5.1 The fractional Tikhonov method
In this section, we apply the fractional Tikhonov method given by Morigi [35]. From now on, we denote
$$ \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) = \xi _{1}E_{\beta ,1} \biggl(-\biggl( \frac{j\pi }{R}\biggr)^{2}T^{\beta } \biggr) + \xi _{2} \int _{0}^{T}E_{ \beta ,1} \biggl(-\biggl( \frac{j \pi }{R} \biggr)^{2}t^{\beta } \biggr) \,dt, $$
we propose the following regularized solution with exact data \((f,G)\):
$$ \mathcal{P}_{\alpha ,\delta } (r) = \sum_{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-1} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{\delta } \langle \Xi ,\psi _{j}\rangle \psi _{j}(r), $$
(5.1)
in which
$$\begin{aligned} \Xi (r) &= \sum_{j=1}^{+\infty } \langle f,\psi _{j}\rangle \psi _{j}(r) \\ &\quad {}- \xi _{1} \sum_{j=1}^{+\infty } \int _{0}^{T} (T-s)^{\beta -1} E_{\beta ,\beta } \biggl(- \biggl(\frac{j\pi }{R} \biggr)^{2}(T-s)^{\beta } \biggr) \bigl\langle G(\cdot ,s),\psi _{j}\bigr\rangle \,ds \psi _{j}(r). \end{aligned}$$
(5.2)
However, if the measured data \(\{f,G\}\) are noised by \(\{f_{\epsilon },G_{\epsilon }\}\), then we get
$$ \mathcal{P}^{\epsilon }_{\alpha ,\delta } (r) = \sum _{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-1} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{\delta } \bigl\langle \Xi ^{\epsilon },\psi _{j}\bigr\rangle \psi _{j}(r), $$
(5.3)
in which
$$\begin{aligned} \Xi ^{\epsilon }(r) &= \sum_{j=1}^{+\infty } \bigl\langle f^{\epsilon },\psi _{j} \bigr\rangle \psi _{j}(r) \\ &\quad {}- \xi _{1} \int _{0}^{T} (T-s)^{\beta -1} E_{\beta ,\beta } \biggl(- \biggl(\frac{j\pi }{R} \biggr)^{2}(T-s)^{\beta } \biggr) \bigl\langle G^{\epsilon }(\cdot ,s), \psi _{j}\bigr\rangle \,ds \psi _{j}(r), \end{aligned}$$
(5.4)
where \(\delta \in (\frac{1}{2};1 ]\) and \(\alpha > 0\). Noting that when \(\delta = 1\), the fractional Tikhonov method becomes a standard Tikhonov regularization.
Lemma 5.1
Let \(\delta \in (\frac{1}{2};1]\), we have
$$ \sup_{j > 0} \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}( \pi ,R,T) \bigr\vert ^{-1} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{\delta } \leq \mathcal{Z}_{\delta }^{\frac{1}{2}} \alpha ^{- \frac{1}{2}}, $$
(5.5)
with \(\mathcal{Z}\) depending on δ.
Proof
The proof of lemma can be found in [36]. □
5.2 An a priori parameter choice rule
Let us consider the operator
$$ \mathcal{Q}_{\beta }(\pi ,R,T) (t)\nu := \sum _{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-1} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{\delta }\langle \nu ,\psi _{j}\rangle \psi _{j}(r), $$
(5.6)
for \(\nu \in L^{2}([0,R];r^{2})\) and \(0 \leq t \leq T\).
By applying the fractional Tikhonov method, we can see that
$$\begin{aligned} &u_{\alpha ,\delta }(r,0) \\ &\quad = \sum_{j=1}^{+\infty } \bigl\vert \Theta _{ \beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-1} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{\delta } \\ &\qquad {}\times \biggl[ \langle f,\psi _{j}\rangle - \xi _{1} \int _{0}^{T} (T-s)^{ \beta -1}E_{\beta ,\beta } \biggl(- \biggl(\frac{j\pi }{R} \biggr)^{2}(T-s)^{ \beta } \biggr) \bigl\langle G(\cdot ,s),\psi _{j}\bigr\rangle \,ds \biggr] \psi _{j}(r). \end{aligned}$$
(5.7)
By choosing the regularization parameter α, the following theorem gives that the choice α is valid by using suitable assumptions. In order to give error estimate, let us assume that \(\|\Xi \|_{\mathcal{H}^{s}([0,R];r^{2})} \leq \mathcal{C}\) for any \(s > 0\), where \(\mathcal{C}\) is a positive constant. Before going to the main theorem, we have auxiliary lemmas as follows.
