Introduction
This paper is concerned with the large-time behavior of nonnegative solutions to the initial-boundary value problem
{ ∂ t α ( u − u 0 ) = Δ u m , x ∈ Ω , t > 0 , u = 0 , x ∈ ∂ Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) ≥ 0 , x ∈ Ω . \begin{cases}
\partial_t^\alpha(u-u_0)=\Delta u^m, & x\in\Omega,\quad t>0,\\
u=0, & x\in\partial\Omega,\quad t>0,\\
u(x,0)=u_0(x)\ge0, & x\in\Omega.
\end{cases} ⎩ ⎨ ⎧ ∂ t α ( u − u 0 ) = Δ u m , u = 0 , u ( x , 0 ) = u 0 ( x ) ≥ 0 , x ∈ Ω , t > 0 , x ∈ ∂ Ω , t > 0 , x ∈ Ω. where m > 0 m>0 m > 0 0 < α < 1 0<\alpha<1 0 < α < 1 N ≥ 3 N\ge3 N ≥ 3 Ω ⊂ R N \Omega\subset\mathbb{R}^N Ω ⊂ R N ∂ t α \partial_t^\alpha ∂ t α α \alpha α
∂ t α v ( t ) : = d d t ( g 1 − α ∗ v ) ( t ) , g 1 − α ( t ) : = t − α Γ ( 1 − α ) . \partial_t^\alpha v(t):=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*v)(t),\qquad g_{1-\alpha}(t):=\frac{t^{-\alpha}}{\Gamma(1-\alpha)}. ∂ t α v ( t ) := d t d ( g 1 − α ∗ v ) ( t ) , g 1 − α ( t ) := Γ ( 1 − α ) t − α . Here ∗ * ∗ ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
( f ∗ g ) ( t ) : = ∫ 0 t f ( t − τ ) g ( τ ) d τ , t ≥ 0. (f*g)(t):=\int_0^t f(t-\tau)g(\tau)\,\mathrm{d}\tau,\qquad t\ge0. ( f ∗ g ) ( t ) := ∫ 0 t f ( t − τ ) g ( τ ) d τ , t ≥ 0. for f , g ∈ L l o c 1 ( [ 0 , ∞ ) ) f,g\in L^1_{\mathrm{loc}}([0,\infty)) f , g ∈ L loc 1 ([ 0 , ∞ ))
Throughout this paper, we use the following notion of an H − 1 H^{-1} H − 1 [1 , 10 ] .
Definition 1.1 We say that u ∈ L l o c 2 ( 0 , ∞ ; H − 1 ( Ω ) ) u\in L^2_{\mathrm{loc}}(0,\infty;H^{-1}(\Omega)) u ∈ L loc 2 ( 0 , ∞ ; H − 1 ( Ω )) H − 1 H^{-1} H − 1
g 1 − α ∗ [ u − u 0 ] ∈ H l o c 1 ( 0 , ∞ ; H − 1 ( Ω ) ) g_{1-\alpha}*[u-u_0]\in H^1_{\mathrm{loc}}(0,\infty;H^{-1}(\Omega)) g 1 − α ∗ [ u − u 0 ] ∈ H loc 1 ( 0 , ∞ ; H − 1 ( Ω )) ( g 1 − α ∗ [ u − u 0 ] ) ( 0 ) = 0 (g_{1-\alpha}*[u-u_0])(0)=0 ( g 1 − α ∗ [ u − u 0 ]) ( 0 ) = 0
u m ∈ L l o c 2 ( 0 , ∞ ; H 0 1 ( Ω ) ) u^m\in L^2_{\mathrm{loc}}(0,\infty;H^1_0(\Omega)) u m ∈ L loc 2 ( 0 , ∞ ; H 0 1 ( Ω ))
u u u
∫ Ω ∂ t α ( u ( x , ⋅ ) − u 0 ( x ) ) ( t ) φ ( x ) d x + ∫ Ω ∇ u m ( x , t ) ⋅ ∇ φ ( x ) d x = 0 , a.e. t > 0 (1) \int_\Omega\partial_t^\alpha(u(x,\cdot)-u_0(x))(t)\varphi(x)\,\mathrm{d}x+\int_\Omega\nabla u^m(x,t)\cdot\nabla\varphi(x)\,\mathrm{d}x=0,\qquad\text{a.e. }t>0 \tag*{(1)} ∫ Ω ∂ t α ( u ( x , ⋅ ) − u 0 ( x )) ( t ) φ ( x ) d x + ∫ Ω ∇ u m ( x , t ) ⋅ ∇ φ ( x ) d x = 0 , a.e. t > 0 (1) for any test function φ ∈ H 0 1 ( Ω ) \varphi\in H^1_0(\Omega) φ ∈ H 0 1 ( Ω )
The existence of solutions to time-fractional nonlinear diffusion equations has been studied extensively (see, e.g., [1 , 5 , 10 , 18 , 29 ] ). In particular, Wittbold, Wolejko, and Zacher [29 ] established the existence of bounded weak solutions for bounded initial data in the degenerate case m > 1 m>1 m > 1 [18 ] extended this result to the singular case 0 < m < 1 0<m<1 0 < m < 1 [1 ] developed the abstract theory for time-fractional gradient flows of subdifferential type in a real Hilbert space H H H
∂ t α ( u − u 0 ) ( t ) + ∂ φ ( u ( t ) ) ∋ 0 in H , t > 0 , \partial_t^\alpha(u-u_0)(t)+\partial\varphi(u(t))\ni0\quad\text{in }H,\qquad t>0, ∂ t α ( u − u 0 ) ( t ) + ∂ φ ( u ( t )) ∋ 0 in H , t > 0 , and proved the existence and uniqueness of strong solutions (see also [5 ] ). Here ∂ φ \partial\varphi ∂ φ φ \varphi φ [1 , 10 , 21 ] , for any nonnegative initial datum u 0 ∈ L m + 1 ( Ω ) ∩ H − 1 ( Ω ) u_0 \in L^{m+1}(\Omega) \cap H^{-1}(\Omega) u 0 ∈ L m + 1 ( Ω ) ∩ H − 1 ( Ω ) H − 1 H^{-1} H − 1 u ∈ L ∞ ( 0 , ∞ ; L m + 1 ( Ω ) ∩ H − 1 ( Ω ) ) ∩ C ( [ 0 , ∞ ) ; H − 1 ( Ω ) ) u \in L^\infty(0,\infty;L^{m+1}(\Omega)\cap H^{-1}(\Omega))\cap C([0,\infty);H^{-1}(\Omega)) u ∈ L ∞ ( 0 , ∞ ; L m + 1 ( Ω ) ∩ H − 1 ( Ω )) ∩ C ([ 0 , ∞ ) ; H − 1 ( Ω ))
∥ u ( t ) ∥ m + 1 ≤ ∥ u 0 ∥ m + 1 , a.e. t > 0 , (2) \|u(t)\|_{m+1} \le\|u_0\|_{m+1}, \qquad\text{a.e. }t>0, \tag*{(2)} ∥ u ( t ) ∥ m + 1 ≤ ∥ u 0 ∥ m + 1 , a.e. t > 0 , (2) where ∥ ⋅ ∥ p = ∥ ⋅ ∥ L p ( Ω ) \|\cdot\|_p=\|\cdot\|_{L^p(\Omega)} ∥ ⋅ ∥ p = ∥ ⋅ ∥ L p ( Ω ) 1 ≤ p ≤ ∞ 1\le p\le\infty 1 ≤ p ≤ ∞
We now turn to the large-time behavior. Time decay estimates for time-fractional diffusion equations have been established by many authors (see, e.g., [12 , 13 , 19 , 20 , 23 , 28 ] ). In particular, Vergara and Zacher [28 ] showed that the L 2 ( Ω ) L^2(\Omega) L 2 ( Ω )
∂ t α ( u − u 0 ) − div ( A ( x , t ) ∇ u ) = 0 \partial_t^\alpha(u-u_0)-\operatorname{div}(A(x,t)\nabla u)=0 ∂ t α ( u − u 0 ) − div ( A ( x , t ) ∇ u ) = 0 decays like t − α t^{-\alpha} t − α [10 ] proved the nonextinction property of nonnegative solutions to problem (P) and established the asymptotic behavior
∥ u ( t ) ∥ p ≍ t − α m as t → ∞ \|u(t)\|_p \asymp t^{-\frac{\alpha}{m}} \quad\text{as }t\to\infty ∥ u ( t ) ∥ p ≍ t − m α as t → ∞ for every p ∈ [ 1 , ∞ ] p\in[1,\infty] p ∈ [ 1 , ∞ ] m = 1 m=1 m = 1
∂ t α ( u − u 0 ) = Δ u , x ∈ R N , t > 0 , u ( x , 0 ) = u 0 ( x ) , x ∈ R N , \partial_t^\alpha(u-u_0)=\Delta u,\quad x\in\mathbb{R}^N,\ t>0,\qquad u(x,0)=u_0(x),\quad x\in\mathbb{R}^N, ∂ t α ( u − u 0 ) = Δ u , x ∈ R N , t > 0 , u ( x , 0 ) = u 0 ( x ) , x ∈ R N , Cortázar, Quirós, and Wolanski [13 ] showed that the rescaled solution t α u ( x , t ) t^\alpha u(x,t) t α u ( x , t ) R N \mathbb{R}^N R N
In the classical case, namely, α = 1 \alpha=1 α = 1 0 < m < 1 0<m<1 0 < m < 1 m > 1 m>1 m > 1 [24 , 26 ] ). After a suitable rescaling, problem (P) is transformed into
{ ∂ τ v = Δ v m + λ v , x ∈ Ω , τ > 0 , v = 0 , x ∈ ∂ Ω , τ > 0 , v ( x , 0 ) = v 0 ( x ) , x ∈ Ω . \begin{cases}
\partial_\tau v=\Delta v^m+\lambda v, & x\in\Omega,\quad\tau>0,\\
v=0, & x\in\partial\Omega,\quad\tau>0,\\
v(x,0)=v_0(x), & x\in\Omega.
\end{cases} ⎩ ⎨ ⎧ ∂ τ v = Δ v m + λ v , v = 0 , v ( x , 0 ) = v 0 ( x ) , x ∈ Ω , τ > 0 , x ∈ ∂ Ω , τ > 0 , x ∈ Ω. Based on the theory developed by Feireisl and Simondon [14 ] , for the fast diffusion range m s < m < 1 m_s<m<1 m s < m < 1 [9 ] proved the convergence of the rescaled solution v v v S S S
− Δ S m = λ S in Ω , S = 0 on ∂ Ω , -\Delta S^m=\lambda S \quad\text{in }\Omega,\qquad S=0 \quad\text{on }\partial\Omega, − Δ S m = λ S in Ω , S = 0 on ∂ Ω , where m s = [ N − 2 ] + / ( N + 2 ) m_s=[N-2]_+/(N+2) m s = [ N − 2 ] + / ( N + 2 ) [25 ] gave the corresponding result for the porous medium range m > 1 m>1 m > 1 [6 ] ). Moreover, related asymptotic results for the fast diffusion equation of the form
∂ t ( ∣ u ∣ q − 2 u ) = Δ u , q > 2 \partial_t(|u|^{q-2}u)=\Delta u,\qquad q>2 ∂ t ( ∣ u ∣ q − 2 u ) = Δ u , q > 2 have also been obtained by energy methods (see, e.g., [2 , 3 , 4 ] and the references therein). Thus, in the classical case, suitably rescaled solutions converge to the spatial profile of a separable solution.
On the other hand, the time-fractional problem exhibits significantly different properties. Due to the time nonlocality of fractional derivatives, the standard chain and product rules, as well as the usual rescaling arguments, are not directly applicable. For instance, for sufficiently smooth functions u u u φ \varphi φ
φ ( t ) ∂ t α u ( t ) = ∂ t α ( φ u ) ( t ) − α Γ ( 1 − α ) ∫ 0 t φ ( t ) − φ ( s ) ( t − s ) α + 1 u ( s ) d s . \varphi(t)\partial_t^\alpha u(t)=\partial_t^\alpha(\varphi u)(t)-\frac{\alpha}{\Gamma(1-\alpha)}\int_0^t\frac{\varphi(t)-\varphi(s)}{(t-s)^{\alpha+1}}u(s)\,\mathrm{d}s. φ ( t ) ∂ t α u ( t ) = ∂ t α ( φ u ) ( t ) − Γ ( 1 − α ) α ∫ 0 t ( t − s ) α + 1 φ ( t ) − φ ( s ) u ( s ) d s . This identity follows from the corresponding product rule for nonlocal derivatives (see Lemma 2.2 below for details). The additional nonlocal term illustrates the difficulty of applying time-dependent rescaling arguments to time-fractional diffusion equations. In particular, the identification of the asymptotic profile for the time-fractional nonlinear problem (P) remains an open problem. The aim of the present paper is to identify a profile S S S S S S
Theorem 1.1 Assume that 0 < α < 1 0<\alpha<1 0 < α < 1 m > 0 m>0 m > 0 N ≥ 3 N\geq3 N ≥ 3 p 0 > 1 p_0>1 p 0 > 1
{ p 0 ≥ m + 1 if m + 1 > N 2 , p 0 > N 2 if N 2 ≥ m + 1. \begin{cases}
p_0\geq m+1 & \text{if } m+1>\dfrac{N}{2},\\
p_0>\dfrac{N}{2} & \text{if } \dfrac{N}{2}\geq m+1.
\end{cases} ⎩ ⎨ ⎧ p 0 ≥ m + 1 p 0 > 2 N if m + 1 > 2 N , if 2 N ≥ m + 1. Let u ∈ L ∞ ( 0 , ∞ ; L m + 1 ( Ω ) ∩ H − 1 ( Ω ) ) ∩ C ( [ 0 , ∞ ) ; H − 1 ( Ω ) ) u\in L^\infty(0,\infty;L^{m+1}(\Omega)\cap H^{-1}(\Omega))\cap C([0,\infty);H^{-1}(\Omega)) u ∈ L ∞ ( 0 , ∞ ; L m + 1 ( Ω ) ∩ H − 1 ( Ω )) ∩ C ([ 0 , ∞ ) ; H − 1 ( Ω )) H − 1 H^{-1} H − 1 u 0 ∈ L p 0 ( Ω ) ∩ H − 1 ( Ω ) u_0\in L^{p_0}(\Omega)\cap H^{-1}(\Omega) u 0 ∈ L p 0 ( Ω ) ∩ H − 1 ( Ω ) S S S
− Δ S m = c α u 0 in Ω , S = 0 on ∂ Ω , (3) -\Delta S^m=c_\alpha u_0\quad\text{in }\Omega,\qquad S=0\quad\text{on }\partial\Omega, \tag*{(3)} − Δ S m = c α u 0 in Ω , S = 0 on ∂ Ω , (3) where c α : = 1 / Γ ( 1 − α ) c_\alpha:=1/\Gamma(1-\alpha) c α := 1/Γ ( 1 − α ) 1 ≤ p < ∞ 1\leq p<\infty 1 ≤ p < ∞
lim t → ∞ ∥ t α / m u ( t ) − S ∥ p = 0. (4) \lim_{t\to\infty}\left\|t^{\alpha/m}u(t)-S\right\|_p=0. \tag*{(4)} t → ∞ lim t α / m u ( t ) − S p = 0. (4) Furthermore, it holds that
lim t → ∞ ∥ t α u m ( t ) − c α ∫ Ω G Ω ( ⋅ , y ) u 0 ( y ) d y ∥ p = 0 (5) \lim_{t\to\infty}\left\|t^\alpha u^m(t)-c_\alpha\int_\Omega G_\Omega(\cdot,y)u_0(y)\,\mathrm{d}y\right\|_p=0 \tag*{(5)} t → ∞ lim t α u m ( t ) − c α ∫ Ω G Ω ( ⋅ , y ) u 0 ( y ) d y p = 0 (5) for every 1 ≤ p < ∞ 1\leq p<\infty 1 ≤ p < ∞ G Ω G_\Omega G Ω − Δ Ω -\Delta_\Omega − Δ Ω
Remark 1.1
(i) Since u 0 ∈ H − 1 ( Ω ) u_0\in H^{-1}(\Omega) u 0 ∈ H − 1 ( Ω )
− Δ V = c α u 0 in Ω , V = 0 on ∂ Ω . -\Delta V=c_\alpha u_0\quad\text{in }\Omega,\qquad V=0\quad\text{on }\partial\Omega. − Δ V = c α u 0 in Ω , V = 0 on ∂ Ω. Moreover, since u 0 ≥ 0 u_0\geq0 u 0 ≥ 0 V ≥ 0 V\geq0 V ≥ 0 S m : = V S^m:=V S m := V
∫ Ω ∇ S m ( x ) ⋅ ∇ φ ( x ) d x = c α ∫ Ω u 0 ( x ) φ ( x ) d x (6) \int_\Omega\nabla S^m(x)\cdot\nabla\varphi(x)\,\mathrm{d}x=c_\alpha\int_\Omega u_0(x)\varphi(x)\,\mathrm{d}x \tag*{(6)} ∫ Ω ∇ S m ( x ) ⋅ ∇ φ ( x ) d x = c α ∫ Ω u 0 ( x ) φ ( x ) d x (6) for any test function φ ∈ H 0 1 ( Ω ) \varphi\in H_0^1(\Omega) φ ∈ H 0 1 ( Ω )
(ii) Since u 0 u_0 u 0 L p 0 ( Ω ) L^{p_0}(\Omega) L p 0 ( Ω ) p 0 > N / 2 p_0>N/2 p 0 > N /2 S m S^m S m
S m ( x ) = c α ∫ Ω G Ω ( x , y ) u 0 ( y ) d y . S^m(x)=c_\alpha\int_\Omega G_\Omega(x,y)u_0(y)\,\mathrm{d}y. S m ( x ) = c α ∫ Ω G Ω ( x , y ) u 0 ( y ) d y . Then it holds that ∥ S ∥ ∞ ≤ C ∥ u 0 ∥ p 0 1 / m \lVert S\rVert_\infty\le C\lVert u_0\rVert_{p_0}^{1/m} ∥ S ∥ ∞ ≤ C ∥ u 0 ∥ p 0 1/ m C > 0 C>0 C > 0 α \alpha α m m m N N N Ω \Omega Ω [10 Lemma 4.5] ).
(iii) When m = 1 m=1 m = 1
We briefly explain the idea of the proof of Theorem 1.1. The proof is based on a nonlinear entropy method introduced in [8 ] . More precisely, we set v ( x , t ) : = t α / m u ( x , t ) v(x,t):=t^{\alpha/m}u(x,t) v ( x , t ) := t α / m u ( x , t )
E [ v ∣ S ] ( t ) : = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x . \mathcal{E}[v|S](t):=\int_\Omega\left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^m(x)(v(x,t)-S(x))\right\}\,\mathrm{d}x. E [ v ∣ S ] ( t ) := ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x . Using the Yosida approximation of the nonlocal time-derivative operator and a nonlocal chain rule inequality, we derive a fractional differential inequality for E [ v ∣ S ] \mathcal{E}[v|S] E [ v ∣ S ] S m S^m S m [13 ] for the linear problem in the whole space.
