A generalization and application of the Michell solution for functionally graded circular planes

Abstract

In this article, a fundamental solution for the field equation of functionally graded (FG) circular planes is developed via generalization of the Michell solution for non-axisymmetric elasticity problems. The FG plane as a special case is assumed to have a constant Poisson’s ratio, while the elasticity modulus varies exponentially in the radial direction from a metal base to a ceramic base. The generalized form of the Frobenius method is extended and employed to solve the governing equations and investigate both plane-stress and plane-strain states. The target structural application of this solution has included circular solid or hollow FG cylinders, in which both inner and outer surfaces may be prescribed tractions. To this end, the stress field is first extracted and validated by available solutions for isotropic plane in the literature. Next, the stress field in an intact annular plane under a constant, as well as an arbitrary traction on its inner and outer borders, is shown. Multiple numerical examples comprising the hollow thick-walled cylinder under uniform internal and external pressure loadings, as well as subjected to normal and shear arbitrary tractions, are presented.

Appendix A: Generalized Michell solution for FG materials

A separable solution form to Eq. (6) is assumed as $\varphi \left( r,\theta \right) =f\left( r \right) e^{b\theta }$ (for example, see [11]). Therefore, $f\left( r \right)$ is governed by the following equation:

$$\begin{aligned}&f^{''''}-\frac{2\left( \kappa r\mathrm {-}1 \right) }{r}f^{'''}+\frac{\kappa ^{2}r^{2}-\left( 2-\nu \right) \kappa r-1+2b^{2}}{r^{2}}f^{''}\nonumber \\&\quad +\frac{1+\kappa r-\kappa ^{2}\nu r^{2}-2b^{2}(1+\kappa r)}{r^{3}}f^{'}+\frac{4+3\kappa r-\kappa ^{2}\nu r^{2}+b^{2}}{r^{4}}b^{2}f=0 \end{aligned}$$

(A.1)

where the prime designates differentiation with respect to the argument. By defining $x=\kappa r$ Eq. (A.1) is rewritten as follows.

$$\begin{aligned}&f^{''''}-\frac{2\left( x\mathrm {-}1 \right) }{x}f^{'''}+\frac{x^{2}-\left( 2-\nu \right) x-1+2b^{2}}{x^{2}}f^{''}+\frac{1+x-\nu x^{2}-2b^{2}\left( 1+x \right) }{x^{3}}f^{'}\nonumber \\&\quad +\frac{4+3x-\nu x^{2}+b^{2}}{x^{4}}b^{2}f=0 \end{aligned}$$

(A.2)

A solution for the above equation is given by $f(x)=\sum \limits _{m=0}^\infty a_{m} x^{m+c}$ where the constant c is obtained using a generalized form of the Frobenius method [35] as follows:

$$\begin{aligned} c\left( c-1 \right) \left( c-2 \right) \left( c-3 \right) +p_{0}c\left( c-1 \right) \left( c-2 \right) +q_{0}c\left( c-1 \right) +r_{0}c+s_{0}=0 \end{aligned}$$

(A.3)

where the coefficients are $p_{0}=2,\, q_{0}=-r_{0}=\mathrm {2}b^{2}-1,$ and $s_{0}=b^{2}\left( 4+b^{2} \right) .$ In fact, Eq. (A.3) is another (indicial) form of Eq. (A.2). To find these coefficients, first the coefficients of $f^{'''},f^{''},f^{'}$ and $f^{'}$ in Eq. (A.2) are multiplied by $x,x^{2},x^{3}$ and $x^{4},$ respectively, and then the limits of them are found when $x\rightarrow 0.$ The roots of Eq. (A.3) are $c=\pm ib, c=2\pm ib.$ In the special case under consideration, namely an asymmetric case $\left( b=0 \right) , \quad \varphi \left( r,\theta \right)$ will be only a function of r. Namely, we deal with an axisymmetric solution which is obtained by changing Eq. (3) as $\frac{\partial }{\partial r}\left( -r^{2}\frac{\partial \epsilon _{\theta \theta }}{\partial r} \right) +r\frac{\partial \epsilon _{rr}}{\partial r}=0$ or $\frac{\partial }{\partial r}\left( r\epsilon _{\theta \theta } \right) -\epsilon _{rr}=0.$ Substituting Eqs. (1) into the latter equation and resubstituting Eqs. (5) into the ensuing equation and considering the state of plane stress, we arrive at: $r\mathrm {[}\frac{\partial ^{3}\varphi }{{\partial r}^{3}}\mathrm {-}\nu \frac{\partial }{\partial r}\left( \frac{\partial \varphi }{r\partial r} \right) ]+\mathrm {(}\nu +1-\kappa r)\frac{\partial ^{2}\varphi }{{\partial r}^{2}}\mathrm {-(}\nu +1-\nu \kappa r)\frac{\partial \varphi }{r\partial r}=0.$ By assuming $x=\kappa r\mathrm {,}\varphi =f(x)$ and $f^{'}\left( x \right) =g\left( x \right) ,$ the following second-order differential equation is obtained:

$$\begin{aligned} x^{2}g^{''}+x(1-x)g^{'}+(\nu x-1)g=0 \end{aligned}$$

(A.4)

Using the Frobenius method, the characteristic equation is reduced to $c^{2}-1=0$ ($c^{2}-1$ is an indicial polynomial.). Therefore, we arrive at $c=1,-1.$ Consequently according to the Frobenius method, for the first solution of Eq. (A.4), we have $g\left( x \right) =\sum \nolimits _{m=0}^\infty b_{m} x^{m+1},$ where $b_{m}$ are unknown coefficients. Substituting of $g\left( x \right) =\sum \nolimits _{m=0}^\infty b_{m} x^{m+1},$ into Eq. (A.4) we arrive at:

$$\begin{aligned} \sum \limits _{m=1}^\infty {[b_{m}m\left( m+2 \right) +b_{m-1}\left( \nu -m \right) ]x^{m+1}} =0 \end{aligned}$$

(A.5)

The above equations are satisfied under the condition $b_{m}=\frac{m-\nu }{m\left( m+2 \right) }b_{m-1}$. Solution to this equation is given by$\, b_{m}=b_{0}\frac{2{(-1)}^{m}}{m!\left( m+2 \right) !}\prod \limits _{j=1}^m \left( \nu -j \right) ,m\ge 1.\,$Using the expansion for $g\left( x \right) =\sum \nolimits _{m=0}^\infty b_{m} x^{m+1}$ we have,

$$\begin{aligned} g\left( x \right) =b_{0}x\left[ 1+\sum \nolimits _{m=1}^\infty {\frac{2\left( -1 \right) ^{m}}{m!\left( m+2 \right) !}\prod \limits _{j=1}^m \left( \nu -j \right) } x^{m} \right] =b_{0}x{}_{1}F_{1}\left( 1-\nu ;3;x \right) \end{aligned}$$

(A.6)

where ${}F_{1}\left( 1-\nu ;3;x \right)$ is the confluent hypergeometric function of the first kind.  To find the other solution of Eq. (A.4), we assume $g\left( x \right) =\sum \nolimits _{m=0}^\infty g_{m} x^{m-1}+hlnx\sum \nolimits _{m=0}^\infty {b_{m}x^{m+1}}$ where $g_{m}$ and h are unknown constants. Substituting this solution form into Eq. (A.4) results in:

$$\begin{aligned}&-\,g_{1}+g_{0}\left( \nu +1 \right) +(\nu g_{1}+2hb_{0})x\nonumber \\&\quad \sum \limits _{m=3}^\infty {\{g_{m}m\left( m-2 \right) +g_{m-1}\left( \nu -m+2 \right) +h[2b_{m-2}\left( m-1 \right) -b_{m-3}]\}x^{m-1}} =0 \end{aligned}$$

(A.7)

Solutions to the above equation are found by:

$$\begin{aligned}&g_{0}=\frac{-2hb_{0}}{\nu \left( \nu +1 \right) },g_{1}=-\frac{2hb_{0}}{\nu },g_{3}=hb_{0}\left( \frac{4\nu -1}{9} \right) -\frac{1}{3}g_{2}\left( \nu -1 \right) \nonumber \\&\quad g_{m}=\frac{1}{m\left( m-2 \right) }\{h\left[ b_{m-3}-2b_{m-2}\left( m-1 \right) \right] -g_{m-1}\left( \nu -m+2 \right) \},m\ge 4 \end{aligned}$$

(A.8)

Finally, we have,

$$\begin{aligned}&g\left( x \right) =\frac{-2}{\nu \left( \nu +1 \right) }hb_{0}E_{0}\left( x,\nu \right) +(hb_{0}lnx+g_{2})x{}_{1}F_{1}\left( 1-\nu ;3;x \right) \end{aligned}$$

