# Toronto Math Forum

## APM346-2022S => APM346--Tests & Quizzes => Quiz 3 => Topic started by: Kexin Wang on February 14, 2022, 07:08:03 PM

Title: Quiz3 1-F
Post by: Kexin Wang 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 $\frac{1}{2}[sech(x+ct)+sech(x-ct)]$ $,x>0$