The rest of the present paper is organized as follows. In Section 2, we collect the tools needed in the proof of the main result. In particular, we recall properties of nonlocal time derivatives. In Section 3, we establish decay estimates for fractional ODEs. In Section 4, we prove Theorem 1.1. We first derive a basic entropy inequality for the rescaled solution. Due to the time nonlocality, this inequality exhibits different behavior depending on whether m > α m>\alpha m > α m = α m=\alpha m = α 0 < m < α 0<m<\alpha 0 < m < α m ≥ 1 m\ge1 m ≥ 1 α < m < 1 \alpha<m<1 α < m < 1 m = α m=\alpha m = α 0 < m < α 0<m<\alpha 0 < m < α m = 1 m=1 m = 1
Preliminaries
In this section, we first recall some properties of the Mittag-Leffler function, which is defined by
E a , b ( z ) = ∑ k = 0 ∞ z k Γ ( a k + b ) , z ∈ R , a , b > 0. E_{a,b}(z)=\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(ak+b)},\qquad z\in\mathbb{R},\quad a,b>0. E a , b ( z ) = k = 0 ∑ ∞ Γ ( ak + b ) z k , z ∈ R , a , b > 0. This function satisfies
1 1 + Γ ( 1 − α ) t ≤ E α , 1 ( − t ) ≤ 1 1 + t Γ ( 1 + α ) , t > 0 (7) \frac{1}{1+\Gamma(1-\alpha)t}\le E_{\alpha,1}(-t)\le\frac{1}{1+\frac{t}{\Gamma(1+\alpha)}},\qquad t>0 \tag*{(7)} 1 + Γ ( 1 − α ) t 1 ≤ E α , 1 ( − t ) ≤ 1 + Γ ( 1 + α ) t 1 , t > 0 (7) for 0 < α < 1 0 < \alpha< 1 0 < α < 1
d d t E α , 1 ( − λ t α ) = − λ t α − 1 E α , α ( − λ t α ) , t > 0 (8) \frac{\mathrm{d}}{\mathrm{d}t}E_{\alpha,1}(-\lambda t^\alpha)=-\lambda t^{\alpha-1}E_{\alpha,\alpha}(-\lambda t^\alpha),\qquad t>0 \tag*{(8)} d t d E α , 1 ( − λ t α ) = − λ t α − 1 E α , α ( − λ t α ) , t > 0 (8) as well as that
1 ( 1 + c α , 1 t ) 2 ≤ Γ ( α ) E α , α ( − t ) ≤ 1 ( 1 + c α , 2 t ) 2 , t ≥ 0 , (9) \frac{1}{(1+c_{\alpha,1}t)^2}\leq\Gamma(\alpha)E_{\alpha,\alpha}(-t)\leq\frac{1}{(1+c_{\alpha,2}t)^2},\qquad t\geq0, \tag*{(9)} ( 1 + c α , 1 t ) 2 1 ≤ Γ ( α ) E α , α ( − t ) ≤ ( 1 + c α , 2 t ) 2 1 , t ≥ 0 , (9) where c α , 1 , c α , 2 c_{\alpha,1},c_{\alpha,2} c α , 1 , c α , 2 α \alpha α [11 ] ). We next recall some basic facts of the time-fractional derivative operator ∂ t α \partial_t^\alpha ∂ t α T > 0 T>0 T > 0 1 ≤ p < ∞ 1\leq p<\infty 1 ≤ p < ∞ X X X B B B
B u = d d t ( g 1 − α ∗ u ) , Bu=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*u), B u = d t d ( g 1 − α ∗ u ) , D ( B ) = { u ∈ L p ( 0 , T ; X ) ; g 1 − α ∗ u ∈ W 1 , p ( 0 , T ; X ) with ( g 1 − α ∗ u ) ( 0 ) = 0 } . D(B)=\left\{u\in L^p(0,T;X);\,g_{1-\alpha}*u\in W^{1,p}(0,T;X)\text{ with }(g_{1-\alpha}*u)(0)=0\right\}. D ( B ) = { u ∈ L p ( 0 , T ; X ) ; g 1 − α ∗ u ∈ W 1 , p ( 0 , T ; X ) with ( g 1 − α ∗ u ) ( 0 ) = 0 } . It is well known that B B B m m m L p ( 0 , T ; X ) L^p(0,T;X) L p ( 0 , T ; X ) [7 , 16 , 17 ] ). Let n ∈ N n\in\mathbb{N} n ∈ N B n B_n B n B B B B n = n B ( n + B ) − 1 B_n=nB(n+B)^{-1} B n = n B ( n + B ) − 1 u ∈ D ( B ) u\in D(B) u ∈ D ( B )
B n u → B u strongly in L p ( 0 , T ; X ) as n → ∞ . (10) B_nu\to Bu\quad\text{strongly in }L^p(0,T;X)\text{ as }n\to\infty. \tag*{(10)} B n u → B u strongly in L p ( 0 , T ; X ) as n → ∞. (10) Moreover, B n B_n B n
B n u = d d t ( g 1 − α , n ∗ u ) , u ∈ L p ( 0 , T ; X ) , B_nu=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*u),\qquad u\in L^p(0,T;X), B n u = d t d ( g 1 − α , n ∗ u ) , u ∈ L p ( 0 , T ; X ) , where g 1 − α , n = n s α , n g_{1-\alpha,n}=ns_{\alpha,n} g 1 − α , n = n s α , n s α , n s_{\alpha,n} s α , n
d d t ( g 1 − α ∗ [ s α , n − s α , n ( 0 ) ] ) + n s α , n = 0 , t > 0 , s α , n ( 0 ) = 1. \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha}*[s_{\alpha,n}-s_{\alpha,n}(0)]\right)+ns_{\alpha,n}=0,\qquad t>0,\qquad s_{\alpha,n}(0)=1. d t d ( g 1 − α ∗ [ s α , n − s α , n ( 0 )] ) + n s α , n = 0 , t > 0 , s α , n ( 0 ) = 1. See, e.g., [27 ] . This problem can also be written as the Volterra integral equation
s α , n ( t ) + n ( s α , n ∗ g α ) ( t ) = 1 , t > 0 , (11) s_{\alpha,n}(t)+n(s_{\alpha,n}*g_\alpha)(t)=1,\qquad t>0, \tag*{(11)} s α , n ( t ) + n ( s α , n ∗ g α ) ( t ) = 1 , t > 0 , (11) and, using the Laplace transform, the solution s α , n s_{\alpha,n} s α , n
s α , n ( t ) = E α , 1 ( − n t α ) , t ≥ 0. s_{\alpha,n}(t)=E_{\alpha,1}(-nt^\alpha),\qquad t\geq0. s α , n ( t ) = E α , 1 ( − n t α ) , t ≥ 0. Therefore, by (2.1)–(2.3), s α , n ∈ W l o c 1 , 1 ( [ 0 , ∞ ) ) s_{\alpha,n}\in W^{1,1}_{\mathrm{loc}}([0,\infty)) s α , n ∈ W loc 1 , 1 ([ 0 , ∞ )) h α , n ∈ L l o c 1 ( [ 0 , ∞ ) ) h_{\alpha,n}\in L^1_{\mathrm{loc}}([0,\infty)) h α , n ∈ L loc 1 ([ 0 , ∞ )) n g α ng_\alpha n g α
h α , n ( t ) + n ( h α , n ∗ g α ) ( t ) = n g α ( t ) , t > 0 (12) h_{\alpha,n}(t)+n(h_{\alpha,n}*g_\alpha)(t)=ng_\alpha(t),\qquad t>0 \tag*{(12)} h α , n ( t ) + n ( h α , n ∗ g α ) ( t ) = n g α ( t ) , t > 0 (12) (see, e.g., [17 ] ). Then it follows from g 1 − α ∗ g α = 1 g_{1-\alpha}*g_\alpha=1 g 1 − α ∗ g α = 1
( g 1 − α ∗ h α , n ) ( t ) + n ( g 1 − α ∗ h α , n ∗ g α ) ( t ) = n , t > 0 , (g_{1-\alpha}*h_{\alpha,n})(t)+n(g_{1-\alpha}*h_{\alpha,n}*g_\alpha)(t)=n,\qquad t>0, ( g 1 − α ∗ h α , n ) ( t ) + n ( g 1 − α ∗ h α , n ∗ g α ) ( t ) = n , t > 0 , which together with (2.5) implies that
g 1 − α , n = n s α , n = g 1 − α ∗ h α , n . g_{1-\alpha,n}=n s_{\alpha,n}=g_{1-\alpha}*h_{\alpha,n}. g 1 − α , n = n s α , n = g 1 − α ∗ h α , n . Note that, for any function f ∈ L p ( 0 , T ; X ) f\in L^p(0,T;X) f ∈ L p ( 0 , T ; X )
h α , n ∗ f → f strongly in L p ( 0 , T ; X ) as n → ∞ . (13) h_{\alpha,n}*f\to f\quad\text{strongly in }L^p(0,T;X)\text{ as }n\to\infty. \tag*{(13)} h α , n ∗ f → f strongly in L p ( 0 , T ; X ) as n → ∞. (13) In fact, set v : = g α ∗ f v:=g_\alpha*f v := g α ∗ f v ∈ D ( B ) v\in D(B) v ∈ D ( B )
d d t ( g 1 − α , n ∗ v ) = B n v → n → ∞ B v = d d t ( g 1 − α ∗ v ) = d d t ( g 1 − α ∗ g α ∗ f ) = f in L p ( 0 , T ; X ) , \frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v)=B_nv\xrightarrow[n\to\infty]{}Bv=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*v)=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*g_\alpha*f)=f\quad\text{in }L^p(0,T;X), d t d ( g 1 − α , n ∗ v ) = B n v n → ∞ B v = d t d ( g 1 − α ∗ v ) = d t d ( g 1 − α ∗ g α ∗ f ) = f in L p ( 0 , T ; X ) , while also
B n v = d d t ( g 1 − α ∗ h α , n ∗ g α ∗ f ) = d d t ( g 1 − α ∗ g α ∗ h α , n ∗ f ) = h α , n ∗ f . B_nv=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*h_{\alpha,n}*g_\alpha*f)=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha}*g_\alpha*h_{\alpha,n}*f)=h_{\alpha,n}*f. B n v = d t d ( g 1 − α ∗ h α , n ∗ g α ∗ f ) = d t d ( g 1 − α ∗ g α ∗ h α , n ∗ f ) = h α , n ∗ f . This implies that (2.7) holds. In particular, putting f = g 1 − α f=g_{1-\alpha} f = g 1 − α X = R X=\mathbb{R} X = R
g 1 − α , n → g 1 − α strongly in L 1 ( 0 , T ) as n → ∞ . (14) g_{1-\alpha,n}\to g_{1-\alpha}\quad\text{strongly in }L^1(0,T)\text{ as }n\to\infty. \tag*{(14)} g 1 − α , n → g 1 − α strongly in L 1 ( 0 , T ) as n → ∞. (14) Next, we recall a chain rule inequality and a product rule for nonlocal time-derivative operators. The following lemmas play a crucial role to give a basic inequality for the entropy functional.
Lemma 2.1 [19 Lemma 6.1] Let T > 0 T>0 T > 0 I ⊂ R I\subset\mathbb{R} I ⊂ R h ∈ W 1 , 1 ( 0 , T ) h\in W^{1,1}(0,T) h ∈ W 1 , 1 ( 0 , T ) H ∈ C 1 ( I ) H\in C^1(I) H ∈ C 1 ( I ) u 0 ∈ I u_0\in I u 0 ∈ I u ∈ L 1 ( 0 , T ) u\in L^1(0,T) u ∈ L 1 ( 0 , T ) u ( t ) ∈ I u(t)\in I u ( t ) ∈ I t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T ) H ( u ) H(u) H ( u ) H ′ ( u ) u H'(u)u H ′ ( u ) u H ′ ( u ) ( h ′ ∗ u ) H'(u)(h'*u) H ′ ( u ) ( h ′ ∗ u ) L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T )
H ′ ( u ( t ) ) d d t ( h ∗ u ( ⋅ ) ) ( t ) = d d t ( h ∗ H ( u ( ⋅ ) ) ) ( t ) + ( − H ( u ( t ) ) + H ′ ( u ( t ) ) u ( t ) ) h ( t ) + ∫ 0 t ( H ( u ( t − s ) ) − H ( u ( t ) ) − H ′ ( u ( t ) ) ( u ( t − s ) − u ( t ) ) ) [ − h ′ ( s ) ] d s , a.e. t ∈ ( 0 , T ) . H'(u(t))\frac{\mathrm{d}}{\mathrm{d}t}(h*u(\cdot))(t)=\frac{\mathrm{d}}{\mathrm{d}t}(h*H(u(\cdot)))(t)+\left(-H(u(t))+H'(u(t))u(t)\right)h(t)+\int_0^t\left(H(u(t-s))-H(u(t))-H'(u(t))(u(t-s)-u(t))\right)\left[-h'(s)\right]\mathrm{d}s,\quad\text{a.e. }t\in(0,T). H ′ ( u ( t )) d t d ( h ∗ u ( ⋅ )) ( t ) = d t d ( h ∗ H ( u ( ⋅ ))) ( t ) + ( − H ( u ( t )) + H ′ ( u ( t )) u ( t ) ) h ( t ) + ∫ 0 t ( H ( u ( t − s )) − H ( u ( t )) − H ′ ( u ( t )) ( u ( t − s ) − u ( t )) ) [ − h ′ ( s ) ] d s , a.e. t ∈ ( 0 , T ) . In particular, if H ∈ C 1 ( R ) H\in C^1(\mathbb{R}) H ∈ C 1 ( R ) H ( 0 ) = 0 H(0)=0 H ( 0 ) = 0 h h h
H ′ ( u ( t ) ) d d t ( h ∗ u ( ⋅ ) ) ( t ) ≥ d d t ( h ∗ H ( u ( ⋅ ) ) ) ( t ) , a.e. t ∈ ( 0 , T ) . (15) H'(u(t))\frac{\mathrm{d}}{\mathrm{d}t}(h*u(\cdot))(t)\ge\frac{\mathrm{d}}{\mathrm{d}t}(h*H(u(\cdot)))(t),\quad\text{a.e. }t\in(0,T). \tag*{(15)} H ′ ( u ( t )) d t d ( h ∗ u ( ⋅ )) ( t ) ≥ d t d ( h ∗ H ( u ( ⋅ ))) ( t ) , a.e. t ∈ ( 0 , T ) . (15) Lemma 2.2 [30 Lemma 2.2] Let T > 0 T>0 T > 0 h ∈ W 1 , 1 ( 0 , T ) h\in W^{1,1}(0,T) h ∈ W 1 , 1 ( 0 , T ) u ∈ L 1 ( 0 , T ) u\in L^1(0,T) u ∈ L 1 ( 0 , T ) φ ∈ C 1 ( [ 0 , T ] ) \varphi\in C^1([0,T]) φ ∈ C 1 ([ 0 , T ])
φ ( t ) d d t ( h ∗ u ) ( t ) = d d t ( h ∗ [ φ u ] ) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s ) ) u ( s ) d s , a.e. t ∈ ( 0 , T ) . \varphi(t)\frac{\mathrm{d}}{\mathrm{d}t}(h*u)(t)=\frac{\mathrm{d}}{\mathrm{d}t}(h*[\varphi u])(t)+\int_0^t h'(t-s)(\varphi(t)-\varphi(s))u(s)\,\mathrm{d}s,\quad\text{a.e. }t\in(0,T). φ ( t ) d t d ( h ∗ u ) ( t ) = d t d ( h ∗ [ φ u ]) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s )) u ( s ) d s , a.e. t ∈ ( 0 , T ) . Remark 2.1. In Lemma 2.2, the assumption φ ∈ C 1 ( [ 0 , T ] ) \varphi\in C^1([0,T]) φ ∈ C 1 ([ 0 , T ])
h ∈ W 1 , 1 ( 0 , T ) , φ ∈ L ∞ ( 0 , T ) , u ∈ L 1 ( 0 , T ) , h \in W^{1,1}(0,T), \qquad\varphi\in L^\infty(0,T), \qquad u \in L^1(0,T), h ∈ W 1 , 1 ( 0 , T ) , φ ∈ L ∞ ( 0 , T ) , u ∈ L 1 ( 0 , T ) , then it follows that
φ ( t ) d d t ( h ∗ u ) ( t ) = d d t ( h ∗ [ φ u ] ) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s ) ) u ( s ) d s (16) \varphi(t)\frac{\mathrm{d}}{\mathrm{d}t}(h \ast u)(t)=\frac{\mathrm{d}}{\mathrm{d}t}(h\ast[\varphi u])(t)+\int_0^t h'(t-s)(\varphi(t)-\varphi(s))u(s)\,\mathrm{d}s \tag*{(16)} φ ( t ) d t d ( h ∗ u ) ( t ) = d t d ( h ∗ [ φ u ]) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s )) u ( s ) d s (16) for a.e. t ∈ ( 0 , T ) t \in(0,T) t ∈ ( 0 , T )
h ∗ f ∈ W 1 , 1 ( 0 , T ) , d d t ( h ∗ f ) ( t ) = ( h ′ ∗ f ) ( t ) + h ( 0 ) f ( t ) , a.e. t ∈ ( 0 , T ) h\ast f\in W^{1,1}(0,T), \qquad\frac{\mathrm{d}}{\mathrm{d}t}(h\ast f)(t)=(h'\ast f)(t)+h(0)f(t), \qquad\text{a.e. }t\in(0,T) h ∗ f ∈ W 1 , 1 ( 0 , T ) , d t d ( h ∗ f ) ( t ) = ( h ′ ∗ f ) ( t ) + h ( 0 ) f ( t ) , a.e. t ∈ ( 0 , T ) for any f ∈ L 1 ( 0 , T ) f\in L^1(0,T) f ∈ L 1 ( 0 , T ) h ∈ W 1 , 1 ( 0 , T ) h\in W^{1,1}(0,T) h ∈ W 1 , 1 ( 0 , T )
φ ( t ) d d t ( h ∗ u ) ( t ) = h ( 0 ) φ ( t ) u ( t ) + φ ( t ) ( h ′ ∗ u ) ( t ) = h ( 0 ) φ ( t ) u ( t ) + ( h ′ ∗ [ φ u ] ) ( t ) − ∫ 0 t h ′ ( t − s ) φ ( s ) u ( s ) d s + ∫ 0 t h ′ ( t − s ) φ ( t ) u ( s ) d s = d d t ( h ∗ [ φ u ] ) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s ) ) u ( s ) d s , a.e. t ∈ ( 0 , T ) . \begin{aligned}
\varphi(t)\frac{\mathrm{d}}{\mathrm{d}t}(h\ast u)(t)
&=h(0)\varphi(t)u(t)+\varphi(t)(h'\ast u)(t)\\
&=h(0)\varphi(t)u(t)+(h'\ast[\varphi u])(t)-\int_0^t h'(t-s)\varphi(s)u(s)\,\mathrm{d}s+\int_0^t h'(t-s)\varphi(t)u(s)\,\mathrm{d}s\\
&=\frac{\mathrm{d}}{\mathrm{d}t}(h\ast[\varphi u])(t)+\int_0^t h'(t-s)(\varphi(t)-\varphi(s))u(s)\,\mathrm{d}s,\qquad\text{a.e. }t\in(0,T).
\end{aligned} φ ( t ) d t d ( h ∗ u ) ( t ) = h ( 0 ) φ ( t ) u ( t ) + φ ( t ) ( h ′ ∗ u ) ( t ) = h ( 0 ) φ ( t ) u ( t ) + ( h ′ ∗ [ φ u ]) ( t ) − ∫ 0 t h ′ ( t − s ) φ ( s ) u ( s ) d s + ∫ 0 t h ′ ( t − s ) φ ( t ) u ( s ) d s = d t d ( h ∗ [ φ u ]) ( t ) + ∫ 0 t h ′ ( t − s ) ( φ ( t ) − φ ( s )) u ( s ) d s , a.e. t ∈ ( 0 , T ) . In particular, this extension is applicable to the regularized kernel g 1 − α , n g_{1-\alpha,n} g 1 − α , n
Finally, we recall some elementary inequalities for the following nonnegative convex function.
Lemma 2.3. Let a , b ≥ 0 a,b\geq0 a , b ≥ 0 θ > 0 \theta>0 θ > 0 F θ F_\theta F θ
F θ ( a , b ) = 1 θ + 1 a θ + 1 − 1 θ + 1 b θ + 1 − b θ ( a − b ) . F_\theta(a,b)=\frac{1}{\theta+1}a^{\theta+1}-\frac{1}{\theta+1}b^{\theta+1}-b^\theta(a-b). F θ ( a , b ) = θ + 1 1 a θ + 1 − θ + 1 1 b θ + 1 − b θ ( a − b ) . Then the following assertions hold.
(i) If θ ≥ 1 \theta\geq1 θ ≥ 1
F θ ( a , b ) ≥ 1 θ + 1 ∣ a − b ∣ θ + 1 . F_\theta(a,b)\geq\frac{1}{\theta+1}|a-b|^{\theta+1}. F θ ( a , b ) ≥ θ + 1 1 ∣ a − b ∣ θ + 1 . (ii) If 0 < θ < 1 0<\theta<1 0 < θ < 1
F θ ( a , b ) ≤ M 1 − θ 2 θ ∣ a θ − b θ ∣ 2 , M = max { a , b } . F_\theta(a,b)\leq\frac{M^{1-\theta}}{2\theta}|a^\theta-b^\theta|^2,\qquad M=\max\{a,b\}. F θ ( a , b ) ≤ 2 θ M 1 − θ ∣ a θ − b θ ∣ 2 , M = max { a , b } . Proof. We first prove assertion (i). Assume that a ≥ b a\geq b a ≥ b
x θ − y θ ≥ ( x − y ) θ , x ≥ y ≥ 0 , θ ≥ 1 , x^\theta-y^\theta\geq(x-y)^\theta,\qquad x\geq y\geq0,\ \theta\geq1, x θ − y θ ≥ ( x − y ) θ , x ≥ y ≥ 0 , θ ≥ 1 , we have
F θ ( a , b ) = ∫ b a ( r θ − b θ ) d r ≥ ∫ b a ( r − b ) θ d r = 1 θ + 1 ( a − b ) θ + 1 . F_\theta(a,b)=\int_b^a (r^\theta-b^\theta)\,dr \ge\int_b^a (r-b)^\theta\,dr=\frac{1}{\theta+1}(a-b)^{\theta+1}. F θ ( a , b ) = ∫ b a ( r θ − b θ ) d r ≥ ∫ b a ( r − b ) θ d r = θ + 1 1 ( a − b ) θ + 1 . The case a < b a<b a < b
Next, let 0 < θ < 1 0<\theta<1 0 < θ < 1 a ≥ b a\ge b a ≥ b s = r θ s=r^\theta s = r θ
F θ ( a , b ) = 1 θ ∫ b θ a θ ( s − b θ ) s 1 θ − 1 d s ≤ a 1 − θ θ ∫ b θ a θ ( s − b θ ) d s = a 1 − θ 2 θ ( a θ − b θ ) 2 . F_\theta(a,b)=\frac{1}{\theta}\int_{b^\theta}^{a^\theta}(s-b^\theta)s^{\frac{1}{\theta}-1}\,ds\le\frac{a^{1-\theta}}{\theta}\int_{b^\theta}^{a^\theta}(s-b^\theta)\,ds=\frac{a^{1-\theta}}{2\theta}(a^\theta-b^\theta)^2. F θ ( a , b ) = θ 1 ∫ b θ a θ ( s − b θ ) s θ 1 − 1 d s ≤ θ a 1 − θ ∫ b θ a θ ( s − b θ ) d s = 2 θ a 1 − θ ( a θ − b θ ) 2 . The case a < b a<b a < b a a a b b b
Decay estimates for fractional ODEs
In this section, we establish decay estimates for fractional ODEs. We first recall an asymptotic decay estimate for a nonlinear fractional ODE with time-dependent coefficients. This result will be applied to the case m > α m>\alpha m > α
Proposition 3.1 [23 Lemma 2.1] Let 0 < α < 1 0<\alpha<1 0 < α < 1 β > − α \beta>-\alpha β > − α γ > 0 \gamma>0 γ > 0 λ > 0 \lambda>0 λ > 0 w ∈ C ( [ 0 , ∞ ) ) w\in C([0,\infty)) w ∈ C ([ 0 , ∞ )) Cauchy problem
∂ t α ( w − w 0 ) + λ t β w γ = 0 , t > 0 , w ( 0 ) = w 0 > 0. \partial_t^\alpha(w-w_0)+\lambda t^\beta w^\gamma=0,\qquad t>0,\qquad w(0)=w_0>0. ∂ t α ( w − w 0 ) + λ t β w γ = 0 , t > 0 , w ( 0 ) = w 0 > 0. Then it holds that
c 0 1 + t α + β γ ≤ w ( t ) ≤ c 1 1 + t α + β γ , t > 0 , \frac{c_0}{1+t^{\frac{\alpha+\beta}{\gamma}}}\le w(t)\le\frac{c_1}{1+t^{\frac{\alpha+\beta}{\gamma}}},\qquad t>0, 1 + t γ α + β c 0 ≤ w ( t ) ≤ 1 + t γ α + β c 1 , t > 0 , where c 0 c_0 c 0 c 1 c_1 c 1 α \alpha α β \beta β γ \gamma γ λ \lambda λ w 0 w_0 w 0
In order to give decay estimates for fractional ODEs, we first develop the comparison principle for inhomogeneous fractional ODEs with time-dependent coefficients. By applying the same arguments as in [28 Lemma 2.6] , we obtain the following lemma.
