1 Introduction

Let \(H\) be a Hilbert space with the inner product \(\langle. \rangle \) and the norm \(\|.\|\) and let \(A\) be a positive self-adjoint operator defined on a dense subspace \(D(A)\subset H\) such that \(-A\) generates a compact contractive semi-group \(\{S(t)\}_{t \ge0}\) on \(H\). We shall consider the backward problem of finding a function \(u:[0,T] \to H\) such that

$$ \left\{ \textstyle\begin{array}{l@{\quad}l} u_{t} + Au= f\bigl(t,u(t)\bigr),& t \in(0,T), \\ u(T)=\varphi, \end{array}\displaystyle \right. $$
(1)

where the data \(\varphi\) is given in \(H\) and the source function \(f\) will be defined later. In practice, the data \(\varphi\in H\) is noisy and is represented by the perturbed data \(\varphi^{\epsilon} \in H\) satisfying

$$\begin{aligned} \bigl\| \varphi^{\epsilon}-\varphi\bigr\| \le\epsilon, \end{aligned}$$
(2)

where the constant \(\epsilon>0\) represents an upper bound on the measurement error. The problem (1) is not well-posed because its solution may not exist and, even if it exists, it does not depend continuously on the “noisy” Cauchy data \(\varphi^{\epsilon}\). Hence, a regularization process is required in order to obtain a stable solution.

The linear homogeneous case \(f=0\) has been studied in many papers, we refer to some interesting ones [1–4, 7, 9, 15, 18, 24]. Although there are many papers on the linear case of the backward parabolic problem, the nonlinear cases has been less investigated [8, 17, 21–23] and it is the purpose of this study to make advances into the semi-linear problem (1). First, the study of backward problem for parabolic are very close to those that have been developed in the context of controllability of parabolic problems. The issues there are also very close to that of solving the equation backwards, although often this appears as a technical preliminary step to deal with more general control problems. We note the reader to the paper [5] of E. Zuazua et al. which some explicit estimates on backward uniqueness for parabolic equations is investigated. The method in [5] can be applied to solve the backward problem and it is a future topic of our work. Next, we give some comments on the development of the backward in time nonlinear parabolic. To solve the nonlinear problem, M. Klibanov and his group apply the new ideas of Carleman estimates. The application of a powerful tool of Carleman estimates leads to much more general and powerful results not only for parabolic equations with backwards time but also for ill-posed Cauchy problems for both linear and quasilinear PDEs of the second order. The first general theory of applications of the quasireversibility method for the case of ill-posed Cauchy problems for linear equations can be found in Chap. 2 of [12]. Carleman estimates ensure Hölder convergence rates of regularized solutions, which is stronger than the logarithmic convergence rate which is in our paper. Moreover, in [11], the authors gave a new globally convergent numerical method for a wide class of ill-posed problems for quasilinear PDEs. Note that Carleman estimate of Lemma 5.3 of [10] combined with the scheme of Sect. 2 of [11] leads to the desired result for backward in time parabolic equation. Furthermore, numerical experiments for the method of [11] can be found in [13]. Therefore, the method of Carleman estimates given in [10–14].

In this paper, we don’t follow the above method. We give a new Fourier method for regularizing the nonlinear problem. The more details of our method can be investigated in next section.

2 Mathematical Analysis

In this paper, we assume that \(A \) admits an orthonormal eigenbasis \(\{ \phi_{n} \}_{n=1}^{\infty}\) in \(H\), associated with the eigenvalues \(\{\lambda_{n}\}_{n=1}^{\infty}\) such that

$$\begin{aligned} 0< \lambda_{1}\leq\lambda_{2}\leq\lambda_{3} \leq\cdots,\quad \text{and}\quad \lim_{n\rightarrow\infty} \lambda_{n} = \infty. \end{aligned}$$

The source function \(f:[0,T] \times H \to H \) satisfies the global Lipschitz condition

$$\begin{aligned} \bigl\| f(t,v_{1})-f(t,v_{2})\bigr\| \le K \|v_{1}-v_{2} \|, \end{aligned}$$
(3)

for some constant \(K>0\) independent of \(t\), \(v_{1}\), \(v_{2}\). As shown in [22], a solution to problem (1) satisfies the following integral equation

$$\begin{aligned} u(t)= \sum_{n=1}^{\infty} \biggl[ e^{(T-t) \lambda_{n} }\varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n}, \end{aligned}$$
(4)

where \(\varphi_{n}= \langle\varphi, \phi_{n} \rangle\), \(f_{n} (u) (s) = \langle f(s,u(s)), \phi_{n} \rangle\). In a few sentences, we present a brief introduction to the Fourier truncated method. From (4), it can be observed that \(e^{(T-t) \lambda_{n} }\) and \(e^{(s-t) \lambda_{n} }\) tend to infinity as \(n\) tends to infinity, so in order to guarantee the convergence of the solution \(u\) given by (4), the coefficient \(\varphi_{n}\) must decay rapidly. But such a decay usually cannot occur for the measured data \(\varphi^{\epsilon}\). Hence, a natural way is to eliminate the high frequencies and consider the solution \(u\) for \(\lambda_{n} \le M_{\epsilon}\), where \(M_{\epsilon}\) is a positive integer; this is the so-called Fourier truncated method, and \(M_{\epsilon}\) plays the role of a regularization parameter. In [17, 20, 22], the authors applied this method to establish a regularized solution of

$$\begin{aligned} w^{\epsilon}(t)= \sum_{\lambda_{n} \le M_{\epsilon}} \biggl[ e^{(T-t) \lambda_{n} } \varphi_{n}^{\epsilon}- \int_{t}^{T} e^{(s-t) \lambda _{n} } f_{n} \bigl(w^{\epsilon}\bigr) (s) ds \biggr] \phi_{n} , \end{aligned}$$
(5)

where \(\varphi_{n}^{\epsilon}= \langle\varphi^{\epsilon}, \phi_{n} \rangle\), \(f_{n} (w^{\epsilon}) (s) = \langle f(s,w ^{\epsilon}(s)), \phi_{n} \rangle\). To get a stability estimate, they required the stronger condition on the exact solution \(u\)

$$\begin{aligned} \sup_{t \in[0,T]} \sum_{n=1}^{\infty}e^{2t \lambda_{n}} \bigl\langle u(t), \phi_{n} \bigr\rangle ^{2} < \infty. \end{aligned}$$
(6)

