# differential equation nondevelopable

I try to solve this differential equation whose solution seems not to be constructable in power series $y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers. If one can help me please?

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Is the coefficient of $y$ supposed to be $x+\frac{a}{x^2} + b$, or $\frac{x+a}{x^2+b}$? The slash notation is ambiguous; please avoid it. – Arturo Magidin Feb 3 '12 at 20:27

Let us solve this problem in detail to show what kind of problems we encounter in here. We insert the ansatz $y(x) = \sum\limits_{n=0}^\infty p_n x^{n+\alpha}$ into the ODE and then equate to zero the coefficients at consecutive powers. We have: \begin{eqnarray} coeff @ x^{\alpha-2} :&& p_0 \alpha (\alpha-1) + a p_0 =0 \\ coeff @ x^{\alpha-1}:&& p_1 (\alpha+1) \alpha + a p_1 = 0 \\ coeff @x^{\alpha} :&& p_2(\alpha+2)(\alpha+1) + a p_2 + b p_0 = 0 \end{eqnarray} This gives $\alpha = \frac{1}{2} \left( 1 \pm \sqrt{1- 4 a} \right)$ and $p_1=0$ and $p_2= -b p_0/(2 (2 \pm \sqrt{1- 4 a}))$. The recursion relation for the coefficients now reads: $$p_{n+2} \left( (n+2+\alpha)(n+1+\alpha) + a \right) + p_n b + p_{n-1}=0$$ for $n=1,2,\dots$. Now we substitute $p_n \rightarrow p_{n+1}$ and we get a following recursion relation: $$p_{n+3} = f_n \left( b \cdot p_{n+1} + p_n \right)$$ for $n=1,2,3,\dots$ with $f_n := - ((n+2)(n+2 \pm\sqrt{1-4 a}))^{-1}$ and subject to $(p_1,p_2)= (1,0)$. By looking at this recursion relation we see that the solution is some polynomial in the variable $b$. Is it possible to find a closed form expression for all coefficients of that polynomial? By starting from a given value of $n$ we can back-propagate this equation following two rules, firstly only steps of length two or three are allowed, secondly a step by two and three is assigned a factor $b$ and a unity respectively. We denote by $(i_l + l-1)$ the positions of the $f_{n-2(i_l-1)-3 \cdot l}$-factors related to movements of length three. Here for $l=1,\dots,s$. All the remaining $f$-factors are related to steps of length two. Since at the end of the back-propagation process we must always hit unity this yields a following constraint: $n - 2(i_{s+1}-1) - 3 s= 1$. Due to the initial conditions he exponent assigned to all the terms in question equals $i_{s+1}-1$. If we now consider two cases of $n$ being even and odd respectively then we easily arrive at the following result: Let $n \notin 2 {\mathbb N}$, ie $n=2 m+1$. Then we have: \begin{eqnarray} p_{n+3} &=& \sum\limits_{j=0}^{\lfloor \frac{m}{3} \rfloor} b^{m- 3 j} \cdot \sum\limits_{1 \le i_1 \le i_2 \le \dots \le i_{2 j} \le m+1- 3j} \left( \prod\limits_{l=1}^{2 j+1} \prod\limits_{\xi=0}^{i_l-i_{l-1}-1} f_{n-2 \cdot i_{l-1} - 3 (l-1) - 2 \xi}\right) \cdot \left(\prod\limits_{l=1}^{2 j} f_{n-2 \cdot (i_l-1)- 3 l} \right)\\ &=& {\mathcal C}_m \sum\limits_{j=0}^{\lfloor \frac{m}{3} \rfloor} b^{m- 3 j} \frac{(-1)^{1-j+m}}{2^{2(1+j+m)}} \pi^{2 j} \sum\limits_{1 < i_1 < i_2 < \dots < i_{2 j} \le m+1- j} \prod\limits_{l=1}^{2 j} \prod\limits_{p=\pm} \binom{\frac{\theta_p-n}{2} + \frac{1}{2} l -2 + i_l}{\frac{1}{2}} \end{eqnarray} where $${\mathcal C}_m := \prod\limits_{p=\pm} \frac{(\frac{\theta_p-n}{2}-1)!}{(\frac{\theta_p-n}{2}+m)!}$$ Likewise let $n \in 2{\mathbb N}$, ie $n=2 m$. Then we have: \begin{eqnarray} p_{n+3} &=& \sum\limits_{j=0}^{\lfloor \frac{(m-2)}{3} \rfloor} b^{m- 3 j-2} \cdot \sum\limits_{1 \le i_1 \le i_2 \le \dots \le i_{2 j+1} \le m-1- 3 j} \left( \prod\limits_{l=1}^{2 j+2} \prod\limits_{\xi=0}^{i_l-i_{l-1}-1} f_{n-2 \cdot i_{l-1} - 3 (l-1) - 2 \xi}\right) \cdot \left(\prod\limits_{l=1}^{2 j+1} f_{n-2 \cdot (i_l-1)- 3 l} \right)\\ &=&{\mathcal D}_m \sum\limits_{j=0}^{\lfloor \frac{(m-2)}{3} \rfloor} b^{m- 3 j-2} \frac{(-1)^{m-j}}{2^{2(1+j+m)}} \pi^{1+2 j} \sum\limits_{1 < i_1 < i_2 < \dots < i_{2 j+1} \le m- j} \prod\limits_{l=1}^{2 j+1} \prod\limits_{p=\pm} \binom{\frac{\theta_p-n}{2} + \frac{1}{2} l -2 + i_l}{\frac{1}{2}} \end{eqnarray} where $${\mathcal D}_m := \prod\limits_{p=\pm} \frac{(\frac{\theta_p-n}{2}-1)!}{(\frac{\theta_p-n}{2}+m-\frac{1}{2})!}$$ and $\theta_\pm = (-2,-2 \pm \sqrt{1-4 a})$. We wrote a simple Mathematica program to verify that those results are correct. Now comes the most interesting part. Is it possible to do the sums over the $i$-indices analytically or otherwise the results above are only some sort of perturbation results in the variable $b$. We leave this question open for the time being.

