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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.

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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).

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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 $C=?$

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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

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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

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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".

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Chapter 1 / Re: Classification of PDEs

« on: January 14, 2022, 02:45:57 AM »

Quote from: Weihan Luo on January 14, 2022, 12:38:03 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. $:)$

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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 $\pm \xi^2$.

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