We deal with the Dirichlet problem in a bounded smooth domain.[omissis

Anedda, Claudia

Cuccu, Fabrizio

Porru, Giovanni

Le Matematiche (Dipartimento di Matematica e Informatica, Università degli Studi di Catania)

LE MATEMATICHEVol. LX (2005) – Fasc. II, pp. 339–352BOUNDARY ESTIMATES FOR SOLUTIONSTO SINGULAR ELLIPTIC EQUATIONSCLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRUWe deal with the Dirichlet problem u+u−γ + g(u) = 0 in a boundedsmooth domain  ⊂ RN with u = 0 on the boundary ∂, where γ > 1and g is a smooth function. Assuming |g(t)| grows like t−ν , 1 < ν < γ , ast → 0, we find optimal estimates of u(x) in terms of the distance of x fromthe boundary ∂.1. Introduction.We investigate the Dirichlet problem(1.1) u + f (u) = 0 in , u = 0 on ∂,where  is a bounded smooth domain in RN and f (t) is a smooth, positive,decreasing function which tends to infinity as t tends to zero. Existence anduniqueness for this problem have been discussed by Crandall, Rabinowitz andTartar in [5]. In particular, in [5] it is proved that there is a classical solutionu ∈C2() ∩ C0() and that, near the boundary ∂, it satisfies(1.2) λ(δ(x )) ≤ u(x ) ≤ (δ(x )),where  = (s) is a positive solution of the problem′′ + f () = 0, (0) = 0,340 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRUδ(x ) is the distance from x to ∂, and λ,  are two suitable positive constants.If f (t) = t−γ , γ > 1, we can take  = φ , with(1.3) φ(s) =( γ + 1√2(γ − 1) s) 2γ+1.In this special case, the estimate (1.2) has been improved in [4], where it isshown that there is a constant B such that(1.4)∣∣u(x )− φ(δ(x ))∣∣ ≤ Bδ(x ).For γ > 3, the last inequality has been made more precise in [13], where theestimate ∣∣u(x )− φ(δ(x ))∣∣ ≤ B(δ(x )) γ+3γ+1has been proved. For 1 < γ < 3, in [3] it is shown that∣∣u(x )− φ(δ(x ))∣∣ ≤ B(δ(x )) 2γγ+1 .The existence of a solutionof the homogeneous Dirichlet problem for moregeneral singular elliptic equations has been recently investigated in [6], [8],[12], [14], [15]. Also the behaviour of the solution near the boundary has beeninvestigated. However, as far as we know, there are not results concerning theprecise evaluation of the second order term in the boundary approximation ofthe solution.In the present paper we investigate problem (1.1) in case f (t) = t−γ +g(t),where γ > 1 and g(t) is a smooth function such that t−γ + g(t) is positive anddecreasing for t > 0. First we assume that |g(t)| grows like the function t−νwithmax[γ − 12, 1]< ν < γ,and prove thatu(x ) = φ(δ(x ))(1+ O(1)δβ),where(1.5) β = 2(γ − ν)γ + 1and O(1) is a bounded quantity. Note that 0 < β < 1.BOUNDARY ESTIMATES FOR SOLUTIONS. . . 