### Recent Posts

Pages: 1 ... 7 8 [9] 10
81
##### 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?
82
##### 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 $\{x=-t, t>0\}$.
83
##### Chapter 2 / Online textbook, Chapter 2.6, example 7
« Last post by Zicheng Ding 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.
84
##### Chapter 1 / Re: HW1 Problem 1 （2）
« Last post by Victor Ivrii 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
85
##### Chapter 1 / HW1 Problem 1 （2）
« Last post by Kexin Wang on January 28, 2022, 06:52:38 PM »
\begin{gather}
u_t+uu_x= 0;\\[2pt]
\end{gather}

Hi Professor, I don't know what the operator for this equation. Does it have an operator? Or is the operator equal to

\begin{gather}

L= \partial_t + u\cdot\partial_x
\end{gather}
86
##### Chapter 1 / Re: HW1 Problem 6 （41）
« Last post by Victor Ivrii on January 28, 2022, 01:41:36 AM »
Yes, it is correct: $u=H(x)+m(y,z)$ where $H$ and $m$ are arbitrary functions.
87
##### Chapter 1 / HW1 Problem 6 （41）
« Last post by Kexin Wang on January 27, 2022, 09:12:08 PM »
\begin{align}
&\left\{\begin{aligned}
&u_{xy}=0,\\[2pt]
&u_{xz}=0;
\end{aligned}\right.\\[2pt]
\end{align}

For this problem I'm not sure if my approach was correct.
\begin{align}
u_{x}=f(x,z)  \textrm{   from the first equation}
\end{align}
\begin{align}
&u_{x}=g(x,y)\textrm{   from the second equation}
\end{align}
\begin{align}
\textrm{Therefore we must have   }\textrm{   }u_{x} = h(x)
\end{align}
\begin{align}
&u=H(x) + m(y,z)\textrm{ } \textrm{where}\textrm{ } H'(x)=h(x)
\end{align}
88
##### Chapter 1 / Re: Question about the posted solution for HW1 3c
« Last post by Victor Ivrii on January 27, 2022, 03:59:38 PM »
It was a misprint; it should be $e^{u(x',y)}$ in the denominator
89
##### Chapter 1 / Question about the posted solution for HW1 3c
« Last post by Yuanhan (Phyllis) Peng on January 27, 2022, 12:39:22 PM »
I am confused about why when we integrate the left hand side Ux/e^u, we used the x' in Ux(x', y) expression, but we didn't change x in e^u(x,y)

The screenshot is attached.

90
##### Chapter 2 / Re: Chapter 2.2 problem 2
« Last post by Zicheng Ding 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.
Pages: 1 ... 7 8 [9] 10