### Author Topic: HA9 Problem 1  (Read 4187 times)

#### Zacharie Leger

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##### HA9 Problem 1
« on: March 24, 2015, 10:56:40 PM »
Find function $u$ harmonic in $\{x^2+y^2+z^2\le 1\}$ and coinciding with $g=z^4$ as $x^2+y^2+z^2=1$.

Hint. According to [Subsection 28.1] solution must be a harmonic polynomial of degree $4$ and it should depend only on $x^2+y^2+z^2$ and $z$ (Explain why). The only way to achive it (and still coincide with $g$ on $\{x^2+y^2+z^2=1\}$) is to find
\begin{equation*}
u= z^4 + az^2(1-x^2-y^2-z^2)+b(1-x^2-y^2-z^2)^2
\end{equation*}
with unknown coefficients $a,b$.

It seems to me that we should have a harmonic polynomial of degree 4 if we want the function to coincide with $g(x)=z^4$ on ${x^2+y^2+z^2=1}$, I'm I missing something?
« Last Edit: March 26, 2015, 02:50:38 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Question 1
« Reply #1 on: March 25, 2015, 07:03:59 AM »
Thanks, corrected.

#### Chaojie Li

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##### Re: HA9 Problem 1
« Reply #2 on: March 26, 2015, 08:06:07 PM »
This is Q1

#### Jessica Chen

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##### Re: HA9 Problem 1
« Reply #3 on: March 26, 2015, 11:26:42 PM »
Q1
From lecture section 28.1, we have

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

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

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

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

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.

« Last Edit: March 28, 2015, 11:28:19 AM by Jessica Chen »

#### Mark Nunez

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##### Re: HA9 Problem 1
« Reply #4 on: March 27, 2015, 12:35:47 AM »
I found different values for a,b.

#### Victor Ivrii

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##### Re: HA9 Problem 1
« Reply #5 on: March 27, 2015, 11:47:05 AM »
All do correct things but do errors in calculations. Actually, it was partially my fault as one needs to look at
\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*}
Calculation of Laplacian applied to first three terms is very easy and we get
$$12 z^2-6a +b (2-2x^2-2y^2-16z^2) = 12 z^2 -6a + 2b(1 -\rho^2 -7 z^2)$$
and for the fourth term $v=c(1-2\rho^2+\rho^4)$ we apply Laplacian in the spherical coordinates
$$v_{\rho\rho}+ 2\rho^{-1}v_\rho= c(-6 + 20\rho^2).$$ So
$$\Delta u= 12 z^2 -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2$$
« Last Edit: March 27, 2015, 12:07:49 PM by Victor Ivrii »

#### Chaojie Li

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• Posts: 21
• Karma: 0
##### Re: HA9 Problem 1
« Reply #6 on: March 28, 2015, 01:25:04 PM »

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

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

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

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

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 &= u_{\xx}+u_{\yy}+u_{zz}\\
&=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.
[/quote]
« Last Edit: March 28, 2015, 01:28:09 PM by Chaojie Li »

#### Chaojie Li

• Full Member
• Posts: 21
• Karma: 0
##### Re: HA9 Problem 1
« Reply #7 on: March 28, 2015, 01:29:47 PM »
most part code is copied from Chen

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

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

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

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

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 &= u_{xx}+u_{yy}+u_{zz}\\
&=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.
[/quote]