Author Topic: HA7 problem 2  (Read 2629 times)

Victor Ivrii

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HA7 problem 2
« on: March 11, 2015, 08:20:22 AM »
a. Try to find the solutions that depend only on $r$ of the equation
    \begin{equation*}
    \Delta u:=u_{xx}+u_{yy}=k^2u,
    \end{equation*}
    where $k$ is a positive constant. What ODE should satisfy $u(r)$?

b.  Try to find the solutions that depend only on $r$ of the equation
    \begin{equation*}
    \Delta u:=u_{xx}+u_{yy}=-k^2u,
    \end{equation*}
    where $k$ is a positive constant. What ODE should satisftfy $u(r)$?

Biao Zhang

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Re: HA7 problem 2
« Reply #1 on: March 12, 2015, 09:20:56 PM »
(a)
 The form of Bessel's function,
 
 $$\frac {\partial^2 y}{\partial x^2}+\frac{1}{x} \frac{\partial y}{\partial x}+(1+\frac{n^2}{x^2})y=0$$
 
we can rewrite the equation:

$$ u_{rr} + \frac{1}{r} u_{r} +(1-\frac{s^2}{r^2})u = 0 $$

$$ \text{Let: } u(r) = v(i k r) \implies u_{r} = i k v_{r}, \phantom{\ } u_{rr} = - k^2 v_{rr} $$

$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} - k^2 u = 0$$

$$- k^2 v_{rr} + \frac{1}{r} i k v_{r} - k^2 v = 0 $$

Since $ k >0$, $ k^2 \ne 0 $, so:

$$ v_{rr} - \frac{i}{k r} v_{r} + v = 0 $$

$ - \frac{i}{k r} = \frac{(-i)}{(1) k r} = \frac{(-i)}{(-i * i) k r} = \frac{\overline{i}}{\overline{i}} \frac{1}{i k r} = \frac{1}{i k r} $ so our ODE in $v(i k r)$, say $v(z)$ where $z = i k r$ is equivalent to:

$$ v_{rr} + \frac{1}{i k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$

Since $v_{rr} + \frac{1}{i k r} v_{r} = -v $ is a linear combination of Bessel funtion of $J_0 \& N_0$

$$ \implies v(z) = v(i k r) = u(r) = A_0 J_0(i k r) + B_0 N_0(i k r) \phantom{\ } $$


(b)
As (a):

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

rewrite in Bessel equation:

$$ u_{zz} + \frac{1}{z} u_{z} +(1-\frac{s^2}{z^2})u = 0 $$

$$ \text{Let: } u(r) = v(k r) \implies u_{r} = k v_{r}, \phantom{\ } u_{rr} = k^2 v_{rr} $$

$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} + k^2 u = 0 arrow k^2 v_{rr} + \frac{1}{r} k v_{r} + k^2 v = 0 $$

Let $z = k r$. Again, $k >0$ so $k^2 \ne 0$ :

$$ v_{rr} + \frac{1}{k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$

Which is clearly a Bessel's differential equation with $s = n = 0$:

$$ v_{zz} + \frac{1}{z} v_{z} + v = v_{zz} + \frac{1}{z} v_{z} +(1-\frac{s^2}{z^2})v = 0 $$

So our solution for $v(z)$ is again $ v(z) = A J_0(z) + B N_0(z) $ for some $ \{A,B\} \in \mathbb{R} $.

$$ \implies v(z) = v(k r) = u(r) = A_0 J_0(k r) + B_0 N_0(k r) \phantom{\ } $$