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### Messages - Zicheng Ding

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##### Chapter 5 / Chapter 5.3 Problem 1.1
« on: March 13, 2022, 02:32:37 PM »
For this problem we have the following equations: $u_{xx} + u_{yy} = 0 \space \space (-\infty < x < \infty , y > 0)$ and $u|_{y=0} = f(x)$.
After doing the Fourier Transformation, the equations become $\hat{u}_{yy} - k^2\hat{u} = 0$ and $\hat{u}(k,0) = \hat{f}(k)$.
In my understanding, we should get a general form as $\hat{u}(k,y) = A(k)e^{ky} + B(k)e^{-ky}$ and drop the first term if $k>0$ since as $y \rightarrow \infty$ the term $e^{ky} \rightarrow \infty$, and if $y<0$ we drop the second one.
On the answer provided by Prof. Kennedy it said $\hat{u}(k,y) = \hat{f}(k)e^{-k|y|}$, so I am a little confused since $\hat{u}(k,y) = \hat{f}(k)e^{-|k|y}$ makes more sense to me.

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##### Chapter 3 / Chapter 3.1 Theorem 4
« on: February 26, 2022, 07:43:05 PM »
In the online text book, chapter 3.1 theorem 4, we have the formula for inhomogeneous heat equations. For the second integral in the formula, since it represents the homogeneous part of the heat equation and the heat equation is linear, should the range be from $-\infty$ to $\infty$ instead of $0$ to $\infty$? The equation we are solving have the range of $-\infty < x < \infty$ and $t > 0$ as usual.

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##### Chapter 4 / Chapter 4.2, Example 6
« on: February 13, 2022, 01:14:16 PM »
In chapter 4.2, example 6, I was a little confused about the notation. My understanding is that $\omega_n$ are the roots for equation $\tan(\omega l) = \frac{(\alpha + \beta)\omega}{\omega^2 - \alpha\beta}$, so should the equation for $X_n$ be $X_n = \omega_n \cos(\omega_n x) + \alpha\sin(\omega_n x)$ instead of $X_n = \omega \cos(\omega_n x) + \alpha\sin(\omega_n x)$ (the first $\omega$ change to $\omega_n$)? I attached a screenshot of the example.

Another minor thing that I noticed is that for homework assignment 6/7, it said problems 1-11 for section 4.1 and 4.2, but there are no problem 11 in neither textbook version, so I guess it was a typo on Quercus.

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##### Chapter 3 / Re: Chapter 3.2 Problem 9
« on: February 06, 2022, 02:19:29 PM »
I am thinking that if we assume maximum is not on the boundary, then in the initially with $u_t = ku_{xx}$ we will have $u_t - ku_{xx} < 0$ as a contradiction, but in this case we will not necessarily have $u_t - xu_{xx} < 0$ since $x$ can switch signs, so we no longer have the contradiction and that breaks down the proof.

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##### Chapter 3 / Homework Assignment Week5
« on: February 06, 2022, 12:55:41 PM »
For week 5 homework assignment, problem 13 of chapter 3.2 is assigned, but there is no problem 13 in the link to the online textbook. I found problem 13 in the pdf version of the textbook, so doing problem 13 in the pdf version is the same?

Another minor typo I noticed is that in online textbook chapter 3.2, theorem 3, I think it should be $\varepsilon > 0$. I marked the error in the attached screenshot.

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##### Chapter 3 / Chapter 3.2 Problem 9
« on: February 06, 2022, 12:45:35 PM »
For question 9 we have $u = - 2xt - x^2$ as a solution for $u_t = xu_{xx}$, and I found the maximum in the closed rectangle {$-2 \leq x \leq 2$, $0 \leq t \leq 1$} at $(x,t) = (-1, 1)$ on the boundary. I notice that at the maximum we have $u_t > 0$ and $u_{xx} < 0$ but since we have an $x$ in the equation, $u_t = xu_{xx}$ is still satisfied. In the proof of the maximum principle with $v = u - \varepsilon t$, the solution for this question also seems valid, so I am a little confused about where in the proof of maximum principle actually breaks down in this example.

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##### Chapter 2 / Online textbook, Chapter 2.6, example 7
« on: January 31, 2022, 09:04:09 PM »
I was a little confused about the region marked in red in this example. For the region on the right we get $\phi(x) = \sin(3x)$ and $\psi(x) = 3\sin(3x)$,  and from the boundary conditions we can get $\phi(x) = \psi(x) = 0$ for $t > 0$, so we have $u = 0$ for the region on the left. For the middle region in red, since $\psi(x - \frac{t}{3})$ is undefined, we can just have it as $\psi(x) = 0$ instead?  The middle region is like a mix of the other two regions so I am a little uncertain.

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##### Chapter 2 / Re: Chapter 2.2 problem 2
« on: January 27, 2022, 08:19:58 AM »
$x^2 - y^2 = C$ is a hyperbolic curve so it has two parts for x. I see now, thank you professor.

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##### Chapter 2 / Chapter 2.2 problem 2
« on: January 26, 2022, 09:22:59 PM »
Chapter 2.2 problem 2 equation (9): $u_t + yu_x + xu_y = x$
and we get $\frac{dt}{1} = \frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}$
By solving $\frac{dx}{y} = \frac{dy}{x}$ first we get $C = x^2 - y^2$
But we solving for either $\frac{dt}{1} = \frac{dx}{y}$ or $\frac{dt}{1} = \frac{dy}{x}$ we need to substitute $x$ for $y$ or the other direction, so in this case we will have a square root function like $x = \pm \sqrt{y^2 + C}$
I am not entirely sure whether or not we should proceed with the $\pm$ sign here or is there another easier approach to this question?

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