Lemma 5.2
For some positive constant, we get
$$ \frac{\alpha \varsigma ^{2-2k}}{\mathcal{N}^{2}+\alpha \varsigma ^{2}} \leq \textstyle\begin{cases} \frac{k^{k}(1-k)^{1-k} }{\mathcal{N}^{2k}} \alpha ^{k},& \textit{if }0 < k < 1, \\ \frac{1}{\mathcal{N}^{2} s^{2k-2} } \alpha ,& \textit{if }k \geq 1. \end{cases} $$
(5.8)
Proof
This lemma is proven similarly [36]. □
Theorem 5.3
Let \(f \in L^{2}([0,R];r^{2})\) and \(G \in L^{\infty }([0,T];L^{2}([0,R];r^{2}))\), inside Θ performed as in the digital formula (5.2), and if we choose the parameter regularization
$$ \alpha = \textstyle\begin{cases} (\frac{\epsilon }{\mathcal{C}} )^{\frac{2}{s+1}},& \textit{if }0 < s < 1, \\ (\frac{\epsilon }{\mathcal{C}} ),& \textit{if }s \geq 1, \end{cases} $$
then it gives:
-
If \(0 < s < 1\), then we have
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \textit{ is of order } \epsilon ^{\frac{s}{s+1}}. $$
(5.9)
-
If \(s \geq 1\), then we have
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \textit{ is of order } \epsilon ^{\frac{1}{2}}. $$
(5.10)
Proof
From the triangle inequality, we have
$$\begin{aligned} \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} &\leq \bigl\Vert u_{\alpha ,\delta }^{\epsilon }( \cdot ,0) - u_{\alpha , \delta }(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad {} + \bigl\Vert u_{\alpha ,\delta }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})}. \end{aligned}$$
(5.11)
First of all, we have the estimate \(\|u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u_{\alpha ,\delta }( \cdot ,0) \|_{L^{2}([0,R];r^{2})}\). Now, using the inequality \((a+b)^{2} \leq 2 ( a^{2} + b^{2} ) \) gives
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} \\ &\quad \leq 2\sum_{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-2} \biggl[ \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{2\delta } \\ &\qquad {}\times \biggl( \bigl\vert \bigl\langle f-f^{\epsilon },\psi _{j}\bigr\rangle \bigr\vert ^{2} \\ &\qquad {}+ \xi ^{2}_{1} \biggl\vert \int _{0}^{T} (T-s)^{\beta -1}E_{\beta , \beta } \biggl(- \biggl(\frac{j\pi }{R} \biggr)^{2}(T-s)^{\beta } \biggr) \bigl\langle G(\cdot ,s)-G^{ \epsilon }(\cdot ,s),\psi _{j}\bigr\rangle \,ds \biggr\vert ^{2} \biggr). \end{aligned}$$
(5.12)
From (5.12), applying Lemma 2.3 and the estimate of (1.3), it is easy to see that
$$\begin{aligned} \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} & \leq 2 \mathcal{Z}_{\delta } \alpha ^{-1} \bigl( \bigl\Vert f-f^{\epsilon } \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} \\ &\quad {} + \xi _{1}^{2} \overline{D}_{5}^{2} \bigl\Vert G^{\epsilon }-G \bigr\Vert ^{2}_{L^{ \infty }(0,T;L^{2}([0,R];r^{2}))} \bigr). \end{aligned}$$
(5.13)
Hence, we conclude that
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0)-u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \leq 2 \mathcal{Z}_{\delta }^{ \frac{1}{2}} \alpha ^{-\frac{1}{2}} \epsilon \bigl[1 + \xi _{1}^{2} \overline{D}_{5}^{2} \bigr]^{\frac{1}{2}}. $$
(5.14)
Moreover, with 〈Ξ, \(\psi _{j}\rangle \) defined in (5.2), we also get
$$\begin{aligned} & \bigl\Vert u_{\alpha , \delta }(\cdot ,0)-u(\cdot ,0) \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} \\ &\quad \leq \sum_{j=1}^{+\infty }|^{2} \bigl\vert \Theta _{\beta ,j}^{\xi _{1}, \xi _{2}}(\pi ,R,T) \bigr\vert ^{-2} \biggl[ 1 - \frac{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr]^{2\delta } \bigl\vert \langle \Xi ,\psi _{j}\rangle \bigr\vert ^{2}. \end{aligned}$$
(5.15)
