Buoyancy Nonlinear second order differential equation

Question

Working on equations of motion of an expandable gaz-filled body immersed uderwater, I have reached the following equation :

$$ \beta^{2} z'' - \alpha(z')^{2} + \beta z z'' - z = 0 $$

Could you give me some advices on how to find and analytical solution, if it exists ? If not, do we have one if we forget the drag, that is if :

$$ \beta^{2} z'' + \beta z z'' - z = 0 $$

Ultimately, what is really of interest for me is not really the entire solutions, but rather stability around the solution $z=0$

Answer 1

First thing we notice is that the equation doesn't contain the independent variable, so we can immediately lower the order by one.

Introducing:

$$z'=u(z) \\ z''=u' z'=u u'$$

We get:

$$\beta^2 u u' - \alpha u^2 + \beta z u u' - z = 0$$

Collecting the derivative:

$$\beta (\beta+z) u u'-\alpha u^2-z=0$$

Let's introduce yet another function:

$$u^2=v(z) \\ v'=2u u'$$

We get:

$$\frac{1}{2} \beta(\beta+z) v'-\alpha v-z=0$$

This is now a linear inhomogeneous ODE, which is easy to solve.

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I hope I haven't made any mistakes, but with ODEs it's always easy to check the final solution by direct substitution, so once you find $v(z), u(z)$ and then $z(x)$ from $z'=u(z)$ you can check.

Answer 2

Since there are no $x$-dependence, you can reduce the order by one.

Recall $z''=\frac12\frac{\mathrm{d}(z')^2}{\mathrm{d}z}$ and we get $$ (\beta^2 + \beta z) \frac{\mathrm{d}(z'^2)}{\mathrm{d}z} - 2\alpha(z')^{2} = 2z $$ Integrating once gives $$ z'^2 = \begin{cases} \frac{\beta^2}{\alpha(\beta-2\alpha)} + \frac{2z}{\beta-2\alpha}+ c (z+\beta)^{2\alpha/\beta} & \alpha\neq 0\\ c+\frac{2 z}{\beta} - 2 \log(z+\beta) & \alpha=0 \end{cases} $$ but I don't think the final integral can be done in closed form.