Error bounds for the SKCWs method
In this subsection, we present error bounds for the SKCWs method. To do this, we define
$$\begin{aligned} &{{ \langle u,v \rangle }_{\omega }} = \int _{0}^{t_{f}} {u(t)v(t){ \omega }(t) \,{d}t},\quad \forall u,v \in L_{\omega }^{2} \bigl([0,{t_{f}} )\bigr), \\ &{ \Vert u \Vert _{\omega }} = { \biggl( \int _{0}^{t_{f}} {{u^{2}}(t){ \omega }(t) \,{d}t } \biggr)^{\frac{1}{2}}}, \quad\forall u \in L_{\omega }^{2} \bigl([0,{t_{f}})\bigr), \end{aligned}$$
which are inner product and norm on the space \(L_{\omega }^{2} ([0,{t_{f}})) \), respectively.
Now we recall the following theorems from our previous work [47].
Theorem 5.1
Let \(f(t)\in L_{\omega }^{2} ([0,{t_{f}}))\) with \(\vert {f''(t)} \vert \le \mathfrak{L}\). The Eq. (7) converges uniformly to \(f(t)\) and the coefficients in (6) explicitly satisfy
$$\begin{aligned} \vert {\hat{f}}_{i,j} \vert < 4\sqrt{2{t_{f}}\pi } \mathfrak{L}\frac{1}{i^{\frac{5}{2}}{(j + 1)}^{2}}, \quad i \ge 1,~ j \ge 0. \end{aligned}$$
Theorem 5.2
Let \(f(t)\in L_{\omega }^{2} ([0,{t_{f}})) \). Then we have
$$\begin{aligned} { \Vert {f - {f_{{2^{k}},M - 1}}} \Vert _{\omega}} < 4 \sqrt{2{t_{f}} \pi } {\mathfrak{L}} { \Biggl( {\sum _{i = 0}^{\infty }{\sum_{j = M}^{\infty }{ \frac{1}{{{{i}^{5}}{{(j + 1)}^{4}}}}} } } + { \sum_{i = {2^{k}} + 1}^{\infty }{ \sum_{j = 0}^{\infty }{ \frac{1}{{{{i}^{5}}{{(j + 1)}^{4}}}}} } } \Biggr)^{\frac{1}{2}}}. \end{aligned}$$
Theorem 5.3
Let \({}_{0}^{C}D_{t}^{\alpha _{l}} f(t) \in L_{\omega }^{2} ([0,{t_{f}})) \) and \(\vert {}_{0}^{C}D_{t}^{\alpha _{l} +2} f(t) \vert \leq \mathfrak{L}_{l}\) for \(l=1,2,\ldots, r \). Then we have
$$\begin{aligned} \bigl\Vert {{}_{0}^{C}D_{t}^{\alpha _{l}} f - {{ \bigl( {{}_{0}^{C}D_{t}^{ \alpha _{l}} f} \bigr)}_{{2^{k}},M - 1}}} \bigr\Vert _{\omega } < 4 \sqrt{2{t_{f}} \pi } {\mathfrak{L}_{l}} \Biggl( \sum_{i = 0}^{\infty }{ \sum_{j = M}^{\infty }{ \frac{1}{{{i^{5}}{{(j + 1)}^{4}}}}} } + \sum_{i = {2^{k}} + 1}^{\infty }{\sum _{j = 0}^{\infty }{ \frac{1}{{{i^{5}}{{(j + 1)}^{4}}}}} } \Biggr)^{\frac{1}{2}}. \end{aligned}$$
Error bounds for the SFOJPs method
Here, we discuss error bounds for the SFOJPs method. To do this, first, we define the following inner product and norm on the weighted space \(L_{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}^{2} ([0,{t_{f}}]) \):
$$\begin{aligned} &{{ \langle u,v \rangle }_{{w_{t_{f}}^{(\lambda,\theta, \vartheta )}}}} = \int _{0}^{t_{f}} {u(t)v(t){w_{t_{f}}^{(\lambda, \theta,\vartheta )}}(t) \,{d}t},\quad \forall u,v \in L_{w_{t_{f}}^{( \lambda,\theta,\vartheta )}}^{2} \bigl([0,{t_{f}}]\bigr), \\ &{ \Vert u \Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}} = { \biggl( \int _{0}^{t_{f}} {{u^{2}}(t){w_{t_{f}}^{(\lambda,\theta, \vartheta )}}(t) \,{d}t } \biggr)^{\frac{1}{2}}},\quad \forall u \in L_{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}^{2} \bigl([0,{t_{f}}]\bigr). \end{aligned}$$
Let
$$\begin{aligned} {{\Lambda }_{N}} = \operatorname{span} \bigl\{ {{ \mathcal{J}}_{i}^{(\lambda,\theta, \vartheta )}(t),~ 0\leq i\leq N } \bigr\} , \end{aligned}$$
(13)
be the fractional-polynomial space of finite dimension.
