I'm trying to find the solution to the equation $ (A^2)'' = 2 \alpha A^2 - 2 A $ on the interval $(0,1)$ where $A'(0) = 0,\, A(1) = 0 $ for $\alpha >0$. Firstly I make the substitution $y = A^2$, this reduces the equation to $y'' = 2 \alpha y -2\sqrt{y}.$ Since the equation is lacking the independent variable $x$, one may make the substitution $y'=v$ which reduces the equation to $ v \frac{dv}{dy} = 2\alpha y - 2\sqrt{y}.$ By seperating variables, and using $y'=v$, one has: $$ y' = \sqrt{\frac{2}{3}}\cdot\sqrt{C+3\alpha y^2 - 4 y^{1.5}}.$$ for some integration constant $C$. This is as far as I've got without rurnning in to problems. I am not able to get this down to an explicit form.
Question 1: Does there exist a solution to the above problem, and if one does can it be written explicitly.
Question 2: If an explicit solution can't be determined, can we learn anything about it by looking at the $(y,y')$ phase plane?
Question 3: Are there any other methods that could be applied to solve this, or to at least find some information about the solution? I'm not sure if you could apply asymptotics as it's a BVP on a finite interval.
EDIT 1: You can use $y'(0) = 0$ to get rid of the $C$ in the above equation. You can the substitute A^2 straight back in to the equation for $y'$ to give: $$ 2A\cdot A' = \sqrt{\frac{2}{3}}\cdot A \cdot \sqrt{3\alpha A^2 - 4A} \implies A' = \frac{1}{\sqrt{6}} \sqrt{3\alpha A^2 - 4A}$$ Which, according to matlab, has the solution: $$A(x) = -\frac{1}{3\alpha}\cdot\bigg\{2\cos\bigg[\frac{\sqrt{3\alpha}}{6}\bigg(C_1+x\sqrt{6}\bigg)\cdot \int_0^x \frac{1}{ln(t)}dt\bigg]-2\bigg\}$$
If anyone would like to point out any floors I've made it would be much appreciative! Also - It would still be interesting to hear the answers to the questions I have previously asked.
You are not allowed to set $C=0$ since you do not know the value of $y(0)$:
$$ y'' = 2\alpha y-2\sqrt{y},\qquad y'(0)=y(1)=0 $$
$$ 2y'y'' = (4\alpha y-4\sqrt{y})y' $$
$$ \int_{0}^{z}2y'y''\,dx = y'^2(z) = \left(2\alpha\, y(z)^2-\frac{8}{3}y(z)^{3/2}\right)\underbrace{-\left(2\alpha\, y(0)^2-\frac{8}{3}y(0)^{3/2}\right)}_{C} $$
$$ \int_{z}^{1}2y'y''\,dx = y'(1)^2-y'^2(z) = -\left(2\alpha\, y(z)^2-\frac{8}{3}y(z)^{3/2}\right) $$
so $C=y'(1)^2$ and the solution is inherently related to the inversion of an elliptic integral.