# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 3 => Topic started by: Thomas Nutz on October 07, 2012, 10:15:42 PM

Title: problem 3
Post by: Thomas Nutz on October 07, 2012, 10:15:42 PM
Could anyone give me a hint on how to integrate problem 3a)? In order to get rid of the absolute value I have to split up the integral and then the -inf to inf formulas don't work any more...

Thanks a lot!
Title: Re: problem 3
Post by: Calvin Arnott on October 08, 2012, 12:18:45 AM
Try completing the square and using the error function. It's a really messy integral.
Title: Re: problem 3
Post by: Victor Ivrii on October 08, 2012, 01:12:46 AM
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).
Title: Re: problem 3
Post by: Peishan Wang on October 08, 2012, 07:43:01 AM
Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!
Title: Re: problem 3
Post by: Victor Ivrii on October 08, 2012, 08:12:31 AM
Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!

For $x>0$ only, I adjusted problem to make it explicit
Title: Re: problem 3
Post by: Shu Wang on October 08, 2012, 07:07:43 PM
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

When you integrate erf(z) it always gives you zero because it's an odd function. When multiplied to any integrated function (and as alpha -> 0), the resulting functions are always 0. Does that make any sense?
Title: Re: problem 3
Post by: Calvin Arnott on October 08, 2012, 07:46:14 PM
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

When you integrate erf(z) it always gives you zero because it's an odd function. When multiplied to any integrated function (and as alpha -> 0), the resulting functions are always 0. Does that make any sense?

The error function itself isn't integrated here, the function G(x,y,t)*g(y) is integrated.

It's true that an odd function will integrate to 0 on a domain symmetric about 0, and anything multiplied by this evaluated integral will be 0, but because we're multiplying G(x,y,t) inside of the integral sign that fact isn't too relevant here.

Consider the analogue to the function f(x) = x. f is an odd function, so the integral on say (-a,a) is 0. But when we multiply inside of the integral sign by another odd function g(x) = x^3, we have g*f(x) = x^4- certainly not 0 on the interval (-a,a) when a != 0.
Title: Re: problem 3
Post by: Bowei Xiao on October 08, 2012, 09:26:55 PM
just wondering is that sounds true if i try to change the integral to some kind of PDF of a normal distribution?
Title: Re: problem 3: OT--Integrating erf(z)
Post by: Victor Ivrii on October 08, 2012, 09:42:55 PM
$\newcommand{\erf}{\operatorname{erf}}$
Integrating by parts
\begin{equation*}
\int _0^z \erf(z)\,dz= z \erf(z)-\int_0^z z\erf '(z)\,dz=z\cdot \erf(z)-\sqrt{\frac{2}{\pi}}\int_0^z  z \cdot e^{-\frac{z^2}{2}}\,dz=
z \cdot \erf(z)-\sqrt{\frac{2}{\pi}}\Bigl[e^{-\frac{z^2}{2}}-1\Bigr]
\end{equation*}
Title: Re: problem 3
Post by: Peishan Wang on October 10, 2012, 10:52:54 PM
Q3 part(a)

(Sorry this part is kind of long so I have to attach 4 pictures. Please let me know if there's anything wrong with the answers.)
Title: Re: problem 3
Post by: Peishan Wang on October 10, 2012, 10:57:53 PM
Q3 part(b) which used the same method as part(a).
Title: Re: problem 3
Post by: Peishan Wang on October 10, 2012, 11:00:01 PM
Q3 part(c)
Title: Re: problem 3
Post by: Peishan Wang on October 10, 2012, 11:03:51 PM
Q3 part(d)

We get the same answer as in part (a) because g(x) is itself an even function, so if we take even reflection we still get g(x).
Title: Re: problem 3
Post by: Zarak Mahmud on October 10, 2012, 11:04:48 PM
Hahah, yeah I think I would have had to spend about 3 hours typing that all up.
Title: Re: problem 3
Post by: Peishan Wang on October 11, 2012, 02:47:33 PM
Yeah my answers are really long but I don't know other ways to solve the problem.... :(
Title: Re: problem 3
Post by: Aida Razi on October 13, 2012, 09:25:45 PM
Question 3 part a: p has to be= ((y-x+2kÎ±t) / (âˆš 2kt)) instead of ((y-x+2kÎ±t) / (âˆš 4kt)) in Peishan's solution.
Title: Re: problem 3
Post by: Peishan Wang on October 14, 2012, 01:56:36 AM
I think it doesn't matter. If you use the error function (the first attachment) provided by our professor, then you let p = ((y-x+2kÎ±t) / (âˆš 2kt)). If you use the other form (the second attachment) which is used in the textbook, then you let p = ((y-x+2kÎ±t) / (âˆš 4kt)).

And in another post, http://forum.math.toronto.edu/index.php?topic=49.0 Julong has shown that these two forms are actually equivalent.

Professor can you provide some feedback to the posted solutions? It seems that you ignored this post completely. Thanks.
Title: Re: problem 3
Post by: Victor Ivrii on October 14, 2012, 04:22:42 AM
I think it doesn't matter. If you use the error function (the first attachment) provided by our professor, then you let p = ((y-x+2kÎ±t) / (âˆš 2kt)). If you use the other form (the second attachment) which is used in the textbook, then you let p = ((y-x+2kÎ±t) / (âˆš 4kt)).

And in another post, http://forum.math.toronto.edu/index.php?topic=49.0 Julong has shown that these two forms are actually equivalent.

Professor can you provide some feedback to the posted solutions? It seems that you ignored this post completely. Thanks.

You are correct.

PS I am not on 24/7 shift
Title: Re: problem 3
Post by: Aida Razi on December 18, 2012, 12:45:51 PM
Solution to question3 part b is attached!
Title: Re: problem 3
Post by: Fanxun Zeng on December 18, 2012, 07:20:40 PM
Thanks Aida for posting Part b in December. As there is still NO solution posted for Part d yet, I spent 1 hour to write part d and post attached same 2 images, first by camera, second by scan.

Thanks professor. Here I used error function in the textbook, instead of that given in the homework, but yes they are same. In part d, I got limit u(x,t)=1 for Neumann condition.