Lemma 3.1 Let f ∈ C ( R ) f\in C(\mathbb{R}) f ∈ C ( R ) nondecreasing function , and r , g ∈ L l o c 1 ( [ 0 , ∞ ) ) ∩ C ( 0 , ∞ ) r,g\in L^1_{\mathrm{loc}}([0,\infty))\cap C(0,\infty) r , g ∈ L loc 1 ([ 0 , ∞ )) ∩ C ( 0 , ∞ ) r r r nonnegative function . Assume that u , v ∈ C ( [ 0 , ∞ ) ) u,v\in C([0,\infty)) u , v ∈ C ([ 0 , ∞ ))
∂ t α ( u − u 0 ) + r ( t ) f ( u ) ≤ g , t > 0 , u ( 0 ) = u 0 , \partial_t^\alpha(u-u_0)+r(t)f(u)\le g,\qquad t>0,\qquad u(0)=u_0, ∂ t α ( u − u 0 ) + r ( t ) f ( u ) ≤ g , t > 0 , u ( 0 ) = u 0 , ∂ t α ( v − v 0 ) + r ( t ) f ( v ) ≥ g , t > 0 , v ( 0 ) = v 0 . \partial_t^\alpha(v-v_0)+r(t)f(v)\ge g,\qquad t>0,\qquad v(0)=v_0. ∂ t α ( v − v 0 ) + r ( t ) f ( v ) ≥ g , t > 0 , v ( 0 ) = v 0 . If u 0 ≤ v 0 u_0\le v_0 u 0 ≤ v 0 then u ( t ) ≤ v ( t ) u(t)\le v(t) u ( t ) ≤ v ( t ) t > 0 t>0 t > 0
The proof of this lemma is very similar to [28 Lemma 2.6] . However, for the reader’s convenience, we write the details.
Proof of Lemma 3.1. Let T > 0 T>0 T > 0 u u u h α , n h_{\alpha,n} h α , n
d d t ( g 1 − α , n ∗ [ u − u 0 ] ) ( t ) + ( h α , n ∗ [ r ( ⋅ ) f ( u ( ⋅ ) ) ] ) ( t ) ≤ ( h α , n ∗ g ) ( t ) , 0 < t < T . (17) \frac{d}{dt}\left(g_{1-\alpha,n}*[u-u_0]\right)(t)+\left(h_{\alpha,n}*\left[r(\cdot)f(u(\cdot))\right]\right)(t)\le(h_{\alpha,n}*g)(t),\qquad0<t<T. \tag*{(17)} d t d ( g 1 − α , n ∗ [ u − u 0 ] ) ( t ) + ( h α , n ∗ [ r ( ⋅ ) f ( u ( ⋅ )) ] ) ( t ) ≤ ( h α , n ∗ g ) ( t ) , 0 < t < T . (17) Indeed, since ( g 1 − α ∗ [ u − u 0 ] ) ( 0 ) = 0 (g_{1-\alpha} * [u-u_0])(0)=0 ( g 1 − α ∗ [ u − u 0 ]) ( 0 ) = 0
( h α , n ∗ [ d d t ( g 1 − α ∗ [ u − u 0 ] ) ] ) ( t ) = d d t ( h α , n ∗ g 1 − α ∗ [ u − u 0 ] ) ( t ) − h α , n ( t ) ( g 1 − α ∗ [ u − u 0 ] ) ( 0 ) = d d t ( g 1 − α , n ∗ [ u − u 0 ] ) ( t ) , 0 < t < T . \left(h_{\alpha,n} * \left[\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha} * [u-u_0])\right]\right)(t)=\frac{\mathrm{d}}{\mathrm{d}t}(h_{\alpha,n} * g_{1-\alpha} * [u-u_0])(t)-h_{\alpha,n}(t)(g_{1-\alpha} * [u-u_0])(0)
=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [u-u_0])(t),\quad0<t<T. ( h α , n ∗ [ d t d ( g 1 − α ∗ [ u − u 0 ]) ] ) ( t ) = d t d ( h α , n ∗ g 1 − α ∗ [ u − u 0 ]) ( t ) − h α , n ( t ) ( g 1 − α ∗ [ u − u 0 ]) ( 0 ) = d t d ( g 1 − α , n ∗ [ u − u 0 ]) ( t ) , 0 < t < T . In the same way, we convolve the inequality for v v v h α , n h_{\alpha,n} h α , n
d d t ( g 1 − α , n ∗ [ u − v ] ) ( t ) + ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ ) ) − f ( v ( ⋅ ) ) ) ] ) ( t ) ≤ g 1 − α , n ( t ) ( u 0 − v 0 ) ≤ 0 \frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [u-v])(t)+\left(h_{\alpha,n} * \left[r(\cdot)\left(f(u(\cdot))-f(v(\cdot))\right)\right]\right)(t)\leq g_{1-\alpha,n}(t)(u_0-v_0)\leq0 d t d ( g 1 − α , n ∗ [ u − v ]) ( t ) + ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ )) − f ( v ( ⋅ )) ) ] ) ( t ) ≤ g 1 − α , n ( t ) ( u 0 − v 0 ) ≤ 0 for 0 < t < T 0<t<T 0 < t < T [ y ] + [y]_+ [ y ] + y ∈ R y\in\mathbb{R} y ∈ R [ u − v ] + [u-v]_+ [ u − v ] + H ( y ) = 1 2 ∣ y ∣ + 2 H(y)=\frac{1}{2}|y|^2_+ H ( y ) = 2 1 ∣ y ∣ + 2
1 2 d d t ( g 1 − α , n ∗ [ u − v ] + 2 ) ( t ) + ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ ) ) − f ( v ( ⋅ ) ) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + ≤ 0 , 0 < t < T . \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [u-v]^2_+)(t)+\left(h_{\alpha,n} * \left[r(\cdot)\left(f(u(\cdot))-f(v(\cdot))\right)\right]\right)(t)[u(t)-v(t)]_+\leq0,\quad0<t<T. 2 1 d t d ( g 1 − α , n ∗ [ u − v ] + 2 ) ( t ) + ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ )) − f ( v ( ⋅ )) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + ≤ 0 , 0 < t < T . Since ( g 1 − α , n ∗ [ u − v ] + 2 ) ( 0 ) = 0 (g_{1-\alpha,n} * [u-v]^2_+)(0)=0 ( g 1 − α , n ∗ [ u − v ] + 2 ) ( 0 ) = 0 0 0 0 T T T
1 2 ( g 1 − α , n ∗ [ u − v ] + 2 ) ( T ) + ∫ 0 T ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ ) ) − f ( v ( ⋅ ) ) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + d t ≤ 0. (18) \frac{1}{2}(g_{1-\alpha,n} * [u-v]^2_+)(T)+\int_0^T\left(h_{\alpha,n} * \left[r(\cdot)\left(f(u(\cdot))-f(v(\cdot))\right)\right]\right)(t)[u(t)-v(t)]_+\,\mathrm{d}t\leq0. \tag*{(18)} 2 1 ( g 1 − α , n ∗ [ u − v ] + 2 ) ( T ) + ∫ 0 T ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ )) − f ( v ( ⋅ )) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + d t ≤ 0. (18) By (2.8), we have
∣ ( g 1 − α , n ∗ [ u − v ] + 2 ) ( t ) − ( g 1 − α ∗ [ u − v ] + 2 ) ( t ) ∣ ≤ ∥ [ u − v ] + 2 ∥ L ∞ ( 0 , T ) ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞ , \left|(g_{1-\alpha,n} * [u-v]^2_+)(t)-(g_{1-\alpha} * [u-v]^2_+)(t)\right|
\leq\|[u-v]^2_+\|_{L^\infty(0,T)}\|g_{1-\alpha,n}-g_{1-\alpha}\|_{L^1(0,T)}
\to0\quad\text{as }n\to\infty, ( g 1 − α , n ∗ [ u − v ] + 2 ) ( t ) − ( g 1 − α ∗ [ u − v ] + 2 ) ( t ) ≤ ∥ [ u − v ] + 2 ∥ L ∞ ( 0 , T ) ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞ , Moreover, since r ∈ L 1 ( 0 , T ) r\in L^1(0,T) r ∈ L 1 ( 0 , T ) f ( u ) ∈ L ∞ ( 0 , T ) f(u)\in L^\infty(0,T) f ( u ) ∈ L ∞ ( 0 , T )
∣ ∫ 0 T ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ ) ) − f ( v ( ⋅ ) ) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + d t − ∫ 0 T r ( t ) ( f ( u ( t ) ) − f ( v ( t ) ) ) [ u ( t ) − v ( t ) ] + d t ∣ ≤ ∥ ( h α , n ∗ [ r ( f ( u ) − f ( v ) ) ] ) − r ( f ( u ) − f ( v ) ) ∥ L 1 ( 0 , T ) ∥ [ u − v ] + ∥ L ∞ ( 0 , T ) → 0 as n → ∞ . \begin{aligned}
&\left|\int_0^T\left(h_{\alpha,n} * \left[r(\cdot)\left(f(u(\cdot))-f(v(\cdot))\right)\right]\right)(t)[u(t)-v(t)]_+\,\mathrm{d}t\right.\\
&\qquad\left.-\int_0^T r(t)\left(f(u(t))-f(v(t))\right)[u(t)-v(t)]_+\,\mathrm{d}t\right|\\
&\leq\left\|\left(h_{\alpha,n} * \left[r\left(f(u)-f(v)\right)\right]\right)-r\left(f(u)-f(v)\right)\right\|_{L^1(0,T)}\|[u-v]_+\|_{L^\infty(0,T)}\\
&\to0\quad\text{as }n\to\infty.
\end{aligned} ∫ 0 T ( h α , n ∗ [ r ( ⋅ ) ( f ( u ( ⋅ )) − f ( v ( ⋅ )) ) ] ) ( t ) [ u ( t ) − v ( t ) ] + d t − ∫ 0 T r ( t ) ( f ( u ( t )) − f ( v ( t )) ) [ u ( t ) − v ( t ) ] + d t ≤ ∥ ( h α , n ∗ [ r ( f ( u ) − f ( v ) ) ] ) − r ( f ( u ) − f ( v ) ) ∥ L 1 ( 0 , T ) ∥ [ u − v ] + ∥ L ∞ ( 0 , T ) → 0 as n → ∞. Therefore, taking the limit n → ∞ n\to\infty n → ∞
1 2 ( g 1 − α ∗ [ u − v ] + 2 ) ( T ) + ∫ 0 T r ( t ) ( f ( u ( t ) ) − f ( v ( t ) ) ) [ u ( t ) − v ( t ) ] + d t ≤ 0. \frac{1}{2}(g_{1-\alpha} * [u-v]^2_+)(T)+\int_0^T r(t)\left(f(u(t))-f(v(t))\right)[u(t)-v(t)]_+\,\mathrm{d}t\leq0. 2 1 ( g 1 − α ∗ [ u − v ] + 2 ) ( T ) + ∫ 0 T r ( t ) ( f ( u ( t )) − f ( v ( t )) ) [ u ( t ) − v ( t ) ] + d t ≤ 0. Since g 1 − α g_{1-\alpha} g 1 − α r r r ( f ( u ) − f ( v ) ) [ u − v ] + ≥ 0 (f(u)-f(v))[u-v]_+\geq0 ( f ( u ) − f ( v )) [ u − v ] + ≥ 0 T > 0 T>0 T > 0 [ u − v ] + = 0 [u-v]_+=0 [ u − v ] + = 0 t > 0 t>0 t > 0
u ( t ) ≤ v ( t ) u(t)\leq v(t) u ( t ) ≤ v ( t ) for all t > 0 t>0 t > 0 □ \square □
We are now ready to establish two decay estimates for fractional ODEs that will be used in the fast diffusion range. First, the following result corresponds to the case 0 < m < α 0<m<\alpha 0 < m < α
Lemma 3.2 Let 0 < α < 1 0<\alpha<1 0 < α < 1 β > 1 − α \beta>1-\alpha β > 1 − α λ , μ > 0 \lambda,\mu>0 λ , μ > 0 y ∈ C ( [ 0 , ∞ ) ) y\in C([0,\infty)) y ∈ C ([ 0 , ∞ ))
∂ t α ( y − y 0 ) + λ t β y = μ t β − 1 , t > 0 , y ( 0 ) = y 0 > 0. (19) \partial_t^\alpha(y-y_0)+\lambda t^\beta y=\mu t^{\beta-1},\qquad t>0,\qquad y(0)=y_0>0. \tag*{(19)} ∂ t α ( y − y 0 ) + λ t β y = μ t β − 1 , t > 0 , y ( 0 ) = y 0 > 0. (19) Then it holds that
y ( t ) ≤ C t − 1 y(t)\le Ct^{-1} y ( t ) ≤ C t − 1 for sufficiently large t > 0 t>0 t > 0 C > 0 C>0 C > 0 α \alpha α β \beta β λ \lambda λ μ \mu μ y 0 y_0 y 0
Proof. We construct a supersolution to problem (3.3), namely, a function w ∈ C ( [ 0 , ∞ ) ) w\in C([0,\infty)) w ∈ C ([ 0 , ∞ ))
∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ 0 , t > 0 , w ( 0 ) = y 0 . (20) \partial_t^\alpha(w-y_0)(t)+\lambda t^\beta w(t)-\mu t^{\beta-1}\ge0,\qquad t>0,\qquad w(0)=y_0. \tag*{(20)} ∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ 0 , t > 0 , w ( 0 ) = y 0 . (20) Define
t 0 α + β : = max { 2 Γ ( α + β ) λ Γ ( β ) , 1 λ Γ ( 1 − α ) ( 2 1 + α + 2 2 + α 1 − α ) } , t_0^{\alpha+\beta}:=\max\left\{\frac{2\Gamma(\alpha+\beta)}{\lambda\Gamma(\beta)},\frac{1}{\lambda\Gamma(1-\alpha)}\left(2^{1+\alpha}+\frac{2^{2+\alpha}}{1-\alpha}\right)\right\}, t 0 α + β := max { λ Γ ( β ) 2Γ ( α + β ) , λ Γ ( 1 − α ) 1 ( 2 1 + α + 1 − α 2 2 + α ) } , and consider the function
w ( t ) = { y 0 + μ Γ ( β ) Γ ( α + β ) t α + β − 1 , 0 ≤ t < t 0 , K t − 1 , t ≥ t 0 . (21) w(t)=
\begin{cases}
y_0+\dfrac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{\alpha+\beta-1}, & 0\le t<t_0,\\
Kt^{-1}, & t\ge t_0.
\end{cases} \tag*{(21)} w ( t ) = ⎩ ⎨ ⎧ y 0 + Γ ( α + β ) μ Γ ( β ) t α + β − 1 , K t − 1 , 0 ≤ t < t 0 , t ≥ t 0 . (21) where K K K
K = y 0 t 0 + μ Γ ( β ) Γ ( α + β ) t 0 α + β . K=y_0t_0+\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}t_0^{\alpha+\beta}. K = y 0 t 0 + Γ ( α + β ) μ Γ ( β ) t 0 α + β . Then we easily see that w ∈ C ( [ 0 , ∞ ) ) ∩ W l o c 1 , 1 ( [ 0 , ∞ ) ) w\in C([0,\infty))\cap W_{\mathrm{loc}}^{1,1}([0,\infty)) w ∈ C ([ 0 , ∞ )) ∩ W loc 1 , 1 ([ 0 , ∞ ))
w ′ ( t ) = { μ Γ ( β ) Γ ( α + β − 1 ) t α + β − 2 , 0 ≤ t < t 0 , − K t − 2 , t ≥ t 0 . w'(t)=
\begin{cases}
\dfrac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta-1)}t^{\alpha+\beta-2}, & 0\le t<t_0,\\
-Kt^{-2}, & t\ge t_0.
\end{cases} w ′ ( t ) = ⎩ ⎨ ⎧ Γ ( α + β − 1 ) μ Γ ( β ) t α + β − 2 , − K t − 2 , 0 ≤ t < t 0 , t ≥ t 0 . For t ∈ [ 0 , t 0 ) t\in[0,t_0) t ∈ [ 0 , t 0 )
∂ t α ( w − y 0 ) ( t ) = ( g 1 − α ∗ w ′ ) ( t ) = μ Γ ( β ) Γ ( α + β − 1 ) Γ ( 1 − α ) ∫ 0 t ( t − s ) − α s α + β − 2 d s = μ t β − 1 . \partial_t^\alpha(w-y_0)(t)=(g_{1-\alpha}*w')(t)
=\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta-1)\Gamma(1-\alpha)}
\int_0^t(t-s)^{-\alpha}s^{\alpha+\beta-2}\,\mathrm{d}s
=\mu t^{\beta-1}. ∂ t α ( w − y 0 ) ( t ) = ( g 1 − α ∗ w ′ ) ( t ) = Γ ( α + β − 1 ) Γ ( 1 − α ) μ Γ ( β ) ∫ 0 t ( t − s ) − α s α + β − 2 d s = μ t β − 1 . This implies that
∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 = μ t β − 1 + λ t β ( y 0 + μ Γ ( β ) Γ ( α + β ) t α + β − 1 ) − μ t β − 1 = λ t β ( y 0 + μ Γ ( β ) Γ ( α + β ) t α + β − 1 ) ≥ 0. \begin{aligned}
\partial_t^\alpha(w-y_0)(t)+\lambda t^\beta w(t)-\mu t^{\beta-1}
&=\mu t^{\beta-1}+\lambda t^\beta\left(y_0+\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{\alpha+\beta-1}\right)-\mu t^{\beta-1}\\
&=\lambda t^\beta\left(y_0+\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{\alpha+\beta-1}\right)\ge0.
\end{aligned} ∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 = μ t β − 1 + λ t β ( y 0 + Γ ( α + β ) μ Γ ( β ) t α + β − 1 ) − μ t β − 1 = λ t β ( y 0 + Γ ( α + β ) μ Γ ( β ) t α + β − 1 ) ≥ 0. for 0 ≤ t < t 0 0 \le t < t_0 0 ≤ t < t 0 t ≥ t 0 t \ge t_0 t ≥ t 0
∂ t α ( w − y 0 ) ( t ) = ( g 1 − α ∗ w ′ ) ( t ) = μ Γ ( β ) Γ ( α + β − 1 ) ∫ 0 t 0 g 1 − α ( t − s ) s α + β − 2 d s − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s ≥ − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s . \begin{aligned}
\partial_t^\alpha(w-y_0)(t)=(g_{1-\alpha}*w')(t)
=\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta-1)}\int_0^{t_0}g_{1-\alpha}(t-s)s^{\alpha+\beta-2}\,\mathrm{d}s-K\int_{t_0}^{t}g_{1-\alpha}(t-s)s^{-2}\,\mathrm{d}s \\
&\ge-K\int_{t_0}^{t}g_{1-\alpha}(t-s)s^{-2}\,\mathrm{d}s.
\end{aligned} ∂ t α ( w − y 0 ) ( t ) = ( g 1 − α ∗ w ′ ) ( t ) = Γ ( α + β − 1 ) μ Γ ( β ) ∫ 0 t 0 g 1 − α ( t − s ) s α + β − 2 d s − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s ≥ − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s . If t 0 ≤ t ≤ 2 t 0 t_0\le t\le2t_0 t 0 ≤ t ≤ 2 t 0
− K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s ≥ − K t 0 − 2 ( t − t 0 ) 1 − α Γ ( 2 − α ) ≥ − K Γ ( 2 − α ) t 0 − 1 − α ≥ − K 2 1 + α Γ ( 2 − α ) t − 1 − α . -K\int_{t_0}^{t}g_{1-\alpha}(t-s)s^{-2}\,\mathrm{d}s\ge-Kt_0^{-2}\frac{(t-t_0)^{1-\alpha}}{\Gamma(2-\alpha)}\ge-\frac{K}{\Gamma(2-\alpha)}t_0^{-1-\alpha}\ge-\frac{K2^{1+\alpha}}{\Gamma(2-\alpha)}t^{-1-\alpha}. − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s ≥ − K t 0 − 2 Γ ( 2 − α ) ( t − t 0 ) 1 − α ≥ − Γ ( 2 − α ) K t 0 − 1 − α ≥ − Γ ( 2 − α ) K 2 1 + α t − 1 − α . By the definition of K K K t 0 t_0 t 0
K ≥ μ Γ ( β ) Γ ( α + β ) t 0 α + β ≥ μ Γ ( β ) Γ ( α + β ) 2 Γ ( α + β ) λ Γ ( β ) = 2 μ λ . K\ge\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}t_0^{\alpha+\beta}\ge\frac{\mu\Gamma(\beta)}{\Gamma(\alpha+\beta)}\frac{2\Gamma(\alpha+\beta)}{\lambda\Gamma(\beta)}=\frac{2\mu}{\lambda}. K ≥ Γ ( α + β ) μ Γ ( β ) t 0 α + β ≥ Γ ( α + β ) μ Γ ( β ) λ Γ ( β ) 2Γ ( α + β ) = λ 2 μ . and then we obtain
∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ − K 2 1 + α Γ ( 2 − α ) t − 1 − α + λ K t β − 1 − λ K 2 t β − 1 ≥ λ K 2 t − α − 1 ( − 2 2 + α λ Γ ( 2 − α ) + t 0 α + β ) ≥ λ K 2 t − α − 1 ( − 2 2 + α λ Γ ( 2 − α ) + 1 λ Γ ( 1 − α ) ( 2 1 + α + 2 2 + α 1 − α ) ) = K 2 α Γ ( 1 − α ) t − α − 1 ≥ 0. \begin{aligned}
\partial_t^\alpha(w-y_0)(t)+\lambda t^\beta w(t)-\mu t^{\beta-1}\ge-\frac{K2^{1+\alpha}}{\Gamma(2-\alpha)}t^{-1-\alpha}+\lambda Kt^{\beta-1}-\frac{\lambda K}{2}t^{\beta-1} \\
&\ge\frac{\lambda K}{2}t^{-\alpha-1}\left(-\frac{2^{2+\alpha}}{\lambda\Gamma(2-\alpha)}+t_0^{\alpha+\beta}\right) \\
&\ge\frac{\lambda K}{2}t^{-\alpha-1}\left(-\frac{2^{2+\alpha}}{\lambda\Gamma(2-\alpha)}+\frac{1}{\lambda\Gamma(1-\alpha)}\left(2^{1+\alpha}+\frac{2^{2+\alpha}}{1-\alpha}\right)\right) \\
&=\frac{K2^\alpha}{\Gamma(1-\alpha)}t^{-\alpha-1}\ge0.