From the definition of \(\|\Xi (\cdot )\|_{\mathcal{H}^{s}([0,R];r^{2})} \leq \mathcal{C}\), for any \(s>0\), \(\delta \in (\frac{1}{2},1]\), we obtain
$$\begin{aligned} \bigl\Vert u_{\alpha , \delta }(\cdot ,0)-u(\cdot ,0) \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} &\leq \sum _{j=1}^{+\infty } \biggl[ \frac{\alpha (j^{2} )^{-2s}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \biggr] \frac{ (j^{2} )^{2s} \vert \langle \Xi ,\psi _{j}\rangle \vert ^{2}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}} \\ &\leq \sup_{j \geq 1} \biggl[ \frac{\alpha (j^{2})^{-2s} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} +\alpha } \biggr] { \mathcal{C}}^{2} \\ &\leq \sup_{j \geq 1} \biggl[ \frac{\alpha (j^{2})^{2-2s} }{ \underline{E}^{2} +\alpha j^{4}} \biggr] { \mathcal{C}}^{2}. \end{aligned}$$
(5.16)
Applying Lemma 5.2 and Lemma 2.4, where we used \(|\Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) |^{2} \geq \frac{\underline{E}^{2}}{j^{4}}\), we get
$$\begin{aligned} \sup_{j \geq 1} \biggl[ \frac{\alpha (j^{2})^{-2s} }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} +\alpha } \biggr] &\leq \sup_{j \geq 1} \biggl[ \frac{\alpha (j^{2})^{2-2s} }{ \underline{E}^{2} +\alpha j^{4}} \biggr] \\ &\leq \textstyle\begin{cases} \frac{s^{s}(1-s)^{1-s} }{\underline{E}^{2s}} \alpha ^{s}, &\text{if }0 < s < 1, \\ \frac{1}{\underline{E}^{2} } \alpha , &\text{if }s \geq 1. \end{cases}\displaystyle \end{aligned}$$
(5.17)
Combining (5.15), (5.16), and (5.17), we obtain
$$ \bigl\Vert u_{\alpha , \delta }(\cdot ,0)-u(\cdot ,0) \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} \leq \textstyle\begin{cases} {\frac{s^{s}(1-s)^{1-s} }{\underline{E}^{2s}}} \mathcal{C}^{2} \alpha ^{s},& \text{if }0 < s < 1, \\ {\frac{1}{\underline{E}^{2} }} \mathcal{C}^{2} \alpha ,& \text{if }s \geq 1. \end{cases} $$
(5.18)
From (5.14) and (5.18), combining the inequality \(\sqrt{a^{2} + b^{2}} \leq a + b\), \(\forall a,b \geq 0\), we deduce that
$$\begin{aligned} \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} &\leq 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} \alpha ^{-\frac{1}{2}} \epsilon \bigl(1 + \xi _{1}^{2} \overline{D}_{5}^{2} \bigr)^{\frac{1}{2}} \\ &\quad {}+ \textstyle\begin{cases} ({\frac{s^{s}(1-s)^{1-s} }{\underline{E}^{2s}}} )^{ \frac{1}{2}} \mathcal{C} \alpha ^{\frac{s}{2}}, &\text{if }0 < s < 1, \\ \frac{1}{\underline{E}} \mathcal{C} \alpha ^{\frac{1}{2}}, &\text{if }s \geq 1. \end{cases}\displaystyle \end{aligned}$$
(5.19)
Choose the regularization parameter α as
$$ \alpha = \textstyle\begin{cases} (\frac{\epsilon }{\mathcal{C}} )^{\frac{2}{s+1}},& \text{if }0 < s < 1, \\ (\frac{\epsilon }{\mathcal{C}} ),& \text{if }s \geq 1, \end{cases} $$
(5.20)
then we have:
-
If \(0 < s < 1\), then we have the following estimate:
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \text{ is of order } \epsilon ^{\frac{s}{s+1}}. $$
(5.21)
-
If \(s \geq 1\), then we have the following estimate:
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \text{ is of order } \epsilon ^{\frac{1}{2}}. $$
(5.22)
The proof is completed. □
5.3 An a posteriori parameter choice rule
In this subsection, considering the choice of the a posteriori regularization parameter in Morozov’s discrepancy principle, [37] we choose the regularization parameter α such that
$$ \bigl\Vert \mathcal{K}\mathcal{P}^{\epsilon }_{\alpha ,\delta } - \Xi ^{ \epsilon } \bigr\Vert _{L^{2}([0,R];r^{2})} = \zeta \epsilon , $$
(5.23)
where \(\|\Xi ^{\epsilon }\|_{L^{2}([0,R];r^{2})}\geq \zeta \epsilon \), \(\zeta > 1\) depends on ϵ.
