, p_{2}) > 0, then p_{2} < N/(N − 2). Let us choose

δ = 2 − N, and α = min{0, 2 − (p_{2} − ε)(N − 2)},

and thus the conclusion follows by taking Γ(q_{2}, p_{2} − ε) > 0.

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Cecilia S. Yarur Departamento de Matematicas Universidad de Santiago de Chile Casilla 307, Correo 2, Santiago, Chile email: cyarur@fermat.usach.cl