Critical Poles and Third-Order Nonlinear Differential Equations

Abstract

The paper deals with the results of a study of a third-order nonlinear differential equation with moving singular points and critical poles. So far, this type of equation cannot be solved in quadratures. The development of the author’s approach in proving the theorem of the existence of moving singular points and solutions in the vicinity of a critical pole, based on a modified Cauchy majorant method, is given. An analytical approximate solution in the vicinity of a moving singular point is obtained, and an expression for the a priori error estimate is presented. A numerical experiment confirming the obtained theoretical results is provided.

1. Introduction

Nonlinear differential equations arise in the study of nonlinear oscillation theory [1,2,3,4,5,6], relaxation oscillations [7], nonlinear diffusion [8], and in the investigation of wave processes in rods and beams [9]. In particular, the authors of work [10] suggest using a third-order nonlinear equation. Unlike linear differential equations, nonlinear differential equations have the distinctive feature of possessing moving singular points, which create complications in solving this category of equations.

It should be noted that this category of equations allows solvability in quadratures if it possesses the P-property (having simple and multiple poles) [11,12,13,14,15,16,17]. The second approach is related to an analytical approximate method of solution, which is applicable to nonlinear equations with moving singular points not only with simple and multiple poles, but also with critical points [18,19,20]. At present, the second approach has also been implemented for a nonlinear equation with a fractional derivative [21], as contrasted with the first one.

This study uses a third-order nonlinear differential equation with a fifth-degree polynomial on the right-hand side in the complex domain, devoid of the P-property. The considered equation cannot be solved in quadratures. This fact confirms the research novelty and the above-mentioned application puts on display the relevance of the presented results. It proves the theorem of existence of a singular point and a solution in the surroundings of this point. An analytical approximate solution is constructed, and an a priori error estimate is obtained. A formula for calculating the domain in which the presented results are valid is obtained. The reliability of the theoretical results is confirmed by numerical experiments.

2. Research Methods

Let us consider the Cauchy problem

$$Y^{‴}\left(z\right)=Y^5\left(z\right)+r\left(z\right),$$

(1)

$$Y\left(z_0\right)=Y_0,\hspace{1em}{Y}^{\prime }\left(z_0\right)=Y_1,\hspace{1em}{Y}^{″}\left(z_0\right)=Y_2,$$

(2)

where $r\left(z\right)$ is an analytical function, $z\in \mathrm{C}$.

Theorem 1.

Let $z^{*}$ be a moving singular point of the Cauchy problem solution (1) and (2) and $r\left(z\right)$ is an analytical function in the domain

$$\left|z^{*}-z\right|<{\rho }_1,$$

where $0<{\rho }_1=const$. Then there is a unique solution (1) and (2) in the domain of a moving singular point $z^{*}$ as follows:

$$Y\left(z\right)={(z^{*}-z)}^{-3/4}\sum _{n=0}^{\infty }{C}_n{(z^{*}-z)}^{n/4},$$

(3)

in the domain

$$\left|z^{*}-z\right|<{\rho }_2,$$

(4)

where

$${\rho }_2=min\left\{{\rho }_1,{\left(2\right(M+1\left)\right)}^{-4/15}\right\},M=max\left\{\underset{n}{sup}{\displaystyle \frac{|\left.r^{\left(n\right)}\left(z^{*}\right)\right|}{n!}}\right\},n=0,1,2,\dots .$$

Proof of Theorem 1.

Let us look up, for Cauchy problem, solution (1) and (2) as follows:

$$Y\left(z\right)={(z^{*}-z)}^{\rho }\sum _{n=0}^{\infty }{C}_n{(z^{*}-z)}^n$$

(5)

where $C_n\in \mathrm{C}$.

Since $z^{*}$ is a regular point for the function $r\left(z\right)$, its conversion into a series is possible

$$r\left(z\right)=\sum _{n=0}^{\infty }{A}_n{(z^{*}-z)}^n,$$

(6)

where $A_n\in \mathrm{C}$.