\end{aligned} ∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ − Γ ( 2 − α ) K 2 1 + α t − 1 − α + λ K t β − 1 − 2 λ K t β − 1 ≥ 2 λ K t − α − 1 ( − λ Γ ( 2 − α ) 2 2 + α + t 0 α + β ) ≥ 2 λ K t − α − 1 ( − λ Γ ( 2 − α ) 2 2 + α + λ Γ ( 1 − α ) 1 ( 2 1 + α + 1 − α 2 2 + α ) ) = Γ ( 1 − α ) K 2 α t − α − 1 ≥ 0. For t ≥ 2 t 0 t\ge2t_0 t ≥ 2 t 0 r = t / t 0 ≥ 2 r=t/t_0\ge2 r = t / t 0 ≥ 2
− K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s = − K Γ ( 1 − α ) t 0 − 1 − α ∫ 1 r ( r − τ ) − α τ − 2 d τ = − K Γ ( 1 − α ) t 0 − 1 − α ( ∫ 1 r / 2 ( r − τ ) − α τ − 2 d τ + ∫ r / 2 r ( r − τ ) − α τ − 2 d τ ) ≥ − K Γ ( 1 − α ) t 0 − 1 − α ( 2 α r − α ∫ 1 ∞ τ − 2 d τ + 4 r − 2 ∫ r / 2 r ( r − τ ) − α d τ ) = − K Γ ( 1 − α ) t 0 − 1 − α ( 2 α r − α + 2 1 + α 1 − α r − 1 − α ) ≥ − K Γ ( 1 − α ) t 0 − 1 − α ( 2 α + 2 1 + α 1 − α ) r β − 1 = − K Γ ( 1 − α ) t 0 − α − β ( 2 α + 2 1 + α 1 − α ) t β − 1 . \begin{aligned}
-K\int_{t_0}^{t}g_{1-\alpha}(t-s)s^{-2}\,\mathrm{d}s
=-\frac{K}{\Gamma(1-\alpha)}t_0^{-1-\alpha}\int_1^r(r-\tau)^{-\alpha}\tau^{-2}\,\mathrm{d}\tau\\
&=-\frac{K}{\Gamma(1-\alpha)}t_0^{-1-\alpha}\left(\int_1^{r/2}(r-\tau)^{-\alpha}\tau^{-2}\,\mathrm{d}\tau+\int_{r/2}^{r}(r-\tau)^{-\alpha}\tau^{-2}\,\mathrm{d}\tau\right) \\
&\ge-\frac{K}{\Gamma(1-\alpha)}t_0^{-1-\alpha}\left(2^\alpha r^{-\alpha}\int_1^\infty\tau^{-2}\,\mathrm{d}\tau+4r^{-2}\int_{r/2}^{r}(r-\tau)^{-\alpha}\,\mathrm{d}\tau\right) \\
&=-\frac{K}{\Gamma(1-\alpha)}t_0^{-1-\alpha}\left(2^\alpha r^{-\alpha}+\frac{2^{1+\alpha}}{1-\alpha}r^{-1-\alpha}\right) \\
&\ge-\frac{K}{\Gamma(1-\alpha)}t_0^{-1-\alpha}\left(2^\alpha+\frac{2^{1+\alpha}}{1-\alpha}\right)r^{\beta-1}
=-\frac{K}{\Gamma(1-\alpha)}t_0^{-\alpha-\beta}\left(2^\alpha+\frac{2^{1+\alpha}}{1-\alpha}\right)t^{\beta-1}.
\end{aligned} − K ∫ t 0 t g 1 − α ( t − s ) s − 2 d s = − Γ ( 1 − α ) K t 0 − 1 − α ∫ 1 r ( r − τ ) − α τ − 2 d τ = − Γ ( 1 − α ) K t 0 − 1 − α ( ∫ 1 r /2 ( r − τ ) − α τ − 2 d τ + ∫ r /2 r ( r − τ ) − α τ − 2 d τ ) ≥ − Γ ( 1 − α ) K t 0 − 1 − α ( 2 α r − α ∫ 1 ∞ τ − 2 d τ + 4 r − 2 ∫ r /2 r ( r − τ ) − α d τ ) = − Γ ( 1 − α ) K t 0 − 1 − α ( 2 α r − α + 1 − α 2 1 + α r − 1 − α ) ≥ − Γ ( 1 − α ) K t 0 − 1 − α ( 2 α + 1 − α 2 1 + α ) r β − 1 = − Γ ( 1 − α ) K t 0 − α − β ( 2 α + 1 − α 2 1 + α ) t β − 1 . Here we used the fact that
r 1 − α ≤ r − α ≤ r β − 1 r^{1-\alpha} \le r^{-\alpha} \le r^{\beta-1} r 1 − α ≤ r − α ≤ r β − 1 for r ≥ 2 r \ge2 r ≥ 2 β − 1 > − α \beta-1>-\alpha β − 1 > − α
∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ − K Γ ( 1 − α ) t 0 − α − β ( 2 α + 2 1 + α 1 − α ) t β − 1 + λ K 2 t β − 1 ≥ λ K 2 t 0 − α − β t β − 1 ( − 1 λ Γ ( 1 − α ) ( 2 α + 1 + 2 2 + α 1 − α ) + t 0 α + β ) ≥ 0. \begin{aligned}
\partial_t^\alpha(w-y_0)(t)+\lambda t^\beta w(t)-\mu t^{\beta-1}
&\ge-\frac{K}{\Gamma(1-\alpha)}t_0^{-\alpha-\beta}\left(2^\alpha+\frac{2^{1+\alpha}}{1-\alpha}\right)t^{\beta-1}+\frac{\lambda K}{2}t^{\beta-1}\\
&\ge\frac{\lambda K}{2}t_0^{-\alpha-\beta}t^{\beta-1}\left(-\frac{1}{\lambda\Gamma(1-\alpha)}\left(2^{\alpha+1}+\frac{2^{2+\alpha}}{1-\alpha}\right)+t_0^{\alpha+\beta}\right)\\
&\ge0.
\end{aligned} ∂ t α ( w − y 0 ) ( t ) + λ t β w ( t ) − μ t β − 1 ≥ − Γ ( 1 − α ) K t 0 − α − β ( 2 α + 1 − α 2 1 + α ) t β − 1 + 2 λ K t β − 1 ≥ 2 λ K t 0 − α − β t β − 1 ( − λ Γ ( 1 − α ) 1 ( 2 α + 1 + 1 − α 2 2 + α ) + t 0 α + β ) ≥ 0. Thus the function w w w (21) satisfies (20) , so we obtain by Lemma 3.1
y ( t ) ≤ w ( t ) , t > 0. y(t)\le w(t),\qquad t>0. y ( t ) ≤ w ( t ) , t > 0. This implies that
y ( t ) ≤ C t − 1 , t ≥ t 0 . y(t)\le Ct^{-1},\qquad t\ge t_0. y ( t ) ≤ C t − 1 , t ≥ t 0 . Hence the proof of Lemma 3.2 is complete. □
Finally, we give the following decay estimate for fractional ODEs with logarithmic term. We will use the following result in the proof of Theorem 1.1 for the critical case m = α m=\alpha m = α
Lemma 3.3 Let 0 < α < 1 0<\alpha<1 0 < α < 1 λ , μ > 0 \lambda,\mu>0 λ , μ > 0 y ∈ C ( [ 0 , ∞ ) ) y\in C([0,\infty)) y ∈ C ([ 0 , ∞ ))
∂ t α ( y − y 0 ) + λ t 1 − α y = μ t − α log ( 1 + t ) , t > 0 , y ( 0 ) = y 0 > 0. \partial_t^\alpha(y-y_0)+\lambda t^{1-\alpha}y=\mu t^{-\alpha}\log(1+t),\qquad t>0,\qquad y(0)=y_0>0. ∂ t α ( y − y 0 ) + λ t 1 − α y = μ t − α log ( 1 + t ) , t > 0 , y ( 0 ) = y 0 > 0. Then it holds that
y ( t ) ≤ C log ( 1 + t ) t y(t)\le C\frac{\log(1+t)}{t} y ( t ) ≤ C t log ( 1 + t ) for sufficiently large t > 0 t>0 t > 0 C > 0 C>0 C > 0 α \alpha α λ \lambda λ μ \mu μ y 0 y_0 y 0
Proof. As in Lemma 3.1, we construct a function z ∈ C ( [ 0 , ∞ ) ) z\in C([0,\infty)) z ∈ C ([ 0 , ∞ ))
∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ 0 , t > 0. (22) \partial_t^\alpha(z-y_0)(t)+\lambda t^{1-\alpha}z(t)-\mu t^{-\alpha}\log(1+t)\ge0,\qquad t>0. \tag*{(22)} ∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ 0 , t > 0. (22) Define
t 0 : = max { e − 1 , exp ( ( 2 − α ) 2 2 + α λ Γ ( 2 − α ) ) − 1 , exp ( 2 α + 1 λ Γ ( 1 − α ) ) − 1 } , t_0:=\max\left\{e-1,\exp\left(\frac{(2-\alpha)2^{2+\alpha}}{\lambda\Gamma(2-\alpha)}\right)-1,\exp\left(\frac{2^{\alpha+1}}{\lambda\Gamma(1-\alpha)}\right)-1\right\}, t 0 := max { e − 1 , exp ( λ Γ ( 2 − α ) ( 2 − α ) 2 2 + α ) − 1 , exp ( λ Γ ( 1 − α ) 2 α + 1 ) − 1 } , and consider the function
z ( t ) = { y 0 + μ Γ ( 2 − α ) t , 0 ≤ t < t 0 , K log ( 1 + t ) t , t ≥ t 0 . (23) z(t)=
\begin{cases}
y_0+\mu\Gamma(2-\alpha)t, & 0\le t<t_0,\\
K\frac{\log(1+t)}{t}, & t\ge t_0.
\end{cases} \tag*{(23)} z ( t ) = { y 0 + μ Γ ( 2 − α ) t , K t l o g ( 1 + t ) , 0 ≤ t < t 0 , t ≥ t 0 . (23) where
K = t 0 ( y 0 + μ Γ ( 2 − α ) t 0 ) log ( 1 + t 0 ) . K=\frac{t_0(y_0+\mu\Gamma(2-\alpha)t_0)}{\log(1+t_0)}. K = log ( 1 + t 0 ) t 0 ( y 0 + μ Γ ( 2 − α ) t 0 ) . Then we observe that z ∈ C ( [ 0 , ∞ ) ) ∩ W l o c 1 , 1 ( [ 0 , ∞ ) ) z\in C([0,\infty))\cap W_{\mathrm{loc}}^{1,1}([0,\infty)) z ∈ C ([ 0 , ∞ )) ∩ W loc 1 , 1 ([ 0 , ∞ ))
z ′ ( t ) = { μ Γ ( 2 − α ) , 0 ≤ t < t 0 , K ( − log ( 1 + t ) t 2 + 1 t ( 1 + t ) ) , t ≥ t 0 . z'(t)=
\begin{cases}
\mu\Gamma(2-\alpha), & 0\le t<t_0,\\
K\left(-\frac{\log(1+t)}{t^2}+\frac{1}{t(1+t)}\right), & t\ge t_0.
\end{cases} z ′ ( t ) = { μ Γ ( 2 − α ) , K ( − t 2 l o g ( 1 + t ) + t ( 1 + t ) 1 ) , 0 ≤ t < t 0 , t ≥ t 0 . For 0 ≤ t < t 0 0\le t<t_0 0 ≤ t < t 0
∂ t α ( z − y 0 ) ( t ) = ( g 1 − α ∗ z ′ ) ( t ) = μ t 1 − α . \partial_t^\alpha(z-y_0)(t)=(g_{1-\alpha}*z')(t)=\mu t^{1-\alpha}. ∂ t α ( z − y 0 ) ( t ) = ( g 1 − α ∗ z ′ ) ( t ) = μ t 1 − α . Hence, since the elementary inequality
log ( t + 1 ) t ≤ 1 , t ≥ 0 , (24) \frac{\log(t+1)}{t}\le1,\qquad t\ge0, \tag*{(24)} t log ( t + 1 ) ≤ 1 , t ≥ 0 , (24) it follows that
∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) = t 1 − α ( μ + λ ( y 0 + μ Γ ( 2 − α ) t ) − μ log ( 1 + t ) t ) ≥ t 1 − α ( μ + λ ( y 0 + μ Γ ( 2 − α ) t ) − μ ) = λ t 1 − α ( y 0 + μ Γ ( 2 − α ) t ) ≥ 0. \begin{aligned}
\partial_t^\alpha(z-y_0)(t)+\lambda t^{1-\alpha}z(t)-\mu t^{-\alpha}\log(1+t)
&=t^{1-\alpha}\left(\mu+\lambda\left(y_0+\mu\Gamma(2-\alpha)t\right)-\mu\frac{\log(1+t)}{t}\right)\\
&\ge t^{1-\alpha}\left(\mu+\lambda\left(y_0+\mu\Gamma(2-\alpha)t\right)-\mu\right)\\
&=\lambda t^{1-\alpha}\left(y_0+\mu\Gamma(2-\alpha)t\right)\ge0.
\end{aligned} ∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) = t 1 − α ( μ + λ ( y 0 + μ Γ ( 2 − α ) t ) − μ t log ( 1 + t ) ) ≥ t 1 − α ( μ + λ ( y 0 + μ Γ ( 2 − α ) t ) − μ ) = λ t 1 − α ( y 0 + μ Γ ( 2 − α ) t ) ≥ 0. For t ≥ t 0 t\ge t_0 t ≥ t 0 g 1 − α > 0 g_{1-\alpha}>0 g 1 − α > 0 z z z
∂ t α ( z − y 0 ) ( t ) = μ Γ ( 2 − α ) ∫ 0 t 0 g 1 − α ( t − s ) d s + K ∫ t 0 t g 1 − α ( t − s ) ( − log ( 1 + s ) s 2 + 1 s ( 1 + s ) ) d s ≥ − K ∫ t 0 t g 1 − α ( t − s ) log ( 1 + s ) s 2 d s . \begin{aligned}
\partial_t^\alpha(z-y_0)(t)
&=\mu\Gamma(2-\alpha)\int_0^{t_0}g_{1-\alpha}(t-s)\,ds\\
&\quad+K\int_{t_0}^{t}g_{1-\alpha}(t-s)\left(-\frac{\log(1+s)}{s^2}+\frac{1}{s(1+s)}\right)\,ds\\
&\ge-K\int_{t_0}^{t}g_{1-\alpha}(t-s)\frac{\log(1+s)}{s^2}\,ds.
\end{aligned} ∂ t α ( z − y 0 ) ( t ) = μ Γ ( 2 − α ) ∫ 0 t 0 g 1 − α ( t − s ) d s + K ∫ t 0 t g 1 − α ( t − s ) ( − s 2 log ( 1 + s ) + s ( 1 + s ) 1 ) d s ≥ − K ∫ t 0 t g 1 − α ( t − s ) s 2 log ( 1 + s ) d s . If t 0 ≤ t ≤ 2 t 0 t_0\le t\le2t_0 t 0 ≤ t ≤ 2 t 0
− K ∫ t 0 t g 1 − α ( t − s ) log ( 1 + s ) s 2 d s ≥ − K t 0 − 1 ∫ t 0 t g 1 − α ( t − s ) d s = − K t 0 − 1 g 2 − α ( t − t 0 ) ≥ − K Γ ( 2 − α ) t 0 − α ≥ − K 2 α Γ ( 2 − α ) t − α . \begin{aligned}
-K\int_{t_0}^{t}g_{1-\alpha}(t-s)\frac{\log(1+s)}{s^2}\,ds
&\ge-Kt_0^{-1}\int_{t_0}^{t}g_{1-\alpha}(t-s)\,ds\\
&=-Kt_0^{-1}g_{2-\alpha}(t-t_0)\\
&\ge-\frac{K}{\Gamma(2-\alpha)}t_0^{-\alpha}
\ge-\frac{K2^\alpha}{\Gamma(2-\alpha)}t^{-\alpha}.
\end{aligned} − K ∫ t 0 t g 1 − α ( t − s ) s 2 log ( 1 + s ) d s ≥ − K t 0 − 1 ∫ t 0 t g 1 − α ( t − s ) d s = − K t 0 − 1 g 2 − α ( t − t 0 ) ≥ − Γ ( 2 − α ) K t 0 − α ≥ − Γ ( 2 − α ) K 2 α t − α . Note that, by the definition of t 0 t_0 t 0 exp ( s ) − 1 ≥ s \exp(s)-1\ge s exp ( s ) − 1 ≥ s s ∈ R s\in\mathbb{R} s ∈ R
t 0 ≥ exp ( ( 2 − α ) 2 1 + α λ Γ ( 2 − α ) ) − 1 ≥ ( 2 − α ) 2 1 + α λ Γ ( 2 − α ) ≥ 2 λ Γ ( 2 − α ) , t_0\ge\exp\left(\frac{(2-\alpha)2^{1+\alpha}}{\lambda\Gamma(2-\alpha)}\right)-1
\ge\frac{(2-\alpha)2^{1+\alpha}}{\lambda\Gamma(2-\alpha)}
\ge\frac{2}{\lambda\Gamma(2-\alpha)}, t 0 ≥ exp ( λ Γ ( 2 − α ) ( 2 − α ) 2 1 + α ) − 1 ≥ λ Γ ( 2 − α ) ( 2 − α ) 2 1 + α ≥ λ Γ ( 2 − α ) 2 , so we get
K ≥ μ Γ ( 2 − α ) t 0 2 log ( 1 + t 0 ) ≥ μ Γ ( 2 − α ) t 0 ≥ μ Γ ( 2 − α ) 2 λ Γ ( 2 − α ) = 2 μ λ . K\ge\frac{\mu\Gamma(2-\alpha)t_0^2}{\log(1+t_0)}
\ge\mu\Gamma(2-\alpha)t_0
\ge\mu\Gamma(2-\alpha)\frac{2}{\lambda\Gamma(2-\alpha)}
=\frac{2\mu}{\lambda}. K ≥ log ( 1 + t 0 ) μ Γ ( 2 − α ) t 0 2 ≥ μ Γ ( 2 − α ) t 0 ≥ μ Γ ( 2 − α ) λ Γ ( 2 − α ) 2 = λ 2 μ . Thus we have for t 0 ≤ t ≤ 2 t 0 t_0 \le t \le2t_0 t 0 ≤ t ≤ 2 t 0
∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ − K 2 α Γ ( 2 − α ) t − α + λ K 2 t − α log ( 1 + t ) ≥ λ K 2 t − α ( − 2 α + 1 λ Γ ( 2 − α ) + log ( 1 + t 0 ) ) ≥ 0. \begin{aligned}
\partial_t^\alpha(z-y_0)(t)+\lambda t^{1-\alpha}z(t)-\mu t^{-\alpha}\log(1+t)
\ge-\frac{K2^\alpha}{\Gamma(2-\alpha)}t^{-\alpha}+\frac{\lambda K}{2}t^{-\alpha}\log(1+t) \\
&\ge\frac{\lambda K}{2}t^{-\alpha}\left(-\frac{2^{\alpha+1}}{\lambda\Gamma(2-\alpha)}+\log(1+t_0)\right)\ge0.
\end{aligned} ∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ − Γ ( 2 − α ) K 2 α t − α + 2 λ K t − α log ( 1 + t ) ≥ 2 λ K t − α ( − λ Γ ( 2 − α ) 2 α + 1 + log ( 1 + t 0 ) ) ≥ 0. For t ≥ 2 t 0 t\ge2t_0 t ≥ 2 t 0 r = t / t 0 ≥ 2 r=t/t_0\ge2 r = t / t 0 ≥ 2
− K ∫ t 0 t g 1 − α ( t − s ) log ( 1 + s ) s 2 d s = − K Γ ( 1 − α ) t 0 α − 1 ∫ 1 r ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ = − K Γ ( 1 − α ) t 0 α − 1 ( ∫ 1 r / 2 ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ + ∫ r / 2 r ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ ) . \begin{aligned}
-K\int_{t_0}^{t}g_{1-\alpha}(t-s)\frac{\log(1+s)}{s^2}\,\mathrm{d}s
=-\frac{K}{\Gamma(1-\alpha)}t_0^{\alpha-1}\int_1^r(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau\\
&=-\frac{K}{\Gamma(1-\alpha)}t_0^{\alpha-1}\left(\int_1^{r/2}(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau+\int_{r/2}^r(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau\right).
