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Messages - Kexin Wang

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1
Test 2 / TT2 Problem1
« on: March 31, 2022, 09:43:26 PM »
Hi Professor,
I did my Term Test 2 today and for problem 1, I don't know if my understanding is correct. I think this question is approachable by both (Inverse) Fourier Transformation and Separation Variable methods. I'm not sure if I have the correct insights, I think both methods work from the information provided.

2
Test 2 / Term Test 2 Variant E problem 2
« on: March 28, 2022, 03:25:10 PM »
Hi, I just want to confirm with someone about the solution for Term test 2 variant E problem2. I think for$\lambda_n$ the n should start from 0, so n should be n = 0, 1, 2 and so on. I don't know if the solution makes a typo or am I missing some insights here.

3
Test 2 / Term Test 2 Variant C Problem4
« on: March 27, 2022, 08:36:36 PM »
Hi Professor,
I was reviewing practice test variant C problem4. I tried to follow your solution but I think where I circled should be the same as the other red circle? I don't know if this is a typo or am I missing some insights here. I've attached a picture.
Thank you

4
Test 2 / Term Test 2 Variant E
« on: March 27, 2022, 03:31:36 PM »
For problem 1 in Practice Test Variant E, I'm wondering if this is a typo because the interval given is from $-\pi$ to $\pi$? (I circled it in red)

5
Quiz 5 / Quiz 5
« on: March 27, 2022, 10:06:51 AM »
My question for Quiz5 was to decompose a function into full Fourier Series [0, $\pi$]. I wonder is it equivalent as decompose into full Fourier Series on [$-\pi$, $\pi$]? I understand it is equivalent when the function is even, but I'm wondering what should I do when the function is odd.

6
Quiz 3 / Quiz3 1-F
« on: February 14, 2022, 07:08:03 PM »

\begin{cases}
u_{tt}- c^2u_{xx} = 0 &\ x>0\\
u\rvert_{t = 0} = sech(x) &\ x>0\\
u_{t}\rvert_{t = 0} = 0 &\ x>0\\
u_{x}\rvert_{x = 0} = 0 &\ t>0\\
\end{cases}

For this problem, my approach was to extend $sech(x)$ as an even function for $x \in R$. Since $sech(x)$ is already an even function so we can write the set of equations as follows.

\begin{cases}
u_{tt}- c^2u_{xx} = 0 &\ x>0\\
u\rvert_{t = 0} = sech(x)\\
u_{t}\rvert_{t = 0} = 0 &\ x>0\\
\end{cases}

And by using D'Alembert Formula $u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]$ we can simplify it as $\frac{1}{2}[sech(x+ct)+sech(x-ct)]$ $,x>0$

7
Chapter 1 / HW1 Problem 1 （2）
« 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}

8
Chapter 1 / HW1 Problem 6 （41）
« 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}

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