Theorem 5.4
Let \({}_{0}^{C}D_{t}^{j \lambda } f(t) \in C([0,{t_{f}}])\), for \(j=0,1,\ldots,N\). If \({f_{N}}(t)\) is the best approximation to \(f(t)\) from \({{\Lambda }_{N}}\), then
$$\begin{aligned} { \Vert {f - {f_{N}}} \Vert _{{w_{t_{f}}^{(\lambda,\theta, \vartheta )}}}} \le \frac{{\mathcal{L}}}{{\Gamma ((N + 1)\lambda + 1)}}\sqrt{ \frac{{t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}} , \end{aligned}$$
(14)
where \(\mathcal{L} \ge \vert {{}_{0}^{C}D_{t}^{(N + 1)\lambda }f(t)} \vert \), for \(t \in [0,{t_{f}}]\).
Proof
Since \({f_{N}}(t)\) is the best approximation to \(f(t)\) from \({{\Lambda }_{N}}\), defined in (13), we have
$$\begin{aligned} { \Vert {f - {f_{N}}} \Vert _{{w_{t_{f}}^{(\lambda,\theta, \vartheta )}}}} \le { \Vert {f - u} \Vert _{{w_{t_{f}}^{( \lambda,\theta,\vartheta )}}}}, \quad\forall u(t) \in {\Lambda _{N}}. \end{aligned}$$
Considering the generalized Taylors formula \(u(t) = \sum_{j = 0}^{N} { \frac{{{t^{j\lambda }}}}{{\Gamma (j\lambda + 1)}} ( {}_{0}^{C}D_{t}^{j \lambda }u ) ({0^{+} })}\) yields
$$\begin{aligned} \bigl\vert {f(t) - u(t)} \bigr\vert = \Biggl\vert {f(t) - \sum_{j = 0}^{N} {\frac{{{t^{j\lambda }}}}{{\Gamma (j\lambda + 1)}} \bigl( {}_{0}^{C}D_{t}^{j \lambda }u \bigr) \bigl({0^{+} }\bigr)} } \Biggr\vert \le \mathcal{L} \frac{{{t^{(N + 1)\lambda }}}}{{\Gamma ((N + 1)\lambda + 1)}}. \end{aligned}$$
(15)
Taking \({L_{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}^{2}}\)-norm in both sides of inequality (15) leads to
$$\begin{aligned} \Vert {f - u} \Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}^{2} & \le \frac{{\mathcal{L}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}} \int _{0}^{{t_{f}}} {{t^{2(N + 1)\lambda }} {w_{t_{f}}^{(\lambda, \theta,\vartheta )}}\,{d}t} \\ &=\frac{{\mathcal{L}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}} \lambda \int _{0}^{{t_{f}}} {{t^{\lambda (2N + 3 + \vartheta ) - 1}} {{ \bigl(t_{f}^{\lambda }- {t^{\lambda }}\bigr)}^{\theta }} \,{d}t}. \end{aligned}$$
(16)
Let \(z = t_{f}^{\lambda }- {t^{\lambda }} \). Then we obtain
$$\begin{aligned} \lambda \int _{0}^{{t_{f}}} {{t^{\lambda (2N + 3 + \vartheta ) - 1}} {{ \bigl(t_{f}^{\lambda }- {t^{\lambda }}\bigr)}^{\theta }} \,{d}t} = \int _{0}^{t_{f}^{\lambda }} {{{ \bigl( {t_{f}^{\lambda }- z} \bigr)}^{2N + 2 + \vartheta }} {z^{\theta }}\,dz}. \end{aligned}$$
(17)
Now setting \(s = \frac{z}{{t_{f}^{\lambda }}} \) gives
