Finding the solution to $V\frac {dV}{dz}=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12}-V)$

96 Views Asked by At

The following equation was found by balancing buoyancy with drag in an engineering field.

$V\frac {dV}{dz}=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12}-V)$

$\gamma$,$\alpha$, $\beta$, and $R$ are all constant, and $0\le z \le R$

Obviously, the equation is nonlinear in V. It almost looks like you can put it in the form $\frac {d}{dz}f(z, V)=\gamma + \alpha(\beta(R^2-z^2)^{-\frac12})$ for some $f(z,V)$ but I have not been able to find one, let alone something separable.

I'm thinking of using Runge–Kutta to do a numerical solution, with the constants able to range over a lot of values I'm worried there might be cases of stange behavior.

Is there a way to simply solve this, numerically or in a closed form, for all constant values?

1

There are 1 best solutions below

2
On

Hint:

Let $V=-\alpha U$ ,

Then $\dfrac{dV}{dz}=-\alpha\dfrac{dV}{dz}$

$\therefore\alpha^2U\dfrac{dU}{dz}=\gamma+\alpha(\beta(R^2-z^2)^{-\frac{1}{2}}+\alpha U)$

$U\dfrac{dU}{dz}=U+\dfrac{\gamma}{\alpha^2}+\dfrac{\beta}{\alpha\sqrt{R^2-z^2}}$

This belongs to an Abel equation of the second kind in the canonical form.

When $\gamma=0$ , this belongs to a special case of Abel equation of the second kind in the canonical form in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=140.

Let $\begin{cases}z=-\dfrac{4RW}{W^2+4}\\U=t-\dfrac{4RW}{W^2+4}\end{cases}$ ,

Then $\pm\dfrac{\beta}{\alpha}\dfrac{dW}{dt}=-\dfrac{tW^2}{4}+RW-t$