Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 51, pp. 1–11. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.51 CONVEXITY OF SOLUTIONS TO ELLIPTIC PDE’S BENYAM MEBRATE, GIOVANNI PORRU Abstract. This article concerns the convexity or concavity of solutions to special second order elliptic partial differential equations in convex domains. We concentrate our investigation to boundary blow up solutions as well as to solutions of particular singular equations. Following a method due to Korevaar and Kennington, we find a new sufficient condition for proving convexity or concavity. This sufficient condition is useful when the semilinear component of the equation is the sum of two or more terms. 1. Introduction The notion of convexity of a function is useful in many branches of Mathematics. In this paper we address the following question: when a solution to a second order elliptic equation in a convex domain is convex? The answer is trivial in dimension n = 1 only. For a discussion of this problem for a general n we refer to the monograph of Kawohl [14] and references therein. It happens that the answer depends on the boundary conditions as well as on the structure of the equation. Brascamp and Lieb [5] proved that, if u(x) is a solution to the eigenvalue problem ∆u+ λu = 0 u > 0 in Ω, u = 0 on ∂Ω, then v(x) = log u(x) is concave provided Ω is convex. Here λ denotes the first eigenvalue. In [16], it is proved that if u(x) is a solution to the problem ∆u+ uq = 0, u > 0 in Ω, u = 0 on ∂Ω, 0 ≤ q < 1, then v(x) = u 1−q 2 (x) is concave provided Ω is convex. This result can be extended (with the same proof) to the case −1 < q < 1. For the general case ∆u+ f(u) = 0 one may find a special function ϕ(t) (depending on f) so that v = ϕ(u(x)) is concave [14]. Concavity results have been found for p-Laplace equations [4, 13, 21] as well as for some fully nonlinear equations [1]. A nice method to study convexity has been established by Korevaar [17] and Kennington [16]. Another powerful method uses the so called constant rank Theo- rem [2, 6, 18]. We also recall a method which uses the concave envelop of a function [8]. For more convexity results we refer to [3, 7, 10, 11, 13] and references therein. In [12] the authors show that if the solution u of ∆u + 1 = 0 in Ω with u = 0 on ∂Ω is such that √ maxu− u(x) is convex then Ω must be an ellipsoid. 2020 Mathematics Subject Classification. 35E10, 35J60, 35B50. Key words and phrases. Convexity; maximum principle; boundary blow-up; singular problems. ©2024. This work is licensed under a CC BY 4.0 license. Submitted March 25, 2024. Published September 6, 2024. 1 2 B. MEBRATE, G. PORRU EJDE-2024/51 In Section 1 we consider the problem ∆u = f(u) in Ω, u → +∞ as x → ∂Ω. This problem has been investigated in the seminal papers [15, 19]. The solution u(x) (if it exists) is named boundary blow-up solution or large solution. It is known that, if the domain Ω is convex and if f satisfies suitable conditions, then u(x) is convex. We are able to prove that, under some additional condition on f , also log u is convex. Next, for proving convexity, we describe a new method which uses a different condition on f . This new condition is very useful when f is the sum of two or more terms. In Section 2 we consider the problem ∆u+ f(u) = 0, u(x) > 0 in Ω, u = 0 on ∂Ω, uν = −∞ on ∂Ω. This kind of problems, named singular problems, are discussed in [9, 20]. If the domain Ω is convex and if f satisfies suitable conditions, the solution u(x) exists and it is concave. Also for this problem, we describe a new method for studying concavity that uses a different condition on f . 