(A.9)

where

$$\begin{aligned}&E_{0}\left( x,\nu \right) =\sum \limits _{m=0}^\infty {r_{m}x^{m-1}} =x^{-1}+\left( \nu +1 \right) -\nu \left( \nu +1 \right) \left( \frac{4\nu -1}{18} \right) x^{2}+\cdots ,\nonumber \\&\quad r_{0}=1,r_{1}=\left( \nu +1 \right) r_{0},r_{2}=0,\, r_{3}=-\nu \left( \nu +1 \right) \left( \frac{4\nu -1}{18} \right) \nonumber \\&\quad r_{m}=\frac{\nu \left( \nu +1 \right) \left( -1 \right) ^{m}\prod \limits _{j=1}^{m-3} \left( \nu -j \right) [2\nu \left( m-1 \right) -\left( m-2 \right) ^{2}]}{m\left( m-2 \right) m!\left( m-2 \right) !}\nonumber \\&\qquad \quad -\frac{\nu -m+2}{m\left( m-2 \right) }r_{m-1} ,m\ge 4 \end{aligned}$$

(A.10)

Viewing Eqs. (A.6) and (A.9), we realize that Eq. (A.4) has two fundamental solutions as $x{}F_{1}\left( 1-\nu ;\right. \left. 3;x \right)$ and $H_{0}\left( x,\nu \right)$ wherein$\mathrm {\, }H_{0}\left( x,\nu \right) =E_{0}\left( x,\nu \right) -\frac{1}{2}\nu \left( \nu +1 \right) xlnx{}_{1}F_{1}\left( 1-\nu ;3;x \right) .$ After adopting some variables of this study with those of [30], the above first solution, namely$\, x{}F_{1}\left( 1-\nu ;3;x \right)$, could be validated. In the latter reference, the elasticity modulus and the Poisson’s ratio are given by the relations $E_{1}\left( r \right) =E_{0}e^{\frac{mr}{r_{out}}},\nu _{1}\left( r \right) =\nu _{0}\left( 1+\frac{nr}{r_{out}} \right) ,$ respectively. For the case $m\ne 0,\, n=0,$ i.e., constant Poisson’s ratio, by comparing the relation $E\left( r \right) =E_{m}e^{\beta (\frac{r-R_{1}}{R_{2}-R_{1}})}=E_{m}e^{\kappa (r-R_{1})}$ with $E_{1}\left( r \right) =E_{0}e^{\frac{mr}{r_{out}}},$ we can conclude that $E_{0}=E_{m}e^{-\kappa R_{1}}\,$and $\kappa =\frac{m}{r_{out}}.$ Using these relations and considering the relation $x=\kappa r,$ Eq. (41) of [30] is similar to Eq. (A.4) of the present study. Solution for Eq. (41) of [30] is given by Eq. (42c) with its first term being $rU\left( 1-\nu _{0};3;\frac{m}{r_{out}}r \right)$. In [30], the confluent hypergeometric function, ${}_{1}F_{1}\left( 1-\nu ;3;x \right)$ was also presented by the function $U\left( 1-\nu _{0};3;\kappa r \right) .$

In fact, for $n=0$ in view of the relation $f^{'}\left( x \right) =g\left( x \right)$ we have three independent solutions for $f\left( x \right) =constant, f\left( x \right) =\int {x{}F_{1}\left( 1-\nu ;3;x \right) \mathrm{d}x}$, and $f\left( x \right) =\int {H_{0}\left( x,\nu \right) \mathrm{d}x} .$ By virtue of Eqs. (5) and the relation $\varphi \left( r,\theta \right) =f\left( r \right) e^{b\theta },$ which is reduced to $\varphi \left( r,\theta \right) =f\left( r \right)$ for $b=0,$ we find the stress components for $b=0$ as $\sigma _{rr}=\frac{\kappa ^{2}}{x}f^{'}\left( x \right) ,\sigma _{\theta \theta }=\kappa ^{2}f^{''}\left( x \right)$and $\sigma _{r\theta }=0.$ For the case of $\kappa =0,$ Eq. (A.4) leads to $r^{2}g^{''}+rg^{'}-g=0.$Therefore, the fundamental solutions of this equation can be given by $g\left( r \right) =r,\frac{1}{r}.$ According to the relations$\, g\left( x \right) =\mathop {\mathrm {lim}}\limits _{{k}\rightarrow {0}}{H_{0}\left( x,\nu \right) }\sim \frac{1}{x}$ and $g\left( x \right) =\mathop {\mathrm {lim}}\limits _{{k}\rightarrow {0}} {x{}F_{1}\left( 1-\nu ;3;x \right) }\sim x,\,$we conclude that the above-mentioned independent solutions for the case of $\kappa =0$ can also be reobtained.

Now, let us focus on the case of $b\ne 0.$ Substituting of $f=\sum \limits _{m=0}^\infty a_{m} x^{m+c}$ into Eq. (A.2) we arrive at:

$$\begin{aligned}&a_{0}\{\left[ b^{2}+c\left( c-2 \right) \right] ^{2}+4b^{2}\}x^{c}+\{a_{1}\left[ b^{2}+c^{2}-1 \right] ^{2}+4b^{2}\}\nonumber \\&\quad +\,[c(-2c^{2}+4c-1)+(3-2c)\{b^{2}+\nu c\left( c-1 \right) ]a_{0}\}x^{1+c}\nonumber \\&\quad \sum \limits _{m=2}^\infty {\{\{\left[ b^{2}+(m+c)\left( m+c-2 \right) \right] ^{2}+4b^{2}\}} a_{m}\nonumber \\&\quad +\,\{[1-2\left( m+c-2 \right) ^{2}]\left( m+c-1 \right) +[5-2\left( m+c \right) ]b^{2}\nonumber \\&\quad +\,\nu \left( m+c-1 \right) \left( m+c-2 \right) \}a_{m-1}\nonumber \\&\quad +\,[\left( m+c-2 \right) \left( m+c-3-\nu \right) -\nu b^{2}]a_{m-2}\}x^{m+c}=0 \end{aligned}$$

(A.11)

The above equation is satisfied for all values of x by setting the coefficients of $x^{c},\, \, x^{1+c}$ and $x^{m+c}$ being zero. This leads to 3 equations in which c are $c=\pm ib, c=2\pm ib.$ We can have the periodic solution by choosing $b=in, n=0,1,\cdots ,$. Therefore, we have,

$$\begin{aligned}&a_{0}\ne 0,\left\{ {\begin{array}{l} a_{1}\left( 1-4n^{2} \right) +a_{0}n\left( n\pm 1 \right) \left( 1+\nu \right) =0 \\ \alpha _{m,\pm n}\left( a_{m-2},a_{m-1},a_{m} \right) =0,m\ge 2 \\ \end{array}} \right. ,\, for\, c=\mp n\nonumber \\&\quad a_{0}\ne 0,\left\{ {\begin{array}{l} 3a_{1}\left( 3+4n\left( n\mp 2 \right) \right) +a_{0}\left( n\mp 1 \right) \left[ n\left( \nu -7 \right) \mp 2\left( \nu -1 \right) \right] =0 \\ \alpha _{m+2,\pm n}\left( a_{m-2},a_{m-1},a_{m} \right) =0,m\ge 2 \\ \end{array}}, \right. \nonumber \\&\quad \, for\, c=2\mp n \end{aligned}$$

(A.12)

where

$$\begin{aligned}&\alpha _{mn}(x,y,z)=x\{2-m+n+\left( m-2-n \right) ^{2}+\nu [2-m+n\left( 1+n \right) ]\}\nonumber \\&\quad +\,zm\left( m-2 \right) \left( m-2n-2 \right) \left( m-2n \right) -y[(\, 4m{-5)n}^{2}+\left( -\, 6m^{2}+20m-15 \right) n\nonumber \\&\quad +\,(m-1)\, \left( 2m^{2}-8m+7 \right) -\left( m-n-1 \right) \left( m-n-2 \right) \nu ] \end{aligned}$$

(A.13)

wherein $\alpha _{m+2n,n}\left( x,y,z \right) =\alpha _{m,-n}\left( x,y,z \right) .$ First, the special cases $n=0,1$ are investigated. Solution to these special cases is given by:

$$\begin{aligned}&for\, n=0,\nonumber \\&\quad \left\{ {\begin{array}{l} a_{0}\ne 0,a_{1}=0,a_{2}\ne 0,a_{m}=\frac{4\left( -1 \right) ^{m}\prod \nolimits _{j=1}^{m-2} \left( \nu -j \right) }{m\left( m-2 \right) !m!}a_{2},for\, m\ge 3,c=0\, \, \, \, \, \, \, \, \, \\ a_{0}\ne 0,a_{m}=\frac{{4\left( -1 \right) }^{m}\prod \nolimits _{j=1}^m \left( \nu -j \right) }{\left( m+2 \right) m!\left( m+2 \right) !}a_{0},\, for\, m\ge 1,c=2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ \, a_{0}\ne 0,a_{m}=0,\, for\, m\ge 1,c=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ \end{array}} \right. \, \nonumber \\&\quad for\, n=1,\nonumber \\&\quad \left\{ {\begin{array}{l} a_{0}=a_{1}=a_{3}=0,a_{2},a_{4}\ne 0,a_{m}=\frac{48\left( -1 \right) ^{m}\prod \nolimits _{j=3}^{m-2} \left( \nu -j \right) }{\left( m-2 \right) \left( m-4 \right) !m!}a_{4},for\, m\ge 5,c=- 1\\ a_{0},a_{1}=0,a_{2}\ne 0,a_{m}=\frac{48\left( -1 \right) ^{m}\prod \nolimits _{j=3}^m \left( \nu -j \right) }{m\left( m-2 \right) !\left( m+2 \right) !}a_{2},\, for\, m\ge 3,c=1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ a_{0}\ne 0,a_{m}=\frac{48\left( -1 \right) ^{m}\prod \nolimits _{j=3}^{m+2} \left( \nu -j \right) }{\left( m+2 \right) m!\left( m+4 \right) !}a_{0},\, for\, m\ge 1,c=3\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ a_{0}\ne 0,a_{m}=0,\, for\, m\ge 1,c=1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ \end{array}} \right. \end{aligned}$$

