Author Topic: HA10 Problem 5  (Read 449 times)

Victor Ivrii

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HA10 Problem 5
« on: March 26, 2015, 03:13:13 PM »
Consider wave equation
\begin{equation}
\rho u_{tt} - (k(x)u_x)_x=0 \label{eq-10.5}
\end{equation}
on $-\infty < x < \infty$ where
$\rho(x)=\left\{\begin{aligned}\rho_-(x)& &&x<0,\\ \rho_+(x)& &&x>0\end{aligned}\right.$ and $k(x)=\left\{\begin{aligned}k_-(x)& &&x<0,\\ k_+(x)& &&x>0.\end{aligned}\right.$

a. Write down equation (\ref{eq-10.5}) for $x< 0$ and $x > 0$ separately.
b. Find out *transmission conditions* (there must be 2 of them) linking $u(-0,t)$, $u(+0,t)$, $u_x(-0,t)$, $u_x(+0,t)$.

Victor Ivrii

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Re: HA10 Problem 5
« Reply #1 on: April 03, 2015, 03:56:08 PM »
a. Equations are easy
\begin{equation}
\rho_\pm u_{tt}- k_\pm u_{xx}=0\qquad \text{as } \pm x>0.
\end{equation}
b. Transmission conditions are a bit more difficult. First $u_x$ must be a function, not a distribution and therefore
\begin{equation}
u_x(-0,t)=u_x(+0,t).
\label{P}
\end{equation}
Next, $\lim _{x\to \pm 0}(k (x)u_x)=k_\pm u_x (\pm 0,t)$ and it must not to have jums so
\begin{equation}
k_-u_x(-0,t)=k_+ u_x(+0,t).
\label{Q}
\end{equation}
(\ref{P}) and (\ref{Q}) are transmission conditions