Erratum to “Note on Some Nonlinear Integral Inequalities and Applications to Differential Equations”

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First, we apologize for the misprints in the original article. Now we correct those misprints. Page 1, line 2: Between equations and The Gronwall-Bellman insert “Among various types of integral inequalities.” Page 1, line 5: Replace diferential by differential. Page 2, line 1: Replace of the situation by to the situation.  Page 2, Lemma 2.1: between For 𝑥∈ℝ₊,  𝑦∈ℝ₊,  1/𝑝+1/𝑞=1, and one has insert “with 𝑝>1.” Page 2, Theorem 2.3: Replace and there by “If there.” Page 2, Theorem 2.3: Replace and 𝑢(𝑡) satisfy by and the function 𝑢(𝑡) satisfies. Page 3, Proof of Theorem 2.3, line 9: Replace it yields by we obtain. Page 4, Remark 2.4, line 2: Replace become by becomes. Page 4, Theorem 2.5, line 1: Replace holds by hold. Page 4, Theorem 2.5, (2.15): Replace ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠. Page 4, Theorem 2.5, (2.16): Replace ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠. Page 5,(2.28): Replace ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑡)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑠) ∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠.  Page 6, (2.29): Replace ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑡)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑠) ∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠.  Page 6, (2.30): Replace ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑡)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀(𝑝^⋆/𝑝−1)𝑏(𝑠) ∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠.  Page 6, (2.31): Replace ∫^𝑡₀𝑏(𝑡)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝^⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠. Page 6, (2.33): Replace ∫^𝑡₀𝑏(𝑡)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝_⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝_⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠.  Page 6, (2.34): Replace ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)(𝑎(𝑡)+𝑝_⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠 by ∫^𝑡₀𝑏(𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝_⋆/𝑝)^{(𝑝⋆/𝑝)−1}𝑑𝑠. Page 6, in Theorem 2.6: Replace holds by hold.  