By putting the expressions (5) and (6) into Equation (1), we obtain

$$\begin{gathered} -\sum _{n=0}^{\infty }(n+\rho )(n+\rho -1)(n+\rho -2)C_n{(z^{*}-z)}^{n+\rho -3} \\ =\sum _{n=0}^{\infty }{C_n^{***}(z^{*}-z)}^{n+5\rho }+\sum _{n=0}^{\infty }{A}_n{(z^{*}-z)}^n, \end{gathered}$$

(7)

where

$$C_n^{***}=\sum _{k=0}^n{C}_k^{**}{C}_{n-k},C_n^{**}=\sum _{k=0}^n{C}_k^{*}{C}_{n-k}^{*},C_n^{*}=\sum _{k=0}^n{C}_k{C}_{n-k}.$$

From the identity conditions in (7), we obtain an equation for determining the parameter $\rho$:

$$n+\rho -3=n+5\rho ,$$

where the second one implies $\rho =-3/4$. Then (5) is configured as (3), and (7) is transformed:

$$\begin{gathered} -\sum _{n=0}^{\infty }({\displaystyle \frac{n-3}{4}})({\displaystyle \frac{n-7}{4}})({\displaystyle \frac{n-11}{4}})C_n{(z^{*}-z)}^{(n-15)/4}= \\ \sum _{n=0}^{\infty }{C_n^{***}(z^{*}-z)}^{(n-15)/4}+\sum _{n=0}^{\infty }{A}_n{(z^{*}-z)}^n, \end{gathered}$$

(8)

Then, from (8), we obtain recurrent ratios for coefficients:

• $-({\displaystyle \frac{n-3}{4}})({\displaystyle \frac{n-7}{4}})({\displaystyle \frac{n-11}{4}})C_n=C_n^{***},$ for n = 0, 14, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 30, 32, …;
• $-({\displaystyle \frac{n-3}{4}})({\displaystyle \frac{n-7}{4}})({\displaystyle \frac{n-11}{4}})C_n=C_n^{***}+A_{(n-15)/4},$ for n = 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, ….

For n = 1–14 values $C_n=0.$ For $n=0,$ it follows from the first recurrent ratio

$$(3\ast 7\ast 11)/4^3{C}_0=C_0^{***},\to (21\ast 11)/4^3{C}_0=C_0^5,\to C_0=\pm \sqrt[4]{231/64}.$$

Let us present the coefficient expressions $C_n$, not equal to zero:

$$\begin{gathered} C_{15}=-{\displaystyle \frac{A_0}{6+5C_0^4}};C_{19}=-{\displaystyle \frac{A_1}{54+5C_0^4}};C_{23}=-{\displaystyle \frac{A_2}{60+5C_0^4}};C_{27}=-{\displaystyle \frac{A_3}{120+5C_0^4}}; \\ {C}_{30}=-{\displaystyle \frac{4C_{15}^2(C_0^4+C_0^3)}{\mathrm{11,799}/64+5C_0^4}};C_{31}=-{\displaystyle \frac{A_4}{210+5C_0^4}};C_{34}=-{\displaystyle \frac{4C_0^3{C}_{15}{C}_{19}}{\mathrm{19,251}/64+5C_0^4}}; \\ {C}_{35}=-{\displaystyle \frac{A_5}{336+5C_0^4}};C_{38}=-{\displaystyle \frac{2C_0^3(6C_{15}{C}_{23}+5C_{19}^2)}{\mathrm{29,295}/64+5C_0^4}};C_{39}=-{\displaystyle \frac{A_6}{504+5C_0^4}}; \\ {C}_{40}=-{\displaystyle \frac{4C_0^3{C}_{15}^2}{\mathrm{35,409}/64+5C_0^4}};C_{42}=-{\displaystyle \frac{16C_0^3(C_{15}{C}_{27}+C_{19}{C}_{23})}{\mathrm{42,315}/64+5C_0^4}};C_{43}=-{\displaystyle \frac{A_7}{720+5C_0^4}}; \\ {C}_{45}=-{\displaystyle \frac{20C_0^3{C}_{15}{C}_{30}+4C_0^3{C}_{15}{C}_{40}+12C_0^2{C}_{15}^3+8C_0^2{C}_{15}^3+8C_0^2{C}_{15}{C}_{30}}{6783/8+9C_0^4}}; \\ {C}_{46}=-{\displaystyle \frac{12C_0^3{C}_{15}{C}_{31}}{\mathrm{58,695}/64+5C_0^4}};C_{47}=-{\displaystyle \frac{A_8}{990+5C_0^4}}; \\ {C}_{49}=-{\displaystyle \frac{24C_0^3(C_{15}{C}_{34}+C_{19}{C}_{30})+26C_0^2{C}_{15}^2{C}_{19}}{9177/8+5C_0^4}}; \\ {C}_{50}=-{\displaystyle \frac{C_0^3(10C_{15}{C}_{35}+12C_{19}{C}_{31}+12C_{23}{C}_{27})}{\mathrm{78,819}/64+5C_0^4}};C_{51}=-{\displaystyle \frac{A_8}{1320+5C_0^4}}. \end{gathered}$$

