I have a PDE of the following form: $$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial f}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2f}{\partial\phi^2} = A\cos\theta\,\max(\cos\phi, 0) + B-Cf^4~.$$
Does anyone know if an analytical solution exists for this equation? We can assume periodic boundary condition such that $f(\theta,0)=f(\theta, 2\pi)$.
Not really sure I am a reputable source, but perhaps my answer will nonetheless prove useful: ;)
I would like to point out at the beginning that the given equation (cf. (6) below) is properly speaking an elliptic partial differential equation, only a heat equation in the sense that it models a static distribution of heat/temperature; note that there are no $t$ (time) derivatives present. This being said,
I assume
$A \ne 0. \tag 0$
I further assume $\phi$, $\theta$ are the usual coordinates on the two-sphere $S^2$, with
$\phi \in [0, 2\pi], \; \theta \in [0, \pi], \tag 1$
where of course we identify the points with coordinates $(\theta, 0)$ and $(\theta, 2\pi)$ for all $\theta \in [0, \pi]$; then any continuous function
$f: S^2 \to \Bbb C \tag 2$
satisfies the stated condition
$f(\theta, 0) = f(\theta, 2\pi); \tag 3$
note that (2) encompasses the case
$f: S^2 \to \Bbb R, \tag 4$
and indeed, (3) may be extended to cover all cases of the form
$f: S^2 \to Y, \tag 5$
where $Y$ is an arbitrary topological space and $f$ is a continuous map; of course this generalization binds by virtue of the fact that the points having coordinates $(\theta, 0)$ and $(\theta, 2\pi)$ are identified for all $\theta \in [0, \pi]$.
I mention these observations since it is not a priori clear that an $f(\theta, \phi)$ which satisfies the given equation
$\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial f}{\partial\theta}\right)+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2f}{\partial\phi^2}$ $= A\cos\theta\,\max(\cos\phi, 0) + B - C f^4 \tag 6$
is meant to be real or complex valued. The case (5) was added as a (nearly) obvious generalization, though it has no direct application here.
Having said these things, we show that a solution $f(\theta, \phi)$ cannot be analytic in the vicinity of any point $p \in S^2$ with
$\phi = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \tag 7$
whose $\theta$ coordinate excludes
$\theta = 0, \dfrac{\pi}{2}, \pi. \tag 8$
The second of these conditions (8) implies that neither
$\cos \theta, \sin \theta = 0 \tag 9$
at $p$, and thus that every coefficient of every derivative of $f$ occurring in (6) is in fact an analytic function of $\theta$, as is also the coefficient $A\cos \theta$ of $\max(\cos \phi, 0)$. Since we have chosen $\theta$ such that
$\cos \theta \ne 0, \tag{10}$
equation (6) may be written in the form
$A\cos\theta\,\max(\cos\phi, 0)$ $= \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial f}{\partial\theta}\right)+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2f}{\partial\phi^2} - B + Cf^4, \tag{11}$
or
$\max(\cos\phi, 0)$ $= (A\cos \theta)^{-1} \left (\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial f}{\partial\theta}\right)+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2f}{\partial\phi^2} - B + Cf^4 \right ), \tag{12}$
which itself expresses $\max(\cos \phi, 0)$ as an analytic function. As such, $\max(\cos \phi, 0)$ must be everywhere differentiable; but since $\cos \phi$ changes sign from positive to negative at $\pi/2$, and from negative to positive at $3\pi/2$, and in fact
$\cos \phi < 0, \; \dfrac{\pi}{2} < \phi < \dfrac{3\pi}{2};\tag{13}$
furthermore,
$(\cos \phi)' = -\sin \phi, \tag{14}$
the derivative of $\max(\cos \phi, 0)$ approaches $-1$ as $\phi$ approaches $\pi/2$ from below, and $1$ as $\phi$ approaches $3\pi/2$ from above, but is $0$ throughout the interval $(\pi/2, 3\pi/2)$; therefore $\max(\cos \phi, 0)$ is non-differentiable at $\pi/2$ and $3\pi/2$; but this contradicts the fact that the right-hand side of (12) is an analytic function; thus no analytic solution of (6) exists in $S^2$.
