# Preface

This section provides an introduction to a very important class of ordinary differential equations and their solutions.

Introduction to Linear Algebra with Mathematica

# Hypergeometric Functions

The differential equation \begin{equation} \label{E66.1} x(1-x) \,y'' + [\gamma - (\alpha + \beta +1)x]\,y' - \alpha\beta\,y =0, %\eqno{(6.1)}$$\end{equation} \index{Hypergeometric equation}% in which \alpha, \beta, and \gamma are parameters, is called the {\bf hypergeometric equation}. Its singular points x=0, x=1, and x=\infty are all regular. For Eq.\;\eqref{E66.1}, suppose we attempt a power series solution of the Frobenius form \index{Frobenius series}% $y(x) = x^m \,\sum_{k=0}^\infty \,a_k x^k = \sum_{k\geq 0} \,a_k x^{k+m} .$ We substitute the power-series form into the differential equation \eqref{E66.1} to obtain$$ \sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m-1} - \sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m}  +\gamma \,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m-1} - (\alpha + \beta + 1)\,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m} -\alpha \beta \,\sum_{k\geq 0} \,a_k x^{k+m} =0. $$In two series that contain powers x^{k+m-1}, we shift the index by setting k-1=j, then k=j+1. When k=0, the index j is -1, and we keep this term out of the sum. This yields$$ a_0 m(m-1) x^{m-1} + \gamma a_0 m x^{m-1} + \sum_{j\geq 0} \,a_{j+1} (j+1+m)(j+m) \,x^{j+m}  + \gamma \,\sum_{j\geq 0} \,a_{j+1} (j+1+m) \,x^{j+m} - \sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m}  - (\alpha + \beta + 1)\,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m} -\alpha \beta \,\sum_{k\geq 0} \,a_k x^{k+m} =0. $$We equate the coefficients of the power m-1 to zero. This leads to a_0 m\, [m-1 +\gamma ] =0. Assuming that a_0 \neq 0, we get$$ m(m-1+\gamma )=0. $$This equation has two roots$$m_1 =0 \quad\mbox{and}\quad m_2 = \gamma -1 .$$We denote the corresponding solution of Eq.\;\eqref{E66.1} by y_1 (x) and y_2 (x). For m= m_1 =0 we have$$ y(x) = \sum_{k\geq 0} \,a_k \, x^k , \quad y' (x) = \sum_{k\geq 1} \,k a_k \, x^{k-1} , \quad y'' (x) = \sum_{k\geq 2} \,k(k-1) a_k \, x^{k-2} . $$We substitute these power series into Eq.\;\eqref{E66.1} to obtain$$ \sum_{k\geq 1} \,a_{k+1} \, k(k+1)\, x^k - \sum_{k\geq 2} \,a_k \, k(k-1)\, x^k + \gamma \,\sum_{k\geq 0} \,a_{k+1} \, (k+1)\, x^k  -(\alpha + \beta +1)\,\sum_{k\geq 1} \,a_k \, k\, x^k - \alpha \beta\, \sum_{k\geq 0} \,a_k \, x^k =0 . $$Next, we equate the coefficients of each power of x to zero. This yields$$ \gamma a_1 - \alpha\beta \,a_0 =0 ,  2 a_2 + \gamma 2 a_2 - (\alpha + \beta +1) a_1 - \alpha \beta a_1 =0\quad\mbox{or}\quad 2 (\gamma +1) a_2 - (\alpha +1)(\beta +1) a_1 =0 ,  3(\gamma +2) \,a_3 - (\alpha +2)(\beta +2)\,a_2 =0 , \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots  (k+1)(\gamma +k) \,a_{k+1} - (\alpha +k)(\beta +k) a_k =0 . $$If \gamma \neq 0, -1, -2, \ldots ,-k, \ldots and a_0 =1 we obtain the following relations: \begin{eqnarray*} a_1 &=& \frac{\alpha\beta}{\gamma} , \\ a_2 &=& \frac{(\alpha +1)(\beta +1)}{2(\gamma +1)} \,a_1 = \frac{\alpha (\alpha +1)\beta (\beta +1)}{1\cdot 2 \gamma (\gamma +1)} , \\ a_3 &=& \frac{(\alpha +2)(\beta +2)}{2(\gamma +2)} \, a_2 = \frac{\alpha (\alpha +1)(\alpha +2)\beta (\beta +1)(\beta +2)}{1\cdot 2 \cdot 3 \gamma (\gamma +1)(\gamma +2)} , \\ &&\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ a_{k+1} &=& \frac{(\alpha +k)(\beta +k)}{(k+1)(\gamma +k)} \,a_k \\ &=& \frac{\alpha (\alpha +1)(\alpha +2)\cdots (\alpha +k)\beta (\beta +1)(\beta +2)\cdots (\beta +k)}{1\cdot 2 \cdot 3 \cdots (k+1)\gamma (\gamma +1)(\gamma +2)\cdots (\gamma +k)} . \end{eqnarray*} The power series with these coefficients, when a_0 is assigned the value 1, is denoted by F(\alpha , \beta ; \gamma ; x) and is called the {\bf hypergeometric function}: \index{Hypergeometric function}% \index{Function!hypergeometric}% \begin{equation} \label{E66.4} F(\alpha , \beta ; \gamma ; x) = 1 + \sum_{k\geq 0} \, \frac{\alpha (\alpha +1)(\alpha +2)\cdots (\alpha +k)\beta (\beta +1)(\beta +2)\cdots (\beta +k)}{(k+1)! \,\gamma (\gamma +1)(\gamma +2)\cdots (\gamma +k)} \, x^{k+1} . %\eqno{(6.4)}$$ \end{equation} Sometimes the hypergeometric function is denoted by $_2F_1 (\alpha , \beta ; \gamma ; x)$ or $F\left(\left.