Author Topic: Problem 2  (Read 8860 times)

Thomas Nutz

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Problem 2
« on: November 12, 2012, 01:13:34 PM »
Hey,

I get an ODE for problem 2, but it is not of the Euler type and I don't know how to solve it. Could anyone give me a hint?

Thanks!

Victor Ivrii

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Re: Problem 2
« Reply #1 on: November 12, 2012, 01:46:02 PM »
Hey,

I get an ODE for problem 2, but it is not of the Euler type and I don't know how to solve it. Could anyone give me a hint?

Thanks!

I am asking you to find solutions, but to try to find them? The crucial question is What ODE should satisfy $u(r)$?

Thomas Nutz

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Re: Problem 2
« Reply #2 on: November 12, 2012, 02:09:39 PM »
Sorry I don't understand. I did find an ODE for $u(r)$, which is solved by Bessel functions. When you ask "try to find solutions", what do you expect me to do?

Thanks again!

Victor Ivrii

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Re: Problem 2
« Reply #3 on: November 12, 2012, 03:05:42 PM »
Sorry I don't understand. I did find an ODE for $u(r)$, which is solved by Bessel functions. When you ask "try to find solutions", what do you expect me to do?

Thanks again!

YES, exactly what you did--found equation--and you actually found that it solved by a class of special functions (which are invented exactly to solve this and more general equation)

Ian Kivlichan

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Re: Problem 2
« Reply #4 on: November 19, 2012, 09:30:00 PM »
Hopeful solutions attached! :)

For 2.b), the ODE is exactly the same, but we switch the sign of k^2 u.

Chen Ge Qu

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Re: Problem 2
« Reply #5 on: November 19, 2012, 09:32:30 PM »
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Kerri, Hu

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Re: Problem 2
« Reply #6 on: November 19, 2012, 09:35:44 PM »
Question 2a)

Calvin Arnott

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Re: Problem 2
« Reply #7 on: November 19, 2012, 09:37:10 PM »
Problem 2

Part a. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u := u_{xx} + u_{yy} = k^2 u    $$

What ODE satisfies $u\left(r\right)$?


Answer:
We have that transforming Laplacian $\Delta_{\left(x,y\right)} := \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ into polar coordinates  $u\left(x,y\right) \rightarrow u\left(r,\theta\right)$ yields us the polar form of the Laplacian:

$$ \Delta_{\left(r,\theta\right)} := \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta ^2} $$

Since we assume a form of $u\left(r,\theta\right)$ which depends only on $r$, we have that $u\left(r\right)$ is constant with respect to $\theta$, so $u_{\theta\theta} = 0 $, and our PDE simplifies to the ODE:

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

We proceed by finding a solution for $u\left(r\right)$ in the form of Bessel's function, which solves the Bessel differential equation of the form:

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

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

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

Moreover, as $ k > 0$, $ k^2 \ne 0 $, so dividing through by $ -k^2 $ yields:

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

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

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

It's clear then that for $ s = n = 0$:

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

Because $n = 0$ is an integer, our solution for $v\left(z\right) $ is given in the form of: $ v\left(z\right) = A J_0\left(z\right) + B N_0\left(z\right) $ for some $ \{A,B\} \in \mathbb{R} $, where $J_0\left(z\right) $ is Bessel's function of the first kind of order $0$, and $N_0\left(z\right)$ is Bessel's function of the second kind, namely, the Neumann function.

$$ J_n\left(z\right) = \sum_{j=0}^{\infty}\frac{\left(-1\right)^j}{\Gamma\left(j+1\right)\Gamma\left(j+n+1\right)}\left(\frac{z}{2}\right)^{2j+n} $$

$$ N_n\left(z\right) = \lim_{s \to n} N_s\left(z\right) = \lim_{s \to n} \frac{J_{s}\left(z\right) \cos\left(\pi s\right) - J_{-s}\left(z\right)}{\sin\left(\pi s\right)}$$

$$ \implies v\left(z\right) = v\left(i k r\right) = u\left(r\right) = A J_0\left(i k r\right) + B N_0\left(i k r\right)  \phantom{\ } \blacksquare $$





Part b. Let $ k > 0 $. Find the solutions that depend only on $r$ of the equation

$$ \Delta u := u_{xx} + u_{yy} = -k^2 u    $$

What ODE satisfies $u\left(r\right)$?

