### Recent Posts

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##### Chapter 3 / Re: Chapter 3.2 Problem 9
« Last post by Zicheng Ding 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 / Re: Homework Assignment Week5
« Last post by Victor Ivrii on February 06, 2022, 01:53:33 PM »
Thanks! Fixed online TB
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##### Chapter 3 / Re: Chapter 3.2 Problem 9
« Last post by Victor Ivrii on February 06, 2022, 01:25:46 PM »
You almost there. Think!
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##### Chapter 3 / Homework Assignment Week5
« Last post by Zicheng Ding 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
« Last post by Zicheng Ding 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 / Re: S2.2 Q1
« Last post by Victor Ivrii on February 02, 2022, 06:25:02 PM »
• If you do not know this integral you need to refresh Calcuus I. one of basic integrals. Or have a table of basic integrals handy.
• Since $x^2+y^2=c^2$ is a circle, you can substitute $x=c\cos(s)$ and $y=c\sin(s)$ and then observe that $s=D-s$. It gives you the answer, less nicely looking than the one you wrote.
• Expressing $x, y$ through $t,c,d$ you can express $C=c\cos(d)$ and $D=c\sin(d)$ through $x,y,t$ which would give you that nice answer.

Write \cos , \sin , \log .... to produce proper (upright) expressions with proper spacing
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##### Chapter 3 / MOVED: S2.2 Q1
« Last post by Victor Ivrii on February 02, 2022, 06:22:00 PM »
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##### Chapter 2 / S2.2 Q1
« Last post by Yifei Hu on February 02, 2022, 04:35:19 PM »
The problem asks for general solution of the equation $U_t+yU_x-xU_y=0; U(0,x,y)=f(x,y)$
I proceed as usual:
$$\frac{dt}{1}=\frac{dx}{y}=\frac{-dy}{x}=\frac{du}{0}$$
Integrate and this gives: $x^2+y^2=C$, $t-\int \frac{1}{\sqrt{c-x^2}}dx=D$
Hence, I conclude that $U=\phi(C,D)=\phi(x^2+y^2,t-\int \frac{1}{\sqrt{c-x^2}}dx)$. However, this involves an integral that I can not calculated by hand, can anyone give me a hint on how to do this integral?
Also, I see that in the solution we can also solve this system with a nice trigonometry form $U=f(xcos(t)-ysin(t),xsin(t)+ycos(t))$but the solution does not specify how to reach that, can anyone shed lights on how the solution is reached?
As $x>0$ it is a correct calculation. However $f(xe^{-t})$ re,mains valid for $x<0$ while $f(t-\ln (x))$ does not.