### Author Topic: Wednesday's quiz  (Read 1828 times)

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Wednesday's quiz
« on: March 29, 2018, 09:20:21 AM »
I was not there but I heard from my friend that they were asked to find harmonic extension in $B_1(0)$ of $g(x,y,z)=x^4+y^4+z^4$ given on $C_1(0)$.

Solution $u$ is sought in the form
\begin{equation}\label{1}u=g-P(x,y,z)(\rho^2-1),\,\rho=\sqrt{x^2+y^2+z^2}.\end{equation}
Where $P$ is a polynomial even and symmetric in $x,y,z$, as does $g$, and $\deg(P)=\deg(g)-2=2$ . Therefore $P=P(\rho)=a\rho^2+b$ for some constant $a,b$, and so \eqref{1} becomes
\begin{equation}\label{2}u=g-(a\rho^4+(b-a)\rho^2-b).\end{equation}
Observe $$\Delta\rho^2=6,\,\Delta\rho^4=20\rho^2,\,\Delta g=12\rho^2.$$ Now set $\Delta u=0$ and \eqref{2} gives
\begin{equation}\label{3}0=12\rho^2-(20a\rho^2+6(b-a)).\end{equation}
Equating both sides of \eqref{3} term by term we find $a=b=\frac{3}{5}$ and so \eqref{2} becomes
$$u=g-\frac{3}{5}(\rho^4-1)=\frac{2}{5}(x^4+y^4+z^4)-\frac{6}{5}(x^2y^2+x^2z^2+y^2z^2)+\frac{3}{5}.$$
« Last Edit: March 29, 2018, 03:15:45 PM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: Wednesday's quiz
« Reply #1 on: March 29, 2018, 10:15:58 AM »
Need to mention that $\Delta g=12\rho^2$.

You should not post before T0101 class (their problem is similar, even not the same)

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: Wednesday's quiz
« Reply #2 on: March 29, 2018, 03:14:47 PM »
Of course I wasn't aware of that. But $\Delta g =12\rho^2$ is really there. I should probably have displayed it.