By using the recurrent ratios and expressions of coefficients $C_n,$ the solution (3) can be transformed:

$$\begin{gathered} Y\left(z\right)={(z^{*}-z)}^{-3/4}\sum _{n=0}^{\infty }{C}_n{(z^{*}-z)}^{n/4}=C_0{(z^{*}-z)}^{-3/4}+\sum _{n=1}^{\infty }{C}_n{(z^{*}-z)}^{n/4} \\ =C_0{(z^{*}-z)}^{-3/4}+\sum _{n=15}^{\infty }{C}_n{(z^{*}-z)}^{(n-15)/4}=C_0{(z^{*}-z)}^{-3/4} \\ +\sum _{i=0}^{14}\sum _{k=1}^{\infty }{C}_{15k+i}{(z^{*}-z)}^{\left(\right(15k-15)+i)/4}. \end{gathered}$$

Thus, in the last expression of the solution, we have 15 series determined by the “i” index. In case of $i=0$, we have a series as follows:

$$\sum _{k=1}^{\infty }{C}_{15k}{(z^{*}-z)}^{(15k-15)/4}.$$

(9)

To estimate coefficients $C_{15k}$ of the series (9), given the expressions of these coefficients, we set the following hypothesis:

$$\left|C_{15k+i}\right|\le {\displaystyle \frac{2^k{(M+1)}^k}{\frac{(15k+i-3)}{4}\frac{(15k+i-7)}{4}\frac{(15k+i-11)}{4}}}=V_{15k+i}.$$

(10)

Let us check this hypothesis. Let us consider the second approach to the recurrent ratios for coefficients $C_{15k}$, as this is more complex. As a result of the checking, we must obtain an estimate as follows:

$$\left|C_{15(k+1)}\right|\le {\displaystyle \frac{2^{k+1}{(M+1)}^{k+1}}{\frac{(15k+12)}{4}\frac{(15k+8)}{4}\frac{(15k+4)}{4}}}=V_{15(k+1)}.$$

(11)

Based on the recurrent ratio 2, we obtain

$$-({\displaystyle \frac{15k+12}{4}})({\displaystyle \frac{15k+8}{4}})({\displaystyle \frac{15k+4}{4}})C_{15(k+1)}=C_{15(k+1)}^{***}+A_{\left(15\right(k+1)-15)/4}.$$

(12)

Let us express the right side of the identity in detail (12).

