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### Messages - Victor Ivrii

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31
##### Chapter 2 / Re: Transport Equation Derivation
« on: January 24, 2022, 05:46:09 AM »
Then your best shot would be
• either to ask during Prof Kennedy's Office hours
• or to formulate the problem and define everything  (that is $u, v, S, \tilde{S}$) here by yourself without screenshots
My guess that it is a continuity condition.

32
##### Chapter 2 / Re: Transport Equation Derivation
« on: January 19, 2022, 05:06:34 AM »
It would be really helpfull if you explained where you took this from (if online TextBook--then section and equation number, if lecture then which lecture and which part).

33
##### Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« on: January 17, 2022, 07:48:38 PM »
Now it is correct $x=Ce^{t}$ and then $C=?$

34
##### Chapter 1 / Re: chapter 1 Problem 4 (1)
« on: January 17, 2022, 01:29:44 AM »
Display formulae are surrounded by double dollars and no empty lines. Multiline formulae use special environments (google LaTeX gather align

35
##### Chapter 1 / Re: home assignment1 Q3(1),(2)
« on: January 16, 2022, 05:47:56 PM »
OK. Remarks:

1. Do not use $*$ as a multiplication sign!
2. Do not use LaTeX for italic text (use markdown of the forum--button I)
3. Escape ln, cos, .... : \ln (x) to produce $\ln (x)$ and so on

36
##### Chapter 1 / Re: Classification of PDEs
« on: January 14, 2022, 01:47:15 PM »
Yes, all linear are also semilinear and all semilinear are also quasilinear. For full mark you need to provide the most precise classification. So, if equation is linear you say "linear", if it is semilinear but not  linear you say "semilinear but not  linear" and so on,... "quasilinear but not  semilinear" and "non-linear and not quasilinear".

37
##### Chapter 1 / Re: Classification of PDEs
« on: January 14, 2022, 02:45:57 AM »
In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?
First, it will be not just quasilinear, but also  semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left

Good job, you mastered some $\LaTeX$ basics. 38
##### Chapter 1 / Re: Second Order canonical Form
« on: January 13, 2022, 07:24:23 PM »
We replace differentiation by $x$, y$by multiplication on$\xi,\eta$. So$\partial_x^2 \mapsto \xi^2$(just square); as a result senior terms like$Au_{xx}+2Bu_{xy}+ Cu_{yy}$are replaced by quadratic form$A\xi^2+2B\xi\eta+C\eta^2$. In the Linear Algebra you studied quadratic forms, right? And you know that • if$AC-B^2 >0$the canonical form is$\pm (\xi^2+\eta^2)$(as$\pm A>0$) • if$AC-B^2 <0$the canonical form is$ (\xi^2-\eta^2)$, • if$AC-B^2 =0$, but at least one of coefficients is not$0$the canonical form is$\pm \xi^2$. 39 ##### Test-2 / Re: Test-2 problem-1 confusion regarding boundary conditions « on: March 07, 2021, 04:12:41 AM » Solution is allowed to be discontinuous. 40 ##### Chapter 3 / Re: Rouché's Theorem « on: December 12, 2020, 06:25:05 PM » You need to indicate that there are no zeroes on$\gamma$41 ##### Chapter 7 / Re: Drawing phase portrait with complex eigenvalues question « on: December 01, 2020, 08:46:13 PM » "How ellipses wouls look like" means the directions and relative size of their semi-axis. See frame 4 of MAT244_W8L3 handout 42 ##### Chapter 4 / Re: Section 4.4 Past Final Exam Question « on: November 22, 2020, 05:19:33 AM » There could be misprints 43 ##### Chapter 4 / Re: General Question related to something in Section 4.1 « on: November 13, 2020, 12:54:44 PM » For higher order equations it is covered in MAT244-LEC0201-W6L2 (see modules). It is mandatory material. 44 ##### Chapter 4 / Re: 4.2 question 28 « on: October 25, 2020, 01:13:50 PM » you need to write it, if you hope for any answer 45 ##### Test 1 / Re: 2020TT1 Deferred Sitting #1 « on: October 15, 2020, 08:17:16 AM » Indeed, instead of$\log(\pm w)$with$+2\pi mi$we write$\log(w)$with$+\pi mi$since$\log (-w)=\log(w)+\i i\$

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