Toronto Math Forum

APM346--2019 => APM346--Lectures & Home Assignments => Home Assignment 5 => Topic started by: Rhamel on February 21, 2019, 05:54:56 PM

Title: Problem 1
Post by: Rhamel on February 21, 2019, 05:54:56 PM
I would like to know if my solution is correct.
for problem 1. 3

I am supposed to solve $$u_{t} = ku_{xx}, u(x, 0) = g(x)$$

where $$g(x) = \exp(-a |x|)$$

My solution:
 $$u(x, t) = \frac{1}{4\sqrt{kt\pi}} \int_{-\inf}^{\inf} \exp(\frac{-(x-y)^2}{4kt}) \exp(-a|y|) dy$$
(*) to get rid of the absolute value:
 $$u(x, t) = \frac{2}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(x-y)^2}{4kt}) \exp(-ay) dy$$
I then complete the square for: $\frac{-(x-y)^2 - 4ktay}{4kt}$
to get:
 $$u(x, t) = 2 \exp(ax-a^2kt) \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$

 I then use:  $$1 =  \frac{1}{4\sqrt{kt\pi}} \int_{-\inf}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$
(**) to write:  $$\frac{1}{2} =  \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$
  $$u(x, t) = \frac{2 \exp(ax-a^2kt)}{2} = \exp(ax-a^2kt) $$

I feel like I might have taken a wrong turn at step (*) and/or (**); could someone let me know if this is right or wrong   
Title: Re: Problem 1
Post by: Yilin Ye on February 22, 2019, 12:41:09 AM
(*)should be

\begin{matrix} \frac{1}{4\sqrt{kt\pi}}\int_{0}^{inf} exp(-(x-y)^2/4kt)exp(-ay)\, dy\end{matrix}+\begin{matrix}\frac{1}{4\sqrt{kt\pi}} \int_{-inf}^{0} exp(-(x-y)^2/4kt)exp(ay)\, dy\end{matrix}