341Next we take γ > 3, suppose that |g(t)| grows like t−ν with 1 < ν ≤(γ − 1)/2, and we prove thatu(x ) = φ(δ(x ))[1+ (N − 1)K3− γ δ + O(1)δβ],where K = K (x ) denotes the mean curvature of the surface {x ∈  : δ(x ) =constant}. The exponent β is defined as in (1.5), but now we have 1 < β < 2for 1 < ν < (γ − 1)/2, and β = 1 for ν = (γ − 1)/2.In case of g ≡ 0 and γ > 3 we prove that(1.6) u(x ) = φ(δ(x ))[1+ (N − 1)K3− γ δ + O(1)δ1+σ],with0 < σ <γ − 3γ + 1 .To prove our results we use the estimate(1.7) limx→∂φ(δ(x ))u(x )= 1.For general f (1.7) follows, for example, by [8]. Indeed, if(1.8) F(t) =∫ ∞tf (τ ) dτ,Theorem 2.6 of [8] assumes that F(t) → ∞ as t → 0 and that, for someM > 1, F(t) < MF(2t) for t small. Both these conditions hold in our situation.We emphasize that when the perturbation |g(t)| grows less than t−γ then,concerning the boundary behaviour of the solution, it does not make effects atthe first level. Furthermore, when |g(t)| grows less than t− γ−12 then it does notmake effects at the first two levels.Estimate (1.6) with o(δ) in place of O(1)δ1+σ has been announced in [2].The influence of the domain geometry in boundary blow-up problems has beendiscussed in [1], [7].342 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRU2. Boundary estimates.Let γ > 1 be a real number and let g(t) be a smooth function satisfying(2.1) g(t) = O(1)t−ν, 1 < ν < γ,where O(1) is a bounded quantity. We also suppose(2.2) t−γ + g(t) > 0 and − γ t−γ−1 + g′(t) < 0 ∀ t > 0.Under conditions (2.1) and (2.2), we investigate the Dirichlet problem(2.3) u + u−γ + g(u) = 0 in , u = 0 on ∂,where  is a bounded smooth domain in RN . The integral function F(t) definedas in (1.8) with f (t) = t−γ + g(t) behaves likeF(t) = t1−γγ − 1 + O(1)t1−ν.Since F(t) → ∞ as t → 0 and since we have, for some M > 1 and for t smallF(t) < M F(2t), we can use the estimate (1.7) for the solution u to problem(2.3). We have the followingTheorem 2.1. Let  ⊂ RN , N ≥ 2, be a bounded smooth domain, let γ > 1be a real number, and let g(t) be a smooth function satisfying (2.2) and (2.1)withmax[γ − 12, 1]< ν < γ.If u(x ) is a solution to problem (2.3) thenu(x ) = φ(δ(x ))[1+ O(1)δβ], β = 2(γ − ν)γ + 1 ,where φ is defined in (1.3), and O(1) is a bounded quantity.Proof. We look for a super-solution of the formw(x ) = φ(δ)(1+ αδβ),where α is a positive real number to be determined. We havewxi = φ′δxi(1+ αδβ) + αβφδβ−1δxi .BOUNDARY ESTIMATES FOR SOLUTIONS. . . 