Preface

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

Hypergeometric Functions

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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

and rewrite it as Applying Adomian's decomposition method, we seek its solution in the form: with the initial condition u₀ = 1. Then applying the inverse operator we find So we get u₁ = αβx/γ, and by induction where is rising factorial of α. In this way, we get the well known solution Example: ???

 

References

 

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