(A.14)

Using the expansion    $f=\sum \nolimits _{m=0}^\infty a_{m} x^{m+c}$and presenting only the fundamental solutions, we have

$$\begin{aligned}&f=\{1,x^{2}\,\,_{2}F_{2}\left( 2,1-\nu ;3,3;x \right) \},n=0\nonumber \\&\quad f=\{x,12x^{-1}\int \int {x^{2}\,\,_{2}F_{2}\left( 2,3-\nu ;3,3;x \right) \mathrm{d}x}\mathrm{d}x\},n=1 \end{aligned}$$

(A.15)

where

$$\begin{aligned}&12x^{-1}\int \int {x^{2}\,\,_{2}F_{2}\left( 2,3-\nu ;3,3;x \right) \mathrm{d}x} \mathrm{d}x\nonumber \\&\quad =x^{3}\left[ 1+\sum \limits _{m=1}^\infty {\frac{48\left( -1 \right) ^{m}\prod \nolimits _{j=3}^{m+2} \left( \nu -j \right) }{\left( m+2 \right) m!\left( m+4 \right) !}x^{m}} \right] \end{aligned}$$

(A.16)

According to the relations $\int {x_{1} F_{1}\left( 1-\nu ;3;x \right) \mathrm{d}x} =\frac{x^{2}}{2}_ {2}F_{2}\left( 2,1-\nu ;3,3;x \right)$ and $f^{'}\left( x \right) =g\left( x \right) ,$ the solution for $n=0,$ given in Eq. (A.6) is obtained again and validated. In other words, the derivative of the solution of f(x) for $n=0$ i.e., $\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{x^{2}}{2}_{2}F_{2}\left( 2,1-\nu ;3,3;x \right) \right\} =x_{1}F_{1}\left( 1-\nu ;3;x \right)$ is a solution for $g\left( x \right) \,$ which is given in Eq. (12). Viewing Eqs. (A.12), we can conclude that in some cases, i.e., for $m=2, m=2n, m=2\left( n-1 \right)$ and $m=2\left( n+1 \right) ,$ the multiplier of $a_{m}$ in the recursive equation (or the coefficient of z in Eq. (A.13)) may be zero and consequently the coefficient $a_{m}$ cannot be computed directly. For the case of $n\ge 2,$ in light of Eqs. (A.12), we can see a relation between the coefficients $a_{0}$ and $a_{1}$. Besides for $m=2$ the recursive equation (i.e., the relation $\alpha _{m,\pm n}\left( a_{m-2},a_{m-1},a_{m} \right) =0$ of Eqs. (A.12)) yields another relation between the coefficient $a_{0}$ and $a_{1}$ which leads to $a_{0}=a_{1}=0.$ Also in the case of $c=-n$, the multipliers of the coefficient of $a_{2n}$ and $a_{2n+2}$ in the recursive equation will be zero and thus there will be a relation between the pair coefficients $(a_{2n-1},a_{2n-2})$ and $(a_{2n+1},a_{2n}).$ Finally according to the linear relationship between $a_{m-2},a_{m-1},a_{m}$, i.e., $\alpha _{m,\pm n}\left( a_{m-2},a_{m-1},a_{m} \right) =\alpha _{m,\pm n}\left( 1,0,0 \right) a_{m-2}+\alpha _{m,\pm n}\left( 0,1,0 \right) a_{m-1}+\alpha _{m,\pm n}\left( 0,0,1 \right) a_{m},$ we find:

$$\begin{aligned}&for\, c=-n\, ,a_{0}=a_{1}=a_{2}=a_{3}=\cdots =a_{2n+1}=0,a_{2n+2}\ne 0,\nonumber \\&\quad a_{2n+3}=-\frac{\alpha _{2n+3,n}\left( 0,1,0 \right) }{\alpha _{2n+3,n}\left( 0,0,1 \right) }a_{2n+2},\nonumber \\&\quad a_{2n+4}=\left[ \frac{\alpha _{2n+4,n}\left( 0,1,0 \right) }{\alpha _{2n+4,n}\left( 0,0,1 \right) }\frac{\alpha _{2n+3,n}\left( 0,1,0 \right) }{\alpha _{2n+3,n}\left( 0,0,1 \right) }-\frac{\alpha _{2n+4,n}\left( 1,0,0 \right) }{\alpha _{2n+4,n}\left( 0,0,1 \right) } \right] a_{2n+2},\cdots \nonumber \\&\quad for\, c=n\, ,a_{0}=a_{1}=,a_{2}\ne 0,a_{3}=-\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }a_{2},\nonumber \\&\quad a_{4}=\left[ \frac{\alpha _{4,-n}\left( 0,1,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) }\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }-\frac{\alpha _{4,-n}\left( 1,0,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) } \right] a_{2},\cdots \nonumber \\&\quad for\, c=2-n\, ,\, a_{0}=a_{1}=a_{2}=a_{3}=\cdots =a_{2n-1}=0,a_{2n}\ne 0,\nonumber \\&\quad a_{2n+1}=-\frac{\alpha _{2n+3,n}\left( 0,1,0 \right) }{\alpha _{2n+3,n}\left( 0,0,1 \right) }a_{2n},\nonumber \\&\quad a_{2n+2}=\left[ \frac{\alpha _{2n+4,n}\left( 0,1,0 \right) }{\alpha _{2n+4,n}\left( 0,0,1 \right) }\frac{\alpha _{2n+3,n}\left( 0,1,0 \right) }{\alpha _{2n+3,n}\left( 0,0,1 \right) }-\frac{\alpha _{2n+4,n}\left( 1,0,0 \right) }{\alpha _{2n+4,n}\left( 0,0,1 \right) } \right] a_{2n},\cdots \nonumber \\&\quad for\, c=2+n,a_{0}\ne 0,a_{1}=-\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }a_{0},\nonumber \\&\quad a_{2}=\left[ \frac{\alpha _{4,-n}\left( 0,1,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) }\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }-\frac{\alpha _{4,-n}\left( 1,0,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) } \right] a_{0},\cdots \end{aligned}$$

(A.17)

According to the relation $\alpha _{m+2n,n}\left( x,y,z \right) =\alpha _{m,-n}\left( x,y,z \right) ,$ the above solutions are the same. In other words, the coefficients for each value of c are the identical multipliers of $a_{2n+2},a_{2},a_{2n},a_{0}$ for the cases $c=-n,n,2-n,2+n,$ respectively. Using the expansion    $f=\sum \nolimits _{m=0}^\infty a_{m} x^{m+c}$ the fundamental solution can be written as $x^{n+2}F_{n}\left( x,\nu \right)$ in which

$$\begin{aligned}&F_{n}\left( x,\nu \right) =1-\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }x+\left[ \frac{\alpha _{4,-n}\left( 0,1,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) }\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }-\frac{\alpha _{4,-n}\left( 1,0,0 \right) }{\alpha _{4,-n}\left( 0,0,1 \right) } \right] x^{2}+\cdots \nonumber \\&\quad =\sum \limits _{m=0}^\infty k_{m} x^{m}=1+\frac{\left( 1+n \right) \left( 7n+2-\left( 2+n \right) \nu \right) }{\left( 2n+3 \right) \left( 2n+1 \right) 3}x+\cdots \end{aligned}$$

(A.18)

