### Author Topic: FE-7  (Read 3630 times)

#### Victor Ivrii

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##### FE-7
« on: December 18, 2015, 07:52:03 PM »
Solve
\begin{align}
&\Delta u=0 && x^2+y^2+z^2<1,\label{7-1}\\
&u=g(x,y,z) && x^2+y^2+z^2=1\label{7-2}
\end{align}
with $g(x,y,z)=z(x^2+y^2)$.

Hint. If $g$ is a polynomial of degree $m$ look for

u=g - P(x,y,z)(x^2+y^2+z^2-R^2)
\label{7-3}

with $P$ a polynomial of degree  $(m-2)$. Here $R$ is the radius of the ball. If $g$ has some rotational symmetry, so $P$ has.

Bonus
Represent $u$ as a sum of homogeneous harmonic polynomials.
« Last Edit: December 21, 2015, 08:06:56 AM by Victor Ivrii »

#### Yiqi Shi

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##### Re: FE-7
« Reply #1 on: December 18, 2015, 10:27:19 PM »
Since g is a polynomial of degree 3, the The degree of $P$ should be 1. Also, notice that $g$ is odd about $z$ and $P$ has the same rotational symmetry as $g$, so we can assume $P=kz$.

$$\Delta \mu=2z+2z-2kz-2kz-6kz=0 \implies k=\frac{2}{5}$$\\
Thus,

\mu=z(x^2+y^2)-\frac{2}{5}z(x^2+y^2+z^2-1)

« Last Edit: December 18, 2015, 10:29:32 PM by Yiqi Shi »

#### Vivian Tan

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##### Re: FE-7
« Reply #2 on: December 18, 2015, 10:29:00 PM »
$g$ is a degree 3 polynomial, so $P$ will be a degree 1 polynomial. $g$ also has rotational symmetry about the z axis ($x^2+y^2=s^2$ in cylindrical coordinates), so $P$ must does as well. No degree 1 polynomial in $x$ and $y$ can have such symmetry, so $P=az$ only, where $a$ is to be determined. Plugging in this expression for $P$ we obtain:
$$\Delta u=2z+2z-2az-2az-6az=0\Rightarrow a=\frac{2}{5}$$
Therefore the answer is, using $R=1$:
$$u=z(x^2 + y^2)-\frac{2}{5}z(x^2+y^2+z^2-1)$$

#### Emily Deibert

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##### Re: FE-7
« Reply #3 on: December 19, 2015, 12:06:50 AM »
$g$ is a degree 3 polynomial, so $P$ will be a degree 1 polynomial. $g$ also has rotational symmetry about the z axis ($x^2+y^2=s^2$ in cylindrical coordinates)

Thank you Vivian for explaining why the symmetry exists!

#### Chi Ma

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##### Re: FE-7
« Reply #4 on: December 19, 2015, 10:57:57 AM »
$u$ can be written as a sum of 3 homogeneous harmonic polynomials.

u = -\frac{1}{5}(z^3-3zx^2) -\frac{1}{5}(z^3-3zy^2) + \frac{2}{5}z

#### Victor Ivrii

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##### Re: FE-7
« Reply #5 on: December 21, 2015, 08:10:18 AM »
\begin{equation*}
u=\underbracket{\frac{3}{5}zx^2+\frac{3}{5}zy^2-\frac{2}{5}z^3}+\underbracket{\frac{2}{5}z}
\end{equation*}
only 2 homogeneous harmonic polynomials as in your decomposition two have the same order of homogeneity and count as one (as they could be joined)

Those who did not use "rotational symmetry and odd with respect to $z$" arguments were forced to use $u=z (x^2+y^2) - (Ax+By+Cz +D) (x^2+y^2+z^2-1)$ rather than $u=z (x^2+y^2) - Cz (x^2+y^2+z^2-1)$ and never performed calculations.
« Last Edit: December 22, 2015, 02:04:31 AM by Victor Ivrii »