\pile{\alpha ,\beta }{\gamma }\right| x\right)$. To determine the domain of convergence of the power series \eqref{E66.4}, we apply the ratio test to obtain $$\lim_{k\rightarrow \infty} \,\left\vert \frac{u_{k+1}}{u_k} \right\vert = \lim_{k\rightarrow \infty} \,\left\vert \frac{(\alpha +k)(\beta +k)}{(1+k)(\gamma +k)} \, x\right\vert = |x| .$$ Therefore the series \eqref{E66.4} converges\footnote{It is proved in advanced books that the power series \eqref{E66.4} converges absolutely when $x=1$ if $\gamma - \alpha -\beta >0$, and diverges if $\gamma - \alpha -\beta \leq 0$. When $x=-1$ it converges absolutely when $\gamma - \alpha -\beta >0$, it converges when $-1 < \gamma - \alpha -\beta \leq 0$, and it diverges when $\gamma - \alpha -\beta \leq -1$.} (absolutely) when $-1< x < 1$ and diverges for $|x| >1$. \qed To find another solution $y_2 (x)$, we make the substitution $y_2 = x^{1-\gamma}\,u$. Then $u$ is a solution of the differential equation $$x(1-x)\, u'' + [(2-\gamma ) -(\alpha + \beta - 2\gamma +3) x]\,u' - (\alpha +1 -\gamma )(\beta +1 -\gamma )\,u =0$$which is a hypergeometric equation. Hence $u= F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x)$ and $$y_2 (x) = x^{1-\gamma} \, F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x) .$$ The functions $y_1 (x)$ and $y_2 (x)$ are linearly independent because the corresponding power series contain different powers. Therefore the general solution of Eq.\;\eqref{E66.1} is their linear combination: \begin{equation} \label{E66.5} y(x) = %C_1 \,y_1 (x) + C_2 \,y_2 (x) = C_1 \,F(\alpha , \beta ; \gamma ; x) + C_2 \,x^{1-\gamma} \, F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x) . \end{equation} The coefficients of the hypergeometric series \eqref{E66.4} can be expressed in terms of the gamma-function as \index{Gamma function}% \index{Function!Gamma}% $F(\alpha , \beta , \gamma , x) = \frac{\Gamma (\gamma )}{\Gamma (\alpha )\Gamma (\beta )} \,\sum_{k=0}^\infty \,\frac{\Gamma (\alpha +k) \Gamma (\beta +k)}{k! \,\Gamma (\gamma +k)} \, x^k , \qquad \Gamma (\nu ) = \int_0^\infty t^{\nu -1} \,e^{-t} \,dt . %\eqno{(6.5)}$ Many elementary and special functions are of the hypergeometric type, as the following list shows \begin{align*} F(\alpha , \beta ; \beta ; x) &= (1-x)^{-\alpha} , &F(-n , \beta ; \beta ; -x) &= F(-n , 1; 1; -x ) = (1+x)^n , \\ F(-n , \beta ; \beta ; 1-x) &= x^n , &F(1,1;2;-x) & = \frac{1}{x} \,\ln (1+x) , \\ F\left( \frac{1}{2} , 1; \frac{3}{2} ; x^2 \right) &= \frac{1}{2x} \,\ln \frac{1+x}{1-x} , &F\left( \frac{1}{2} , \frac{1}{2} ; \frac{3}{2} ; x^2 \right) &= \frac{1}{x} \,\ln \arcsin x , \\ F\left( \frac{1}{2} , 1; \frac{3}{2} ; - x^2 \right) &= \frac{1}{x} \,\arctan x , &F\left( \frac{1}{2} , \frac{1}{2} ; \frac{1}{2} ; x^2 \right) &= (1- x^2 )^{1/2} = \sqrt{1- x^2} , \\ F\left( \frac{k}{2} , -\frac{k}{2} ; \frac{1}{2} ; x^2 \right) &= \cos (k\arcsin x) , &\lim_{\beta \rightarrow \infty} \, F\left( 1, \beta ; 1 ; \frac{x}{\beta} \right) &= e^x , \\ \lim_{\alpha \rightarrow \infty} \, F\left( \alpha , \beta ; \beta ; \frac{x}{\alpha} \right) &= e^x , &\lim_{\alpha \rightarrow \infty} \, F\left( \alpha, \alpha ; \frac{1}{2} ; -\frac{x^2}{4\alpha^2} \right) &= \cos x , \\ \lim_{\alpha \rightarrow \infty} \, F\left( \alpha, \alpha ; \frac{3}{2} ; -\frac{x^2}{4\alpha^2} \right) &= \frac{1}{x}\,\sin x , &\lim_{\alpha ,\beta\rightarrow \infty} \, F\left( \alpha, \beta ; \frac{1}{2} ; \frac{x^2}{4\alpha\beta} \right) &= \cosh x . \end{align*}

