### Lemma 3.1

*The function*
\(U(t)\)
*satisfies the following vector ordinary differential equation*:

$$ \frac{dU(t)}{dt}=aU(t)+\frac{0.25a^{3}(t-nh)^{2}}{[1-0.5a(t-nh)]^{2}}Y _{n}, \quad t\in \bigl[nh,(n+1)h\bigr], n=0,1,2,\dots , $$

*and initial conditions*
\(U(nh)=Y_{n}\). *Furthermore*,

$$ U\bigl((n+1)h\bigr)=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \biggr]Y_{n}. $$

### Proof

Firstly, it follows from the expression of function \(U(t)\) that

$$ \bigl(U(t)\bigr)_{n=n}=\frac{1+0.5a(t-nh)}{1-0.5a(t-nh)}Y_{n}, \quad nh\leq t \leq (n+1)h. $$

The derivative of function \(U(t)\) is given by

$$ \begin{aligned} \frac{dU(t)}{dt}&=\frac{a}{[1-0.5a(t-nh)]^{2}}Y_{n} \\ &= \frac{a[1-0.25a^{2}(t-nh)^{2}]-a[1-0.25a^{2}(t-nh)^{2}]+a}{[1-0.5a(t-nh)]^{2}}Y _{n} \\ &=a\frac{1-0.5a(t+nh)}{1-0.5a(t-nh)}Y_{n}+ \frac{0.25a^{3}(t-nh)^{2}}{[1-0.5a(t-nh)]^{2}}Y_{n} \\ &=aU(t)+\frac{0.25a^{3}(t-nh)^{2}}{[1-0.5a(t-nh)]^{2}}Y_{n}. \end{aligned} $$

Secondly, we have \(U(nh)=Y_{n}\).

Lastly, we integrate the derivative \(\frac{dU(t)}{dt}\) and obtain

$$ \int _{nh}^{(n+1)h}\frac{dU(t)}{dt}\,dt= \int _{nh}^{(n+1)h}aU(t)\,dt+ \int _{nh}^{(n+1)h}\frac{0.25a^{3}(t-nh)^{2}}{[1-0.5a(t-nh)]^{2}}Y_{n}\,dt. $$

We make the transform \(t-nh=\tau \) and have

$$ U\bigl((n+1)h\bigr)-U(nh)= \int _{0}^{h}aU(t)\,dt+\frac{1}{4}a^{3} \int _{0}^{h}\frac{ \tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau Y_{n}. $$

Therefore, we can obtain the claim of Lemma 3.1 as follows:

$$ U\bigl((n+1)h\bigr)=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{\tau ^{2}}{(1-0.5a \tau )^{2}}\,d\tau \biggr]Y_{n}. $$

This completes our proof. □

### Lemma 3.2

*Let the local error be*
\(E_{n}=Y_{n}-Y(nh)\). *Then it satisfies the following equality*:

$$ E_{n+1}=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \biggr]E_{n}+\frac{1}{4}a^{3}e^{(n+1)ha} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \mathit{II}. $$

(7)

### Proof

If *t* satisfies the condition \((n+1)h\leq t\leq (n+2)h\), we have

$$ U(t)=\frac{1+0.5a(t-(n+1)h)}{1-0.5a(t-(n+1)h)}Y_{n+1}. $$

Then we can obtain \(U((t+1)h)=Y_{n+1}\). It follows from Lemma 3.1 that

$$ Y_{n+1}=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \biggr]Y_{n}. $$

By the vector differential equation \(Y'=aY\), \(Y(0)=\mathit{II}\), its solution is \(Y(t)=e^{at}\mathit{II}\). By the definition of local error \(E_{n}\), we have

$$ \begin{aligned} E_{n+1}&=Y_{n+1}-Y\bigl((n+1)h\bigr) \\ &=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a \tau )^{2}}\,d\tau \biggr]Y_{n}-e^{(n+1)ah}\mathit{II} \\ &=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a \tau )^{2}}\,d\tau \biggr] \bigl(E_{n}+Y(nh)\bigr)-e^{(n+1)ah}\mathit{II} \\ &=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a \tau )^{2}}\,d\tau \biggr]E_{n}\\ &\quad {}+e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e ^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \biggr]e^{nha}\mathit{II}-e^{(n+1)ah} \mathit{II} \\ &=e^{ah}\biggl[1+\frac{1}{4}a^{3} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a \tau )^{2}}\,d\tau \biggr]E_{n}+\frac{1}{4}a^{3}e^{(n+1)ha} \int _{0}^{h}\frac{e ^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \mathit{II}. \end{aligned} $$

This completes our proof. □

By the conclusions of Lemmas 3.1 and 3.2, we obtain the following error control theorem.

### Theorem 3.1

*If*
\(a<0\), *the ECI algorithm* (1) *satisfies the following error propagation inequality*:

$$ \Vert E_{n+1} \Vert \leq e^{ah}\biggl[1+\frac{1}{4} \vert a \vert ^{3} \int _{0}^{h} e^{-\tau a} \tau ^{2} \,d\tau \biggr] \Vert E_{n} \Vert +\frac{1}{4} \vert a \vert ^{3}e^{(n+1)ha} \int _{0}^{h}e ^{-\tau a}\tau ^{2}\,d \tau . $$

### Proof

When \(a<0\), it follows from the condition \(0<\tau <h\) that we have the conclusion \(1-0.5a\tau >1\). By the properties of the integral, we obtain

$$ \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \leq \int _{0}^{h}e^{-\tau a}\tau ^{2}\,d \tau . $$

Therefore, by Lemma 3.2 and the triangle inequality of the norm, we obtain

$$ \begin{aligned} \Vert E_{n+1} \Vert &\leq e^{ah}\biggl[1+ \frac{1}{4} \vert a \vert ^{3} \int _{0}^{h}\frac{e^{- \tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d\tau \biggr] \Vert E_{n} \Vert +\frac{1}{4} \vert a \vert ^{3}e ^{(n+1)ha} \int _{0}^{h}\frac{e^{-\tau a}\tau ^{2}}{(1-0.5a\tau )^{2}}\,d \tau \\ &\leq e^{ah}\biggl[1+\frac{1}{4} \vert a \vert ^{3} \int _{0}^{h} e^{-\tau a}\tau ^{2} \,d \tau \biggr] \Vert E_{n} \Vert +\frac{1}{4} \vert a \vert ^{3}e^{(n+1)ha} \int _{0}^{h}e^{-\tau a} \tau ^{2}\,d \tau . \end{aligned} $$

Therefore, the conclusion of Theorem 3.1 follows from Lemma 3.1. □

### Remark 2

The advantages of this method are not only in its convergence, that is, the iterated error being limited in a small interval, but also in its ability to simulate the solutions of ODEs continuously and explicitly, which can help simulate the true solutions more accurately.