$$\begin{aligned} \int _{0}^{t_{f}^{\lambda }} {{{ \bigl( {t_{f}^{\lambda }- z} \bigr)}^{2N + 2 + \vartheta }} {z^{\theta }}\,dz} & = t_{f}^{(2N + 3 + \vartheta + \theta )\lambda } \int _{0}^{1} {{{ ( {1 - s} )}^{2N + 2 + \vartheta }} {s^{\theta }}\,ds} \\ & = t_{f}^{(2N + 3 + \vartheta + \theta )\lambda } \frac{{\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}. \end{aligned}$$
(18)
By substituting the above relation into (17) and from (16), we can write
$$\begin{aligned} \Vert {f - u} \Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}^{2} \le \frac{{\mathcal{L}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}}t_{f}^{(2N + 3 + \vartheta + \theta )\lambda } \frac{{\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}. \end{aligned}$$
(19)
By taking the square root of (19), we obtain inequality (14). □
Theorem 5.5
Let \(_{0}^{C}D_{t}^{j \lambda + \alpha _{l}} f(t) \in C([0,{t_{f}}])\). If \({ ( {{}_{0}^{C}D_{t}^{\alpha _{l}} f} )}_{N} (t)\) is the best approximation to \({{}_{0}^{C}D_{t}^{\alpha _{l}} f}(t)\) from \({{\Lambda }_{N}}\), then
$$\begin{aligned} & \bigl\Vert {{}_{0}^{C}D_{t}^{\alpha _{l}} f - {{ \bigl( {{}_{0}^{C}D_{t}^{ \alpha _{l}} f} \bigr)}_{N}}} \bigr\Vert _{{w_{t_{f}}^{(\lambda, \theta,\vartheta )}}} \\ &\quad \leq \frac{{{\mathcal{L}_{l} }}}{{\Gamma ((N + 1)\lambda + 1)}} \sqrt{ \frac{{t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}} , \end{aligned}$$
(20)
where \({\mathcal{L}_{l} } \ge \vert {{}_{0}^{C}D_{t}^{(N + 1)\lambda + { \alpha _{l}} }f(t)} \vert \), for \(l=1,2,\ldots,r\) and \(t \in [0,{t_{f}}]\).
Proof
Since \({{ ( {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f} )}_{N}}(t)\) is the best approximation to \({{{{}_{0}^{C}D_{t}^{{\alpha _{l}}}} }}f(t)\) from \({{\Lambda }_{N}}\), we have
$$\begin{aligned} { \bigl\Vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f - {{ \bigl( {{}_{0}^{C}D_{t}^{{ \alpha _{l}}}f} \bigr)}_{N}}} \bigr\Vert _{{w_{t_{f}}^{(\lambda, \theta,\vartheta )}}}} \le { \bigl\Vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f - _{0}^{C}D_{t}^{{\alpha _{l}}}u} \bigr\Vert _{{w_{t_{f}}^{(\lambda, \theta,\vartheta )}}}},\quad \forall u(t) \in {\Lambda _{N}}. \end{aligned}$$
Considering the generalized Taylor formula \({}_{0}^{C}D_{t}^{{\alpha _{l}}}u(t) = \sum_{j = 0}^{N} { \frac{{{t^{j\lambda }}}}{{\Gamma (j\lambda + 1)}} ( {}_{0}^{C}D_{t}^{j \lambda + {\alpha _{l}}}u ) ({0^{+} })}\) yields
$$\begin{aligned} \bigl\vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f(t) - _{0}^{C}D_{t}^{{\alpha _{l}}}u(t)} \bigr\vert & = \Biggl\vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f(t) - \sum_{j = 0}^{N} { \frac{{{t^{j\lambda }}}}{{\Gamma (j\lambda + 1)}} \bigl( {}_{0}^{C}D_{t}^{j \lambda + {\alpha _{l}}}u \bigr) \bigl({0^{+} }\bigr)} } \Biggr\vert \\ & \le {\mathcal{L}_{l}} \frac{{{t^{(N + 1)\lambda }}}}{{\Gamma ((N + 1)\lambda + 1)}}. \end{aligned}$$
