Erratum to “Positive Solution to a Fractional Boundary Value Problem”

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In the paper entitled “Positive solution to a fractional boundray value problems,” the following problem (P1) is studied: ^𝑐𝐷^𝑞₀₊𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),^𝑐𝐷^𝜎₀⁺𝑢(𝑡()𝑢(0)),0<𝑡<1,1.1=𝑢^′′(0)=0,𝑢^′(1)=𝛼𝑢^′′(1),(1.2) where 𝑓∶[0,1]×ℝ×ℝ→ℝ is a given function, 2<𝑞<3, and 1<𝜎<2. Remarking that all the calculuses in this paper are done for 0<𝜎<1 and that if we take 1<𝜎<2, then ^𝑐𝐷^𝜎₀⁺𝑇𝑢=(1/Γ(2−𝜎))∫^𝑡₀((𝑇𝑢)^′′(𝑠)/(𝑡−𝑠)^𝜎−1)𝑑𝑠 and the second derivative with respect to 𝑡 of 𝐺(𝑡,𝑠) is discontinuous for 𝑠=𝑡, consequently we cannot apply this method to establish the existence and positivity of solution. For this reason, we correct the study of problem (P1) by taking 0<𝜎<1, and then the following corrections are needed.

(1) In page 3, in Lemma 2.3, we should correct ^𝑐𝐷^𝛼₀⁺𝑡^𝛽−1=(Γ(𝛽)/Γ(𝛽−𝛼))𝑡^{𝛽−𝛼−1}, 𝛽>𝑛.

(2) Equation (2.6) must be 𝑢(𝑡)=1Γ𝑞−2¹₀1(1−𝑠)^3−𝑞𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠.(2.6)

The Green function in (2.7) is 𝐺()=()𝑡,𝑠1−𝑠^3−𝑞(𝑡−𝑠)^𝑞−1𝑡()𝑞−1𝑞−2+𝛼𝑡−1−𝑠𝑡()𝑞−2,𝑠<𝑡,𝛼𝑡−1−𝑠()𝑞−2,𝑡≤𝑠.2.7

(3) Equation (2.11) becomes 𝑢(𝑡)=1Γ×𝑞−2_𝑡0(𝑡−𝑠)^𝑞−1+𝑞−1𝑞−2𝛼𝑡(1−𝑠)^3−𝑞−𝑡(1−𝑠)^𝑞−2(𝑠)+1𝑞−2×𝑦𝑑𝑠Γ𝑞−2¹_𝑡𝛼𝑡(1−𝑠)^3−𝑞−𝑡(1−𝑠)^𝑞−2𝑦(𝑠)()𝑞−2𝑑𝑠.2.11

Equation (2.12) must be 𝑢(𝑡)=1Γ𝑞−2¹₀1(1−𝑠)^3−𝑞𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠.(2.12)

(4) Equation (3.1) must be (𝑡)=1𝑇𝑢Γ×𝑞−2₁₀1(1−𝑠)^3−𝑞𝐺(𝑡,𝑠)𝑓𝑠,𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠())𝑑𝑠.3.1

In Theorem 3.2, the condition (3.5) must be 𝐶_𝑔+𝐶_ℎ<12,𝐴_𝑔+𝐴_ℎ<Γ(2−𝜎)2.(3.5)

(5) Equation (3.12) must be ||<1𝑇𝑢−𝑇𝑣2‖𝑢−𝑣‖,(3.12) and (3.13) becomes ^𝑐𝐷^𝜎₀⁺𝑇𝑢−^𝑐𝐷^𝜎₀⁺1𝑇𝑣=Γ(1−𝜎)^𝑡₀(𝑇𝑢)^′(𝑠)−(𝑇𝑣)^′(𝑠)(𝑡−𝑠)^𝜎𝑑𝑠.(3.13)

(6) Equation (3.14) is 𝐺₁()=()𝑡,𝑠𝜕𝐺𝑡,𝑠=()𝜕𝑡1−𝑠^3−𝑞(𝑡−𝑠)^𝑞−2()𝑞−2+𝛼−1−𝑠()𝑞−2,𝑠<𝑡,𝛼−1−𝑠()𝑞−2,𝑡≤𝑠.3.14