\end{aligned} − K ∫ t 0 t g 1 − α ( t − s ) s 2 log ( 1 + s ) d s = − Γ ( 1 − α ) K t 0 α − 1 ∫ 1 r ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ = − Γ ( 1 − α ) K t 0 α − 1 ( ∫ 1 r /2 ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ + ∫ r /2 r ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ ) . For τ ∈ ( 1 , r / 2 ) \tau\in(1,r/2) τ ∈ ( 1 , r /2 ) τ ≥ 1 \tau\ge1 τ ≥ 1 1 + t 0 ≥ e 1+t_0\ge e 1 + t 0 ≥ e
log ( 1 + t 0 τ ) ≤ log ( ( 1 + t 0 ) τ ) \log(1+t_0\tau)\le\log((1+t_0)\tau) log ( 1 + t 0 τ ) ≤ log (( 1 + t 0 ) τ ) and that log ( ( 1 + t 0 ) τ ) ≥ 1 \log((1+t_0)\tau)\ge1 log (( 1 + t 0 ) τ ) ≥ 1
∫ 1 r / 2 ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ ≤ 2 α r − α ∫ 1 r / 2 log ( 1 + t 0 τ ) τ 2 d τ \int_1^{r/2}(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau
\le2^\alpha r^{-\alpha}\int_1^{r/2}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau ∫ 1 r /2 ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ ≤ 2 α r − α ∫ 1 r /2 τ 2 log ( 1 + t 0 τ ) d τ ≤ 2 α r − α ( ∫ 1 r / 2 log ( ( 1 + t 0 ) τ ) − 1 τ 2 d τ + ∫ 1 r / 2 τ − 2 d τ ) . (25) \le2^\alpha r^{-\alpha}\left(\int_1^{r/2}\frac{\log((1+t_0)\tau)-1}{\tau^2}\,\mathrm{d}\tau+\int_1^{r/2}\tau^{-2}\,\mathrm{d}\tau\right). \tag*{(25)} ≤ 2 α r − α ( ∫ 1 r /2 τ 2 log (( 1 + t 0 ) τ ) − 1 d τ + ∫ 1 r /2 τ − 2 d τ ) . (25) Note that
0 ≤ log ( ( 1 + t 0 ) τ ) − 1 τ 2 = − d d τ ( log ( ( 1 + t 0 ) τ ) τ ) , 0\le\frac{\log((1+t_0)\tau)-1}{\tau^2}=-\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\log((1+t_0)\tau)}{\tau}\right), 0 ≤ τ 2 log (( 1 + t 0 ) τ ) − 1 = − d τ d ( τ log (( 1 + t 0 ) τ ) ) , we have
∫ 1 r / 2 log ( ( 1 + t 0 ) τ ) − 1 τ 2 d τ ≤ − ∫ 1 ∞ d d τ ( log ( ( 1 + t 0 ) τ ) τ ) d τ = log ( 1 + t 0 ) . \int_1^{r/2}\frac{\log((1+t_0)\tau)-1}{\tau^2}\,\mathrm{d}\tau
\le-\int_1^\infty\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\log((1+t_0)\tau)}{\tau}\right)\,\mathrm{d}\tau
=\log(1+t_0). ∫ 1 r /2 τ 2 log (( 1 + t 0 ) τ ) − 1 d τ ≤ − ∫ 1 ∞ d τ d ( τ log (( 1 + t 0 ) τ ) ) d τ = log ( 1 + t 0 ) . Moreover, we also observe that
∫ 1 r / 2 τ − 2 d τ ≤ ∫ 1 ∞ τ − 2 d τ = 1. \int_1^{r/2}\tau^{-2}\,\mathrm{d}\tau\le\int_1^\infty\tau^{-2}\,\mathrm{d}\tau=1. ∫ 1 r /2 τ − 2 d τ ≤ ∫ 1 ∞ τ − 2 d τ = 1. These together with (3.9) imply that
∫ 1 r / 2 ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ ≤ 2 α r − α log ( 1 + t 0 ) + 2 α r − α . \int_1^{r/2}(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau
\le2^\alpha r^{-\alpha}\log(1+t_0)+2^\alpha r^{-\alpha}. ∫ 1 r /2 ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ ≤ 2 α r − α log ( 1 + t 0 ) + 2 α r − α . Since log ( ( 1 + t 0 ) r ) / r \log((1+t_0)r)/r log (( 1 + t 0 ) r ) / r 1 + t 0 ≥ e 1+t_0 \ge e 1 + t 0 ≥ e r ≥ 1 r \ge1 r ≥ 1
log ( ( 1 + t 0 ) r ) r ≤ log ( 1 + t 0 ) , r ≥ 1. \frac{\log((1+t_0)r)}{r} \le\log(1+t_0), \qquad r \ge1. r log (( 1 + t 0 ) r ) ≤ log ( 1 + t 0 ) , r ≥ 1. Thus, for τ ∈ ( r / 2 , r ) \tau\in(r/2,r) τ ∈ ( r /2 , r )
∫ r / 2 r ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ ≤ 4 r − 2 log ( ( 1 + t 0 ) r ) ∫ r / 2 r ( r − τ ) − α d τ ≤ 2 1 + α 1 − α r − α log ( 1 + t 0 ) . \begin{aligned}
\int_{r/2}^{r}(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau
\le4r^{-2}\log((1+t_0)r)\int_{r/2}^{r}(r-\tau)^{-\alpha}\,\mathrm{d}\tau\\
&\le\frac{2^{1+\alpha}}{1-\alpha}r^{-\alpha}\log(1+t_0).
\end{aligned} ∫ r /2 r ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ ≤ 4 r − 2 log (( 1 + t 0 ) r ) ∫ r /2 r ( r − τ ) − α d τ ≤ 1 − α 2 1 + α r − α log ( 1 + t 0 ) . Hence, since r = t / t 0 r=t/t_0 r = t / t 0 t 0 ≥ 1 t_0\ge1 t 0 ≥ 1
∂ t α ( z − y 0 ) ( t ) ≥ − K Γ ( 1 − α ) t 0 − α − 1 ∫ 1 r ( r − τ ) − α log ( 1 + t 0 τ ) τ 2 d τ ≥ − K Γ ( 1 − α ) t 0 − α − 1 ( 2 α r − α log ( 1 + t 0 ) + 2 α r − α + 2 1 + α 1 − α r − α log ( 1 + t 0 ) ) = − K Γ ( 1 − α ) t − α ( 2 α log ( 1 + t 0 ) t 0 + 2 α t 0 + 2 1 + α 1 − α log ( 1 + t 0 ) t 0 ) ≥ − K Γ ( 1 − α ) t − α ( 2 α + 2 α + 2 α + 1 1 − α ) = − K ( 2 − α ) Γ ( 2 − α ) 2 1 + α t − α . \begin{aligned}
\partial_t^\alpha(z-y_0)(t)
\ge-\frac{K}{\Gamma(1-\alpha)}t_0^{-\alpha-1}\int_1^r(r-\tau)^{-\alpha}\frac{\log(1+t_0\tau)}{\tau^2}\,\mathrm{d}\tau\\
&\ge-\frac{K}{\Gamma(1-\alpha)}t_0^{-\alpha-1}\left(2^\alpha r^{-\alpha}\log(1+t_0)+2^\alpha r^{-\alpha}+\frac{2^{1+\alpha}}{1-\alpha}r^{-\alpha}\log(1+t_0)\right) \\
&= -\frac{K}{\Gamma(1-\alpha)}t^{-\alpha}\left(2^\alpha\frac{\log(1+t_0)}{t_0}+\frac{2^\alpha}{t_0}+\frac{2^{1+\alpha}}{1-\alpha}\frac{\log(1+t_0)}{t_0}\right) \\
&\ge-\frac{K}{\Gamma(1-\alpha)}t^{-\alpha}\left(2^\alpha+2^\alpha+\frac{2^{\alpha+1}}{1-\alpha}\right) \\
&= -\frac{K(2-\alpha)}{\Gamma(2-\alpha)}2^{1+\alpha}t^{-\alpha}.
\end{aligned} ∂ t α ( z − y 0 ) ( t ) ≥ − Γ ( 1 − α ) K t 0 − α − 1 ∫ 1 r ( r − τ ) − α τ 2 log ( 1 + t 0 τ ) d τ ≥ − Γ ( 1 − α ) K t 0 − α − 1 ( 2 α r − α log ( 1 + t 0 ) + 2 α r − α + 1 − α 2 1 + α r − α log ( 1 + t 0 ) ) = − Γ ( 1 − α ) K t − α ( 2 α t 0 log ( 1 + t 0 ) + t 0 2 α + 1 − α 2 1 + α t 0 log ( 1 + t 0 ) ) ≥ − Γ ( 1 − α ) K t − α ( 2 α + 2 α + 1 − α 2 α + 1 ) = − Γ ( 2 − α ) K ( 2 − α ) 2 1 + α t − α . Furthermore, we get
∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ − K ( 2 − α ) Γ ( 2 − α ) 2 1 + α t − α + λ K 2 t − α log ( 1 + t ) ≥ λ K 2 t − α ( − 2 − α λ Γ ( 2 − α ) 2 2 + α + log ( 1 + t 0 ) ) \begin{aligned}
\partial_t^\alpha(z-y_0)(t)+\lambda t^{1-\alpha}z(t)-\mu t^{-\alpha}\log(1+t)
\ge-\frac{K(2-\alpha)}{\Gamma(2-\alpha)}2^{1+\alpha}t^{-\alpha}+\frac{\lambda K}{2}t^{-\alpha}\log(1+t) \\
&\ge\frac{\lambda K}{2}t^{-\alpha}\left(-\frac{2-\alpha}{\lambda\Gamma(2-\alpha)}2^{2+\alpha}+\log(1+t_0)\right)
\end{aligned} ∂ t α ( z − y 0 ) ( t ) + λ t 1 − α z ( t ) − μ t − α log ( 1 + t ) ≥ − Γ ( 2 − α ) K ( 2 − α ) 2 1 + α t − α + 2 λ K t − α log ( 1 + t ) ≥ 2 λ K t − α ( − λ Γ ( 2 − α ) 2 − α 2 2 + α + log ( 1 + t 0 ) ) for t ≥ 2 t 0 t\ge2t_0 t ≥ 2 t 0 z z z (23) satisfies (22) . Then, by Lemma 3.1, it holds that
y ( t ) ≤ z ( t ) , t > 0. y(t)\le z(t),\qquad t>0. y ( t ) ≤ z ( t ) , t > 0. Thus we conclude that
y ( t ) ≤ C log ( 1 + t ) t y(t)\le C\frac{\log(1+t)}{t} y ( t ) ≤ C t log ( 1 + t ) for t ≥ t 0 t\ge t_0 t ≥ t 0
Proof of Theorem 1.1
In this section, we prove Theorem 1.1.
A basic entropy inequality
We first prove the following entropy estimate.
Lemma 4.1 Assume the same conditions as in Theorem 1.1. Let T > 0 T>0 T > 0 v ( x , t ) : = t α m u ( x , t ) v(x,t):=t^{\frac{\alpha}{m}}u(x,t) v ( x , t ) := t m α u ( x , t ) E [ v ∣ S ] \mathcal{E}[v|S] E [ v ∣ S ]
E [ v ∣ S ] ( t ) : = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x . \mathcal{E}[v|S](t):=\int_{\Omega}\left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^m(x)(v(x,t)-S(x))\right\}\,\mathrm{d}x. E [ v ∣ S ] ( t ) := ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x . with
E 0 : = E [ v ∣ S ] ( 0 ) = ∫ Ω { − S m + 1 ( x ) m + 1 + S m + 1 ( x ) } d x = m m + 1 ∫ Ω S m + 1 ( x ) d x . \mathcal{E}_0:=\mathcal{E}[v|S](0)=\int_{\Omega}\left\{-\frac{S^{m+1}(x)}{m+1}+S^{m+1}(x)\right\}\,\mathrm{d}x=\frac{m}{m+1}\int_{\Omega}S^{m+1}(x)\,\mathrm{d}x. E 0 := E [ v ∣ S ] ( 0 ) = ∫ Ω { − m + 1 S m + 1 ( x ) + S m + 1 ( x ) } d x = m + 1 m ∫ Ω S m + 1 ( x ) d x . Then the following assertions hold.
(a) E : = E [ v ∣ S ] \mathcal{E}:=\mathcal{E}[v|S] E := E [ v ∣ S ]
d d t ( g 1 − α , n ∗ [ E − E 0 ] ) ( t ) + t α ( 1 m − 1 ) ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 ≤ F ( t ) + I n ( t ) + J n ( t ) (26) \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*[\mathcal{E}-\mathcal{E}_0]\right)(t)+t^{\alpha\left(\frac{1}{m}-1\right)}\left\|\nabla(v^m(t)-S^m)\right\|_2^2\leq F(t)+I_n(t)+J_n(t) \tag*{(26)} d t d ( g 1 − α , n ∗ [ E − E 0 ] ) ( t ) + t α ( m 1 − 1 ) ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 ≤ F ( t ) + I n ( t ) + J n ( t ) (26) for a.e. 0 < t < T 0<t<T 0 < t < T
F ( t ) : = C ∫ 0 t ( t − s ) − α − 1 ( t α m − s α m ) ∥ u ( s ) ∥ m + 1 d s , F(t):=C\int_0^t(t-s)^{-\alpha-1}\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\|u(s)\|_{m+1}\,\mathrm{d}s, F ( t ) := C ∫ 0 t ( t − s ) − α − 1 ( t m α − s m α ) ∥ u ( s ) ∥ m + 1 d s , and C > 0 C>0 C > 0 m m m α \alpha α N N N Ω \Omega Ω ∥ u 0 ∥ p 0 \|u_0\|_{p_0} ∥ u 0 ∥ p 0 I n , J n → 0 I_n,J_n\to0 I n , J n → 0 L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T ) n → ∞ n\to\infty n → ∞
(b) F F F
F ( t ) ≤ { ν g 1 − α ( t ) , if m > α , μ t − α log ( 1 + t ) + ν g 1 − α ( t ) , if m = α , μ t β − 1 + ν g 1 − α ( t ) , if 0 < m < α , 0 < t < T , (27) F(t)\leq
\begin{cases}
\nu g_{1-\alpha}(t), & \text{if }m>\alpha,\\
\mu t^{-\alpha}\log(1+t)+\nu g_{1-\alpha}(t), & \text{if }m=\alpha,\\
\mu t^{\beta-1}+\nu g_{1-\alpha}(t), & \text{if }0<m<\alpha,
\end{cases}
\qquad0<t<T, \tag*{(27)} F ( t ) ≤ ⎩ ⎨ ⎧ ν g 1 − α ( t ) , μ t − α log ( 1 + t ) + ν g 1 − α ( t ) , μ t β − 1 + ν g 1 − α ( t ) , if m > α , if m = α , if 0 < m < α , 0 < t < T , (27) where β = α ( 1 / m − 1 ) \beta=\alpha(1/m-1) β = α ( 1/ m − 1 ) μ , ν \mu,\nu μ , ν m m m α \alpha α N N N Ω \Omega Ω ∥ u 0 ∥ L p 0 ( Ω ) \|u_0\|_{L^{p_0}(\Omega)} ∥ u 0 ∥ L p 0 ( Ω )
The estimates (27) imply that the fast diffusion range 0 < m < 1 0<m<1 0 < m < 1 α < m < 1 \alpha<m<1 α < m < 1 m = α m=\alpha m = α 0 < m < α 0<m<\alpha 0 < m < α m = α m=\alpha m = α H − 1 H^{-1} H − 1
Proposition 4.1 [10 Theorem 1.7] Assume the same conditions as in Theorem 1.1. Then, for every 1 ≤ p ≤ ∞ 1\leq p\leq\infty 1 ≤ p ≤ ∞
c 0 1 + t α m ≤ ∥ u ( t ) ∥ p ≤ c 1 t α m , t > 0 , (28) \frac{c_0}{1+t^{\frac{\alpha}{m}}}\leq\|u(t)\|_p\leq\frac{c_1}{t^{\frac{\alpha}{m}}},\qquad t>0, \tag*{(28)} 1 + t m α c 0 ≤ ∥ u ( t ) ∥ p ≤ t m α c 1 , t > 0 , (28) where c 0 , c 1 > 0 c_0,c_1>0 c 0 , c 1 > 0 N N N α \alpha α p p p m m m Ω \Omega Ω ∥ u 0 ∥ p 0 \|u_0\|_{p_0} ∥ u 0 ∥ p 0
We are now ready to prove Lemma 4.1.
Proof of assertion (a) of Lemma 4.1. Let h α , n h_{\alpha,n} h α , n (12) . Convolving (1.1) with the kernel h α , n h_{\alpha,n} h α , n
( h α , n ∗ [ d d t ( g 1 − α ∗ [ u − u 0 ] ) ] ) ( t ) = d d t ( g 1 − α , n ∗ [ u − u 0 ] ) ( t ) , a.e. 0 < t < T , \left(h_{\alpha,n} * \left[\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha} * [u-u_0])\right]\right)(t)=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [u-u_0])(t),\qquad\text{a.e. } 0<t<T, ( h α , n ∗ [ d t d ( g 1 − α ∗ [ u − u 0 ]) ] ) ( t ) = d t d ( g 1 − α , n ∗ [ u − u 0 ]) ( t ) , a.e. 0 < t < T , we obtain
∫ Ω d d t ( g 1 − α , n ∗ u ( x , ⋅ ) ) ( t ) φ ( x ) d x + ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = 0 (29) \begin{aligned}
&\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*u(x,\cdot))(t)\varphi(x)\,\mathrm{d}x\\
&\quad+\int_{\Omega}\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\cdot\nabla\varphi(x)\,\mathrm{d}x-g_{1-\alpha,n}(t)\int_{\Omega}u_0(x)\varphi(x)\,\mathrm{d}x=0 \tag*{(29)}
\end{aligned} ∫ Ω d t d ( g 1 − α , n ∗ u ( x , ⋅ )) ( t ) φ ( x ) d x + ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = 0 (29) for a.e. t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T ) g 1 − α , n ∈ W l o c 1 , 1 ( [ 0 , ∞ ) ) g_{1-\alpha,n}\in W_{\mathrm{loc}}^{1,1}([0,\infty)) g 1 − α , n ∈ W loc 1 , 1 ([ 0 , ∞ )) u ∈ L ∞ ( 0 , ∞ ; H − 1 ( Ω ) ) u\in L^\infty(0,\infty;H^{-1}(\Omega)) u ∈ L ∞ ( 0 , ∞ ; H − 1 ( Ω ))
∫ Ω d d t ( g 1 − α , n ∗ u ( x , ⋅ ) ) ( t ) φ ( x ) d x = ∫ Ω ( ( g 1 − α , n ′ ∗ u ( x , ⋅ ) ) ( t ) + g 1 − α , n ( 0 ) u ( x , t ) ) φ ( x ) d x = ( g 1 − α , n ′ ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + g 1 − α , n ( 0 ) ∫ Ω u ( x , t ) φ ( x ) d x = d d t ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) , a.e. 0 < t < T . \begin{aligned}
\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*u(x,\cdot))(t)\varphi(x)\,\mathrm{d}x
&=\int_{\Omega}\left((g'_{1-\alpha,n}*u(x,\cdot))(t)+g_{1-\alpha,n}(0)u(x,t)\right)\varphi(x)\,\mathrm{d}x\\
&=\left(g'_{1-\alpha,n}*\left[\int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)+g_{1-\alpha,n}(0)\int_{\Omega}u(x,t)\varphi(x)\,\mathrm{d}x\\
&=\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*\left[\int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t),\qquad\text{a.e. } 0<t<T.
\end{aligned} ∫ Ω d t d ( g 1 − α , n ∗ u ( x , ⋅ )) ( t ) φ ( x ) d x = ∫ Ω ( ( g 1 − α , n ′ ∗ u ( x , ⋅ )) ( t ) + g 1 − α , n ( 0 ) u ( x , t ) ) φ ( x ) d x = ( g 1 − α , n ′ ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + g 1 − α , n ( 0 ) ∫ Ω u ( x , t ) φ ( x ) d x = d t d ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) , a.e. 0 < t < T . Multiplying (4.4) by t α m t^{\frac{\alpha}{m}} t m α
t α m d d t ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + t α m ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − t α m g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = 0 (30) \begin{aligned}
&t^{\frac{\alpha}{m}}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*\left[\int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)\\
&\quad+t^{\frac{\alpha}{m}}\int_{\Omega}\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\cdot\nabla\varphi(x)\,\mathrm{d}x
-t^{\frac{\alpha}{m}}g_{1-\alpha,n}(t)\int_{\Omega}u_0(x)\varphi(x)\,\mathrm{d}x=0 \tag*{(30)}
\end{aligned} t m α d t d ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + t m α ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − t m α g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = 0 (30) for a.e. t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T )
∫ Ω u ( x , ⋅ ) φ ( x ) d x ∈ L 1 ( 0 , T ) , \int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\in L^1(0,T), ∫ Ω u ( x , ⋅ ) φ ( x ) d x ∈ L 1 ( 0 , T ) , for the first term on the left-hand side, by using (2.10), we get
t α m d d t ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) = d d t ( g 1 − α , n ∗ [ t α m ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α m − s α m ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s = d d t ( g 1 − α , n ∗ [ ∫ Ω v ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α m − s α m ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s \begin{aligned}
t^{\frac{\alpha}{m}}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*\left[\int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)
&=\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*\left[t^{\frac{\alpha}{m}}\int_{\Omega}u(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)\\
&\quad+\int_0^t g'_{1-\alpha,n}(t-s)\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\left\{\int_{\Omega}u(x,s)\varphi(x)\,\mathrm{d}x\right\}\,\mathrm{d}s\\
&=\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*\left[\int_{\Omega}v(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)\\
&\quad+\int_0^t g'_{1-\alpha,n}(t-s)\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\left\{\int_{\Omega}u(x,s)\varphi(x)\,\mathrm{d}x\right\}\,\mathrm{d}s
\end{aligned} t m α d t d ( g 1 − α , n ∗ [ ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) = d t d ( g 1 − α , n ∗ [ t m α ∫ Ω u ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + ∫ 0 t g 1 − α , n ′ ( t − s ) ( t m α − s m α ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s = d t d ( g 1 − α , n ∗ [ ∫ Ω v ( x , ⋅ ) φ ( x ) d x ] ) ( t ) + ∫ 0 t g 1 − α , n ′ ( t − s ) ( t m α − s m α ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s for a.e. 0 < t < T 0<t<T 0 < t < T
d d t ( g 1 − α , n ∗ [ ∫ Ω v ( x , ⋅ ) φ ( x ) d x ] ) ( t ) = ∫ Ω d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) φ ( x ) d x . \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n} * \left[\int_{\Omega} v(x,\cdot)\varphi(x)\,\mathrm{d}x\right]\right)(t)
=
\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*v(x,\cdot)\right)(t)\varphi(x)\,\mathrm{d}x. d t d ( g 1 − α , n ∗ [ ∫ Ω v ( x , ⋅ ) φ ( x ) d x ] ) ( t ) = ∫ Ω d t d ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) φ ( x ) d x . for a.e. 0 < t < T 0<t<T 0 < t < T (30) imply
∫ Ω d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) φ ( x ) d x + t α m ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − t α m g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α m − s α m ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s , a.e. 0 < t < T . \begin{aligned}
&\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*v(x,\cdot)\right)(t)\varphi(x)\,\mathrm{d}x
+t^{\frac{\alpha}{m}}\int_{\Omega}\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\cdot\nabla\varphi(x)\,\mathrm{d}x\\
&\qquad-t^{\frac{\alpha}{m}}g_{1-\alpha,n}(t)\int_{\Omega}u_{0}(x)\varphi(x)\,\mathrm{d}x\\
&=-\int_{0}^{t}g'_{1-\alpha,n}(t-s)\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\left\{\int_{\Omega}u(x,s)\varphi(x)\,\mathrm{d}x\right\}\,\mathrm{d}s,\qquad\text{a.e. }0<t<T.
