### Author Topic: S2.3 problem2(17)(18)  (Read 2500 times)

#### Yilin Ye

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##### S2.3 problem2(17)(18)
« on: January 30, 2019, 11:49:08 PM »
Do we need to consider all situations when calculating
\begin{matrix}\frac{1}{2} \int_{x-ct}^{x+ct} h(x)\, dx\end{matrix}

Like
(1) x-ct,x+ct >1
(2)x+ct>1, x-ct<1
(3) -1<x-ct<x+ct<1
(4)-1<x+ct<1,x-ct<-1
(5)x-ct,x+ct<-1

#### Victor Ivrii

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##### Re: S2.3 problem2(17)(18)
« Reply #1 on: January 31, 2019, 03:47:08 AM »
duplicate removed.

Yes, you need to consider all cases and it is recommended to draw a plane and different regions there

#### Mengmeng Shang

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##### Re: S2.3 problem2(17)(18)
« Reply #2 on: January 31, 2019, 04:53:26 AM »
Dear professor,

do we always assume that x-ct < x+ct? or we have to take cases? thanks

#### Victor Ivrii

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##### Re: S2.3 problem2(17)(18)
« Reply #3 on: January 31, 2019, 07:19:41 AM »
Since in this problem nothing is said about $t>0$ the complete solution should cover all cases. However, since in the problems either $g(x)=0$, or $h(x)=0$ then $u(x,t)$ is odd or even with respect to $t$, respectively. At least in some problems you can observe that solution must be even or odd with respect to $x$ as well.

In such problem, as (17), we have several regions. But we need to work out only some of them and extend to the rest by above arguments.

#### Wanying Zhang

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##### Re: S2.3 problem2(17)(18)
« Reply #4 on: January 31, 2019, 08:47:07 AM »
Professor, may I ask how to derive all the five conditions mentioned above? Why it's not necessary to consider the condition like ${x+ct>1, x-ct<-1}$ or ${x-ct>1, x+ct,-1}$ something like this?

#### Victor Ivrii

No, you draw lines according to conditions: $x=1+ct$, $x=-1+ct$, $x=1-ct$ and $x=-1-ct$.
And normally one need to consider all 9 domains. However, our problems have two symmetries and it is sufficient to consider only 4 domains intersecting with the 1st quadrant and extend solution to the remaining 5, using the fact that solution is either odd or even with respect to $t$ (you need to understand, what is the case), and also solution is either odd or even with respect to $x$ (you need to understand, what is the case).