(7) Equation (3.15) is as follows: ^𝑐𝐷^𝜎₀⁺𝑇𝑢−^𝑐𝐷^𝜎₀⁺=1𝑇𝑣ΓΓ()×𝑞−21−𝜎_𝑡010(𝑡−𝑠)^−𝜎1(1−𝑟)^3−𝑞𝐺₁()×𝑓(𝑟),𝑠,𝑟𝑟,𝑢^𝑐𝐷^𝜎₀⁺𝑢(𝑟)−𝑓𝑟,𝑣(𝑟),^𝑐𝐷^𝜎₀⁺𝑣(𝑟)()𝑑𝑟𝑑𝑠.3.15

Equation (3.16) is as follows: ^𝑐𝐷^𝜎₀⁺𝑇𝑢−^𝑐𝐷^𝜎₀⁺≤||𝑇𝑣max𝑢−𝑣ΓΓ()×𝑞−21−𝜎_𝑡010(𝑡−𝑠)^−𝜎1(1−𝑟)^3−𝑞𝐺₁(𝑔(𝑟)+𝑠,𝑟)𝑑𝑟𝑑𝑠max^𝑐𝐷^𝜎₀⁺𝑢−^𝑐𝐷^𝜎₀⁺𝑣ΓΓ()×𝑞−21−𝜎_𝑡010(𝑡−𝑠)^−𝜎1(1−𝑟)^3−𝑞𝐺₁(ℎ(𝑟)()𝑠,𝑟)𝑑𝑟𝑑𝑠.3.16

Equation (3.17) is as follows: ¹₀1(1−𝑟)^3−𝑞𝐺₁(𝑠,𝑟)𝑔(𝑟)𝑑𝑟≤Γ𝑞−22𝐼₀^𝑞−1₊𝑔(1)+|𝛼|𝐼₀^𝑞−2₊𝑔(1).(3.17)

(8) Equation (3.18) is as follows: ^𝑐𝐷^𝜎₀⁺𝑇𝑢−^𝑐𝐷^𝜎₀⁺≤‖‖1𝑇𝑣𝑢−𝑣Γ()𝐴2−𝜎^𝑔+𝐴_ℎ.(3.18)

Equation (3.19) becomes ^𝑐𝐷^𝜎₀⁺𝑇𝑢−^𝑐𝐷^𝜎₀⁺≤1𝑇𝑣2‖𝑢−𝑣‖.(3.19)

Equation (3.21) is as follows: a.e(𝑡,𝑥,𝑥)∈[0,1]×ℝ².(3.21)

Equation (3.22) is as follows: 𝜓(𝑟)(𝑟)𝐶+𝜙+1¹Γ+𝐶𝑞−2Γ(²)()2−𝜎<𝑟.3.22

Equation (3.29) becomes ^𝑐𝐷^𝜎₀⁺≤𝐶𝑇𝑢²Γ()𝜓(𝑟)(𝑟).()2−𝜎+𝜙+13.29

Equation (3.30) is ‖‖=𝜓(𝑟)(𝑟)𝐶𝑇𝑢+𝜙+1¹Γ+𝐶𝑞−2Γ(²).()2−𝜎3.30

Equation (3.32) is as follows: 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂≤𝐶Γ×𝑞−2_𝑡10𝑡₂−𝑠^𝑞−1−𝑡₁−𝑠^𝑞−1+𝑡𝑞−1𝑞−2₂−𝑡₁|𝛼|(1−𝑠)^3−𝑞−(1−𝑠)^𝑞−2+𝑞−2𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝑞−1+𝑡𝑞−1𝑞−2₂−𝑡₁|𝛼|(1−𝑠)^3−𝑞−(1−𝑠)^𝑞−2+𝑞−2𝑑𝑠_1𝑡2𝑡₁−𝑡₂|𝛼|(1−𝑠)^3−𝑞−(1−𝑠)^𝑞−2()𝑞−2𝑑𝑠.3.32

