Author Topic: Problem 1  (Read 4957 times)

Fanxun Zeng

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Problem 1
« on: November 28, 2012, 09:27:32 PM »
Problem 1 a b required c d bonus Solutions posting after 21:30

Is it a flood? You could at least post a link

http://www.math.toronto.edu/courses/apm346h1/20129/HA8.html#problem-8.1

And guys, don't start new topics for the same question-I merged all. V.I.
« Last Edit: November 29, 2012, 05:00:18 AM by Victor Ivrii »

Aida Razi

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Problem1
« Reply #1 on: November 28, 2012, 09:30:03 PM »
Here is the solution!

Ian Kivlichan

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Problem 1
« Reply #2 on: November 28, 2012, 09:30:04 PM »
Hopeful solutions to parts a), b), c), and d) attached!

Fanxun Zeng

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Re: Problem 1
« Reply #3 on: November 28, 2012, 09:31:54 PM »
Problem 1 part a solution attached

Fanxun Zeng

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Re: Problem 1
« Reply #4 on: November 28, 2012, 09:32:25 PM »
Problem 1 part b solution attached

Calvin Arnott

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Re: Problem 1
« Reply #5 on: November 28, 2012, 09:47:14 PM »
Problem 1

Part a. Find the solutions that depend only on $r$ of the equation
$$ \Delta u\left(x,y\right) := u_{xx} + u_{yy} = 0    $$

Answer:
We derived that transforming Laplacian $\Delta := \partial_x^2 + \partial_y^2$ into polar coordinates  $u\left(x,y\right) \rightarrow u\left(r,\theta\right)$ yields us the polar form of the Laplacian:
$$ \Delta := \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta ^2 $$
$u\left(r\right)$ is constant with respect to $\theta$, $u_{\theta\theta} = 0 $, so we have:
$$ \Delta u\left(x,y\right) := u_{xx} + u_{yy} = 0 $$
$$ \implies \Delta u\left(r\right) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = 0  $$
$$ \implies r u_{rr} + u_r = 0 \implies \partial_r \left(r u_r\right) = 0 \implies r u_r = A \implies u_r = \frac{A}{r} $$
$$ \implies u\left(r\right) =  A \log\left(r\right) + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$


Part b. Find the solutions that depend only on $\rho$ of the equation
$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = 0    $$

Answer:
We derived that transforming Laplacian $\Delta := \partial_x^2 + \partial_y^2 + \partial_z^2$ into spherical coordinates  $u\left(x,y,z\right) \rightarrow u\left(\rho,\theta,\phi\right)$ yields us the spherical form of the Laplacian:
$$ \Delta := \partial_\rho^2 + \frac{2}{\rho} \partial_\rho + \frac{1}{\rho^2}\left( \partial_\theta ^2 + \cot\left(\theta\right) \partial_\theta + \frac{1}{\sin\left(\theta\right)^2} \partial_{\phi \phi}\right) $$
$u\left(\rho\right)$ is constant with respect to $\theta$, $u_{\theta} = u_{\theta\theta} = u_{\phi \phi} = 0 $, and we have:
$$ \Delta u\left(x,y,z\right) := u_{xx} + u_{yy} + u_{zz} = 0 $$
$$ \implies \Delta u\left(\rho\right) := u_{\rho \rho} + \frac{2}{\rho} u_{\rho} = 0  $$
$$ \implies \rho^2 u_{\rho \rho} + 2 \rho u_{\rho} = 0 \implies \partial_\rho \left(\rho^2 u_{\rho}\right) = 0  $$
$$ \implies \rho^2 u_{\rho} = -A \implies u_{\rho} = -\frac{A}{\rho ^2} \implies u_{\rho} = - \frac{A}{\rho ^2} $$
$$ \implies u\left(\rho\right) =  A \rho^{-1} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$


Part c. In the $n$-dimensional case, prove that for $u=u\left(r\right)$, with $r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}} $:
\begin{equation}
\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0 \label{1.c.1}
\end{equation}

