This equation shouldn't be so hard, and yet I'm stymied.
$$ \left( \frac{dw}{dz} \right )^2 + \alpha \frac{dw}{dz} + w \beta = 0 $$ with $w(0) = w_0>0$ $w(L) = 0$ for some known L and $\alpha(z)>0$ and $\beta(z)>0$ known. $\alpha$ and $\beta$ actually begin life as functions of known $w$, so I dont worry about existence of solution, what I want to prove is that I can invert back from $\alpha$ and $\beta$ to give unique $w$. $d \alpha / dz = \beta + C$ for some constant $C$ if it helps.
I can solve for some special cases but I'm interested in a general expression, or at least to prove uniqueness. Interestingly, those special cases are not unique if I just put constraint at $z=0$ but become so with the $z=L$ constraint.
Start with $$\left( \frac{dw}{dz} \right )^2 + \alpha(z) \frac{dw}{dz} + w \beta(z) = 0$$
We obtain:
$$\frac{dw}{dz}=-\frac{1}{2}\alpha(z)+\frac{1}{2}\left(\alpha^2(z)-4w \beta(z)\right)^{1/2}......(1)$$
$$\frac{dw}{dz}=-\frac{1}{2}\alpha(z)-\frac{1}{2}\left(\alpha^2(z)-4w \beta(z)\right)^{1/2}......(2)$$
You might be integrate (1) and (2) to find $w(z)$.
EDIT:
Not quite yet.
$w(z)$ appeared in the square root.