Other coefficients of the above function are computed using the recursive equation $k_{m}=-\frac{\alpha _{m+2,-n}\left( 1,0,0 \right) }{\alpha _{m+2,-n}\left( 0,0,1 \right) } k_{m-2}-\frac{\alpha _{m+2,-n}\left( 0,1,0 \right) }{\alpha _{m+2,-n}\left( 0,0,1 \right) }k_{m-1},m\ge 2$ in which $k_{0}=1,k_{1}=-\frac{\alpha _{3,-n}\left( 0,1,0 \right) }{\alpha _{3,-n}\left( 0,0,1 \right) }.$ Comparing the $F_{0}\left( x,\nu \right)$ with the solution for $n=0$, i.e., function ${}_{2}F_{2}\left( 2,1-\nu ;3,3;x \right)$ shows that they are exactly identical. Also comparing $x^{3}F_{1}\left( x,\nu \right)$ with the solution for $n=1,$ that is, Eqs. (A.15) and (A.16) prove the correctness of the above solution for special cases. It is worth mentioning that for case $\kappa =0$ by assuming $f=\sum \limits _{m=0}^\infty a_{m} r^{m+c}$and substituting into Eq. (A.1) we find that all coefficients $a_{m}$ are vanished except for $a_{0}\ne 0$ and also we have$\left[ -n^{2}+c\left( c-2 \right) \right] ^{2}-4n^{2}=0.$ Therefore, the fundamental solutions are given by $f=\left\{ r^{n},r^{-n},r^{2+n},r^{2-n} \right\} .$ To obtain the other solutions of Eq. (A.2) for $b=in,n\ne 0,$ we assume $f\left( x \right) =V(x)x^{n+2}F_{n}\left( x,\nu \right)$. In fact, we find the unknown function V(x) as a multiplier of the former solution by constructing a new differential equation which its order is less than the governing equation for $f\left( x \right) .$ This technique will be used through this text repeatedly. By substituting $f\left( x \right) =V(x)x^{n+2}F_{n}\left( x,\nu \right)$ into Eq. (A.1), we arrive at:

$$\begin{aligned}&x^{3}F_{n}\left( x,\nu \right) U^{'''}+2x^{2}2x\left[ F_{n}\left( x,\nu \right) \right] ^{'}+\left( 1+2\left( n+2 \right) -x \right) F_{n}\left( x,\nu \right) U^{''}\nonumber \\&\quad +\,x6x^{2}\left[ F_{n}\left( x,\nu \right) \right] ^{''}+\left( -6x+12n+30 \right) x\left[ F_{n}\left( x,\nu \right) \right] ^{'}\nonumber \\&\quad +\,\{x^{2}-\left( 6n+14-\nu \right) x+(4n^{2}+24n+23)\}F_{n}\left( x,\nu \right) U^{'}\nonumber \\&\quad +\,\{4x^{3}\left[ F_{n}\left( x,\nu \right) \right] ^{'''}+\left( -6x+12n+30 \right) x^{2}\left[ F_{n}\left( x,\nu \right) \right] ^{''}\nonumber \\&\quad +\,2[x^{2}-\left( 6n+14-\nu \right) x+{4n}^{2}+24n+23]x\left[ F_{n}\left( x,\nu \right) \right] ^{'}\nonumber \\&\quad +\,\{{\left( 2n+4-\nu \right) x}^{2}+\left( {-4n}^{2}-22n-19+2\nu \left( n+2 \right) \right) x\nonumber \\&\quad +\,3(4n^{2}+8n+3)\}F_{n}\left( x,\nu \right) U=0 \quad \end{aligned}$$

(A.19)

where $V^{'}=U.$ By substituting the expansion of $F_{n}\left( x,\nu \right) =\sum \nolimits _{m=0}^\infty k_{m} x^{m}$ into the above equation we arrive at $x^{3}\sum \nolimits _{l=0}^\infty {k_{l}x^{l}} U^{'''}+x^{2}\sum \nolimits _{l=0}^\infty {\theta _{l}x}^{l} U^{''}+x\sum \nolimits _{l=0}^\infty {\rho _{l}x}^{l} U^{'}+\sum \nolimits _{l=0}^\infty {\delta _{l}x}^{l} U=0$ where

$$\begin{aligned}&\theta _{0}=2\left( 2n+5 \right) ,\theta _{1}=2\left[ \left( 2n+7 \right) k_{1}-1 \right] ,\theta _{l}=2\left[ \left( 2n+5+2l \right) k_{l}-k_{l-1} \right] ,l\ge 2\nonumber \\&\rho _{0}=4n^{2}+24n+23,\rho _{1}=\left( 4n^{2}+36n+53 \right) k_{1}-\left( 6n+14-\nu \right) \nonumber \\&\rho _{2}=\left( 4n^{2}+48n+95 \right) k_{2}-\left( 20+6n-\nu \right) k_{1}+1\nonumber \\&\rho _{l}=\left[ \left( 6l+12n+24 \right) l+4n^{2}+24n+23 \right] k_{l}-\left( 6l+6n+8-\nu \right) k_{l-1}+k_{l-2},\nonumber \\&l\ge 3\nonumber \\&\delta _{0}=3\left( 4n^{2}+8n+3 \right) \nonumber \\&\delta _{1}={-4n}^{2}-22n-19+2\nu \left( n+2 \right) +(20n^{2}+72n+55)k_{1}\nonumber \\&\delta _{2}=2n+4-\nu +\left( 28n^{2}+144n+161 \right) k_{2}+\left[ {-4n}^{2}-34n-47+2\nu \left( n+3 \right) \right] k_{1}\nonumber \\&\delta _{l}=\left[ l^{2}\left( 4l+12n+18 \right) +\left( 8n^{2}+36n+24 \right) l+3\left( 4n^{2}+8n+3 \right) \right] k_{l}+[-6l^{2}\nonumber \\&\qquad +\,2\left( \nu -6n-5 \right) l{-4n}^{2}-10n-3+2\nu \left( n+1 \right) ]k_{l-1}+\left( 2n+2l-\nu \right) k_{l-2},l\ge 3 \end{aligned}$$

(A.20)

Solution to the above equation is assumed to be in the form of $U\left( x \right) =\sum \nolimits _{m=0}^\infty w_{m} x^{m+p}.$ Substituting this solution into the above equation yields:

$$\begin{aligned}&\sum \limits _{q=0}^\infty \sum \limits _{m=0}^q {{\{[k}_{m}\left( q-m+p-2 \right) +\theta _{m}]\left( q-m+p \right) \left( q-m+p-1 \right) } \nonumber \\&\quad +\,\rho _{m}\left( q-m+p \right) +\delta _{m}\}{w_{q-m}x}^{q}=0 \end{aligned}$$

(A.21)

The above relation holds for each value of x under these conditions:

$$\begin{aligned}&{[k}_{0}\left( p-2 \right) +\theta _{0}]p\left( p-1 \right) +\rho _{0}p+\delta _{0}=0\nonumber \\&\quad w_{q}=-\sum \limits _{m=1}^q {{\{[k}_{m}\left( q-m+p-2 \right) +\theta _{m}]\left( q-m+p \right) \left( q-m+p-1 \right) } \nonumber \\&\quad +\,\rho _{m}\left( q-m+p \right) +\delta _{m}\}w_{q-m}/\{{[k}_{0}\left( q+p-2 \right) +\theta _{0}]\left( q+p \right) \left( q+p-1 \right) \nonumber \\&\quad +\,\rho _{0}\left( q+p \right) +\delta _{0}\},q\ge 1\, \nonumber \\&\quad when\, {[k}_{0}\left( q+p-2 \right) +\theta _{0}]\left( q+p \right) \left( q+p-1 \right) +\rho _{0}\left( q+p \right) +\delta _{0}\ne 0 \end{aligned}$$

(A.22)

Another noteworthy point is that when the term ${[k}_{0}\left( q+p-2 \right) +\theta _{0}]\left( q+p \right) \left( q+p-1 \right) +\rho _{0}\left( q+p \right) +\delta _{0}$ is zero, we can have $w_{q}\ne 0.$ The first equation of (A.22) is solved as $p=-3,p=-1-2n,p=-3-2n.$ Substituting p into the second equation of (A.22), we arrive at:

$$\begin{aligned}&w_{q}=\sum \limits _{m=1}^q \gamma _{qmn} w_{q-m},\, q\ge 1,p=-3\nonumber \\&w_{q}=\sum \limits _{m=1}^q \gamma _{q-2n+2,mn} w_{q-m},\, q\ge 1,p=-1-2n\nonumber \\&w_{q}=\sum \limits _{m=1}^q {\gamma _{q-2n,mn}w_{q-m}} ,\, q\ge 1,p=-3-2n \end{aligned}$$

(A.23)

where

$$\begin{aligned} \gamma _{qmn}=&-{\{[k}_{m}\left( q-m-5 \right) +\theta _{m}]\left( q-m-3 \right) \left( q-m-4 \right) +\rho _{m}\left( q-m-3 \right) \nonumber \\&+\,\delta _{m}\}/q(q+2n-2)(q+2n) \end{aligned}$$

(A.24)

As it can be seen for $p=-1-2n,$ the dominator of $\gamma _{q-2n+2,mn}$ or the coefficient of $w_{2n-2}$ in Eq. (A.24) is zero; therefore, we have $w_{2n-2}\ne 0.$ According to Eq. (A.24), the dominator of $\gamma _{q-2n+2,mn}\,$is $(q-2n+2)q(q+2)$ which tends to zero for $q=2n-2.$ Also, for $p=-3-2n$ we conclude that $w_{2},w_{2n}\ne 0.$ To find the new solutions, let us first compare the above solutions with each other. By shifting q for the case$\, p=-3,$ to $q-2n+2,$ we arrive at $w_{q-2n+2}=\sum \limits _{m=1}^q \gamma _{q-2n+2,mn} w_{q-2n+2-m}.$ This equation is similar to the second equation of (A.23) which shows that the $\left( q-2n+2 \right)$th coefficient of case $p=-1-2n$, i.e., $w_{q-2n+2}$ is similar to the qth coefficient of case $p=-3,$ which is, $w_{q}.\,$By setting $w_{0}=w_{1}=\cdots =w_{2n-3}=0$ in the case of $p=-1-2n$ and choosing $w_{2n-2}$ to be equal with $w_{0}$ of case $p=-3,$ we conclude that $w_{q+2n-2}$ of the case $p=-1-2n$ is equal to $w_{q}$ of case $p=-3.$ In other word, these cases give similar solutions with the above explained conditions. Therefore, for $p=-1-2n$ we have no new solution distinct from the solution for $p=-3.$ Similarly, for the case $p=-3-2n$ the solution is similar to that of $p=-3.$ Finally, the new solutions for f(x) by considering $U\left( x \right) =V^{'}(x)$ and $V\left( x \right) =w_{0}E_{n}\left( x,\nu \right)$is given by $f(x)=w_{0}E_{n}\left( x,\nu \right) x^{n+2}F_{n}\left( x,\nu \right) ,\, for\, p=-3$ where

$$\begin{aligned}&E_{n}\left( x,\nu \right) =-\frac{1}{2}x^{-2}-h_{1n}x^{-1}+h_{2n}lnx+\sum \limits _{m=3}^\infty {\frac{h_{mn}}{m-2}x^{m-2}} ,\nonumber \\&\quad h_{0n}=1,h_{qn}=\sum \limits _{m=1}^q \gamma _{qmn} h_{q-m,n},\, q\ge 1,p=-3 \end{aligned}$$