2. Blow-up problems Let f(t) be smooth, positive and increasing for t > 0, with f(0) = 0. If∫ ∞ 1 (F (t))−1/2dt < ∞, F (t) = ∫ t 0 f(τ)dτ, then, the boundary blow-up problem ∆u(x) = f(u(x)) in Ω, u = +∞ on ∂Ω, (2.1) has a positive solution [15, 19]. In this article, Ω is assumed to be bounded and convex. To investigate the convexity of u(x), define the concavity function c(x, y) = 2u(z)− u(x)− u(y), x, y ∈ Ω, z = x+ y 2 . Clearly, u(x) is convex if and only if c(x, y) ≤ 0 in Ω× Ω. By the condition u(x) = +∞ on ∂Ω, we have c(x, y) < 0 on ∂(Ω × Ω). To show that c(x, y) ≤ 0 in Ω × Ω, it is enough to prove that c(x, y) cannot have any positive maximum in Ω× Ω. The following result is well known (see [16, 17]). Theorem 2.1 (Korevaar-Kennington). Let f(t) be smooth and positive for t > 0, with f(0) = 0. We also assume f(t) to be strictly increasing and harmonic concave. If u ∈ C2(Ω) is a solution to Problem (2.1), then the corresponding concavity function c(x, y) cannot have any positive maximum in Ω× Ω. If f satisfies some additional conditions, we can improve Theorem 2.1 as follows. Theorem 2.2. Let f(t) be smooth and positive for t > 0, with f(0) = 0. We also assume that f(et) et is strictly increasing and harmonic concave. If u ∈ C2(Ω) is a solution to Problem (2.1), then the function v(x) = log u(x) is convex. Recall that f(et)/et is harmonic concave if and only if et/f(et) is convex, that is, if f2(t)− 3tf ′(t)f(t)− t2f ′′(t)f(t) + 2t2(f ′(t))2 ≥ 0. (2.2) The latter condition is satisfied by f(t) = tp with p > 1. EJDE-2024/51 CONVEXITY OF SOLUTIONS TO ELLIPTIC PDE’S 3 Proof of theorem 2.2. Putting v = log u, our problem reads as ∆v = −|Dv|2 + f(ev) ev in Ω, v = +∞ on ∂Ω. Clearly, the corresponding concavity function c(x, y) = 2v(z)− v(x)− v(y), z = x+ y 2 , is negative on ∂(Ω×Ω). Let us prove that c(x, y) cannot have any positive maximum in Ω × Ω. Arguing by contradiction, suppose c(x, y) has a positive maximum at some point (x, y) ∈ Ω× Ω. At this point, we have Dxc(x, y) = Dv(z)−Dv(x) = 0, Dyc(x, y) = Dv(z)−Dv(y) = 0. Moreover, we find that Dxxc(x, y) = 1 2 ∆v(z)−∆v(x), Dxyc(x, y) = 1 2 ∆v(z), Dyyc(x, y) = 1 2 ∆v(z)−∆v(y). (2.3) Following [16], define Lc(x, y) = r2Dxxc(x, y) + 2rsDxyc(x, y) + s2Dyyc(x, y). (2.4) We suppose that r2 + s2 > 0. The corresponding coefficient (2n× 2n) matrix is[ r2I rsI rsI s2I ] , where I is the n×n unitary matrix. The eigenvalues of this matrix are 0 (computed n times) and r2 + s2 (computed n times), hence, it is non-negative. Inserting the values of Dxxc(x, y), Dxyc(x, y) and Dyyc(x, y) given in (2.3) into (2.4), we have Lc(x, y) = r2 (1 2 ∆v(z)−∆v(x) ) + rs∆v(z) + s2 (1 2 ∆v(z)−∆v(y) ) = (r + s)2 2 ∆v(z)− r2∆v(x)− s2∆v(y). On using our equation and recalling that Dv(x) = Dv(z) = Dv(y), we have Lc(x, y) = −|Dv(z)|2 [ (r + s)2 2 − r2 − s2 ] + (r + s)2 2 f(ev(z)) ev(z) − r2 f(ev(x)) ev(x) − s2 f(ev(y)) ev(y) . On noting that − [ (r + s)2 2 − r2 − s2 ] = (r − s)2 2 ≥ 0, we find Lc(x, y) ≥ (r + s)2 2 f(ev(z)) ev(z) − r2 f(ev(x)) ev(x) − s2 f(ev(y)) ev(y) . 