### Author Topic: Thursday's quiz  (Read 2298 times)

#### Jingxuan Zhang

• Elder Member
• Posts: 106
• Karma: 20
##### Thursday's quiz
« on: March 29, 2018, 03:21:42 PM »
It was question 3.3 as of
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter8/S8.P.html

Since $g$ is already harmonic, it's harmonic extension in unit ball has the same formula! Moreover $g(kx)=k^3g(x)$ so this formula is in fact a sum of homogeneous harmonic polynomial consisting of one term! So
$$u(x,y,z)=xyz,x^2+y^2+z^2\leq 1$$ or $$\tilde{u}(\rho,\theta,\phi)=\rho^3\cos^2\phi\sin\phi\cos\theta\sin\theta,0\leq\rho\leq1,0\leq\theta\leq 2\pi, 0\leq\phi\leq\pi.$$
« Last Edit: March 29, 2018, 03:23:16 PM by Jingxuan Zhang »

#### Zhongnan Wu

• Jr. Member
• Posts: 5
• Karma: 2
##### Re: Thursday's quiz
« Reply #1 on: April 02, 2018, 07:28:17 PM »
u=xyz-P(x, y, z)(x2+y2+z2-R2)
u=-[Pxx(x2+y2+z2-R2)+2xPx+2xPx+2P+Pyy(x2+y2+z2-R2)+2yPy+2yPy+2P+ Pzz(x2+y2+z2-R2)+2zPz+2zPz+2P]=0
Rearranging it we can get:
P(x2+y2+z2-R2)+4(xPx+yPy+zPz)+6P=0
As P is rotational symmetric, we first consider for P(x, x, x)=mx+n
Then 3Pxx(3x2-R2)+12Pxx+6P=0
Plug in P=mx+n, Px=m, Pxx=0, we can get m=n=0
So P(x, y, z)=0
Then u=xyz