Answer:
As the Laplacian term of this PDE is identical to part a. we follow the same derivation by first changing to polar coordinates and assuming a radial solution. Then we have an ODE in $u\left(r\right)$:

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

We wish to again factor this into a Bessel differential equation:

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

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

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

Let $z = k r$. Again, $k > 0$ so $k^2 \ne 0$ and dividing through gives us:

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

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

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

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

$$ \implies v\left(z\right) = v\left(k r\right) = u\left(r\right) = A J_0\left(k r\right) + B N_0\left(k r\right)  \phantom{\ } \blacksquare $$

Ian Kivlichan

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Re: Problem 2
« Reply #8 on: November 19, 2012, 09:48:54 PM »
Calvin: I wasn't really sure that for 2.b) you could have a Bessel function of a complex variable, or if you'd need to do something slightly different?

I read something about this in the textbook (Strauss), but don't have it with me and might not remember correctly. ;P

Victor Ivrii

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Re: Problem 2
« Reply #9 on: November 20, 2012, 05:48:17 AM »
Calvin: I wasn't really sure that for 2.b) you could have a Bessel function of a complex variable, or if you'd need to do something slightly different?

I read something about this in the textbook (Strauss), but don't have it with me and might not remember correctly. ;P

Actually Bessel functions are analytic functions.


If you consider $n$-dimensional problem and not necessarily spherically symmetric solution then
\begin{equation}
\Delta =\partial_r^2 + (n-1)r^{-1}\partial_r +\r^{-2}\Lambda
\label{eq-1}
\end{equation}
where $\Lambda$ is Laplace-Beltrami on the sphere ($(n-1)$-dimensional sphere $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$).

If we consider equation $\Delta u=0$ and try to solve it by separation of variables we get
\begin{equation}
\frac{r^2 R''+(n-1)r R'}{R}+\frac{\Lambda V}{V}=0
\label{eq-2}
\end{equation}
where $V$ contains spherical variables (except $r$) and therefore
\begin{align}
&\Lambda V=-\lambda V,\label{eq-3}\\
&r^2 R''+(n-1)r R'=\lambda R.\label{eq-4}
\end{align}
One can prove that such solutions $U=VR$ are polynomials of degree $m=0,1,2,\ldots$ with respect to $x$ (assuming solutions are not singular at $0$) and therefore $R=r^m$ (don't care about coefficient) and therefore $\lambda_m= m(m+n-2)$ but their multiplicities are large: the most interesting case is $n=3$ when $\lambda_m=m(m+1)$ of multiplicity $2m+1$ (application to theory of Hydrogen atom).

Solutions of (\ref{eq-3}) are spherical functions which I don't describe. Let us return to $k\ne 0$. In this case separating variables we get
\begin{equation}
\Bigl[\frac{r^2 R''+(n-1)r R'}{R}\mp k^2r^2\Bigr]+\frac{\Lambda V}{V}=0
\label{eq-5}
\end{equation}
and therefore we get (\ref{eq-3}) implying $\lambda=m(m+n-2)$ and instead of (\ref{eq-4}) we get
\begin{equation}
r^2 R''+(n-1)r R' \mp k^2r^2R -\lambda R=0
\label{eq-6}
\end{equation}
which is Bessel equation. By scaling $r_{new}=k r_{old}$ we can get rid off $k$ (and even off sign making complex scaling).

I just want to note that Bessel functions depend on $n$ and $m$. However for odd $n$ (in particular as $n=3$) and $\lambda=0$ they are elementary functions (see problem 1 for $n=3$)