This condition appears in some other papers, for example [6, 17, 19, 20]. One can further remark that there are not too many functions \(u\) which satisfy condition (6) and moreover, in practice, such condition is difficult to be checked. Motivated by this reason, in our study, we first develop a new Fourier truncated solution \(u^{\epsilon}\) by modifying the right-hand side of (5) as follows

$$\begin{aligned} u^{\epsilon}(t)&= \sum_{\lambda_{n} \le M_{\epsilon}} \biggl[ e^{(T-t) \lambda_{n} } \varphi_{n}^{\epsilon}- \int_{t}^{T} e^{(s-t) \lambda _{n} } f_{n} \bigl(u^{\epsilon}\bigr) (s) ds \biggr] \phi_{n} \\ &\quad{}+ \sum _{\lambda_{n} > M_{\epsilon}} \biggl[ \int_{0}^{t} e^{(s-t) \lambda _{n} }f_{n} \bigl(u^{\epsilon}\bigr) (s) ds \biggr] \phi_{n}, \end{aligned}$$
(7)

where \(f_{n} (u^{\epsilon}) (s) = \langle f(s,u^{\epsilon}(s)), \phi_{n} \rangle\). To obtain the error estimate between the exact solution \(u\) and the regularized solution \(u^{\epsilon}\), we only require a weak assumption

$$\begin{aligned} \bigl\| u(0)\bigr\| _{H^{r}} \le E,\quad r\ge0, \end{aligned}$$
(8)

or another condition

$$ \bigl\| u(t)\bigr\| _{H} \le B, \quad\text{for all }0 \le t\le T, $$
(9)

where \(E>0\) is a positive real number. Here \(H^{r}\), \(r\ge0\) is a Hilbert scale space defined by

$$\begin{aligned} H^{r} = \Biggl\{ h \in H: \sum_{n=1}^{\infty} \lambda_{n}^{2r} \langle h, \phi_{n} \rangle^{2} < \infty \Biggr\} ,\quad \text{(see [16])}. \end{aligned}$$
(10)

It is obvious that our conditions (8) or (9) is much weaker than (6). To the best of our knowledge, there are not related references in which the Fourier truncated method was used to solve the semi-linear problem (1) under the assumption (8).

Our main results are stated in the two following theorem. In Theorem 1, we present a new Fourier method for solving the problem (1) in the case that

$$ 0< KT< 1, $$
(11)

In Theorem 2, we establish another regularized solution for solving (1) for any \(K>0\).

Theorem 1

The problem (7) has a unique solution \(u^{\epsilon}\in C([0,T];H)\). Assume that problem (1) has a weak solution \(u \in C([0,T];H)\) which satisfies (8). Choose \(M_{\epsilon}>0\) such that \(\lim_{\epsilon\to0} M_{\epsilon}^{-1}= \lim_{\epsilon\to0} e^{T M_{\epsilon}} \epsilon=0\). Then we have the following estimate

$$\begin{aligned} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| \le R(p) \sqrt{e^{2T M_{\epsilon}} \epsilon^{2} + M_{\epsilon}^{-2r} E^{2} } e^{-tM_{\epsilon}},\quad \textit{for all } t \in[0,T], \end{aligned}$$
(12)

where \(r \ge0\) and

$$\begin{aligned} R( p)= \sqrt{\frac{2 ( 1+\frac{1}{p} ) }{ 1- (1+p) K^{2} T ^{2} }}, \end{aligned}$$
(13)

for any \(0 < p < \frac{1}{K^{2}T^{2}}-1\).

Corollary 1

Let us choose \(M_{\epsilon}=\frac{1}{T+\alpha} \ln (\frac{1}{ \epsilon} )\) for \(\alpha>0\) then

$$\begin{aligned} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| \le R(p) \sqrt{ \epsilon^{\frac{2\alpha}{T+\alpha}}+ \biggl( \frac{1}{T+\alpha} \ln \biggl( \frac{1}{\epsilon} \biggr) \biggr) ^{-2r} E^{2} }\; \epsilon^{\frac{t}{T+\alpha}},\quad \textit{for all } t \in[0,T]. \end{aligned}$$
(14)

Remark 1

In Theorem 1, our method has a drawback that the condition for \(K \in [0, \frac{1}{T})\) is still too strong which may restrict the class of equations. We are trying to remove this constraint in the next theorem.

Theorem 2

Assume that problem (1) has a weak solution \(u \in C([0,T];H)\) which satisfies (9). Let us define a sequence \(\{ T_{i} \}\), \(i=0,1,\ldots2m\) such that

$$\begin{aligned} T_{0}=0 < T_{1}= \overline{h} T < T_{2}=2 \overline{h} T< \cdots < T_{2m}=2m \overline{h} T=T, \end{aligned}$$
(15)

where \(\overline{h}=\frac{1}{2m}\). Let \(m\) be a positive integer number such that \(m>KT\). We denote

$$\begin{aligned} \alpha_{k}(\epsilon) = \frac{m}{T 2^{2m-k}} \ln \biggl( \frac{1}{\epsilon} \biggr) , \quad\textit{for all } 2 \le k \le 2m, \quad \textit{and} \quad\alpha_{1}= \frac{m}{T2^{2m-1}}\ln \biggl( \frac{1}{\epsilon} \biggr). \end{aligned}$$

Define the regularized solution \(\mathbf{U^{\epsilon}}\) as follows

$$ {\mathbf{U}^{\epsilon}}(t) = \mathbf{V}_{T_{2m-i-2},T_{2m-i}}^{\alpha _{2m-i}} \bigl( \mathbf{U}^{\epsilon}(T_{2m-i}) \bigr) (t), \quad\textit{if } T_{2m-i-1} \le t \le T_{2m-i},\ i=\overline{0, 2m-2} $$
(16)

and

$$ {\mathbf{U}^{\epsilon}}(t) = \mathbf{V}_{T_{0},T_{1}}^{\alpha _{1}} \bigl( \mathbf{U}^{\epsilon}(T_{1}) \bigr) (t), \quad\textit{if } 0 \le t \le T_{1}. $$
(17)

