Author Topic: Lecture 20 Notes  (Read 9930 times)

Chiara Moraglia

  • Jr. Member
  • **
  • Posts: 9
  • Karma: 1
    • View Profile
Lecture 20 Notes
« on: November 11, 2012, 03:28:11 PM »
Hi, I was just wondering what the symbol O stands for in the notes of lecture 20, for example in formula (10). Thanks!

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Lecture 20 Notes
« Reply #1 on: November 11, 2012, 05:18:44 PM »
Hi, I was just wondering what the symbol O stands for in the notes of lecture 20, for example in formula (10). Thanks!

Too bad that in Calculus I you were not given the very standard math notations:

$f=O(g)$ means that $\frac{f}{g}$ is bounded;

$f=o(g)$ means that $\frac{f}{g}$ tends to $0$;

$f\sim g$ means that $\frac{f}{g}$ tends to $1$

and a bit less traditional

$f \asymp g$ means that $c^{-1}\le |\frac{f}{g}|\le c$ (or equivalently $f=O(g)$ and $g=O(f)$).

Thomas Nutz

  • Full Member
  • ***
  • Posts: 26
  • Karma: 1
    • View Profile
Re: Lecture 20 Notes
« Reply #2 on: November 13, 2012, 12:24:36 PM »
In the section \textbf{Laplace equation in half-plane} it says

"The problem $u_{yy}+u_{xx}=0$, $y>0$,  $u(x,0)=g(x)$ [...] is not uniquely solvable". As an example the function $u(x,y)=y$ is given, which satisfies the Laplace equation. But it does obviously not satisfy $u(x,0)=g(x)$, so I don't see how we can conclude immediately that only bounded solutions are unique...

Can anybode help me with this?
Thanks!

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Lecture 20 Notes
« Reply #3 on: November 13, 2012, 03:28:17 PM »
In the section \textbf{Laplace equation in half-plane} it says

"The problem $u_{yy}+u_{xx}=0$, $y>0$,  $u(x,0)=g(x)$ [...] is not uniquely solvable". As an example the function $u(x,y)=y$ is given, which satisfies the Laplace equation. But it does obviously not satisfy $u(x,0)=g(x)$, so I don't see how we can conclude immediately that only bounded solutions are unique...

Can anybode help me with this?
Thanks!

It satisfies $u(x,0)=0$; so does trivial solution $u=0$.