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Is there analytical solution to this heat equation?

Mathematics Asked by titanium on November 2, 2021

I have a PDE of the following form:
$$frac{1}{sintheta}frac{partial}{partialtheta}left(sinthetafrac{partial f}{partialtheta}right)+frac{1}{sin^2theta}frac{partial^2f}{partialphi^2} = Acostheta,max(cosphi, 0) + B-Cf^4~.$$

Does anyone know if an analytical solution exists for this equation? We can assume periodic boundary condition such that $f(theta,0)=f(theta, 2pi)$.

One Answer

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

$phi in [0, 2pi], ; theta in [0, pi], tag 1$

where of course we identify the points with coordinates $(theta, 0)$ and $(theta, 2pi)$ for all $theta in [0, pi]$; then any continuous function

$f: S^2 to Bbb C tag 2$

satisfies the stated condition

$f(theta, 0) = f(theta, 2pi); tag 3$

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 $(theta, 2pi)$ 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

$dfrac{1}{sintheta}dfrac{partial}{partialtheta}left(sinthetadfrac{partial f}{partialtheta}right)+dfrac{1}{sin^2theta}dfrac{partial^2f}{partialphi^2}$ $= Acostheta,max(cosphi, 0) + B - C f^4 tag 6$

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

$phi = dfrac{pi}{2}, dfrac{3pi}{2}, tag 7$

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 $Acos theta$ 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

$Acostheta,max(cosphi, 0)$ $= dfrac{1}{sintheta}dfrac{partial}{partialtheta}left(sinthetadfrac{partial f}{partialtheta}right)+dfrac{1}{sin^2theta}dfrac{partial^2f}{partialphi^2} - B + Cf^4, tag{11}$

or

$max(cosphi, 0)$ $= (Acos theta)^{-1} left (dfrac{1}{sintheta}dfrac{partial}{partialtheta}left(sinthetadfrac{partial f}{partialtheta}right)+dfrac{1}{sin^2theta}dfrac{partial^2f}{partialphi^2} - B + Cf^4 right ), tag{12}$

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 $3pi/2$, and in fact

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

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 $3pi/2$ from above, but is $0$ throughout the interval $(pi/2, 3pi/2)$; therefore $max(cos phi, 0)$ is non-differentiable at $pi/2$ and $3pi/2$; but this contradicts the fact that the right-hand side of (12) is an analytic function; thus no analytic solution of (6) exists in $S^2$.

Answered by Robert Lewis on November 2, 2021

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