Now let us assume that $0< b \ll 1$. Then clearly the solutions read: $$y(x) = y^{(0)}(x) + y^{(1)}(x) b^1 + O(b^2)$$ We analyze now the coefficients in that expansion.

Zeroth order approximation:

Now, we select the coefficients at $b^0$. The exponent is zero if and only if, firstly $n=2 m+1$ and $m=3 \tilde{m}$ and $j=\tilde{m}$ and secondly $n=2 m$ and $m=3\tilde{m}+2$ and $j=\tilde{m}$. Both choices lead to $i_l=l+1$ for $l=1,\dots,2 j$ in the first case and $l=1,\dots,2 j+1$ in the second case respectively. After fairly simple manipulations we get: $$p_{n+3} = \frac{(-1)}{9^{2 j}} \prod\limits_{p=\pm} \frac{1}{(\theta_p - 1)} \frac{1}{ (\frac{(4 - \theta_p)}{3})^{(2 j)}}$$ for $n=2 m+1 = 6 j+1$ for $j=0,1,2,\dots,$ and $$p_{n+3} = \frac{(+1)}{9^{2 j+1}} \prod\limits_{p=\pm} \frac{1}{(\theta_p - 1)} \frac{1}{ (\frac{(4 - \theta_p)}{3})^{(2 j+1)}}$$ for $n=2 m=6 j+4$ for $j=-1,0,1,2,\dots$. Putting both cases together we get the solution: \begin{eqnarray} y^{(0)}(x) &=& \sum\limits_{j=0}^\infty p_{6 j+4} x^{6 j+3 + \alpha} + \sum\limits_{j=-1}^\infty p_{6 j+7} x^{6 j+6 + \alpha} \nonumber \\ &=& \sum\limits_{j=-1}^\infty \frac{(-1)^{j+1}}{9^j} \left( \prod\limits_{p=\pm} \frac{1}{(\theta_p - 1)} \frac{1}{ (\frac{(4 - \theta_p)}{3})^{(2 j+1)}} \right) \cdot x^{3 j+3+\alpha} \nonumber \\ &=& x^{\frac{1}{2} (1\pm \sqrt{1-4 a})} F_{0,1}\left[1 \pm \frac{1}{3} \sqrt{1-4 a}; -\frac{x^3}{9}\right] = {\mathcal A}_\pm \sqrt{x} J_{\pm \frac{1}{3} \sqrt{1- 4 a}} \left( \frac{2}{3} x^{3/2}\right) \end{eqnarray} where ${\mathcal A}_\pm = (\pm \frac{1}{3} \sqrt{1- 4 a})! 3^{\pm \frac{1}{3} \sqrt{1- 4 a}}$.

