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### Messages - Andrew Lee Chung

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##### HA3 / Re: HA3-P6
« on: October 08, 2015, 03:19:59 PM »
Why do we have a rectangle as domain of dependence as compared to the one in the book?
How do we set the limits for the double integral?

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##### Textbook errors / Re: Section 3.1 error in equation 9
« on: October 07, 2015, 11:24:01 PM »

(9) Should be:
\begin{equation} u(x,t) = \frac{1}{2\sqrt{\pi kt}}e^{-\frac{x^2}{4kt}}   \end{equation}

Indeed

Error in change of variable
\begin{equation} z = x/ \sqrt{2kt} \end{equation}

(12) Shouldn't upper limit be
\begin{equation}  \frac{x}{ \sqrt{4kt}}\end{equation}

You can change variables in a different way and get different expressions; it looks like with $e^{-z^2}$ rather than $e^{-z^2/2}$ it became more standard; so I change it (but it will take time to deal with all instances in forthcoming sections, so be vigilant)

Error function shouldn't have variables in the limits
\begin{equation}erf(x) =   \sqrt{ \frac{2}{\pi} }   \int_{0}^{x} \ e^{- z^{2}/2 }dz\end{equation}

It can have variable limits or we can get constant limit but then we need integrand depending on $x$

It's actually equivalent to the wikipedia version:
\begin{equation}erf(x) = \frac{2}{ \sqrt{\pi} } \int_{0}^{x} \ e^{- t^{2} }  dt \end{equation}

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##### HA3 / Re: HA3-P5
« 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)?

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