\end{aligned} ∫ Ω d t d ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) φ ( x ) d x + t m α ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ φ ( x ) d x − t m α g 1 − α , n ( t ) ∫ Ω u 0 ( x ) φ ( x ) d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t m α − s m α ) { ∫ Ω u ( x , s ) φ ( x ) d x } d s , a.e. 0 < t < T . We choose φ = v m ( t ) − S m ∈ H 0 1 ( Ω ) \varphi=v^{m}(t)-S^{m}\in H_{0}^{1}(\Omega) φ = v m ( t ) − S m ∈ H 0 1 ( Ω )
∫ Ω d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) ( v m ( x , t ) − S m ( x ) ) d x + t α m ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − t α m g 1 − α , n ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α m − s α m ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x ) ) d x } d s . (31) \begin{aligned}
&\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*v(x,\cdot)\right)(t)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&\quad+t^{\frac{\alpha}{m}}\int_{\Omega}\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\cdot\nabla\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&\quad-t^{\frac{\alpha}{m}}g_{1-\alpha,n}(t)\int_{\Omega}u_{0}(x)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&=-\int_{0}^{t}g'_{1-\alpha,n}(t-s)\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\left\{\int_{\Omega}u(x,s)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\right\}\,\mathrm{d}s. \tag*{(31)}
\end{aligned} ∫ Ω d t d ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) ( v m ( x , t ) − S m ( x ) ) d x + t m α ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − t m α g 1 − α , n ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t m α − s m α ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x ) ) d x } d s . (31) for a.e. t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T )
t α m ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x = t α m ∫ Ω ∇ u m ( x , t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − t α m ∫ Ω { ∇ u m ( x , t ) − ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) } ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x = : t α m ∫ Ω ∇ u m ( x , t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − I n ( t ) , \begin{aligned}
&t^{\frac{\alpha}{m}}\int_{\Omega}\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\cdot\nabla\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&=t^{\frac{\alpha}{m}}\int_{\Omega}\nabla u^{m}(x,t)\cdot\nabla\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&\quad-t^{\frac{\alpha}{m}}\int_{\Omega}\left\{\nabla u^{m}(x,t)-\left(h_{\alpha,n}*\nabla u^{m}(x,\cdot)\right)(t)\right\}\cdot\nabla\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&=:t^{\frac{\alpha}{m}}\int_{\Omega}\nabla u^{m}(x,t)\cdot\nabla\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x-I_{n}(t),
\end{aligned} t m α ∫ Ω ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x = t m α ∫ Ω ∇ u m ( x , t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − t m α ∫ Ω { ∇ u m ( x , t ) − ( h α , n ∗ ∇ u m ( x , ⋅ ) ) ( t ) } ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x =: t m α ∫ Ω ∇ u m ( x , t ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x − I n ( t ) , and
t α m g 1 − α , n ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = t α m g 1 − α ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x − t α m ( g 1 − α ( t ) − g 1 − α , n ( t ) ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = t α ( 1 m − 1 ) Γ ( 1 − α ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x − J n ( t ) . \begin{aligned}
&t^{\frac{\alpha}{m}}g_{1-\alpha,n}(t)\int_{\Omega}u_{0}(x)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&=t^{\frac{\alpha}{m}}g_{1-\alpha}(t)\int_{\Omega}u_{0}(x)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&\quad-t^{\frac{\alpha}{m}}\left(g_{1-\alpha}(t)-g_{1-\alpha,n}(t)\right)\int_{\Omega}u_{0}(x)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x\\
&=\frac{t^{\alpha\left(\frac{1}{m}-1\right)}}{\Gamma(1-\alpha)}\int_{\Omega}u_{0}(x)\left(v^{m}(x,t)-S^{m}(x)\right)\,\mathrm{d}x-J_{n}(t).
\end{aligned} t m α g 1 − α , n ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = t m α g 1 − α ( t ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x − t m α ( g 1 − α ( t ) − g 1 − α , n ( t ) ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = Γ ( 1 − α ) t α ( m 1 − 1 ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x − J n ( t ) . Then it holds that
I n , J n → 0 strongly in L 1 ( 0 , T ) as n → ∞ . I_n,J_n \to0 \quad\text{strongly in } L^1(0,T) \text{ as } n\to\infty. I n , J n → 0 strongly in L 1 ( 0 , T ) as n → ∞. Indeed, by Hölder’s inequality and (13) , we obtain
∫ 0 T ∣ I n ( t ) ∣ d t ≤ T α m ∥ ∇ v m − ( h α , n ∗ ∇ v m ) ∥ L 2 ( 0 , T ; L 2 ( Ω ) ) ∥ ∇ ( v m − S m ) ∥ L 2 ( 0 , T ; L 2 ( Ω ) ) → 0 as n → ∞ . \int_0^T |I_n(t)|\,\mathrm{d}t \le T^{\frac{\alpha}{m}}\|\nabla v^m-(h_{\alpha,n}\ast\nabla v^m)\|_{L^2(0,T;L^2(\Omega))}\|\nabla(v^m-S^m)\|_{L^2(0,T;L^2(\Omega))}
\to0 \quad\text{as } n\to\infty. ∫ 0 T ∣ I n ( t ) ∣ d t ≤ T m α ∥∇ v m − ( h α , n ∗ ∇ v m ) ∥ L 2 ( 0 , T ; L 2 ( Ω )) ∥∇ ( v m − S m ) ∥ L 2 ( 0 , T ; L 2 ( Ω )) → 0 as n → ∞. Note that, by (28) in Proposition 4.1 and assertion (ii) of Remark 1.1, it holds that
∥ v ( t ) ∥ ∞ = t α m ∥ u ( t ) ∥ ∞ ≤ C , ∥ S ∥ ∞ ≤ C ∥ u 0 ∥ L 1 p 0 . \|v(t)\|_\infty=t^{\frac{\alpha}{m}}\|u(t)\|_\infty\le C,\qquad\|S\|_\infty\le C\|u_0\|_{L^{\frac{1}{p_0}}}. ∥ v ( t ) ∥ ∞ = t m α ∥ u ( t ) ∥ ∞ ≤ C , ∥ S ∥ ∞ ≤ C ∥ u 0 ∥ L p 0 1 . Since Ω \Omega Ω u 0 ∈ L 1 ( Ω ) u_0\in L^1(\Omega) u 0 ∈ L 1 ( Ω ) (14) imply that
∫ 0 T ∣ J n ( t ) ∣ d t ≤ C T α m ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞ . \int_0^T |J_n(t)|\,\mathrm{d}t\le CT^{\frac{\alpha}{m}}\|g_{1-\alpha,n}-g_{1-\alpha}\|_{L^1(0,T)}\to0\quad\text{as }n\to\infty. ∫ 0 T ∣ J n ( t ) ∣ d t ≤ C T m α ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞. Moreover, by (6) with φ = v m ( t ) − S m \varphi=v^m(t)-S^m φ = v m ( t ) − S m
t α ( 1 m − 1 ) Γ ( 1 − α ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x ) ) d x = t α ( 1 m − 1 ) ∫ Ω ∇ S m ( x ) ⋅ ∇ ( v m ( x , t ) − S m ( x ) ) d x \frac{t^{\alpha\left(\frac{1}{m}-1\right)}}{\Gamma(1-\alpha)}\int_\Omega u_0(x)(v^m(x,t)-S^m(x))\,\mathrm{d}x
=t^{\alpha\left(\frac{1}{m}-1\right)}\int_\Omega\nabla S^m(x)\cdot\nabla(v^m(x,t)-S^m(x))\,\mathrm{d}x Γ ( 1 − α ) t α ( m 1 − 1 ) ∫ Ω u 0 ( x ) ( v m ( x , t ) − S m ( x )) d x = t α ( m 1 − 1 ) ∫ Ω ∇ S m ( x ) ⋅ ∇ ( v m ( x , t ) − S m ( x )) d x for a.e. t > 0 t>0 t > 0 u m = t − α v m u^m=t^{-\alpha}v^m u m = t − α v m (31) as
∫ Ω d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) ( v m ( x , t ) − S m ( x ) ) d x + t α ( 1 m − 1 ) ∫ Ω ∣ ∇ ( v m ( x , t ) − S m ( x ) ) ∣ 2 d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α m − s α m ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x ) ) d x } d s + I n ( t ) + J n ( t ) (32) \begin{aligned}
&\int_\Omega\frac{\mathrm{d}}{\mathrm{d}t}\bigl(g_{1-\alpha,n}\ast v(x,\cdot)\bigr)(t)(v^m(x,t)-S^m(x))\,\mathrm{d}x
+t^{\alpha\left(\frac{1}{m}-1\right)}\int_\Omega|\nabla(v^m(x,t)-S^m(x))|^2\,\mathrm{d}x\\
&=-\int_0^t g'_{1-\alpha,n}(t-s)\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)
\left\{\int_\Omega u(x,s)(v^m(x,t)-S^m(x))\,\mathrm{d}x\right\}\,\mathrm{d}s+I_n(t)+J_n(t) \tag*{(32)}
\end{aligned} ∫ Ω d t d ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) ( v m ( x , t ) − S m ( x )) d x + t α ( m 1 − 1 ) ∫ Ω ∣∇ ( v m ( x , t ) − S m ( x )) ∣ 2 d x = − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t m α − s m α ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x )) d x } d s + I n ( t ) + J n ( t ) (32) for a.e. t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T ) (15) for H ( y ) = 1 m + 1 y m + 1 H(y)=\frac{1}{m+1}y^{m+1} H ( y ) = m + 1 1 y m + 1
1 m + 1 d d t ( g 1 − α , n ∗ v m + 1 ( x , ⋅ ) ) ( t ) ≤ v m ( x , t ) d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) , a.e. x ∈ Ω , t ∈ ( 0 , T ) . \frac{1}{m+1}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}\ast v^{m+1}(x,\cdot))(t)\le v^m(x,t)\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}\ast v(x,\cdot))(t),\quad\text{a.e. }x\in\Omega,\ t\in(0,T). m + 1 1 d t d ( g 1 − α , n ∗ v m + 1 ( x , ⋅ )) ( t ) ≤ v m ( x , t ) d t d ( g 1 − α , n ∗ v ( x , ⋅ )) ( t ) , a.e. x ∈ Ω , t ∈ ( 0 , T ) . Since
E ( t ) − E 0 = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x − ∫ Ω { − S m + 1 ( x ) m + 1 + S m + 1 ( x ) } d x = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m ( x ) v ( x , t ) } d x , \begin{aligned}
\mathcal{E}(t)-\mathcal{E}_0
&=\int_\Omega\left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^m(x)(v(x,t)-S(x))\right\}\,\mathrm{d}x\\
&\quad-\int_\Omega\left\{-\frac{S^{m+1}(x)}{m+1}+S^{m+1}(x)\right\}\,\mathrm{d}x\\
&=\int_\Omega\left\{\frac{v^{m+1}(x,t)}{m+1}-S^m(x)v(x,t)\right\}\,\mathrm{d}x,
\end{aligned} E ( t ) − E 0 = ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x − ∫ Ω { − m + 1 S m + 1 ( x ) + S m + 1 ( x ) } d x = ∫ Ω { m + 1 v m + 1 ( x , t ) − S m ( x ) v ( x , t ) } d x , we have
d d t ( g 1 − α , n ∗ [ E − E 0 ] ) ( t ) = ∫ Ω { 1 m + 1 d d t ( g 1 − α , n ∗ v m + 1 ( x , ⋅ ) ) ( t ) − S m ( x ) d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) } d x ≤ ∫ Ω { v m ( x , t ) d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) − S m ( x ) d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) } d x = ∫ Ω d d t ( g 1 − α , n ∗ v ( x , ⋅ ) ) ( t ) ( v m ( x , t ) − S m ( x ) ) d x , a.e. 0 < t < T . \begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [\mathcal{E}-\mathcal{E}_0])(t)
&=\int_{\Omega}\left\{\frac{1}{m+1}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v^{m+1}(x,\cdot))(t)-S^m(x)\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v(x,\cdot))(t)\right\}\,\mathrm{d}x\\
&\le\int_{\Omega}\left\{v^m(x,t)\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v(x,\cdot))(t)-S^m(x)\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v(x,\cdot))(t)\right\}\,\mathrm{d}x\\
&=\int_{\Omega}\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*v(x,\cdot))(t)(v^m(x,t)-S^m(x))\,\mathrm{d}x,\quad\text{a.e. }0<t<T.
\end{aligned} d t d ( g 1 − α , n ∗ [ E − E 0 ]) ( t ) = ∫ Ω { m + 1 1 d t d ( g 1 − α , n ∗ v m + 1 ( x , ⋅ )) ( t ) − S m ( x ) d t d ( g 1 − α , n ∗ v ( x , ⋅ )) ( t ) } d x ≤ ∫ Ω { v m ( x , t ) d t d ( g 1 − α , n ∗ v ( x , ⋅ )) ( t ) − S m ( x ) d t d ( g 1 − α , n ∗ v ( x , ⋅ )) ( t ) } d x = ∫ Ω d t d ( g 1 − α , n ∗ v ( x , ⋅ )) ( t ) ( v m ( x , t ) − S m ( x )) d x , a.e. 0 < t < T . This together with (4.7) implies
d d t ( g 1 − α , n ∗ [ E − E 0 ] ) ( t ) + t α ( 1 m − 1 ) ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 ≤ − ∫ 0 t g 1 − α , n ′ ( t − s ) ( t α / m m − s α / m m ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x ) ) d x } d s (33) \frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*[\mathcal{E}-\mathcal{E}_0])(t)+t^{\alpha\left(\frac{1}{m}-1\right)}\|\nabla(v^m(t)-S^m)\|_2^2
\le-\int_0^t g'_{1-\alpha,n}(t-s)\left(\frac{t^{\alpha/m}}{m}-\frac{s^{\alpha/m}}{m}\right)\left\{\int_{\Omega}u(x,s)(v^m(x,t)-S^m(x))\,\mathrm{d}x\right\}\,\mathrm{d}s \tag*{(33)} d t d ( g 1 − α , n ∗ [ E − E 0 ]) ( t ) + t α ( m 1 − 1 ) ∥∇ ( v m ( t ) − S m ) ∥ 2 2 ≤ − ∫ 0 t g 1 − α , n ′ ( t − s ) ( m t α / m − m s α / m ) { ∫ Ω u ( x , s ) ( v m ( x , t ) − S m ( x )) d x } d s (33) for a.e. t ∈ ( 0 , T ) t\in(0,T) t ∈ ( 0 , T ) − g 1 − α , n ′ ≥ 0 -g'_{1-\alpha,n}\ge0 − g 1 − α , n ′ ≥ 0 n n n
− g 1 − α , n ′ ( t ) = − n s α , n ′ ( t ) = n 2 t α − 1 E α , α ( − n t α ) ≤ 1 Γ ( α ) n 2 t 2 α − 1 ( 1 + c α , 2 n t α ) 2 = t − α − 1 ( n t α ) 2 Γ ( α ) ( 1 + c α , 2 n t α ) 2 ≤ C α t − α − 1 , 0 < t < T , \begin{aligned}
-g'_{1-\alpha,n}(t)&=-ns'_{\alpha,n}(t)=n^2t^{\alpha-1}E_{\alpha,\alpha}(-nt^\alpha)\le\frac{1}{\Gamma(\alpha)}\frac{n^{2}t^{2\alpha-1}}{(1+c_{\alpha,2}nt^\alpha)^2}\\
&=t^{-\alpha-1}\frac{(nt^\alpha)^2}{\Gamma(\alpha)(1+c_{\alpha,2}nt^\alpha)^2}\le C_\alpha t^{-\alpha-1},\quad0<t<T,
\end{aligned} − g 1 − α , n ′ ( t ) = − n s α , n ′ ( t ) = n 2 t α − 1 E α , α ( − n t α ) ≤ Γ ( α ) 1 ( 1 + c α , 2 n t α ) 2 n 2 t 2 α − 1 = t − α − 1 Γ ( α ) ( 1 + c α , 2 n t α ) 2 ( n t α ) 2 ≤ C α t − α − 1 , 0 < t < T , where C α C_\alpha C α α \alpha α Ω \Omega Ω v v v S S S
∣ ∫ 0 t [ − g 1 − α , n ′ ( t − s ) ] ( t α / m m − s α / m m ) { ∫ Ω ( v m ( x , t ) − S m ( x ) ) u ( x , s ) d x } d s ∣ ≤ C ∫ 0 t [ − g 1 − α , n ′ ( t − s ) ] ( t α / m m − s α / m m ) ∥ u ( s ) ∥ m + 1 d s ≤ C ∫ 0 t ( t − s ) − α − 1 ( t α / m m − s α / m m ) ∥ u ( s ) ∥ m + 1 d s , a.e. 0 < t < T . \begin{aligned}
\left|\int_0^t[-g'_{1-\alpha,n}(t-s)]\left(\frac{t^{\alpha/m}}{m}-\frac{s^{\alpha/m}}{m}\right)\left\{\int_{\Omega}(v^m(x,t)-S^m(x))u(x,s)\,\mathrm{d}x\right\}\,\mathrm{d}s\right|\\
\le C\int_0^t[-g'_{1-\alpha,n}(t-s)]\left(\frac{t^{\alpha/m}}{m}-\frac{s^{\alpha/m}}{m}\right)\|u(s)\|_{m+1}\,\mathrm{d}s\\
\le C\int_0^t(t-s)^{-\alpha-1}\left(\frac{t^{\alpha/m}}{m}-\frac{s^{\alpha/m}}{m}\right)\|u(s)\|_{m+1}\,\mathrm{d}s,\quad\text{a.e. }0<t<T.
\end{aligned} ∫ 0 t [ − g 1 − α , n ′ ( t − s )] ( m t α / m − m s α / m ) { ∫ Ω ( v m ( x , t ) − S m ( x )) u ( x , s ) d x } d s ≤ C ∫ 0 t [ − g 1 − α , n ′ ( t − s )] ( m t α / m − m s α / m ) ∥ u ( s ) ∥ m + 1 d s ≤ C ∫ 0 t ( t − s ) − α − 1 ( m t α / m − m s α / m ) ∥ u ( s ) ∥ m + 1 d s , a.e. 0 < t < T . This together with (4.8) yields (4.1). Therefore, assertion (a) of Lemma 4.1 follows. □ \square □
Proof of assertion (b) of Lemma 4.1. Combining (4.3) in Proposition 4.1 and (1.2), we have
∥ u ( t ) ∥ m + 1 ≲ ( 1 + t ) − α / m , t > 0 , \|u(t)\|_{m+1}\lesssim(1+t)^{-\alpha/m},\quad t>0, ∥ u ( t ) ∥ m + 1 ≲ ( 1 + t ) − α / m , t > 0 , where a ≲ b a\lesssim b a ≲ b a ≤ C b a\le Cb a ≤ C b C C C a a a b b b m > α m>\alpha m > α α / m < 1 \alpha/m<1 α / m < 1 ∥ u ( t ) ∥ m + 1 ≲ t − α / m \|u(t)\|_{m+1}\lesssim t^{-\alpha/m} ∥ u ( t ) ∥ m + 1 ≲ t − α / m t − α / m ∈ L l o c 1 ( [ 0 , ∞ ) ) t^{-\alpha/m}\in L^1_{\mathrm{loc}}([0,\infty)) t − α / m ∈ L loc 1 ([ 0 , ∞ )) t − α = Γ ( 1 − α ) g 1 − α ( t ) t^{-\alpha}=\Gamma(1-\alpha)g_{1-\alpha}(t) t − α = Γ ( 1 − α ) g 1 − α ( t )
F ( t ) = C ∫ 0 t / 2 ( t − s ) − α − 1 ( t α m − s α m ) ∥ u ( s ) ∥ m + 1 d s + C ∫ t / 2 t ( t − s ) − α − 1 ( t α m − s α m ) ∥ u ( s ) ∥ m + 1 d s ≲ t − α − 1 + α m ∫ 0 t / 2 s − α m d s + t α m − 1 ∫ t / 2 t ( t − s ) − α s − α m d s ≲ g 1 − α ( t ) . \begin{aligned}
F(t)={}&C\int_{0}^{t/2}(t-s)^{-\alpha-1}\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\lVert u(s)\rVert_{m+1}\,ds\\
&+C\int_{t/2}^{t}(t-s)^{-\alpha-1}\left(t^{\frac{\alpha}{m}}-s^{\frac{\alpha}{m}}\right)\lVert u(s)\rVert_{m+1}\,ds\\
&\lesssim t^{-\alpha-1+\frac{\alpha}{m}}\int_{0}^{t/2}s^{-\frac{\alpha}{m}}\,ds+t^{\frac{\alpha}{m}-1}\int_{t/2}^{t}(t-s)^{-\alpha}s^{-\frac{\alpha}{m}}\,ds\lesssim g_{1-\alpha}(t).
\end{aligned} F ( t ) = C ∫ 0 t /2 ( t − s ) − α − 1 ( t m α − s m α ) ∥ u ( s ) ∥ m + 1 d s + C ∫ t /2 t ( t − s ) − α − 1 ( t m α − s m α ) ∥ u ( s ) ∥ m + 1 d s ≲ t − α − 1 + m α ∫ 0 t /2 s − m α d s + t m α − 1 ∫ t /2 t ( t − s ) − α s − m α d s ≲ g 1 − α ( t ) . On the other hand, when 0 < m ≤ α 0<m\leq\alpha 0 < m ≤ α ( 1 + s ) − α / m (1+s)^{-\alpha/m} ( 1 + s ) − α / m m = α m=\alpha m = α
F ( t ) ≲ t − α ∫ 0 t / 2 ( 1 + s ) − 1 d s + t − 1 ∫ t / 2 t ( t − s ) − α d s ≲ t − α log ( 1 + t ) + g 1 − α ( t ) . \begin{aligned}
F(t)&\lesssim t^{-\alpha}\int_{0}^{t/2}(1+s)^{-1}\,ds+t^{-1}\int_{t/2}^{t}(t-s)^{-\alpha}\,ds\\
&\lesssim t^{-\alpha}\log(1+t)+g_{1-\alpha}(t).
\end{aligned} F ( t ) ≲ t − α ∫ 0 t /2 ( 1 + s ) − 1 d s + t − 1 ∫ t /2 t ( t − s ) − α d s ≲ t − α log ( 1 + t ) + g 1 − α ( t ) . If 0 < m < α 0<m<\alpha 0 < m < α
β : = α ( 1 m − 1 ) > 1 − α , \beta:=\alpha\left(\frac{1}{m}-1\right)>1-\alpha, β := α ( m 1 − 1 ) > 1 − α , we similarly obtain
F ( t ) ≲ t − α − 1 + α m ∫ 0 t / 2 ( 1 + s ) − α m d s + t α m − 1 ∫ t / 2 t ( t − s ) − α ( 1 + s ) − α m d s ≲ t β − 1 + g 1 − α ( t ) . \begin{aligned}
F(t)&\lesssim t^{-\alpha-1+\frac{\alpha}{m}}\int_{0}^{t/2}(1+s)^{-\frac{\alpha}{m}}\,ds+t^{\frac{\alpha}{m}-1}\int_{t/2}^{t}(t-s)^{-\alpha}(1+s)^{-\frac{\alpha}{m}}\,ds\\
&\lesssim t^{\beta-1}+g_{1-\alpha}(t).