The special functions are extremely useful tools for obtaining closed form as well as series solutions to a variety of problems arising in science and engineering we tryed to reobtain the known results by the new method.

Although the Adomian’s goal is to find a method to unify linear and nonlinear, ordinary or partial differential equations for solving initial and boundary value problems, we shall deal in the following only with linear second order differential equations.

Because the solutions to many mathematical physics problems are expressed by the hypergeometric functions, we consider first this equation whose standard form is

$x \left( 1 - x \right) f'' (x) + \left[ \gamma - \left( \alpha + \beta + 1 \right) x \right] f' (x) - \alpha\beta\, f(x) =0$
and rewrite it as
\begin{align*} L\,f(x) &= x\, f'' (x) + \gamma \,f' (x) \equiv x^{1-\gamma} \frac{\text d}{{\text d}x} \left( x^{\gamma} \frac{{\text d}f}{{\text d}x} \right) \\ &= x^2 f'' (x) + \left[ \left( \alpha + \beta + 1 \right) x \right] f' (x) - \alpha\beta\, f(x) . \end{align*}
Applying Adomian's decomposition method, we seek its solution in the form:
$f(x) = u_0 (x) + u_1 (x) + u_2 (x) + \cdots ,$
with the initial condition u0 = 1. Then applying the inverse operator
$\left( L^{-1} f\right) (x) = \int_0^x x^{-\gamma} {\text d}x \int_0^x x^{\gamma -1} f(x) \,{\text d}x - f(0) ,$
we find
\begin{align*} u_n (x) &= L^{-1} \left[ x^2 u''_{n-1} (x) + \left( \alpha + \beta + 1 \right) x \,u'_{n-1} (x) + \alpha\beta\, u_{n-1} (x) \right] \\ &= \int_0^x x^{-\gamma} {\text d}x \int_0^x {\text d}x \,x^{\gamma -1} \left[ x^2 u''_{n-1} (x) + \left( \alpha + \beta + 1 \right) x \,u'_{n-1} (x) + \alpha\beta\, u_{n-1} (x) \right] {\text d}x . \end{align*}
So we get u1 = αβx/γ, and by induction
$u_n (x) = \frac{\alpha^{\overline{n}} \,\beta^{\overline{n}}{\gamma^{\overline{n}} \, \frac{x^n}{n!} , \qquad n=1,2,\ldots ;$
where
$\alpha^{\overline{n}} = \alpha \left( \alpha +1 \right) \cdots \left( \alpha + n -1 \right)$
is rising factorial of α. In this way, we get the well known solution
$f(x) = _2F_1 \left( \alpha , \beta; \gamma; x \right) = 1 + \sum_{n\ge 1} \frac{\alpha^{\overline{n}} \,\beta^{\overline{n}}{\gamma^{\overline{n}} \, \frac{x^n}{n!} .$

Example:

1. Dită, P. and Grama, N., On Adamian's decomposition method for solving differential equations, Institute of Atomic Physics, Bucharest, 2008.