Toronto Math Forum

Hi I have facing difficulty with solving the 10th part of the Problem 4 in Chapter 3.2

This equation is quite different from others in Problem 4, since I am really not sure with my solution, can anyone check for me? My friend said he will solve it in a different way and he thinks my solution is wrong but he did not actually solve it. I have attached the question and my solution. It is really near the quiz and I have struggled on this problem for two days, please help me!

It is the same answer for $x>ct$ and $x-ct$. Explain why

\begin{equation}
    \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}       
\end{equation}

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{equation}
    \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}       
\end{equation}

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

Indeed, fixed. For consistency added index $_n$ to similar places of Example 4.2.7


Please post in the appropriate subforum

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.