(A.25)

Equation (A.2) is a forth order differential equation with 4 fundamental solutions for $n\ne 0.$ Until now, we have found two fundamental solutions as $x^{n+2}F_{n}\left( x,\nu \right)$ and $x^{n+2}E_{n}\left( x,\nu \right) F_{n}\left( x,\nu \right) .$ For simplification we show $h_{mn}$ with $h_{m}$ in the rest of the paper. To find two additional solutions, we assume $U\left( x \right) =v(x)\sum \nolimits _{m=0}^\infty {h_{m}x^{m-3}}$and substituting this relation into the equation $x^{3}\sum \nolimits _{l=0}^\infty {k_{l}x^{l}} U^{'''}+x^{2}\sum \nolimits _{l=0}^\infty {\theta _{l}x}^{l} U^{''}+x\sum \nolimits _{l=0}^\infty {\rho _{l}x}^{l} U^{'}+\sum \nolimits _{l=0}^\infty {\delta _{l}x}^{l} U=0$ and defining $v^{'}\left( x \right) =u\left( x \right) ,$ we arrive at:

$$\begin{aligned}&u^{''}\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}x^{l+m}} +\left[ 3\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}\left( m-3 \right) x^{l+m-1}} +\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\theta _{l}h_{m}x^{l+m}} \right] u^{'}\nonumber \\&\quad +\,[3\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}\left( m-3 \right) \left( m-4 \right) x^{l+m-2}}\nonumber \\&\quad +2\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\theta _{l}h_{m}\left( m-3 \right) x^{l+m-1}} \nonumber \\&\quad +\,\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\rho _{l}h_{m}x^{l+m}} ]u=0 \end{aligned}$$

(A.26)

As it can be seen again, the order of equation $x^{3}\sum \nolimits _{l=0}^\infty {k_{l}x^{l}} U^{'''}+x^{2}\sum \nolimits _{l=0}^\infty {\theta _{l}x}^{l} U^{''}+x\sum \nolimits _{l=0}^\infty {\rho _{l}x}^{l} U^{'}+\sum \nolimits _{l=0}^\infty {\delta _{l}x}^{l} U=0$ is reduced from 3 to 2 in Eq. (A.26). The above double series appeared as the coefficients of $u^{''},u^{'}$ and u are transformed to single series as:

$$\begin{aligned}&\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}x^{l+m}} =\sum \limits _{s=0}^\infty {\sigma _{s}x^{s}} \nonumber \\&\quad 3\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}\left( m-3 \right) x^{l+m-1}} +\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\theta _{l}h_{m}x^{l+m}} ={{-9h}_{0}k}_{0}x^{-1}+\sum \limits _{s=0}^\infty \tau _{s} x^{s},\nonumber \\&\quad 3\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {k_{l}h_{m}\left( m-3 \right) \left( m-4 \right) x^{l+m-2}} +2\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\theta _{l}h_{m}\left( m-3 \right) x^{l+m-1}} \nonumber \\&\quad +\,\sum \limits _{l=0}^\infty \sum \limits _{m=0}^\infty {\rho _{l}h_{m}x^{l+m}} ={36k_{0}h}_{0}x^{-2}+6\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] x^{-1}+\sum \limits _{s=0}^\infty \varphi _{s} x^{s} \end{aligned}$$

(A.27)

where

$$\begin{aligned} {\sigma _{s}}=&\sum \limits _{l=0}^s k_{l} {h}_{s-l},\, \, \tau _{s}={-9h}_{0}k_{s+1}+\sum \limits _{l=0}^s {{[3h}_{s-l+1}\left( s-l-2 \right) k_{l}+h_{s-l}\theta _{l}]} \nonumber \\ \varphi _{s}=&6\left( 6{k_{s+2}h}_{0}+3{k_{s+1}h}_{1}-{\theta _{s+1}h}_{0} \right) \nonumber \\&+\,\sum \limits _{l=0}^s {[{3k_{l}h}_{s-l+2}\left( s-l-1 \right) \left( s-l-2 \right) +2{\theta _{l}h}_{s-l+1}\left( s-l-2 \right) +{\rho _{l}h}_{s-l}]} \end{aligned}$$

(A.28)

By substituting $u(x)=\sum \limits _{e=0}^\infty u_{e} x^{e+q}$into the above equation, we find:

$$\begin{aligned}&\sum \limits _{e=0}^\infty \sum \limits _{s=0}^\infty {\sigma _{s}u_{e}\left( e+q \right) \left( e+q-1 \right) x^{s+e}} +\sum \limits _{s=0}^\infty \sum \limits _{e=0}^\infty {\tau _{s}u_{e}\left( e+q \right) x^{s+e+1}} \nonumber \\&\quad +\,\sum \limits _{s=0}^\infty {\sum \limits _{e=0}^\infty {u_{e}\varphi _{s}} x^{s+e+2}} +{9k_{0}h}_{0}\sum \limits _{e=0}^\infty {u_{e}\left( 4-e-q \right) x^{e}} \nonumber \\&\quad +\,6\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] \sum \limits _{e=0}^\infty u_{e} x^{e+1}=0 \end{aligned}$$

(A.29)

By defining $e+s=j$ and synchronizing the above series to start at the same index value, Eqs. (A.29) are satisfied under the following conditions:

$$\begin{aligned}&q\left( q-10 \right) +36=0,u_{1}=-\Bigg \{\frac{6\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] +\sigma _{1}q(q-1)+\tau _{0}q}{\sigma _{0}\left[ q\left( q-8 \right) +27 \right] }\Bigg \}u_{0}\nonumber \\&u_{j}=\Bigg \{-\sum \limits _{s=1}^j {\sigma _{s}u_{j-s}\left( j-s+q \right) \left( j-s+q-1 \right) } -\sum \limits _{s=0}^{j-1} {\tau _{s}u_{j-s-1}\left( j-s+q-1 \right) } \nonumber \\&\quad -\,\sum \limits _{s=0}^{j-2} {u_{j-s-2}\varphi _{s}} -6\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] u_{j-1}\Bigg \}/\sigma _{0}\left[ \left( j+q \right) \left( j+q-10 \right) +36 \right] ,\nonumber \\&\quad j\ge 2 \end{aligned}$$

(A.30)

Roots of the indicial polynomial $q\left( q-10 \right) +36$ are two complex numbers as $q=5\pm \sqrt{11} i,$ in which $i=\sqrt{-1} .$ Substituting these values into the above equations, we arrive at:

$$\begin{aligned}&u_{1}=-\,\frac{2\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] (1\mp 2\sqrt{11} i)+3[(23\mp \sqrt{11} i)\sigma _{1}+(3\mp \sqrt{11} i)\tau _{0}]}{15\sigma _{0}}u_{0},\nonumber \\&u_{j}=\bigg \{-\sum \limits _{s=1}^j {\sigma _{s}u_{j-s}\left( j-s+5\pm \sqrt{11} i \right) \left( j-s+4\pm \sqrt{11} i \right) } \nonumber \\&\quad -\,\sum \limits _{s=0}^{j-1} {\tau _{s}u_{j-s-1}\left( j-s+4\pm \sqrt{11} i \right) } -\sum \limits _{s=0}^{j-2} {u_{j-s-2}\varphi _{s}} \nonumber \\&\quad -\,6\left[ 3\left( {k_{0}h}_{1}+2{k_{1}h}_{0} \right) -{\theta _{0}h}_{0} \right] u_{j-1}\bigg \}/\sigma _{0}j\left( j\pm 2\sqrt{11} i \right) ,j\ge 2 \end{aligned}$$

(A.31)

