### Author Topic: Section2.3 equation 2  (Read 1453 times)

#### Yunheng Chen

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• • Posts: 4
• Karma: 0 ##### Section2.3 equation 2
« on: October 18, 2015, 02:00:18 PM »
For equation 2 we have (âˆ‚^2tâˆ’c^2âˆ‚^2x)=(âˆ‚tâˆ’câˆ‚x)(âˆ‚t+câˆ‚x)u=0  (2)
Shouldn't we have a u on the left side?

#### Yunheng Chen

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• • Posts: 4
• Karma: 0 ##### Re: Section2.3 equation 2
« Reply #1 on: October 18, 2015, 02:10:17 PM »
Also in the following line for equation2
We have "w=(âˆ‚tâˆ’câˆ‚x)u=utâˆ’cux"
Shouldn't it be "w=(âˆ‚tâˆ’câˆ‚x)u=wt - wux"

I just found out that I was wrong about it.
Thanks for Emily's explanation
« Last Edit: October 18, 2015, 05:27:27 PM by Yunheng Chen »

#### Emily Deibert ##### Re: Section2.3 equation 2
« Reply #2 on: October 18, 2015, 04:50:37 PM »
Also in the following line for equation2
We have "w=(âˆ‚tâˆ’câˆ‚x)u=utâˆ’cux"
Shouldn't it be "w=(âˆ‚tâˆ’câˆ‚x)u=wt - wux"

But why would the w come into the equation here? I think we are saying that \begin{equation}
w = (\partial_t - c\partial_x)u = \partial_tu - c\partial_xu \end{equation}
As in, we are expanding what is in the brackets so that it acts on u, and thus this is what w is equal to. Maybe I am misunderstanding?

Edit: I agree with the original post though, I think a u is missing.

#### Victor Ivrii ##### Re: Section2.3 equation 2
« Reply #3 on: October 19, 2015, 03:23:28 AM »
Indeed, $u$  was missing