# Toronto Math Forum

## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Yifei Hu on February 19, 2022, 08:10:02 PM

Title: Derivation of D' Alembert formula under Characteristic Coordinate
Post by: Yifei Hu 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?
Title: Re: Derivation of D' Alembert formula under Characteristic Coordinate
Post by: Victor Ivrii on February 20, 2022, 10:01:28 AM
Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.
Title: Re: Derivation of D' Alembert formula under Characteristic Coordinate
Post by: Yifei Hu 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?
Title: Re: Derivation of D' Alembert formula under Characteristic Coordinate
Post by: Victor Ivrii 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.