The comparison of the above equations for $q=5\pm \sqrt{11} i$ shows that the coefficients $u_{j},j=1,2,\cdots$for $q=5+\sqrt{11} i$ are conjugates of those for $q=5-\sqrt{11} i.$ According to the previous relations, we know $V^{'}\left( x \right) =U\left( x \right) =v\left( x \right) \sum \nolimits _{m=0}^\infty {h_{m}x^{m-3}} =\sum \nolimits _{s=0}^\infty \frac{u_{s}}{s+q+1} x^{s+q+1}\sum \nolimits _{m=0}^\infty {h_{m}x^{m-3}} .$ By integration of the above relation for $V^{'}\left( x \right)$ with respect to x and using the relation $f\left( x \right) =V\left( x \right) x^{n+2}F_{n}\left( x,\nu \right) ,$ we arrive at:

$$\begin{aligned}&f\left( x \right) =x^{n+2}F_{n}\left( x,\nu \right) \sum \limits _{s=0}^\infty \sum \limits _{m=0}^\infty \frac{h_{m}u_{s}x^{s+q+m-1}}{\left( s+q+1 \right) \left( s+q+m-1 \right) } \nonumber \\&\quad =x^{6\pm \sqrt{11} i+n}F_{n}\left( x,\nu \right) \sum \limits _{e=0}^\infty \sum \limits _{m=0}^e \frac{h_{m}u_{e-m}x^{e}}{\left( e-m+6\pm \sqrt{11} i \right) \left( e+4\pm \sqrt{11} i \right) } \end{aligned}$$

(A.32)

Finally, we conclude that two additional fundamental solution will be $K_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right)$ and $H_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right)$ where

$$\begin{aligned}&K_{n}\left( x,\nu \right) =\sum \limits _{e=0}^\infty \sum \limits _{m=0}^e {h_{m}Real\bigg \{\frac{u_{e-m}x^{e\pm \sqrt{11} i}}{\left( e-m+6\pm \sqrt{11} i \right) \left( e+4\pm \sqrt{11} i \right) }\bigg \}} \nonumber \\&\quad H_{n}\left( x,\nu \right) =\sum \limits _{e=0}^\infty \sum \limits _{m=0}^e {h_{m}Im\bigg \{\frac{u_{e-m}x^{e\pm \sqrt{11} i}}{\left( e-m+6\pm \sqrt{11} i \right) \left( e+4\pm \sqrt{11} i \right) }\bigg \}} \end{aligned}$$

(A.33)

Since the coefficients $u_{j},j=1,2,\cdots$for $q=5+\sqrt{11} i$ are conjugates of those for $q=5-\sqrt{11} i$, both the positive $(+)$ or negative $\left( - \right) \,$signs in the above relations lead to the same fundamental functions. At the end, we conclude that the fundamental solutions of Eq. (A.2) for$\, n\ne 0$ are $x^{n+2}F_{n}\left( x,\nu \right) ,E_{n}\left( x,\nu \right) x^{n+2}F_{n} \left( x,\nu \right) ,K_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right)$ and $H_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right) .$ For $n=1,$ we have two solutions as x and $x^{3}F_{1}\left( x,\nu \right) \mathrm {=}x^{3}\left[ 1+\sum \nolimits _{m=1}^\infty {\frac{48\left( -1 \right) ^{m}\prod \nolimits _{j=3}^{m+2} \left( \nu -j \right) }{\left( m+2 \right) m!\left( m+4 \right) !}x^{m}} \right] .$ For $x^{3}F_{1}\left( x,\nu \right)$ or in general for $x^{n+2}F_{n}\left( x,\nu \right)$ we found three new solutions as $E_{n}\left( x,\nu \right) x^{n+2}F_{n}\left( x,\nu \right) ,K_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right)$ and $H_{n}\left( x,\nu \right) x^{n+6}F_{n}\left( x,\nu \right) .$ But for $n=1$ we did not investigate the solution x. Therefore, similar to the procedure used for $n\ne 1$ to obtain the rest of new solutions, we can assume $f\left( x \right) =xW\left( x \right) .$ Substituting this solution into Eq. (A.2) we arrive at:

$$\begin{aligned}&x^{3}U^{'''}+\left( 6-2x \right) x^{2}U^{''}+\left[ x^{2}+\left( \nu -8 \right) x+3 \right] xU^{'}\nonumber \\&\quad +\,[\left( 2-\nu \right) x^{2}+\left( 2\nu -1 \right) x-3]U=0 \end{aligned}$$

(A.34)

where $U\left( x \right) =W^{'}(x)$is found by replacing $U\left( x \right) =W^{'}(x)=\sum \nolimits _{m=0}^\infty z_{m} x^{m+c}$ in the above equation. Subsequently,

$$\begin{aligned}&z_{0}\left( c-1 \right) \left( c+1 \right) \left( c+3 \right) x^{c}+\{z_{1}c\left( c+2 \right) \left( c+4 \right) \nonumber \\&\quad +\,z_{0}\left\{ c\left[ \nu -6-2c \right] +\left( 2\nu -1 \right) \right\} \}x^{1+c}\nonumber \\&\quad +\,\sum \limits _{m=2}^\infty {\{z_{m}\left( m+c-1 \right) \left( m+c+1 \right) \left( m+c+3 \right) } \nonumber \\&\quad +\,z_{m-1}\left\{ \left( m+c-1 \right) \left[ \nu -6-2\left( m-1+c \right) \right] +\left( 2\nu -1 \right) \right\} +z_{m-2}\left( m+c-\nu \right) \}x^{m+c} =0 \end{aligned}$$

(A.35)

The above equation is satisfied for $c=1,-1,-3$, and the solutions for $U\left( x \right)$ are, respectively, written as:

$$\begin{aligned}&U\left( x \right) =x\left[ 1-\left( \frac{\nu -3}{5} \right) x+\frac{\left( \nu -3 \right) \left( \nu -4 \right) }{60}x^{2}\cdots \right] \times \left\{ {\begin{array}{l} \, z_{2}\, \, \, \, for\, c=-1 \\ z_{0}\, \, \, \, \, \, \, for\, c=1 \\ \end{array}} \right. ,\nonumber \\&\quad U\left( x \right) =z_{4}(x-\frac{\nu -3}{5}x^{2}+\cdots )\nonumber \\&\quad +\,z_{0}\left[ x^{-3}+\frac{\nu +1}{3}x^{-2}+\frac{2\nu -1}{6}x^{-1}-\frac{1}{6}\left( 1-\nu \right) +\cdots \right] ,\, for\, c=-3, \end{aligned}$$

(A.36)

By integrating the above relations, we find:

$$\begin{aligned}&W\left( x \right) =\frac{1}{2}x^{2}\left[ 1-2\left( \frac{\nu -3}{15} \right) x+\frac{\left( \nu -3 \right) \left( \nu -4 \right) }{120}x^{2}+\cdots \right] \times \left\{ {\begin{array}{l} \, z_{2}\, \, \, \, for\, c=-1 \\ z_{0}\, \, \, \, \, \, \, for\, c=1 \\ \end{array}} \right. \nonumber \\&\quad =\frac{1}{2}x^{2}F_{1}\left( x,\nu \right) \times \left\{ {\begin{array}{l} \, z_{2}\, \, \, \, for\, c=-1 \\ z_{0}\, \, \, \, \, \, \, for\, c=1 \\ \end{array}} \right. \nonumber \\&\quad W\left( x \right) =\frac{1}{2}z_{4}x^{2}F_{1}\left( x,\nu \right) +z_{0}\left[ -\frac{1}{2}x^{-2}-\frac{\nu +1}{3}x^{-1}+\frac{2\nu -1}{6}lnx-\frac{1-\nu }{6}x+\cdots \right] \nonumber \\&\quad =\frac{1}{2}z_{4}x^{2}F_{1}\left( x,\nu \right) +z_{0}L_{1}\left( x,\nu \right) ,\, for\, c=-3 \end{aligned}$$

(A.37)

where

$$\begin{aligned}&L_{1}\left( x,\nu \right) =t_{2}lnx+\sum \limits _{m=0,m\ne 2}^\infty \frac{t_{m}}{m-2} x^{m-2},\nonumber \\&\quad t_{0}=1,t_{1}=\frac{\nu +1}{3},t_{2}=\frac{2\nu -1}{6},t_{3}=-\frac{1}{6}\left( 1-\nu \right) ,t_{4}=0,\nonumber \\&\quad t_{m}=\frac{t_{m-1}\left[ \left( m-4 \right) \left( 2m-2-\nu \right) +\left( 1-2\nu \right) \right] +t_{m-2}\left( 3-m+\nu \right) }{\left( m-4 \right) \left( m-2 \right) m},m\ge 5 \end{aligned}$$

(A.38)

The solution for $c=-1$ and $c=1$, i.e., $f(x)=xW\left( x \right) =x^{3}F_{1}\left( x,\nu \right)$ is not a new solution, while the solution for $c=-3,$ that is, $f\left( x \right) =x(t_{2}lnx+\sum \nolimits _{m=0,m\ne 2}^\infty \frac{t_{m}}{m-2} x^{m-2})\,$is a new solution. To find the final solution for $n=1$, we presume that $U\left( x \right) =w\left( x \right) \sum \nolimits _{m=0}^\infty t_{m} x^{m-3}$ and substitute it into Eq. (A.34) and find an equation in terms of $w^{'}\left( x \right) .$ By solving the ensuing equation with similar previously used procedure, w(x) and consequently $U\left( x \right) , \quad W\left( x \right)$ and $f\left( x \right)$ can be found. Then, the equation for w(x) is given by:

$$\begin{aligned}&x^{2}w^{'''}\sum \limits _{m=0}^\infty t_{m} x^{m}+xw^{''}\left\{ \sum \limits _{m=1}^\infty {[t_{m}\left( 3m-3 \right) {-2t_{m-1}]x}^{m}} -3t_{0} \right\} \nonumber \\&\quad +\,w^{'}\{3t_{0}+\left[ t_{0}\left( \nu +4 \right) -3t_{1} \right] x\nonumber \\&\quad +\,\sum \limits _{m=2}^\infty {{\{3t}_{m}[m\left( m-3 \right) {+1]+t_{m-1}\left( \nu +8-4m \right) +t_{m-2}\}x}^{m}}\} =0 \end{aligned}$$