343Recall that [9]N∑i=1δxi δxi = 1,N∑i=1δxi xi = −(N − 1)K ,where K = K (x ) is the mean curvature of the surface {x ∈  : δ(x ) =constant}. Putting (N − 1)K = H we find(2.4) w = φ′′(1+ αδβ) − φ′H (1+ αδβ) + 2αβδβ−1φ′ ++ φαβ(β − 1)δβ−2 − αβφH δβ−1.The function φ = φ(s) satisfies the equations(2.5) φ′′ = −φ−γ , φ′ = φ−γ γ + 1γ − 1s, φ = φ−γ (γ + 1)22(γ − 1)s2.By (2.4) and (2.5) we obtain(2.6) −w = φ−γ[1+ αδβ + γ + 1γ − 1δH (1+ αδβ)− 2αβ γ + 1γ − 1δβ− αβ(β − 1) (γ + 1)22(γ − 1)δβ + αβ (γ + 1)22(γ − 1)δβ+1H].By (2.6) we get(2.7) −w > φ−γ[1+αδβ(1−2γ + 1γ − 1β−(γ + 1)22(γ − 1)β(β−1)−C1δ)−C2δ],where Ci , i = 1, 2, · · ·, here and in the sequel, denote suitable positive constantsindependent of α. We will take α and δ0 > 0 such that(2.8) αδβ0 <12,and consider points x ∈  such that δ(x ) < δ0. Using condition (2.1) andTaylor’s expansion we findw−γ + g(w) = φ−γ [(1+ αδβ)−γ + O(1)φγ−ν](2.9)< φ−γ[1− γαδβ + C3(αδβ)2 + C4δβ].344 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRUBy (2.7) and (2.9) we find that(2.10) −w > w−γ + g(w)provided1+ αδβ(1− 2γ + 1γ − 1β −(γ + 1)22(γ − 1)β(β − 1)− C1δ)− C2δ> 1− γαδβ + C3(αδβ)2 + C4δβ.After simplification we find(2.11) C4 +C2δ1−β < α[ (γ + 1)22(γ − 1) (β + 1)(2(γ − 1)γ + 1 − β)−C1δ −C3αδβ].Since β < 2(γ − 1)/(γ + 1), we can take α and δ0 so that (in addition to (2.8))C1δ0 + C3αδβ0 <(γ + 1)22(γ − 1)(β + 1)(2(γ − 1)γ + 1 − β).Since (γ − 1)/2 < ν , we have β < 1. Increase α (and decrease δ0) until (2.11)holds for δ < δ0.Let us show now that we can choose δ0 and α so that u(x ) < w(x ) forδ(x ) = δ0. Let q = αδβ0 , where α and δ0 are as in above. Since by (1.7)limx→∂φ(δ(x ))u(x )= 1,we can decrease δ0 (increasing α according to αδβ0 = q) untilφ(δ(x ))u(x )>11+ qfor δ(x ) ≤ δ0. Multiplying by(1+ αδβ) we findw(x )u(x )>11+ q(1+ αδβ).Then w(x ) > u(x ) for δ(x ) = δ0. On ∂ we have w(x ) = u(x ) = 0.Therefore, since t−γ + g(t) is decreasing, by (2.10) we find that w(x ) ≥ u(x )on {x ∈ : δ(x ) < δ0}.BOUNDARY ESTIMATES FOR SOLUTIONS. . . 345Now we seek for a sub-solution of the typev(x ) = φ(δ)(1− αδβ).We find an equation similar to (2.6) with −α in place of α, that is− v = φ−γ[1− αδβ + γ + 1γ − 1δH (1− αδβ)+ 2αβ γ + 1γ − 1δβ+ αβ(β − 1) (γ + 1)22(γ − 1)δβ − αβ (γ + 1)22(γ − 1)δβ+1H].From this equation we find(2.12) −v < φ−γ[1−αδβ(1−2γ + 1γ − 1β−(γ + 1)22(γ − 1)β(β−1)−C5δ)+C6δ].Let δ0 and α satisfy (2.8). For δ(x ) < δ0, using condition (2.1) and Taylor’sexpansion we find(2.13) v−γ + g(v) = φ−γ [(1− αδβ)−γ + O(1)φγ−ν]> φ−γ[1− γαδβ − C7δβ].By (2.12) and (2.13) we find that(2.14) −v < v−γ + g(v)provided1−αδβ(1−2γ + 1γ − 1β −(γ + 1)22(γ − 1)β(β −1)−C5 δ)+C6δ < 1−γαδβ −C7δβ.After simplification we find(2.15) C7 + C6δ1−β < α[ (γ + 1)22(γ − 1) (β + 1)(2(γ − 1)γ + 1 − β)− C5δ].Inequality (2.15) (which is similar to (2.11) but easier) holds for δ < δ0 with δ0small enough and α large enough.346 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRULet us show that we can choose δ0 and α so that v(x ) < u(x ) for δ(x ) = δ0.Let q = αδβ0 , with α and δ0 as in above. Since we havelimx→∂φ(δ(x ))u(x )= 1,we can decrease δ0 (increasing α according to αδβ0 = q) untilφ(δ(x ))u(x )<11− qfor δ(x ) ≤ δ0. As a consequence we havev(x )u(x )<11− q(1− αδβ).Then v(x ) < u(x ) for δ(x ) = δ0. By (2.14) it follows that v(x ) ≤ u(x ) on{x ∈ : δ(x ) < δ0}.We have proved that for x near ∂ and α large enough we haveφ(δ)(1− αδβ) ≤ u(x ) ≤ φ(δ)(1+ αδβ).Equivalently we haveu(x ) = φ(δ)[1+ O(1)δβ].The theorem is proved. Recall that δ = −(N − 1)K , where K is the mean curvature of thesurface {x ∈ : δ(x ) = constant}.We state now our main result.Theorem 2.2. Let  ⊂ RN , N ≥ 2, be a bounded smooth domain, let γ > 3be a real number, and let g(t) be a smooth function satisfying (2.2) and (2.1)with1 < ν ≤ γ − 12.If u(x ) is a solution to problem (2.3) thenu(x ) = φ(δ(x ))[1+ (N − 1)K3− γ δ + O(1)δβ], β = 2(γ − ν)γ + 1 ,where φ is defined in (1.3), and O(1) is a bounded quantity. Furthermore, ifg(t) ≡ 0 then we haveu(x ) = φ(δ(x ))[1+ (N − 1)K3− γ δ + O(1)δ1+σ],where σ is any positive real number such that σ < (γ − 3)/(γ + 1).BOUNDARY ESTIMATES FOR SOLUTIONS. . . 347Proof. Define(2.16) A = H3− γ , H = (N − 1)K .We look for a super-solution of the kindw(x ) = φ(δ)(1+ Aδ + αδ1+σ ),where σ is a positive real number such that σ < (γ − 3)/(γ + 1) and α is apositive real number to be determined. We havewxi = φ′δxi(1+ Aδ + αδ1+σ) + φ(Axi δ + Aδxi + α(1+ σ )δσ δxi ).Since δ = −H we findw = φ′′(1+ Aδ + αδ1+σ)− φ′H (1+ Aδ + αδ1+σ )+ 2φ′(∇A · ∇δ δ + A + α(1+ σ )δσ )+ φ(A δ + 2∇A · ∇ δ − AH + α(1+ σ )σδσ−1 − α(1+ σ )H δσ).Using equations (2.5) we find−w = φ−γ[1+ Aδ + αδ1+σ + γ + 1γ − 1δH(1+ Aδ + αδ1+σ)− 2γ + 1γ − 1δ(∇A · ∇δ δ + A + α(1+ σ )δσ )− (γ + 1)22(γ − 1)δ2(A δ + 2∇A · ∇ δ − AH + α(1+ σ )σδσ−1 − α(1+ σ )H δσ)].Denoting with Ci , i = 1, 2, · · · , positive constants independent of α we get(2.17) −w > φ−γ[1+ δ(A + γ + 1γ − 1H − 2γ + 1γ − 1 A)+ αδ1+σ(1− 2γ + 1γ − 1 (1+ σ ) −(γ + 1)22(γ − 1) (1+ σ )σ − C1δ)− C2δ2].For α fixed, we consider δ0 > 0 such that for δ(x ) < δ0 we have(2.18) −12< Aδ + αδ1+σ < 1.348 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRUUsing condition (2.1) and Taylor’s expansion we have(2.19) w−γ + g(w) = φ−γ[(1+ Aδ + αδ1+σ)−γ + O(1)φγ−ν]< φ−γ[1− γ Aδ − γαδ1+σ + C3δ2 + C4δαδ1+σ + C5(αδ1+σ )2 + C6δβ].Observe that we can take C6 = 0 when g(t) ≤ 0. By (2.17) and (2.19) we findthat(2.20) −w > w−γ + g(w)provided(2.21) 1+(A + γ + 1γ − 1H − 2γ + 1γ − 1 A)δ+ αδ1+σ(1− 2γ + 1γ − 1 (1+ σ ) −(γ + 1)22(γ − 1)(1+ σ )σ − C1δ)− C2δ2> 1− γ Aδ − γαδ1+σ + C3δ2 + C4δαδ1+σ + C5(αδ1+σ )2 + C6δβ.Using (2.16) we find(2.22) A + γ + 1γ − 1H − 2γ + 1γ − 1 A = −γ A.Hence, (2.21) can be rewritten as(2.23) αδ1+σ(1− 2γ + 1γ − 1(1+ σ )−(γ + 1)22(γ − 1)(1+ σ )σ − C1δ)− C2δ2> −γαδ1+σ + C3δ2 + C4δαδ1+σ + C5(αδ1+σ )2 + C6δβ.Rearranging we have(2.24) (C2 + C3)δ2 + C6δβ< αδ1+σ[ (γ + 1)22(γ − 1)(γ − 3γ + 1 − σ)(σ + 2)− C5αδ1+σ − (C1 + C4)δ].If we take 1+ σ = β we find(2.25) (C2 + C3)δ2−β + C6BOUNDARY ESTIMATES FOR SOLUTIONS. . . 