Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment 3 => Topic started by: Kun Guo 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?

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

I think there is still problem 2/ sqrt(pi) instead of sqrt(2/pi), is this true?

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

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,

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).

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{eqErf}
\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*}

(a)
\begin{equation}
G(x,y,t) = \frac{1}{2 \sqrt{k \pi t}} e^{\frac{(xy)^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{(xy)^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{(xy)^2}{4kt}} + e^{\frac{(x+y)^2}{4kt}} \right] g(y)\,dy.
\end{equation*}

There is a missing g(y) in the final solution for both conditions.

There is a missing g(y) in the final solution for both conditions.
Right, I fixed them. Thanks!

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

Zarak,
You don't need to surround
\begin{equation} ....\end{equation}
by double dollars as equation, align, gather (and their * versions) are LaTeX environments, and multline (and its * version) is AMSLaTeX 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.