Second order ODE with non-integral exponent

Question

Consider the following 2nd-order nonlinear ODE with non-integral exponent. It arises from a diffusion problem in electrochemistry with a concentration-dependent sink term:

$y''-cy^\alpha=0$

with $c>0$ and $0<\alpha<1$. I'm looking for the real-valued solution that should be non-negative, since $y$ represents a concentration. Of particular interest are the special cases $\alpha=1/4$ and $\alpha=1/2$.

Answer 1

Here is the closed form solution using Inverse Beta Regularized $\text I^{-1}_s(a,b)$ and the Incomplete Beta function Leaving off from @JJacquelin’s solution using $\alpha=a$:

$$x=\pm \frac y{\sqrt{c_1}}\:_2\text{F}_1\left(\frac{1}{2}\:,\: \frac{1}{a+1}\:;\: \frac1{a+1}+1\:;\:-\frac{2c\,y^{a+1}}{c_1(a +1)}\right)+c_2= \pm \frac y{\sqrt{c_1}}\frac1{a+1}\left(-\frac{2c\,y^{a+1}}{c_1(a+1)}\right)^{-\frac1{a+1}}\text B_{-\frac{2c\,y^{a+1}}{c_1(a+1)}}\left(\frac1{a+1},\frac12\right)+c_2= c_2\pm \sqrt[a+1]{\frac{c_1(a+1)}{2c}}\frac{\text B_{\frac{2c\,y^{a+1}}{c_1(a+1)}}\left(\frac1{a+1},\frac12\right)}{\sqrt{c_1}(a+1)}$$

Now invert:

$$y’’=cy^a\implies y=\sqrt[a+1]{-\frac{(a+1)c_1}{2c}\text I^{-1}_\frac{\sqrt[a+1]{-\frac{2c}{c_1 (a+1)}}\sqrt{c_1}(a+1)(x-c_2) }{\text B\left(\frac1{a+1},\frac12\right)}\left(\frac1{a+1},\frac12\right)};0\le\frac{\sqrt[a+1]{-\frac{2c}{c_1 (a+1)}}\sqrt{c_1}(a+1)(x-c_2) }{\text B\left(\frac1{a+1},\frac12\right)}\le1,a>0$$

which works when seeing the first link’s “Expanded form assuming” section and simplifying the coefficients with the second link

Special cases include specific cases of the Jacobi Amplitude and Weierstrass P

Answer 2

$$y''-cy^\alpha=0$$ $$2y''y'-2cy^\alpha y'=0 \quad\to\quad y'^2-\frac{2c}{\alpha+1} y^{\alpha+1}=c_1$$ $$\frac{dy}{dx}=y'=\pm\sqrt{c_1+\frac{2c}{\alpha+1} y^{\alpha+1}}$$ $$x=\pm\int \frac{dy}{\sqrt{c_1+\frac{2c}{\alpha+1} y^{\alpha+1}}}+c_2$$ $$x=\pm \frac{y}{\sqrt{c_1}}\:_2\text{F}_1\left(\frac{1}{2}\:,\: \frac{1}{\alpha+1}\:;\: \frac{\alpha+2}{\alpha+1}\:;\:-\frac{2c\,y^{\alpha+1}}{c_1(\alpha+1)}\right)+c_2$$ where $_2$F$_1$ is the Gauss hypergeometric function.

This gives $x$ as a function of $y$. It is doubtful that a closed form exists for the inverse function $y(x)$ in the general case, that is for any $\alpha$ in the specified range.