APM346-2015F > Final Exam

FE-7

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Victor Ivrii:
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
\begin{equation}
u=g - P(x,y,z)(x^2+y^2+z^2-R^2)
\label{7-3}
\end{equation}
 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.

Yiqi Shi:
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,
\begin{equation}
\mu=z(x^2+y^2)-\frac{2}{5}z(x^2+y^2+z^2-1)
\end{equation}

Vivian Tan:
$g$ 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:

--- Quote from: Vivian Tan 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)

--- End quote ---

Thank you Vivian for explaining why the symmetry exists!

Chi Ma:
$u$ can be written as a sum of 3 homogeneous harmonic polynomials.
\begin{equation}
u = -\frac{1}{5}(z^3-3zx^2) -\frac{1}{5}(z^3-3zy^2) + \frac{2}{5}z
\end{equation}

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