I have the following differential equation $$y''-xy=0\,.$$ By using $y=\sum\limits_{n=0}^{\infty} a_n x^n$, I show that $$(n+2)(n+1)a_{n+2}=a_{n-1}$$ for $n=1,2,...$, and $a_2=0$. (I differentiate the sum and by plugging it into the differential equation which is zero I get that result.) My question is how can I show that solutions converge for all $x \in \mathbb{R} $ since I cannot apply the ratio test.

  • $\begingroup$ note that if $a_2=0$ then $a_4=0=a_6=a_8....=0$ $\endgroup$ – Isham Jul 30 '18 at 14:03
  • $\begingroup$ Let $b_n = a_{2n-1}x^{2n-1}$. Since $a_{2n} = 0$, your sum is equivalent to $\sum b_n$ which you can apply the ratio test to. $\endgroup$ – AlexanderJ93 Jul 30 '18 at 14:03
  • $\begingroup$ Is your recurrence relation correct? $\endgroup$ – Nosrati Jul 30 '18 at 14:07
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    $\begingroup$ Your recurrence seems not right…? 'Cause for $y''$, the coefficient of $x^n$ is $(n+2) (n+1) a_{n+2}$; for $xy$, the corresponding coefficient is $a_{n-1}$, so… $\endgroup$ – xbh Jul 30 '18 at 14:08
  • $\begingroup$ For ratio test, maybe you could solve the recurrence first then test it? $\endgroup$ – xbh Jul 30 '18 at 14:10

There are two linearly independent solutions $f_0$ and $f_1$. We assume that the first one $f_0(x)$ satisfies the conditions $f_0(0)=1$ and $f_0'(0)=0$, whilst the second one $f_1(x)$ satisfies the conditions $f_1(0)=0$ and $f_1'(0)=1$. Any solution $y=f(x)$ of this differential equation is written as the sum $$f(x)=f(0)\,f_0(x)+f'(0)\,f_1(x)\text{ for all }x\in\mathbb{R}\,.$$ If we show that $f_0$ and $f_1$ are analytic on the whole $\mathbb{R}$, then we are done.

Now, from the result you obtain regarding the relations of the coefficients, you have that $$f_0(x)=\sum_{k=0}^\infty\,u_{k}\,x^{3k}\text{ and }f_1(x)=\sum_{k=0}^\infty\,v_{k}\,x^{3k+1}\,,$$ where $$u_k:=\frac{1}{\prod\limits_{r=1}^k\,\big((3k)(3k-1)\big)}=\frac{1}{(3k)_{k,3}\,(3k-1)_{k,3}}$$ and $$v_k:=\frac{1}{\prod\limits_{r=1}^k\,\big((3k+1)(3k)\big)}=\frac{1}{(3k+1)_{k,3}\,(3k)_{k,3}}\,.$$ Here, the falling factorial power $(x)_{n,h}$ for $x,h\in\mathbb{C}$ and $n\in\mathbb{Z}_{\geq 0}$ is defined by $$(x)_{n,h}:=x\,(x-h)\,(x-2h)\,\ldots\,\big(x-(n-1)h\big)\,,$$ with $(x)_{0,h}:=1$.

To show that the radii of convergence of $f_0$ and $f_1$ are both infinite, we need to verify that $$\limsup_{k\to\infty}\,\sqrt[3k]{u_k}=0\text{ and }\limsup_{k\to\infty}\,\sqrt[3k+1]{v_k}=0\,.$$ However, this follows from the observation that $$(3k+1)_{k,3}\geq k!\,,\,\,(3k)_{k,3}\geq k!\,,\text{ and }(3k-1)_{k,3}\geq k!$$ for all $k=0,1,2,\ldots$, and the well known fact that $$\sqrt[k]{k!}\to\infty\text{ as }k\to\infty\,.$$

It turns out that $$f_0(x)=\frac{\sqrt[3]{3}^2\,\Gamma\left(\frac{2}{3}\right)}{2}\,\text{Ai}(x)+\frac{\sqrt[6]{3}\,\Gamma\left(\frac{2}{3}\right)}{2}\,\text{Bi}(x)$$ and $$f_1(x)=-\frac{\sqrt[3]{3}\,\Gamma\left(\frac{1}{3}\right)}{2}\,\text{Ai}(x)+\frac{\Gamma\left(\frac{1}{3}\right)}{2\,\sqrt[6]{3}}\,\text{Bi}(x)\,,$$ where $\text{Ai}$ and $\text{Bi}$ are Airy functions, and $\Gamma$ the usual gamma function. Alternatively, $$\text{Ai}(x)=\frac{1}{\sqrt[3]{3}^2\,\Gamma\left(\frac23\right)}\,f_0(x)-\frac{1}{\sqrt[3]{3}\,\Gamma\left(\frac13\right)}\,f_1(x)$$ and $$\text{Bi}(x)=\frac{1}{\sqrt[6]{3}\,\Gamma\left(\frac23\right)}\,f_0(x)+\frac{\sqrt[6]{3}}{\Gamma\left(\frac13\right)}\,f_1(x)\,.$$


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