Toronto Math Forum
APM3462018S => APM346––Home Assignments => Topic started by: Jingxuan Zhang on March 20, 2018, 08:24:05 AM

I am referring to Q2.3 on
http://www.math.toronto.edu/courses/apm346h1/20181/PDEtextbook/Chapter8/S8.P.html
So how do I actually solve
$$\sin^2\phi \Phi''+\sin\phi\cos\phi \Phi' (l(l+1)\sin^2\phim^2)\Phi=0$$
which, suppose it's correctly derived, has cost me an entire afternoon? I remember I certain remark
in lecture that there should be constrain on $m$ and some thing like $(x+iy)^m$, but that part of my note
is very much blurred.
I heuristically plugged in $\sin,\cos,\sin^2,\cos^2,\sin\cos$ but there does not seem to be a good cancellation.

$m\le l$ and both are integers

But what then is meant by $(x+iy)^m$? are these the solutions? apparently it doesn't seem to be.

In case of 2 variables $(x\pm iy)^{m}$ are solutions (harmonic functions)

But in the hint you say the solution should be in the form of trig polyn? and the ODE should have one variable? how am I supposed to interpret this $(x\pm iy)^{m}$?

We are talking about different things, $(x\pm yi)^{m}$ is a homogeneous harmonic polynomial of two variables.
Let me give you example. Consider harmonic polynomial of order $3$, containing $z^3$ (there is only one of them, the rest differ by polynomials, which do not contain $z^3$). First of all find it: We need only $z^1$ (as we separate odd and even with respect to $z$; $zxy$ is harmonic by its own, so we take $z^3 az(x^2+y^2)$ for symmetry. Obviously, this is harmonic if $a=\frac{3}{4}$. So, consider
$$z^3\frac{3}{4}z(x^2+y^2).$$
Plugging $z=\rho\cos(\phi)$, $x=\rho\sin(\phi)\cos(\theta)$, $y=\rho\sin(\phi)\sin(\theta)$ we get
$$
\rho^3\Bigl(\cos^3(\phi) \frac{3}{4}\cos(\phi)\sin^2(\phi)\Bigr)=\rho^3\Bigl(\frac{7}{4}\cos^3(\phi)\frac{3}{4}\Bigr).
$$
Here $m=0$, obviously.
Consider harmonic polynomial of order $3$, containing $z^2$. Well it must contain $z^2x $ or $z^2y$, or, better $z^2(x\pm iy)$. This is not harmonic polynomial, to make it correct to $z^2(x\pm yi) + a(x^2+y^2)(x\pm yi)$. Obviously, this is harmonic if $a=\frac{1}{2}$.
$$
\bigl(z^2\frac{1}{2}(x^2+y^2)\bigr)(x\pm yi) = \rho^3 \bigl(\cos^2(\phi)\frac{1}[2}\sin^2(\phi)\bigl)\sin(\phi) e^{\pm i\theta}.
$$
Here $m=\pm 1$.
To get to $m=\pm 2$ consider this way $z(x\pm yi)^2$. $m=\pm 3$ consider this way $(x\pm yi)^3$. All four are harmonic without corrections.