4 B. MEBRATE, G. PORRU EJDE-2024/51 Putting H(t) := f(et)/et, we find that Lc(x, y) ≥ (r + s)2 2 H(v(z))− r2H(v(x))− s2H(v(y)). Since we are assuming that u(z) > u(x) + u(y) 2 and since H(t) is strictly increasing, we have Lc(x, y) > (r + s)2 2 H (v(x) + v(y) 2 ) − r2H(v(x))− s2H(v(y)). With a = v(x) and b = v(y), we choose r = H(b), s = H(a). After some computations, we find that Lc(x, y) > (r + s) [H(a) +H(b) 2 H (a+ b 2 ) −H(a)H(b) ] . Finally, since H(t) is harmonic concavity, Lc(x, y) > 0. But, at (x, y) (point of maximum) we have Lc(x, y) ≤ 0, a contradiction. The proof is complete. □ By Theorem 2.2, if u(x) is a blow-up solution to ∆u = up, p > 1, the function v(x) = log u(x) is convex. In case u(x) ≥ 1, we may ask if also z = ( log u(x) )β for β ∈ (0, 1) is convex. The answer is negative in general. Indeed, for n > 2 consider the problem ∆u = n(n− 2)u n+2 n−2 , u = +∞ on ∂Ω. If Ω is the unit ball centered at the origin, a (radial) solution is u(r) = (1− r2) 2−n 2 , r = |x|. The function v = log u = 2−n 2 log(1− r2), for r near to 0, behaves like n−2 2 r2, and r2β is not convex for 0 < β < 1 2 . Theorem 2.2 cannot be applied when f(t) = t logα(t + 1), α > 2, because this function does not satisfy condition (2.2). The following is a weaker result. Theorem 2.3. Let 0 < β < 1. Let f(t) be smooth and positive for t > 0, with f(0) = 0, and such that t β−1 β f(t 1 β ) is strictly increasing and harmonic concave. If u ∈ C2(Ω) is a positive solution to Problem (2.1), then the function v(x) = uβ(x) is convex. Note that t β−1 β f(t 1 β ) is harmonic concave if and only if t 1−β β ( f(t 1 β ) )−1 is convex, that is, if (1− β)(1− 2β)f2(t)− 3(1− β)tf ′(t)f(t)− t2f ′′(t)f(t) + 2t2(f ′(t))2 ≥ 0. (2.5) The latter condition is satisfied by f(t) = t logα(t + 1) with α > 2 and β = 1/3. Indeed, with β = 1/3 and f as in above we find (1− β)(1− 2β)f2(t)− 3(1− β)tf ′(t)f(t)− t2f ′′(t)f(t) + 2t2(f ′(t))2 = t2 log2α−2(t+ 1) [2 9 log2(t+ 1) + αt2 (t+ 1)2 log(t+ 1) + α(α+ 1)t2 (t+ 1)2 ] ≥ 0. EJDE-2024/51 CONVEXITY OF SOLUTIONS TO ELLIPTIC PDE’S 5 Proof of theorem 2.3. The proof is similar to that of Theorem 2.2. Putting v = uβ , our problem reads as ∆v = β − 1 β |Dv|2 v + βv β−1 β f(v 1 β ) in Ω, v = +∞ on ∂Ω. Clearly, the corresponding concavity function c(x, y) = 2v(z)− v(x)− v(y), z = x+ y 2 is negative on ∂(Ω×Ω). Let us prove that c(x, y) cannot have any positive maximum in Ω × Ω. Arguing by contradiction, suppose c(x, y) has a positive maximum at some point (x, y) ∈ Ω×Ω. If L is defined as in (2.4) and if we replace the expressions for Dxxc(x, y), Dxyc(x, y) and Dyyc(x, y) given in (2.3), we find (as in the proof of Theorem 2.2) Lc(x, y) = (r + s)2 2 ∆v(z)− r2∆v(x)− s2∆v(y). On using our equation for v and recalling that (at the point of maximum) Dv(x) = Dv(z) = Dv(y), we find that Lc(x, y) = β − 1 β |Dv(z)|2 [ (r + s)2 2v(z) − r2 v(x) − s2 v(y) ] +β [ (r + s)2 2 v β−1 β (z)f(v 1 β (z)) − r2v β−1 β (x)f(v 1 β (x))− s2v β−1 β (y)f(v 1 β (y)) ] . Since we are assuming c(x, y) > 0, we have 1 2v(z) < 1 v(x) + v(y) and β − 1 β |Dv(z)|2 [ (r + s)2 2v(z) − r2 v(x) − s2 v(y) ] ≥ β − 1 β |Dv(z)|2 [ (r + s)2 v(x) + v(y) − r2 v(x) − s2 v(y) ] = 1− β β |Dv(z)|2 (rv(y)− sv(x))2( v(x) + v(y) ) v(x)v(y) ≥ 0. Therefore, Lc(x, y) ≥ β [ (r + s)2 2 v β−1 β (z)f(v 1 β (z))− r2v β−1 β (x)f(v 1 β (x))− s2v β−1 β (y)f(v 1 β (y)) ] . With K(t) := t β−1 β f(t 1 β ), we have Lc(x, y) ≥ β [ (r + s)2 2 K(v(z))− r2K(v(x))− s2K(v(y)) ] . Since K(t) is strictly increasing and v(z) > v(x)+v(y) 2 , we have K(v(z)) > K (v(x) + v(y) 2 ) . 6 B. MEBRATE, G. PORRU EJDE-2024/51 Therefore, Lc(x, y) > β [ (r + s)2 2 K (u(x) + u(y) 2 ) − r2K(u(x))− s2K(u(y)) ] . With a = v(x) and b = v(y), we choose r = K(b) and s = K(a). Then Lc(x, y) > β(r + s) [K(a) +K(b) 2 K (a+ b 2 ) −K(a)K(b) ] . Finally, since K(t) is harmonic concavity, we find Lc(x, y) > 0. But, at (x, y) (point of maximum) we have Lc(x, y) ≤ 0, a contradiction. The proof is complete. □ Let us recall that Theorem 2.1 has been extended to the problem ∆u(x) = f(u(x)) ( 1 + k|Du|2 ) in Ω, u = +∞ on ∂Ω, where k is a positive constant. We shall prove the following version. Theorem 2.4. Let φ(t) be a positive function such that φ1/2(t) is convex. Let f(t) and g(t) be positive, increasing smooth functions with f(t) strictly increasing and with f(0) = g(0) = 0. We suppose that φ(t)f(t) and φ(t)g(t) are concave. If u ∈ C2(Ω) is a solution to ∆u(x) + k1|Du|2 = f(u(x)) + k2|Du(x)|2g(u(x)), k1, k2 ≥ 0, (2.6) in a convex domain Ω then the corresponding concavity function c(x, y) cannot have a positive maximum in Ω× Ω. Proof. By contradiction, suppose c(x, y) has a positive maximum at some point (x, y) ∈ Ω × Ω. If L is defined as in (2.4), by using (2.3) with u in place of v we have Lc(x, y) = (r + s)2 2 ∆u(z)− r2∆u(x)− s2∆u(y), z = x+ y 2 . On using our equation (2.6) and recalling that Du(x) = Du(z) = Du(y) we have Lc(x, y) = k1 (r − s)2 2 |Du(z)|2 + (r + s)2 2 f(u(z))− r2f(u(x))− s2f(u(y)) + k2|Du(z)|2 [ (r + s)2 2 g(u(z))− r2g(u(x))− s2g(u(y)) ] . Since we are assuming that u(z) > u(x) + u(y) 2 , and since f(t) is strictly increasing and g(t) is increasing, we find that f(u(z)) > f (u(x) + u(y) 2 ) , g(u(z)) ≥ g (u(x) + u(y) 2 ) . Therefore, Lc(x, y) > (r + s)2 2 f (u(x) + u(y) 2 ) − r2f(u(x))− s2f(u(y)) + k2|Du(z)|2 [ (r + s)2 2 g (u(x) + u(y) 2 ) − r2g(u(x))− s2g(u(y)) ] . EJDE-2024/51 CONVEXITY OF SOLUTIONS TO ELLIPTIC PDE’S 7 Put u(x) = a, u(y) = b, and choosing r = φ1/2(a) and s = φ1/2(b), we find that Lc(x, y) > 1 2 ( φ1/2(a) + φ1/2(b) )2 f (a+ b 2 ) − φ(a)f(a)− φ(b)f(b) + k2|Du(z)|2 [1 2 ( φ1/2(a) + φ1/2(b) )2 g (a+ b 2 ) − φ(a)g(a)− φ(b)g(b) ] . Since φ1/2(t) is convex, we have φ1/2(a) + φ1/2(b) 2 ≥ φ1/2 (a+ b 2 ) , which can be written as 1 2 ( φ1/2(a) + φ1/2(b) )2 ≥ 2φ (a+ b 2 ) . On using this inequality, we find 1 2 ( φ1/2(a) + φ1/2(b) )2 f (a+ b 2 ) − φ(a)f(a)− φ(b)f(b) ≥ 2φ (a+ b 2 ) f (a+ b 2 ) − φ(a)f(a)− φ(b)f(b) ≥ 0, and 1 2 ( φ1/2(a) + φ1/2(b) )2 g (a+ b 2 ) − φ(a)g(a)− φ(b)g(b) ≥ 2φ (a+ b 2 ) g (a+ b 2 ) − φ(a)g(a)− φ(b)g(b) ≥ 0, where the concavity of φ(t)f(t) and φ(t)g(t) have been used. Hence, Lc(x, y) > 