I am solving a differential equation related to fluid mechanics, a rigid air bubble rising towards the surface of a liquid. Doing all the calculations, I have come to this differential equation, which I have not been able to solve, by far.

$$\frac{d^2y}{dt^2}=A(P_{atm}-\rho g y)^{2/3}\cdot \left(\frac{dy}{dt}\right)^2+By+C \tag{1}$$ Where $A,B,C, P_{atm}, \rho$ and $g$ are all constants.

I have created an alternative differential equation by substituting $v=y'$, as it seems like a legitimate method, however I am not sure how I will solve this, as this equation also seems unsolvable:

$$v\frac{dv}{dy}=A(P_{atm}-\rho g y)^{2/3}\cdot v^2+By+C$$

$$\frac{dv}{dy}=A(P_{atm}-\rho g y)^{2/3}\cdot v+\frac{By+C}{v}\ \tag{2}$$

In conclusion, I would like some guidance to obtain a solution for $y(t)$. I am not sure whether if I solved $(2)$, if it would still be possible to get $y(t)$ from $v(y)$.


The idea to substitute $v = \frac{\text{d} y}{\text{d} t}$ is a good one. To start with your last equation, we can multiply on both sides with $v$ to obtain \begin{equation} v \frac{\text{d} v}{\text{d} y} = A (P- \rho g\, y)^{\frac{2}{3}} v^2 + B y + C. \tag{1} \end{equation} Now, we recognise the left hand side as the derivative of $\frac{1}{2} v^2$. Therefore, writing $w = v^2$, we can write $(1)$ in terms of $w$ as \begin{equation} \frac{\text{d} w}{\text{d} y} = 2A (P- \rho g\, y)^{\frac{2}{3}} w + 2B y + 2C, \tag{2} \end{equation} which is a linear first order ODE for $w$. Now, I would suggest that you rescale $w$ and $y$ to simplify $(2)$ a bit. A good choice would be to introduce \begin{equation} \eta = \frac{P - \rho g y}{(\frac{\rho g}{2A})^{\frac{3}{5}}} \;\Leftrightarrow \; y = \frac{P - (\frac{\rho g}{2A})^{\frac{3}{5}}\eta}{\rho g} \end{equation} and \begin{equation} z = \frac{\rho g A}{B} (\frac{2A}{\rho g})^{\frac{1}{5}} w\;\Leftrightarrow\;w = \frac{B}{\rho g A} (\frac{\rho g}{2A})^{\frac{1}{5}} z \end{equation} to reduce $(2)$ to \begin{equation} \frac{\text{d} z}{\text{d} \eta} = - \eta^{\frac{2}{3}} z + \eta - D, \tag{3} \end{equation} where \begin{equation} D = \frac{B P +C \rho g}{B (\frac{\rho g}{2A})^{\frac{3}{5}}}. \end{equation} You can solve $(3)$ explicitly in terms of the exponential integral.

  • $\begingroup$ Thank you for the effort. Would this mean that I would have to find an integrating factor, and get something approximately looking like the Exponential integral? $\endgroup$ – projectilemotion Mar 16 '16 at 23:08
  • $\begingroup$ Well, this being a linear first order differential equation, finding an integrating factor is equivalent to solving the homogeneous ODE $z' = -\eta^{\frac{2}{3}} \$. You can then find the solution to the original, inhomogeneous ODE using variation of parameters. Standard stuff, in a way: see en.wikipedia.org/wiki/… . $\endgroup$ – Frits Veerman Mar 17 '16 at 9:25
  • $\begingroup$ The variation of parameters method seemed to work. One question - Where did you get the substitutions \eta and z from, how did you get the 1/5 and 3/5 powers? $\endgroup$ – projectilemotion Mar 17 '16 at 17:02
  • 1
    $\begingroup$ Well, I tried linear rescalings of $y$ and $w$, so $y \to \alpha y + \beta$ and $w \to \gamma w + \delta$. It seemed useful to call the quantity within brackets '$\eta$', but then I realised that actually, using the substitution $\eta = \alpha (P - \rho g y)$ allows you to choose $\alpha$ conveniently such that a lot of parameters are scaled away. Same story with $w$: it didn't seem useful to introduce an offset $\delta$, so I just substituted $z = \gamma w$ and chose $\gamma$ such that as many prefactors as possible became 1. $\endgroup$ – Frits Veerman Mar 17 '16 at 17:48

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