(A.39)

By assuming an expansion as $w^{'}(x)=\sum \nolimits _{l=0}^\infty p_{l} x^{l+c}$ and substituting it into the above equation, we find:

$$\begin{aligned}&t_{0}p_{0}\left( c-1 \right) \left( c-3 \right) +\{t_{0}p_{1}c\left( c-2 \right) +p_{0}\left\{ t_{1}\left[ c\left( c-1 \right) -3 \right] +t_{0}\left( \nu +4-2c \right) \right\} \}x\nonumber \\&\quad \sum \limits _{s=2}^\infty {\{\sum \limits _{l=0}^s {t_{s-l}p_{l}\left( l+c \right) \left( l+c-1 \right) } +\sum \limits _{l=0}^{s-1} {p_{l}\left( l+c \right) \left[ t_{s-l}\left( 3\left( s-l \right) -3 \right) -2t_{s-l-1} \right] } } \nonumber \\&\quad +\,\sum \limits _{l=0}^{s-2} {p_{l}{\{3t}_{s-l}[(s-l)\left( s-l-3 \right) +1]+t_{s-l-1}\left( \nu +8-4\left( s-l \right) \right) +t_{s-l-2}\}} \nonumber \\&\quad +\,3t_{0}p_{s}\left( 1-s-c \right) +\left[ t_{0}\left( \nu +4 \right) -3t_{1} \right] p_{s-1}\}x^{s}=0 \end{aligned}$$

(A.40)

Solution to the above equation is given by:

$$\begin{aligned}&p_{1}=\left\{ \frac{\left[ 3-c\left( c-1 \right) \right] t_{1}+t_{0}\left[ 2c-\left( \nu +4 \right) \right] }{t_{0}c\left( c-2 \right) } \right\} p_{0},\nonumber \\&\quad t_{0}\left( s+c-1 \right) \left( s+c-3 \right) p_{s}\nonumber \\&\quad =\{-\left[ t_{0}\left( \nu +4 \right) -3t_{1} \right] p_{s-1}-\sum \limits _{l=0}^{s-1} {p_{l}\left( l+c \right) \left[ t_{s-l}\left( c+3s-2l-4 \right) -2t_{s-l-1} \right] } \nonumber \\&\quad -\,\sum \limits _{l=0}^{s-2} {p_{l}{\{3t}_{s-l}[(s-l)\left( s-l-3 \right) +1]+t_{s-l-1}\left( \nu +8-4\left( s-l \right) \right) +t_{s-l-2}\}} ,s\ge 2 \end{aligned}$$

(A.41)

where $c=1$ or $c=3.$ At the end, for both values of $c=1$ $c=3,$ the function w(x) is as $p_{2}x^{4}Y_{1}\left( x,\nu \right)$ and $p_{0}x^{4}Y_{1}\left( x,\nu \right) ,$ respectively, where

$$\begin{aligned}&Y_{1}\left( x,\nu \right) =\frac{1}{4}+\frac{(1-2\nu )}{15}x+\frac{1}{144}\left( 7\nu ^{2}-13\nu +10 \right) x^{2}+\cdots =\sum \limits _{l=0}^\infty \frac{q_{l}}{l+4} x^{l} \end{aligned}$$

(A.42)

where $q_{l}=\frac{p_{l}}{p_{0}}.$ Since $f\left( x \right) =x\int {w\left( x \right) \sum \nolimits _{m=0}^\infty t_{m} x^{m-3}} \mathrm{d}x=x\int {x^{4}Y_{1}\left( x,\nu \right) \sum \nolimits _{m=0}^\infty t_{m} x^{m-3}} \mathrm{d}x$ the new possibly fundamental solution for $n=1$ can be as follows:

$$\begin{aligned} f\left( x \right) =\frac{x^{3}}{2}F_{1}\left( x,\nu \right) =\frac{x^{3}}{2}\{1-2\frac{\nu -3}{15}x+\frac{1}{120}(\nu -3)(\nu -4)x^{2}+\cdots \} \end{aligned}$$

(A.43)

But for $n=1,$ the solution $x^{3}F_{1}\left( x,\nu \right)$ is not an independent solution. To find the last solution to Eq. (A.39) for $n=1,$ for which one of the independent solutions is $w\left( x \right) =x^{4}Y_{1}\left( x,\nu \right) =\sum \nolimits _{l=0}^\infty \frac{q_{l}}{l+4} x^{l+4}$ or $w^{'}\left( x \right) =\sum \nolimits _{l=0}^\infty q_{l} x^{l+3},$ let us assume a new solution as $w^{'}\left( x \right) =p\left( x \right) \sum \nolimits _{l=0}^\infty q_{l} x^{l+3}.$ By substituting this relation into Eq. (A.39), we arrive at:

$$\begin{aligned} q^{'}\left( x \right) \sum \limits _{m=5}^\infty \omega _{m} x^{m}+q\left( x \right) \sum \limits _{m=4}^\infty \mu _{m} x^{m}=0 \end{aligned}$$

(A.44)

where $q\left( x \right) =p^{'}\left( x \right)$ and

$$\begin{aligned}&\sum \limits _{m=5}^\infty \omega _{m} x^{m}=\sum \limits _{l=0}^\infty q_{l} x^{l+5}\sum \limits _{m=0}^\infty t_{m} x^{m}=\sum \limits _{s=5}^\infty \sum \limits _{l=0}^{s-5} {q_{l}t_{s-l-5}x^{s}} \nonumber \\&\quad \sum \limits _{m=4}^\infty \mu _{m} x^{m}=2\sum \limits _{l=0}^\infty q_{l} \left( l+3 \right) x^{l+4}\sum \limits _{m=0}^\infty t_{m} x^{m}+\sum \limits _{l=0}^\infty q_{l} x^{l+4}\{-3t_{0}\nonumber \\&\quad +\,\sum \limits _{m=1}^\infty {[3t_{m}\left( m-1 \right) {-2t_{m-1}]x}^{m}}\} =\sum \limits _{s=4}^\infty {[2\sum \limits _{l=0}^{s-4} {q_{l}t_{s-l-4}\left( l+3 \right) } -3t_{0}q_{s-4}]x^{s}} \nonumber \\&\quad +\,\sum \limits _{s=5}^\infty \sum \limits _{l=0}^{s-5} {q_{l}[t_{s-l-4}\left( 3\left( s-l-4 \right) -3 \right) {-2t_{s-l-5}]x}^{s}} \end{aligned}$$

(A.45)

Therefore, the coefficients $\omega _{m}$ and $\mu _{m}$ are computed by the following relations:

$$\begin{aligned}&\omega _{s}=\sum \limits _{l=0}^{s-5} {q_{l}t_{s-l-5}} ,s\ge 5,\, \, \mu _{4}=3q_{0}t_{0}=3\nonumber \\&\quad \mu _{s}=2\sum \limits _{l=0}^{s-4} {q_{l}t_{s-l-4}\left( l+3 \right) } -3t_{0}q_{s-4}\nonumber \\&\quad +\,\sum \limits _{l=0}^{s-5} {q_{l}[t_{s-l-4}\left( 3\left( s-l-4 \right) -3 \right) -2t_{s-l-5}],s\ge 5} \end{aligned}$$

(A.46)

To solve Eq. (A.44), we assume that $\frac{\sum \nolimits _{m=4}^\infty {\mu _{m}x^{m}} }{\sum \nolimits _{m=5}^\infty {\omega _{m}x^{m}} }=\sum \nolimits _{l=-1}^\infty {\vartheta _{l}x^{l}} .$ Thus, we can easily find that $\mu _{s}=\sum \nolimits _{l=-1}^{s-5} \vartheta _{l} \omega _{s-l},s\ge 4.$ By solving these algebraic equations, the coefficients $\vartheta _{l}$ are found and consequently the solution to Eq. (A.44) is given by $q\left( x \right) =e^{-[\vartheta _{-1}lnx+\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} ]}.$

Division of two infinite power series $\sum \nolimits _{m=4}^\infty {\mu _{m}x^{m}}$ and $\sum \nolimits _{m=5}^\infty {\omega _{m}x^{m}}$ yields an infinite power series like $\sum \nolimits _{l=?}^\infty {\vartheta _{l}x^{l}} .$ The mathematical operations like addition, subtraction, multiplication and division can be applied to the infinite power series of x, and the result is a new infinite power series of x. The lower limit of the $\sum \nolimits _{l=?}^\infty {\vartheta _{l}x^{l}}$ should be -1 because the multiplication of two series $\sum \nolimits _{m=5}^\infty {\omega _{m}x^{m}}$ and $\sum \nolimits _{l=?}^\infty {\vartheta _{l}x^{l}}$ is $\sum \nolimits _{m=4}^\infty {\mu _{m}x^{m}} .$ In fact, multiplication of the first terms of these two series, i.e., $x^{5}$and $x^{-1}$, leads to the first term of series $\sum \nolimits _{m=4}^\infty {\mu _{m}x^{m}},$ i.e., $x^{4}.$ The coefficients of $\vartheta _{l}$ can be computed by solving the equations $\mu _{s}=\sum \nolimits _{l=-1}^{s-5} \vartheta _{l} \omega _{s-l},s\ge 4.$ This equation is similar to the equations given in [36] to compute multiplication of two infinite power series.

