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Chapter 1 / Re: Classification of PDEs
« Last post by Weihan Luo on January 14, 2022, 11:28:53 AM »
Thank you for your response.

Does it mean that all linear PDEs are also quasilinear/or semilinear? If so, on a quiz, I should classify those PDEs as linear right?
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Chapter 1 / Re: Classification of PDEs
« Last post by Victor Ivrii 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. :)
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Chapter 1 / Classification of PDEs
« Last post by Weihan Luo on January 14, 2022, 12:38:03 AM »
I am a little bit confused about the classifications of PDES. Namely, I have trouble distinguishing between linear equations versus quasi-linear equations.

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?
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Chapter 1 / Re: Second Order canonical Form
« Last post by Victor Ivrii 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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Chapter 1 / Second Order canonical Form
« Last post by Yifei Hu on January 13, 2022, 02:37:35 PM »
What is the definition here (when classifying the second order PDEs) for the second order canonical form? what are the Xi and Eta here? Is the operation here defined as taking derivative? e.g: Eta^2 = second derivative of Eta?
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Quiz-6 / Qz6-TWO-C
« Last post by duoyizhang on March 25, 2021, 01:45:34 PM »
Question:Find Fourier transformation of the function $$e^{-\alpha x^2/2}sin(\beta x)$$
with $$\alpha>0,\beta>0$$
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Quiz-6 / Section 0201 quiz6-question 2B
« Last post by Houze Xu on March 18, 2021, 01:09:00 PM »
f(x) = e^−α|x|sin(βx)
Answer: f1(x) = e^−α|x|
f1^(k) = (2/pi)^1/2(α/α^2 + k^2)
f^(k) = F(f(x)) = -i/2(F(e^−α|x|e^iβx)-F(e^−α|x|e^-iβx))
F(e^−α|x|e^iβx) = (2/pi)^1/2(α/α^2 +(k^2- β^2)^2)
F(e^−α|x|e^-iβx))= (2/pi)^1/2(α/α^2 +(k^2+β^2)^2)
f^(k) = -i/(2pi)^1/2(α/α^2 +(k^2- β^2)^2 - α/α^2 +(k^2+β^2)^2 )
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Quiz-5 / Qz5-THREE-E
« Last post by duoyizhang on March 16, 2021, 01:27:02 AM »
The question is:Decompose f(x) = xcos(x)  into full Fourier series on interval [0, pi].
My confusion is how to decompose it on an interval like  [0,𝑙] rather than[-l,l]
Firstly,I compute f(x) into  full Fourier series on interval [-pi, pi] by the formula,Which is $$f1=f(x)=-\frac{sinx}{2}+\sum_{n=2}^\infty\frac{2n(-1)^{n}}{n^{2}-1}$$
Then what should we do to compute f(x) on[0,pi],I tried to use the property that f(x)is an odd function but it seems to be wrong.
Any help will be appreciated!
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Test-2 / Re: Test-2 problem-1 confusion regarding boundary conditions
« Last post by Victor Ivrii on March 07, 2021, 04:12:41 AM »
Solution is allowed to be discontinuous.
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Test-2 / Test-2 problem-1 confusion regarding boundary conditions
« Last post by aryakim on March 06, 2021, 10:07:36 AM »
I had a question about problem $1$ of my Test$2$. (My version was called alternative-F, night section).There was something confusing about this problem that I realized during the test.
Here is the problem:

\begin{equation}
\nonumber
\left\{ \begin{aligned}
& u_{tt}-u_{xx}=0, &&0< t < \pi, 0 < x < \pi, &(1.1) \\\
&u|_{t=0}= 2\cos (x),   && 0< x < \pi, &(1.2)\\
&u_t|_{t=0}= 0,  && 0< x < \pi,  &(1.3)\\
&u|_{x=0}= u|_{x= \pi}=0, && 0< t < \pi.  &(1.4)
\end{aligned}
\right.
\end{equation}

So, on the $t-x$ diagram, the lines $t = 0$ and $x = \pi$ intersect at ($t=0,x=\pi$), which will be a boundary point of the region where $0<t<x<\pi$.
If this point (i.e. ($t=0,x=\pi$)) on the diagram is approached by the line $t = 0$, equation $(1.2)$ is used to conclude that the value of $u(x,t)$ approaches $-2$.
On the other hand, if the point is approached from the line $x = \pi$,  $u(x,t)$ should become zero, which is in contradiction with the other boundary condition. It seems this would make it impossible to incorporate conditions $(1.2)$ and $(1.4$) at the same time. I was hoping someone could clear my confusion regarding this problem.


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