Toronto Math Forum
APM3462019 => APM346Lectures & Home Assignments => Home Assignment 5 => Topic started 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:
$$u(x, t) = \frac{1}{4\sqrt{kt\pi}} \int_{\inf}^{\inf} \exp(\frac{(xy)^2}{4kt}) \exp(ay) dy$$
(*) to get rid of the absolute value:
$$u(x, t) = \frac{2}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{(xy)^2}{4kt}) \exp(ay) dy$$
I then complete the square for: $\frac{(xy)^2  4ktay}{4kt}$
to get:
$$u(x, t) = 2 \exp(axa^2kt) \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{(y+2katx)^2}{4kt}) dy$$
I then use: $$1 = \frac{1}{4\sqrt{kt\pi}} \int_{\inf}^{\inf} \exp(\frac{(y+2katx)^2}{4kt}) dy$$
(**) to write: $$\frac{1}{2} = \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{(y+2katx)^2}{4kt}) dy$$
$$u(x, t) = \frac{2 \exp(axa^2kt)}{2} = \exp(axa^2kt) $$
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

（*）should be
𝑢(𝑥,𝑡)=
\begin{matrix} \frac{1}{4\sqrt{kt\pi}}\int_{0}^{inf} exp((xy)^2/4kt)exp(ay)\, dy\end{matrix}+\begin{matrix}\frac{1}{4\sqrt{kt\pi}} \int_{inf}^{0} exp((xy)^2/4kt)exp(ay)\, dy\end{matrix}