Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment 3 => Topic started 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!

Try completing the square and using the error function. It's a really messy integral.

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 variablesnot $\erf$ but many others).

Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!

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

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 variablesnot $\erf$ 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?

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 variablesnot $\erf$ 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.

just wondering is that sounds true if i try to change the integral to some kind of PDF of a normal distribution?

$\newcommand{\erf}{\operatorname{erf}}$
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*}

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.)

Q3 part(b) which used the same method as part(a).

Q3 part(c)

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).

Hahah, yeah I think I would have had to spend about 3 hours typing that all up.

Yeah my answers are really long but I don't know other ways to solve the problem.... :(

Question 3 part a: p has to be= ((yx+2kÎ±t) / (âˆš 2kt)) instead of ((yx+2kÎ±t) / (âˆš 4kt)) in Peishan's solution.

I think it doesn't matter. If you use the error function (the first attachment) provided by our professor, then you let p = ((yx+2kÎ±t) / (âˆš 2kt)). If you use the other form (the second attachment) which is used in the textbook, then you let p = ((yx+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.

I think it doesn't matter. If you use the error function (the first attachment) provided by our professor, then you let p = ((yx+2kÎ±t) / (âˆš 2kt)). If you use the other form (the second attachment) which is used in the textbook, then you let p = ((yx+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

Solution to question3 part b is attached!

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.