(21)
Taking \({L_{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}^{2}}\)-norm in both sides of inequality (21) leads to
$$\begin{aligned} \bigl\Vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f - _{0}^{C}D_{t}^{{\alpha _{l}}}u} \bigr\Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}^{2} & \le \frac{{\mathcal{L}_{l}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}} \int _{0}^{{t_{f}}} {{t^{2(N + 1)\lambda }} {w_{t_{f}}^{(\lambda, \theta,\vartheta )}}\,{d}t} \\ &= \frac{{\mathcal{L}_{l}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}} \lambda \int _{0}^{{t_{f}}} {{t^{\lambda (2N + 3 + \vartheta ) - 1}} {{ \bigl(t_{f}^{\lambda }- {t^{\lambda }}\bigr)}^{\theta }} \,{d}t}. \end{aligned}$$
(22)
From (17), (18), and (22), we can write
$$\begin{aligned} \bigl\Vert {{}_{0}^{C}D_{t}^{{\alpha _{l}}}f - _{0}^{C}D_{t}^{{\alpha _{l}}}u} \bigr\Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}^{2} & \le \frac{{\mathcal{L}_{l}^{2}}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}}} \int _{0}^{{t_{f}}} {{t^{2(N + 1)\lambda }} {w_{t_{f}}^{(\lambda, \theta,\vartheta )}}\,{d}t} \\ & = \frac{{\mathcal{L}_{l}^{2} t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{{{(\Gamma ((N + 1)\lambda + 1))}^{2}}\Gamma (4 + 2N + \theta + \vartheta )}}. \end{aligned}$$
(23)
Now we take the square root of both sides of (23), and therefore inequality (20) can be obtained. □
Theorem 5.6
Let \({}_{0}^{C}D_{t}^{j \lambda } f(t)\), \({}_{0}^{C}D_{t}^{j \lambda + q} f(t)\), \({}_{0}^{C}D_{t}^{j \lambda + \alpha _{l}} f(t) \in C([0,{t_{f}}]) \). Suppose that \(\vert {}_{0}^{C}D_{t}^{j \lambda } f(t) \vert \leq \mathcal{L}\), \(\vert {}_{0}^{C}D_{t}^{j \lambda + q} f(t) \vert \leq \mathcal{L}_{q}\), \(\vert {}_{0}^{C}D_{t}^{j \lambda + \alpha _{l} } f(t) \vert \leq \mathcal{L}_{l}\), for \(l=1,2,\ldots,r\). In Eq. (1), let \(G_{1} \) be Lipschitz with the constant μ. Therefore, the error bound of the SFOJPs method for the modified equation is
$$\begin{aligned} { \bigl\Vert {E_{{k}}^{M}} \bigr\Vert _{{w_{t_{f}}^{(\lambda,\theta, \vartheta )}}}} \leq \frac{ ( {\mu } (\beta - \alpha ) + (r+1){\kappa } )\rho }{\Gamma ((N + 1)\lambda + 1)} \sqrt{ \frac{{t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}} , \end{aligned}$$
where \(\kappa = \mathop{\max }_{j = 0, \ldots,r} \{ { \kappa _{j}} \} \) and \(\rho = \mathop{\max }_{l = 1, \ldots,r} \{ \mathcal{L},{\mathcal{L}_{q}},{\mathcal{L}_{l}} \} \).
