ABSTRACT
Consider the following problem D 0 + α ( t k D 0 + β u ) = t γ Δ u + | u | p + 1 , t > 0 , x ∈ R N , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_1.jpg"/> lim t → 0 u ( x , t ) = u ( x , 0 ) , x ∈ R N , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_2.jpg"/> where 0 < α, β < 1, p > 0, γ ∈ R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_1.jpg"/> , k ∈ R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_2.jpg"/> . We denote with D 0 + α https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_3.jpg"/> and D 0 + β https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_4.jpg"/> the left Riemann-Liouville fractional derivatives in t, Δ = ∑ i = 1 n ∂ 2 ∂ x i 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_5.jpg"/> is the Laplacian in the space variables. For convenience, we restrict our investigations for the case N = 1. In this case, problem (3.1) and (3.2) can be written as follows. D 0 + α ( t k D 0 + β u ) = t γ u x x + | u | p + 1 , t > 0 , x ∈ R , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_3.jpg"/> lim t → 0 u ( x , t ) = u ( x , 0 ) , x ∈ R . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_4.jpg"/>
Let α, β ∈ (0, 1), γ ≥ 0, k ≥ β and p > 0 be fixed. Then problem (3.3), (3.4) has a solution u ∈ C 1 ( ( 0 , + ∞ ) , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_6.jpg"/> .
Let b > 3 and Q > 0 be fixed. We set A 1 = max { 1 , sup x ∈ R | ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b d w d y | , sup x ∈ R | ∫ 0 x 1 ( w 2 + 1 ) b d w | , sup x ∈ R | ∫ 0 x ∫ 0 y | 1 − ( 1 + 2 b ) w 2 | ( w 2 + 1 ) b + 2 d w d y | , sup x ∈ R | ∫ 0 x | 1 − ( 1 + 2 b ) w 2 | ( w 2 + 1 ) b + 2 d w | } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_1.jpg"/> 138Because b > 3, we have that A 1 < ∞. We choose a constant a > 3 such that a − γ > 5 , a > k + 4 − α − β , 11 a > k + 4 − α − β . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_5.jpg"/> Let ε, q ∈ (0, 1) be chosen so that { ϵ Q + A 1 Q ( a ( 6 − 3 β ) + 2 − β ) ( 1 − q ) ( a 2 − 1 ) ( 1 − α ) ( 1 − β ) Γ ( 1 − α ) Γ ( 1 − β ) + ( 1 − q ) Q a − γ − 1 ( 1 + 6 b A 1 ) + ( 1 − q ) A 1 Q p + 1 ≤ Q , ϵ Q + 2 ( 1 − q ) A 1 Q ( 1 − α ) Γ ( 1 − α ) Γ ( 1 − β ) + ( 30 − 12 β ) a ( 1 − q ) A 1 Q ( a − 1 ) ( 1 − α ) ( 1 − β ) Γ ( 1 − α ) Γ ( 1 − β ) + ( 40 − 20 β ) a 2 ( 1 − q ) A 1 Q ( a 2 − 1 ) ( 1 − α ) ( 1 − β ) Γ ( 1 − α ) Γ ( 1 − β ) + ( 1 − q ) Q + 10 a ( 1 − q ) Q a − γ − 1 + 4 b ( 1 − q ) A 1 Q + 40 a b ( 1 − q ) A 1 Q a − γ − 1 + 2 b ( 1 − q ) A 1 Q + 20 a b A 1 Q a − γ − 1 + 2 ( 1 − q ) A 1 Q p + 1 ≤ Q . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math3_6.jpg"/>
Let C 1 ( [ 0 , ∞ ) , C 2 ( R 1 ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_7.jpg"/> be endowed with the norm | | u | | = max { sup t ∈ [ 0 , ∞ ) , x ∈ R | u ( x , t ) | , sup t ∈ [ 0 , ∞ ) , x ∈ R | u x ( x , t ) | , sup t ∈ [ 0 , ∞ ) , x ∈ R | u t ( x , t ) | , sup t ∈ [ 0 , ∞ ) , x ∈ R | u x x ( x , t ) | } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_2.jpg"/> With N ~ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_8.jpg"/> we will denote the set of all equi-continuous families in C 1 ( [ 0 , ∞ ) , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_9.jpg"/> , i.e., if F ∈ N ~ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_10.jpg"/> is a family of elements of C 1 ( [ 0 , ∞ ) , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_11.jpg"/> , then for every ε > 0 there exists δ = δ(ε) > 0 such that | v ( x 1 , t 1 ) − v ( x 2 , t 2 ) | < ϵ , | v t ( x 1 , t 1 ) − v t ( x 2 , t 2 ) | < ϵ , | v x ( x 1 , t 1 ) − v x ( x 2 , t 2 ) | < ϵ , | v x x ( x 1 , t 1 ) − v x x ( x 2 , t 2 ) | < ϵ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_3.jpg"/> for every v ∈ F https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_12.jpg"/> , whenever |t 1 − t 2| < δ, |x 1 − x 2| < δ, t 1, t 2 ∈ [0, ∞), x 1 , x 2 ∈ R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_13.jpg"/> . Let N 1 = N ~ ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_14.jpg"/> , where N ~ ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_15.jpg"/> is the closure of N ~ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_16.jpg"/> , and N = { u ∈ N 1 : | | u | | ≤ Q } , N 1 = { u ∈ N 1 : | | u | | ≤ ( 1 + ϵ ) Q } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_4.jpg"/> Note that the set N is a compact subset of N 1.
