Author Topic: Q2-T0101  (Read 1900 times)

Victor Ivrii

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Q2-T0101
« on: February 02, 2018, 07:17:32 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}
            &1 &&|x| < 1,\\
            &0 &&|x| \ge 1.
            \end{aligned}\right.
&&h(x)=0 \tag*{Part (a)}
\\[5pt]     
&g(x)=0, &&h(x)=\left\{\begin{aligned}
            &\cos (x) &&|x| < \pi/2,\\
            &0 &&|x| \ge \pi/2.
            \end{aligned}\right.
            \tag*{Part (b)}
\end{align}

Jilong Bi

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Re: Q2-T0101
« Reply #1 on: February 03, 2018, 01:13:09 PM »
For the general solution :   \begin{equation} u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]+\frac{1}{2c}
  \int_{x-ct}^{x+ct}h(y)dy
\end{equation}
$$\\
$$for part(a):
\begin{equation}
 u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]
\end{equation}
$$
u(x,t) =
\begin{cases}
0, & \text{$x-ct >1 , x+ct \geq1$}  \\
\frac{1}{2}, & \text{$\vert x-ct \vert <1 , x+ct \geq1$}\\
1, & \text{$\vert x-ct \vert <1 , \vert x+ct \vert<1$}\\
1, & \text{$x-ct <-1 , x+ct \geq1$}\\
\end{cases}
$$for part(b):
\begin{equation}
 u(x,t) =\frac{1}{2c}
  \int_{x-ct}^{x+ct}h(y)dy
\end{equation}
$$
u(x,t) =
\begin{cases}
0, & \text{$x-ct >\frac{\pi}{2} , x+ct >\frac{\pi}{2}$}  \\
\frac{1}{2c}(1-\sin(x-ct), & \text{$\vert x-ct \vert <\frac{\pi}{2} ,x+ct \geq\frac{\pi}{2}$}\\
\frac{1}{2c}(\sin(x+ct)-\sin(x-ct), & \text{$\vert x-ct \vert <\frac{\pi}{2} , \vert x+ct \vert<\frac{\pi}{2}$}\\
\frac{1}{c}, & \text{$x-ct <-\frac{\pi}{2}  , x+ct \geq\frac{\pi}{2} $}\\
\end{cases}
$$


Xinrui Wang

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Re: Q2-T0101
« Reply #2 on: February 03, 2018, 01:44:39 PM »
The attachment is my solution, the (x,t) plane has been divided into 9 regions for both parts.

Victor Ivrii

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Re: Q2-T0101
« Reply #3 on: February 04, 2018, 09:25:47 AM »
Convert images to png –– those are displayed right on forum, no need to download