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### Messages - Victor Ivrii

Pages: 1 [2] 3 4 ... 155
16
##### Chapter 3 / Re: Chapter 3.1 Theorem 4
« on: February 27, 2022, 07:52:52 PM »
Indeed, to be corrected. Thanks

17
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« 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.

18
##### Chapter 2 / Re: Text Book 2.4 Example 2.1
« on: February 20, 2022, 10:04:48 AM »
We integrate from $t=0$ because for $t=0$ initial conditions are done. $-1<t$ is a domain where $f(x,t)$ is defined

19
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« on: February 20, 2022, 10:01:28 AM »
Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.

20
##### Quiz 3 / Re: LEC0101 Quiz3-1c
« on: February 15, 2022, 12:46:54 PM »
It is the same answer for $x>ct$ and $x-ct$. Explain why

21
##### Chapter 4 / Re: Chapter 4.2, Example 6
« on: February 14, 2022, 07:12:14 AM »
Indeed, fixed. For consistency added index $_n$ to similar places of Example 4.2.7

Please post in the appropriate subforum

22
##### Chapter 3 / MOVED: Chapter 4.2, Example 6
« on: February 14, 2022, 07:10:42 AM »

23
##### Chapter 3 / Re: Chapter 3.2 Problem 9
« on: February 06, 2022, 02:21:30 PM »
Indeed. You got it!

24
##### Chapter 3 / Re: Homework Assignment Week5
« on: February 06, 2022, 01:53:33 PM »
Thanks! Fixed online TB

25
##### Chapter 3 / Re: Chapter 3.2 Problem 9
« on: February 06, 2022, 01:25:46 PM »
You almost there. Think!

26
##### Chapter 2 / Re: S2.2 Q1
« on: February 02, 2022, 06:25:02 PM »
• If you do not know this integral you need to refresh Calcuus I. one of basic integrals. Or have a table of basic integrals handy.
• Since $x^2+y^2=c^2$ is a circle, you can substitute $x=c\cos(s)$ and $y=c\sin(s)$ and then observe that $s=D-s$. It gives you the answer, less nicely looking than the one you wrote.
• Expressing $x, y$ through $t,c,d$ you can express $C=c\cos(d)$ and $D=c\sin(d)$ through $x,y,t$ which would give you that nice answer.

Write \cos , \sin , \log .... to produce proper (upright) expressions with proper spacing

27
##### Chapter 3 / MOVED: S2.2 Q1
« on: February 02, 2022, 06:22:00 PM »

28
##### Chapter 2 / Re: Ut+xUx=0
« on: February 01, 2022, 12:16:37 PM »
As $x>0$ it is a correct calculation. However $f(xe^{-t})$ re,mains valid for $x<0$ while $f(t-\ln (x))$ does not.

29
##### Chapter 2 / Re: f(x) in Method of Characteristics
« on: February 01, 2022, 12:12:52 PM »
Yes, there are many answers which are correct because they include arbitrary functions (and in ODE arbitrary constants)

30
##### Chapter 2 / Re: Online textbook, Chapter 2.6, example 7
« on: February 01, 2022, 05:56:58 AM »
For the middle region in red, since $\psi(x - \frac{t}{3})$ is undefined,
It is defined, because on the line $\{x=-t, t>0\}$ we have not 1 but 2 boundary conditions, so in the domain $\{x>-t, t>0\}$ we have essentially a Cauchy problem with the date on the line consisting of two rays: $\{x>0,t=0\}$ and $\{x=-t, t>0\}$.

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