Equation (3.33) is as follows: 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂≤𝐶Γ×𝑡𝑞−2₂−𝑡₁×^𝑡1₀1+|𝛼|𝑞−2(1−𝑠)^3−𝑞+(1−𝑠)^𝑞−2+𝑞−2𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝑞−1+𝑡𝑞−2𝑞−1₂−𝑡₁|𝛼|(1−𝑠)^3−𝑞+(1−𝑠)^𝑞−2+𝑡𝑞−2𝑑𝑠₁−𝑡₂¹_𝑡2|𝛼|(1−𝑠)^3−𝑞+(1−𝑠)^𝑞−2.()𝑞−2𝑑𝑠3.33

Equation (3.34) is as follows: 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂≤𝐶𝑡₂−𝑡₁Γ|𝛼|𝑞−24+3+𝐶𝑞−2Γ𝑞−2^𝑡2_𝑡1𝑡₂−𝑠^𝑞−1()𝑞−2𝑞−1𝑑𝑠.3.34

Equation (3.35) is as follows: ^𝑐𝐷^𝜎₀⁺𝑇𝑢𝑡₁−^𝑐𝐷^𝜎₀⁺𝑇𝑢𝑡₂≤1Γ(1−𝜎)^𝑡1₀𝑡₁−𝑠^−𝜎−𝑡₂−𝑠^−𝜎×(𝑇𝑢(𝑠))^′+1𝑑𝑠Γ(1−𝜎)^𝑡2_𝑡1𝑡₂−𝑠^−𝜎(𝑇𝑢(𝑠))^′()𝑑𝑠.3.35

Equation (3.37) is as follows: ^𝑐𝐷^𝜎₀⁺𝑇𝑢𝑡₁−^𝑐𝐷^𝜎₀⁺𝑇𝑢𝑡₂≤𝜓(𝑟)+𝜙(𝑟)+1×𝐶₂2𝑡₂−𝑡₁^1−𝜎+𝑡₂^1−𝜎−𝑡₁^1−𝜎×(Γ(2−𝜎))⁻¹.(3.37)

Equation (3.39) is as follows: ^𝑐𝐷^𝜎₀⁺𝑢(𝑡≤𝐶)²Γ()𝜓(𝑟)(𝑟).()2−𝜎+𝜙+13.39

Equation (3.40) is as follows: ‖𝑢‖≤𝜓(𝑟)(𝑟)𝐶+𝜙+1¹Γ+𝐶𝑞−2Γ(²)()2−𝜎<𝑟.3.40

(9) (H2) 0<∫¹₀(1/(1−𝑠)^3−𝑞)𝐺(𝑠,𝑠)𝑎(𝑠)𝑑𝑠<∞.

In Lemma 4.1, we have 𝛼≥1/(𝑞−2) and (4.1) becomes:

If 𝑡,𝑠∈[𝜏,1], 𝜏>0, then ()()≤20<𝜏𝐺𝑠,𝑠≤𝐺𝑡,𝑠𝜏²𝐺(𝑠,𝑠),0<𝐺₁(𝑠,𝑠)≤𝐺₁()≤2𝑡,𝑠𝜏𝐺₁().()𝑠,𝑠4.1

Equation (4.2) should be 𝛼𝑡(1−𝑠)^3−𝑞−𝑡(1−𝑠)^𝑞−2=𝑡𝑞−2𝑞−2(1−𝑠)^3−𝑞()𝑞−2𝛼−1+𝑠4.2 which is positive if 𝛼≥1/(𝑞−2).

Equation (4.3) : let 𝑡,𝑠∈[𝜏,1]; it is easy to see that 𝐺(𝑠,𝑠)≠0, and then we have 𝐺(𝑡,𝑠)𝐺()=()𝑠,𝑠𝑡−𝑠^𝑞−1(1−𝑠)^3−𝑞𝑠+𝑡𝑞−1𝑞−2𝛼−1+𝑠𝑠≤1+(1−𝑠)²𝑠²≤2𝜏²𝐺(),0<𝜏≤𝑠≤𝑡≤1,𝑡,𝑠𝐺()=𝑡𝑠,𝑠𝑠≤2𝜏≤2𝜏²(),0<𝜏≤𝑡≤𝑠≤1.4.3 Finally, since 𝐺(𝑠,𝑠) is nonnegative, we obtain 0<𝜏𝐺(𝑠,𝑠)≤𝐺(𝑡,𝑠)≤(2/𝜏²)𝐺(𝑠,𝑠).

