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 $f(\theta,0)=f(\theta, 2\pi)$.

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    $\begingroup$ Are $A$, $B$, $C$ constants? $\endgroup$ Jul 17 '20 at 2:27
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    $\begingroup$ @RobertLewis Yes they are constants. $\endgroup$
    – titanium
    Jul 17 '20 at 6:16
  • $\begingroup$ Do you want an explicit formula for all solutions? Say, what the answer for your question would you expect for the standard heat equation $u_t=u_{xx}$? $\endgroup$
    – Andrew
    Jul 18 '20 at 8:42
  • $\begingroup$ No, I am only interested in steady state solution, so I don't care about time dependence. $\endgroup$
    – titanium
    Jul 18 '20 at 19:58
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    $\begingroup$ @titanium: I answered your question, hope it helps! Cheers! $\endgroup$ Jul 21 '20 at 16:21

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

$A\cos\theta\,\max(\cos\phi, 0)$ $= \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial f}{\partial\theta}\right)+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2f}{\partial\phi^2} - B + Cf^4, \tag{11}$


$\max(\cos\phi, 0)$ $= (A\cos \theta)^{-1} \left (\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial f}{\partial\theta}\right)+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2f}{\partial\phi^2} - B + Cf^4 \right ), \tag{12}$

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

$\cos \phi < 0, \; \dfrac{\pi}{2} < \phi < \dfrac{3\pi}{2};\tag{13}$


$(\cos \phi)' = -\sin \phi, \tag{14}$

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 $S^2$.

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    $\begingroup$ This is very useful. I was trying to obtain the numerical solution of this equation using Python. But, I keep getting a very unstable solution. Do you have any idea on numerical implementation for the case of $A=C$ and $B=0$, for example? $\endgroup$
    – titanium
    Jul 21 '20 at 19:18
  • $\begingroup$ @titanium: Thanks for the kind words, I suspect implementing a numerical solution will have to be handled very carefully to avoid instabilities. You might consider pulling the equation back to a rectangular coordinate patch in the plane, bearing in mind that this might add some new variable coefficients due to non-flat geometry of the sphere; then use a fine gridding of the planar patch to keep the truncation errors down; then write out the equations in terms of the point variables which arise from discretization. $\endgroup$ Jul 22 '20 at 0:20
  • $\begingroup$ @titanium: this will give a big nonlinear algebraic system, to which if you are lucky you can apply Newton's method. Such things have been known to work, but not without some effort. Letting $B = 0$ and $A = C$ probably won't simplify things too much. Solving nonlinear PDEs on nonlinear spaces is a nasty business. Anyway, if you really find my answer and remarks helpful, you might consider giving my answer formal acceptance. Best of luck with it. Cheers! $\endgroup$ Jul 22 '20 at 0:24
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    $\begingroup$ @titanium: hey, muchas gracias for the bounty! Cheers! $\endgroup$ Jul 24 '20 at 17:46

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