Author Topic: Q2-T5102  (Read 2029 times)

Victor Ivrii

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Q2-T5102
« on: February 02, 2018, 07:18:39 AM »
Solve IVP
\begin{equation*}
\left\{\begin{aligned}
&u_{tt}-c^2u_{xx}=0,
\\
&u|_{t=0}=g(x),\\
&u_t|_{t=0}=h(x)
\end{aligned}\right.
\end{equation*}
with
\begin{align}     
&g(x)=\left\{\begin{aligned}
            &\cos (x) &&|x| < \pi/2,\\
            &0 &&|x| \ge \pi/2 ,
            \end{aligned}\right.
&&h(x)=0\tag*{Part (a)}\\[5pt]
&g(x)=0, &&h(x)=\left\{\begin{aligned}
            &0 &&x <  0,\\
            &1 &&x \ge 0.
            \end{aligned}\right.
            \tag*{Part (b)}
\end{align}

Jingxuan Zhang

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Re: Q2-T5102
« Reply #1 on: February 02, 2018, 09:23:39 PM »
Plug in D'Alembert's instantly we obtain:

\begin{align}     
&u=\left\{\begin{aligned}
            &0 && x+ct\leq -\pi/2 \text{ or } x-ct \geq \pi/2\\
            &\cos(x-ct)/2 && x-ct <\pi/2 \leq x+ct\\
            &\cos(x)\cos(ct) && -\pi/2 < x-ct < x+ct < \pi/2\\
            &\cos(x+ct)/2 && x-ct \leq -\pi/2 < x+ct\\
            \end{aligned}\right.
\tag*{Part (a)}\\[5pt]
&u=\left\{\begin{aligned}
            &0 && x+ct< 0\\
            &x/2c+t/2 && x-ct <0 \leq x+ct\\
            &t && 0 \leq x-ct\\
            \end{aligned}\right.
\tag*{Part (b)}
\end{align}