# Is there analytical solution to this heat equation?

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

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

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

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 $$S^2$$.

• 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? Jul 21 '20 at 19:18
• @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. Jul 22 '20 at 0:20
• @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! Jul 22 '20 at 0:24
• @titanium: hey, muchas gracias for the bounty! Cheers! Jul 24 '20 at 17:46