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**Chapter 1 / Re: HW1 Problem 1 （2）**

« **on:**January 29, 2022, 05:35:01 AM »

You should ask everybody, not just me. And the question was a bit different anyway: to classify an equation

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You should ask everybody, not just me. And the question was a bit different anyway: to classify an equation

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Yes, it is correct: $u=H(x)+m(y,z)$ where $H$ and $m$ are arbitrary functions.

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It was a misprint; it should be $e^{u(x',y)}$ in the denominator

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Both signs. What are curves $x^2-y^2=C$?

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Please post non-mathematical questions in Quercus discussions.

Please read description of Quiz 1.

Please read description of Quiz 1.

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Then your best shot would be *continuity condition*.

- 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

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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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Now it is correct $x=Ce^{t}$ and then $C=?$

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Display formulae are surrounded by double dollars and no empty lines. Multiline formulae use special environments (google LaTeX gather align

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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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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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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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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

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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Solution is allowed to be discontinuous.

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You need to indicate that there are no zeroes on $\gamma$