Hence, BVPHDEF (1) has a maximal solution on J.
Let V be a maximal solution on J 0 and we denote by β the upper bound of J 0. We suppose that β < + ∞.
for all t ∈ J. Taking the limit as n → ∞ in above Eq. (5.5) yields r ( t ) = f ( t, r ( t ) ) + 1 Γ ( q ) ∫ t 0 t ( t − s ) q − 1 g ( s, r ( s ) ) d s. for all t ∈ J. Thus, the function r is a solution of FHDE (2.1) on J. Finally, from inequality (5.3), it follows that u ( t ) ≤ r ( t ) for all t ∈ J. Hence, FHDE (2.1) has a maximal solution on J.
Then u ( t ) ≤ r ( t ) (6.2). for all t ∈ J, where r is a maximal solution of FHDE (2.1) on J. Proof Let ε > 0 be arbitrary small.
} t in J, afrac{u(0)}{f(0,u(0))}+bfrac{u(T)}{f(T,u(T))}leq c, end{cases} (20) then u(t) leq r(t) (21) for all (t in J ), where r is a maximal solution of BVPHDEF (1) on J. Let (varepsilon> 0) be arbitrarily small.
Let be a solution of (1.2) in, then is said to be a maximal solution of (1.2), if for every solution of (1.2) existing on, the inequality,, holds.
Then (r(t)) is said to be a maximal solution of (2.2) if, for every solution (u(t)) of (2.2) existing on ((0, +infty)), the inequality (u(t)leq r(t)), (tin 0, +infty)) holds.
Let r ( x ) be the maximal solution of D β u = g ( x, u ), u 0 ≥ 0, existing on [ x 0, x 0 + η ].
By Theorem 5.2, (r t, varepsilon)) is a maximal solution of BVPHDEF (16) so that the limit r(t) =lim_{varepsilonrightarrow0} r t,varepsilon) (22) is uniform on J and the function r is a maximal solution of BVPHDEF (1) on J. Hence, we obtain textstylebegin{cases} D^{alpha} (frac{r t,varepsilon)}{f t,r t,varepsilon))} ) =g t,r t,varepsilon)) + varepsilon quad mbox{a.e.e
Finally, we prove that is a weakly maximal solution of.
Suppose that is not a weakly maximal solution of.
