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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?

2
Textbook errors / Re: Section 3.1 error in equation 9
« on: October 07, 2015, 11:24:01 PM »
I mentioned about this error during lecture today

(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}



3
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)?

4
HA1 / Re: HA1-P4
« on: September 30, 2015, 04:53:04 PM »
@Chi
Seems good

@Emily

\begin{equation} ln(u_{x}) = u + f(x)\end{equation}

Then proceed as Chi did

5
HA1 / Re: HA1-P3
« on: September 30, 2015, 04:44:59 PM »
@Emily

 d)
\begin{equation}
u_{xy} = 2u_{x}+e^{x+y} \\
Let \ v = u_{x} \\
 v_{y} = 2v +e^{x+y}  \\
v-2vy=e^{x+y} +f_{1}(x) \\
 v=u_{x}= \frac{e^{x+y}+f_{1}(x)}{1-2y} \\
u= \frac{e^{x+y}+f_{2}(x)}{1-2y} +g(y) \\where \
f_{2}(x) = \int f_{1}(x)dx
\end{equation}

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HA1 / Re: Problem 5
« on: September 30, 2015, 04:21:26 PM »
My take on problem 5:
For b) shouldn't it be:
\begin{equation}
u = \int \int f(x,y)dydx + \int g(x,z)dx + h(y,z)
\end{equation}

Also is it safe to assume \begin{equation} \int \int f(x,y)dydx =  f_{2}(x,y) \end{equation}
for convenience?

So that we can rewrite it as:
\begin{equation} u =  f_{2}(x,y) + g_{1}(x,z) + h(y,z)\end{equation}

The rest seems fine

7
HA1 / Re: Problem 2
« on: September 30, 2015, 04:08:31 PM »
@Yaodong
Lu is a partial differential expression linear with respect to unknown function u
It could be the 0 order wrt u

From the book, the example given had this form
\begin{equation} Lu:= a_{11} u_{xx} +  2a_{12} u_{xy}+ a_{22}  u_{yy} + a_{1}  u_{x} + a_{2} u_{y} + au = f(x) \end{equation}
So it does include u

8
HA1 / Re: Problem 2
« on: September 30, 2015, 03:50:16 PM »
My take on qu2:
@Emily
e) 4th order linear inhomogeneous (sin(x) is not a dependent variable, u, nor any of its partial derivatives)
g) 4th order linear homogeneous (u occurs linearly in the equation)
h) 4th order semi linear (sin(x)sin(u) is a function f(x,u))

9
HA1 / Re: Problem 1
« on: September 30, 2015, 03:34:51 PM »
Similar answers except for:
e) semi linear is always inhomogeneous?
shouldn't h) be linear homogeneous?

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