$$\begin{gathered} C_{15(k+1)}^{***}+A_{\left(15\right(k+1)-15)/4}=\sum _{l=0}^{15(k+1)}{C}_l^{**}{C}_{15(k+1)-l}+A_{15k/4}=C_0^{**}{C}_{15(k+1)}+C_0{C}_{15(k+1)}^{**} \\ +\sum _{l=1}^{15(k+1)-1}{C}_l^{**}{C}_{15(k+1)-l}+A_{15k/4}=C_0^4{C}_{15(k+1)}+C_0\sum _{m=0}^{15(k+1)}{C}_m^{*}{C}_{15(k+1)-m}^{*} \\ +\sum _{l=1}^{15(k+1)-1}(\sum _{m=0}^l{C}_m^{*}{C}_{l-m}^{*})C_{15(k+1)-l}+A_{15k/4}=C_0^4{C}_{15(k+1)}+2C_0{C}_0^{**}{C}_{15(k+1)}^{*} \\ +C_0\sum _{m=1}^{15(k+1)-1}{C}_m^{*}{C}_{15(k+1)-m}^{*}+\sum _{l=1}^{15(k+1)-1}\left[\sum _{m=0}^l(\sum _{q=0}^m{C}_q{C}_{m-q})(\sum _{p=0}^{l-m}{C}_p{C}_{l-m-p})C_{15(k+1)-l}\right] \\ +A_{15k/4}=5C_0^4{C}_{15(k+1)}+2C_0^3\sum _{m=1}^{15(k+1)-1}{C}_m{C}_{15(k+1)-m}+C_0\sum _{m=1}^{15(k+1)-1}[(\sum _{q=0}^m{C}_q{C}_{m-q}) \\ \times (\sum _{p=0}^{15(k+1)-m}{C}_p{C}_{15(k+1)-m-p})]+\sum _{l=1}^{15(k+1)-1}\left[2C_0^{*}{C}_l^{*}+\sum _{m=1}^{l-1}{C}_m^{*}{C}_{l-m}^{*}\right]C_{15(k+1)-l}+A_{15k/4}. \end{gathered}$$

Or, considering that $C_n$ = 0, n = 1–14, we have

$$\begin{gathered} C_{15(k+1)}^{***}+A_{\left(15\right(k+1)-15)/4}=5C_0^4{C}_{15(k+1)}+2C_0^3\sum _{m=15}^{15k}{C}_m{C}_{15(k+1)-m} \\ +C_0\sum _{m=15}^{15k}\left[(2C_0{C}_m+\sum _{q=15}^{m-15}{C}_q{C}_{m-q})(2C_0{C}_{15(k+1)-m}+\sum _{p=15}^{15k-m}{C}_p{C}_{15(k+1)-m-p})\right] \\ +\sum _{l=15}^{15k}\left[2C_0^2(2C_0{C}_l+\sum _{p=15}^{l-15}{C}_p{C}_{l-p})+\sum _{m=15}^{l-15}(2C_0{C}_m+\sum _{p=15}^{m-15}{C}_p{C}_{m-p})\right.(2C_0{C}_{l-m} \\ +\left.\sum _{q=15}^{l-m-15}{C}_q{C}_{l-m-q})\right]C_{15(k+1)-l}+A_{15k/4}. \end{gathered}$$

Taking into account the last expression, from (12) we obtain

$$\begin{gathered} C_{15(k+1)}=-{\displaystyle \frac{1}{(15k+12)(15k+8)(15k+4)/64}}\left(2C_0^3\sum _{m=15}^{15k}{C}_m{C}_{15(k+1)-m}\right) \\ +C_0\sum _{m=15}^{15k}\left[(2C_0{C}_m+\sum _{q=15}^{m-15}{C}_q{C}_{m-q}\left)\right(2C_0{C}_{15(k+1)-m}+\sum _{p=15}^{15k-m}{C}_p{C}_{15(k+1)-m-p})\right] \\ +\sum _{l=15}^{15k}[2C_0^2(2C_0{C}_l+\sum _{p=15}^{l-15}{C}_p{C}_{l-p}) \\ +\sum _{m=15}^{l-15}(2C_0{C}_m+\sum _{p=15}^{m-15}{C}_p{C}_{m-p})(2C_0{C}_{l-m}+\sum _{q-15}^{l-m-15}{C}_q{C}_{l-m-q})]C_{15(k+1)-l}+A_{15k/4}. \end{gathered}$$