Page 6, in Proof: Replace for by For.  Page 7, in Remark 2.7: Replace if by If.  Page 7, in Theorem 2.8: Replace holds by hold.  Page 7, in Theorem 2.8 line 2: Replace nonnegative by positive.  Page 8, in Theorem 2.9: Replace 𝜕/𝜕𝑠𝑘(𝑡,𝑠) by (𝜕/𝜕𝑠)𝑘(𝑡,𝑠).  Page 12, line 3: Replace tacking account by Taking into account.  Page 12, in Remark 2.10: Replace if by If. Page 11, (2.63): Replace ∫^𝑡₀(𝜕/𝜕𝑠)𝑘(𝑡,𝑠)(𝑎(𝑠)+𝑝^⋆/𝑝+𝑏(𝑠)𝑣(𝑠))^𝑝⋆/𝑝 by ∫^𝑡₀(𝜕/𝜕𝑠)𝑘(𝑡,𝑠)∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑠)(𝑎(𝑠)+ 𝑝^⋆/𝑝+𝑏(𝑠)𝑣(𝑠))^𝑝⋆/𝑝.  Page 12, (3.1): Insert 𝑡≥𝑡₀.  Page 12 in (3.2): Replace {𝑐+∫^𝑡_𝑡0𝑓(𝜏)[((𝑞/𝑝)𝑐+(𝑝−𝑞)/ 𝑝)exp∫^𝑡_𝑡0𝐴^∗(𝑠)𝑑𝑠+∫^𝑡_𝑡0𝐵^∗(𝑠)(exp∫^𝑡_𝑠𝐴^∗(𝜎)𝑑𝜎)𝑑𝑠]𝑑𝜏}^1/𝑝 by {𝑐+∫^𝑡_𝑡0𝑓(𝜏)[((𝑞/𝑝)𝑐+(𝑝−𝑞)/ 𝑝)exp∫^𝜏_𝑡0𝐴^∗(𝑠)𝑑𝑠+∫^𝜏_𝑡0𝐵^∗(𝑠)(exp∫^𝜏_𝑠𝐴^∗(𝜎)𝑑𝜎)𝑑𝑠]𝑑𝜏}^1/𝑝,  𝑡≥𝑡₀.  Page 13, Proof of Theorem 3.1, in (3.4), (3.5), (3.6), and (3.7): Replace 𝑍(𝑡) by 𝑧(𝑡).  Page 13 between (3.7) and (3.8): Replace its follow by it follows.  Page 14, in application, line 2: Replace equation by equations.  Page 14, in end of the Section 3: Replace then the result required is found by: using (3.13) in (3.14), we get the required inequality in (3.2).  Page 14 in (3.13): Replace 𝑧(𝑡)≤𝑐+∫^𝑡_𝑡0𝑓(𝜏)[((𝑞/𝑝)𝑐+(𝑝−𝑞)/𝑝)exp∫^𝑡_𝑡0𝐴^∗(𝑠)𝑑𝑠+ ∫^𝑡_𝑡0𝐵^∗(𝑠)(exp∫^𝑡_𝑠𝐴^∗(𝜎)𝑑𝜎)𝑑𝑠]𝑑𝜏, by 𝑧(𝑡)≤𝑐+∫^𝑡_𝑡0𝑓(𝜏)[((𝑞/𝑝)𝑐+(𝑝−𝑞)/𝑝)exp∫^𝜏_𝑡0𝐴^∗(𝑠)𝑑𝑠+∫^𝜏_𝑡0𝐵^∗(𝑠)(exp ∫^𝜏_𝑠𝐴^∗(𝜎)𝑑𝜎)𝑑𝑠]𝑑𝜏.  Page 14, in (4.1) insert 𝑡>0.  Page 14, in (4.2): Replace ∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)𝑢^𝑝𝑖(𝑡) by ∑^𝑖=𝑛_𝑖=1ℎ_𝑖(𝑡)|𝑢|^𝑝𝑖.  Page 14, in Example 4.1: Replace 𝑝,  𝑝_𝑖(𝑖=1,…,𝑛)≥0 by 𝑝,  𝑝_𝑖(𝑖=1,…,𝑛)>0.  Page 15, line 2: we estimate the solution 𝑢(𝑡) of (4.1).  Page 15 in Proof of Theorem 4.3: Replace (4.5) by this formula 𝑢(𝑡)≤{|𝑐|^𝑝+∫^𝑡_𝑡0𝑓(𝜏)[((𝑞/𝑝)|𝑐|^𝑝+(𝑝−𝑞)/𝑝)exp∫^𝜏_𝑡0𝐴^∗(𝑠)𝑑𝑠+∫^𝜏_𝑡0𝐵^∗(𝑠)(exp∫^𝜏_𝑠𝐴^∗(𝜎)𝑑𝜎)𝑑𝑠]𝑑𝜏}^1/𝑝.  Page 15 in Proof of Theorem 4.3: Replace 𝑢^𝑝(𝑡)=𝑐^𝑝+∫^𝑡_𝑡0𝑓(𝜏)[𝑢^𝑞(𝜏)+∫^𝑡_𝑡0𝑘(𝜏,𝑠)𝑢^𝑟(𝑠)𝑑𝑠]𝑑𝜏 by 𝑢^𝑝(𝑡)=𝑐^𝑝+∫^𝑡_𝑡0𝑓(𝜏)[𝑢^𝑞(𝜏)+∫^𝜏_𝑡0𝑘(𝜏,𝑠)𝑢^𝑟(𝑠)𝑑𝑠]𝑑𝜏.  Page 15, Proof of Theorem 4.3: Replace (4.7) by |𝑢(𝑡)|^𝑝≤|𝑐|^𝑝+∫^𝑡_𝑡0𝑓(𝜏)[|𝑢(𝜏)|^𝑞+∫^𝜏_𝑡0𝑘(𝜏,𝑠)|𝑢(𝜏)|^𝑟𝑑𝑠]𝑑𝜏,  𝑡₀≤𝑠≤𝜏≤𝑡.  End of the corrigendum.