Author Topic: Problem 1  (Read 5719 times)

Kun Guo

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Problem 1
« on: October 04, 2012, 05:47:45 PM »
 I noticed the given error function(Erf) in the problem set sheet is different with what's given in WolframAlpha. Which one should I use? Also, are we expected to write our final answers in form of Erf?

Victor Ivrii

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Re: Problem 1
« Reply #1 on: October 04, 2012, 05:52:05 PM »
I noticed the given error function(Erf) in the problem set sheet is different with what's given in WolframAlpha. Which one should I use? Also, are we expected to write our final answers in form of Erf?

It was a misprint in HA3, which I just fixed
« Last Edit: October 04, 2012, 06:00:50 PM by Victor Ivrii »

Kun Guo

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Re: Problem 1
« Reply #2 on: October 05, 2012, 12:22:45 AM »
I think there is still problem 2/ sqrt(pi) instead of sqrt(2/pi), is this true?

Victor Ivrii

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Re: Problem 1
« Reply #3 on: October 05, 2012, 07:03:41 AM »
I think there is still problem 2/ sqrt(pi) instead of sqrt(2/pi), is this true?

$\sqrt{\frac{2}{\pi}}$ as it should be

Aida Razi

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Re: Problem 1
« Reply #4 on: October 07, 2012, 03:50:06 PM »
I was wondering if first we need to find IVP solution to heat equation by separation of variable and then apply IBVP OR
we can use IVP solution from lecture note and just apply IBVP?

Thank you,

Victor Ivrii

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Re: Problem 1
« Reply #5 on: October 07, 2012, 04:47:44 PM »
I was wondering if first we need to find IVP solution to heat equation by separation of variable and then apply IBVP OR
we can use IVP solution from lecture note and just apply IBVP?

Thank you,

So far we have not studied separation of variables for heat equation (at least not on infinite or semiinfinite interval).

Jinlong Fu

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Re: Problem 1
« Reply #6 on: October 09, 2012, 10:59:24 PM »
The two forms are actually the same,

\begin{equation*}
erf(z)=\sqrt{\frac{2}{\pi}}\int_0^ze^{-y^2/2}\,dy
\tag{Erf}\label{eq-Erf}
\end{equation*}

let $\frac{y}{\sqrt{2}}= t$, you can transform the above formula into the form in Wolfram


\begin{equation*}
erf(z)=\frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}\,dt
\end{equation*}
« Last Edit: October 10, 2012, 12:04:15 AM by Jinlong Fu »

Zarak Mahmud

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Re: Problem 1
« Reply #7 on: October 10, 2012, 09:31:36 PM »
(a)

\begin{equation}
G(x,y,t) = \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-y)^2}{4kt}}\
\end{equation}

\begin{equation*}
u(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g(y)dy \\
 = \int_{-\infty}^{0} G(x,y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy \\
\end{equation*}
Let $-y = z$,
$$ = \int_{\infty}^{0} G(x,-z,t)g(-z)(-dz) + \int_{0}^{\infty} G(x,y,t)g(y)dy$$

We have Dirichlet boundary condition so g(z) is odd;$ g(-z) = -g(z).$

\begin{equation*}
= -\int_{\infty}^{0} G(x,-z,t)g(z)dz + \int_{0}^{\infty} G(x,y,t)g(y)dy\\
 = -\int_{\infty}^{0} G(x,-y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy\\
 = \int_{0}^{\infty}\big[G(x,y,t)g(y) - G(x,-y,t)g(y)\big]dy\\
 = \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} - e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.\\
\end{equation*}


(b)
Neumann boundary conditions: $g(y)$ is even $\implies g(x) = g(-x)$
Making the same substitution as in (a), we have:

\begin{equation*}
u(x,t) = \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} + e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.
\end{equation*}
« Last Edit: October 14, 2012, 08:18:19 AM by Zarak Mahmud »

Aida Razi

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Re: Problem 1
« Reply #8 on: October 13, 2012, 02:17:05 PM »
There is a missing g(y) in the final solution for both conditions.

Victor Ivrii

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Re: Problem 1
« Reply #9 on: October 13, 2012, 02:47:52 PM »
There is a missing g(y) in the final solution for both conditions.

Right, I fixed them. Thanks!

Victor Ivrii

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Re: Problem 1
« Reply #10 on: October 13, 2012, 02:51:37 PM »
Zarak,

You don't need to surround
Code: [Select]
\begin{equation} ....\end{equation}by double dollars as equation, align, gather (and their * versions) are LaTeX environments, and multline  (and its * version) is AMS-LaTeX environment, all recognizable by MJ. Basically DD are  deprecated in LaTeX.

Zarak Mahmud

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Re: Problem 1
« Reply #11 on: October 14, 2012, 08:15:42 AM »
Zarak,

You don't need to surround
Code: [Select]
\begin{equation} ....\end{equation}by double dollars as equation, align, gather (and their * versions) are LaTeX environments, and multline  (and its * version) is AMS-LaTeX environment, all recognizable by MJ. Basically DD are  deprecated in LaTeX.

Oh I see. I edited them out of the code for this post and will dispense with them in future posts. Now that I think about it, the double dollars do seem redundant when using equation, align, gather etc.