Then, we have the error estimation

$$\begin{aligned} &\bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| \le(2m-k) \varPhi^{2m-k}(m,K,p_{0}) (1+B)\epsilon^{\frac {1}{2^{2m-k}}}, \\ & \quad t \in[T_{k}, T_{k+1}],\ k=\overline{1, 2m-1} \end{aligned}$$
(18)

and

$$\begin{aligned} \bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| \le m \varPhi^{2m}(m,K,p_{0}) (1+B)\epsilon^{\frac{mt}{2^{2m-1}}}, \quad t\in[0,T_{1}]. \end{aligned}$$
(19)

Here

$$\begin{aligned} \varPhi(m,K,p_{0})= \frac{m\sqrt{2 ( 1+\frac{1}{p_{0}} ) }}{m-{T K \sqrt{1+p_{0}}} } \end{aligned}$$

with \(p_{0}\) is any number satisfies \(0< p_{0} < \frac{m^{2}}{T^{2}K^{2}}-1\).

3 Proof of Main Theorem

3.1 Proof of Theorem 1

The proof of Theorem 1 consists of two steps. In Step 1, we prove the existence and the uniqueness of a solution to problem (7). In Step 2, the error between the exact solution and the regularized solution is obtained.

Step 1. The existence and the uniqueness of a solution of (7).

Let us define on \(C([0, T];H) \) the following Bielecki norm

$$\begin{aligned} \|g\|_{1}= \sup_{0 \le t \le T} e^{t M_{\epsilon}} \bigl\| g(t)\bigr\| ,\quad \text{for all } g \in C\bigl([0, T];H\bigr) . \end{aligned}$$
(20)

It is easy to show that \(\|.\|_{1}\) is a norm of \(C([0, T];H) \). For \(w \in C([0, T];H)\), we consider the following function

$$\begin{aligned} J(w) (t) &=\sum_{\lambda_{n} \le M_{\epsilon}} \biggl[ e^{(T-t) \lambda _{n} } \varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(w) (s) ds \biggr] \phi_{n} \\ &\quad{} + \sum_{\lambda_{n} > M_{\epsilon}} \biggl[ \int_{0} ^{t} e^{(s-t) \lambda_{n} } f_{n}(w) (s) ds \biggr] \phi_{n}, \end{aligned}$$
(21)

where \(f_{n} (w) (s) = \langle f(s,w(s)), \phi_{n} \rangle \). We shall prove that, for every \(w_{1},w_{2} \in C([0, T];H)\),

$$\begin{aligned} \bigl\| J(w_{1})-J(w_{2})\bigr\| _{1} \le KT \|w_{1}-w_{2}\|_{1}. \end{aligned}$$
(22)

First, by using the Hölder inequality and (3), we have the following estimates for all \(t\in[0,T]\)

$$\begin{aligned} &\sum_{\lambda_{n} \le M_{\epsilon}} \biggl( \int_{t}^{T}e^{(s-t) \lambda_{n} } \bigl[f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr] ds \biggr)^{2} \\ &\quad \le(T-t) \sum_{\lambda_{n} \le M_{\epsilon}} \int_{t}^{T} \bigl| e ^{(s-t) \lambda_{n} } \bigl[ f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr] \bigr|^{2}ds \\ &\quad\le(T-t) \sum_{\lambda_{n} \le M_{\epsilon}} \int_{t}^{T} e^{2(s-t)M _{\epsilon}} \bigl|f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr|^{2}ds \\ &\quad\le K^{2} (T-t) \int_{t}^{T} e^{2(s-t)M_{\epsilon}} \bigl\| w_{1}(s)-w _{2}(s)\bigr\| ^{2}ds \\ &\quad \le e^{-2tM_{\epsilon}} K^{2} (T-t)^{2} \sup _{0\le s \le T} e^{2sM _{\epsilon}} \bigl\| w_{1}(s)-w_{2}(s) \bigr\| ^{2} \\ &\quad=e^{-2tM_{\epsilon}} K^{2} (T-t)^{2} \|w_{1}-w_{2} \|_{1}^{2}, \end{aligned}$$
(23)

and

$$\begin{aligned} &\sum_{\lambda_{n} > M_{\epsilon}} \biggl( \int_{0}^{t} e^{(s-t) \lambda_{n} } \bigl[ f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr] ds \biggr)^{2} \\ &\quad\le t \sum_{\lambda_{n} > M_{\epsilon}} \int_{0}^{t} \bigl| e^{(s-t) \lambda_{n} } \bigl[f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr] \bigr|^{2}ds \\ &\quad\le t \sum_{\lambda_{n} > M_{\epsilon}} \int_{0}^{t} e^{2(s-t)M _{\epsilon}} \bigl|f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr|^{2}ds \\ &\quad\le K^{2} t \int_{0}^{t} e^{2(s-t)M_{\epsilon}} \bigl\| w_{1}(s)-w_{2}(s) \bigr\| ^{2}ds \\ &\quad \le e^{-2tM_{\epsilon}} K^{2} t^{2} \sup _{0\le s \le T} e^{2sM _{\epsilon}} \bigl\| w_{1}(s)-w_{2}(s) \bigr\| ^{2} \\ &\quad=e^{-2tM_{\epsilon}} K^{2} t^{2} \|w_{1}-w_{2} \|_{1}^{2}. \end{aligned}$$
(24)