And finally, we have $p\left( x \right) =\int {x^{-\vartheta _{-1}}e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }\mathrm{d}x} .$ After that using the relation $w^{'}\left( x \right) =p\left( x \right) \sum \nolimits _{l=0}^\infty q_{l} x^{l+3}$ we get $w^{'}\left( x \right) =\sum \nolimits _{l=0}^\infty q_{l} x^{l+3}\int {x^{-\vartheta _{-1}}e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }\mathrm{d}x} .$ Employing the relation $W^{'}\left( x \right) =U\left( x \right) =w\left( x \right) \sum \limits _{m=0}^\infty t_{m} x^{m-3}$ and integration with respect to $x,\,$the function $U\left( x \right)$ and subsequently $f\left( x \right)$ is provided as

$$\begin{aligned}&f\left( x \right) =x\sum \limits _{m=0}^\infty t_{m} \sum \limits _{l=0}^\infty q_{l} \int {\left( x^{m-3}\left( \int {x^{l+3}\left( \int {x^{-\vartheta _{-1}}e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }\mathrm{d}x} \right) \mathrm{d}x} \right) \right) \mathrm{d}x} \nonumber \\&\quad =xZ_{1}\left( x,\nu \right) \end{aligned}$$

(A.47)

To compute the function $Z_{1}\left( x,\nu \right)$, we consider the following Taylor’s expansion

$$\begin{aligned} e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }=\sum \limits _{s=0}^\infty \psi _{s} x^{s} \end{aligned}$$

(A.48)

where

$$\begin{aligned} \psi _{s}=-\sum \limits _{r=0}^{s-1} {\frac{\vartheta _{s-r-1}}{s}\psi _{r}} ,s\ge 1,\psi _{0}=1 \end{aligned}$$

(A.49)

Replacing Eq. (A.48) into Eq. (A.47) and knowing that $\vartheta _{-1}=3,$ we arrive

$$\begin{aligned}&Z_{1}\left( x,\nu \right) =\sum \limits _{m=0}^\infty t_{m} \sum \limits _{l=0}^\infty {q_{l}\{\psi _{2}[lnx-\frac{1}{m+l+2}-\frac{1}{l+4}]\frac{x^{m+l+2}}{\left( l+4 \right) \left( m+l+2 \right) }} \nonumber \\&\quad +\,\sum \limits _{s=0,s\ne 2}^\infty \psi _{s} \frac{x^{m+l+s}}{\left( s-2 \right) \left( m+l+s \right) \left( l+s+2 \right) }\} \end{aligned}$$

(A.50)

For proof of (A.48), please see Appendix B. Finally, for $n=1$ we conclude that the fundamental solutions are $x,x^{3}F_{1}\left( x,\nu \right) ,\, {xL}_{1}\left( x,\nu \right)$ and $xZ_{1}\left( x,\nu \right) .$ It can be shown that the solution $x^{3}F_{1}\left( x,\nu \right)$ produces the single-valued displacement components, while the solutions ${xL}_{1}\left( x,\nu \right)$ and $xZ_{1}\left( x,\nu \right)$ provide the multiple-valued displacement components.

The final point can be realized by viewing Eqs. (A.3). By sorting this equation in terms of b as $b^{4}+2\left[ c\left( c-2 \right) +2 \right] b^{2}+\left[ c\left( c-2 \right) \right] ^{2}=0,$ we conclude that the repeated roots for b is found for ${[c\left( c-2 \right) +2]}^{2}-\left[ c\left( c-2 \right) \right] ^{2}=0.$ In other words, for $c=1$ we have two repeated roots for b as $b=+i,+i,-i,-i.$ Also for $c=0$ or $c=2$ the values $b=0,0$ can be repeated roots for b. According to the relation $\varphi \left( r,\theta \right) =f\left( r \right) e^{b\theta },$ for $b=0,0$ we consider the solutions $\varphi =f\left( r \right)$ and $\varphi =f\left( r \right) \theta$ (the term $\theta$ is considered because of repeated root for b). Moreover, for $b=+i,+i,-i,-i$ viewing the relation $\varphi \left( r,\theta \right) =f\left( r \right) e^{b\theta },$ we consider the solutions $\varphi =f\left( r \right) \cos \theta ,f\left( r \right) \sin \theta ,f\left( r \right) \theta \cos \theta$ and $\varphi =f\left( r \right) \theta \sin \theta .$ By substituting these relations into Eq. (6), two equations are attained including Eq. (A.1) and the equation $2r^{2}\frac{\partial ^{2}f}{{\partial r}^{2}}-2r(1+\kappa r)\frac{\partial f}{\partial r}+(2+3\kappa r-\kappa ^{2}\nu r^{2})f=0.$ For $\kappa \ne 0,\,$there is no common solution for Eq. (A.1) and this equation. But for $\kappa =0,$ the common solutions are r and lnr,  respectively. Therefore, for $\kappa =0,$ we have four new solutions as $\varphi =r\theta \cos \theta ,\varphi =r\theta \sin \theta ,\varphi =\theta \cos \theta rlnr$ and $\varphi =\theta \sin \theta rlnr.$ Finally, viewing Eqs. (A.9), (A.15), the solution $E_{n}\left( x,\nu \right) x^{n+2}F_{n}\left( x,\nu \right) ,$ Eqs. (A.33), (A.38) and (A.47) the complete Michell solution to the functionally graded materials is given by Eq. (7).

Appendix B: Proof of Eq. (A.48)

“The Taylor’s expansion” of the function $e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }$ is written as $=\sum \nolimits _{s=0}^\infty \psi _{s} x^{s}$where $\psi _{s}=\frac{1}{s!}\left. \frac{\mathrm{d}^{s}}{\mathrm{d}x^{s}}\left( \mathrm {\, }e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} } \right) \right| _{x=0}.$ The s-th derivative of the function $e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }$ can be found by using “Leibniz’s rule” as:

$$\begin{aligned}&\left. \frac{\mathrm{d}^{s}}{\mathrm{d}x^{s}}\left( \mathrm {\, }e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} } \right) \right| _{x=0}=-\left. \frac{\mathrm{d}^{s-1}}{\mathrm{d}x^{s-1}}\left( e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} }\mathrm {\, }\sum \limits _{l=0}^\infty {\vartheta _{l}x^{l}} \right) \right| _{x=0} \\&\quad =-\sum \limits _{r=0}^{s-1} {\left( {\begin{array}{l} s-1 \\ r \\ \end{array}} \right) \left. \frac{\mathrm{d}^{s-1-r}}{\mathrm{d}x^{s-1-r}}\left( \mathrm {\, }\sum \limits _{l=0}^\infty {\vartheta _{l}x^{l}} \right) \right| _{x=0}\left. \frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}\left( \mathrm {\, \, }e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} } \right) \right| _{x=0}} \\&\quad =-\sum \limits _{r=0}^{s-1} {\left( {\begin{array}{l} s-1 \\ r \\ \end{array}} \right) \left. \left( \mathrm {\, }\sum \limits _{l=0}^\infty {\vartheta _{l}\frac{\mathrm{d}^{s-1-r}}{\mathrm{d}x^{s-1-r}}x^{l}} \right) \right| _{x=0}r!\psi _{r}} \\&\quad =-\sum \limits _{r=0}^{s-1} {\left( {\begin{array}{l} s-1 \\ r \\ \end{array}} \right) \left. \left( \mathrm {\, }\sum \limits _{l=0}^\infty {\vartheta _{l}\frac{l!}{\left( l-(s-1-r) \right) !}x^{l-(s-1-r)}} \right) \right| _{x=0}r!\psi _{r}} \\&\quad =-\sum \limits _{r=0}^{s-1} {\left( {\begin{array}{l} s-1 \\ r \\ \end{array}} \right) \vartheta _{s-1-r}\left( s-1-r \right) !r!\psi _{r}} =-\left( s-1 \right) !\sum \limits _{r=0}^{s-1} {\vartheta _{s-1-r}\psi _{r}} \end{aligned}$$

By substituting the above relation into the relation $\psi _{s}=\frac{1}{s!}\left. \frac{\mathrm{d}^{s}}{\mathrm{d}x^{s}}\left( \mathrm {\, }e^{-\sum \nolimits _{l=0}^\infty {\frac{\vartheta _{l}}{l+1}x^{l+1}} } \right) \right| _{x=0},$ relation (A.49) is obtained and accordingly Eq. (A.48) is proven.