"Mathematical Physics by Carl Bender"의 두 판 사이의 차이

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==lecture 2==
 
==lecture 2==
 
* $y''+Q(x)y=0$ Schrodinger equation
 
* $y''+Q(x)y=0$ Schrodinger equation
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* {{수학노트|url=이계_선형_미분방정식}}
  
  

2014년 3월 1일 (토) 08:35 판

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lecture 1 perturbation method

  • solve $x^5+x=1$

method 1

  • try $x^5+\epsilon x=1$
  • find $x(\epsilon)$ satisfying $x(\epsilon)^5+\epsilon x(\epsilon)=1$
  • answer

$$x(\epsilon)=1-\frac{\epsilon }{5}-\frac{\epsilon ^2}{25}-\frac{\epsilon ^3}{125}+\frac{21 \epsilon ^5}{15625}+\frac{78 \epsilon ^6}{78125}+\cdots$$

  • Setting $\epsilon=1$ gives numerical value $0.75\cdots$


weak coupling approach

  • use the similar idea to Feynman diagrams
  • try $\epsilon x^5+ x=1$
  • we get

$$ x(\epsilon)=1-\epsilon +5 \epsilon ^2-35 \epsilon ^3+285 \epsilon ^4-2530 \epsilon ^5+23751 \epsilon ^6+\cdots $$

  • can we get a meaningful number out of this?
  • yes, for example, Pade summation can be used


asymptotics

  • $f\sim g\, \quad (x\to x_0)$ iff $$\lim_{x\to x_0}\frac{f(x)}{g(x)}=1$$
  • apply the method of dominant balance to $\epsilon x^5+ x=1$
  • $x^4\sim -1/\epsilon\, \quad (\epsilon \to 0)$ and thus

$$x\sim \frac{\omega}{\epsilon^{1/4}}\, \quad (\epsilon \to 0)$$ where $\omega^4=-1$

  • this is the first order approximation and we can have more terms


lecture 2


computational resource


books

  • Bender, Carl M., and Steven A. Orszag. 1999. Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory. Springer.