Consider a collapsing spherical cavity in a large pond of water. The initial pressure of water in the pond is $P_0$, the initial radius of the cavity is $R_0$. Determine the speed of the edge of the cavity when its radius reaches $r_0$ ($r_0 < R_0$). The water density is $\rho$.
Note: The problem statement is translated (and slightly modified) from the original text in Savchenko et al 1981.
Discussion
I did not plan on writing up this problem as it seemed trivial to me at first. I thought, ok there is some kind of differential equation to compose and solve.
I tried to use the Newton's law and the energy conservation equation approaches but could not really get it.
I asked ChatGPT, and it came up with a solution based on the energy balance equation, the solution did resemble the correct one but it was off by a constant factor. ChatGPT mentioned the Rayleigh-Plesset equation which helped me to come up with my solution below. But here I decided to base the solution on the Newton's equation of motion to be able to present a self-contained solution, at least from my stand point.
Solution
Let's consider a water parcel at a distance $x$ from the center of the cavity $C$ (see the figure below).
Figure 1: Cross-section of collapsing cavity in a large pond. Dashed circles show positions of the cavity's edge when its sizes are $R_0, r, r_0$.
The pressure gradient will act to accelerate the parcel towards the point $C$.
\begin{equation}
\rho \frac{dv}{dt} = -\frac{\partial P}{\partial x}
\end{equation}
where $v=v(x, t)$ - is velocity of the water parcel at the distance $x$ from the center of the collapsing hollow cavity at time $t$.
We can express the derivative of $v(x(t), t)$ in the above equation by assuming that the parcell travels along a tragectory $x(t)$ and employing the compound function differentiation formula:
\begin{align*}
\frac{dv}{dt} &= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} \frac{d x}{d t} \\
&= \frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x}
\end{align*}
Therefore, the equation of motion takes the following form (this is also known as an Euler form of the equation of motion):
\begin{align}
\boxed{\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} = -\frac{1}{\rho}\frac{\partial P}{\partial x}}
\end{align}
Let's consider the continuity equation for the spheres with radii $x$ and $R_0$ and centered around the point $C$:
\begin{align*}
& v(x, t) \cdot 4\pi x^2 = v(R_0, t) \cdot 4\pi R_0^2 & \Rightarrow \\
& v(x, t) \cdot x^2 = v(R_0, t) \cdot R_0^2 = F(t) & \Rightarrow \\
& \boxed{v(x, t) = \frac{F(t)}{x^2}} &
\end{align*}
Now let's plug the above expression for $v(x, t)$ into the equation of motion:
\begin{align*}
& \frac{1}{x^2} \frac{d F(t)}{d t} + \frac{F(t)^2}{x^2}\left(-\frac{2}{x^3} \right) = -\frac{1}{\rho}\frac{\partial P}{\partial x} & \Rightarrow \\
& \frac{1}{x^2} \frac{d F(t)}{d t} - \frac{2F(t)^2}{x^5} = -\frac{1}{\rho}\frac{\partial P}{\partial x}
\end{align*}
Let's consider the above equation at a fixed moment in time when the radius of the hollow cavity is $r(t)$. Then integrating the above equation from $r(t)$ to $+\infty$ and using that $P(r) = 0$ and $P(+\infty)=P_0$ we get:
\begin{align*}
& \frac{d F(t)}{d t}\int\limits_r^{+\infty}\frac{dx}{x^2} - 2F(t)^2\int\limits_r^{+\infty}\frac{dx}{x^5} = -\frac{1}{\rho}\int\limits_r^{+\infty}\frac{\partial P}{\partial x} dx & \Rightarrow \\
& \frac{1}{r}\frac{d F(t)}{d t} - \frac{1}{2} \frac{F(t)^2}{r^4} = -\frac{P_0}{\rho} &
\end{align*}
We can express $F(t)$ using flux continuity at the edge of the cavity as follows:
\begin{align*}
F(t) = u(t)r^2
\end{align*}
where $u(t) = dr/dt=\dot{r}$ - is the speed of the edge of the cavity at time $t$ when its radius is $r=r(t)$.
After plugging in the above expression for $F(t)$ into the integrated equation of motion, we get:
\begin{align*}
\boxed{\dot{u}r + \frac{3}{2} u^2 = -\frac{P_0}{\rho}}
\end{align*}
The above is a particular case of an existing named equation (Rayleigh-Plesset equation: $\ddot{r} r + \frac{3}{2} \dot{r}^2 = -P_0/\rho$, where $\dot{r}=u$).
Finally, to solve the equation we will invert the $r(t)$ function and will consider the speed of the edge of the cavity as a function of the radius of the cavity.
\begin{align*}
\dot{u} = \frac{d u}{d r} \frac{1}{\dot{r}}
\end{align*}
Replacing the temporal derivative $\dot{u}$, using the above formula, in the initial differential equation we get an easily separable first order differential equation with respect to $u=u(r)$:
\begin{align*}
\frac{du}{dr}ur + \frac{3}{2} u^2 = -\frac{P_0}{\rho}
\end{align*}
Further, we can separate $du$ and $dr$ and integrate the equation from $R_0$ to $r_0$, taking into account that $u(R_0) = 0$ (i.e. the velocity of the cavity edge was 0 in the beginning when its radius was $R_0$):
\begin{align*}
& -\int\limits_{u(R_0)}^{u(r_0)} \frac{u du}{\frac{P_0}{\rho} + \frac{3}{2}u^2} = \int\limits_{R_0}^{r_0}\frac{dr}{r} & \Rightarrow \\
& \ln\left(\frac{\frac{P_0}{\rho} + \frac{3}{2}u_(r_0)^2}{\frac{P_0}{\rho}}\right) = -3 \ln\left(\frac{r_0}{R_0}\right) & \Rightarrow \\
& \frac{P_0}{\rho} + \frac{3}{2}u(r_0)^2 = \frac{P_0}{\rho}\left(\frac{R_0}{r_0}\right)^3 & \Rightarrow \\
& \boxed{\left| u(r_0)\right| = \sqrt{\frac{2}{3}\frac{P_0}{\rho}\frac{R_0^3 - r_0^3}{r_0^3}} } &
\end{align*}
So the above expression is the formula for the speed of the edge of the collapsing cavity at the moment when its radius reaches $r_0$ from $R_0$. It is interesting to note that there is a singularity at $r_0 \rightarrow 0$, the speed of the cavity edge increases to infinity.
