### Author Topic: Example 8b  (Read 2758 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

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