Lemma 5.4
For some positive constants k, α, \(\mathcal{N}\), ς, we get
$$ \frac{\alpha \varsigma ^{-k+1}}{\mathcal{N}^{2}+\alpha \varsigma ^{2}} \leq \textstyle\begin{cases} \frac{2^{-1} (k+1)^{\frac{k+1}{2}} }{\mathcal{N}^{k+1}} \alpha ^{ \frac{k+1}{2}},& \textit{if }0 < k < 1, \\ \frac{1}{\mathcal{N}^{2} \varsigma ^{1-k} } \alpha ,& \textit{if }k \geq 1. \end{cases} $$
(5.24)
Proof
This lemma is proven similarly [36]. □
Lemma 5.5
From (5.23), if we can find that ζ is satisfied, then we have the estimate of α as follows:
$$ \alpha ^{-1} \leq \textstyle\begin{cases} ( \frac{\mathcal{V}_{3}(\overline{E},\underline{E},s)}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{2}{s+1}} (\frac{\mathcal{C}}{\epsilon } )^{ \frac{2}{s+1}} ,& \textit{if }0 < s < 1, \\ \frac{\mathcal{V}_{4}(\overline{E},\underline{E})}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } (\frac{\mathcal{C}}{\epsilon } ),& \textit{if }s \geq 1. \end{cases} $$
(5.25)
Proof
From (5.23), it gives
$$\begin{aligned} \zeta \epsilon &= \bigl\Vert \mathcal{K} \mathcal{P}^{\epsilon }_{\alpha , \delta }- \Xi ^{\epsilon } \bigr\Vert _{L^{2}([0,R];r^{2})} \leq \Biggl\Vert \sum _{j=1}^{+\infty } \frac{\alpha }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \bigl\langle \Xi ^{\epsilon },\psi _{j}\bigr\rangle \psi _{j} \Biggr\Vert _{L^{2}([0,R];r^{2})} \\ &\leq \underbrace{ \bigl\Vert \Xi ^{\epsilon } - \Xi \bigr\Vert _{L^{2}([0,R];r^{2})}}_{: \mathcal{X}_{1}} + \underbrace{ \Biggl\Vert \sum _{j=1}^{+\infty }\frac{\alpha }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \langle \Xi , \psi _{j}\rangle \psi _{j} \Biggr\Vert _{L^{2}([0,R];r^{2})}}_{:= \mathcal{X}_{2}}. \end{aligned}$$
(5.26)
We have the estimate ζϵ through two steps as follows, one by one.
Step 1: Estimate of \(\mathcal{X}_{1}\), to do this, we recall Ξ, \(\Xi ^{\epsilon }\) from expressions (5.2) and (5.4), and we have
$$\begin{aligned} \mathcal{X}_{1}^{2} & \leq 2 \bigl\Vert f^{\epsilon }-f \bigr\Vert ^{2}_{L^{2}([0,R];r^{2})} + 2 \xi _{1}^{2}\overline{D}_{5}^{2} \bigl\Vert G^{\epsilon }-G \bigr\Vert ^{2}_{L^{ \infty }(0,T;L^{2}([0,T];r^{2}))} \\ &\leq 2\epsilon ^{2} \bigl(1 + \xi _{1}^{2} \overline{D}_{5}^{2} \bigr). \end{aligned}$$
(5.27)
Step 2: Estimate of \(\mathcal{X}_{2}\), using again the a priori bound condition of Ξ, we obtain
$$\begin{aligned} \mathcal{X}_{2} &= \biggl\Vert \frac{\alpha \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert (j^{2})^{-s}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \frac{(j^{2})^{s}\langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert } \biggr\Vert _{L^{2}([0,R];r^{2})} \\ &\leq \sup_{j \geq 1} \biggl( \frac{\alpha (j^{2})^{-s} \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2} + \alpha } \biggr) \mathcal{C}. \end{aligned}$$
(5.28)
From (5.28), using Lemma 2.4 and Lemma 5.4 implies that
$$ \frac{ \alpha (j^{2})^{-s} \frac{\overline{E}}{j^{2}} }{ \frac{\underline{E}^{2}}{j^{4}} + \alpha } \leq \overline{E} \frac{\alpha (j^{2})^{-s-1}}{ \underline{E}^{2} + \alpha j^{4} } \leq \textstyle\begin{cases} \underbrace{\overline{E} \tfrac{(s+1)^{s+1}}{2\underline{E}^{s+1}}}_{:= \mathcal{V}_{1}(\overline{E},\underline{E},s)} \alpha ^{ \frac{s+1}{2}},& \text{if }0 < s < 1, \\ \underbrace{ \tfrac{\overline{E}}{\underline{E}^{2}}}_{\mathcal{V}_{2}( \overline{E},\underline{E})} \alpha ,& \text{if }s \geq 1. \end{cases} $$
