A Singular Initial-Value Problem for Second-Order Differential Equations

and Applied Analysis 3 Now let us calculate the derivatives x󸀠(t) and x󸀠󸀠(t) from (6) by using the Leibniz rule:

In the case  = 0 the existence of the solution for the problem (2) has been studied in [19], where the authors demonstrated the importance of the condition  = 0 for the existence.We find the conditions for () and (, ()) to guarantee the existence of the solution for  ̸ = 0.
In this section, we generalize the existence theorem of solutions in [19] (see also, [20]).
Theorem 1.Let  and  satisfy the following conditions: Then a solution to the initial-value problem (2) with  = 0 exists.
In [5] the authors demonstrated the importance of the condition  = 0 for the existence.
To overcome the difficulties in the case  ̸ = 0 we consider a generalization of Theorem 1 and show that the statement of the theorem is true without condition (D3) and with weaker conditions on (, ).Theorem 2. Suppose that () is integrable on the interval [, ] for all  > 0 and  and  satisfy the following conditions: (D4 * ) there exist ,  with  <  < ,  > 0, and an integrable (improper, in general) () such that Then a solution to the initial-value problem (2) exists for all  ∈  such that where () ∈ [0, 1] is a solution of the problem That is we suppose the existence of solution of the problem (4) for some ().For the problems with  = 0, the initialvalue problem (4) always has a solution () = , for () ≡ 0. So Theorem 1 corresponds to the cases () = 0 and () = .
One of the advantages of Theorem 2 is that the problem (4) always has a solution for some appropriate (); for example, for () = −(), the problem (4) has a solution () =  + .The conclusion of the theorem remains valid for all solutions of (4).
It is also clear from the conclusion of Theorem 2 that the interval [0, 1] can be taken as [0,  0 ] for some small enough  0 > 0.
Proof of Theorem 2. For  ∈ (0, 1], we define the functions The function ℎ() is a bounded function which is continuous for  ∈ (0, 1].It is continuous or has a removable discontinuity at  = 0 and is differentiable a.e. We will show that the problem (2) is equivalent to the following integral equation: First, let us show the existence of the integral in (6) It follows from  ≥  on the set In like manner we obtain So the right-hand side of (6) makes sense for any () ≥ 0 and |(, ()) − ()| ≤  and lim Abstract and Applied Analysis 3 Now let us calculate the derivatives   () and   () from ( 6) by using the Leibniz rule: It follows from (12) that That is, the problem (2) is equivalent to (6).Let us define the recurrence relations where () is a solution of the problem (4).It follows from ( 9), (10), and ( 14) that  <   () <  for  <  −1 () <  and for small enough  ∈ [0,  0 ).Now, for  1 ,  2 ∈ [0,  0 ), we have from ( 9) and ( 10) that for some constant  1 .Thus, the sequence   () is uniformly bounded and uniformly continuous and, by Ascoli-Arzela lemma, there exists a continuous () such that    () → () uniformly on [0, ], for any fixed  ∈ [0,  0 ).Without loss of generality, say   () → ().Then using the Lebesgue dominated convergence theorem.
Note that the positivity condition of the function () can be weakened.
The positivity of () has been used in the proof of Theorem 2 to show the (removable) continuity of the function ℎ() at 0. Now assuming that the following condition holds: (C2) || is integrable on [, ] for any fixed ,  ∈ (0, 1],  < , and  ≤ ∫    ()  < +∞; for some fixed  (18) we can prove a similar theorem.

Theorem 3. The conclusion of Theorem 2 remains valid if condition (D2) is replaced by (C2).
Proof.We need to make some modifications to the proof of Theorem 2; for example, instead of the inequality for  ≥ , we will have for small enough  and .
The inequalities of the type ( 7)-( 10) can be easily established for the function (, ) with where () is absolutely integrable function, and the more general theorem can be stated as follows.

Theorem 4. The conclusion of Theorem 2 remains valid if the condition (D4 * c) is replaced by (D4 * d).
The more applicable version of the existence theorems can be received from Theorems 2, 3, and 4 if the function () is replaced by (, ).For example, Theorem 2 can be improved as follows.
Remark 7. Biles et al. [19] give an example which satisfies conditions of Theorem 1 except condition  ≥ 0. They considered the problem with the family of solutions  =  2 +1, where  is an arbitrary constant.

Applications
Now we can find wide classes of IVPs with corresponding existence and uniqueness criteria.The class of solvable problems can be extended by adding a function () to the function (, ), where () is taken from the equation of the type (4) with a solution.
Let us rephrase the main conclusion of Theorem 5 as follows.If the (singular) problem has a solution () =  + .Then existence of solution of (32) follows from Theorem 2.

Since
ln   → 0 as  → 0, for any  > 0 and  ≥ 0, we have that (ln  )  is integrable and so the problem has a solution.For the approximate solution of the problems like (36) see [22].

Concluding Remarks
We extended the class of solvable second-order singular IVPs.We established that the difficulties related to the singularity can be overcome for the problems of the type (2) with  ≥ 0 or  ≤ ∫    ()  < +∞; for some fixed .
The problem of the existence of a solution is reduced to the finding of a solution of some more easy problems like (4).The approach used here can be useful for the problems on the existence of solutions of boundary value problems [23][24][25][26].The authors in [23,24]

Theorem 6 .
Suppose the conditions of Theorem 2 or Theorem 3 hold and, in addition, suppose that  is Lipschitz in  on [, ].Then the IVP (2) has a unique solution.