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If $u(\xi=0+, \eta)=u(\xi=0-,\eta)$

Does this mean

$\lim \limits_{\xi \to 0+}u(\xi,\eta)=\lim \limits_{\xi \to 0-}u(\xi,\eta)$ ?

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usainlightning
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1 Answers1

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Yes, that would be the meaning, if I should take a guess - especially if it makes sense in whatever context you've seen it in.

Mankind
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  • I'm not sure if it does make sense. The next line states: $\frac{\partial u}{\partial \eta}(0+,\eta$)=$\frac{\partial u}{\partial \eta}(0-,\eta$). We have assumed $u$ is continuous at $(0,\eta)$. – usainlightning Apr 01 '15 at 19:17
  • That next line looks like you have also assumed that the partial derivative wrt. $\eta$ is continuous at $(0,\eta)$. Is this so? – Mankind Apr 01 '15 at 19:25
  • The only extra information I have is that $u$ is smooth for $\xi \neq 0$. Does this not mean that the partial derivatives exist everywhere except $\xi=0$? – usainlightning Apr 01 '15 at 21:45
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    @usainlightning yes, it does. So I don't know why your source writes up the definition of continuity at $(0,\eta)$, but there's probably a reason given in the context. – Mankind Apr 02 '15 at 18:14