In this section, we provide the error analysis of the spectral collocation method for SVIDE. We will particularly determine the spectral accuracy for the numerical solution \(y^{N}(t)\) [26]. Let us consider
$$\begin{aligned} U_{T, 1}^{N}(t)=I_{T, N}\frac{d}{dt}Y(t)- \frac{d}{dt}I_{T, N}Y(t) \end{aligned}$$
and
$$\begin{aligned} U_{T, 2}^{N}(t)={} & f \biggl(y^{N}(t), \int _{0}^{t_{i}}G(t, s)y^{N}(s)\,ds, t \biggr)-f \biggl(I_{T, N}Y(t), I_{T, N} \int _{0}^{t_{i}}G(t, s)Y(s)\,ds, t \biggr) \\ &{}\vee g \biggl(y^{N}(t), \int _{0}^{t_{i}}H(t, s)y^{N}(s)\,ds, t \biggr) \\ &{}-g \biggl(I_{T, N}Y(t), I_{T, N} \int _{0}^{t_{i}}H(t, s)Y(s)\,ds, t \biggr). \end{aligned}$$
(15)
Then we have from Eq. (1)
$$\begin{aligned} \frac{d}{dt}I_{T, N}Y(t) &=f \biggl(Y(t), \int _{0}^{t_{i}}G(t, s)Y(s)\,ds, t \biggr) \\ &\quad{} \vee g \biggl(Y(t), \int _{0}^{t_{i}}H(t, s)Y(s)\,ds, t \biggr)-W_{T, 1} ^{N}(t), \quad t\in \varLambda _{N}. \end{aligned}$$
(16)
Furthermore, let \(E^{N}(t)= y^{N}(t)-I_{N, T}Y(t)\), then we have from Eq. (6) and Eq. (16) that
$$ \textstyle\begin{cases} \frac{d}{dt}E^{N}(t)=U_{T, 1}^{N}(t)+U_{T, 2}^{N}(t), \quad t\in\varLambda _{N}, \\ E^{N}(0)=\varPhi (0)-I_{T, N}\varPhi (0)=0. \end{cases} $$
(17)
Lemma 3.1
Using the condition Eq. (17), the following inequality holds:
$$\begin{aligned} \bigl(E^{N}(T)\bigr)^{2}\leq 2 \bigl\Vert E^{N} \bigr\Vert _{T, N} \bigl( \bigl\Vert U_{T, 1}^{N} \bigr\Vert _{T, N}+ \bigl\Vert U _{T, 2}^{N} \bigr\Vert _{T, N} \bigr). \end{aligned}$$
(18)
Proof
Since \(E^{N}(0)=0\), multiplying Eq. (17) by \(2E^{N}(t _{T, k}^{N})\omega _{T, k}^{N}\) and having the resulting equation summed up, \(1\leq k\leq N\), we obtain
$$\begin{aligned} 2\biggl(E^{N}, \frac{d}{dt}E^{N} \biggr)_{T, N}=2\bigl(U_{T, 1}^{N}, E^{N} \bigr)_{T, N}+2\bigl(U _{T, 2}^{N}, E^{N} \bigr)_{T, N}. \end{aligned}$$
(19)
Since \(\frac{d}{dt}E^{N}(t)\in P_{N-1}(0, T)\), then using the result, if \(\psi _{1}, \psi _{2} \in P_{2N-1}(0, T)\) then \((\psi _{1}, \psi _{2})_{T} = (\psi _{1}, \psi _{2})_{T, N}\). Using this, then
$$\begin{aligned} 2\biggl(E^{N}, \frac{d}{dt}E^{N} \biggr)_{T, N}=2\biggl(E^{N}, \frac{d}{dt}E^{N} \biggr)_{T}=\bigl(E ^{N}(T)\bigr)^{2}. \end{aligned}$$
(20)
Put Eq. (20) into Eq. (19), and using a Cauchy–Schwartz inequality, we obtain Eq. (18). □
Next we estimate, \(U_{T, 1}^{N}\). For this purpose, the following lemma is given; one may refer to [27].