From the definition of \(J\) in (21), we have

$$\begin{aligned} J(w_{1}) (t)- J(w_{2}) (t) & = \sum _{\lambda_{n} \le M_{\epsilon}} \biggl( - \int_{t}^{T}e^{(s-t) \lambda_{n} } \bigl[f_{n}(w_{1}) (s)- f _{n}(w_{2}) (s) \bigr] ds \biggr) \phi_{n} \\ &\quad{}+ \sum_{\lambda_{n} > M_{\epsilon}} \biggl( \int_{0}^{t} e^{(s-t) \lambda_{n} } \bigl[ f_{n}(w_{1}) (s)- f_{n}(w_{2}) (s) \bigr] ds \biggr) \phi_{n}. \end{aligned}$$
(25)

Combining (23), (24), (25) and using the inequality \((a + b)^{2} \le(1 + m)a^{2} + ( 1 + \frac{1}{m} ) b^{2}\) for any real numbers \(a\), \(b\) and \(m>0\), we get the following estimate for all \(t \in(0,T)\)

$$\begin{aligned} \textstyle\begin{array}[b]{rl} \bigl\| J(w_{1}) (t)- J(w_{2}) (t) \bigr\| ^{2} &\le e^{-2tM_{\epsilon}} K^{2} (1+m)t ^{2} \|w_{1}-w_{2} \|_{1}^{2} \\ &\quad{} + e^{-2tM_{\epsilon}} K^{2} \biggl( 1+\frac{1}{m} \biggr) (T-t)^{2} \|w _{1}-w_{2}\|_{1}^{2}. \end{array}\displaystyle \end{aligned}$$
(26)

By choosing \(m=\frac{T-t}{t}\), we obtain

$$\begin{aligned} e^{2tM_{\epsilon}} \bigl\| J(w_{1}) (t)- J(w_{2}) (t) \bigr\| ^{2} \le K^{2} T^{2} \|w_{1}-w_{2} \|^{2}_{1}, \quad \text{for all } t \in(0,T). \end{aligned}$$
(27)

On other hand, letting \(t=T\) in (24), we deduce

$$\begin{aligned} e^{2TM_{\epsilon}} \bigl\| J(w_{1}) (T)- J(w_{2}) (T) \bigr\| ^{2} \le K^{2} T^{2} \|w_{1}-w_{2} \|^{2}_{1}. \end{aligned}$$
(28)

By letting \(t=0\) in (23), we have

$$\begin{aligned} \bigl\| J(w_{1}) (0)- J(w_{2}) (0) \bigr\| ^{2} \le K^{2} T^{2} \|w_{1}-w_{2} \|^{2} _{1}. \end{aligned}$$
(29)

Combining (27), (28) and (29), we obtain

$$\begin{aligned} e^{tM_{\epsilon}} \bigl\| J(w_{1}) (t)- J(w_{2}) (t) \bigr\| \le KT \|w_{1}-w_{2} \|_{1}, \quad 0\le t \le T, \end{aligned}$$

which leads to (22). Since \(KT<1\), we can conclude that \(J\) is a contraction; by the Banach fixed point theorem, it follows that the equation \(J(w) = w\) has a unique solution \(w \in C([0, T];H)\).

Step 2. The error estimates between the exact solution \(u\) and the regularized solution \(u^{\epsilon}\).

From (4), a solution to problem (1) can be rewritten as

$$\begin{aligned} u(t) &= \sum_{n=1}^{\infty} \biggl[ e^{(T-t) \lambda_{n} } \varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} \\ &= \sum_{\lambda_{n} \le M_{\epsilon}} \biggl[e^{(T-t) \lambda_{n} } \varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} \\ &\quad{} + \sum_{\lambda_{n} > M_{\epsilon}} \biggl[e^{(T-t) \lambda_{n} } \varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} \\ &= \sum_{\lambda_{n} \le M_{\epsilon}} \biggl[e^{(T-t) \lambda_{n} } \varphi_{n} - \int_{t}^{T} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} \\ &\quad{} + \sum_{\lambda_{n} > M_{\epsilon}} \biggl[ \int_{0}^{t} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} + \sum_{\lambda_{n} >M_{\epsilon}} e^{-t \lambda_{n} } u_{n}(0) \phi _{n}, \end{aligned}$$
(30)

where we have used the equality \(\varphi_{n}= e^{-T \lambda_{n}} u _{n}(0)+ \int_{0}^{T} e^{(s-T)\lambda_{n}} f_{n}(u)(s)ds\). Subtracting (7) from (30) and taking the inner product in \(H\) yield the following estimate

$$\begin{aligned} \bigl| \bigl\langle u^{\epsilon}(t)-u(t), \phi_{n} \bigr\rangle \bigr| &\le e^{(T-t) \lambda_{n} } \bigl| \varphi^{\epsilon}_{n}- \varphi_{n} \bigr| \qquad ( \lambda_{n} \le M_{\epsilon} )\\ &\quad{}+ e^{-t \lambda_{n} } \lambda_{n}^{-r} \lambda_{n}^{r} \bigl| u _{n}(0) \bigr| \qquad ( \lambda_{n} > M_{\epsilon} )\\ &\quad{}+ \int_{t}^{T} e^{(s-t) \lambda_{n} } \bigl|f_{n}\bigl(u^{\epsilon}\bigr)(s)- f _{n}(u)(s) \bigr| ds \qquad ( \lambda_{n} \le M_{\epsilon} )\\ &\quad{}+ \int_{0}^{t} e^{(s-t) \lambda_{n} } \bigl|f_{n}\bigl(u^{\epsilon}\bigr)(s)- f _{n}(u)(s) \bigr| ds \qquad ( \lambda_{n} > M_{\epsilon} )\\ &\le e^{(T-t)M_{\epsilon}} \bigl| \varphi^{\epsilon}_{n}-\varphi _{n} \bigr|+e^{-t M_{\epsilon}} M_{\epsilon}^{-r} \lambda_{n}^{r} \bigl| u_{n}(0) \bigr|\\ &\quad{} + \int_{0}^{T} e^{(s-t) M_{\epsilon}} \bigl|f_{n}\bigl(u^{\epsilon}\bigr)(s)- f _{n}(u)(s) \bigr| ds. \end{aligned}$$