(10) In Lemma 4.3, put 𝛼≥1/(𝑞−2) and inequality (4.5) becomes min_{𝑡∈[𝜏,1]}𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺𝑢(𝑡)≥𝜏³2‖𝑢‖.(4.5)

Equation (4.7) is as follows: 𝑢(𝑡)≤2𝜏²Γ×𝑞−2₁₀1(1−𝑠)^3−𝑞𝐺(𝑠,𝑠)𝑎(𝑠)𝑓₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠())𝑑𝑠.4.7

Equation (4.8) is as follows: ^𝑐𝐷^𝜎₀⁺𝑢(𝑡)=1ΓΓ()×𝑞−21−𝜎_𝑡010(𝑡−𝑠)^−𝜎1(1−𝑟)^3−𝑞𝐺₁()(𝑟)𝑓𝑠,𝑟×𝑎₁𝑢(𝑟),^𝑐𝐷^𝜎₀⁺𝑢(𝑟)≤2𝑑𝑠𝑑𝑟Γ()×𝜏Γ𝑞−22−𝜎₁₀1(1−𝑟)^3−𝑞𝐺₁(𝑟,𝑟)𝑎(𝑟)×𝑓₁𝑢(𝑟),^𝑐𝐷^𝜎₀⁺𝑢(𝑟)()𝑑𝑟.4.8

Equation (4.9) is as follows: ‖𝑢‖≤2𝜏²Γ×𝑞−2₁₀1(1−𝑠)^3−𝑞𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()(𝑠)𝑓2−𝜎×𝑎₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠)()𝑑𝑠.4.9

Equation (4.10) is as follows: ¹₀1(1−𝑠)^3−𝑞𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()(𝑠)𝑓2−𝜎×𝑎₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠≥𝜏)𝑑𝑠₂Γ𝑞−22‖𝑢‖.(4.10)

Equation (4.11) : in view of the left hand side of (4.1) , we obtain for all 𝑡∈[𝜏,1] 𝑢(𝑡)≥𝜏Γ×𝑞−2₁₀1(1−𝑠)^3−𝑞𝐺(𝑠,𝑠)𝑎(𝑠)𝑓₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠)()𝑑𝑠.4.11

Equation (4.12) is as follows: ^𝑐𝐷^𝜎₀⁺𝑢(𝑡)≥1ΓΓ()×𝑞−22−𝜎₁₀1(1−𝑟)^3−𝑞𝐺₁(𝑟,𝑟)𝑎(𝑟)×𝑓₁𝑢(𝑟),^𝑐𝐷^𝜎₀⁺𝑢(𝑟)()𝑑𝑟.4.12

Equation (4.13) is as follows: min_{𝑡∈[𝜏,1]}𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺𝑢(𝑡≥𝜏)Γ𝑞−2¹₀1(1−𝑠)^3−𝑞×𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()(𝑠)𝑓2−𝜎×𝑎₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠())𝑑𝑠.4.13

Equation (4.14) is as follows: min_{𝑡∈[𝜏,1]}𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺𝑢(𝑡)≥𝜏³2‖𝑢‖.(4.14)

Equation (4.17) is as follows: 𝐾=𝑢∈𝐸⁺,min_{𝑡∈[𝜏,1]}𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺𝑢(𝑡)≥𝜏³2‖𝑢‖.(4.17)

Equation (4.19) is as follows: (𝑡)=1𝑇𝑢Γ×𝑞−2₁₀1(1−𝑠)^3−𝑞×𝐺(𝑡,𝑠)𝑎(𝑠)𝑓₁𝑢(𝑠),^𝑐𝐷^𝜎₀⁺𝑢(𝑠)≤‖𝑢‖𝑑𝑠2𝜀𝜏²Γ𝑞−2¹₀1(1−𝑠)^3−𝑞𝐺()𝑎(𝑠)()𝑠,𝑠𝑑𝑠.4.19