Lemma 3.2
For the integers
\(1\leq r\leq N\),
$$\begin{aligned} \bigl\Vert U_{T, 1}^{N} \bigr\Vert _{T} \leq cTN^{-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(21)
where
\(R_{T, 1}^{r}(\varPhi )\)
is finite.
Proof
For this purpose, using Lemma 3.1. for integers \(1 \leq r \leq N+1\), we have
$$\begin{aligned} \biggl\Vert I_{T, N}\frac{d}{dt}Y- \frac{d}{dt}Y \biggr\Vert _{T}\leq c_{1}TN^{-r}R_{T, 1} ^{r}(Y). \end{aligned}$$
(22)
Again we have, \(0 \leq r\leq N\),
$$\begin{aligned} \biggl\Vert \frac{d}{dt}(I_{T, N}Y-Y) \biggr\Vert _{T}\leq c_{1}TN^{-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(23)
Now by Eq. (22) and Eq. (23), we have
$$\begin{aligned} \bigl\Vert U_{T, 1}^{N} \bigr\Vert \leq \biggl\Vert I_{T, N}\frac{d}{dt}Y-\frac{d}{dt}Y \biggr\Vert _{T}+ \biggl\Vert \frac{d}{dt}(I_{T, N}Y-Y) \biggr\Vert _{T}\leq cTN^{-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(24)
Here \(2c_{1}=c\), which completes the proof. □
Now, to analyze the numerical error of functions f and g, we consider the following, for a positive number \(\beta < 1/4\).
Theorem 3.1
([27])
If conditions Eq. (2)–Eq. (5) hold, \(R_{T, 1}^{r}(\varPhi )\)
is a finite function for integers
\(1\leq r\leq N\), and
$$\begin{aligned} T\sqrt{2+N^{-1}}\bigl(1+\sqrt{2+N^{-1}}\bigr) \bigl(K_{1} + K_{2}(1-\lambda )\bigr) \leq \beta < \frac{1}{4}, \end{aligned}$$
(25)
where
\(0\leq \lambda <1\), then
$$\begin{aligned}& \bigl\Vert Y-y^{N} \bigr\Vert _{T}\leq c_{\beta }T^{2}N^{-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(26)
$$\begin{aligned}& \bigl\vert Y(T)-y^{N}(T) \bigr\vert \leq c_{\beta }T^{\frac{3}{2}}N^{-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(27)
and, in particular,
$$\begin{aligned} \max_{t\in [0, T]} \bigl\vert Y(t)-y^{N}(t) \bigr\vert \leq c_{\beta }TN^{1-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(28)
where the constant
\(c_{\beta }>0\), depending on β.
