### Author Topic: Problem 1  (Read 1232 times)

#### Rhamel

• Jr. Member
• Posts: 7
• Karma: 0
##### Problem 1
« 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

« Last Edit: February 21, 2019, 06:28:51 PM by Rhamel »
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#### Yilin Ye

• Newbie
• Posts: 3
• Karma: 0
##### Re: Problem 1
« Reply #1 on: February 22, 2019, 12:41:09 AM »
（*）should be
𝑢(𝑥,𝑡)=

\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}