First order correction:

Now, we select the coefficients at $b^1$. The exponent is zero if and only if, firstly $n=2 m+1$ and $m=3 \tilde{m}+1$ and $j=\tilde{m}$ and secondly $n=2 m$ and $m=3\tilde{m}+3$ and $j=\tilde{m}$. Both cases lead to $i_l=l+1+1_{l \ge \xi}$ for $l=1,\dots,2 j$ and $\xi=1,\dots,2 j+1$ in the first case and $l=1,\dots,2 j+1$ and $\xi=1,\dots,2 j+2$ in the second case respectively. After some manipulations we get for the odd case: \begin{eqnarray} p_{n+3} = \frac{3^4}{4^4} \frac{(-1)^{4 j}}{ 9^{2 j}} {\mathcal D}^{o}_j \cdot \sum\limits_{\xi=1}^{2 j+1} % \prod\limits_{p=\pm} % % \frac{1} {\left(\frac{1}{2} (-6 j+\theta_p+3 \xi -6)\right)^{(2)}} % \frac{\left(\frac{1}{3} (-6 j+\theta_p+3 \xi-4)\right)!} {\left(\frac{1}{3} (-6 j+\theta_p+3 \xi -9)\right)!} \end{eqnarray} for $n=2 m+1=6j+3$ for $j=0,1,2,\dots$. Likewise in the even case we get: \begin{eqnarray} p_{n+3} = \frac{3^4}{4^4} \frac{(-1)^{4 j+1}}{ 9^{2 j+1}} {\mathcal D}^{e}_j \cdot \sum\limits_{\xi=1}^{2 j+2} % \prod\limits_{p=\pm} % \frac{1}{ \left(\frac{1}{2} (-6 j+\theta_p+3 \xi -9)\right)^{(2)}} \frac{ \left(\frac{1}{3} (-6 j+\theta_p+3 \xi -7)\right)!}{ \left(\frac{1}{3} (-6 j+\theta_p+3 \xi -12)\right)!} \end{eqnarray} for $n=2 m=6j+6$ for $j=-1,0,1,2,\dots$. The prefactors read: \begin{eqnarray} {\mathcal D}^{o}_j &:=& \prod\limits_{p=\pm} % \frac{\left(\frac{1}{2} (-6 j+\theta_p-5)\right)!}{\left(\frac{1}{2} (\theta_p-1)\right)!} % \frac{\left(\frac{1}{2} (\theta_p-3)\right)! \left(\frac{1}{3} (-6 j+\theta_p-3)\right)!}{\left(\frac{1}{3} (\theta_p-4)\right)! \left(\frac{1}{2} (-6 j+\theta_p-3)\right)!}\\ {\mathcal D}^{e}_j &:=& \prod\limits_{p=\pm} % \frac{\left(\frac{1}{2} (-6 j+\theta_p-8)\right)!}{\left(\frac{1}{2} (\theta_p-1)\right)!} % \frac{\left(\frac{1}{2} (\theta_p-3)\right)! \left(\frac{1}{3} (-6 j+\theta_p-6)\right)!}{\left(\frac{1}{3} (\theta_p-4)\right)! \left(\frac{1}{2} (-6 j+\theta_p-6)\right)!} \end{eqnarray} Puting things together the solution reads: $$y^{(1)}(x) = \sum\limits_{j=0}^\infty p_{6 j+6} x^{6 j+5 + \alpha} + \sum\limits_{j=-1}^\infty p_{6 j+9} x^{6 j+8 + \alpha} = \dots$$