139For u ∈ N 1, we define the operators. T ( u ) ( x , t ) = ( 1 + ϵ ) u ( x , t ) , S ( u ) ( x , t ) = − ϵ u ( x , t ) + ( 1 − q ) Γ ( 1 − α ) Γ ( 1 − β ) 1 ( t + 1 ) 10 a × ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t 1 ( τ + 1 ) a ∫ 0 τ s k ( τ − s ) α d d s ∫ 0 s u ( w , z ) ( s − z ) β d z d s d τ d w d y + a ( 1 − q ) Γ ( 1 − α ) Γ ( 1 − β ) 1 ( t + 1 ) 10 a × ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t ∫ 0 τ 1 1 ( τ 2 + 1 ) a + 1 ∫ 0 τ 2 s k ( τ 2 − s ) α d d s ∫ 0 s u ( w , z ) ( s − z ) β d z d s d τ 2 d τ 1 d w d y − ( 1 − q ) ( x 2 + 1 ) b 1 ( t + 1 ) 10 a ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( x , τ 2 ) d τ 2 d τ 1 − 4 b ( 1 − q ) 1 ( t + 1 ) 10 a ∫ 0 x y ( y 2 + 1 ) b + 1 ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( y , τ 2 ) d τ 2 d τ 1 d y + 2 b ( 1 − q ) 1 ( t + 1 ) 10 a ∫ 0 x ∫ 0 y 1 − ( 1 + 2 b ) w 2 ( w 2 + 1 ) b + 2 ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( w , τ 2 ) d τ 2 d τ 1 d w d y − ( 1 − q ) 1 ( t + 1 ) 10 a ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t ∫ 0 τ 1 1 ( τ 2 + 1 ) a | u ( w , τ 2 ) | p + 1 d τ 2 d τ 1 d w d y , R ( u ) ( x , t ) = T ( u ) ( x , t ) + S ( u ) ( x , t ) , x ∈ R , t ∈ [ 0 , ∞ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_5.jpg"/> Our first observation is as follows.
The fixed points of the operator R are solutions of problem (3.3) and (3.4).
Let u ∈ N 1 be such that R(u)(x, t) = u(x, t), x ∈ R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_17.jpg"/> , t ≥ 0. Then 1 Γ ( 1 − α ) Γ ( 1 − β ) ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t 1 ( τ + 1 ) a ∫ 0 τ s k ( τ − s ) α d d s ∫ 0 s u ( w , z ) ( s − z ) β d z d s d τ d w d y https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_6.jpg"/> + a Γ ( 1 − α ) Γ ( 1 − β ) × ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t ∫ 0 τ 1 1 ( τ 2 + 1 ) a + 1 ∫ 0 τ 2 s k ( τ 2 − s ) α d d s ∫ 0 s u ( w , z ) ( s − z ) β d z d s d τ 2 d τ 1 d w d y − 1 ( x 2 + 1 ) b ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( x , τ 2 ) d τ 2 d τ 1 − 4 b ∫ 0 x y ( y 2 + 1 ) b + 1 ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( y , τ 2 ) d τ 2 d τ 1 d y + 2 b ∫ 0 x ∫ 0 y 1 − ( 1 + 2 b ) w 2 ( w 2 + 1 ) b + 2 ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u ( w , τ 2 ) d τ 2 d τ 1 d w d y − ∫ 0 x ∫ 0 y 1 ( w 2 + 1 ) b ∫ 0 t ∫ 0 τ 1 | u ( w , τ 2 ) | p + 1 ( τ 2 + 1 ) a d τ 2 d τ 1 d w d y = 0 , x ∈ R , t ≥ 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_7.jpg"/> 140The last equation we differentiate twice in x and we get 1 Γ ( 1 − α ) Γ ( 1 − β ) ∫ 0 t 1 ( τ + 1 ) a ∫ 0 τ s k ( τ − s ) α d d s ∫ 0 s u ( x , z ) ( s − z ) β d z d s d τ + a Γ ( 1 − α ) Γ ( 1 − β ) ∫ 0 t ∫ 0 τ 1 1 ( τ 2 + 1 ) a + 1 ∫ 0 τ 2 s k ( τ 2 − s ) α d d s ∫ 0 s u ( x , z ) ( s − z ) β d z d s d τ 2 d τ 1 − ∫ 0 t ∫ 0 τ 1 τ 2 γ ( τ 2 + 1 ) a u x x ( x , τ 2 ) d τ 2 d τ 1 − ∫ 0 t ∫ 0 τ 1 | u ( x , τ 2 ) | p + 1 ( τ 2 + 1 ) a d τ 2 d τ 1 = 0 , x ∈ R , t ≥ 0 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_8.jpg"/> which we differentiate twice in t and we find 1 Γ ( 1 − α ) Γ ( 1 − β ) d d t ∫ 0 t s k ( t − s ) α d d s ∫ 0 s u ( x , z ) ( s − z ) β d z d s − t γ u x x ( x , t ) − | u ( x , t ) | p + 1 = 0 , x ∈ R , t ≥ 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_9.jpg"/> Therefore, u satisfies equation (3.3). Since u ∈ N 1, we have u ∈ C 1 ( [ 0 , + ∞ ) , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math3_18.jpg"/> and hence lim t → 0 u ( x , t ) = u ( x , 0 ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath3_10.jpg"/>
