Solving a 2D partial differential equation from an old science paper

Question

I am reading an atmospheric science paper (B. J. Hoskins, 1985) and stumble on a second-order partial differential equation that after some simplification roughly reads \begin{align*} \frac{\partial}{\partial r} (\frac{1}{r} \frac{\partial (rv)}{\partial r}) + \frac{\partial^2 v}{\partial t^2} &= \frac{\partial P}{\partial r} \\ \frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2} + \frac{\partial^2 v}{\partial t^2} &= \frac{\partial P}{\partial r} \end{align*} where $v(r, t)$ is some function of $r$ and $t$. $P$ is a quantity that has a value of $1+\epsilon$ inside the circular region $r^2 + t^2 = C^2$, and and $1$ outside the circle, where $\epsilon$ is a very small number. The boundary conditions should be $v = 0$, when $r \to 0$ and $r \to \infty$. The solution given in the paper is \begin{align*} v &= \frac{1}{3} \epsilon r & r^2 + t^2 < C^2 \\ v &= \frac{1}{3} \epsilon r (\frac{C^2}{r^2 + t^2})^{3/2} & r^2 + t^2 > C^2 \end{align*} The paper claims that this is a straight forward theoretical exercise, but I have no idea how to arrive at the given solution. I have tried Separation of Variables but I can at most deduce that $v \propto r$ inside the circular region. Laplace Transform and Fourier Transform seems futile. I have tried to use Duhamel's Principle but my answer is wrong. I am thinking if some transformations can be applied. Can anyone give me some suggestions? Any help would be appreciated.

Answer 1

I have found a way (although not perfect) to derive the solution in a reasonable way. First make a transformation $Q = v/r$ to make the equation simpler, leaving three terms in the Equation. Then, by observation, one solution for $Q$ is a constant. Another solution for $Q$ is motivated by the circular boundary. Let $z = r^2 + t^2$ and assume that the solution is solely a function of $z$. By Chain Rule, the equation can be transformed into an ordinary equation in $z$ and it is possible to get another solution $1/z^{3/2}$. Due to the physics, we know that $Q = \alpha$ inside the circle, and $Q = \beta/z^{3/2}$ outside the circle. Finally, we consider the difference in P across the interface. By integrating over a small neighborhood at the interface, $\alpha$ and $\beta$ can be determined.