Then, given the estimate (10) it follows from the last one that

$$\begin{gathered} \left|C_{15(k+1)}\right|\le {\displaystyle \frac{1}{(15k+12)(15k+8)(15k+4)/64}}(2C_0^3\sum _{m=15}^{15k}{\displaystyle \frac{{\left(2(M+1)\right)}^{m/15}}{(m-3)(m-7)(m-11)/64}} \\ \times {\displaystyle \frac{{\left(2(M+1)\right)}^{k+1-m/15}}{(15k+12-m)(15k+8)(15k+4)/64}}+C_0\sum _{m=15}^{15k}\left[(2C_0{\displaystyle \frac{{\left(2(M+1)\right)}^{m/15}}{(m-3)(m-7)(m-11)/64}}\right. \\ +\sum _{q=15}^{m-15}{\displaystyle \frac{{\left(2(M+1)\right)}^{q/15}}{(q-3)(q-7)(q-11)/64}}{\displaystyle \frac{{\left(2(M+1)\right)}^{(m-q)/15}}{(m-q-3)(m-q-7)(m-q-11)/64}}) \\ \times (C_0{\displaystyle \frac{{\left(2(M+1)\right)}^{k+1-m/15}}{(15k+12-m)(15k+8-m)(15k+4-m)/64}} \\ +\sum _{p=15}^{15k-m}{\displaystyle \frac{{\left(2(M+1)\right)}^{p/15}}{(p-3)(p-7)(p-11)/64}} \\ \times \left.\left.{\displaystyle \frac{{\left(2(M+1)\right)}^{k+1-(m+p)/15}}{(15k+12-m-p)(15k+8-m-p)(15k+4-m-p)/64}}\right)\right] \\ +\sum _{l=15}^{15k}[2C_0^2(2C_0{\displaystyle \frac{{\left(2(M+1)\right)}^{l/15}}{(l-3)(l-7)(l-11)/64}}+\sum _{p=15}^{l-15}{\displaystyle \frac{{\left(2(M+1)\right)}^{p/15}}{(p-3)(p-7)(p-11)/64}} \\ \times {\displaystyle \frac{{\left(2(M+1)\right)}^{(l-p)/15}}{(l-p-3)(l-p-7)(l-p-11)/64}}+\sum _{m=15}^{l-15}\left(2C_0{\displaystyle \frac{{\left(2(M+1)\right)}^{m/15}}{(m-3)(m-7)(m-11)/64}}\right. \\ +\sum _{p=15}^{m-15}{\displaystyle \frac{{\left(2(M+1)\right)}^{p/15}}{(p-3)(p-7)(p-11)/64}}{\displaystyle \frac{{\left(2(M+1)\right)}^{(m-p)/15}}{(m-p-3)(m-p-7)(m-p-11)/64}})(2C_0 \\ \times {\displaystyle \frac{{\left(2(M+1)\right)}^{(l-m)/15}}{(l-m-3)(l-m-7)(l-m-11)/64}}+\sum _{q-15}^{l-m-15}{\displaystyle \frac{{\left(2(M+1)\right)}^{q/15}}{(q-3)(q-7)(q-11)/64}} \\ \times \left.{\displaystyle \frac{{\left(2(M+1)\right)}^{(l-m-q)/15}}{(l-m-q-3)(l-m-q-7)(l-m-q-11)/64}}\right)] \\ \times {\displaystyle \frac{{\left(2(M+1)\right)}^{k+1-l/15}}{(15k+12-l)(15k+8-l)(15k+4-l)/64}}+M). \end{gathered}$$

Simplifying the last one, we obtain

$$\begin{gathered} \left|C_{15(k+1)}\right|\le {\displaystyle \frac{{(M+1)}^{k+1}{2}^{k-1}+{(M+1)}^{k+1}{2}^{k-1}+{(M+1)}^{k+1}{2}^{k-1}+M}{(15k+12)(15k+8)(15k+4)/64}} \\ \le {\displaystyle \frac{2^{k+1}{(M+1)}^{k+1}}{(15k+12)(15k+8)(15k+4)/64}}. \end{gathered}$$

The estimates for the coefficients of the 14 next series are proved in a similar way:

$$\sum _{k=1}^{\infty }{C}_{15k+i}{(z^{*}-z)}^{(15k-15+i)/4},$$

where the index takes values i = 1–14.