Answer:
$$ \text{Let: } r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right). $$
$$\implies \Delta u = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(r\right) = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right) = 0 $$
$$ \implies \sum_{i=1}^{n} [ \partial_{x_i}^2 u\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) ] = \sum_{i=1}^{n} [ \partial_{x_i}\frac{ x_i u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} ]$$
$$ = \sum_{i=1}^{n} [ \frac{ u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} - \frac{ x_{i}^2 u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{ x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}] $$
$$ = \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}}          +     \frac{x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}  ]$$
$$ = \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} ]         +    \frac{\left(\sum_{j=1}^{n}  x_{j}^2\right) u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\sum_{j=1}^{n}  x_{j}^2}  $$
$$ = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)         +   \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} \sum_{i=1}^{n} \left(    \frac{ \left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2 }{ \sum_{j=1}^{n} x_j^2} \right)   $$
$$ \text{Notice that: } \sum_{i=1}^{n} [    \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) }{ \left(\sum_{j=1}^{n} x_j^2\right)} ] = \sum_{i=1}^{n} \left(    \frac{ \sum_{j = 1}^{n} x_j^2 }{ \sum_{j=1}^{n} x_j^2} \right) - \sum_{i=1}^{n} \left(    \frac{ x_{i}^2}{ \sum_{j=1}^{n} x_j^2} \right)  $$
$$ = n -    \frac{ \sum_{i=1}^{n} x_{i}^2}{ \sum_{j=1}^{n} x_j^2} = n -1 \text{ so we have:}$$
$$ \implies \Delta u = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)         +  \left(n-1\right) \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} $$
$$ =  u_{rr} + \frac{n-1}{r}u_r = 0 \text{, as needed. } \blacksquare $$


Part d. In the $n$-dimensional case, $\left(n \ne 2\right)$, prove that $u = u\left(r\right)$ satisfies the Laplace equation for $r \ne 0 \iff u = A r^{2-n} + B$, $\{ A, B \} \in \mathbb{R}$.

Answer:
$$ \text{Let: } n \in  \mathbb{N} \setminus 2, \phantom{O} \{ x_1 \dots x_n \} \in \mathbb{R}^n, \phantom{O} r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right). $$
By part c. we have that the Laplacian of $u\left(r\right)$ satisfies \eqref{1.c.1}:
$$ \Delta u = u_{rr} + \frac{n-1}{r}u_r = 0 $$
If $r \ne 0$, $u\left(r\right) = A r^{2-n} + B$, $u_{r} = A\left(2-n\right) r^{1-n}$, $u_{rr} = A\left(1-n\right)\left(2-n\right) r^{-n}$ and clearly:
$$ u_{rr} + \frac{n-1}{r}u_r = A \left(1-n\right) \left(2 - n\right) r^{-n} +  \frac{n-1}{r} A \left(2 - n\right) r^{1 - n} $$
$$ = A \left(1 - n\right) \left(2 - n\right) r^{-n} - A \left(1 - n\right) \left(2 - n\right) r^{-n} = 0 \phantom{O} \square$$
Thus $u$ satisfies Laplace's equation in $r$. Conversely, if $u\left(r\right)$ satisfies Laplace's equation in $r$ \eqref{1.c.1} for $r\ne 0$, then:
$$u_{rr} + \frac{n-1}{r}u_r = 0 \implies r^{n-1} u_{rr} + \left(n-1\right)r^{n-2}u_r = 0 \implies \partial_r\left(r^{n-1} u_{r}\right)= 0 $$
$$\implies r^{n-1} u_{r}= \left(2-n\right) A \implies u_{r}= \left(2-n\right)\frac{A}{r^{n-1}} $$
$$ \implies u\left(r\right) = A r^{2-n} + B \phantom{O} \square$$
Thus we have $u = u\left(r\right)$ satisfies Laplace's equation in $r \ne 0$, $n \in  \mathbb{N} \setminus 2$,
$$\Delta u\left(r\right) = 0 \iff u\left(r\right) = A r^{2-n} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$
« Last Edit: November 28, 2012, 10:00:43 PM by Calvin Arnott »

Jinchao Lin

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question a-c
« Reply #6 on: November 28, 2012, 10:33:54 PM »
Question (a)-(c)

Jinchao Lin

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Re: Problem1
« Reply #7 on: November 28, 2012, 10:37:51 PM »
part(d)


Jinchao--are we supposed to rotate a monitor? Sure one can download and rotate an image but methinks it would be simpler for you to rotate it before posting. Also, why all these weird colours--it is a home work in math, not in visual arts :D V.I.
« Last Edit: November 29, 2012, 05:08:45 AM by Victor Ivrii »

Fanxun Zeng

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Re: Problem 1
« Reply #8 on: November 28, 2012, 11:59:50 PM »
Thanks Calvin's solutions at 21:47, especially parts c and d bonus, which are not required. I only posted parts a and b at 21:32, which are all required work for Problem 1. By the way, note in my part b, r should be replaced by ρ. Part a posted t 21:31 is OK.