Author Topic: Q2-T0101  (Read 1900 times)

Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
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

• Jr. Member
• Posts: 8
• Karma: 9
Re: Q2-T0101
« Reply #1 on: February 03, 2018, 01:13:09 PM »
For the general solution :    u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]+\frac{1}{2c}
\int_{x-ct}^{x+ct}h(y)dy

$$\\$$for part(a):

u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]

$$u(x,t) = \begin{cases} 0, & \text{x-ct >1 , x+ct \geq1} \\ \frac{1}{2}, & \text{\vert x-ct \vert <1 , x+ct \geq1}\\ 1, & \text{\vert x-ct \vert <1 , \vert x+ct \vert<1}\\ 1, & \text{x-ct <-1 , x+ct \geq1}\\ \end{cases}$$for part(b):

u(x,t) =\frac{1}{2c}
\int_{x-ct}^{x+ct}h(y)dy

$$u(x,t) = \begin{cases} 0, & \text{x-ct >\frac{\pi}{2} , x+ct >\frac{\pi}{2}} \\ \frac{1}{2c}(1-\sin(x-ct), & \text{\vert x-ct \vert <\frac{\pi}{2} ,x+ct \geq\frac{\pi}{2}}\\ \frac{1}{2c}(\sin(x+ct)-\sin(x-ct), & \text{\vert x-ct \vert <\frac{\pi}{2} , \vert x+ct \vert<\frac{\pi}{2}}\\ \frac{1}{c}, & \text{x-ct <-\frac{\pi}{2} , x+ct \geq\frac{\pi}{2} }\\ \end{cases}$$

Xinrui Wang

• Newbie
• Posts: 1
• Karma: 1
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.