Author Topic: Derivation of D' Alembert formula under Characteristic Coordinate  (Read 342 times)

Yifei Hu

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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 »

Victor Ivrii

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Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Reply #3 on: February 21, 2022, 04:34:15 AM »
Quote
But how does this qualify us to replace the indefinite integral with the definite one?

Did you take Calculus I? Then you must know that if the preimitive (indefinite integral) is a set of definite integrals which differ by an arbitrary constant.