Hence, using Parseval’s identity and the inequality \((c_{1}+c_{2}+c _{3})^{2} \le2 ( 1+\frac{1}{p} ) c_{1}^{2}+ 2 ( 1+ \frac{1}{p} ) c_{2}^{2}+(1+p)c_{3}^{2} \) for any real numbers \(c_{1}\), \(c_{2}\), \(c_{3}\) and \(p>0\) yields

$$\begin{aligned} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| ^{2} &= \sum_{n=1}^{\infty} \bigl| \bigl\langle u^{\epsilon}(t)-u(t), \phi_{n} \bigr\rangle \bigr|^{2} \\ &\le2 \biggl( 1+\frac{1}{p} \biggr) e^{2(T-t)M_{\epsilon}}\sum _{n=1} ^{\infty} \bigl| \varphi^{\epsilon}_{n}- \varphi_{n} \bigr|^{2} \\ &\quad{}+ 2 \biggl( 1+ \frac{1}{p} \biggr) e^{-2tM_{\epsilon}} M_{\epsilon}^{-2r} \sum _{n=1} ^{\infty}\lambda_{n}^{2r} \bigl|u_{n}(0) \bigr|^{2} \\ &\quad{}+ (1+p) \sum_{n=1}^{\infty} \biggl[ \int_{0}^{T} e^{(s-t) M_{\epsilon }} \bigl|f_{n} \bigl(u^{\epsilon}\bigr) (s)- f_{n}(u) (s) \bigr| ds \biggr]^{2} \\ &\le2 \biggl( 1+\frac{1}{p} \biggr) e^{2(T-t)M_{\epsilon}} \bigl\| \varphi^{\epsilon}-\varphi\bigr\| ^{2}+ 2 \biggl( 1+\frac{1}{p} \biggr) e^{-2tM _{\epsilon}} M_{\epsilon}^{-2r}\bigl\| u(0) \bigr\| _{H^{r}}^{2} \\ &\quad{}+ (1+p) T e^{-2tM_{\epsilon}} \int_{0}^{T} e^{2sM_{\epsilon}} \bigl\| f\bigl(s,u ^{\epsilon}(s)\bigr) -f\bigl(s,u(s)\bigr)\bigr\| ^{2}ds. \end{aligned}$$
(31)

Multiplying by \(e^{2tM_{\epsilon}} \) both sides and using (3), we obtain

$$\begin{aligned} e^{2tM_{\epsilon}} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| ^{2} &\le2 \biggl( 1+ \frac{1}{p} \biggr) e^{2T M_{\epsilon}} \epsilon^{2} + 2 \biggl( 1+ \frac{1}{p} \biggr) M_{\epsilon}^{-2r} \bigl\| u(0) \bigr\| _{H^{r}}^{2} \\ &\quad{}+ (1+p) K^{2} T \int_{0}^{T} e^{2sM_{\epsilon}} \bigl\| u^{\epsilon}(s)-u(s) \bigr\| ^{2} ds. \end{aligned}$$
(32)

Set \(Q(t)= e^{2tM_{\epsilon}} \|u^{\epsilon}(t)-u(t)\|^{2} \) for all \(t \in[0,T]\). Since \(u^{\epsilon}, u \in C([0, T];H) \), the function \(Q\) is continuous on \([0,T]\) and attains over there its maximum \(P\) at some point \(t_{0} \in[0,T]\). Therefore

$$\begin{aligned} \int_{0}^{T} e^{2sM_{\epsilon}} \bigl\| u^{\epsilon}(s)-u(s) \bigr\| ^{2} ds \le \int_{0}^{T} \sup_{0 \le s \le T }e^{2sM_{\epsilon}} \bigl\| u^{\epsilon }(s)-u(s)\bigr\| ^{2} ds= P \int_{0}^{T} ds=TP. \end{aligned}$$
(33)

Combining the latter estimate with (32), we deduce

$$\begin{aligned} e^{2tM_{\epsilon}} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| ^{2} \le2 \biggl( 1+ \frac{1}{p} \biggr) e^{2T M_{\epsilon}} \epsilon^{2} + 2 \biggl( 1+ \frac{1}{p} \biggr) M_{\epsilon}^{-2r} \bigl\| u(0) \bigr\| _{H^{r}}^{2}+ (1+p) K ^{2} T^{2} P . \end{aligned}$$
(34)

Choosing \(t=t_{0}\) on the left-hand side of the latter inequality, we get

$$\begin{aligned} P \le2 \biggl( 1+\frac{1}{p} \biggr) e^{2T M_{\epsilon}} \epsilon^{2} + 2 \biggl( 1+\frac{1}{p} \biggr) M_{\epsilon}^{-2r} \bigl\| u(0)\bigr\| _{H^{r}}^{2} + (1+p) K^{2} T^{2} P. \end{aligned}$$
(35)

Or

$$\begin{aligned} \bigl[ 1- (1+p) K^{2} T^{2} \bigr] P \le2 \biggl( 1+ \frac{1}{p} \biggr) \bigl[ e^{2T M_{\epsilon}} \epsilon^{2} + M_{\epsilon}^{-2r} \bigl\| u(0) \bigr\| _{H^{r}}^{2} \bigr]. \end{aligned}$$
(36)

Since \(0 < p < \frac{1}{K^{2}T^{2}}-1\), it follows that the left hand-side bracket is positive. This implies that for all \(t \in[0,T]\)

$$\begin{aligned} e^{2tM_{\epsilon}} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| ^{2} \le \frac{2 ( 1+ \frac{1}{p} ) [ e^{2T M_{\epsilon}} \epsilon^{2} + M_{ \epsilon}^{-2r} \|u(0)\|_{H^{r}}^{2} ]}{ 1- (1+p) K^{2} T^{2} }. \end{aligned}$$
(37)

Hence

$$\begin{aligned} \bigl\| u^{\epsilon}(t)-u(t)\bigr\| ^{2} \le\frac{2 ( 1+\frac{1}{p} ) [ e^{2T M_{\epsilon}} \epsilon^{2} + M_{\epsilon}^{-2r} \|u(0) \|_{H^{r}}^{2} ]}{ 1- (1+p) K^{2} T^{2} }e^{-2tM_{\epsilon}}. \end{aligned}$$
(38)

This completes the proof of the theorem.

