Author Topic: P3 Night  (Read 1106 times)

Victor Ivrii

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P3 Night
« on: February 15, 2018, 08:21:20 PM »
Find   solution to
\begin{align}
&u_{tt}-u_{xx}=0, &&   t>0,\; x>  2-2\sqrt{t+1} \tag{1}\\
&u|_{t=0}=0, && x>0,\tag{2}\\
&u_t|_{t=0}=0, && x>0,\tag{3}\\
&u|_{x=2-2\sqrt{t+1}}= t, &&t>0.\tag{4}
\end{align}

Victor Ivrii

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Re: P3 Night
« Reply #1 on: February 22, 2018, 07:38:07 AM »
(a) Solution to equation is
\begin{equation}
u(x,t)=\phi(x+t)+\psi(x-t)
\label{eq-3-5}
\end{equation}
with unknown functions $\phi$ and $\psi$. Plugging into initial conditions we get
\begin{gather*}
\phi(x)+\psi(x)=0, \quad \phi'(x)-\psi'(x)=0\implies \phi(x)=\psi(x)=0 \qquad x>0
\end{gather*}
and $u(x,t)=\sin(x+3t)$ as $x>t$.

(b) Plugging into boundary condition  we get $\psi (2-2\sqrt{t+1}-t)= t$ as $t>0$. For $t>0$
$2-2\sqrt{t+1}-t$ is a monotone decreasing function, from $0$ to $-\infty$. Solve:
$$
x=-2-2\sqrt{t+1}-t\implies (\sqrt{t+1}+1)^2 = -x\implies  \sqrt{t+1}=\sqrt{1-x}-1\implies
t=-1+(\sqrt{1-x}-1)^2,
$$
and therefore $\psi(x)=-1+(\sqrt{1-x}-1)^2$ for $x<0$ and finally
$$
u(x,t)=-1+(\sqrt{1-x+t}-1)^2\qquad x<t.
$$


« Last Edit: February 26, 2019, 04:14:11 PM by Victor Ivrii »