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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?
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##### Chapter 2 / Re: Ut+xUx=0
« Last post by Victor Ivrii on February 01, 2022, 12:16:37 PM »
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.
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##### Chapter 2 / Re: f(x) in Method of Characteristics
« Last post by Victor Ivrii on February 01, 2022, 12:12:52 PM »
Yes, there are many answers which are correct because they include arbitrary functions (and in ODE arbitrary constants)
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##### Chapter 2 / Impose Initial Condition on Inhomogeneous Equations
« Last post by Yifei Hu on February 01, 2022, 11:34:15 AM »
Consider problem: $U_x+3U_y=xy, U(0,y)=0$, I proceed as follow:
$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$
Integrate on first two terms:
$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$
Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but  connected by (*).
$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D$$
D must be constant along the integral curve hence $D=\phi(x-1/3y)$
Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.
Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.

Update:
Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?
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##### Chapter 2 / Ut+xUx=0
« Last post by Yifei Hu on February 01, 2022, 10:08:42 AM »
When solving this problem, I proceed as follow:
$$\frac{dt}{1}=\frac{dx}{x}=\frac{du}{0}$$
Hence, U does not depend on x and t, integrate on first part of equation:
$$t=ln(x)+C$$
I did not take exponential on both sides to get $e^t=Cx$ but I directly use $C=t-ln(x)$ and got $U=f(t-ln(x))$. Can anyone help me identify why this calculation is wrong?
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##### Chapter 2 / f(x) in Method of Characteristics
« Last post by Yifei Hu on February 01, 2022, 09:47:20 AM »
In solving the problem $2U_t+3U_x=0, U(0,x)=f(x)$, tutorial gave answer $U=f(x-\frac{3}{2}t)$ while I have $U=f(3t-2x)$. Are both answers valid? We can use different form of C as long as it represent the same family of characteristics right?
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##### Chapter 2 / Re: Online textbook, Chapter 2.6, example 7
« Last post by Victor Ivrii on February 01, 2022, 05:56:58 AM »
For the middle region in red, since $\psi(x - \frac{t}{3})$ is undefined,
It is defined, because on the line $\{x=-t, t>0\}$ we have not 1 but 2 boundary conditions, so in the domain $\{x>-t, t>0\}$ we have essentially a Cauchy problem with the date on the line consisting of two rays: $\{x>0,t=0\}$ and $\{x=-t, t>0\}$.
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