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The differential equation $$y''+ \left(x+ \frac{a}{x^2}+b \right)y=0$$ has a regular singular point at $x=0$. In such a case, it is not always possible to construct a power series solution. However, it is always possible to find a solution of the form $$y = x^\alpha p(x)$$ with $\alpha \in \mathbb{C}$ and $p(x) = \sum_{n=0} p_n x^n$.

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And what if OP really meant $$y''+{x+a\over x^2+b}y=0$$ – Gerry Myerson Mar 4 '12 at 23:51
@Gerry: then he should be able to develop $y(x)$ in a power series around $x=0$. – Fabian Mar 5 '12 at 3:11

Hint:

$y''+\left(x+\dfrac{a}{x^2}+b\right)y=0$

$x^2y''+(x^3+bx^2+a)y=0$

Let $y=x^ku$ ,

Then $y'=x^ku'+kx^{k-1}u$

$y''=x^ku''+kx^{k-1}u'+kx^{k-1}u'+k(k-1)x^{k-2}u=x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u$

$\therefore x^2(x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u)+(x^3+bx^2+a)x^ku=0$

$x^{k+2}u''+2kx^{k+1}u'+(x^3+bx^2+k(k-1)+a)x^ku=0$

$x^2u''+2kxu'+(x^3+bx^2+k(k-1)+a)u=0$

Choose $k(k-1)+a=0$ , i.e. $k=\dfrac{1\pm\sqrt{1-4a}}{2}$ , the ODE becomes

$x^2u''+(1\pm\sqrt{1-4a})xu'+(x^3+bx^2)u=0$

$xu''+(1\pm\sqrt{1-4a})u'+(x^2+bx)u=0$

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Another way of solving this equation will be a series expansion in the parameter $b$. We know that for $b=0$ the two independent solutions can be easily found by means of the series expansion method.Those solutions are related to Bessel functions. Let us therefore assume that the whole solution reads: $$y(x) = \sum\limits_{j=0}^\infty b^j y^{(j)}(x)$$ We call the $j=0$ function the basic solution and the $j>0$ functions the corrections of order $j$. Inserting the ansatz into the ODE we get: $$y^{(0)}_\pm(x) = \sqrt{x} J_{\pm \frac{1}{3} \sqrt{1-4 a}} \left(\frac{2}{3} x^{\frac{3}{2}}\right) = \frac{x^\alpha}{{\mathcal A}_\pm} F_{0,1}[1\pm \frac{1}{3} \sqrt{1-4 a};-\frac{x^3}{9}]$$ where $\alpha=1/2(1\pm \sqrt{1-4 a})$ and ${\mathcal A}_\pm = 3^{\frac{\pm}{3} \sqrt{1-4 a} } \left(\frac{\pm}{3} \sqrt{1-4 a} \right)!$. We have: $$\left[\frac{d^2}{d x^2} + (x + \frac{a}{x^2})\right] y^{(j)}(x) = -y^{(j-1)}(x)$$ for $j=1,2,\dots$. The above equation can be solved by means of Greens functions. The solution reads: $$y^{(j)}(x) = \int\limits_0^x \left(\frac{y^{(0)}_-(x) y^{(0)}_+(\xi) - y^{(0)}_+(x) y^{(0)}_-(\xi)}{{\mathcal W}\left[y^{(0)}_+,y^{(0)}_-\right](\xi)}\right) \cdot (-) y^{(j-1)}(\xi) d \xi$$ Here ${\mathcal W}[y^{(0)}_+,y^{(0)}_-]$is the Wronskian. It is a constant as a function of $\xi$ and it reads: $${\mathcal W}[y^{(0)}_+,y^{(0)}_-](\xi) = - \frac{\sin\left[\frac{\pi}{3} \sqrt{1- 4 a}\right]}{\frac{\pi}{3}}$$ Writing the above equation in a compact way we get: $$y^{(J)}(x) = \frac{1}{{\mathcal W}^J} \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot y^{(0)}(\xi) d \xi$$ where $${\mathcal K}^{(J)}(x,\xi) := \int\limits_\xi^x {\mathcal K}^{(J-1)}(x,\eta) \cdot {\mathcal K}(\eta,\xi) d \eta$$ for $J=2,3,\dots$ and $${\mathcal K}^{(1)}(x,\xi) := \left| \begin{array}{rr} y_+^{(0)}(x) & y_-^{(0)}(x) \\ y_+^{(0)}(\xi) & y_-^{(0)}(\xi) \end{array} \right|$$ Now, in order to construct the higher order corrections we compute a following quantity(to be termed moments): $${\mathcal M}_{J,l}(x) := \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot \xi^{l+\alpha} d \xi$$ where $J=1,2,\dots$ and $l=0,1,2,\dots$. Integrating by parts twice we obtain a following recursion relation for the moments: $$(l+2)(l+2\pm \sqrt{1-4 a}){\mathcal M}_{J,l} + {\mathcal M}_{J,l+3} = -{\mathcal W}\cdot\left({\mathcal M}_{J-1,l+2} 1_{J>1} + \delta_{J,1} x^{l+2+\alpha}\right)$$ for $J=1,2,\dots$. The solutions to the recursion relations are pretty straightforward and read: \begin{eqnarray} {\mathcal M}_{1,l} &=& {\mathcal W} \cdot x^{l+\alpha-1} \cdot \left\{ F_{1,2}\left[\begin{array}{rr} 1 \\ \frac{l+2}{3} & \frac{l+2\pm\sqrt{1-4 a}}{3}\end{array};-\frac{x^3}{3^2}\right]-1\right\} \\ {\mathcal M}_{J,l} &=& {\mathcal W} \sum\limits_{j=0}^\infty \left(\frac{-1}{3^2}\right)^{j+1} \cdot \frac{{\mathcal M}_{J-1,l+3 j+2}}{\left(\frac{l+2}{3}\right)^{(j+1)} \left(\frac{l+2\pm \sqrt{1-4 a}}{3}\right)^{(j+1)}} \end{eqnarray} for $J>1$. Using the expressions for the moments along with the expression for the correction of order $j$ we obtain the first order correction: \begin{eqnarray} && y_\pm^{(1)}(x) = \frac{x^{\alpha+2}}{(-9)^1 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^2 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)! \sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{2}{3}+j)!(\frac{2\pm\sqrt{1-4 a}}{3}+j)!}\\ &&\sum\limits_{j_1=0}^j \binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}} \end{eqnarray} Likewise the second order correction reads: \begin{eqnarray} &&y_\pm^{(2)}(x) = \frac{x^{\alpha+4}}{(-9)^2 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^4 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)! \sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{4}{3}+j)!(\frac{4\pm\sqrt{1-4 a}}{3}+j)!}\\ &&\sum\limits_{0\le j_1 \le j_2 \le j}^j \binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}} \binom{j_2+\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_2+\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}} \end{eqnarray} It is now easy to see what is the pattern for all higher order corrections. As such the problem is in principle solved. It would be nice to reduce the multiple sums to single sums though.

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