Proof
Since \(U_{T, 2}^{N}(0)=0\), we have from Eq. (15)
$$\begin{aligned} \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N} &= \bigl\Vert f\bigl(y^{N}, v_{1}^{N}, \cdot \bigr)-f(I_{T, N}Y, I_{T, N}V_{1}, \cdot ) \bigr\Vert _{T, N} \\ &\quad{} \vee \bigl\Vert g\bigl(y^{N}, v_{2}^{N}, \cdot \bigr)-g(I_{T, N}Y, I_{T, N}V_{2}, \cdot ) \bigr\Vert _{T, N}, \quad t\in \varLambda _{N}, \end{aligned}$$
(29)
where \(v_{1}^{N}(t)=\int _{0}^{t_{i}}G(t, s)y^{N}(s)\,ds\), \(v_{2}^{N}(t)= \int _{0}^{t_{i}}H(t, s)y^{N}(s)\,ds \in P_{N}(0, T)\). For simplicity, we denote \(K_{N}=\sqrt{2+N^{-1}}\), then, by using Eq. (2) and Eq. (3), we see that Eq. (29) takes the form
$$\begin{aligned} \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N} &\leq \bigl\Vert f\bigl(y^{N}, v_{1}^{N}, \cdot \bigr)-f \bigl(I_{T, N}Y, v _{1}^{N}, \cdot \bigr) \bigr\Vert _{T, N} \\ &\quad {}+ \bigl\Vert f\bigl(I_{T, N}Y, v_{1}^{N}, \cdot \bigr)-f(I_{T, N}Y, I_{T, N}V_{1}, \cdot ) \bigr\Vert _{T, N} \\ &\quad {}\vee \bigl\Vert g\bigl(y^{N}, v_{1}^{N}, \cdot \bigr)-g\bigl(I_{T, N}Y, v_{1}^{N}, \cdot \bigr) \bigr\Vert _{T, N} \\ &\quad {}+ \bigl\Vert g\bigl(I _{T, N}Y, v_{1}^{N}, \cdot \bigr)-g(I_{T, N}Y, I_{T, N}V_{1}, \cdot ) \bigr\Vert _{T, N} \\ &\leq K_{1} \bigl\Vert E^{N} \bigr\Vert _{T, N}+K_{2} \bigl\Vert v^{N}-I_{T, N}V \bigr\Vert _{T, N} \\ &\leq K_{N} \bigl(K_{1} \bigl\Vert E^{N} \bigr\Vert _{T}+K_{2} \bigl\Vert v^{N}-I_{T, N}V \bigr\Vert _{T} \bigr). \end{aligned}$$
(30)
Furthermore, using Eq. (21) we obtain
$$\begin{aligned} \bigl\Vert v^{N}-I_{T, N}V \bigr\Vert _{T} \leq{}& \bigl\Vert v^{N}-V \bigr\Vert _{T}+ \Vert V-I_{T, N}V \Vert _{T} \\ \leq{}& (1-\lambda )^{-\frac{1}{2}} \bigl\Vert Y-y^{N} \bigr\Vert _{T}+cT^{2}N^{-r-1}R _{T, 1}^{r} (Y). \end{aligned}$$
(31)
Now by Eq. (25), Eq. (30), and Eq. (31), we get
$$\begin{aligned} \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N}\leq K_{N} \bigl(K_{1} \bigl\Vert E^{N} \bigr\Vert _{T}+K_{2}(1- \lambda )^{-\frac{1}{2}} \bigl\Vert Y-y^{N} \bigr\Vert _{T} \bigr)+cK_{N}K_{2}T^{2}N^{-r-1}R _{T, 1}^{r}(Y). \end{aligned}$$
(32)
Now for any \(t\in [0, T]\) (cf. [28])
$$\begin{aligned} \bigl(E^{N}(t)\bigr)^{2}\leq \biggl(T^{\frac{1}{2}}\bigl(E^{N}(T)\bigr)^{2}+2 \bigl\Vert E^{N} \bigr\Vert _{T} \biggl\Vert t^{\frac{1}{2}} \frac{d}{dt}E^{N} \biggr\Vert _{T} \biggr)t^{-\frac{1}{2}}. \end{aligned}$$
(33)
Integrating Eq. (33) with respect to t,
$$\begin{aligned} \bigl\Vert E^{N} \bigr\Vert _{T}^{2} \leq 2T\bigl(E^{N}(T)\bigr)^{2}+4T^{\frac{1}{2}} \bigl\Vert E^{N} \bigr\Vert _{T} \biggl\Vert t^{\frac{1}{2}} \frac{d}{dt}E^{N} \biggr\Vert _{T}. \end{aligned}$$
(34)
Now since \(\frac{d}{dt}E^{N}(t)\in P_{N-1}(0, T)\) and \(\vert U_{T, 1}^{N}(0)+U _{T, 2}^{N}(0) \vert < \infty \), we have