(5.29)
From the analytics assessment on the side, we get
$$ \epsilon \bigl( \zeta - \sqrt{2} \bigl(1 + \xi _{1}^{2} \overline{D}_{5}^{2}\bigr)^{ \frac{1}{2}} \bigr) \leq \textstyle\begin{cases} \mathcal{V}_{3}(\overline{E},\underline{E},s) \mathcal{C} \alpha ^{ \frac{s+1}{2}}, &\text{if }0 < s < 1, \\ \mathcal{V}_{4}(\overline{E},\underline{E}) \mathcal{C} \alpha , &\text{if }s \geq 1. \end{cases} $$
(5.30)
This yields
$$ \alpha ^{-1} \leq \textstyle\begin{cases} ( \frac{\mathcal{V}_{3}(\overline{E},\underline{E},s)}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{2}{s+1}} (\frac{\mathcal{C}}{\epsilon } )^{ \frac{2}{s+1}} , &\text{if }0 < s < 1, \\ \frac{\mathcal{V}_{4}(\overline{E},\underline{E})}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } (\frac{\mathcal{C}}{\epsilon } ), &\text{if }s \geq 1. \end{cases} $$
(5.31)
□
Theorem 5.6
Assume that (1.3) holds, recalling the α in Lemma 5.5, then we have the following estimate:
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq \underline{E}^{-\frac{s}{s+1}} (1+\zeta )^{\frac{s}{s+1}} \epsilon ^{\frac{s}{s+1}} \mathcal{C}^{\frac{1}{s+1}} \\ &\qquad {}+ \textstyle\begin{cases} 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{3}(\overline{E},\underline{E},s)}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{s+1}} \mathcal{C}^{\frac{1}{s+1}}\epsilon ^{ \frac{s}{s+1}} ,& \textit{if }0 < s < 1, \\ 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{4}(\overline{E},\underline{E})}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{2}} \mathcal{C}^{\frac{1}{2}}\epsilon ^{\frac{1}{2}},& \textit{if }s \geq 1. \end{cases}\displaystyle \end{aligned}$$
(5.32)
Proof
From the triangle inequality, we get
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq\bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} + \bigl\Vert u_{\alpha ,\delta }( \cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})}. \end{aligned}$$
(5.33)
From (5.14), we obtain
$$ \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0)-u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \leq 2 \mathcal{Z}_{\delta }^{ \frac{1}{2}} \alpha ^{-\frac{1}{2}} \epsilon \bigl( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} \bigr)^{\frac{1}{2}}. $$
(5.34)
Substituting (5.31) into above equation, one has
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0)-u_{\alpha ,\delta }( \cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq \textstyle\begin{cases} 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{3}(\overline{E},\underline{E},s)}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{s+1}} \mathcal{C}^{\frac{1}{s+1}}\epsilon ^{ \frac{s}{s+1}} , &\text{if }0 < s < 1, \\ 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{4}(\overline{E},\underline{E})}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{2}} \mathcal{C}^{\frac{1}{2}}\epsilon ^{\frac{1}{2}}, &\text{if }s \geq 1. \end{cases}\displaystyle \end{aligned}$$
(5.35)
Next, we give the estimate \(\| u_{\alpha ,\delta }(\cdot ,0) - u(\cdot ,0) \|_{L^{2}([0,R];r^{2})}\) as follows:
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq \Biggl\Vert \sum_{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}( \pi ,R,T) \bigr\vert ^{-1} \frac{\alpha \langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \Biggr\Vert _{L^{2}([0,R];r^{2})}. \end{aligned}$$
