I am reading both Griffiths and Gasiorowicz and I can't get either of them to tell me why the continuity of the derivative of the natural log of the amplitude(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{d(ln(u(x)))}{dx}=\frac{1}{u(x)}\frac{du(x)}{dx}[/tex]

or put a different way

[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]

at a boundary a implies that

[tex]u(a^{-})=u(a^{+})[/tex]

and

[tex]\frac{du(a^{-})}{dx}=\frac{du(a^{+})}{dx}[/tex]

Why is this true?

It looks to me that there still is some ambiguity (as in the top equation is neccessary but not sufficient) because

[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]

can be reconfigured to

[tex]\frac{u(a^{+})}{u(a^{-})}=\frac{\frac{du(a^{+})}{dx}}{\frac{du(a^{-})}{dx}}[/tex]

Which just means that the ratios have to be equal. But say for instance that [tex]u(a^{+})[/tex] is twice as big as [tex]u(a^{-})[/tex] so that the ratio is 2. Then all that means is the ratio of the slopes at a has to be 2. So, I don't see where they are limited to be identical.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Continuity of Function and Derivative at boundary.

Loading...

Similar Threads for Continuity Function Derivative |
---|

B Do systems continually evolve? |

A Classical Mechanics: Continuous or Discrete universe |

I Disctrete vs continuous spectrum |

I Understanding Continuous Variable QKD |

B Inner product of functions of continuous variable |

**Physics Forums | Science Articles, Homework Help, Discussion**