### Author Topic: HA3-P5  (Read 3302 times)

#### Victor Ivrii ##### HA3-P5
« on: September 28, 2015, 01:05:56 PM »

#### Zaihao Zhou

• Full Member
•   • Posts: 29
• Karma: 0 ##### Re: HA3-P5
« Reply #1 on: October 06, 2015, 06:08:14 PM »
The general solution is
\begin{equation} u(x,t) = \frac{1}{2c}\int_0^t{\int_{x-c(t-t')}^{x+c(t-t')} f(x',t')dx'dt'}   \end{equation}
For problem: \begin{equation}u_{tt} - c^2u_{xx} = f(x,t)  \quad and \quad g(x)=0,\ h(x)=0\end{equation}
To solve problems in P1 (4) - (7), substitute the f(x,t) to corresponding equations.

Below are my results, not sure if they are right.
a)
\begin{equation} u(x,t) = \frac{1}{2\alpha^2c^2}[sin(\alpha x + \alpha ct) - sin(\alpha x - \alpha ct)]\end{equation}

b):  \begin{equation} u(x,t) = \frac{sin(\alpha x)}{\beta^2 - \alpha^2c^2}[sin(\beta t) - sin(\alpha ct)] \end{equation}

c):  \begin{equation} u(x,t) = \frac{1}{2c^2}[F(x+ct)-F(x-ct)] \end{equation}

d): stuck at  \begin{equation} u(x,t) = \frac{1}{2c}\int^t_0[F''(x+ct-ct')-F''(x-ct+ct')]t'dt' \end{equation}

Kinda feel we can still do the integration but don't know how after this.

#### Chi Ma

• Full Member
•   • Posts: 16
• Karma: 0 ##### Re: HA3-P5
« Reply #2 on: October 06, 2015, 11:02:06 PM »
This is what I got for c:

\begin{align} u(x,t) = \frac{1}{2c^2}[F(x+ct)+F(x-ct)-2F(x)] \end{align}

For d, apply integration by part, which is why we need the third derivative $f(x)=F'''(x)$.

\begin{align}
u(x,t) &= \frac{1}{2c}\int^t_0[t'F''(x+c(t-t'))-t'F''(x-c(t-t'))]dt' \\
&= -\frac{1}{2c^2} \left\{  \int^t_0 t'dF'(x+c(t-t')) + \int^t_0 t'dF'(x-c(t-t')) \right\} \\
&= -\frac{1}{c^2}tF'(x) + \frac{1}{2c^3} [F(x+ct)+F(x-ct)]
\end{align}
« Last Edit: October 07, 2015, 04:37:31 PM by Victor Ivrii »

#### Andrew Lee Chung

• Jr. Member
•  • Posts: 9
• Karma: 0 ##### Re: HA3-P5
« Reply #3 on: October 07, 2015, 04:03:32 PM »
@Zaihao

I got this for
a)
\begin{equation} u(x,t) = \frac{1}{2\alpha^2c^2}[2\sin(\alpha x) - \sin(\alpha (x - ct))- \sin(\alpha (x + ct))] \end{equation}

b)
\begin{equation} u(x,t) = \frac{sin(\alpha x)}{c \alpha(\beta^2 - \alpha^2c^2)}[(\beta) sin(\alpha ct) - (\alpha c) sin(\beta t)] \end{equation}

c), d) Same as Chi Ma

Could someone else check a) and b)?
« Last Edit: October 08, 2015, 10:01:00 AM by Andrew Lee Chung »

#### Zaihao Zhou

• Full Member
•   • Posts: 29
• Karma: 0 ##### Re: HA3-P5
« Reply #4 on: October 08, 2015, 08:57:34 AM »
Sorry my fault. Lost one of negative signs in the process throughout. Thank you for correcting me.
Confirmed all, you guys are correct
« Last Edit: October 08, 2015, 09:45:58 AM by Zaihao Zhou »