\end{aligned} F ( t ) ≲ t − α − 1 + m α ∫ 0 t /2 ( 1 + s ) − m α d s + t m α − 1 ∫ t /2 t ( t − s ) − α ( 1 + s ) − m α d s ≲ t β − 1 + g 1 − α ( t ) . Thus, assertion (b) of Lemma 4.1 follows, and the proof of Lemma 4.1 is complete. □ \square □
The porous medium and linear cases m ≥ 1 m\geq1 m ≥ 1
We first prove Theorem 1.1 for the case m ≥ 1 m\geq1 m ≥ 1
λ E m ( t ) ≤ ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T , (34) \lambda\mathcal{E}^{m}(t)\leq\lVert\nabla(v^{m}(t)-S^{m})\rVert_{2}^{2},\qquad\text{a.e. }0<t<T, \tag*{(34)} λ E m ( t ) ≤ ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T , (34) where λ > 0 \lambda>0 λ > 0 N N N α \alpha α m m m Ω \Omega Ω ∥ u 0 ∥ p 0 \lVert u_{0}\rVert_{p_{0}} ∥ u 0 ∥ p 0 c ∈ ( a , b ) c\in(a,b) c ∈ ( a , b )
a m + 1 m + 1 − b m + 1 m + 1 − b m ( a − b ) = m 2 c m − 1 ( a − b ) 2 . \frac{a^{m+1}}{m+1}-\frac{b^{m+1}}{m+1}-b^{m}(a-b)=\frac{m}{2}c^{m-1}(a-b)^{2}. m + 1 a m + 1 − m + 1 b m + 1 − b m ( a − b ) = 2 m c m − 1 ( a − b ) 2 . Hence there exists a nonnegative function η ∈ L ∞ ( Ω × ( 0 , ∞ ) ) \eta\in L^{\infty}(\Omega\times(0,\infty)) η ∈ L ∞ ( Ω × ( 0 , ∞ ))
0 ≤ η ( x , t ) ≤ M : = max { ess sup t > 0 ∥ v ( t ) ∥ ∞ , ∥ S ∥ ∞ } < ∞ 0\leq\eta(x,t)\leq M:=\max\left\{\operatorname*{ess\,sup}_{t>0}\lVert v(t)\rVert_{\infty},\lVert S\rVert_{\infty}\right\}<\infty 0 ≤ η ( x , t ) ≤ M := max { t > 0 ess sup ∥ v ( t ) ∥ ∞ , ∥ S ∥ ∞ } < ∞ and
v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) = m 2 η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 . \frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^{m}(x)(v(x,t)-S(x))=\frac{m}{2}\eta^{m-1}(x,t)\lvert v(x,t)-S(x)\rvert^{2}. m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) = 2 m η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 . This together with m ≥ 1 m \ge1 m ≥ 1
E ( t ) = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x = m 2 ∫ Ω η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≤ m M m − 1 2 ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ 2 d x , a.e. 0 < t < T . \begin{aligned}
\mathcal{E}(t) &= \int_{\Omega} \left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^m(x)(v(x,t)-S(x))\right\}\,dx \\
&= \frac{m}{2}\int_{\Omega}\eta^{m-1}(x,t)|v(x,t)-S(x)|^2\,dx \\
&\le\frac{mM^{m-1}}{2}\int_{\Omega}|v(x,t)-S(x)|^2\,dx,\qquad\text{a.e. }0<t<T.
\end{aligned} E ( t ) = ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x = 2 m ∫ Ω η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≤ 2 m M m − 1 ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ 2 d x , a.e. 0 < t < T . Furthermore, by using the elementary inequality
∣ a − b ∣ ≤ ∣ a m − b m ∣ 1 m , a , b ≥ 0 , m ≥ 1 , |a-b|\leq|a^m-b^m|^{\frac{1}{m}},\qquad a,b\geq0,\quad m\geq1, ∣ a − b ∣ ≤ ∣ a m − b m ∣ m 1 , a , b ≥ 0 , m ≥ 1 , Hölder’s inequality and the Poincaré inequality
∫ Ω ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≤ ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ 2 m d x ≤ ( ∫ Ω d x ) 1 − 1 m ( ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ 2 d x ) 1 m = C ∥ v m ( t ) − S m ∥ 2 m 2 m ≤ C ∥ ∇ ( v m ( t ) − S m ) ∥ 2 m 2 m , a.e. 0 < t < T . \begin{aligned}
\int_{\Omega}|v(x,t)-S(x)|^2\,dx
&\leq\int_{\Omega}|v^m(x,t)-S^m(x)|^{\frac{2}{m}}\,dx \\
&\leq\left(\int_{\Omega}dx\right)^{1-\frac{1}{m}}\left(\int_{\Omega}|v^m(x,t)-S^m(x)|^2\,dx\right)^{\frac{1}{m}} \\
&= C\|v^m(t)-S^m\|_{\frac{2}{m}}^{\frac{2}{m}} \\
&\leq C\|\nabla(v^m(t)-S^m)\|_{\frac{2}{m}}^{\frac{2}{m}},\qquad\text{a.e. }0<t<T.
\end{aligned} ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≤ ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ m 2 d x ≤ ( ∫ Ω d x ) 1 − m 1 ( ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ 2 d x ) m 1 = C ∥ v m ( t ) − S m ∥ m 2 m 2 ≤ C ∥∇ ( v m ( t ) − S m ) ∥ m 2 m 2 , a.e. 0 < t < T . Thus we get
E ( t ) ≤ C ∥ ∇ ( v m ( t ) − S m ) ∥ 2 m 2 m , a.e. 0 < t < T , \mathcal{E}(t)\leq C\|\nabla(v^m(t)-S^m)\|_{\frac{2}{m}}^{\frac{2}{m}},\qquad\text{a.e. }0<t<T, E ( t ) ≤ C ∥∇ ( v m ( t ) − S m ) ∥ m 2 m 2 , a.e. 0 < t < T , that is, (4.9). Combining (4.1), (4.9) and the first estimate in (4.2), we have
d d t ( g 1 − α , n ∗ [ E − E 0 ] ) ( t ) + λ t − α ( 1 − 1 m ) E m ( t ) ≤ ν g 1 − α ( t ) + I n ( t ) + J n ( t ) , a.e. 0 < t < T . \frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n} * [\mathcal{E}-\mathcal{E}_0])(t)+\lambda t^{-\alpha(1-\frac{1}{m})}\mathcal{E}^m(t)\leq\nu g_{1-\alpha}(t)+I_n(t)+J_n(t),\qquad\text{a.e. }0<t<T. d t d ( g 1 − α , n ∗ [ E − E 0 ]) ( t ) + λ t − α ( 1 − m 1 ) E m ( t ) ≤ ν g 1 − α ( t ) + I n ( t ) + J n ( t ) , a.e. 0 < t < T . Moreover, since
g 1 − α , n ( t ) = d d t ( g 1 − α , n ∗ 1 ) ( t ) , g_{1-\alpha,n}(t)=\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*1)(t), g 1 − α , n ( t ) = d t d ( g 1 − α , n ∗ 1 ) ( t ) , we can write
ν g 1 − α ( t ) = ν g 1 − α , n ( t ) + ν ( g 1 − α ( t ) − g 1 − α , n ( t ) ) = ν d d t ( g 1 − α , n ∗ 1 ) ( t ) + ν ( g 1 − α ( t ) − g 1 − α , n ( t ) ) = : ν d d t ( g 1 − α , n ∗ 1 ) ( t ) + K n ( t ) . \begin{aligned}
\nu g_{1-\alpha}(t)=\nu g_{1-\alpha,n}(t)+\nu\bigl(g_{1-\alpha}(t)-g_{1-\alpha,n}(t)\bigr) \\
&=\nu\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*1)(t)+\nu\bigl(g_{1-\alpha}(t)-g_{1-\alpha,n}(t)\bigr) \\
&=:\nu\frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*1)(t)+K_n(t).
\end{aligned} ν g 1 − α ( t ) = ν g 1 − α , n ( t ) + ν ( g 1 − α ( t ) − g 1 − α , n ( t ) ) = ν d t d ( g 1 − α , n ∗ 1 ) ( t ) + ν ( g 1 − α ( t ) − g 1 − α , n ( t ) ) =: ν d t d ( g 1 − α , n ∗ 1 ) ( t ) + K n ( t ) . Thus we obtain
d d t ( g 1 − α , n ∗ [ E − ( E 0 + ν ) ] ) ( t ) + λ t − α ( 1 − 1 m ) E m ( t ) ≤ L n ( t ) , a.e. 0 < t < T . (35) \frac{\mathrm{d}}{\mathrm{d}t}(g_{1-\alpha,n}*[\mathcal{E}-(\mathcal{E}_0+\nu)])(t)+\lambda t^{-\alpha(1-\frac{1}{m})}\mathcal{E}^m(t)\leq L_n(t),\qquad\text{a.e. }0<t<T. \tag*{(35)} d t d ( g 1 − α , n ∗ [ E − ( E 0 + ν )]) ( t ) + λ t − α ( 1 − m 1 ) E m ( t ) ≤ L n ( t ) , a.e. 0 < t < T . (35) where L n : = I n + J n + K n L_n:=I_n+J_n+K_n L n := I n + J n + K n L n → 0 L_n\to0 L n → 0 L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T ) n → ∞ n\to\infty n → ∞ Z ∈ C ( [ 0 , ∞ ) ) Z\in C([0,\infty)) Z ∈ C ([ 0 , ∞ ))
∂ t α ( Z − Z 0 ) + λ t − α ( 1 − 1 m ) Z m = 0 , t > 0 , Z ( 0 ) = Z 0 = E 0 + ν > 0. \partial_t^\alpha(Z-Z_0)+\lambda t^{-\alpha(1-\frac{1}{m})}Z^m=0,\qquad t>0,\qquad Z(0)=Z_0=\mathcal{E}_0+\nu>0. ∂ t α ( Z − Z 0 ) + λ t − α ( 1 − m 1 ) Z m = 0 , t > 0 , Z ( 0 ) = Z 0 = E 0 + ν > 0. Convolving the equation for Z Z Z h α , n h_{\alpha,n} h α , n (12) and taking the difference with (35) , we obtain
d d t ( g 1 − α , n ∗ [ E − Z ] ) ( t ) + λ { t − α ( 1 − 1 m ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − 1 m ) Z m ( ⋅ ) ] ) ( t ) } ≤ L n ( t ) (36) \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n} * [\mathcal{E}-Z]\right)(t)+\lambda\left\{t^{-\alpha\left(1-\frac{1}{m}\right)}\mathcal{E}^{m}(t)-\left(h_{\alpha,n} * \left[t^{-\alpha\left(1-\frac{1}{m}\right)}Z^{m}(\cdot)\right]\right)(t)\right\}\leq L_n(t) \tag*{(36)} d t d ( g 1 − α , n ∗ [ E − Z ] ) ( t ) + λ { t − α ( 1 − m 1 ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − m 1 ) Z m ( ⋅ ) ] ) ( t ) } ≤ L n ( t ) (36) for a.e. 0 < t < T 0<t<T 0 < t < T E ( t ) ≤ Z ( t ) \mathcal{E}(t)\leq Z(t) E ( t ) ≤ Z ( t ) 0 < t < T 0<t<T 0 < t < T E ( t ) > Z ( t ) \mathcal{E}(t)>Z(t) E ( t ) > Z ( t ) 0 < t < T 0<t<T 0 < t < T (36) by [ E − Z ] + ∈ L ∞ ( 0 , ∞ ) [\mathcal{E}-Z]_+\in L^\infty(0,\infty) [ E − Z ] + ∈ L ∞ ( 0 , ∞ ) H ( y ) = 1 2 [ y ] + 2 H(y)=\frac{1}{2}[y]_+^2 H ( y ) = 2 1 [ y ] + 2
1 2 d d t ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( t ) + λ { t − α ( 1 − 1 m ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − 1 m ) Z m ( ⋅ ) ] ) ( t ) } [ E ( t ) − Z ( t ) ] + ≤ L n ( t ) [ E ( t ) − Z ( t ) ] + , a.e. 0 < t < T . \begin{aligned}
\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n} * [\mathcal{E}-Z]_+^2\right)(t)
&+\lambda\left\{t^{-\alpha\left(1-\frac{1}{m}\right)}\mathcal{E}^{m}(t)-\left(h_{\alpha,n} * \left[t^{-\alpha\left(1-\frac{1}{m}\right)}Z^{m}(\cdot)\right]\right)(t)\right\}[\mathcal{E}(t)-Z(t)]_+\\
&\leq L_n(t)[\mathcal{E}(t)-Z(t)]_+,\qquad\text{a.e. }0<t<T.
\end{aligned} 2 1 d t d ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( t ) + λ { t − α ( 1 − m 1 ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − m 1 ) Z m ( ⋅ ) ] ) ( t ) } [ E ( t ) − Z ( t ) ] + ≤ L n ( t ) [ E ( t ) − Z ( t ) ] + , a.e. 0 < t < T . Since ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( 0 ) = 0 (g_{1-\alpha,n} * [\mathcal{E}-Z]_+^2)(0)=0 ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( 0 ) = 0 0 0 0 T T T
1 2 ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( T ) + λ ∫ 0 T { t − α ( 1 − 1 m ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − 1 m ) Z m ( ⋅ ) ] ) ( t ) } [ E ( t ) − Z ( t ) ] + d t ≤ ∫ 0 T L n ( t ) [ E ( t ) − Z ( t ) ] + d t . (37) \begin{aligned}
\frac{1}{2}(g_{1-\alpha,n} * [\mathcal{E}-Z]_+^2)(T)+\lambda\int_0^T\left\{t^{-\alpha\left(1-\frac{1}{m}\right)}\mathcal{E}^{m}(t)-\left(h_{\alpha,n} * \left[t^{-\alpha\left(1-\frac{1}{m}\right)}Z^{m}(\cdot)\right]\right)(t)\right\}[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t \\
&\leq\int_0^T L_n(t)[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t. \tag*{(37)}
\end{aligned} 2 1 ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( T ) + λ ∫ 0 T { t − α ( 1 − m 1 ) E m ( t ) − ( h α , n ∗ [ t − α ( 1 − m 1 ) Z m ( ⋅ ) ] ) ( t ) } [ E ( t ) − Z ( t ) ] + d t ≤ ∫ 0 T L n ( t ) [ E ( t ) − Z ( t ) ] + d t . (37) Furthermore, by (14) it holds that
∣ ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( T ) − ( g 1 − α ∗ [ E − Z ] + 2 ) ( T ) ∣ ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) 2 ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞ . \left|(g_{1-\alpha,n} * [\mathcal{E}-Z]_+^2)(T)-(g_{1-\alpha} * [\mathcal{E}-Z]_+^2)(T)\right|
\leq\|[\mathcal{E}-Z]_+\|_{L^\infty(0,T)}^2\|g_{1-\alpha,n}-g_{1-\alpha}\|_{L^1(0,T)}\to0\quad\text{as }n\to\infty. ( g 1 − α , n ∗ [ E − Z ] + 2 ) ( T ) − ( g 1 − α ∗ [ E − Z ] + 2 ) ( T ) ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) 2 ∥ g 1 − α , n − g 1 − α ∥ L 1 ( 0 , T ) → 0 as n → ∞. Since Z ∈ L ∞ ( 0 , ∞ ) Z\in L^\infty(0,\infty) Z ∈ L ∞ ( 0 , ∞ ) t − α ( 1 − 1 m ) ∈ L 1 ( 0 , T ) t^{-\alpha\left(1-\frac{1}{m}\right)}\in L^1(0,T) t − α ( 1 − m 1 ) ∈ L 1 ( 0 , T ) t − α ( 1 − 1 m ) Z m ∈ L 1 ( 0 , T ) t^{-\alpha\left(1-\frac{1}{m}\right)}Z^m\in L^1(0,T) t − α ( 1 − m 1 ) Z m ∈ L 1 ( 0 , T )
∣ ∫ 0 T ( h α , n ∗ [ t − α ( 1 − 1 m ) Z m ( ⋅ ) ] ) ( t ) [ E ( t ) − Z ( t ) ] + d t − ∫ 0 T t − α ( 1 − 1 m ) Z m ( t ) [ E ( t ) − Z ( t ) ] + d t ∣ ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) ∥ ( h α , n ∗ [ t − α ( 1 − 1 m ) Z m ] − t − α ( 1 − 1 m ) Z m ) ∥ L 1 ( 0 , T ) → 0 as n → ∞ . \begin{aligned}
\left|\int_0^T\left(h_{\alpha,n} * \left[t^{-\alpha\left(1-\frac{1}{m}\right)}Z^m(\cdot)\right]\right)(t)[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t-\int_0^Tt^{-\alpha\left(1-\frac{1}{m}\right)}Z^m(t)[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t\right|\\
\leq\|[\mathcal{E}-Z]_+\|_{L^\infty(0,T)}\left\|\left(h_{\alpha,n} * \left[t^{-\alpha\left(1-\frac{1}{m}\right)}Z^m\right]-t^{-\alpha\left(1-\frac{1}{m}\right)}Z^m\right)\right\|_{L^1(0,T)}\\
\to0\quad\text{as }n\to\infty.
\end{aligned} ∫ 0 T ( h α , n ∗ [ t − α ( 1 − m 1 ) Z m ( ⋅ ) ] ) ( t ) [ E ( t ) − Z ( t ) ] + d t − ∫ 0 T t − α ( 1 − m 1 ) Z m ( t ) [ E ( t ) − Z ( t ) ] + d t ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) ( h α , n ∗ [ t − α ( 1 − m 1 ) Z m ] − t − α ( 1 − m 1 ) Z m ) L 1 ( 0 , T ) → 0 as n → ∞. Moreover, since L n → 0 L_n\to0 L n → 0 L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T ) n → ∞ n\to\infty n → ∞
∣ ∫ 0 T L n ( t ) [ E ( t ) − Z ( t ) ] + d t ∣ ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) ∥ L n ∥ L 1 ( 0 , T ) → 0 as n → ∞ . \left|\int_0^T L_n(t)[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t\right|
\leq\|[\mathcal{E}-Z]_+\|_{L^\infty(0,T)}\|L_n\|_{L^1(0,T)}\to0\quad\text{as }n\to\infty. ∫ 0 T L n ( t ) [ E ( t ) − Z ( t ) ] + d t ≤ ∥ [ E − Z ] + ∥ L ∞ ( 0 , T ) ∥ L n ∥ L 1 ( 0 , T ) → 0 as n → ∞. Thus, passing to the limit n → ∞ n\to\infty n → ∞ (37) , we obtain
1 2 ( g 1 − α ∗ [ E − Z ] + 2 ) ( T ) + λ ∫ 0 T t − α ( 1 − 1 m ) ( E m ( t ) − Z m ( t ) ) [ E ( t ) − Z ( t ) ] + d t ≤ 0. \frac{1}{2}(g_{1-\alpha} * [\mathcal{E}-Z]_+^2)(T)+\lambda\int_0^Tt^{-\alpha\left(1-\frac{1}{m}\right)}(\mathcal{E}^{m}(t)-Z^{m}(t))[\mathcal{E}(t)-Z(t)]_+\,\mathrm{d}t\leq0. 2 1 ( g 1 − α ∗ [ E − Z ] + 2 ) ( T ) + λ ∫ 0 T t − α ( 1 − m 1 ) ( E m ( t ) − Z m ( t )) [ E ( t ) − Z ( t ) ] + d t ≤ 0. Since ( E m − Z m ) [ E − Z ] + ≥ 0 (\mathcal{E}^{m}-Z^{m})[\mathcal{E}-Z]_{+}\ge0 ( E m − Z m ) [ E − Z ] + ≥ 0 g 1 − α g_{1-\alpha} g 1 − α T > 0 T>0 T > 0 [ E ( t ) − Z ( t ) ] + = 0 [\mathcal{E}(t)-Z(t)]_{+}=0 [ E ( t ) − Z ( t ) ] + = 0 t > 0 t>0 t > 0
E ( t ) ≤ Z ( t ) , a.e. t > 0. \mathcal{E}(t)\le Z(t),\qquad\text{a.e. }t>0. E ( t ) ≤ Z ( t ) , a.e. t > 0. Therefore, applying Proposition 3.1 for β = − α ( 1 − 1 / m ) > − α \beta=-\alpha(1-1/m)>-\alpha β = − α ( 1 − 1/ m ) > − α γ = m \gamma=m γ = m
E ( t ) ≤ C 1 + t α / m 2 , t > 0. \mathcal{E}(t)\le\frac{C}{1+t^{\alpha/m^{2}}},\qquad t>0. E ( t ) ≤ 1 + t α / m 2 C , t > 0. Furthermore, by assertion (i) of Lemma 2.3, we see that
E ( t ) = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x ≥ 1 m + 1 ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ m + 1 d x = 1 m + 1 ∥ v ( t ) − S ∥ m + 1 m + 1 , a.e. t > 0. \begin{aligned}
\mathcal{E}(t)
&=\int_{\Omega}\left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^{m}(x)(v(x,t)-S(x))\right\}\,dx\\
&\ge\frac{1}{m+1}\int_{\Omega}|v(x,t)-S(x)|^{m+1}\,dx\\
&=\frac{1}{m+1}\|v(t)-S\|_{m+1}^{m+1},\qquad\text{a.e. }t>0.