Based on the sufficient D’Alembert criterion for the convergence of power series, we obtain the domain

$$\left|z^{*}-z\right|<{\left(2\right(M+1\left)\right)}^{-4/15}.$$

Choosing the value ${\rho }_2=min\left\{{\rho }_1{\left(2\right(M+1\left)\right)}^{-4/15}\right\}$, we finish the proof of Theorem 1, of the existence of both the moving singular point and the analytical solution with respect to the vicinity of this point. □

The proven theorem allows construction of an analytical approximate solution in the vicinity of a moving singular point.

$$Y_N\left(z\right)={(z^{*}-z)}^{-3/4}\sum _{n=0}^N{C}_n{(z^{*}-z)}^{n/4}.$$

(13)

Theorem 2.

For the approximate solution (13) of the Cauchy problem (1) and (2), where $r\left(z\right)$ is the analytical function, in the vicinity

$$\left|z^{*}-z\right|<{\rho }_2,$$

the following error estimate is fair:

$$\begin{gathered} \Delta y_{N+1}\left(z\right)\le {\displaystyle \frac{{\left(2(M+1)\right)}^{\left[(N+1)/15\right]-1}{\left|z^{*}-z\right|}^{15(\left[(N+1)/15\right]-1)/4}}{(N-17)(N-21)(N-25)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}} \\ \times \sum _{i=N+1-15\left[(N+1)/15\right]}^{14}{\left|z^{*}-z\right|}^{i/4}+{\displaystyle \frac{{\left(2(M+1)\right)}^{\left[(N+1)/15\right]}{\left|z^{*}-z\right|}^{15\left[(N+1)/15\right]/4}}{(N-2)(N-6)(N-10)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}} \\ \times \sum _{i=0}^{N+1-15\left[(N+1)/15\right]-1}{\left|z^{*}-z\right|}^{i/4}, \end{gathered}$$

where $M,N>25$ and ${\rho }_2$ are established in Theorem 1.

Proof of Theorem 2.

By definition,

$$\Delta Y_{N+1}\left(z\right)=\left|Y\left(z\right)-Y_N\left(z\right)\right|=\left|\sum _{n=N+1}^{\infty }{C}_n{(z^{*}-z)}^{(n-15)/4}\right|.$$

Let us set the integer index j:

$$j=\left[{\displaystyle \frac{N+1}{15}}\right].$$

(14)

Taking into consideration the coefficients $C_n$, we obtain

$$\begin{gathered} \Delta Y_{N+1}\left(z\right)=\left|\sum _{n=N+1}^{\infty }{C}_n{(z^{*}-z)}^{(n-15)/4}\right|=\left|\sum _{k=j}^{\infty }\sum _{i=N+1-15l}^{14}{C}_{15(k-1)+i}{(z^{*}-z)}^{\left(15\right(k-1)+i)/4}\right. \\ +\left.\sum _{k=j+1}^{\infty }\sum _{i=0}^{N+1-15j-1}{C}_{15(k-1)+i}{(z^{*}-z)}^{\left(15\right(k-1)+i)/4}\right| \\ =\left|\sum _{i=N+1-15j}^{14}\sum _{k=j}^{\infty }{C}_{15(k-1)+i}{(z^{*}-z)}^{\left(15\right(k-1)+i)/4}\right. \\ +\left.\sum _{i=0}^{N+1-15j-1}\sum _{k=j+1}^{\infty }{C}_{15(k-1)+i}{(z^{*}-z)}^{\left(15\right(k-1)+i)/4}\right|. \end{gathered}$$

Then, taking into account the estimate (10) for the coefficients $C_n$, the last one implies that