Equation (4.20) is as follows: ^𝑐𝐷^𝜎₀⁺(𝑡)=1𝑇𝑢ΓΓ()×𝑞−21−𝜎_𝑡0101(1−𝑟)^3−𝑞(𝑡−𝑠)^−𝜎𝐺₁()(𝑟)𝑓𝑠,𝑟×𝑎₁𝑢(𝑟),^𝑐𝐷^𝜎₀⁺𝑢(𝑟)≤2𝑑𝑠𝑑𝑟Γ()×𝜏Γ𝑞−22−𝜎₁₀1(1−𝑟)^3−𝑞𝐺₁()𝑎(𝑟)×|𝑢(𝑟+𝑟,𝑟)|^𝑐𝐷^𝜎₀⁺𝑢(𝑟≤‖𝑢‖)𝑑𝑟2𝜀𝜏²Γ𝑞−2Γ(2−𝜎)¹₀1(1−𝑠)^3−𝑞𝐺₁()𝑎(𝑠)()𝑠,𝑠𝑑𝑠.4.20

Equation (4.21) is as follows: ‖‖≤‖𝑢‖𝑇𝑢2𝜀𝜏²Γ𝑞−2¹₀1(1−𝑠)^3−𝑞×𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()𝑎(𝑠)()2−𝜎𝑑𝑠.4.21

Equation (4.22) is as follows: 𝜀≤𝜏²Γ×2𝑞−2₁₀1(1−𝑠)^3−𝑞×𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()𝑎(𝑠)2−𝜎𝑑𝑠⁻¹.(4.22)

Let 𝑅=max2𝑅₁,2𝑅₂𝜏³.(1)

Equation (4.23) is as follows: min_{𝑡∈[𝜏,1]}𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺𝑢(𝑡)≥𝜏³2‖𝑢‖=𝜏³2𝑅≥𝑅₂.(4.23)

Using the left hand side of (4.1) and Lemma 4.1, we obtain (4.24) : (𝑡)≥𝜏𝑇𝑢⁴𝑀‖𝑢‖2Γ𝑞−2¹₀1(1−𝑠)^3−𝑞𝐺(𝑠,𝑠)𝑎(𝑠)𝑑𝑠.(4.24)

Equation (4.25) is as follows: ^𝑐𝐷^𝜎₀⁺(𝑡)≥𝜏𝑇𝑢³𝑀‖𝑢‖Γ()×2Γ𝑞−22−𝜎₁₀1(1−𝑠)^3−𝑞𝐺₁()𝑎(𝑠)()𝑠,𝑠𝑑𝑠.4.25

Equation (4.26) is as follows: 𝑇𝑢(𝑡)+^𝑐𝐷^𝜎₀⁺(𝑡)≥𝜏𝑇𝑢⁴𝑀‖𝑢‖×2Γ𝑞−2₁₀1(1−𝑠)^3−𝑞𝐺()+𝐺𝑠,𝑠¹(𝑠,𝑠)Γ()(𝑠)()2−𝜎×𝑎𝑑𝑠.4.26

Equation (4.27) is as follows: ×𝜏𝑀≥2Γ𝑞−2₄₁₀1/(1−𝑠)^3−𝑞𝐺(𝑠,𝑠)+𝐺₁(𝑠,𝑠)/Γ(2−𝜎)𝑎(𝑠)𝑑𝑠⁻¹.(4.27)

(11) In Example 4.6, if we choose 𝜎=1/4<1; then we get the same results with 𝐶_𝑔+𝐶_ℎ1=0.49821<2,𝐴_𝑔+𝐴_ℎΓ()=0.42552<2−𝜎2=0.459.(2)

In Example 4.7, choose 𝜎=1/5, 𝜓(𝑥)=(𝑥/10)²+1, and 𝜙(𝑥)=ln(1+𝑥²)/9+1; then we get the same results.

Remark 1. One can study the problem (P1) for 1<𝜎<2 and the function 𝑓 depending only on 𝑡 and 𝑢 instead of 𝑓(𝑡,𝑢(𝑡),^𝑐𝐷^𝜎₀⁺𝑢(𝑡)).