\end{aligned} E ( t ) = ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x ≥ m + 1 1 ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ m + 1 d x = m + 1 1 ∥ v ( t ) − S ∥ m + 1 m + 1 , a.e. t > 0. This implies that
∥ v ( t ) − S ∥ m + 1 m + 1 ≤ C 1 + t α / m 2 , t > 0. \|v(t)-S\|_{m+1}^{m+1}\le\frac{C}{1+t^{\alpha/m^{2}}},\qquad t>0. ∥ v ( t ) − S ∥ m + 1 m + 1 ≤ 1 + t α / m 2 C , t > 0. For 1 ≤ p < m + 1 1\le p<m+1 1 ≤ p < m + 1 Ω \Omega Ω
∥ v ( t ) − S ∥ p ≤ ∣ Ω ∣ 1 p − 1 m + 1 ∥ v ( t ) − S ∥ m + 1 → 0 as t → ∞ . \|v(t)-S\|_{p}\le|\Omega|^{\frac{1}{p}-\frac{1}{m+1}}\|v(t)-S\|_{m+1}\to0\quad\text{as }t\to\infty. ∥ v ( t ) − S ∥ p ≤ ∣Ω ∣ p 1 − m + 1 1 ∥ v ( t ) − S ∥ m + 1 → 0 as t → ∞. Moreover, for m + 1 < p < ∞ m+1<p<\infty m + 1 < p < ∞ v v v S S S
∥ v ( t ) − S ∥ p ≤ ∥ v ( t ) − S ∥ ∞ 1 − m + 1 p ∥ v ( t ) − S ∥ m + 1 m + 1 p ≤ C ∥ v ( t ) − S ∥ m + 1 m + 1 p → 0 as t → ∞ . \|v(t)-S\|_{p}\le\|v(t)-S\|_{\infty}^{1-\frac{m+1}{p}}\|v(t)-S\|_{m+1}^{\frac{m+1}{p}}\le C\|v(t)-S\|_{m+1}^{\frac{m+1}{p}}\to0\quad\text{as }t\to\infty. ∥ v ( t ) − S ∥ p ≤ ∥ v ( t ) − S ∥ ∞ 1 − p m + 1 ∥ v ( t ) − S ∥ m + 1 p m + 1 ≤ C ∥ v ( t ) − S ∥ m + 1 p m + 1 → 0 as t → ∞. This implies that (1.4) holds for m ≥ 1 m\ge1 m ≥ 1
∣ a m − b m ∣ ≤ C m ( a m − 1 + b m − 1 ) ∣ a − b ∣ |a^{m}-b^{m}|\le C_{m}(a^{m-1}+b^{m-1})|a-b| ∣ a m − b m ∣ ≤ C m ( a m − 1 + b m − 1 ) ∣ a − b ∣ for a , b ≥ 0 a,b\ge0 a , b ≥ 0 m ≥ 1 m\ge1 m ≥ 1
∥ v m ( t ) − S m ∥ p ≤ C ∥ v ( t ) − S ∥ p → 0 as t → ∞ . \|v^{m}(t)-S^{m}\|_{p}\le C\|v(t)-S\|_{p}\to0\quad\text{as }t\to\infty. ∥ v m ( t ) − S m ∥ p ≤ C ∥ v ( t ) − S ∥ p → 0 as t → ∞. Therefore, by assertion (ii) of Remark 1.1, we obtain the convergence (1.4) and (1.5) for m ≥ 1 m\ge1 m ≥ 1
The fast diffusion case 0 < m < 1 0<m<1 0 < m < 1
Next, we prove Theorem 1.1 for the fast diffusion case 0 < m < 1 0<m<1 0 < m < 1
λ E ( t ) ≤ ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T . (38) \lambda\mathcal{E}(t)\le\|\nabla(v^{m}(t)-S^{m})\|_{2}^{2},\qquad\text{a.e. }0<t<T. \tag*{(38)} λ E ( t ) ≤ ∥∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T . (38) where λ > 0 \lambda> 0 λ > 0 α \alpha α m m m N N N Ω \Omega Ω ∥ u 0 ∥ L p 0 ( Ω ) \|u_0\|_{L^{p_0}(\Omega)} ∥ u 0 ∥ L p 0 ( Ω ) Lemma 2.3 and the Poincaré inequality, we deduce that
E ( t ) = ∫ Ω { v m + 1 ( x , t ) m + 1 − S m + 1 ( x ) m + 1 − S m ( x ) ( v ( x , t ) − S ( x ) ) } d x ≤ C ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ 2 d x ≤ C ∥ ∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T . \begin{aligned}
\mathcal{E}(t)=\int_{\Omega}\left\{\frac{v^{m+1}(x,t)}{m+1}-\frac{S^{m+1}(x)}{m+1}-S^m(x)(v(x,t)-S(x))\right\}\,dx \\
&\leq C\int_{\Omega}|v^m(x,t)-S^m(x)|^2\,dx \\
&\leq C\|\nabla(v^m(t)-S^m)\|_2^2,\quad\text{a.e. }0<t<T.
\end{aligned} E ( t ) = ∫ Ω { m + 1 v m + 1 ( x , t ) − m + 1 S m + 1 ( x ) − S m ( x ) ( v ( x , t ) − S ( x )) } d x ≤ C ∫ Ω ∣ v m ( x , t ) − S m ( x ) ∣ 2 d x ≤ C ∥∇ ( v m ( t ) − S m ) ∥ 2 2 , a.e. 0 < t < T . The case α < m < 1 \alpha< m < 1 α < m < 1
Since α < m < 1 \alpha< m < 1 α < m < 1 (26) , the first estimate in (27) , and (38) , we obtain
d d t ( g 1 − α , n ∗ [ E − ( E 0 + ν ) ] ) ( t ) + λ t α ( 1 m − 1 ) E ( t ) ≤ L n ( t ) , a.e. 0 < t < T . \frac{d}{dt}\left(g_{1-\alpha,n}*\left[\mathcal{E}-(\mathcal{E}_0+\nu)\right]\right)(t)+\lambda t^{\alpha\left(\frac{1}{m}-1\right)}\mathcal{E}(t)\leq L_n(t),\quad\text{a.e. }0<t<T. d t d ( g 1 − α , n ∗ [ E − ( E 0 + ν ) ] ) ( t ) + λ t α ( m 1 − 1 ) E ( t ) ≤ L n ( t ) , a.e. 0 < t < T . Let W ∈ C ( [ 0 , ∞ ) ) W\in C([0,\infty)) W ∈ C ([ 0 , ∞ ))
∂ t α ( W − W 0 ) + λ t α ( 1 m − 1 ) W = 0 , t > 0 , W ( 0 ) = W 0 = E 0 + ν > 0. \partial_t^\alpha(W-W_0)+\lambda t^{\alpha\left(\frac{1}{m}-1\right)}W=0,\quad t>0,\qquad W(0)=W_0=\mathcal{E}_0+\nu>0. ∂ t α ( W − W 0 ) + λ t α ( m 1 − 1 ) W = 0 , t > 0 , W ( 0 ) = W 0 = E 0 + ν > 0. Repeating the same argument as in the case m ≥ 1 m\geq1 m ≥ 1
E ( t ) ≤ W ( t ) , a.e. t > 0. \mathcal{E}(t)\leq W(t),\quad\text{a.e. }t>0. E ( t ) ≤ W ( t ) , a.e. t > 0. By Proposition 3.1 for β = α ( 1 / m − 1 ) \beta=\alpha(1/m-1) β = α ( 1/ m − 1 ) γ = 1 \gamma=1 γ = 1
E ( t ) ≤ C 1 + t α m , t > 0. (39) \mathcal{E}(t)\leq\frac{C}{1+t^{\frac{\alpha}{m}}},\quad t>0. \tag*{(39)} E ( t ) ≤ 1 + t m α C , t > 0. (39) Furthermore, since m < 1 m<1 m < 1
E ( t ) = m 2 ∫ Ω η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≥ m M m − 1 2 ∥ v ( t ) − S ∥ 2 2 , a.e. t > 0 , (40) \mathcal{E}(t)=\frac{m}{2}\int_{\Omega}\eta^{m-1}(x,t)|v(x,t)-S(x)|^2\,dx\geq\frac{mM^{m-1}}{2}\|v(t)-S\|_2^2,\quad\text{a.e. }t>0, \tag*{(40)} E ( t ) = 2 m ∫ Ω η m − 1 ( x , t ) ∣ v ( x , t ) − S ( x ) ∣ 2 d x ≥ 2 m M m − 1 ∥ v ( t ) − S ∥ 2 2 , a.e. t > 0 , (40) where
M = max { ess sup t > 0 ∥ v ( t ) ∥ ∞ , ∥ S ∥ ∞ } . M=\max\left\{\mathop{\operatorname{ess\,sup}}_{t>0}\|v(t)\|_\infty,\|S\|_\infty\right\}. M = max { ess sup t > 0 ∥ v ( t ) ∥ ∞ , ∥ S ∥ ∞ } . This together with (39) implies that
∥ v ( t ) − S ∥ 2 2 ≤ C 1 + t α m , t > 0. \|v(t)-S\|_2^2\leq\frac{C}{1+t^{\frac{\alpha}{m}}},\quad t>0. ∥ v ( t ) − S ∥ 2 2 ≤ 1 + t m α C , t > 0. Therefore, using again boundness of v v v S S S Ω \Omega Ω α < m < 1 \alpha<m<1 α < m < 1
∣ a m − b m ∣ ≤ C m ∣ a − b ∣ m , a , b ≥ 0 , 0 < m < 1 , |a^m-b^m|\leq C_m|a-b|^m,\quad a,b\geq0,\quad0<m<1, ∣ a m − b m ∣ ≤ C m ∣ a − b ∣ m , a , b ≥ 0 , 0 < m < 1 , we also obtain
∥ v ( t ) − S ∥ p m ≤ C m ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ p m d x ≤ C m ( ∫ Ω d x ) 1 − m ( ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ p d x ) m = C ∥ v ( t ) − S ∥ p p m \begin{aligned}
\|v(t)-S\|_p^m \le C_m \int_{\Omega} |v(x,t)-S(x)|^{pm}\,dx \\
&\le C_m \left(\int_{\Omega} dx\right)^{1-m}\left(\int_{\Omega}|v(x,t)-S(x)|^p\,dx\right)^m \\
&= C\|v(t)-S\|_p^{pm}
\end{aligned} ∥ v ( t ) − S ∥ p m ≤ C m ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ p m d x ≤ C m ( ∫ Ω d x ) 1 − m ( ∫ Ω ∣ v ( x , t ) − S ( x ) ∣ p d x ) m = C ∥ v ( t ) − S ∥ p p m → 0 as t → ∞ . \to0 \quad\text{as } t\to\infty. → 0 as t → ∞. This implies that (5) also holds for α < m < 1 \alpha<m<1 α < m < 1
The case 0 < m < α 0<m<\alpha 0 < m < α
We next consider the case 0 < m < α 0<m<\alpha 0 < m < α α / m > 1 \alpha/m>1 α / m > 1 (26) , the last estimate in (27) with β = α ( 1 / m − 1 ) > 1 − α \beta=\alpha(1/m-1)>1-\alpha β = α ( 1/ m − 1 ) > 1 − α (38) , we also obtain
d d t ( g 1 − α , n ∗ [ E − ( E 0 + ν ) ] ) ( t ) + λ t β E ( t ) ≤ μ t β − 1 + L n ( t ) \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*[\mathcal{E}-(\mathcal{E}_0+\nu)]\right)(t)+\lambda t^\beta\mathcal{E}(t)\le\mu t^{\beta-1}+L_n(t) d t d ( g 1 − α , n ∗ [ E − ( E 0 + ν )] ) ( t ) + λ t β E ( t ) ≤ μ t β − 1 + L n ( t ) for a.e. t > 0 t>0 t > 0 Z ~ ∈ C ( [ 0 , ∞ ) ) \widetilde{Z}\in C([0,\infty)) Z ∈ C ([ 0 , ∞ ))
∂ t α ( Z ~ − Z ~ 0 ) + λ t β Z ~ = μ t β − 1 , t > 0 , Z ~ ( 0 ) = Z ~ 0 = E 0 + ν . \partial_t^\alpha(\widetilde{Z}-\widetilde{Z}_0)+\lambda t^\beta\widetilde{Z}=\mu t^{\beta-1},\quad t>0,\qquad\widetilde{Z}(0)=\widetilde{Z}_0=\mathcal{E}_0+\nu. ∂ t α ( Z − Z 0 ) + λ t β Z = μ t β − 1 , t > 0 , Z ( 0 ) = Z 0 = E 0 + ν . Then, using regularization arguments and Lemma 2.1 again, we have
E ( t ) ≤ Z ~ ( t ) , t > 0. \mathcal{E}(t)\le\widetilde{Z}(t),\quad t>0. E ( t ) ≤ Z ( t ) , t > 0. Thus, applying Lemma 3.2 for β = α ( 1 / m − 1 ) \beta=\alpha(1/m-1) β = α ( 1/ m − 1 )
E ( t ) ≤ C t − 1 \mathcal{E}(t)\le Ct^{-1} E ( t ) ≤ C t − 1 for sufficiently large t > 0 t>0 t > 0 (40) implies that
∥ v ( t ) − S ∥ 2 2 ≤ C t − 1 \|v(t)-S\|_2^2\le Ct^{-1} ∥ v ( t ) − S ∥ 2 2 ≤ C t − 1 for sufficiently large t>0. Thus the convergence (4) and (5) also hold for $0<m<\alpha. □
The critical case m = α m=\alpha m = α
Finally, we consider the critical case m = α m=\alpha m = α (26) , the second estimate in (27) , and (38) , we have
d d t ( g 1 − α , n ∗ [ E − ( E 0 + ν ) ] ) ( t ) + λ t 1 − α E ( t ) ≤ μ t − α log ( 1 + t ) + L n ( t ) \frac{\mathrm{d}}{\mathrm{d}t}\left(g_{1-\alpha,n}*[\mathcal{E}-(\mathcal{E}_0+\nu)]\right)(t)+\lambda t^{1-\alpha}\mathcal{E}(t)\le\mu t^{-\alpha}\log(1+t)+L_n(t) d t d ( g 1 − α , n ∗ [ E − ( E 0 + ν )] ) ( t ) + λ t 1 − α E ( t ) ≤ μ t − α log ( 1 + t ) + L n ( t ) for a.e. 0 < t < T 0<t<T 0 < t < T W ~ ∈ C ( [ 0 , ∞ ) ) \widetilde{W}\in C([0,\infty)) W ∈ C ([ 0 , ∞ ))
∂ t α ( W ~ − W ~ 0 ) + λ t 1 − α W ~ = μ t − α log ( 1 + t ) , t > 0 , W ~ ( 0 ) = W ~ 0 = E 0 + ν . \partial_t^\alpha(\widetilde{W}-\widetilde{W}_0)+\lambda t^{1-\alpha}\widetilde{W}=\mu t^{-\alpha}\log(1+t),\quad t>0,\qquad\widetilde{W}(0)=\widetilde{W}_0=\mathcal{E}_0+\nu. ∂ t α ( W − W 0 ) + λ t 1 − α W = μ t − α log ( 1 + t ) , t > 0 , W ( 0 ) = W 0 = E 0 + ν . Then we obtain
E ( t ) ≤ W ~ ( t ) , t > 0. \mathcal{E}(t) \le\widetilde{W}(t), \quad t>0. E ( t ) ≤ W ( t ) , t > 0. By Lemma 3.3 and (4.15), we have
∥ v ( t ) − S ∥ 2 2 ≤ C log ( 1 + t ) t . \lVert v(t)-S\rVert_2^2 \le C\frac{\log(1+t)}{t}. ∥ v ( t ) − S ∥ 2 2 ≤ C t log ( 1 + t ) . for sufficiently large t > 0 t>0 t > 0 m = α m=\alpha m = α □ \square □
An alternative proof in the linear case m = 1 m=1 m = 1
In this section, we explain assertion (iii) of Remark 1.1. In the linear case m = 1 m=1 m = 1
u ( x , t ) = ∑ k = 1 ∞ E α , 1 ( − λ k t α ) u ^ 0 , k ϕ k ( x ) . u(x,t)=\sum_{k=1}^{\infty}E_{\alpha,1}(-\lambda_k t^\alpha)\widehat{u}_{0,k}\phi_k(x). u ( x , t ) = k = 1 ∑ ∞ E α , 1 ( − λ k t α ) u 0 , k ϕ k ( x ) . where ( λ k , ϕ k ) (\lambda_k,\phi_k) ( λ k , ϕ k )
− Δ ϕ k = λ k ϕ k in Ω , ϕ k = 0 on ∂ Ω , -\Delta\phi_k=\lambda_k\phi_k \quad\text{in }\Omega,\qquad\phi_k=0 \quad\text{on }\partial\Omega, − Δ ϕ k = λ k ϕ k in Ω , ϕ k = 0 on ∂ Ω , and 0 < λ 1 ≤ λ 2 ≤ ⋯ 0<\lambda_1\le\lambda_2\le\cdots 0 < λ 1 ≤ λ 2 ≤ ⋯ [22 ] . Moreover, the Fourier coefficients u ^ 0 , k \widehat{u}_{0,k} u 0 , k
u ^ 0 , k = ∫ Ω ϕ k ( x ) u 0 ( x ) d x , k = 1 , 2 , ⋯ . \widehat{u}_{0,k}=\int_{\Omega}\phi_k(x)u_0(x)\,\mathrm{d}x,\qquad k=1,2,\cdots. u 0 , k = ∫ Ω ϕ k ( x ) u 0 ( x ) d x , k = 1 , 2 , ⋯ . On the other hand, S S S
S ( x ) = 1 Γ ( 1 − α ) ( − Δ ) − 1 u 0 ( x ) = 1 Γ ( 1 − α ) ∑ k = 1 ∞ λ k − 1 u ^ 0 , k ϕ k ( x ) . S(x)=\frac{1}{\Gamma(1-\alpha)}(-\Delta)^{-1}u_0(x)=\frac{1}{\Gamma(1-\alpha)}\sum_{k=1}^{\infty}\lambda_k^{-1}\widehat{u}_{0,k}\phi_k(x). S ( x ) = Γ ( 1 − α ) 1 ( − Δ ) − 1 u 0 ( x ) = Γ ( 1 − α ) 1 k = 1 ∑ ∞ λ k − 1 u 0 , k ϕ k ( x ) . Thus, since { ϕ k } \{\phi_k\} { ϕ k } L 2 ( Ω ) L^2(\Omega) L 2 ( Ω )
∥ t α u ( t ) − S ∥ 2 2 = ∑ k = 1 ∞ ∥ u ^ 0 , k ( t α E α , 1 ( − λ k t α ) − 1 Γ ( 1 − α ) λ k ) ϕ k ∥ 2 2 = ∑ k = 1 ∞ ∣ u ^ 0 , k ∣ 2 ∣ t α E α , 1 ( − λ k t α ) − 1 Γ ( 1 − α ) λ k ∣ 2 . \begin{aligned}
\lVert t^\alpha u(t)-S\rVert_2^2
=\sum_{k=1}^{\infty}\left\lVert\widehat{u}_{0,k}\left(t^\alpha E_{\alpha,1}(-\lambda_k t^\alpha)-\frac{1}{\Gamma(1-\alpha)\lambda_k}\right)\phi_k\right\rVert_2^2 \\
&=\sum_{k=1}^{\infty}|\widehat{u}_{0,k}|^2\left|t^\alpha E_{\alpha,1}(-\lambda_k t^\alpha)-\frac{1}{\Gamma(1-\alpha)\lambda_k}\right|^2.
\end{aligned} ∥ t α u ( t ) − S ∥ 2 2 = k = 1 ∑ ∞ u 0 , k ( t α E α , 1 ( − λ k t α ) − Γ ( 1 − α ) λ k 1 ) ϕ k 2 2 = k = 1 ∑ ∞ ∣ u 0 , k ∣ 2 t α E α , 1 ( − λ k t α ) − Γ ( 1 − α ) λ k 1 2 . Note that
E α , 1 ( − τ ) ∼ 1 Γ ( 1 − α ) τ as τ → ∞ E_{\alpha,1}(-\tau)\sim\frac{1}{\Gamma(1-\alpha)\tau}\quad\text{as }\tau\to\infty E α , 1 ( − τ ) ∼ Γ ( 1 − α ) τ 1 as τ → ∞ (see, e.g., [11 ] ). Hence, for each k ∈ N k\in\mathbb{N} k ∈ N
∣ t α E α , 1 ( − λ k t α ) − 1 Γ ( 1 − α ) λ k ∣ → 0 as t → ∞ . \left|t^\alpha E_{\alpha,1}(-\lambda_k t^\alpha)-\frac{1}{\Gamma(1-\alpha)\lambda_k}\right|\to0\quad\text{as }t\to\infty. t α E α , 1 ( − λ k t α ) − Γ ( 1 − α ) λ k 1 → 0 as t → ∞. Furthermore, by (2.1), we obtain the following estimate uniformly in t > 0 t>0 t > 0
∑ k = 1 ∞ ∣ u ~ 0 , k ∣ 2 ∣ t α E α , 1 ( − λ k t α ) − 1 Γ ( 1 − α ) λ k ∣ 2 ≤ ∑ k = 1 ∞ ∣ u ~ 0 , k ∣ 2 ∣ C λ k − 1 Γ ( 1 − α ) λ k ∣ 2 ≤ ∑ k = 1 ∞ ∣ u ~ 0 , k ∣ 2 C λ k 2 ≤ C λ 1 2 ∑ k = 1 ∞ ∣ u ~ 0 , k ∣ 2 ≤ C λ 1 2 ∥ u 0 ∥ 2 2 , \begin{aligned}
\sum_{k=1}^{\infty}|\widetilde{u}_{0,k}|^2\left|t^\alpha E_{\alpha,1}(-\lambda_k t^\alpha)-\frac{1}{\Gamma(1-\alpha)\lambda_k}\right|^2
&\le\sum_{k=1}^{\infty}|\widetilde{u}_{0,k}|^2\left|\frac{C}{\lambda_k}-\frac{1}{\Gamma(1-\alpha)\lambda_k}\right|^2 \le\sum_{k=1}^{\infty}|\widetilde{u}_{0,k}|^2\frac{C}{\lambda_k^2}\\
&\le\frac{C}{\lambda_1^2}\sum_{k=1}^{\infty}|\widetilde{u}_{0,k}|^2\le\frac{C}{\lambda_1^2}\|u_0\|_2^2,
\end{aligned} k = 1 ∑ ∞ ∣ u 0 , k ∣ 2 t α E α , 1 ( − λ k t α ) − Γ ( 1 − α ) λ k 1 2 ≤ k = 1 ∑ ∞ ∣ u 0 , k ∣ 2 λ k C − Γ ( 1 − α ) λ k 1 2 ≤ k = 1 ∑ ∞ ∣ u 0 , k ∣ 2 λ k 2 C ≤ λ 1 2 C k = 1 ∑ ∞ ∣ u 0 , k ∣ 2 ≤ λ 1 2 C ∥ u 0 ∥ 2 2 , where we used Bessel’s inequality in the last inequality. Thus, by the dominated convergence theorem for series, we have
∥ t α u ( t ) − S ∥ 2 2 → 0 as t → ∞ . \|t^\alpha u(t)-S\|_2^2\to0 \quad\text{as }t\to\infty. ∥ t α u ( t ) − S ∥ 2 2 → 0 as t → ∞. This completes the proof in the linear case. □
Acknowledgements
Acknowledgements. The author would like to express his sincere gratitude to Professor Tatsuki Kawakami for his helpful comments on this study.