349< α[ (γ + 1)22(γ − 1)(β + 1)(2(γ − 1)γ + 1 − β)− C5αδβ − (C1 + C4)δ].Take α and δ0 > 0 so thatC5αδβ0 + (C1 + C4)δ0 <(γ + 1)22(γ − 1) (β + 1)(2(γ − 1)γ + 1 − β).Decrease δ0 again and increase α so that (2.25) and (2.18) (with 1 + σ = β )hold for δ(x ) < δ0. In this situation, inequality (2.20) holds.As in the proof of the previous theorem, let us show that we can choose δ0and α so that u(x ) < w(x ) for δ(x ) = δ0. Let q = αδβ0 , with α and δ0 as inabove. Since we havelimx→∂φ(δ(x ))u(x )= 1,we can decrease δ0 (increasing α according to αδβ0 = q) untilφ(δ(x ))u(x )>22+ qfor δ(x ) ≤ δ0. As a consequence we havew(x )u(x )>22+ q(1+ Aδ + αδβ).Decrease δ0 again and increase α in order to have αδβ0 = q and Aδ0 > −q/2.Then w(x ) > u(x ) for δ(x ) = δ0. Since w(x ) = u(x ) on ∂, and sincet−γ + g(t) is decreasing, by (2.20) it follows that w(x ) ≥ u(x ) on {x ∈  :δ(x ) < δ0}.To finish, we look for a sub-solution of the kindv(x ) = φ(δ)(1+ Aδ − αδ1+σ).We find an inequality similar to (2.17) with −α in place of α, that is(2.26) −v < φ−γ[1+ δ(A + γ + 1γ − 1H − 2γ + 1γ − 1 A)− αδ1+σ(1− 2γ + 1γ − 1 (1+ σ ) −(γ + 1)22(γ − 1) (1+ σ )σ − C7δ)+ C8δ2].350 CLAUDIA ANEDDA - FABRIZIO CUCCU - GIOVANNI PORRUTake α and δ0 > 0 such that for δ(x ) < δ0 we have(2.27) −12< Aδ − αδ1+σ < 1.Using condition (2.1) and Taylor’s expansion we have(2.28) v−γ + g(v) = φ−γ[(1+ Aδ − αδ1+σ)−γ + O(1)φγ−ν]> φ−γ[1− γ Aδ + γαδ1+σ − C9δβ].In case of g(t) ≥ 0 we can take C9 = 0. By (2.26) and (2.28) we find that(2.29) −v < v−γ + g(v)when1+ δ(A + γ + 1γ − 1H − 2γ + 1γ − 1 A)− αδ1+σ(1− 2γ + 1γ − 1 (1+ σ ) −(γ + 1)22(γ − 1)(1+ σ )σ − C7δ)+ C8δ2< 1− γ Aδ + γαδ1+σ − C9δβ.Using (2.22), after simplification we find the following inequality (similar to(2.24) but easier):(2.30) C8δ2 + C9δβ < αδ1+σ[ (γ + 1)22(γ − 1)(γ − 3γ + 1 − σ)(σ + 2)− C7δ].With 1+ σ = β we have(2.31) C8δ2−β + C9 < α[ (γ + 1)22(γ − 1) (β + 1)(2(γ − 1)γ + 1 − β)− C7δ].Now let us take α and δ0 so that (2.31) and (2.27) (with 1 + σ = β ) hold forδ < δ0.With α and δ0 as in above, let q = αδβ0 . Since we havelimx→∂φ(δ(x ))u(x )= 1,BOUNDARY ESTIMATES FOR SOLUTIONS. . . 