Q1

From lecture section 28.1, we have

\begin{equation}

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +

\frac{1}{\rho^2}\Lambda

\end{equation}

with

\begin{equation}

\Lambda:=

\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.

\end{equation}

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$: \begin{equation}

P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

\end{equation}

which could be rewritten as

\begin{equation}

\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+

\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0

\end{equation}

and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:

\begin{align}

\rho^2 P'' +2\rho P' = \lambda P,\\

\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).

\end{align}

The first equation is of Euler type and it has solutions $P:=\rho^l$ iff

$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$

Then suppose that

\begin{align}

u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2) + c(1-x^2-y^2-z^2) ^2\\\\

=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2

\end{align}

\begin{align}

\Delta u &= uxx+uyy+uzz\\

&=12 z^2 -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\

&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\

&=12z^2-6a-12bz^2-6c+20c-20cz^2\\

&=(12-12b-20c)z^2-6a-6c=0

\end{align}

Then we have

\begin{align}

12-12b-20c=0\\

-6a-6c=0

\end{align}

Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.