### Author Topic: question 1 (a)-(b)  (Read 2570 times)

#### Yiyun Liu

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##### question 1 (a)-(b)
« on: March 19, 2015, 09:10:40 PM »
1.  Find the solutions that depend only on $r$ of the equation
\begin{equation*}
\Delta u:=u_{xx}+u_{yy}=0.
\end{equation*}
2.  Find the solutions that depend only on $\rho$ of the equation
\begin{equation*}
\Delta u:=u_{xx}+u_{yy}+u_{zz}=0.
\end{equation*}
3.  (bonus) In $n$-dimensional case prove that if $u=u(r)$ with  $r=(x_1^2+x_2^2+\ldots+x_n^2)^{\frac{1}{2}}$ then

\Delta u = u_{rr}+ \frac{n-1}{r}u_r=0.
\label{equ-H8.1}
4.  (bonus) In $n$-dimensional case prove ($n\ne 2$) that  $u=u(r)$ satisfies Laplace equation as $x\ne 0$ iff  $u=Ar^{2-n}+B$.

$\begin{array}{l} part(a):\\ \Delta u = {u_{rr}} + \frac{1}{r}{u_r} + \frac{1}{{{r^2}}}{u_{\theta \theta }} = 0\\ \Delta u = {u_{rr}} + \frac{1}{r}{u_r} = 0\\ \frac{\partial }{{\partial r}}(r{u_r}) = {u_r} + r{u_{rr}} = r({u_{rr}} + \frac{1}{r}{u_r}) = 0\\ \frac{\partial }{{\partial r}}(r{u_r}) = C\\ u = D\ln (r) + E\\ \\ part(b):\\ \Delta u = {u_{\rho \rho }} + \frac{2}{\rho }{u_\rho } + \frac{1}{{{\rho ^2}}}({u_{\theta \theta }} + \cot (\theta ){u_\theta } + \frac{1}{{{{\sin }^2}(\theta )}}{u_{\theta \theta }}) = 0\\ \Delta u = {u_{\rho \rho }} + \frac{2}{\rho }{u_\rho } = 0\\ \frac{\partial }{{\partial \rho }}({\rho ^2}{u_\rho }) = 2\rho {u_\rho } + {\rho ^2}{u_{\rho \rho }} = {\rho ^2}({u_{\rho \rho }} + \frac{2}{\rho }{u_\rho }) = 0\\ {\rho ^2}{u_\rho } = C,cons\tan ts\\ u = D\frac{1}{\rho } + E \end{array}$
« Last Edit: March 20, 2015, 07:38:30 AM by Victor Ivrii »

#### Mark Nunez

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##### Re: question 1 (a)-(b)
« Reply #1 on: March 19, 2015, 09:17:37 PM »
c,d.

« Last Edit: March 20, 2015, 01:10:46 AM by Mark Nunez »

#### Chaojie Li

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##### Re: question 1 (a)-(b)
« Reply #2 on: March 19, 2015, 09:19:32 PM »
Part c.
$$\text{Let: } r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right).$$
$$\implies \Delta u = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(r\right) = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right) = 0$$
$$\implies \sum_{i=1}^{n} [ \partial_{x_i}^2 u\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) ] = \sum_{i=1}^{n} [ \partial_{x_i}\frac{ x_i u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} ]$$
$$= \sum_{i=1}^{n} [ \frac{ u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} - \frac{ x_{i}^2 u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{ x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}]$$
$$= \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)} ]$$
$$= \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} ] + \frac{\left(\sum_{j=1}^{n} x_{j}^2\right) u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\sum_{j=1}^{n} x_{j}^2}$$
$$= u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) + \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} \sum_{i=1}^{n} \left( \frac{ \left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2 }{ \sum_{j=1}^{n} x_j^2} \right)$$
$$\text{Notice that: } \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) }{ \left(\sum_{j=1}^{n} x_j^2\right)} ] = \sum_{i=1}^{n} \left( \frac{ \sum_{j = 1}^{n} x_j^2 }{ \sum_{j=1}^{n} x_j^2} \right) - \sum_{i=1}^{n} \left( \frac{ x_{i}^2}{ \sum_{j=1}^{n} x_j^2} \right)$$
$$= n - \frac{ \sum_{i=1}^{n} x_{i}^2}{ \sum_{j=1}^{n} x_j^2} = n -1 \text{ so we have:}$$
$$\implies \Delta u = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) + \left(n-1\right) \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}}$$
$$= u_{rr} + \frac{n-1}{r}u_r = 0 \text{, as needed. } \blacksquare$$

Part d.
$$\text{Let: } n \in \mathbb{N} \setminus 2, \phantom{O} \{ x_1 \dots x_n \} \in \mathbb{R}^n, \phantom{O} r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right).$$
By part c. we have that the Laplacian of $u\left(r\right)$ satisfies part c equation named (*):
$$\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0$$
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(*) 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: March 19, 2015, 09:24:52 PM by Chaojie Li »