3.2 Proof of Theorem 2

Now, we consider the following lemma

Lemma 1

Let \(0 \le T_{h} \le T_{p} \le T\). For \(\theta\in C([T_{h}, T_{p}];H)\), we consider the following function

$$\begin{aligned} {\mathbf{V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (t)&= \sum _{\lambda_{n} \le\alpha(\epsilon) } \biggl[ e^{(T_{p}-t) \lambda _{n} } < \theta, \phi_{n} > - \int_{t}^{T_{p}} e^{(s-t) \lambda_{n} } f_{n} \bigl( \mathbf{V}_{T_{h},T_{p}}^{\alpha(\epsilon )}(\theta) \bigr) (s) ds \biggr] \phi_{n} \\ &\quad{}+ \sum_{\lambda_{n} > \alpha(\epsilon) } \biggl[ \int_{T_{h}}^{t} e^{(s-t) \lambda_{n} } f_{n} \bigl( \mathbf{V}_{T_{h},T_{p}}^{\alpha (\epsilon)}(\theta) \bigr) (s) ds \biggr] \phi_{n}. \end{aligned}$$
(39)

Assume that \(K (T_{p}-T_{h}) <1\). Then Problem (39) has unique solution \(\mathbf{V}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta ) \in C([T_{h}, T_{p}]; H)\).

Proof

The existence and uniqueness of solution of problem (39) is proved by a similar way as in theorem. Hence, we omit it here. Next, we prove the estimate between \(\mathbf{V}_{T_{h},T_{p}}^{\alpha(\epsilon )}(\theta)(t)\) and \(u(t)\). It is easy to see that

$$\begin{aligned} u(t) &= \sum_{\lambda_{n} \le\alpha(\epsilon) } \biggl[e^{(T_{p}-t) \lambda_{n} } < u(T_{p}), \phi_{n} > - \int_{t}^{T_{p}} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} \\ &\quad{}+ \sum_{\lambda_{n} > \alpha(\epsilon )} \biggl[ \int_{T_{h}}^{t} e^{(s-t) \lambda_{n} } f_{n}(u) (s) ds \biggr] \phi_{n} + \sum_{\lambda_{n} >M_{\epsilon}} e^{-(t-T_{h}) \lambda_{n} } u_{n}(T_{h}) \phi_{n}, \end{aligned}$$
(40)

for all \(t \in[T_{h}, T_{p}]\). By a similar way, we get for all \(t \in[T_{h}, T_{p}]\)

$$\begin{aligned} & \bigl\| {\mathbf{V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta ) (t)-u(t) \bigr\| ^{2} \\ &\quad= \sum_{n=1}^{\infty} \bigl| \bigl\langle {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (t)-u(t), \phi_{n} \bigr\rangle \bigr|^{2} \\ &\quad\le2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{2(T_{p}-t)\alpha(\epsilon )}\sum _{n=1}^{\infty} \bigl| \bigl\langle \theta-u(T_{p}), \phi_{n} \bigr\rangle \bigr|^{2} \\ &\qquad{}+ 2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{-2(t-T_{h})\alpha(\epsilon)} \sum _{n=1}^{\infty} \bigl|u_{n}(T_{h}) \bigr|^{2} \\ &\qquad{}+ (1+p_{0}) \sum_{n=1}^{\infty} \biggl[ \int_{T_{h}}^{T_{p}} e^{(s-t) \alpha(\epsilon) } \bigl| f_{n} \bigl( \mathbf {V}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) \bigr) (s) - f_{n}(u) (s) \bigr| ds \biggr]^{2} \\ &\quad\le2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{2(T_{p}-t) \alpha(\epsilon)} \bigl\| \theta- u(T_{p})\bigr\| ^{2}+ 2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{-2(t-T_{h}) \alpha(\epsilon)} \bigl\| u(T_{h})\bigr\| ^{2} \\ &\qquad{}+ (1+p_{0}) (T_{p}-T_{h}) e^{-2(t-T_{h})\alpha(\epsilon)} \\ &\qquad{}\times\int _{T_{h}}^{T_{p}} e^{2(s-T_{h})\alpha(\epsilon)} \bigl\| f \bigl( s, \mathbf{V}_{T_{h},T_{p}}^{\alpha(\epsilon)}\bigl(\theta(s)\bigr) \bigr) -f \bigl(s,u(s)\bigr) \bigr\| ^{2}ds. \end{aligned}$$
(41)

Multiplying by \(e^{2(t-T_{h})\alpha(\epsilon)} \) both sides and using (3), we obtain

$$\begin{aligned} &e^{2(t-T_{h})\alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (t)-u(t) \bigr\| ^{2} \\ &\quad\le2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{2(T_{p}-T_{h}) \alpha (\epsilon)} \bigl\| \theta- u(T_{p})\bigr\| ^{2}+ 2 \biggl( 1+ \frac{1}{p_{0}} \biggr) \bigl\| u(T_{h})\bigr\| ^{2} \\ &\qquad{}+ K^{2} (1+p_{0}) (T_{p}-T_{h}) \int_{T_{h}}^{T_{p}} e^{2(s-T_{h})\alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (s)-u(s) \bigr\| ^{2} ds. \end{aligned}$$
(42)

Set \(\overline{Q}(t)= e^{2(t-T_{h})\alpha(\epsilon)} \| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta)(t)-u(t) \|^{2} \) for all \(t \in[T_{h},T_{p}]\). Since \(\mathbf{V}_{T_{h},T_{p}}^{\alpha (\epsilon)}(\theta)(t), u \in C([T_{h}, T_{p}];H) \), the function \(\overline{Q}\) is continuous on \([T_{h},T_{p}]\) and attains over there its maximum \(\overline{P}\) at some point \(t_{1} \in[T_{h},T_{p}]\). Therefore

$$\begin{aligned} &\int_{T_{h}}^{T_{p}} e^{2(s-T_{h}) \alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (s)-u(s) \bigr\| ^{2}ds \\ &\quad\le \int_{T_{h}}^{T_{p}} \sup_{T_{h} \le s \le T_{p} } e^{2(s-T_{h})\alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (s)-u(s) \bigr\| ^{2} ds = \overline{P} \int_{T_{h}}^{T_{p}} ds=\overline{P}(T_{p}-T_{h}). \end{aligned}$$
(43)