351we can decrease δ0 (increasing α according to αδβ0 = q) untilφ(δ(x ))u(x )<22− qfor δ(x ) ≤ δ0. As a consequence we havev(x )u(x )<22− q(1+ Aδ − αδβ).Decrease δ0 again and increase α in order to have αδβ0 = q and Aδ0 < q/2.Then v(x ) < u(x ) for δ(x ) = δ0. By (2.29) it follows that v(x ) ≤ u(x ) on{x ∈ : δ(x ) < δ0}.Therefore, for x near ∂ and α large enough we haveφ(δ)(1+ Aδ − αδβ) ≤ u(x ) ≤ φ(δ)(1+ Aδ + αδβ).Equivalently we haveu(x ) = φ(δ)(1+ Aδ + O(1)δβ).Recalling the definition of A given in (2.16), the first assertion of the theoremfollows.Now let g ≡ 0. Looking for a super-solution of the formw(x ) = φ(δ)(1+ Aδ + αδ1+σ ),we find that−w > w−γprovided inequality (2.24) with C6 = 0 holds, that is, provided(2.32) (C2 + C3)δ1−σ< α[ (γ + 1)22(γ − 1)(γ − 3γ + 1 − σ)(σ + 2)− C5αδ1+σ − (C1 + C4)δ].Since 0 < σ < (γ − 3)/(γ + 1), for α fixed (2.32) holds with δ small enough.By the same argument used in the proof of the first part of this theorem, we canincrease α and decrease δ0 so that w(x ) > u(x ) for δ(x ) = δ0. It follows thatw(x ) is a super-solution on {x ∈ : δ(x ) < δ0}.By a similar argument, using (2.30) with C9 = 0, we find that the functionv(x ) = φ(δ)(1+ Aδ − αδ1+σ),is a sub-solution on {x ∈ : δ(x ) < δ0}. 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Letelier, The influence of domain geometry in boundary blow-upelliptic problems, Nonlinear Anal. TMA, 48 (2002), pp. 897-904.[8] E. Giarrusso - G. Porru, Boundary behaviour of solutions to nonlinear ellipticsingular problems, Appl. Mathematics in the Golden Age (J.C. Misra ed.), Narosa,New Delhi, 2003, pp. 163-178.[9] D. Gilbarg - N.S. Trudinger, Elliptic Partial Differential Equations of SecondOrder, Springer Verlag, Berlin, 1977.[10] B. Kawohl, On a class of singular elliptic equations, Pitman Res. Notes Math.Ser., 266 (1992), pp. 156-163.[11] A.C. Lazer - P.J. McKenna, Asymptotic behaviour of solutions of boundary blow-up problems, Differential and Integral Equations, 7 (1994), pp. 1001-1019.[12] A.C. Lazer - P.J. McKenna, On a singular nonlinear elliptic boundary valueproblem, Proc. American. Math. Soc., 111 (1991), pp. 721-730.[13] P.J. McKenna - W. Reichel, Sign-changing solutions to singular second-orderboundary value problems, Adv. Differential Equations, 6 (2001), pp. 441-460.[14] Z. Zhang - J. Yu, On a singular nonlinear Dirichlet problem with a convectionterm, SIAM J. Math. Anal., 32 (2000), pp. 916-927.[15] Z. Zhang - J. Cheng, Existence and optimal estimates of solutions for singularnonlinear Dirichlet problems, Nonlinear Analysis, 57 (2004), pp. 473-484.Dipartimento di Matematica e Informatica,Universita` di Cagliari,Via Ospedale 72, 09124 Cagliari (ITALY)e-mail: porru@unica.it

Boundary estimates for solutions to singular elliptic equations

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Boundary estimates for solutions to singular elliptic equations

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