$$\begin{aligned} \begin{aligned}[b] \biggl\Vert t^{\frac{1}{2}}\frac{d}{dt}E^{N} \biggr\Vert _{T}&= \biggl\Vert t^{\frac{1}{2}} \frac{d}{dt}E^{N} \biggr\Vert _{T, N}= \bigl\Vert t^{\frac{1}{2}} \bigl(U_{T, 1}^{N} + U_{T, 2}^{N} \bigr) \bigr\Vert _{T, N}\\ &\leq T^{\frac{1}{2}} \bigl\Vert U_{T, 1}^{N} \bigr\Vert _{T, N}+T^{ \frac{1}{2}} \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N}.\end{aligned} \end{aligned}$$
(35)
We use Eq. (34) and Eq. (35) to derive that
$$\begin{aligned} \bigl\Vert E^{N} \bigr\Vert _{T}^{2} \leq 2T\bigl(E^{N}(T)\bigr)^{2}+4T \bigl\Vert E^{N} \bigr\Vert _{T} \bigl( \bigl\Vert W_{T, 1} ^{N} \bigr\Vert _{T, N}+ \bigl\Vert W_{T, 2}^{N} \bigr\Vert _{T, N} \bigr), \end{aligned}$$
(36)
which implies
$$\begin{aligned} \bigl(E^{N}(T)\bigr)^{2}\geq \frac{1}{2T} \bigl\Vert E^{N} \bigr\Vert _{T}^{2}-2 \bigl\Vert E^{N} \bigr\Vert _{T} \bigl( \bigl\Vert U_{T, 1}^{N} \bigr\Vert _{T, N}+ \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N} \bigr). \end{aligned}$$
(37)
Combining Eq. (18) and Eq. (37) gives
$$\begin{aligned} &\frac{1}{2T} \bigl\Vert E^{N} \bigr\Vert _{T}^{2}-2 \bigl\Vert E^{N} \bigr\Vert _{T} \bigl( \bigl\Vert U_{T, 1}^{N} \bigr\Vert _{T, N}+ \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N} \bigr) \\ &\quad \leq 2 \bigl\Vert E^{N} \bigr\Vert _{T, N} \bigl\Vert U_{T,1}^{N} \bigr\Vert _{T, N}+2 \bigl\Vert E^{N} \bigr\Vert _{T, N} \bigl\Vert U _{T,2}^{N} \bigr\Vert _{T, N}, \end{aligned}$$
(38)
which together with Eq. (21), Eq. (25) and Eq. (32) yields
$$\begin{aligned} \bigl\Vert E^{N} \bigr\Vert _{T} &\leq 4T(1+K_{N}) \bigl( \bigl\Vert U_{T,1}^{N} \bigr\Vert _{T, N}+ \bigl\Vert U_{T,2} ^{N} \bigr\Vert _{T, N} \bigr) \\ &\leq 4TK_{N}(1+K_{N}) \bigl(K_{1} \bigl\Vert E^{N} \bigr\Vert _{T}+K_{2}(1-\lambda )^{- \frac{1}{2}} \bigl\Vert Y-y^{N} \bigr\Vert _{T} \bigr)+cT^{2}N^{-r}R_{T, 1}^{r} (Y). \end{aligned}$$
(39)
Here
$$\begin{aligned} \bigl\Vert Y-y^{N} \bigr\Vert _{T}\leq \bigl\Vert E^{N} \bigr\Vert _{T}+ \Vert Y-I_{T, N}Y \Vert _{T}\leq + \bigl\Vert E^{N} \bigr\Vert _{T}+cT^{2}N^{-r-1}R_{T, 1}^{r} (Y). \end{aligned}$$
(40)
Putting Eq. (40) into Eq. (39), we have
$$\begin{aligned} & \bigl(1-4TK_{N}(1+K_{N}) \bigl(K_{1}+K_{2}(1- \lambda )^{-\frac{1}{2}} \bigr) \bigr) \bigl\Vert E^{N} \bigr\Vert _{T} \\ &\quad \leq cK_{2}K_{N}(1+K_{N}) (1-\lambda )^{-\frac{1}{2}}T^{3}N^{-r-1}R _{T, 1}^{r}(Y)+cT^{2}N^{-r}R_{T, 1}^{r}(Y) \\ &\quad \leq cT^{2}N^{-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(41)
combining Eq. (41) with Eq. (25), we have
$$\begin{aligned} \bigl\Vert E^{N} \bigr\Vert _{T}\leq c_{\beta }T^{2}N^{-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(42)
Equation (42) with Eq. (40) gives
$$\begin{aligned} \bigl\Vert Y-y^{N} \bigr\Vert _{T}\leq c_{\beta }T^{2}N^{-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(43)
which leads to Eq. (26).