From above observations, we obtain

$$\begin{aligned} &e^{2(t-T_{h})\alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (t)-u(t) \bigr\| ^{2} \\ &\quad\le2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{2(T_{p}-T_{h}) \alpha (\epsilon)} \bigl\| \theta- u(T_{p})\bigr\| ^{2}+ 2 \biggl( 1+ \frac{1}{p_{0}} \biggr) \bigl\| u(T_{h})\bigr\| ^{2} \\ &\qquad{}+ K^{2}(1+p_{0}) \overline{P} (T_{p}-T_{h})^{2}. \end{aligned}$$
(44)

Choosing \(t=t_{1}\) on the left-hand side of the latter inequality, we get

$$\begin{aligned} \bigl[ 1- K^{2} (1+p_{0}) (T_{p}-T_{h})^{2} \bigr]\overline{P} &\le 2 \biggl( 1+\frac{1}{p_{0}} \biggr) e^{2(T_{p}-T_{h}) \alpha(\epsilon)} \bigl\| \theta- u(T_{p})\bigr\| ^{2} \\ &\quad{}+ 2 \biggl( 1+\frac{1}{p_{0}} \biggr) \bigl\| u(T_{h}) \bigr\| ^{2}. \end{aligned}$$
(45)

Since \(0 < p_{0} < \frac{1}{K^{2} (T_{p}-T_{h})^{2}}-1\), it follows that the left hand-side bracket is positive. This implies that for all \(t \in[T_{h},T_{p}]\)

$$\begin{aligned} e^{2(t-T_{h})\alpha(\epsilon)} \bigl\| {\mathbf {V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta) (t)-u(t) \bigr\| ^{2} \le \frac{2 ( 1+\frac{1}{p_{0}} ) ( e^{2(T_{p}-T_{h}) \alpha (\epsilon)} \| \theta- u(T_{p})\|^{2}+ \|u(T_{h})\|^{2} )}{ 1- K^{2}(1+p_{0}) (T_{p}-T_{h})^{2} }. \end{aligned}$$
(46)

Hence

$$\begin{aligned} \bigl\| {\mathbf{V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta ) (t)-u(t) \bigr\| ^{2} \le\frac{2 ( 1+\frac{1}{p_{0}} ) ( e^{2(T_{p}-T_{h}) \alpha(\epsilon)} \| \theta- u(T_{p})\|^{2}+ \| u(T_{h})\|^{2} )}{ 1- K^{2}(1+p_{0}) (T_{p}-T_{h})^{2} } e^{2(T_{h}-t) \alpha(\epsilon)} . \end{aligned}$$
(47)

This implies that

$$\begin{aligned} &\bigl\| {\mathbf{V}}_{T_{h},T_{p}}^{\alpha(\epsilon)}(\theta ) (t)-u(t) \bigr\| \\ &\quad\le \frac{\sqrt{2 ( 1+\frac{1}{p_{0}} ) } }{ 1- K \sqrt{1+p_{0}} (T_{p}-T_{h}) } \bigl( e^{(T_{p}-T_{h}) \alpha(\epsilon)} \bigl\| \theta- u(T_{p})\bigr\| + \bigl\| u(T_{h})\bigr\| \bigr) e^{(T_{h}-t)\alpha(\epsilon)}, \end{aligned}$$
(48)

for all \(t \in[T_{h}, T_{p}]\). □

Now, we return the proof of Theorem 2.

  • If \(t \in[T_{2m-1}, T]\) then since \(\alpha_{2m}(\epsilon)=\frac {m}{T} \ln(\frac{1}{\epsilon}) \), we get

    $$\begin{aligned} &\bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| \\ &\quad= \bigl\| {\mathbf {V}}_{T_{2m-2},T_{2m}}^{\alpha_{2m}(\epsilon)} \bigl(\varphi^{\epsilon }\bigr) (t)-u(t) \bigr\| \\ &\quad\le\frac{m\sqrt{2 ( 1+\frac{1}{p_{0}} ) } }{m-{T K \sqrt {1+p_{0}}} } \bigl[ e^{(T_{2m}-T_{2m-2}) \alpha_{2m}(\epsilon)} \bigl\| \varphi^{\epsilon}-\varphi \bigr\| + \bigl\| u(T_{2m-2})\bigr\| \bigr] e^{(T_{2m-2}-t) \alpha_{2m}(\epsilon)} \\ &\quad\le\frac{m\sqrt{2 ( 1+\frac{1}{p_{0}} ) } }{m-{T K \sqrt {1+p_{0}}} } ( 1+B )\epsilon^{\frac{1}{2}}= \varPhi(m,K,p_{0}) ( 1+B )\epsilon^{\frac{1}{2}} , \end{aligned}$$
    (49)

    which we note that \(e^{(T_{2m-2}-t) \alpha_{2m}(\epsilon)} \le e^{(T_{2m-2}-T_{2m-1}) \alpha_{2m}(\epsilon)}=\epsilon^{\frac {1}{2}}\) and

    $$\begin{aligned} \varPhi(m,K,p_{0})=\max \biggl\{ 1, \frac{m\sqrt{2 ( 1+\frac{1}{p_{0}} ) } }{m-{T K \sqrt{1+p_{0}}}} \biggr\} ,\quad\mbox{then } \varPhi(m,K,p_{0}) \geq1. \end{aligned}$$

