Author Topic: Derivation of D' Alembert formula under Characteristic Coordinate  (Read 1310 times)

Yifei Hu

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Derivation of D' Alembert formula under Characteristic Coordinate
« on: February 19, 2022, 08:10:02 PM »
In text book 2.4, we had one line in the derivation: $\tilde{u_{\xi}}=-\frac{1}{4c^2} \int^\xi f(\xi,\eta')d\eta'=-\frac{1}{4c^2} \int_\xi^\eta\tilde f(\xi,\eta')d\eta' + \phi'(\xi)$.
Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit?
« Last Edit: February 20, 2022, 07:59:23 PM by Yifei Hu »

Victor Ivrii

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Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Reply #1 on: February 20, 2022, 10:01:28 AM »
Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.

Yifei Hu

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Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Reply #2 on: February 20, 2022, 08:12:28 PM »
I understand that I can show $\phi'(\xi)=0$ by:

1) when t = 0, $\xi = x+ct = x = x-ct = \eta$
2) $u_\xi = u_t \frac{dt}{d\xi} + u_x \frac{dx}{d\xi}$ by chain rule.
3)By initial condition: $u_t|_{t=0} = u_x|_{t=0}$ = 0, we must have $u_\xi=0$
4) $u_\xi = \phi'(x)$ hence $\phi'(\xi)=0$

But how does this qualify us to replace the indefinite integral with the definite one?
« Last Edit: February 20, 2022, 08:14:08 PM by Yifei Hu »