### Author Topic: An ODE  (Read 1539 times)

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### An ODE
« on: March 20, 2018, 08:24:05 AM »
I am referring to Q2.3 on
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter8/S8.P.html

So how do I actually solve
$$\sin^2\phi \Phi''+\sin\phi\cos\phi \Phi' -(l(l+1)\sin^2\phi-m^2)\Phi=0$$
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.

« Last Edit: March 27, 2018, 07:56:32 PM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: An ODE
« Reply #1 on: March 20, 2018, 08:49:19 AM »
$|m|\le l$ and both are integers

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: An ODE
« Reply #2 on: March 20, 2018, 01:12:50 PM »
But what then is meant by $(x+iy)^m$? are these the solutions? apparently it doesn't seem to be.

#### Victor Ivrii ##### Re: An ODE
« Reply #3 on: March 20, 2018, 01:34:59 PM »
In case of 2 variables $(x\pm iy)^{|m|}$ are solutions (harmonic functions)

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: An ODE
« Reply #4 on: March 21, 2018, 08:28:35 AM »
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 $(x\pm iy)^{|m|}$?

#### Victor Ivrii ##### Re: An ODE
« Reply #5 on: March 21, 2018, 10:55:58 AM »
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.