(5.36)
From (5.36), applying the Hölder inequality, we get
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq \underbrace{ \Biggl\Vert \sum_{j=1}^{+\infty } \bigl\vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \bigr\vert ^{-1} \frac{\alpha \langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \Biggr\Vert ^{\frac{s}{s+1}}_{L^{2}([0,R];r^{2})}}_{:= \mathcal{I}_{1}} \\ &\qquad {} \times \underbrace{ \Biggl\Vert \sum _{j=1}^{+\infty }\frac{\alpha }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \frac{\langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert } \Biggr\Vert ^{\frac{1}{s+1}}_{L^{2}([0,R];r^{2})} }_{:= \mathcal{I}_{2}}. \end{aligned}$$
(5.37)
Because of \(\|\Xi \|_{\mathcal{H}^{s}([0,R];r^{2})} \leq \mathcal{C}\), we obtain
$$ \mathcal{I}_{2} \leq \Biggl\Vert \sum _{j=1}^{+\infty }\bigl(j^{2} \bigr)^{-s} \frac{(j^{2})^{s}\langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert } \Biggr\Vert ^{\frac{1}{s+1}} \leq \sup_{j \geq 1} \bigl(j^{2}\bigr)^{-\frac{s}{s+1}} \mathcal{C}^{\frac{1}{s+1}}. $$
(5.38)
Next, using Lemma 2.4, \(\mathcal{I}_{1}\) can be bounded as follows:
$$\begin{aligned} \mathcal{I}_{1} &\leq \sup_{j \geq 1} \bigl\vert \Theta _{\beta ,j}^{\xi _{1}, \xi _{2}}(\pi ,R,T) \bigr\vert ^{-\frac{s}{s+1}} \Biggl\Vert \sum_{j=1}^{+ \infty } \frac{\alpha \langle \Xi ,\psi _{j}\rangle \psi _{j}}{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \Biggr\Vert ^{\frac{s}{s+1}}_{L^{2}([0,R];r^{2})} \\ &\leq \sup_{j \geq 1} \bigl(j^{2} \bigr)^{\frac{s}{s+1}} \underline{E}^{- \frac{s}{s+1}} \Biggl( \bigl\Vert \Xi - \Xi ^{\epsilon } \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad {} + \Biggl\Vert \sum_{j=1}^{+\infty } \frac{\alpha }{ \vert \Theta _{\beta ,j}^{\xi _{1},\xi _{2}}(\pi ,R,T) \vert ^{2}+\alpha } \bigl\langle \Xi ^{\epsilon },\psi _{j} \bigr\rangle \psi _{j}(\cdot ) \Biggr\Vert _{L^{2}([0,R];r^{2})} \Biggr)^{\frac{s}{s+1}} \\ &\leq \sup_{j \geq 1} \bigl(j^{2} \bigr)^{\frac{s}{s+1}} \underline{E}^{- \frac{s}{s+1}} (\epsilon + \zeta \epsilon )^{\frac{s}{s+1}}. \end{aligned}$$
(5.39)
Combining (5.37) to (5.39), we conclude that
$$ \bigl\Vert u_{\alpha ,\delta }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \leq \underline{E}^{-\frac{s}{s+1}} (1+\zeta )^{\frac{s}{s+1}} \epsilon ^{\frac{s}{s+1}} \mathcal{C}^{\frac{1}{s+1}}. $$
(5.40)
From (5.33) and (5.40), we know that
$$\begin{aligned} & \bigl\Vert u_{\alpha ,\delta }^{\epsilon }(\cdot ,0) - u(\cdot ,0) \bigr\Vert _{L^{2}([0,R];r^{2})} \\ &\quad \leq \underline{E}^{-\frac{s}{s+1}} (1+\zeta )^{\frac{s}{s+1}} \epsilon ^{\frac{s}{s+1}} \mathcal{C}^{\frac{1}{s+1}} \\ &\qquad {}+ \textstyle\begin{cases} 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{3}(\overline{E},\underline{E},s)}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{s+1}} \mathcal{C}^{\frac{1}{s+1}}\epsilon ^{ \frac{s}{s+1}} , &\text{if }0 < s < 1, \\ 2 \mathcal{Z}_{\delta }^{\frac{1}{2}} ( 1 + \xi _{1}^{2} \overline{D}_{5}^{2} )^{\frac{1}{2}} ( \frac{\mathcal{V}_{4}(\overline{E},\underline{E})}{ ( \zeta - \sqrt{2} (1 + \xi _{1}^{2}\overline{D}_{5}^{2})^{\frac{1}{2}} ) } )^{\frac{1}{2}} \mathcal{C}^{\frac{1}{2}}\epsilon ^{\frac{1}{2}}, &\text{if }s \geq 1. \end{cases}\displaystyle \end{aligned}$$
(5.41)
This ends the proof of this theorem. □