$$\begin{gathered} \Delta Y_{N+1}\left(z\right)\le \left|\sum _{i=N+1-15j}^{14}{(z^{*}-z)}^{i/4}\sum _{k=j}^{\infty }{\displaystyle \frac{{\left(2(M+1)\right)}^{k-1}{(z^{*}-z)}^{15(k-1)/4}}{(15k-18+i)(15k-22+i)(15k-26+i)/64}}\right| \\ +\left|\sum _{i=0}^{N+1-15j-1}{(z^{*}-z)}^{i/4}\sum _{k=j+1}^{\infty }{\displaystyle \frac{{\left(2(M+1)\right)}^{k-1}{(z^{*}-z)}^{15(k-1)/4}}{(15k-18+i)(15k-22+i)(15k-26+i)/64}}\right| \\ \le {\displaystyle \frac{{\left(2(M+1)\right)}^{j-1}{\left|z^{*}-z\right|}^{15(j-1)/4}}{(15j-18)(15j-22)(15j-26)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}}\sum _{i=N+1-15j}^{14}{\left|z^{*}-z\right|}^{i/4} \\ +{\displaystyle \frac{{\left(2(M+1)\right)}^j{\left|z^{*}-z\right|}^{15j/4}}{(15j-3)(15j-7)(15j-11)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}}\sum _{i=0}^{N+1-15j-1}{\left|z^{*}-z\right|}^{i/4}. \end{gathered}$$

Or, taking into account Formula (14) and $N>25$, we finally obtain

$$\begin{gathered} \Delta y_{N+1}\left(z\right)\le {\displaystyle \frac{{\left(2(M+1)\right)}^{\left[(N+1)/15\right]-1}{\left|z^{*}-z\right|}^{15(\left[(N+1)/15\right]-1)/4}}{(N-17)(N-21)(N-25)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}} \\ \times \sum _{i=N+1-15\left[(N+1)/15\right]}^{14}{\left|z^{*}-z\right|}^{i/4}+{\displaystyle \frac{{\left(2(M+1)\right)}^{\left[(N+1)/15\right]}{\left|z^{*}-z\right|}^{15\left[(N+1)/15\right]/4}}{(N-2)(N-6)(N-10)/64(1-2(M+1){\left|z^{*}-z\right|}^{15/4})}} \\ \times \sum _{i=0}^{N+1-15\left[(N+1)/15\right]-1}{\left|z^{*}-z\right|}^{i/4}. \end{gathered}$$

□

3. Discussion

The Cauchy problem (1) and (2)

$$r\left(z\right)=z^2;\hspace{1em}{z}_0=0.5;\hspace{1em}Y\left(z_0\right)=i;\hspace{1em}{Y}^{\prime }\left(z_0\right)=0.5;\hspace{1em}{Y}^{″}\left(z_0\right)=0.5+0.5i$$

has a moving singular point $z^{*}=3.378175$. The calculations for the approximate analytical solution to (13), where $N=34;$ $Y_{34}\left(z\right)$, were obtained by means of the program complex MATLAB R2024a and are presented in

Table 1. Numerical characteristics for approximate analytical solution $Y_{34}\left(z_1\right)$.

${\mathit{z}}_1$	${\mathit{Y}}_{34}\left({\mathit{z}}_1\right)$	${\mathbf{\Delta }}_1$	${\mathbf{\Delta }}_2$
3.15 + 0.2i	3.14653856 + 1.88920102i	0.01057528	0.0005

. To carry out the calculations, the expressions for the coefficients $C_n$ mentioned above were used.

Definition 1.

$z_1$—the argument value; $Y_{34}\left(z_1\right)$—the value of an approximate analytical solution (13); ${\Delta }_1$—the value of the a priori error estimate; ${\Delta }_2$—the value of the a posteriori error estimate.

For the a posteriori error, the value of the index N for the analytical approximate solution (13) should be 49. All terms in the structure (13) from 35 to 49, when checked, together do not exceed the value ${\Delta }_2$. Therefore, the value $Y_{34}\left(z_1\right)$ has the accuracy $\epsilon =5\ast {10}^{-4}$.

4. Conclusions

For a third-order nonlinear differential equation that is not solvable in quadratures, a theorem has been proven on the existence of a moving critical pole, as well as the existence and uniqueness of the solution in the vicinity of this singular point. In the course of proving the existence theorem, a formula was obtained for calculating the domain in which the results of the study are valid. An analytical approximate solution is proposed in the vicinity of this singular point, and an a priori error estimate is obtained. Optimization of the a priori error is carried out by using the a posteriori error. The theoretical results are confirmed by numerical experiment calculations.