Author Topic: HA7 Problem 1  (Read 2689 times)

Victor Ivrii

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HA7 Problem 1
« on: March 11, 2015, 08:18:47 AM »
a.  Find the solutions that depend only on $r$ of the equation
    \begin{equation*}
    \Delta u:=u_{xx}+u_{yy}+u_{zz}=k^2u,
    \end{equation*}
    where $k$ is a positive constant. (Hint:  Substitute $u=v/r$)

b.  Find the solutions that depend only on $r$ of the equation
    \begin{equation*}
    \Delta u:=u_{xx}+u_{yy}+u_{zz}=-k^2u,
    \end{equation*}
    where $k$ is a positive constant. (Hint: Substitute $u=v/r$)

Biao Zhang

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Re: HA7 Problem 1
« Reply #1 on: March 12, 2015, 09:20:08 PM »
(a)
in the polar variables:\\

$$ \Delta u(r,\theta,\phi) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}(u_{\theta\theta} +\cot(\theta)u_\theta + \frac{1}{\sin(\theta)^2}u_{\phi\phi}) = k^2 u $$

Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:

$$ \Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = k^2 u $$

Substituting in $ u(r) = \frac{v(r)}{r} $ then:

$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $$,
 $$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3} $$
 
$$\implies$$

$$  \Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = k^2  \frac{v}{r}  $$

$$\implies$$

$$ \frac{v_{rr}}{r}=k^2\frac{v}{r}\implies v_{rr} = k^2 v $$

Since we have that $ k >0 $, $k^2 >0$,  :

$$ v(r) = c_1 e^{k r} + c_2 e^{-k r} = c_3 \cosh(k r) + c_4 \sinh(k r) $$

$$ \implies u(r) = \frac{1}{r} v(r) = c_1\frac{e^{k r}}{r}+c_2\frac{e^{-k r}}{r} =c_3\frac{\cosh(k r)}{r}  + c_4 \frac{\sinh(k r)}{r}\phantom{\ } $$

where $ \{c_1,c_2,c_3,c_4\} \in \mathbb{R} $

(b)

Similar to a:\\

$$ \Delta u(r,\theta,\phi) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}(u_{\theta\theta} +\cot(\theta)u_\theta + \frac{1}{\sin(\theta)^2}u_{\phi\phi}) = -k^2 u $$

Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:

$$ \Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = -k^2 u $$

Substituting in $ u(r) = \frac{v(r)}{r} $ then:

$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2} $$,
 $$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3} $$
 
$$\implies$$

$$  \Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = -k^2  \frac{v}{r}  $$

$$\implies$$

$$ v_{rr} = - k^2 v $$

Since we have that $ k >0 $, $k^2 >0$,  :

$$ v(r) = c_1 \sin (k r) + c_2 \cos (k r) $$

$$ \implies u(r) = \frac{v}{r} = c_1\frac{\sin (k r)}{r}+c_2\frac{\cos (k r)}{r} $$

where $ \{c_1,c_2\} \in \mathbb{R} $