Author Topic: Example 8b  (Read 2489 times)

Thomas Nutz

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Example 8b
« on: September 28, 2012, 05:02:12 PM »
Dear all,

in the 8th lecture notes in the example b) we end up at
$$u(x,t)=\frac{1}{2}(sin(x+tt)+sin(x-t))$$ for $x>t$
and $$u(x,t)=\frac {1}{2}(sin(x+t)-sin(x-t))$$ for 0<x<t. But isn't the latter $u$ valid on $-t<x<t$, and can't we even say that
$$u(x,t)=\frac{1}{2}(-sin(x+t)-sin(x-t))$$ for $x<-t$?

Thomas Nutz

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Re: Example 8b
« Reply #1 on: September 28, 2012, 05:15:08 PM »
... and in example 8d): Why should $u$ be given by eq. (9)? Shouldn't that be eq. (9) with all sines exchanged by cosines?

Victor Ivrii

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Re: Example 8b
« Reply #2 on: September 28, 2012, 05:36:19 PM »
If we are looking for $t>0$ then we are interested in $0<x<ct$ and $x>ct$ because the original problem is dealing with $x>0$. Continuation is the method to reduce it to IVP but we need to come back

Transitions in (9), (12) from the middle to the r.h. expression is due to the trigonometric formulae. Everything is ok

PS. Please use \sin,  \cos etc instead of sin, cos in LaTeX code as those are predefined macros.

PPS. In Lecture 9 we use \erf but as this is not a predefined macro, we define it by ourselves