Next by using Eq. (32), Eq. (42), and Eq. (43), we have
$$\begin{aligned} \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N}\leq c_{\beta }T^{2}N^{-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(44)
Furthermore, using Eq. (18), Eq. (21), Eq. (42), and Eq. (44), we deduce that
$$\begin{aligned} \bigl\vert Y(T)-y^{N}(T) \bigr\vert ^{2}&= \bigl(E^{N}(T)\bigr)^{2} \leq 2K_{N} \bigl\Vert E^{N} \bigr\Vert _{T} \bigl( \bigl\Vert U _{T, 1}^{N} \bigr\Vert _{T, N}+ \bigl\Vert U_{T, 2}^{N} \bigr\Vert _{T, N} \bigr) \\ &\leq c_{\beta }T^{3}N^{-2r}\bigl(R_{T, 1}^{r}(Y) \bigr)^{2}. \end{aligned}$$
(45)
This proves Eq. (27).
Again using Eq. (23), and Eq. (42), we deduce that
$$\begin{aligned} \biggl\Vert \frac{d}{dt}\bigl(Y-y^{N}\bigr) \biggr\Vert _{T} &\leq \biggl\Vert \frac{d}{dt}E^{N} \biggr\Vert _{T}+ \biggl\Vert \frac{d}{dt}(Y-I_{T, N}Y) \biggr\Vert _{T} \\ &\leq cT^{-1}N^{2} \bigl\Vert E^{N} \bigr\Vert _{T}+ \biggl\Vert \frac{d}{dt}(Y-I_{T, N}Y) \biggr\Vert _{T} \\ &\leq c_{\beta }TN^{2-r}R_{T, 1}^{r}(Y)+c TN^{-r}R_{T, 1}^{r}(Y) \\ &\leq c_{\beta }TN^{2-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(46)
This together with Eq. (43) implies that
$$\begin{aligned} \bigl\Vert Y-y^{N} \bigr\Vert _{H^{1}(0, T)}\leq c_{\beta }TN^{2-r}R_{T, 1}^{r}(Y). \end{aligned}$$
(47)
For this purpose, according to the Sobolev inequality [26],
$$\begin{aligned} \Vert u \Vert _{L^{\infty }(0, T)}\leq \sqrt{2+T^{-1}} \Vert u \Vert _{L^{2}(0, T)} ^{\frac{1}{2}} \Vert u \Vert _{H^{1}(0, T)}^{\frac{1}{2}}. \end{aligned}$$
(48)
Hence by Eq. (26) and Eq. (46), we obtain
$$\begin{aligned} \max_{t\in [0, T]} \bigl\vert Y(t)-y^{N}(t) \bigr\vert & \leq \sqrt{2+T^{-1}} \bigl\Vert Y-y^{N} \bigr\Vert _{L^{2}(0, T)}^{\frac{1}{2}} \bigl\Vert Y-y^{N} \bigr\Vert _{H^{1}(0, T)}^{\frac{1}{2}} \\ &\leq c_{\beta }TN^{1-r}R_{T, 1}^{r}(Y), \end{aligned}$$
(49)
which is the proof of Eq. (28). □