0. But, at (x, y) (point of maximum) we have Lc(x, y) ≤ 0, a contradiction. The proof is complete. □ Examples. If f(t) = tp and g(t) = tp−γ with p > 1 and 0 < γ ≤ 1, we can choose φ(t) = t1−p. If f(t) = tet and g(t) = tγet with 0 < γ ≤ 1, we can choose φ(t) = e−t. 3. Singular problems First we recall the following result. Proposition 3.1. Let Ω ⊂ Rn be a bounded domain with a smooth boundary. Let f(t) be positive for t > 0, smooth, decreasing, and such that lim t→0+ ∫ 1 t f(τ)dτ = ∞. Then the boundary value problem ∆u(x) + f(u(x)) = 0 in Ω, u = 0 on ∂Ω, has a positive solution. Moreover we have uν = −∞ on ∂Ω, where uν denotes external normal derivative on ∂Ω. Proof. For a proof of existence we refer to [9]. We claim that uν = −∞. Following the proof of [9, Theorem 2.2], we consider the ordinary differential equation p′′(s) + f(p(s)) = 0, p(0) = 0, p(s) > 0. 8 B. MEBRATE, G. PORRU EJDE-2024/51 Integrating this equation we find, with some a > 0, (p′(a))2 − (p′(s))2 2 + ∫ p(a) p(s) f(τ)dτ = 0. On using our assumption on f we find lim s→0+ (p′(s))2 = +∞. By [9, Theorem 2.2] we have that for some λ > 0, λp(d(x)) ≤ u(x), where d(x) denotes the distance of x from ∂Ω. Recalling that p(0) = 0 and that u = 0 on ∂Ω, it follows that uν = −∞ on ∂Ω. The claim is proved. Our aim is to prove that u(x) is concave. To show this, we shall prove that the corresponding concavity function c(x, y) satisfies c(x, y) ≥ 0 in Ω×Ω. Let us show first that c(x, y) cannot have a minimum at a point (x, y) with y ∈ ∂Ω. Indeed, suppose a minimum occurs at a point (x, y) with y ∈ ∂Ω. If we take the derivative of c(x, y) with respect to ν (external normal) and compute it for x ∈ Ω and y ∈ ∂Ω, we find cν(x, y) = 2uν(z)− uν(x) +∞ = +∞, z = x+ y 2 . But, if c(x, y) has a minimum then cν(x, y) ≤ 0, a contradiction. □ Now, to prove that c(x, y) ≥ 0 it is sufficient to show that c(x, y) cannot have a negative minimum in Ω× Ω. The following result is well-known [16]. Theorem 3.2 (Korevaar-Kennington). Let f(t) be a function smooth, strictly de- creasing and harmonic concave. If u ∈ C2(Ω) is a solution to ∆u(x) + f(u(x)) ( 1 + k|Du(x)|2 ) = 0, k ≥ 0, then, the corresponding concavity function c(x, y) cannot have any negative mini- mum in Ω× Ω. Let us prove a different version of Theorem 3.2. Theorem 3.3. Let φ(t) be a positive function such that φ1/2(t) is convex. Let f(t) and g(t) be smooth, decreasing functions for t > 0, with f(t) strictly decreasing. Suppose that φ(t)f(t) and φ(t)g(t) are concave. If u ∈ C2(Ω) is a solution to ∆u(x) + f(u(x)) + k|Du(x)|2g(u(x)) = 0, k ≥ 0, (3.1) in a convex domain Ω then, the corresponding concavity function c(x, y) cannot have any negative minimum in Ω× Ω. Proof. The proof is similar to that of Theorem 2.4. By contradiction, suppose c(x, y) has a negative minimum at (x, y) ∈ Ω × Ω. If L is defined as in (2.4), by using (2.3) with u in place of v, we find Lc(x, y) = (r + s)2 2 ∆u(z)− r2∆u(x)− s2∆u(y). Since (x, y) is a point of minimum for c(x, y), we have Du(x) = Du(z) = Du(y). Hence, on using (3.1), we find that Lc(x, y) = − (r + s)2 2 f(u(z)) + r2f(u(x)) + s2f(u(y)) EJDE-2024/51 CONVEXITY OF SOLUTIONS TO ELLIPTIC PDE’S 9 + k|Du(z)|2 [ − (r + s)2 