Proof
Since \(G_{1} \) is Lipschitz with the constant μ, we can write
$$\begin{aligned} { \vert {E_{N}} \vert } ={}& \biggl\vert { \int _{\alpha }^{\beta }{{G_{1}} \bigl(q,{{ \bigl( {{}_{0}^{C}D_{t}^{q} f} \bigr)}_{N}}(t) \bigr)\,dq}} \\ &{} + {G_{2}} \bigl(t,{f_{N}}(t),{{ \bigl( {{}_{0}^{C}D_{t}^{ \alpha _{1}} f} \bigr)}_{N}}(t), \ldots,{{ \bigl( {{}_{0}^{C}D_{t}^{ \alpha _{r}} f} \bigr)}_{N}}(t) \bigr) - g(t) \biggr\vert \\ ={}& \biggl\vert \int _{\alpha }^{\beta }{{G_{1}} \bigl(q,{{ \bigl( {{}_{0}^{C}D_{t}^{q} f} \bigr)}_{N}}(t) \bigr)\,dq} + {G_{2}} \bigl(t,{f_{N}}(t),{{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{1}} f} \bigr)}_{N}}(t), \ldots, {{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{r}} f} \bigr)}_{N}}(t) \bigr) \\ &{} - \int _{\alpha }^{\beta }{{G_{1}} \bigl(q,{{}_{0}^{C}D{}_{t}^{q}}f(t) \bigr)\,dq} - {G_{2}} \bigl(t,f(t),{{}_{0}^{C}D{}_{t}^{{\alpha _{1}}}}f(t), \ldots,{{}_{0}^{C}D{}_{t}^{{\alpha _{r}}}}f(t) \bigr) \biggr\vert \\ \le{} & {\mu } \int _{\alpha }^{\beta }{{{ \bigl\vert {{{}_{0}^{C}D{}_{t}^{q}}f(t) - {{ \bigl( {{}_{0}^{C}D_{t}^{q} f} \bigr)}_{N}}(t)} \bigr\vert }}\,dq} + { \kappa } { \bigl\vert {f(t) - {f_{N}}(t)} \bigr\vert } \\ &{}+ {\kappa } { \bigl\vert {{{}_{0}^{C}D{}_{t}^{{\alpha _{1}}}}f(t) - {{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{1}} f} \bigr)}_{N}}(t)} \bigr\vert } + \cdots + {\kappa } { \bigl\vert {{{}_{0}^{C}D{}_{t}^{{\alpha _{r}}}}f(t) - {{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{r}} f} \bigr)}_{N}}(t)} \bigr\vert } \\ \le {}& {\mu } \int _{\alpha }^{\beta }{{{ \bigl\Vert {{{}_{0}^{C}D{}_{t}^{q}}f - {{ \bigl( {{}_{0}^{C}D_{t}^{q} f} \bigr)}_{N}}} \bigr\Vert }_{w_{t_{f}}^{( \lambda,\theta,\vartheta )}}}\,dq} + {\kappa } { \Vert {f - {f_{N}}} \Vert _{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}} \\ &{}+ {\kappa } { \bigl\Vert {{{}_{0}^{C}D{}_{t}^{{\alpha _{1}}}}f - {{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{1}} f} \bigr)}_{N}}} \bigr\Vert _{w_{t_{f}}^{( \lambda,\theta,\vartheta )}}} + \cdots + {\kappa } { \bigl\Vert {{{}_{0}^{C}D{}_{t}^{{ \alpha _{r}}}}f - {{ \bigl( {{}_{0}^{C}D_{t}^{\alpha _{r}} f} \bigr)}_{N}}} \bigr\Vert _{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}. \end{aligned}$$
By using Theorems 5.4 and 5.5, we obtain
$$\begin{aligned} { \Vert {E_{N}} \Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}} \leq \frac{{\mu } (\beta - \alpha ) {\mathcal{L}_{q}} + {\kappa } ( {\mathcal{L}} + {\mathcal{L}_{1}} + \cdots + {\mathcal{L}_{r}} )}{{\Gamma ((N + 1)\lambda + 1)}} \sqrt{ \frac{{t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}} . \end{aligned}$$
Let
$$\begin{aligned} \rho = \mathop{\max } _{l = 1, \ldots,r} \{ \mathcal{L},{\mathcal{L}_{q}},{ \mathcal{L}_{l}} \}. \end{aligned}$$
Therefore, we get
$$\begin{aligned} { \Vert {E_{N}} \Vert _{{w_{t_{f}}^{(\lambda,\theta,\vartheta )}}}} \leq \frac{ ( {\mu } (\beta - \alpha ) + (r+1){\kappa } )\rho }{\Gamma ((N + 1)\lambda + 1)} \sqrt{ \frac{{t_{f}^{(2N + 3 + \vartheta + \theta )\lambda }\Gamma (1 + \theta )\Gamma (2N + 3 + \vartheta )}}{{\Gamma (4 + 2N + \theta + \vartheta )}}} . \end{aligned}$$
□