  • If \(t \in[T_{2m-2}, T_{2m-1}]\) then since \(\alpha _{2m-1}(\epsilon)=\frac{m}{2T} \ln(\frac{1}{\epsilon}) \), we get

    $$\begin{aligned} \bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| &= \bigl\| {\mathbf {V}}_{T_{2m-3},T_{2m-1}}^{\alpha_{2m-1}(\epsilon)} \bigl( \mathbf {U}^{\epsilon}(T_{2m-1}) \bigr) (t)-u(t) \bigr\| \\ &\le\varPhi(m,K,p_{0}) \bigl[ e^{(T_{2m-1}-T_{2m-3}) \alpha _{2m-1}(\epsilon)} \bigl\| { \mathbf{U}}^{\epsilon }(T_{2m-1})-u(T_{2m-1}) \bigr\| \\ &\quad{}+ \bigl\| u(T_{2m-3})\bigr\| \bigr] e^{(T_{2m-3}-t) \alpha_{2m-1}(\epsilon)} \\ &\le\varPhi(m,K,p_{0}) \bigl[ \epsilon^{-1/2} \varPhi(m,K,p_{0}) ( 1+B )\epsilon^{\frac{1}{2}} +B \bigr] \epsilon^{\frac{1}{4}} \\ &\le\varPhi(m,K,p_{0}) \bigl[ \varPhi(m,K,p_{0}) ( 1+B )+B \bigr]\epsilon^{\frac{1}{4}} \\ &\le\varPhi(m,K,p_{0}) \bigl[ \bigl(1+\varPhi(m,K,p_{0}) \bigr) ( 1+B ) \bigr]\epsilon^{\frac{1}{4}} \\ &\le2 \varPhi^{2}(m,K,p_{0}) (1+B)\epsilon^{\frac{1}{4}} , \end{aligned}$$
    (50)

    which we note that \(e^{(T_{2m-3}-t) \alpha_{2m-1}(\epsilon)} \le e^{(T_{2m-3}-T_{2m-2}) \alpha_{2m-1}(\epsilon) } =\epsilon^{\frac{1}{4}}\).

  • If \(t \in[T_{2m-3}, T_{2m-2}]\) then since \(\alpha_{2m-2} (\epsilon)=\frac{m}{4T} \ln(\frac{1}{\epsilon}) \), we get

    $$\begin{aligned} \bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| &= \bigl\| {\mathbf {V}}_{T_{2m-4},T_{2m-2}}^{\alpha_{2m-2}(\epsilon)} \bigl( \mathbf {U}^{\epsilon}(T_{2m-2}) \bigr) (t)-u(t) \bigr\| \\ &\le\varPhi(m,K,p_{0}) \bigl[ e^{(T_{2m-2}-T_{2m-4}) \alpha_{2m-2}} \bigl\| { \mathbf{U}}^{\epsilon}(T_{2m-2})-u(T_{2m-2}) \bigr\| \\ &\quad{}+ \bigl\| u(T_{2m-4})\bigr\| \bigr] e^{(T_{2m-4}-t) \alpha_{2m-2}(\epsilon)} \\ &\le\varPhi(m,K,p_{0}) \bigl[ \epsilon^{-1/4}2 \varPhi^{2}(m,K,p_{0}) ( 1+B )\epsilon^{\frac{1}{4}} +B \bigr] \epsilon^{\frac{1}{8}} \\ &\le\varPhi(m,K,p_{0}) \bigl[ 2 \varPhi^{2}(m,K,p_{0}) ( 1+B )+B \bigr]\epsilon^{\frac{1}{8}} \\ & \le\varPhi(m,K,p_{0}) \bigl[ \bigl( 2 \varPhi^{2}(m,K,p_{0}) + 1 \bigr) ( 1+B ) \bigr]\epsilon^{\frac{1}{8}} \\ &\le3 \varPhi^{3}(m,K,p_{0}) (1+B)\epsilon^{\frac{1}{8}}. \end{aligned}$$
    (51)

    By induction, we prove that

  • If \(t \in[T_{1}, T_{2}]\) then since \(\alpha_{2}(\epsilon)=\frac {m}{2^{2m-2}T} \ln(\frac{1}{\epsilon}) \), we get

    $$\begin{aligned} \bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| &= \bigl\| {\mathbf {V}}_{T_{0},T_{2}}^{\alpha_{2}} \bigl( \mathbf{U}^{\epsilon}(T_{2}) \bigr) (t)-u(t) \bigr\| \\ &\le\varPhi(m,K,p_{0}) \bigl[ e^{(T_{0}-T_{2}) \alpha_{2}} \bigl\| { \mathbf{U}}^{\epsilon}(T_{2})-u(T_{2}) \bigr\| + \bigl\| u(T_{0})\bigr\| \bigr] e^{(T_{0}-t) \alpha_{2}} \\ &\le(2m-1) \varPhi^{2m-1}(m,K,p_{0}) (1+B)\epsilon^{\frac{1}{2^{2m-1}}}. \end{aligned}$$
    (52)

  • If \(t \in[0, T_{1}]\) then since \(\alpha_{1}(\epsilon)=\frac {m}{2^{2m-1}T} \ln(\frac{1}{\epsilon}) \), we get

    $$\begin{aligned} &\bigl\| {\mathbf{U}}^{\epsilon}(t)-u(t) \bigr\| \\ &\quad= \bigl\| {\mathbf {V}}_{T_{0},T_{1}}^{\alpha_{2}} \bigl( \mathbf{U}^{\epsilon}(T_{1}) \bigr) (t)-u(t) \bigr\| \\ &\quad\le\varPhi(m,K,p_{0}) \bigl[ e^{(T_{0}-T_{1}) \alpha_{1}} \bigl\| { \mathbf{U}}^{\epsilon}(T_{1})-u(T_{1}) \bigr\| + \bigl\| u(T_{0})\bigr\| \bigr] e^{(T_{0}-t) \alpha_{1}} \\ &\quad\le\varPhi(m,K,p_{0}) \bigl[ \epsilon^{\frac{-1}{2^{2m-1}}}(2m-1) \varPhi^{2m-1}(m,K,p_{0}) (1+B)\epsilon^{\frac{1}{2^{2m-1}}} +B \bigr] \epsilon^{\frac{mt}{2^{2m-1}}} \\ &\quad\le m \varPhi^{2m}(m,K,p_{0}) (1+B)\epsilon^{\frac{mt}{2^{2m-1}}}. \end{aligned}$$
    (53)