2 g(u(z)) + r2g(u(x)) + s2g(u(y)) ] . Since f(t) is strictly decreasing, g(t) is decreasing and v(z) < v(x)+v(y) 2 , we have −f(u(z)) < −f (u(x) + u(y) 2 ) and −g(u(z)) ≤ −g (u(x) + u(y) 2 ) . Therefore, Lc(x, y) < − (r + s)2 2 f (u(x) + u(y) 2 ) + r2f(u(x)) + s2f(u(y)) + k|Du(z)|2 [ − (r + s)2 2 g (u(x) + u(y) 2 ) + r2g(u(x)) + s2g(u(y)) ] . Puting u(x) = a, u(y) = b, and choosing r = φ1/2(a) and s = φ1/2(b), we find that Lc(x, y) < −1 2 ( φ1/2(a) + φ1/2(b) )2 f (a+ b 2 ) + φ(a)f(a) + φ(b)f(b) + k|Du(z)|2 [ −1 2 ( φ1/2(a) + φ1/2(b) )2 g (a+ b 2 ) + φ(a)g(a) + φ(b)g(b) ] . Since φ1/2(t) is convex, we have φ1/2(a) + φ1/2(b) 2 ≥ φ1/2 (a+ b 2 ) , which can be written as −1 2 ( φ1/2(a) + φ1/2(b) )2 ≤ −2φ (a+ b 2 ) . On using this inequality, we find − 1 2 ( φ1/2(a) + φ1/2(b) )2 f (a+ b 2 ) + φ(a)f(a) + φ(b)f(b) ≤ −2φ (a+ b 2 ) f (a+ b 2 ) + φ(a)f(a) + φ(b)f(b) ≤ 0, and − 1 2 ( φ1/2(a) + φ1/2(b) )2 g (a+ b 2 ) + φ(a)g(a) + φ(b)g(b) ≤ −2φ (a+ b 2 ) g (a+ b 2 ) + φ(a)g(a) + φ(b)g(b) ≤ 0, where the concavity of φ(t)f(t) and φ(t)g(t) have been used. Hence, Lc(x, y) < 0. But, at (x, y) (point of minimum) we have Lc(x, y) ≥ 0, a contradiction. The proof is complete. □ Examples. If f(t) = 1/tγ and g(t) = 1/tγ+1 with γ ≥ 1, we can choose φ(t) = tγ+1. If f(t) = 1 tet and g(t) = 1 t2et , we can chose φ(t) = t2et. Recall that Ω is assumed to be convex. By Theorem 3.2, the solution u to the problem ∆u+ u−γ = 0 in Ω, u = 0 on ∂Ω, with γ > 1 concave in Ω. Can we say that uα is concave for some α > 1? Let us prove the following results. Recall that Ω is a bounded convex domain with a smooth boundary. 10 B. MEBRATE, G. PORRU EJDE-2024/51 Proposition 3.4. Let γ > 1 and a ≥ γ−1 2 . If u is a positive solution to ∆u+ a |Du|2 u + u−γ = 0 in Ω, u = 0 on ∂Ω, then the function v(x) = u γ+1 2 (x) is concave. Proof. The equation in terms of v reads as follows ∆v + λ |Dv|2 v + γ + 1 2v = 0, λ = 2 γ + 1 ( a− γ − 1 2 ) . Clearly, u = 0 implies v = 0. Moreover, the condition a ≥ γ−1 2 implies λ ≥ 0. One proves (for example by using Theorem 3.3 with φ(t) = t2) that the corre- sponding concavity function c(x, y) cannot have a negative minimum in Ω× Ω. To conclude that c(x, y) ≥ 0 in Ω× Ω, it suffices to show that vν = −∞ on ∂Ω. To prove this, we note that from ∆v + γ + 1 2v < 0 in Ω, u = 0 on ∂Ω and ∆z + γ + 1 2z = 0 in Ω, z = 0 on ∂Ω it follows that v(x) ≥ z(x) in Ω and |vν | ≥ |zν | on ∂Ω. By Proposition 3.1 we have zν = −∞ on ∂Ω. It follows that vν = −∞ on ∂Ω. The proof is complete. □ Sometimes it is convenient to transform a blow-up problem into a singular prob- lem. Proposition 3.5. Let a > 0 and q > 0. 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Sakaguchi; Concavity properties of solutions to some degenerate quasilinear elliptic Dirich- let problems, Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4e série, tome 14, no. 3 (1987), 403–421. Benyam Mebrate Department of Mathematics, Wollo University, Dessie, Ethiopia Email address: benyam134@gmail.com Giovanni Porru Department of Mathematics and Informatics, University of Cagliari, Cagliari, Italy Email address: gporru856@gmail.com 1. Introduction 2. Blow-up problems Examples 3. Singular problems Examples References